CN104156353A - Computer-based method and device for analyzing natural language syntactic structures - Google Patents

Abstract

The invention discloses a computer-based method and device for analyzing natural language syntactic structures. According to the method and device, the natural language syntactic structures are analyzed by building a matrix model and a linear model and constructing a recursive function through the mathematical thought of a composite function according to mathematical principles of subjects including the abstract algebra, the set theory, the combinatorial mathematics, the computability theory, the computational linguistics and the like and corresponding computer technologies; meanwhile, methods such as the mathematical induction are comprehensively applied to proving important conclusions. By means of the method and device, one set of brand new mathematic models are built for sentences of the natural language, and the thought is basically different from that of a conventional traditional method; the two overall plug-in methods including the single-side same-direction order preserving method and the single-side same-direction non-order-preserving method are creatively provided, and a parallel syntactic constituent generating and processing method of a set family is creatively applied; the rules of the mathematic subjects and the computer subjects are sufficiently used, and the method is high in accuracy and large in operation amount, and has certain technological difficulty.

Description

The method and apparatus that a kind of computer based natural language syntactic structure is resolved

Technical field

The present invention relates to field of computer data processing, be specifically related to the method and apparatus that a kind of computer based natural language syntactic structure is resolved.

Background technology

Natural language processing is an important directions in computer science and artificial intelligence field.Its research can realize between people and computing machine uses natural language to carry out various theories and the method for efficient communication.

It is an importance of natural language processing that syntactic structure is resolved, and it carries out automatically dividing to assist the further processing for statement to the sentence element of natural language statement by computing machine.In existing syntactic structure analytic technique, conventionally adopt probability context without bounding algorithm (Probabilistic Context Free Grammars, PCFG), it has the feature of complicated nested property based on natural language, the rule match probability of computing statement and syntactic structure analysis result, chooses the syntax analysis result of maximum probability as final syntactic structure.

But the method complexity is high, and, for the also urgently further raising of parsing accuracy of combined type sentence structure.

Summary of the invention

In view of this, the invention provides the method and apparatus that a kind of computer based natural language syntactic structure is resolved, design is unique, method is ingenious, it is full and accurate to prove, take full advantage of the rule of mathematics and Computer Subject, described method accuracy is higher, operand is very large, has certain technical difficulty.

The invention provides a kind of method that computer based natural language syntactic structure is resolved, comprising:

S1, read pretreated phrase data structure to be resolved, the conjunctive word arranged side by side unit, subordinate conjunctive word unit, predicate verb unit, the noun pronoun unit that in described pretreated phrase data structure, only comprise statement, and each word unit is numbered according to the order in described pretreated statement, and marks type;

S2, to each predicate verb unit, generate corresponding leading question element, subject element, predicate element and object element;

The possible value of described leading question element is less than one of the conjunctive word arranged side by side unit of corresponding predicate verb element number or subordinate conjunctive word unit for numbering, or be less than the conjunctive word arranged side by side unit of corresponding predicate verb element number and one by a numbering and be adjacent and number and be less than one of the conjunctive word mix vector that corresponding predicate verb element number and numbering are greater than the subordinate conjunctive word cell formation of this conjunctive word element number arranged side by side, or dummy cell;

The possible value of described subject element is less than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of major term unit is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or at one of syntax vector corresponding to the predicate element of front appearance, or dummy cell;

Described predicate element is corresponding described predicate verb unit;

The possible value numbering of described object element is greater than corresponding predicate verb element number and is less than one of noun pronoun unit of the adjacent predicate verb element number in rear appearance, or the numbering of minimum word unit is greater than corresponding predicate verb element number and is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of the adjacent predicate verb element number in rear appearance, or at one of syntax vector corresponding to the predicate element of rear appearance, or dummy cell;

S3, according to the possible value of described leading question element, subject element, predicate element and object element, the institute that obtains syntax vector that each predicate verb unit is corresponding is value likely, and described syntax vector comprises leading question element, subject element, predicate element and object element;

S4, according to institute's value likely of all syntax vectors, generating at least one syntactic structure may matrix solution, described syntactic structure may matrix solution by forming according to the tactic syntax vector of predicate verb element number;

Whether the statement that S5, checking obtain according to syntactic structure possibility matrix solution is identical with described pretreated statement, if identical, using each syntax vector in this syntactic structure possibility matrix solution as one of syntactic structure analysis result;

Wherein, S5 comprises in order the following operation of execution successively, gets rid of ineligible syntactic structure feasible solution:

If S5.1 exists the sequence valve not occurring in this syntactic structure possibility matrix solution, get rid of this syntactic structure possibility matrix solution;

If S5.2 occurs identical sequence valve or occurs identical syntax vector in different syntax vectors, get rid of this syntactic structure possibility matrix solution;

S5.3, in each may matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there is the intersection contradiction of two syntax vectors after equivalent substitution, get rid of this syntactic structure possibility matrix solution;

S5.4, in each may matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there are two sequence valves that position is converse after equivalent substitution, get rid of this syntactic structure possibility matrix solution;

S5.5, in any one may matrix solution, if there is no the syntax vector of mutual substitution relation between existence and other syntax vectors, carry out plug hole and operate to obtain the corresponding possibility of all these possibility matrix solutions syntax analytic structure, and whether the statement that checking obtains according to described possibility syntax analytic structure is identical with described pretreated statement, it further comprises:

S5.5.1, first to existing each other the syntax vector of substitution relation to carry out equivalent substitution in this possibility matrix solution, thereby this possibility matrix solution is converted into one group of syntax vector that does not have each other a substitution relation syntax vector in possibility matrix solution is called to first kind syntax vector, the syntax that conversion is obtained vector be called Equations of The Second Kind syntax vector;

S5.5.2, appoint and to get an Equations of The Second Kind syntax vector according to predetermined direction, mark one by one in the sequence valve of each syntax elements; After the sequence valve of mark syntax elements, appoint and get in i syntax elements, only in the first side of this syntax elements, construct unique room; After making sky, appoint and get a syntax vector equations of The Second Kind syntax vector in addition mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, this new syntax vector is designated as and the syntax that whole plug hole is obtained vector, be referred to as the 3rd class syntax vector;

S5.5.3, vectorial to the 3rd class syntax according to predetermined direction to from vector in first syntax elements of the first side start to vector in the vector that comprises first syntax elements of the second side till each syntax elements, all mark sequence valve; Be positioned at vector in the vector that comprises the element of the first side, does not mark sequence valve; By vector first syntax elements of the second side be designated as will be according to aforementioned manner to vector the syntax vector part of mark, is designated as whipping syntax vector after mark sequence valve, appoint and get a j syntax elements in aforesaid whipping vector, only in this element first side, construct unique room; After making sky, appoint and get an original Equations of The Second Kind syntax vector mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, newly-generated syntax vector is designated as or

The 3rd class syntax vector according to predetermined direction, distich normal vector in each syntax elements all mark sequence valve; After the sequence valve of mark syntax elements, appoint and get one in t syntax elements, in the first side of this syntax elements, construct unique room; After making sky, appoint and get use Equations of The Second Kind syntax vector in the mode of whole plug hole by this vector insert the room of constructing above, and then generate a new vector, this new vector is designated as

S5.5.4, repeat S5.5.3, when the last time makes empty and plug hole step and finishes, the empty and plug hole made carrying out is next time operated, until all Equations of The Second Kind syntaxes are vectorial through last the 3rd class syntax vector of making empty and plug hole step and obtaining all plug hole is complete, finally obtains the 3rd class syntax vector of a single file, and described the 3rd class syntax vector finally obtaining is called to final single file vector;

If all there are two sequence valves that position is converse in may the syntax analytic structure corresponding all described final single file vector of one of S5.5.5, get rid of this possibility syntax analytic structure;

S5.5.6, repeat S5.5.2 to S5.5.5 until all may syntax analytic structure be traversed.

Further, S2 comprises generation coordinate noun pronoun mix vector family:

S2.1 chooses unduplicated two noun pronoun unit:

If A does not have other word unit between these two noun pronoun unit, using these two noun pronoun unit as a coordinate noun pronoun mix vector, and retain this coordinate noun pronoun mix vector;

If there is other word unit in B between these two noun pronoun unit, check between each the word unit between these two noun pronoun unit: if any one the word unit between these two noun pronoun unit, all noun pronoun unit or conjunctive word arranged side by side unit, using selected two noun pronoun unit and all words unit between these two noun pronoun unit as a coordinate noun pronoun mix vector, and retain this coordinate noun pronoun mix vector; Otherwise, do not generate coordinate noun pronoun mix vector;

The capable S2.1 of S2.2 retry, until the array mode of all noun pronoun unit is traversed, generates all coordinate noun pronoun mix vectors that obtain;

If there is coordinate noun pronoun mix vector in this possibility syntax analytic structure of S2.3, all coordinate noun pronoun mix vectors are divided, thereby form several coordinate noun pronoun mix vector families, make: in each coordinate noun pronoun mix vector family, each the coordinate noun pronoun mix vector comprising in this coordinate noun pronoun mix vector family has all comprised two common noun pronoun unit.

S2.4, in each noun pronoun mix vector family, chooses the word unit of the numbering maximum comprising in all noun pronoun mix vectors, and the major term unit as this noun pronoun mix vector family, is used during in order to follow-up generation subject; Choose the word unit of the numbering minimum comprising in all noun pronoun mix vectors, as the minimum word unit of this noun pronoun mix vector family, during in order to follow-up generation object, use.

Further, generating corresponding subject element comprises:

When corresponding predicate verb element number is minimum predicate verb element number, the possible value of described subject element is less than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of its major term unit is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or dummy cell.

When corresponding predicate verb element number is not minimum predicate verb element number, the possible value of described subject element is less than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of major term unit is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or at one of syntax vector corresponding to the predicate verb unit of front appearance, or dummy cell.

Further, generating corresponding object element comprises:

When corresponding predicate verb element number is maximum predicate verb element number, the possible value of described object element is greater than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of its minimum word unit is greater than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or dummy cell.

When corresponding predicate verb element number is not maximum predicate verb element number, the possible value of described object element is greater than corresponding predicate verb element number and is less than one of noun pronoun unit of the adjacent predicate verb element number in rear appearance for numbering, or the numbering of its minimum word unit is greater than corresponding predicate verb element number and is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of the adjacent predicate verb element number in rear appearance, or at one of syntax vector corresponding to the predicate verb unit of rear appearance, or dummy cell.

Further, in S4 and two steps of S5, utilize with syntactic structure and may linear representation solution substitute described syntactic structure possibility matrix solution;

Described syntactic structure possibility linear representation solution and described syntactic structure possibility matrix solution are of equal value;

Described syntactic structure may linear representation solution comprise by forming according to the tactic syntax vector expression of predicate verb element number; Described in each, syntax vector expression is the expression formula that leading question element, subject element, predicate element, the object element of corresponding syntax vector added up in order item by item partially.

Further, described method also comprises:

Each syntax vector and corresponding syntax structural relationship in syntactic structure analysis result are shown in human-computer interaction interface with tree structure.

The device that the present invention also provides a kind of computer based natural language syntactic structure to resolve, comprising:

Fetch unit, for reading pretreated phrase data structure to be resolved, the conjunctive word arranged side by side unit, subordinate conjunctive word unit, predicate verb unit, the noun pronoun unit that in described pretreated phrase data structure, only comprise statement, and each word unit is numbered according to the order in described pretreated statement, and marks type;

Element generation parts, for to each predicate verb unit, generate corresponding leading question element, subject element, predicate element and object element;

Wherein, the possible value of described leading question element is less than one of the conjunctive word arranged side by side unit of corresponding predicate verb element number or subordinate conjunctive word unit for numbering, or be less than the conjunctive word arranged side by side unit of corresponding predicate verb element number and one by a numbering and be adjacent and number and be less than one of the conjunctive word mix vector that corresponding predicate verb element number and numbering are greater than the subordinate conjunctive word cell formation of this conjunctive word element number arranged side by side, or dummy cell;

The possible value of described subject element is less than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of major term unit is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or at one of syntax vector corresponding to the predicate element of front appearance, or dummy cell;

Described predicate element is corresponding described predicate verb unit;

The possible value numbering of described object element is greater than corresponding predicate verb element number and is less than one of noun pronoun unit of the adjacent predicate verb element number in rear appearance, or the numbering of minimum word unit is greater than corresponding predicate verb element number and is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of the adjacent predicate verb element number in rear appearance, or at one of syntax vector corresponding to the predicate element of rear appearance, or dummy cell;

Vector generates parts, be used for according to the possible value of described leading question element, subject element, predicate element and object element, the institute that obtains syntax vector that each predicate verb unit is corresponding is value likely, and described syntax vector comprises leading question element, subject element, predicate element and object element;

Matrix generates parts, and for according to the institute of all syntax vectors value likely, generating at least one syntactic structure may matrix solution, described syntactic structure may matrix solution by forming according to the tactic syntax vector of predicate verb element number;

Decider, whether identical with described pretreated statement for verifying the statement obtaining according to syntactic structure possibility matrix solution, if identical, using each syntax vector in this syntactic structure possibility matrix solution as one of syntactic structure analysis result;

Wherein, described decider is got rid of ineligible syntactic structure feasible solution by following module operation:

First row, except module, if there is the sequence valve not occurring in this syntactic structure possibility matrix solution, is got rid of this syntactic structure possibility matrix solution;

Second row, except module, if occur identical sequence valve or occur identical syntax vector in different syntax vectors, is got rid of this syntactic structure possibility matrix solution;

The 3rd gets rid of module, in each possibility matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there is the intersection contradiction of two syntax vectors after equivalent substitution, getting rid of this syntactic structure may matrix solution;

The 4th gets rid of module, in each possibility matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there are two sequence valves that position is converse after equivalent substitution, getting rid of this syntactic structure may matrix solution;

The 5th gets rid of module, in any one possibility matrix solution, if there is no the syntax vector of mutual substitution relation between existence and other syntax vectors, carry out plug hole and operate to obtain the corresponding possibility of all these possibility matrix solutions syntax analytic structure, and whether the statement that checking obtains according to described possibility syntax analytic structure is identical with described pretreated statement, it further comprises:

The first submodule, first to existing each other the syntax vector of substitution relation to carry out equivalent substitution in this possibility matrix solution, thereby is converted into one group of syntax vector that does not have each other substitution relation by this possibility matrix solution syntax vector in possibility matrix solution is called to first kind syntax vector, the syntax that conversion is obtained vector be called Equations of The Second Kind syntax vector;

The second submodule, appoint and to get an Equations of The Second Kind syntax vector according to predetermined direction, mark one by one in the sequence valve of each syntax elements; After the sequence valve of mark syntax elements, appoint and get in i syntax elements, only in the first side of this syntax elements, construct unique room; After making sky, appoint and get a syntax vector equations of The Second Kind syntax vector in addition mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, this new syntax vector is designated as and the syntax that whole plug hole is obtained vector, be referred to as the 3rd class syntax vector;

The 3rd submodule, to the 3rd class syntax vector according to predetermined direction to from vector in first syntax elements of the first side start to vector in the vector that comprises first syntax elements of the second side till each syntax elements, all mark sequence valve; Be positioned at vector in the vector that comprises the element of the first side, does not mark sequence valve; By vector first syntax elements of the second side be designated as will be according to aforementioned manner to vector the syntax vector part of mark, is designated as whipping syntax vector after mark sequence valve, appoint and get a j syntax elements in aforesaid whipping vector, only in this element first side, construct unique room; After making sky, appoint and get an original Equations of The Second Kind syntax vector mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, newly-generated syntax vector is designated as or

The 3rd class syntax vector according to predetermined direction, distich normal vector in each syntax elements all mark sequence valve; After the sequence valve of mark syntax elements, appoint and get one in t syntax elements, in the first side of this syntax elements, construct unique room; After making sky, appoint and get use Equations of The Second Kind syntax vector in the mode of whole plug hole by this vector insert the room of constructing above, and then generate a new vector, this new vector is designated as

The 4th submodule, repeat the operation of the 3rd submodule, when the last time makes empty and plug hole step and finishes, the empty and plug hole made carrying out is next time operated, until all Equations of The Second Kind syntaxes are vectorial through last the 3rd class syntax vector of making empty and plug hole step and obtaining all plug hole is complete, finally obtains the 3rd class syntax vector of a single file, and described the 3rd class syntax vector finally obtaining is called to final single file vector;

The 5th submodule, if one may syntax all has two sequence valves that position is converse in all described final single file vector corresponding to analytic structure, gets rid of this possibility syntax analytic structure;

The 6th submodule, the operation that repeats to call the second submodule to the five submodules is until all may being traversed by syntax analytic structure.

Further, also comprise:

Result display unit, carries out each syntax vector and corresponding syntax structural relationship in syntactic structure analysis result to show on human-computer interaction interface with tree structure.

Accompanying drawing explanation

By the description to the embodiment of the present invention referring to accompanying drawing, above-mentioned and other objects of the present invention, feature and advantage will be more clear, in the accompanying drawings:

Fig. 1 is the process flow diagram of the computer based natural language syntactic structure of the embodiment of the present invention method of resolving;

Fig. 2 is the schematic diagram of the computer based natural language syntactic structure of the embodiment of the present invention device of resolving.

Embodiment

Based on preferred embodiment, present invention is described below, but the present invention is not restricted to these embodiment.In below details of the present invention being described, detailed some specific detail sections of having described.Do not have for a person skilled in the art the description of these detail sections can understand the present invention completely yet.For fear of obscuring essence of the present invention, known method, flow process, element and circuit do not describe in detail.

Partial ordering relation on A Partial Semigroup and additive operation partially

A1 part natural language is freely the one semigroup in vocabulary and punctuation mark set

According to the theory of Abstract Algebra and computational linguistics, natural language is freely the one semigroup in vocabulary and punctuation mark set.Below in English for example describes, still one of ordinary skill in the art will readily recognize that method of the present invention is also applicable to other natural languages.

A given set A, the symbol string on A be element in A in abutting connection with forming, in abutting connection with time can repeat, form a time-limited linear array.For example: from set, { c}, can form symbol string acbaab for a, b.What this symbol string comprised a occurs for three times, twice appearance of b, and the once appearance of c, it is different from symbol string acaabb.Although the occurrence number of each symbol is identical, their order is different.Visible, symbol string is orderly.Especially, the symbol string that length is 0 is 0 symbol string, is designated as e.Thus, for given limited assemble of symbol A, the symbol string that the upper length of A is n is exactly mapping: a f:N → A from natural manifold N to A.

From two symbol strings, we can form by the way in abutting connection with them new symbol string.For example, at the right-hand member of symbol string abac, in abutting connection with symbol string bbac, just formed new symbol string abacbbac.

This computing in abutting connection with symbol string is called: adjoin computing, referred to as adjoining.

Given length is the symbol string φ of n and the symbol string ψ that length is m, wherein:

φ＝{(1，x ₁)，(2，x ₂)，(3，x ₃)，……，(n-1，x _n-1)，(n，x _n)}；

ψ＝{(1，y ₁)，(2，y ₂)，(3，y ₃)，……，(m-1，y _m-1)，(m，y _m)}；

Adjoining of φ and ψ is designated as: φ ^ ψ.It is length be n+m and by set { (1, x ₁), (2, x ₂), (3, x ₃) ..., (n-1, x _n-1), (n, x _n), (n+1, y ₁), (n+2, y ₂) ..., (n+m, y _m) symbol string that provides.So, adjoining is a kind of dyadic operation being defined in symbol string, and the result of computing is to obtain a new symbol string.

φ and ψ adjoin, and also can omit and adjoin mark ^, and simplification is designated as: φ ψ.

Have: φ ^ ψ=φ ψ.

It is combinative adjoining computing, because for any symbol string φ, and ψ, ω, has:

φ^(ψ^ω)＝(φ^ψ)^ω

Existing each English word and english punctuation mark are defined as a symbol, so the set A of all words and punctuation mark={ a in S ₁, a ₂, a ₃..., a _n(n ∈ N) be exactly a glossary of symbols.

Appoint a time-limited symbol string b who is formed by English word and english punctuation mark who gives ₁b ₂... b _k(k ∈ N), is called word unit or continuous word string.For appointing a word unit a=b who gives ₁b ₂... b _m(m ∈ N), claims that a is the word unit that element forms in A, and and if only if, b ₁, b ₂..., b _m∈ A.

Length is that unique word unit of 0 is called dummy cell, is designated as e.

The set of note all words unit that element forms in A (word string continuously) is A ^s, establish statement S=a ₁a ₂a ₃... a _n, wherein, a _nfor forming the word unit of statement.Algebra system (A ^s, ^, e) and be freely the one semigroup in English word and punctuation mark set A.

According to it, the order in statement is arranged in order in each word unit, under it, is designated as serial number, and note T (α) is the numbering of word unit α in sentence S.

Construct a syntactic constituent Sequential Mapping ω, the condition of ω is as follows:

(1) ω: { a ₁, a ₂, a ₃..., a _n} → N, N is nature manifold;

(2) to any one a _i, a _i∈ S, has: ω (a _i)=T (a _i).

Obviously, ω is a single mapping.

A2 partly defines a kind of partial ordering relation

Meanwhile, for algebra system (A ^s, ^, e), definition binary relation < _□:

For A ^sin word unit α arbitrarily, β ∈ A ^s, claim α < _□β, and if only if α, the code T of β (α), T (β) meets: T (α) < T (β).

According to definition, binary relation < _□meet following condition:

(1) appoint the A to a ∈ ^s, have a ≮ _□a;

(2) for any a, b, c ∈ A ^sif, a < _□b, b ≮ _□a;

(3) for any a, b, c ∈ A ^sif, a < _□b and b < _□c, a < _□c.

The definition of the strict partial ordering relation of foundation, binary relation < _□it is strict partial ordering relation.

A3 partly defines inclined to one side additive operation and syntax sequence valve

Meanwhile, at algebra system (A ^s, ^, e) and upper, define a new dyadic operation+<.Title+< is for being defined in A ^sin strict partial ordering relation < _□on inclined to one side additive operation, be called for short and partially add, it meets following characteristic: for any a, b ∈ A ^sif, a < _□b, has a+ < b=a^b=ab.

We can determine: for any a, and b ∈ A ^sif, a < _□b, has inclined to one side additive operation+< and adjoins computing ^ of equal value.Additive operation+<, can be regarded as and be limited in strict partial ordering relation < partially _□on adjoin computing.

The statement S of arbitrary natural language can regard as by each word unit according to strict partial ordering relation < _□the word string formula being formed by connecting, that is: S=a ₁+ < a ₂+ < a ₃+ < ... + < a _n.This feature, for launching, mathematics manipulation is highly beneficial.

In former S, according to order from left to right, once from beginning of the sentence to a sentence tail, be the individual word string α continuously of n adjacent in full sentence ₁, α ₂..., α _nmark serial number: 1,2 ... .., n.

Once determine, in mark as above, remember that the serial number of an any given continuous word string α is τ (α), claim τ (α) for the sequence valve from left to right of α.That is, appoint to the syntax elements γ in a former sentence S, the syntax sequence valve by syntax elements γ in former sentence S is designated as τ (γ).

B part technical scheme describes in detail

The preliminary classification of B1 part to language message

In the present invention, the word unit a of statement will be formed _iregard as constant.Word unit a _ithere is its linguistic property.The word unit that forms kernel sentence structure can be divided into 4 types of conjunctive word arranged side by side unit, subordinate conjunctive word unit, predicate verb unit, noun pronoun unit.Each word unit comprises at least one natural language vocabulary, its can be the phrase of word, ad hoc structure or a plurality of same attribute words side by side.

For conjunctive word arranged side by side unit, it can be the coordinating conjunction and that connects compound sentence and syntactic constituent arranged side by side, but, or, so, yet etc.

For subordinate conjunctive word unit, it can be the guiding conjunctional pronoun of subordinate clause or the conjunctive phrase of conjunctive adverbs and guiding subordinate clause, for typical introducer, be listed below: that, what, which, who, whom, wherever, whenever, whose, where, when, why, how, whoever, whichever, while, whether, because, before, after, whatever, whomever, as, if, once, until, though, unless, although, no matter what, no matter who, no matter whom, no matter which, in that, in order that, as though, as if, even though, even if, so that etc.It mainly comprises: by word, served as the conjunctive word unit that guides subordinate clause, served as the conjunctive word unit that guides subordinate clause by phrase, connect the conjunctive word unit of compound sentence and compound sentence.

For predicate verb unit, it can be also verb or verb phrase, for example, and can do, do.Predicate is defined as the main action language in a natural sentences in English.In structure, conventionally by two parts, formed: auxiliary verb+notional verb (except principal series list structure).Predicate has the call format of tense and voice, with calculating philological formula, is defined as follows:

For noun pronoun unit, can be: pure noun phrase (not being included in the noun phrase in guest's Jie phrase), (the verb phrase definition of noun: there is verb phrase noun character, that can serve as subject or this class name part of speech syntactic constituent of object of the verb phrase of noun, comprise: infinitive phrase and the large class of gerund phrase two), the pronoun that can use separately.Noun pronoun unit is exemplified below: food, wolf, the men, me, it, this, to do etc.

The verb phrase of noun has call format, with calculating philological formula, is defined as follows:

1	To+VB	7	RB+To+VB
				2	To+VB+VBN	8	RB+To+VB+VBN
3	To+VB+VBN+VBN	9	RB+To+VB+VBN+VBN
				4	VBG	10	RB+VBG
5	VBG+VBN	11	RB+VBG+VBN
				6	VBG+VBN+VBN	12	RB+VBG+VBN+VBN

{ explanation of important sign }

r	Predicate verb unit
		k	The Ser.No. of the current predicate verb unit of processing
Lead	Subordinate conjunctive word unit
		NPI	Pure noun unit
Conj	Conjunctive word arranged side by side unit
		VNP	The verb unit of noun character
NOMP	Nominative pronoun unit
		OBJP	Objective case pronoun unit
NP	The general designation of noun pronoun unit

In above-mentioned word unit list, between the set of word unit, there is following relation:

{NP}＝{NPI}∪{VNP}∪{NOMP}∪{OBJP}。

B2 partly defines important concept

Illustrate: the subordinate sentence of natural language statement is defined as follows: subordinate sentence is exactly simple sentence, i.e. the most basic sentence formula of natural language.A subordinate sentence is exactly a set of subject-predicate matching structure.The trunk of above three class word cell formation natural language statement subordinate sentences, wherein, predicate verb unit serves as predicate, and noun pronoun unit serves as subject or object.

In the present invention, defining variable is x, y, and z, wherein x is leading question element, and y is subject element, and z is object element, and meanwhile, note r is predicate element, the subject-predicate matching structure in each statement can be expressed as:

f(x，y，r，z)＝x+＜Λ+＜y+＜σ+＜r+＜ρ+＜z+＜μ

Λ, σ, ρ, μ represents respectively x, y, r, any composition or punctuation mark outside z, referred to as impurity, can remove impurity by existing statement preconditioning technique.Can be by function f (x, y, r, the z)=x+ < y+ < r+ < z removing after impurity.By the mode of vector (x, y, r, z), represent.

Leading question element x is a composition of simple sentence: when simple sentence is subordinate clause, and the conjunctive phrase of the conjunctional pronoun that leading question element is guiding subordinate clause or conjunctive adverbs, guiding subordinate clause; When simple sentence is compound sentence, leading question element is the coordinating conjunction that this compound sentence is connected with preceding other compound sentences.That is, in a simple sentence, leading question element x be by conjunctive word cell formation, for the syntactic constituent of direct subsequent simple sentence.

If a current just function f in treatment S, remembers that this current function f is f _k; The Ser.No. of remembering the current predicate verb unit of processing is k.(k ∈ N, N is nature manifold, k≤n)

B3 partly generates three crucial set: { x _k, { y _k, { z _k}

B3.1 partly generates { x _k}

[B3.1.1] preparation work: be defined as follows subclass:

1)Lead _k＝{Lead|Lead＜ _□r _k}；

2)conj _k＝{conj|conj＜ _□r _k}；

3)(conj _kοLead _k)＝

{R _k|R _k＝conj+＜Lead，conj＜ _□r _k，Lead＜ _□r _k，τ(Lead)＝τ(conj)+1}；

[B3.1.2] { x _kgenerating algorithm:

{x _k}＝Lead _k∪conj _k∪(conj _kοLead _k)∪{e}。

B3.2 part is about the special explanation (subject arranged side by side of take is example with object arranged side by side) of the generation method of syntactic constituent arranged side by side

[B3.2.1] directviewing description

Illustrate: in narration below, for the ease of expressing, by the formula Ф of continuous word string ^tor in comprise syntax elements be designated as $\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\&SubsetEqual;{\mathrm{\&Phi;}}^t,\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\&SubsetEqual;{\mathrm{\&Phi;}}_{\mathrm{NP}}^t.$

The 1st step

Get the phrase of all noun character in former sentence, it is a set that the phrase of all noun character in former sentence is compiled, and is designated as set Ψ={ α ₁..., α _m-1, α _m, m ∈ N, m is the number of the element in set Ψ.

The 2nd step

According to mode, get set Ψ={ α ₁..., α _m-1, α _min whole combinations of any two elements, $\{{\mathrm{\&alpha;}}_i^1,{\mathrm{\&alpha;}}_j^1\}\&Element;\mathrm{\&Psi;},\{{\mathrm{\&alpha;}}_i^2,{\mathrm{\&alpha;}}_j^2\}\&Element;\mathrm{\&Psi;},......,\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\&Element;\mathrm{\&Psi;}$ If set $C_{\mathrm{\&Psi;}}^2=\left\{\right\{{\mathrm{\&alpha;}}_i^1,{\mathrm{\&alpha;}}_j^1\},\{{\mathrm{\&alpha;}}_i^2,{\mathrm{\&alpha;}}_j^2\},$ $......,\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\}.$

The 3rd step

To appointing one that gives according to element the arrangement from small to large of the syntax sequence valve in former sentence S.Might as well establish can obtain ordered pair set up the formula of a continuous word string ${\mathrm{\&Phi;}}^t={\mathrm{\&alpha;}}_i^t+<{\mathrm{\&beta;}}_k+<...+<{\mathrm{\&beta;}}_l+<{\mathrm{\&alpha;}}_j^t,$ Wherein ${\mathrm{\&alpha;}}_i^t,{\mathrm{\&beta;}}_k,......,{\mathrm{\&beta;}}_l,{\mathrm{\&alpha;}}_j^t$ Be in former sentence S from arrive one group of adjacent continuous word string or empty word string.The ordered pair that limit is such and continuously word string formula.

The 4th step

To formula Ф ^tcheck, if for formula Ф ^tin appoint give between element with between element γ, have: the phrase of γ or noun character, or arranged side by side conjunction, or empty string, by Ф ^tmark change into be called Ф ^tgenerate if set ?

The 5th step

Appoint and get set $\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\&Element;C_{\mathrm{\&Psi;}}^2,{\mathrm{\&alpha;}}_i^p\&NotEqual;{\mathrm{\&alpha;}}_j^p,$ If set existence correspondence define a set family, this set family is by comprising set all set form.This set family is designated as to following expression-form: $I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)=\left\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\right|\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\&SubsetEqual;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;{\mathrm{\&Omega;}}_{\mathrm{NP}}\}.$

The 6th step

If set there is corresponding set family by any one set family in the syntax elements that has under its command of each set all take out, classify set as $P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\left\{\mathrm{\&alpha;}\right|\mathrm{\&alpha;}\&Element;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;$ $I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\}.$

The 7th step

Take out respectively set $P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\left\{\mathrm{\&alpha;}\right|\mathrm{\&alpha;}\&Element;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\}$ In the minimum and maximum element of syntax sequence valve in former sentence S.

Note: the method also can be used for generating the composition arranged side by side of other types, for example, generate adjective phrase arranged side by side.As long as by all NPI in the method, all VNP, all NOMP phrases, all NPI in former sentence, all VNP, all NOMP phrases are replaced to corresponding syntactic constituent, can obtain.

[B3.2.2] formal definitions

Definition: function of a single variable A (S), A (S) represents to take out all NPI phrases in former sentence S, all VNP phrases, all NOMP phrases, meanwhile, classify all NPI phrases in former sentence, all VNP phrases, all NOMP phrases as a set, this set is designated as to Ψ={ α ₁..., α _m-1, α _m, m ∈ N, m is the number of the element in set Ψ.A (S)=Ψ={ α ₁..., α _m-1, α _m.

Definition: function of a single variable B (Ψ), B (Ψ) represent according to mode get set Ψ={ α ₁..., α _m-1, α _min whole combinations of any two elements, $\{{\mathrm{\&alpha;}}_i^1,{\mathrm{\&alpha;}}_j^1\}\&SubsetEqual;\mathrm{\&Psi;},\{{\mathrm{\&alpha;}}_i^2,{\mathrm{\&alpha;}}_j^2\}\&SubsetEqual;\mathrm{\&Psi;},......,\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\&SubsetEqual;\mathrm{\&Psi;},$ If set $C_{\mathrm{\&Psi;}}^2=$ $\left\{\right\{{\mathrm{\&alpha;}}_i^1,{\mathrm{\&alpha;}}_j^1\},\{{\mathrm{\&alpha;}}_i^2,{\mathrm{\&alpha;}}_j^2\},...,\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right\}.$ ? $B\left(\mathrm{\&Psi;}\right)=C_{\mathrm{\&Psi;}}^2=\left\{\right\{{\mathrm{\&alpha;}}_i^1,{\mathrm{\&alpha;}}_j^1\},\{{\mathrm{\&alpha;}}_i^2,{\mathrm{\&alpha;}}_j^2\},...,\{{\mathrm{\&alpha;}}_i^p,$ ${\mathrm{\&alpha;}}_j^p\left\}\right\}.$ To appoint one that gives $\{{\mathrm{\&alpha;}}_i^t,{\mathrm{\&alpha;}}_j^t\}\&Element;C_{\mathrm{\&Psi;}}^2$ Be designated as ? ?

Definition: binary function K (α, β), K (α, β) represents the result to function of a single variable B (Ψ), to appointing one that gives according to element the arrangement from small to large of the syntax sequence valve in former sentence S.Might as well establish $\mathrm{\&tau;}\left({\mathrm{\&alpha;}}_i^t\right)\&le;\mathrm{\&tau;}\left({\mathrm{\&alpha;}}_j^t\right),$ Can obtain ordered pair $<{\mathrm{\&alpha;}}_i^t,{\mathrm{\&alpha;}}_j^t>.$ If set $\mathrm{\&pi;}(<{\mathrm{\&alpha;}}_i^t,{\mathrm{\&alpha;}}_j^t>)=\left\{{\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\right|\mathrm{\&tau;}\left({\mathrm{\&alpha;}}_i^t\right)<\mathrm{\&tau;}\left({\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\right)<$ $\mathrm{\&tau;}\left({\mathrm{\&alpha;}}_j^t\right),{\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\&Element;S\},$ And then set up a continuous word string formula ${\mathrm{\&Phi;}}^t={\mathrm{\&alpha;}}_i^t+<{\mathrm{\&beta;}}_k+<...+<{\mathrm{\&beta;}}_l+<{\mathrm{\&alpha;}}_j^t,$ Wherein ${\mathrm{\&alpha;}}_i^t,{\mathrm{\&beta;}}_k,......,{\mathrm{\&beta;}}_l,{\mathrm{\&alpha;}}_j^t$ Be in former sentence S from arrive one group of adjacent continuous word string or empty word string, and $\mathrm{\&pi;}(<$ ${\mathrm{\&alpha;}}_i^t,{\mathrm{\&alpha;}}_j^t>)=\{{\mathrm{\&beta;}}_k,...,{\mathrm{\&beta;}}_l\}.$ ? $K(\mathrm{\&alpha;},\mathrm{\&beta;})={\mathrm{\&Phi;}}^t={\mathrm{\&alpha;}}_i^t+<{\mathrm{\&beta;}}_k+<...+<{\mathrm{\&beta;}}_l+<{\mathrm{\&alpha;}}_j^t.$

Definition: function of a single variable H (Ф ^t), H (Ф ^t) represent the binary function K (α, β) to generate check: if to appoint to element γ ∈ Ф ^t, and and have: γ=NPI or γ=VNP or γ=NOMP or γ=CONJ or γ=e, by Ф ^tmark change into be called Ф ^tgenerate if set ?

Definition: binary function M (α, β), M (α, β) represents a set of getting for appointing if set existence correspondence define a set family, this set family is by comprising set all set form, this set family is designated as $I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)=\left\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\right|\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\&SubsetEqual;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;{\mathrm{\&Omega;}}_{\mathrm{NP}}\}.$ ? $M(\mathrm{\&alpha;},\mathrm{\&beta;})=$ $\left\{I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right|\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\&Element;C_{\mathrm{\&Psi;}}^2\}.$

Definition: binary function N (α, β), N (α, β) represents the result to binary function M (α, β) for appointing, get set $\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\&Element;C_{\mathrm{\&Psi;}}^2,{\mathrm{\&alpha;}}_i^p\&NotEqual;{\mathrm{\&alpha;}}_j^p,$ If set there is corresponding set family construct a new set as follows $P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\left\{\mathrm{\&alpha;}\right|\mathrm{\&alpha;}\&Element;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\}.$ ? $N(\mathrm{\&alpha;},\mathrm{\&beta;})=$ $P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\left\{\mathrm{\&alpha;}\right|\mathrm{\&alpha;}\&Element;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\}.$

Definition: function of a single variable U (α), U (α) represents the result to binary function N (α, β) get $P^{\max }\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ If $P^{\max }\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\mathrm{\&delta;},\mathrm{\&delta;}\&Element;P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ To appointing the element γ giving, $\mathrm{\&gamma;}\&Element;P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ There is τ (γ)≤τ (δ).? $U\left(\mathrm{\&alpha;}\right)=P^{\max }\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right].$

Definition: function of a single variable V (β), V (β) represents the result to binary function N (α, β) get $P^{\min }\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ If $P^{\min }\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\mathrm{\&delta;},\mathrm{\&delta;}\&Element;P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ To appointing the element γ giving, $\mathrm{\&gamma;}\&Element;P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ There is τ (δ)≤τ (γ).? $V\left(\mathrm{\&beta;}\right)=P^{\min }\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right].$

[B3.2.3] be the generating algorithm of subject side by side:

[B3.2.4] be the generating algorithm of object side by side:

[B3.2.4] illustrates the generating algorithm of subject arranged side by side and object arranged side by side

Illustrate: word order list is:

Former sentence phrase	Phrase type	Serial number
			After	Subordinate conjunctive word unit	1
Jack	Noun pronoun unit	2
			Mary	Noun pronoun unit	3
and	Conjunctive word arranged side by side unit	4
			Linda	Noun pronoun unit	5
left	Predicate verb unit	6
			I	Noun pronoun unit	7
gave	Predicate verb unit	8
			my?son	Noun pronoun unit	9
a?book	Noun pronoun unit	10

Generating subject element set { y ₁, { y ₂process in, move subject generating algorithm arranged side by side as follows:

1. A (S) takes out all NPI phrases in former sentence, all VNP phrases, all NOMP phrases, and classifies all NPI phrases in former sentence, all VNP phrases, all NOMP phrases as a set, and this set is designated as to Ψ={ Jack, Mary, Linda, I, my son, a book}={2,3,5,7,9,10}.

2. B (Ψ) represent according to mode get set Ψ={ in 2,3,5,7,9,10}, whole combinations of any two elements, establish set $C_{\mathrm{\&Psi;}}^2=\left\{\right\{\mathrm{2,3}\},\{\mathrm{2,5}\},\{\mathrm{2,7}\},\{\mathrm{2,9}\},\{\mathrm{2,10}\},\{\mathrm{3,5}\},\{\mathrm{3,7}\},\{\mathrm{3,9}\},\{\mathrm{3,10}\},$ $\left\{\mathrm{5,7}\right\},\left\{\mathrm{5,9}\right\},\left\{\mathrm{5,10}\right\},\left\{\mathrm{7,9}\right\},\left\{\mathrm{10,7}\right\},\left\{\mathrm{10,9}\right\}\}.$ B (Ψ)={ { 2,3}, { 2,5}, { 2,7}, { 2,9}, { 2,10}, { 3,5}, { 3,7}, { 3,9}, { 3,10}, { 5,7}, { 5,9}, { 5,10}, { 7,9}, { 10,7}, { 10,9}}.

3. the result of K (α, β) to function of a single variable B (Ψ), to appointing one that gives according to element the arrangement from small to large of the syntax sequence valve in former sentence S.Might as well establish can obtain ordered pair ? the ordered pair generating is:

{<2，3>，<2，5>，<2，7>，<2，9>，<2，10>，<3，5>，<3，7>，<3，9>，<3，10>，<5，7>，<5，9>，<5，10>，<7，9>，<7，10>，<9，10>}；

If set $\mathrm{\&pi;}(\&lang;{\mathrm{\&alpha;}}_i^t,{\mathrm{\&alpha;}}_j^t\&rang;)=\left\{{\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\right|\mathrm{\&tau;}\left({\mathrm{\&alpha;}}_i^t\right)<\mathrm{\&tau;}\left({\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\right)<\mathrm{\&tau;}\left({\mathrm{\&alpha;}}_j^t\right),{\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\&Element;S\},$ And then set up a continuous word string formula ${\mathrm{\&Phi;}}^t={\mathrm{\&alpha;}}_i^t<{\mathrm{\&beta;}}_k+<...+<{\mathrm{\&beta;}}_l+<{\mathrm{\&alpha;}}_j^t,$ Wherein be in former sentence S from arrive one group of adjacent continuous word string or empty word string, and $\mathrm{\&pi;}(\&lang;{\mathrm{\&alpha;}}_i^t,{\mathrm{\&alpha;}}_j^t\&rang;)=\{{\mathrm{\&beta;}}_k,...,{\mathrm{\&beta;}}_l\}.$ ? $K(\mathrm{\&alpha;},\mathrm{\&beta;})={\mathrm{\&Phi;}}^t={\mathrm{\&alpha;}}_i^t+<{\mathrm{\&beta;}}_k+<...$ $+<{\mathrm{\&beta;}}_l+<{\mathrm{\&alpha;}}_j^t.$ Φ ¹=2+ < e+ < 3, Φ ²=2+ < 3+ < 4+ < 5, Φ ³=2+ < 3+ < 4+ < 5+ < 6+ < 7, Φ ⁴=2+ < 3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9, Φ ⁵=2+ < 3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9+ < 10, Φ ⁶=3+ < 4+ < 5, Φ ⁷=3+ < 4+ < 5+ < 6+ < 7, Φ ⁸=3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9, Φ ⁹=3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9+ < 10, Φ ¹⁰=5+ < 6+ < 7, Φ ¹¹=5+ < 6+ < 7+ < 8+ < 9, Φ ¹²=5+ < 6+ < 7+ < 8+ < 9+ < 10, Φ ¹³=7+ < 8+ < 9, Φ ¹⁴=7+ < 8+ < 9+ < 10, Φ ¹⁵=9+ < e+ < 10.

4. H (Φ ^t) binary function K (α, β) is generated check: if to appoint to element γ ∈ Φ ^t, and and have: γ=NPI or γ=VNP or γ=NOMP or γ=CONJ or γ=e, by Φ ^tmark change into be called Φ ^tgenerate if set set ${\mathrm{\&Omega;}}_{\mathrm{NP}}=\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6,{\mathrm{\&Phi;}}_{\mathrm{NP}}^{15}\}.$ ? ${H(\mathrm{\&alpha;},\mathrm{\&beta;})=\mathrm{\&Omega;}}_{\mathrm{NP}}=\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6,{\mathrm{\&Phi;}}_{\mathrm{NP}}^{15}\}.$

5. M (α, β) represents a set of getting for appointing if set existence correspondence define a set family, this set family is by comprising set all set form, this set family is designated as $I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)=\left\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\right|\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\&SubsetEqual;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;{\mathrm{\&Omega;}}_{\mathrm{NP}}\}.$ ? $I_1\left(\right\{\mathrm{2,3}\left\}\right)=\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2\},$ $I_2\left(\right\{\mathrm{3,5}\left\}\right)=$ $\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6\},$ m (α, β)={ I ₁(2,3}), I ₂(3,5}), I ₃({ 9,10}) }.

6. the result of N (α, β) to binary function M (α, β) for appointing, get set if set there is corresponding set family construct a new set as follows $P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\left\{\mathrm{\&alpha;}\right|\mathrm{\&alpha;}\&Element;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},\mathrm{\&alpha;}\&NotEqual;e,{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\}.$ Can obtain P[I ₁(2,3})]={ 2,3,4,5}, P[I ₂(3,5})]={ 2,3,4,5}, P[I ₃(9,10})]={ 9,10}.

7. the result of U (α) to binary function N (α, β) get if $P^{\max }\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\mathrm{\&delta;},\mathrm{\&delta;}\&Element;P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ To appointing the element γ giving, $\mathrm{\&gamma;}\&Element;P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ There is τ (γ)≤τ (δ).P ^max[I ₁(2,3})]=5, P ^max[I ₂(3,5})]=5, P ^max[I ₃(9,10})]=10.

Generating object element set { z ₁, { z ₂process in, move object generating algorithm arranged side by side as follows:

1. A (S) takes out all NPI phrases in former sentence, all VNP phrases, all NOMP phrases, and classifies all NPI phrases in former sentence, all VNP phrases, all NOMP phrases as a set, and this set is designated as to Ψ={ Jack, Mary, Linda, I, my son, a book}={2,3,5,7,9,10}.

2. B (Ψ) represent according to mode get set Ψ={ in 2,3,5,7,9,10}, whole combinations of any two elements, establish set $C_{\mathrm{\&Psi;}}^2=\left\{\right\{\mathrm{2,3}\},\{\mathrm{2,5}\},\{\mathrm{2,7}\},\{\mathrm{2,9}\},\{\mathrm{2,10}\},\{\mathrm{3,5}\},\{\mathrm{3,7}\},\{\mathrm{3,9}\},\{\mathrm{3,10}\},$ $\left\{\mathrm{5,7}\right\},\left\{\mathrm{5,9}\right\},\left\{\mathrm{5,10}\right\},\left\{\mathrm{7,9}\right\},\left\{\mathrm{10,7}\right\},\left\{\mathrm{10,9}\right\}\}.$ B (Ψ)={ { 2,3}, { 2,5}, { 2,7}, { 2,9}, { 2,10}, { 3,5}, { 3,7}, { 3,9}, { 3,10}, { 5,7}, { 5,9}, { 5,10}, { 7,9}, { 10,7}, { 10,9}}.

3. the result of K (α, β) to function of a single variable B (Ψ), to appointing one that gives according to element the arrangement from small to large of the syntax sequence valve in former sentence S.Might as well establish can obtain ordered pair ? the ordered pair generating is:

{<2，3>，<2，5>，<2，7>，<2，9>，<2，10>，<3，5>，<3，7>，<3，9>，<3，10>，<5，7>，<5，9>，<5，10>，<7，9>，<7，10>，<9，10>}；

If set $\mathrm{\&pi;}(\&lang;{\mathrm{\&alpha;}}_i^t,{\mathrm{\&alpha;}}_j^t\&rang;)=\left\{{\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\right|\mathrm{\&tau;}\left({\mathrm{\&alpha;}}_i^t\right)<\mathrm{\&tau;}\left({\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\right)<\mathrm{\&tau;}\left({\mathrm{\&alpha;}}_j^t\right),{\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\&Element;S\},$ And then set up a continuous word string formula ${\mathrm{\&Phi;}}^t={\mathrm{\&alpha;}}_i^t<{\mathrm{\&beta;}}_k+<...+<{\mathrm{\&beta;}}_l+<{\mathrm{\&alpha;}}_j^t,$ Wherein be in former sentence S from arrive one group of adjacent continuous word string or empty word string, and $\mathrm{\&pi;}(\&lang;{\mathrm{\&alpha;}}_i^t,{\mathrm{\&alpha;}}_j^t\&rang;)=\{{\mathrm{\&beta;}}_k,...,{\mathrm{\&beta;}}_l\}.$ ? $K(\mathrm{\&alpha;},\mathrm{\&beta;})={\mathrm{\&Phi;}}^t={\mathrm{\&alpha;}}_i^t+<{\mathrm{\&beta;}}_k+<...$ $+<{\mathrm{\&beta;}}_l+<{\mathrm{\&alpha;}}_j^t.$ Φ ¹=2+ < e+ < 3, Φ ²=2+ < 3+ < 4+ < 5, Φ ³=2+ < 3+ < 4+ < 5+ < 6+ < 7, Φ ⁴=2+ < 3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9, Φ ⁵=2+ < 3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9+ < 10, Φ ⁶=3+ < 4+ < 5, Φ ⁷=3+ < 4+ < 5+ < 6+ < 7, Φ ⁸=3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9, Φ ⁹=3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9+ < 10, Φ ¹⁰=5+ < 6+ < 7, Φ ¹¹=5+ < 6+ < 7+ < 8+ < 9, Φ ¹²=5+ < 6+ < 7+ < 8+ < 9+ < 10, Φ ¹³=7+ < 8+ < 9, Φ ¹⁴=7+ < 8+ < 9+ < 10, Φ ¹⁵=9+ < e+ < 10.

4. H (Φ ^t) binary function K (α, β) is generated check: if to appoint to element γ ∈ Φ ^t, and and have: γ=NPI or γ=VNP or γ=NOMP or γ=CONJ or γ=e, by Φ ^tmark change into be called Φ ^tgenerate if set set ${\mathrm{\&Omega;}}_{\mathrm{NP}}=\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6,{\mathrm{\&Phi;}}_{\mathrm{NP}}^{15}\}.$ ? ${H(\mathrm{\&alpha;},\mathrm{\&beta;})=\mathrm{\&Omega;}}_{\mathrm{NP}}=\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6,{\mathrm{\&Phi;}}_{\mathrm{NP}}^{15}\}.$

5. M (α, β) represents a set of getting for appointing if set existence correspondence define a set family, this set family is by comprising set all set form, this set family is designated as $I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)=\left\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\right|\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\&SubsetEqual;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;{\mathrm{\&Omega;}}_{\mathrm{NP}}\}.$ ? $I_1\left(\right\{\mathrm{2,3}\left\}\right)=\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2\},$ $I_2\left(\right\{\mathrm{3,5}\left\}\right)=$ $\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6\},$ m (α, β)={ I ₁(2,3}), I ₂(3,5}), I ₃({ 9,10}) }.

6. the result of N (α, β) to binary function M (α, β) for appointing, get set if set there is corresponding set family construct a new set as follows $P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\left\{\mathrm{\&alpha;}\right|\mathrm{\&alpha;}\&Element;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},\mathrm{\&alpha;}\&NotEqual;e,{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\}.$ Can obtain P[I ₁(2,3})]={ 2,3,4,5}, P[I ₂(3,5})]={ 2,3,4,5}, P[I ₃(9,10})]={ 9,10}.

7. V (β) represents the result to binary function N (α, β) get if $P^{\min }\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\mathrm{\&delta;},\mathrm{\&delta;}\&Element;P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ To appointing the element γ giving, $\mathrm{\&gamma;}\&Element;P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ There is τ (δ)≤τ (γ).P ^min[I ₁(2,3})]=2, P ^min[i ₂(3,5})]=2, P ^min[I ₃(9,10})]=9.

B3.3 part { y _kgeneration method

[B3.3.1] preparation work: be defined as follows subclass:

1)NPI _yk＝{NPI|NPI＜ _□r _k}。

2)VNP _yk＝{VNP|VNP＜ _□r _k}。

3)NOMP _k＝{NOMP|NOMP＜ _□r _k}。

4)

Wherein:

5)ry _k＝{r _α|α＜k，α∈N}。(N is nature manifold)

6)fy _k＝{f _α|α＜k，α∈N}。(N is nature manifold)

[B3.3.2] { y _kgenerating algorithm

2. convert to: when there being r _k-1time: { y _k}=NPI _yk∪ VNP _yk∪ NOMP _k∪ G _k∪ fy _k∪ e}, above formula is converted into:

B3.4 part { z _kgeneration method

[B3.4.1] preparation work: be defined as follows subclass:

1)

2)

3)

4)

Wherein:

5)rz _k＝{r _α|k＜α，α∈N}。(N is nature manifold)

6)fz _k＝{f _α|k＜α，α∈N}。(N is nature manifold)

[B3.4.2] { z _kgenerating algorithm

2. convert to: when there being r _k+1time: { z _k}=NPI _zk∪ VNP _zk∪ OBJP _k∪ H _k∪ fz _k∪ e}, above formula is converted into:

B3.5 part matrix expression formula and linear representation

[B3.5.1] matrix expression

And then statement S can express with matrix form, that is:

When a function f _jserve as another function f _ksubject element or for example, during object element: work as f _k=x+ < y+ < r+ < f _jor f _k=x+ < f _jduring+< r+ < y, claim f _kto obtain through compound operation.Compound operation is designated as f (f) in the present invention.

Due to function f, see on the whole it is also word unit, so partially add computing, be applicable to function.If function f _i, f _jmeet f _i< _□f _j, and another function f _kcan be expressed as f _iand f _jpartially to add be f _k=f _i+ < f _j, claim f _kthrough partially adding computing, obtain.

Each English statement S that does not omit predicate verb can regard as by n function f ₁..., f _n(n equals predicate verb element number) passes through the compound of limited number of time and partially adds computing and obtain.Accordingly, the English statement S that any one can not omitted to predicate is designated as:

$S={\mathrm{\&Phi;}}_{+<}^{f\left(f\right)}(f_1,f_2,......,f_n).$

Also, any one English statement that does not omit predicate by the vector that comprises leading question element, subject element, predicate element or object element through compound or partially add computing and obtain.Next, just face the problem of choosing a kind of reasonable expression formula for English natural sentences S.This expression formula, must show rightly to comprise in S all are compound and partially add computing.Matrix form possesses such condition just, and it can embody the compound operation of function with the position of element in a certain row vector, for example: f _k(f _j)=f _k(x _k, f _j, r _k, z _k), just show f _kwith f _jcompound operation relation between the two; Meanwhile, do not destroy again the relation that partially adds between element: f _k=x _k+ < f _j+ < r _k+ < z _k.To sum up, in order accurately, intuitively, clearly to express English natural sentences S, in order to disclose better the inherent mathematical and physical structure of natural sentences S, we adopt matrix as the primary expression formula of natural sentences S.

[B3.5.2] linear representation

Meanwhile, can also utilize linear forms to express statement S, that is:

Lay special stress on:

What 1. each linear representation that does not omit the English natural sentences S of predicate had comprised limited number of time adds computing and compound operation partially.Adopt linear representation as the supplementary expression formula of natural sentences S herein.

2. between matrix expression of the present invention and linear representation, be relation of equivalence.

3. the linear representation of an English natural sentences S, be natively also simultaneously one with function f ₁..., f _n(n equals predicate verb element number) is the system of linear equations of unknown quantity, and therefore, ensuing process of trying to achieve syntactic structure analysis result by Substitution method, also naturally can be regarded as and solve this with function f herein ₁..., f _nthe process of the system of linear equations that (n equals predicate verb element number) is unknown quantity.

The substitution solver of B3.6 part matrix

The 1st step

If there is the sequence valve not occurring in this syntactic structure possibility matrix solution, get rid of this syntactic structure possibility matrix solution;

For example, for following possible matrix solution

Be numbered 4 not appearance of word unit, get rid of.

The 2nd step

If occur identical sequence valve or occur identical syntax vector in different syntax vectors, get rid of this syntactic structure possibility matrix solution;

For example, for following possible matrix solution

Be numbered 5 word unit and occurred twice, get rid of.

The 3rd step

In each may matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there is the intersection contradiction of two syntax vectors equivalent substitution after, get rid of the possible matrix solution of this syntactic structure;

For example, for following possible matrix solution

Above-mentioned matrix is carried out to substitution, f ₂and f ₃the substitution intersection contradiction that has occurred function.Substitution obtains: f ₂=3+ < e+ < 6+ < (4+ < f ₂+ < 7+ < e).There is f in two ends, equation left and right simultaneously ₂, this just occurred () logical contradiction.Get rid of.

The 4th step

In each may matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there are two sequence valves that position is converse equivalent substitution after, get rid of the possible matrix solution of this syntactic structure; This is both fundamental requirement of mathematics manipulation, is to be also defined in strict partial ordering relation < _□on the essential requirement of inclined to one side additive operation.

For example, for following possible matrix solution

It is carried out to substitution, f ₂=4+ < 5+ < 6+ < 3+ < e+ < 7+ < e, obtains order for (4,5,6,3, e, 7, e), there is the sequence valve that position is converse, get rid of.

The 5th step

In any one possibility matrix solution, if if there is no the syntax vector of mutual substitution relation between existence and other syntax vectors, carry out plug hole and operate to obtain the corresponding possibility of all these possibility matrix solutions syntax analytic structure, and whether the statement that checking obtains according to described possibility syntax analytic structure is identical with described pretreated statement, it further comprises:

5.5.1, first to existing each other the syntax vector of substitution relation to carry out equivalent substitution in this possibility matrix solution, thereby this possibility matrix solution is converted into one group of syntax vector that does not have each other a substitution relation syntax vector in possibility matrix solution is called to first kind syntax vector, the syntax that conversion is obtained vector be called Equations of The Second Kind syntax vector;

5.5.2, appoint and get an Equations of The Second Kind syntax vector according to predetermined direction, mark one by one in the sequence valve of each syntax elements; After the sequence valve of mark syntax elements, appoint and get in i syntax elements, only in the first side of this syntax elements, construct unique room; After making sky, appoint and get a syntax vector equations of The Second Kind syntax vector in addition mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, this new syntax vector is designated as and the syntax that whole plug hole is obtained vector, be referred to as the 3rd class syntax vector;

5.5.3, to the 3rd class syntax vector according to predetermined direction to from vector in first syntax elements of the first side start to vector in the vector that comprises first syntax elements of the second side till each syntax elements, all mark sequence valve; Be positioned at vector in the vector that comprises the element of the first side, does not mark sequence valve; By vector first syntax elements of the second side be designated as will be according to aforementioned manner to vector the syntax vector part of mark, is designated as whipping syntax vector after mark sequence valve, appoint and get a j syntax elements in aforesaid whipping vector, only in this element first side, construct unique room; After making sky, appoint and get an original Equations of The Second Kind syntax vector mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, newly-generated syntax vector is designated as or

The 3rd class syntax vector according to predetermined direction, distich normal vector in each syntax elements all mark sequence valve; After the sequence valve of mark syntax elements, appoint and get one in t syntax elements, in the first side of this syntax elements, construct unique room; After making sky, appoint and get use Equations of The Second Kind syntax vector in the mode of whole plug hole by this vector insert the room of constructing above, and then generate a new vector, this new vector is designated as

5.5.4, repeat 5.5.3, when the last time makes empty and plug hole step and finishes, the empty and plug hole made carrying out is next time operated, until all Equations of The Second Kind syntaxes are vectorial through last the 3rd class syntax vector of making empty and plug hole step and obtaining all plug hole is complete, finally obtains the 3rd class syntax vector of a single file, and described the 3rd class syntax vector finally obtaining is called to final single file vector;

If 5.5.5 one may syntax all exists two sequence valves that position is converse in all described final single file vector corresponding to analytic structure, get rid of this possibility syntax analytic structure;

5.5.6, repeat 5.5.2 to 5.5.5 until all may being traversed by syntax analytic structure.

B3.7 part matrix revision program

If desired, proceed to revision program, to plural syntactic structure analysis result is revised, specifically comprise following operation:

(1) noun pronoun unit serves as heavily inspection and the choice of subject and object.

(2) use language rule to check syntactic structure.For example:

1. according to English syntactic structure rule, the introducer of subject clause can not omit.

That of guiding subject clause can not omit;

2. according to English syntactic structure rule, subject it is called and quantitatively will be consistent with predicate;

3. according to verb and thing and not as good as properties, judge and whether connect thereafter object.

(3) structural ambiguity reexamines inspection and gets rid of.

(4) upside-down mounting, omission, there be treat as special case.

(5) composition of extraction is put back to.

(6) generate and export final solution.

By revising, can overcome the nonstandard problem raising of part sentence structure parsing accuracy.

Preferably, can syntactic structure be formed to syntax tree data structure according to analysis result.

The special explanation of B3.8 part to two kinds of plug hole methods

The common principle of [B3.8.1] two kinds of different plug hole methods:

The sequence valve ordered series of numbers 1 of former sentence, 2, ..., k, can be regarded as through the equivalent substitution of the syntax vector of finding clear and definite position in possibility matrix solution and obtains through the whole plug hole of the limited number of time between the syntax vector that can not find clear and definite position in possibility matrix solution.That is, the initial syntax vector corresponding with former sentence can be regarded as first the equivalent substitution through the syntax vector of finding clear and definite position in may matrix solution, then obtain through the whole plug hole of limited number of time between the syntax vector that can not find clear and definite position in may matrix solution.Various not identical plug hole situations are exactly arrangement and the combination in combinatorics in essence.

[B3.8.2] the 1st kind of plug hole method:

In any one possibility matrix solution, if all there is no the syntax vector of clear and definite substitution relation between existence and other any syntax vectors, first in this possibility matrix solution and exist the syntax vector of substitution relation all to carry out equivalent substitution between other syntax vectors, with season this possibility matrix solution in and between other syntax vectors, do not exist the syntax vector of substitution relation all to remain unchanged, comprehensive aforementioned two aspects, are converted into one group of syntax vector that does not all have each other substitution relation by this possibility matrix solution original syntax vector f in possible matrix solution ₁, f ₂..., f _δbe referred to as first kind syntax vector; After aforesaid equivalent substitution, by the one group of syntax vector that does not have each other substitution relation that transforms out according to aforementioned manner be referred to as Equations of The Second Kind syntax vector; Emphasize, Equations of The Second Kind syntax vector is all the syntax vector that does not have each other substitution relation.When θ >=2, whole plug hole is meaningful; Following discussion is default θ >=2 all.

Next, carry out the whole plug hole of one-sided order-preserving in the same way, also can be called the whole plug hole of one-sided forward order-preserving: appoint and get an Equations of The Second Kind syntax vector according to direction from right to left (also can from left to right), mark one by one syntax vector in the sequence valve of each syntax elements.After the sequence valve of mark syntax elements, appoint and get one in syntax elements, might as well establish this syntax elements and be in several i the elements in the right, only on the right side of this syntax elements, (also can only in left side) constructs unique room; After making sky, appoint and get one except syntax vector equations of The Second Kind syntax vector in addition mode with whole plug hole is vectorial by syntax insert the room of constructing above, and then generate a new syntax vector, this new vector is designated as the whole plug hole of every process and the syntax vector that obtains is referred to as the 3rd class syntax vector, it is the 3rd class syntax vector; For appointing the vectorial α of two syntaxes giving and β, if vectorial β is inserted to several i corresponding rooms of syntax elements, the right of syntax vector α in the mode of whole plug hole, obtained a 3rd new class syntax vector, the 3rd class syntax vector of this new acquisition has been designated as to [α] ⁱ+ < β; Emphasize, the 3rd class syntax vector is all the syntax vector that does not have each other substitution relation.Make empty and plug hole step is complete for the 1st.

Proceed to the 2nd and make empty and plug hole step.Vectorial to the 3rd class syntax of making empty and plug hole step and obtaining through the 1st according to direction from right to left (also can be from left to right, but will to mark the direction that order chooses identical with the last time, with last time mark order in the same side), to from vector in the right count first syntax elements and start to vector in the vector that comprises the left side count each syntax elements till first syntax elements, all mark sequence valve; Be positioned at vector in the vector that comprises the syntax elements in left side, does not mark sequence valve; By vector the left side count first syntax elements and be designated as will be according to aforementioned manner to vector the syntax vector part of middle mark order, is designated as: syntax vector claim this syntax vector to be: whipping vector.After mark sequence valve, appoint and get an element in aforesaid whipping vector, might as well establish this element is whipping vector in several j the elements in the right, only on this element right side, (also can be only in left side, but will to make the direction that sky chooses identical with the last time, make empty in the same side with the last time), constructs unique room; After making sky, appoint and get one except used syntax vector in the 1st is made empty and plug hole step with equations of The Second Kind syntax vector in addition mode with whole plug hole is vectorial by syntax insert the room of constructing above, and then generate a new syntax vector, newly-generated syntax vector is designated as for appointing syntax vector α and the β giving, first syntax elements is counted in the left side of vectorial β and be designated as λ (β), if there is syntax vector [α] ⁱ+ < β, according to aforementioned manner to vector [α] ⁱ+ < β marks, and by the syntax vector part according to aforementioned manner mark, is designated as: vector [α _kλ (β _k-1)], claim this vector to be: whipping vector.Make empty and plug hole step is complete for the 2nd.

According to preceding method, the 3rd class syntax vector obtaining making sky and plug hole step through the last time is chosen whipping vector, and according to aforesaid method to selected whipping vector mark sequence valve, but will to mark the direction that order chooses identical with the last time, with last time mark order in the same side; After mark sequence valve, appoint and to get a syntax elements in this whipping vector, according to preceding method, construct unique one-sided room, but will to make the direction that sky chooses identical with the last time, make empty in the same side with the last time; After making sky, appoint get one except previous make empty and plug hole step in Equations of The Second Kind syntax vector used syntax vector, the room of this Equations of The Second Kind syntax vector insertion being constructed above in the mode of whole plug hole, and then new syntax of generation is vectorial; Repeat aforesaid operation: when the last time makes empty and plug hole step and finishes, all according to aforesaid method, the 3rd class syntax vector obtaining making sky and plug hole step through the last time carries out the empty and plug hole operation of making next time, until by Equations of The Second Kind syntax vector all plug hole is complete, finally obtains the 3rd class syntax vector of a single file; The 3rd class syntax vector finally obtaining is called to final single file vector.

By aforesaid vectorial from choosing first Equations of The Second Kind syntax to an entire flow that generates final single file vector as a concrete scheme, thereby the aforesaid namely step in concrete scheme of empty and plug hole step of making each time.

By whole possibility situations of exhaustive each step, exhaustive whole schemes.Check each final single file vector of exhaustive generation: the final single file vector of deleting the converse sequence valve in two positions of appearance.

If there are two sequence valves that position is converse in each final single file vector of exhaustive generation, violate the rules of nature, get rid of all final single file vector, and then get rid of this syntactic structure possibility matrix solution.

Every final single file vector that does not occur the sequence valve that two positions are converse, all meets the natural law, is all reasonably final single file vector; Retain reasonably final single file vector as one of correct result, and retain this syntactic structure possibility matrix solution as one of correct result, in order to generating the use of syntax tree.

[B3.8.3] the 2nd kind of plug hole method:

In any one possibility matrix solution, if all there is no the syntax vector of clear and definite substitution relation between existence and other any syntax vectors, first in this possibility matrix solution and exist the syntax vector of substitution relation all to carry out equivalent substitution between other syntax vectors, with season this possibility matrix solution in and between other syntax vectors, do not exist the syntax vector of substitution relation all to remain unchanged, comprehensive aforementioned two aspects, are converted into one group of syntax vector that does not all have each other substitution relation by this possibility matrix solution by original those syntax vector f in this possibility matrix solution ₁, f ₂..., f _δbe referred to as first kind syntax vector; After aforesaid equivalent substitution, by the one group of syntax vector that does not have each other substitution relation that transforms out according to aforementioned manner be referred to as Equations of The Second Kind syntax vector; Emphasize, Equations of The Second Kind syntax vector is all the syntax vector that does not have each other substitution relation.When θ >=2, whole plug hole is meaningful; Following discussion is default θ >=2 all.

Next, carry out the whole plug hole of one-sided not order-preserving in the same way, also can be called the whole plug hole of one-sided forward not order-preserving: appoint and get an Equations of The Second Kind syntax vector according to direction from right to left (also can from left to right) distich normal vector in each syntax elements mark one by one sequence valve.After the sequence valve of mark syntax elements, appoint and get one in syntax elements, might as well establish this syntax elements and be in several m the elements in the right, only on the right side of this syntax elements, (also can only in left side) constructs unique room; After making sky, appoint and get one except syntax vector equations of The Second Kind syntax vector in addition in the mode of whole plug hole by vector insert the room of constructing above, and then generate a new syntax vector, newly-generated syntax vector is designated as the whole plug hole of every process and the syntax vector that obtains is referred to as the 3rd class syntax vector; For appointing the vectorial α of two syntaxes giving and β, if vectorial β is inserted to several m corresponding rooms of syntax elements, the right of vectorial α in the mode of whole plug hole, obtained a 3rd new class syntax vector, the 3rd class syntax vector of this new acquisition has been designated as to (α) ^m+ < β.Make empty and plug hole step is complete for the 1st.

Proceed to the 2nd and make empty and plug hole step.Vectorial for the 3rd class syntax of making empty and plug hole step and obtaining through the 1st according to direction from right to left (also can be from left to right, but will to mark the direction that order chooses identical with the last time, with last time mark order in the same side), distich normal vector in each syntax elements all mark sequence valve; After the sequence valve of mark syntax elements, appoint and get one in syntax elements, might as well establish this element and be in several t the syntax elements in the right, only on this syntax elements right side, (also can be only in left side, but will to make the direction that sky chooses identical with the last time, make empty in the same side with the last time), constructs unique room; After making sky, appoint and get one except used syntax vector in the 1st is made empty and plug hole step with equations of The Second Kind syntax vector in addition in the mode of whole plug hole by this vector insert the room of constructing above, and then generate a new vector, this new vector is designated as make empty and plug hole step is complete for the 2nd.

According to aforesaid method, to through last the 3rd class syntax vector mark sequence valve of making empty and plug hole step and obtaining, but will to mark the direction that order chooses identical with the last time, mark order in the same side with the last time; After mark sequence valve, appoint and to get a syntax elements in the 3rd class syntax vector, according to the unique one-sided room of aforesaid method construct, but will to make the direction that sky chooses identical with the last time, make empty in the same side with the last time; After making sky, appoint get one except previous make empty and plug hole step in Equations of The Second Kind syntax vector used syntax vector, the room of this Equations of The Second Kind syntax vector insertion being constructed above in the mode of whole plug hole, and then new syntax of generation is vectorial; Repeat aforesaid operation: when the last time makes empty and plug hole step and finishes, all according to aforesaid method, the 3rd class syntax vector obtaining making sky and plug hole step through the last time carries out the empty and plug hole operation of making next time, until by Equations of The Second Kind syntax vector all plug hole is complete, finally obtains the 3rd class syntax vector of a single file; The 3rd class syntax vector finally obtaining is called to final single file vector.

By aforesaid vectorial from choosing first Equations of The Second Kind syntax to an entire flow that generates final single file vector as a concrete scheme, thereby the aforesaid namely step in concrete scheme of empty and plug hole step of making each time.

By whole possibility situations of exhaustive each step, exhaustive whole schemes.Check each final single file vector of exhaustive generation: under the prerequisite of the e on the diverse location in distinguishing possibility matrix solution, two or more identical final single file vectors are retained to one, the identical final single file of Delete superfluous is vectorial, and then deletes the final single file vector of the converse sequence valve in two positions of appearance.

After the final single file vector duplicating of Delete superfluous, if two sequence valves that position is converse appear in each final single file vector, violate the rules of nature, get rid of all final single files vector, and then get rid of this syntactic structure possibility matrix solution.

After the identical final single file vector of Delete superfluous, every final single file vector that does not occur the sequence valve that two positions are converse, all meets the natural law, is all reasonably final single file vector; Retain reasonably final single file vector as one of correct result, and retain this syntactic structure possibility matrix solution as one of correct result, in order to generating the use of syntax tree.

[B3.8.4] is to the Features Summarization of two kinds of methods and comparison:

Can prove by mathematical method: the whole schemes in aforementioned two methods and Overall Steps are all limited, fixing, provable, and can provide the number computing formula of whole concrete schemes and Overall Steps, can also provide by the number computing formula of whole final single file vectors of exhaustive generation.Aforementioned two methods have all been constructed corresponding mapping and have entirely been arranged set and corresponding recursive function as its mathematical model.Each link in aforementioned two methods has strict Mathematics Proof and tight mathematic(al) argument.Aforementioned two methods are the methods that meet the natural law completely.

Aforesaid two methods are all launched for following principle, are all the specific embodiments of following principle:

The sequence valve ordered series of numbers 1 of former sentence, 2, ..., k, can be regarded as through the equivalent substitution of the syntax vector of finding clear and definite position in possibility matrix solution and obtains through the whole plug hole of the limited number of time between the syntax vector that can not find clear and definite position in possibility matrix solution.That is, the initial syntax vector corresponding with former sentence can be regarded as first the equivalent substitution through the syntax vector of finding clear and definite position in may matrix solution, then obtain through the whole plug hole of limited number of time between the syntax vector that can not find clear and definite position in may matrix solution.Various not identical plug hole situations are exactly arrangement and the combination in combinatorics in essence.

Aforesaid two methods have all met the requirement of mentioned above principle, and the net result of two methods is on all four.As can be seen here, aforesaid two methods are equivalent methods.

Choosing for making aspect empty syntax elements: aforesaid the 1st kind of method is conditional for the selection of making sky element, is actually the sequencing that requires to keep plug hole syntax vector; Aforesaid the 2nd kind of method is hard-core for the selection of making sky element, is actually and do not require the sequencing that keeps plug hole syntax vector.

Aspect final single file vector: distinguishing under the prerequisite of the e on the diverse location in possibility matrix solution, the final single file vector that aforementioned the 1st kind of method generates is all syntax vector different between two, and duplicating may appear in the final single file vector that aforesaid the 2nd kind of method generates, therefore want the identical final single file vector of Delete superfluous.

Below, will respectively aforementioned two kinds of methods be illustrated herein.

[B3.8.5] illustrates the 1st kind of plug hole method

[B3.8.5.1] structure is as the mapping of scheme mode

Note: coordinate noun pronoun mix vector and conjunctive word mix vector are all regarded an integral body as, can not be by the whole plug hole of other syntax vectors.

The sequence valve ordered series of numbers 1 of former sentence, 2, ..., k, can be regarded as through the equivalent substitution of the syntax vector of finding clear and definite position in possibility matrix solution and obtains through the whole plug hole of the limited number of time between the syntax vector that can not find clear and definite position in possibility matrix solution.That is, the initial syntax vector corresponding with former sentence can be regarded as first the equivalent substitution through the syntax vector of finding clear and definite position in may matrix solution, then obtain through the whole plug hole of limited number of time between the syntax vector that can not find clear and definite position in may matrix solution.Various not identical plug hole situations are exactly arrangement and the combination in combinatorics in essence.

Introduce in detail the whole plug hole method of aforesaid one-sided order-preserving below.The method can accurately depict each situation of the whole plug hole of limited number of time between the syntax vector that can not find clear and definite position in possibility matrix solution.

By former, may be converted into θ new syntax vector that can not find clear and definite position by matrix solution be designated as: this θ syntax vector arranged entirely, and according to the relative theory of combinatorics, so full rank results is individual; Pass through so full arrangement, obtain altogether θ! The orderly group of individual θ unit.By the θ obtaining through so full arrangement! The set that the orderly group of individual θ unit forms is designated as (θ>=2)

Structure j is not identical mapping ρ j of θ unit between two, j ∈ N, 1≤j≤θ! Making each mapping ρ j of θ unit is from set { t ₁, t ₂..., t _θto set mapping; Set { t ₁, t ₂..., t _θonly be used for representing the field of definition of mapping ρ j, and then the θ of auxiliary calibration set Φ unit is complete arranges, and there is no other physical meaning.Structure mapping is as follows: j ∈ N, a 1≤j≤θ! ; To appointing the j giving ₁and j ₂, j ₁∈ N, j ₂∈ N, 1≤j ₁≤ θ! , 1≤j ₂≤ θ! If, j ₁≠ j ₂, ρ j ₁≠ ρ j ₂.(θ≥2)

For the mapping ρ j of θ unit, have following conclusion to set up: obviously, to any one ρ j (t _k), all exist 1≤δ≤θ, makes be any one ρ j (t _k) all demarcated set in a syntax vector (θ>=2)

Known according to aforesaid structure: by θ! The finite aggregate Ω that the individual θ mapping ρ j of unit forms=and ρ 1, and ρ 2, and ρ 3, and ρ 4, and ρ 5, and ρ 6 ..., a ρ θ! Just portrayed finite aggregate set full arrangement of θ unit.(θ≥2)

Set π=and ρ 1, and ρ 2, and ρ 3, and ρ 4, and ρ 5, and ρ 6 ..., a ρ θ! Tabulate: (θ >=2)

Definition 1.1: will be aforesaid from choosing Equations of The Second Kind syntax vector at one to any one that generates whole θ Equations of The Second Kind syntax vectors using in the complete operation flow process of final single file vector, sequentially arrange, as a scheme mode.Lifting an example describes in detail: if as scheme mode, represent that the 1st step is to choose vector with and by vector mode with the whole plug hole of one-sided order-preserving is inserted vector the 2nd step is to choose and will in the mode of the whole plug hole of one-sided order-preserving, insert in the new syntax vector of the 1st step generation ..., θ step is to choose vector and will in the mode of the whole plug hole of one-sided order-preserving, insert in the new syntax vector of (θ-1) individual step generation.Obviously visible, any one mapping ρ j of θ unit is a scheme mode, and all set of the mapping ρ of θ unit j be exactly all set of scheme mode, all the sum of scheme mode is θ! Individual.(θ≥2)

Definition 1.2: by the operation of any plug hole carrying out according to any one scheme mode the new vector of generation, as a step of this scheme mode.To any one scheme mode ρ j, k the step of carrying out according to ρ j is designated as [n _k] represent that k step always has n _kplant alternative situation.

Definition 1.3: each step of carrying out according to any one scheme mode ρ j is chosen to any one concrete condition, then each step is all joined together, as a concrete scheme.Any one concrete scheme is designated as ρ j represents the scheme mode of this concrete scheme institute foundation, i _krepresent that k step choose the i in this step _kthe situation of kind.

[B3.8.5.2] structure plug hole recursive function

Next, to, according to any one scheme mode ρ j, construct a plug hole recursive algorithm herein by this recursive algorithm, just can portray the specific operation process of the whole plug hole of aforesaid one-sided order-preserving each time.Before structure plug hole recursive algorithm, following 5 definition of given first, as pre-knowledge:

The plug hole recursive algorithm that will construct below be exactly aforesaid k the step of carrying out according to scheme mode ρ j.K wherein represents plug hole recursive algorithm the number of times of operation, carries out the number of times that the whole plug hole of aforesaid one-sided order-preserving operates.

Definition 1.4: appoint to a syntax vector α, function of a single variable W represents to take out and mark-up syntax vector α.W (α)=α _krepresent to take out syntax vector α, and syntax vector α is labeled as to α _k, claim α _kfor: input vector.

Definition 1.5: appoint to a syntax vector β, function of a single variable Q represents to take out and mark-up syntax vector β.Q (β)=β _krepresent to take out syntax vector β, and syntax vector β is labeled as to β _k, claim β _kfor: plug hole vector.At operation plug hole recursive algorithm in process, be by syntax vector β _kinsert syntax vector α _kin.

Definition 1.6: binary function Z represents distich normal vector α _kmark sequence valve, by syntax vector α _kin from right several the 1st syntax elements mark sequence valve 1, then mark successively from right to left sequence valve 2,3 ..., until mark is to vectorial α _kin the vectorial β that comprises _k-1in till left the 1st syntax elements several.By vectorial β _k-1in from left the 1st syntax elements several, be designated as λ (β _k-1), by λ (β _k-1) sequence valve be designated as n _k, n _kit is the maximum sequence valve of aforementioned mark.This process portray for: establish syntax vector α _k=b...... λ (β _k-1) ... b ₂b ₁, element λ (β _k-1) represent vectorial β _k-1in from left the 1st element several.In k step, to α _koperation binary function Z _k(α, β): λ (β _k-1) <n _k>......b ₂<2>b ₁<1>, is designated as this result: vector [α _kλ (β _k-1)], claim this vector to be: whipping vector.Mark represent: to whipping vector [α _kλ (β _k-1)] mark maximum sequence valve be n _k.Operation binary function Z obtains $Z_k(\mathrm{\&alpha;},\mathrm{\&beta;})=[{\mathrm{\&alpha;}}_k\backslash$ $\mathrm{\&lambda;}\left({\mathrm{\&beta;}}_{k-1}\right){]}^{n_k}.$

Definition 1.7: binary function T represents whipping vector to carry out whole plug hole operation after binary function Z operation is complete, is chosen from the i of the right number on whipping vector _kindividual syntax elements, and at i _kunique room is constructed on the right side of individual syntax elements, then by vectorial β _kin the mode of whole plug hole, insert this room.This process portray for: establish syntax vector α _k=b...... λ (β _k-1) ... b ₂b ₁, element λ (β _k-1) represent vectorial β _k-1in from left several the 1st element.Whipping vector is [α _kλ (β _k-1)]=λ (β _k-1) <n _k>......b ₂<2>b ₁<1>.In k step, to vector [α _kλ (β _k-1)] and vectorial β _koperation binary function T _k(α, β), by β _kmode with whole plug hole is inserted [α _kλ (β _k-1)] the several i in the right _kthe corresponding room of individual syntax elements obtains: $......b_{i_k}<{\mathrm{\&beta;}}_k>b_{i_{(k-1)}}......b_2<2>b_1<1>.$ The new vector obtaining through aforesaid plug hole is designated as ${[{\mathrm{\&alpha;}}_k\backslash \mathrm{\&lambda;}\left({\mathrm{\&beta;}}_{k-1}\right)]}^{i_k}+<{\mathrm{\&beta;}}_k.$ Claim vector ${[{\mathrm{\&alpha;}}_k\backslash \mathrm{\&lambda;}\left({\mathrm{\&beta;}}_{k-1}\right)]}^{i_k}+<{\mathrm{\&beta;}}_k.$ For: output vector. $({[{\mathrm{\&alpha;}}_k\backslash \mathrm{\&lambda;}\left({\mathrm{\&beta;}}_{k-1}\right)]}^{i_k}+<{\mathrm{\&beta;}}_k=.......b_{i_k}<{\mathrm{\&beta;}}_k>b_{i_{(k-1)}}......b_2<2>b_1<1>)$

Definition 1.8: appoint to a syntax vector α, the number of the syntax elements comprising in α is designated as to σ [α].If comprise n syntax elements in syntax vector α, n ∈ N, obviously has: n=σ [α].

Note: in the definition of plug hole recursive algorithm below, can see such equation:

$W_k\left(\mathrm{\&alpha;}\right)=T_{k-1}(\mathrm{\&alpha;},\mathrm{\&beta;});T_k(\mathrm{\&alpha;},\mathrm{\&beta;})={[{\mathrm{\&alpha;}}_k\backslash \mathrm{\&lambda;}\left({\mathrm{\&beta;}}_{k-1}\right)]}^{i_k}+<{\mathrm{\&beta;}}_k.$

α wherein and β are the implications of independent variable, are abstract marks, can wide in range value.Therefore, above-mentioned notation contradiction not.

Below, according to the aforesaid mapping ρ j of this paper and 4 functions, definition plug hole recursive algorithm as follows:

Note: appoint and get a scheme mode ρ j; Set

Lay special stress on: when k=1, plug hole recursive algorithm starting condition be:

Recursive algorithm feature be: the whole plug hole operation of aforesaid one-sided order-preserving has been resolved into four processes: 1. get and make blank vector; 2. get plug hole vector; 3. to making specific syntax elements in blank vector, mark sequence valve, and intercept whipping vector; 4. choose arbitrarily a syntax elements in whipping vector, and construct unique room on its right side, the room of structure before then plug hole vector being inserted in the mode of whole plug hole.

[B3.8.5.3] plug hole recursive algorithm operation for example

(distinguishing the e on diverse location)

Order set θ=4, θ unequal to 24, ρ j=ρ 2.

Order ?

Make n _εthe number of the syntax elements in the whipping vector chosen of ε step, i _εit is the order of elements number that the double-void of ε step structure is answered.

The pattern that carries into execution a plan ρ 2, will be by plug hole recursive algorithm move 3 times.Choose a concrete scheme ${\{W~Q~Z~T\}}_{\mathrm{\&rho;}2}^k<\mathrm{3,4,2}>:$

move as follows: (n ₁∈ N, i ₁∈ N, 1≤i ₁≤ n ₁)

N is easy to get ₁=4; Get i ₁=3, obtain: T ₁(α, β)=(7e _c(e _a5 6e _b) 8e _d)

move as follows: (n ₂∈ N, i ₂∈ N, 1≤i ₂≤ n ₂)

N is easy to get ₂=6; Get i ₂=4, obtain:

T ₂(α，β)＝(7e _C(e _A?5?6(1?2?3?4)e _B)8e _D)

move as follows: (n ₃∈ N, i ₃∈ N, 1≤i ₃≤ n ₃)

N is easy to get ₃=7; Get i ₃=2, obtain:

T ₃(α，β)＝(7e _C(e _A?5?6(1?2?3?4)e _B)8(e _E9?10e _F)e _D)

[B3.8.5.4] explanation to this method exhaustive and heterogeneite

Conclusion 1.1: the full arrangement set of the first mapping of aforesaid θ with aforesaid plug hole recursive algorithm limit the whole of the whole plug hole of limited number of time between the aforesaid syntax vector that can not find clear and definite position may.

Conclusion 1.2: according to aforesaid plug hole recursive algorithm definition known, in any two concrete schemes of this paper method, all there are respectively mutually different two steps, any two concrete schemes are not identical.Therefore, distinguishing under the prerequisite of the e on diverse location, the whole final single file vector that whole concrete schemes of this paper method generate is all syntax vector different between two.

[B3.8.5.5] first group of computing formula and relevant proof

Note: coordinate noun pronoun mix vector and conjunctive word mix vector are all regarded an integral body as, can not be by the whole plug hole of other syntax vectors.

Note: following lemma and theorem are all at given any one scheme mode ρ j and plug hole recursive algorithm in this paper method prerequisite under set off a discussion; All to set off a discussion under the prerequisite of the e on the diverse location of distinguishing in may matrix solution.

Definition 1.9: operation plug hole recurrence is calculated the number that is located at the whipping vector intercepting in k step is n, and might as well be located at n the whipping vector intercepting in k step is α ₁, α ₂..., α _n, in k step, each in these vectors is carried out to whole plug hole, the number sum of these vectorial plug hole situations is designated as to τ [∑ (k)].

The definition of lemma 1.1:(σ, referring to definition 1.8)

$\left\{\begin{array}{c}\left(1\right)---\mathrm{\&sigma;}[{\mathrm{\&alpha;}}_k\backslash \mathrm{\&lambda;}\left({\mathrm{\&beta;}}_{k-1}\right)]=\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_{k-1}\right]+(i_{k-1}-1)(k\&GreaterEqual;2)\\ \left(2\right)---\mathrm{\&sigma;}[{\mathrm{\&alpha;}}_k\backslash \mathrm{\&lambda;}\left({\mathrm{\&beta;}}_{k-1}\right)]=\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right](k=1)\end{array}\right.$

Card: as follows:

(1) k >=2 o'clock, known according to aforesaid definition 1.6, the whipping vector of k step $[{\mathrm{\&alpha;}}_k\backslash \mathrm{\&lambda;}\left({\mathrm{\&beta;}}_{k-1}\right)]=<{\mathrm{\&beta;}}_{k-1}>$ $b_{i_{(k-1)}}<k-1>......b_2<2>b_1<1>,$ Syntax elements wherein represent vectorial α _kin the element that comprises, < β _k-1> represents vectorial β _k-1at vectorial α _kin element on corresponding room.According to whipping vector [α _kλ (β _k-1)] expression formula, in this syntax vector, the number of element obviously: σ [β _k]+(i _k-1).Conclusion to be proved is set up.

(2) during k=1, according to aforesaid plug hole recursive algorithm definition directly obtain.

Card is finished

Lemma 1.2: appoint to a k, k ∈ N and k >=1, to the maximum sequence valve of the syntax elements of the whipping vector mark intercepting in k step be:

$\left\{\begin{array}{c}\left(1\right)---n_k=\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_{k-1}\right]+(i_{k-1}-1)(k\&GreaterEqual;2)\\ \left(2\right)---n_1=\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right](k=1)\end{array}\right.$

Card: according to algorithm in function Z _kthe recursive definition of (α, β), to the whipping vector [α intercepting in k step _kλ (β _k-1)] the maximum sequence valve of syntax elements of mark is:

$\left\{\begin{array}{c}\left(1\right)---n_k=\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_k\backslash \mathrm{\&lambda;}\left({\mathrm{\&beta;}}_{k-1}\right)\right])(k\&GreaterEqual;2)\\ \left(2\right)---n_1=\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right](k=1)\end{array}\right.$

Whipping vector [α according to lemma 1.1, a k step _kλ (β _k-1)] number of the syntax elements that comprises is:

$\left\{\begin{array}{c}\left(1\right)---\mathrm{\&sigma;}[{\mathrm{\&alpha;}}_k\backslash \mathrm{\&lambda;}\left({\mathrm{\&beta;}}_{k-1}\right)]=\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_{k-1}\right]+(i_{k-1}-1)(k\&GreaterEqual;2)\\ \left(2\right)---\mathrm{\&sigma;}[{\mathrm{\&alpha;}}_1\backslash \mathrm{\&lambda;}\left({\mathrm{\&beta;}}_0\right)]=\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right](k=1)\end{array}\right.$

To the maximum sequence valve of the syntax elements of the whipping vector mark intercepting in k step, be:

$\left\{\begin{array}{c}\left(1\right)---n_k=\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_{k-1}\right]+(i_{k-1}-1)(k\&GreaterEqual;2)\\ \left(2\right)---n_1=\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right](k=1)\end{array}\right.$

Card is finished

Lemma 1.3: plug hole recursive algorithm ${\{W~Q~Z~T\}}_{\mathrm{\&rho;j}}^k\left[n_k\right]$ The concrete scheme generating ${\{W~Q~Z~T\}}_{\mathrm{\&rho;}2}^k<i_1,i_2,...,i_k>$ Number be: (definition of σ, referring to definition 1.8) (k ∈ N, k >=1)

$\left\{\begin{array}{c}\left(1\right)---{\mathrm{\&Sigma;}}_{i_1=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right]}\{{\mathrm{\&Sigma;}}_{i_2=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_1\right]+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{k-1}=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_{k-2}\right]+(i_{k-2}-1)}\left\{\mathrm{\&sigma;}\right[{\mathrm{\&beta;}}_{k-1}]+(i_{k-1}-1)\}\}...\}\\ (k\&GreaterEqual;2)\\ \left(2\right)---\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right](k=1)\end{array}\right.$

Card: according to lemma 1.2, appoint to a k, k ∈ N and k >=1, to the maximum sequence valve of the syntax elements of the whipping vector mark intercepting in k step be:

$\left\{\begin{array}{c}\left(1\right)---n_k=\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_{k-1}\right]+(i_{k-1}-1),(k\&GreaterEqual;2)\\ \left(2\right)---n_1=\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right],(k=1)\end{array}\right.$

According to algorithm in plug hole function T _kthe recursive definition of (α, β), i ₁span be: 1≤i ₁≤ n ₁, i.e. 1≤i ₁≤ σ [α ₁].

According to algorithm in plug hole function T _kthe recursive definition of (α, β), k>=3 o'clock i _k-1span be: 1≤i _k-1≤ n _k-1, i.e. 1≤i _k-1≤ σ [β _k-2]+(i _k-2-1).

Might as well establish β ₁=λ (β ₁) ... b ₂b ₁, according to aforesaid result and plug hole recursive algorithm definition, the plug hole situation sum of the 1st step is σ [α ₁], i.e. β ₁there is σ [α ₁] individual insertion α ₁plug hole situation.

Might as well establish β ₂=λ (β ₂) ... c ₂c ₁, $[{\left[{\mathrm{\&alpha;}}_1\right]}^{i_1}+<{\mathrm{\&beta;}}_1\backslash \mathrm{\&lambda;}\left({\mathrm{\&beta;}}_1\right)]=\mathrm{\&lambda;}\left({\mathrm{\&beta;}}_1\right)...b_2{b}_1{a}_{i_1-1}...a_1,$ According to aforesaid result and algorithm in plug hole function T _kthe recursive definition of (α, β), to an any given i ₁, 1≤i ₁≤ σ [α ₁], the number of the plug hole situation of the 2nd step is all σ [β ₁]+(i ₁-1); Due to i ₁obtaining value method be traversal real number interval [1, σ [α ₁]] interior all natural numbers, for i ₁whole values, according to the mode of summation ∑, count, can calculate: the 2nd step plug hole situation sum be be β ₂have the plug hole situation of individual insertion whipping vector.

Might as well establish β ₃=λ (β ₃) ... d ₂d ₁, $[{\left[{\mathrm{\&alpha;}}_2\right]}^{i_2}+<{\mathrm{\&beta;}}_2\backslash \mathrm{\&lambda;}\left({\mathrm{\&beta;}}_2\right)]=\mathrm{\&lambda;}\left({\mathrm{\&beta;}}_2\right)...c_1{a}_{i_2-1}...a_1$ Or $[{\left[{\mathrm{\&alpha;}}_2\right]}^{i_2}+<$ ${\mathrm{\&beta;}}_2\backslash \mathrm{\&lambda;}\left({\mathrm{\&beta;}}_2\right)]=\mathrm{\&lambda;}\left({\mathrm{\&beta;}}_2\right)...c_1{b}_{i_2-1-\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right]}...b_1{a}_{\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right]}...a_1,$ According to aforesaid result and algorithm in plug hole function T _kthe recursive definition of (α, β), to an any given i ₂, 1≤i ₂≤ σ [β ₁]+(i ₁-1), the plug hole situation number of the 3rd step is all σ [β ₂]+(i ₂-1); Due to i ₂obtaining value method be traversal real number interval [1, σ [β ₂]+(i ₂-1) all natural numbers], i ₁obtaining value method be traversal real number interval [1, σ [α ₁]] interior all natural numbers, for i ₂whole values, according to the mode of summation ∑, count and adds up, can calculate: the 3rd step plug hole situation sum is be β ₃have ${\mathrm{\&Sigma;}}_{i_1=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right]}\left\{{\mathrm{\&Sigma;}}_{i_2=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_1\right]+(i_1-1)}\right\{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_2\right]+(i_2-1)\left\}\right\}$ The plug hole situation of individual insertion whipping vector.

By mathematical induction, prove below: in k step, all whipping vectors are carried out to whole plug hole, the total τ of the plug hole situation of acquisition [∑ (k)] is: (k >=2)

${\mathrm{\&Sigma;}}_{i_1=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right]}\{{\mathrm{\&Sigma;}}_{i_2=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_1\right]+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{k-1}=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_{k-2}\right]+(i_{k-2}-1)}\left\{\mathrm{\&sigma;}\right[{\mathrm{\&beta;}}_{k-1}]+(i_{k-1}-1)\}\}...\}.$

Suppose: in k step, all whipping vectors are carried out to whole plug hole, the total τ of the plug hole situation of acquisition [∑ (k)] is: (k >=2) ${\mathrm{\&Sigma;}}_{i_1=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right]}\{{\mathrm{\&Sigma;}}_{i_2=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_1\right]+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{k-1}=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_{k-2}\right]+(i_{k-2}-1)}\left\{\mathrm{\&sigma;}\right[{\mathrm{\&beta;}}_{k-1}]+(i_{k-1}-1)\}\}...\}.$

According to algorithm in plug hole function T _kthe recursive definition of (α, β), k>=2 o'clock i _kspan be: 1≤i _k≤ n _k, i.e. 1≤i _k≤ σ [β _k-1]+(i _k-1-1).

According to algorithm in plug hole function T _kthe recursive definition of (α, β), to an any given i _k, 1≤i _k≤ σ [β _k-1]+(i _k-1-1), the plug hole situation number of (k+1) individual step is all σ [β _k]+(i _k-1); Due to i _kobtaining value method be traversal real number interval [1, σ [β _k-1]+(i _k-1-1) all natural numbers], for an any given σ [β _k-1]+(i _k-1-1) value, counts and adds up according to the mode of summation ∑, traversal i _kwhole values, easily calculate: by the add up sum of plug hole situation of (k+1) individual step of obtaining of number, be for appointing a σ [β who gives _k-1]+(i _k-1-1) value, β _k+1have the plug hole situation of individual insertion whipping vector.

Further, inductive assumption provides formula σ [β _k-1]+(i _k-1-1) expression formula, can be determined σ [β by inductive assumption that is _k-1]+(i _k-1-1) whole values.So, according to the mode of summation ∑, count and add up, from formula set out, according to inductive assumption formula ${\mathrm{\&Sigma;}}_{i_1=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right]}\{{\mathrm{\&Sigma;}}_{i_2=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_1\right]+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{k-1}=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_{k-2}\right]+(i_{k-2}-1)}\left\{\mathrm{\&sigma;}\right[{\mathrm{\&beta;}}_{k-1}]+(i_{k-1}-1)\}\}...\},$ Traversal i _k-1..., i ₂, i ₁whole values, thereby cancellation parameter i _k-1..., i ₂, i ₁, directly calculate: the plug hole situation sum of (k+1) individual step is:

${\mathrm{\&tau;}\left[\mathrm{\&Sigma;}(k+1)\right]=\mathrm{\&Sigma;}}_{i_1=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right]}\{{\mathrm{\&Sigma;}}_{i_2=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_1\right]+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{k-1}=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_{k-1}\right]+(i_{k-1}-1)}\left\{\mathrm{\&sigma;}\right[{\mathrm{\&beta;}}_k]+(i_{k-1}-1)\}\}...\}.$

Be β _k+1total total ${\mathrm{\&Sigma;}}_{i_1=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right]}\{{\mathrm{\&Sigma;}}_{i_2=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_1\right]+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{k-1}=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_{k-1}\right]+(i_{k-1}-1)}\left\{\mathrm{\&sigma;}\right[{\mathrm{\&beta;}}_k]+(i_{k-1}-1)\}\}...\}$ The plug hole situation of individual insertion whipping vector.Conclusion to be proved is set up, and mathematical induction proof is complete.

Comprehensive above-mentioned result, the total τ of the plug hole situation of k step [∑ (k)] is as follows: (k >=1)

$\left\{\begin{array}{c}\left(1\right)---{\mathrm{\&Sigma;}}_{i_1=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right]}\{{\mathrm{\&Sigma;}}_{i_2=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_1\right]+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{k-1}=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_{k-2}\right]+(i_{k-2}-1)}\left\{\mathrm{\&sigma;}\right[{\mathrm{\&beta;}}_{k-1}]+(i_{k-1}-1)\}\}...\}\\ (k\&GreaterEqual;2)\\ \left(2\right)---\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right],(k=1)\end{array}\right.$

According to plug hole recursive algorithm definition, the total τ of the plug hole situation of k step [∑ (k)], namely algorithm the total τ [∑ (k)] of plug hole situation of last step, with algorithm the concrete scheme generating <i ₁, i ₂..., i _kthe number of > equates, comprehensive above-mentioned result, plug hole recursive algorithm the concrete scheme generating <i ₁, i ₂..., i _kthe number of > is: (definition of σ, referring to definition 1.8) (k ∈ N, k>=1)

$\left\{\begin{array}{c}\left(1\right)---{\mathrm{\&Sigma;}}_{i_1=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right]}\{{\mathrm{\&Sigma;}}_{i_2=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_1\right]+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{k-1}=1}^{\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_{k-2}\right]+(i_{k-2}-1)}\left\{\mathrm{\&sigma;}\right[{\mathrm{\&beta;}}_{k-1}]+(i_{k-1}-1)\}\}...\}\\ (k\&GreaterEqual;2)\\ \left(2\right)---\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_1\right],(k=1)\end{array}\right.$

Card is finished

Theorem 1.1: the concrete scheme that any one the scheme mode ρ j in this paper method is generated <i ₁, i ₂..., i _kthe number of > is designated as Ω [ρ j], has: Ω [ρ j]=(formula is as follows) (θ>=2)

$\left\{\begin{array}{c}\left(1\right)---{\mathrm{\&Sigma;}}_{i_1=1}^{\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_1\right)\right]}\{{\mathrm{\&Sigma;}}_{i_2=1}^{\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_2\right)\right]+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{\mathrm{\&theta;}-2}=1}^{\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_{\mathrm{\&theta;}-2}\right)\right]+(i_{\mathrm{\&theta;}-3}-1)}\left\{\mathrm{\&sigma;}\right[\mathrm{\&rho;j}\left(t_{\mathrm{\&theta;}-1}\right)]+(i_{\mathrm{\&theta;}-2}-1)\}\}...\}\\ (\mathrm{\&theta;}\&GreaterEqual;3)\\ \left(2\right)---\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_1\right)\right],(\mathrm{\&theta;}=2)\end{array}\right.$

Card: according to plug hole recursive algorithm definition, then apply mechanically lemma 1.3 in conjunction with this definition, conclusion to be demonstrate,proved is obviously set up.Card is finished

Theorem 1.2: the number of the final single file vector that any one the scheme mode ρ j in this paper method is generated is designated as Ω [ρ j], has: Ω [ρ j]=(formula is as follows) (θ >=2)

$\left\{\begin{array}{c}\left(1\right)---{\mathrm{\&Sigma;}}_{i_1=1}^{\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_1\right)\right]}\{{\mathrm{\&Sigma;}}_{i_2=1}^{\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_2\right)\right]+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{\mathrm{\&theta;}-2}=1}^{\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_{\mathrm{\&theta;}-2}\right)\right]+(i_{\mathrm{\&theta;}-3}-1)}\left\{\mathrm{\&sigma;}\right[\mathrm{\&rho;j}\left(t_{\mathrm{\&theta;}-1}\right)]+(i_{\mathrm{\&theta;}-2}-1)\}\}...\}\\ (\mathrm{\&theta;}\&GreaterEqual;3)\\ \left(2\right)---\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_1\right)\right],(\mathrm{\&theta;}=2)\end{array}\right.$

Card: according to the definition of final single file vector sum concrete scheme, the final number of single file vector and the number of concrete scheme equate.According to theorem 1.1, can obtain again.Card is finished

[B3.8.5.6] gives an example to the operation of the 1st kind of plug hole method

Get set θ=4, θ unequal to 24, ρ j=ρ 2.

Get ?

?

Number Ω [ρ 2]=(formula is as follows) of the final single file vector that scheme mode ρ 2 generates:

Work as i ₁=1 o'clock:

Work as i ₁=2 o'clock:

Work as i ₁=3 o'clock:

Work as i ₁=4 o'clock:

The number of the final single file vector that scheme mode ρ 2 generates is: Ω [ρ 2]=140.

The number of the final single file vector that whole scheme modes of this paper method generate is: ${\mathrm{\&Sigma;}}_{j=1}^{\mathrm{\&theta;}!}\left(\mathrm{\&Omega;}\right[\mathrm{\&rho;j}\left]\right)=140+140+140+$ $140+140+140=840.$

[B3.8.5.7] second group of computing formula and relevant proof

By former, may be converted into θ new syntax vector that can not find clear and definite position by matrix solution be designated as: by the newly-generated syntax vector that can not find clear and definite position be referred to as Equations of The Second Kind syntax vector.The syntax vector that any one is eliminated in aforesaid equivalent substitution process is called predecessor's syntax vector.For any one newly-generated Equations of The Second Kind syntax vector will be in aforesaid equivalent substitution process quilt the number of the predecessor's syntax vector f replacing is designated as u ε.? through ε equivalent substitution of u, obtain.

For example: by vector f ₂=e+ < f ₃+ < 7+ < f ₅and f ₃=3+ < e+ < 4+ < e and f ₅=8+ < e+ < 9+ < 10 has generated an Equations of The Second Kind vector through equivalent substitution obvious known u ₁=2, through 2 equivalent substitutions, obtain.

Theorem 1.3: appoint to an Equations of The Second Kind syntax vector will in the number of the syntax elements that comprises be designated as by syntax vector the number of predecessor's syntax vector f of cancellation is designated as u _ε, meet recursion formula:

Card: use mathematical induction proof as follows:

(1), if u _ε=0, syntax is vectorial the number of predecessor's syntax vector f of cancellation is 0, the syntax vector that does not pass through equivalent substitution in former possibility matrix solution, syntax vector in the syntax elements number that comprises be 4, obvious formula now set up.

(2), suppose to work as u _εduring=k set up, now the number of predecessor's syntax vector f of cancellation is k, in the element number that comprises be 3k+4; Work as u _εduring=k+1, can regard as k predecessor's syntax vector f of first cancellation, then on this basis, when deducting self syntax elements, introduce again Liao Yige predecessor's syntax vector f, when deducting self 1 syntax elements, 4 elements have been introduced again.Now in the element number that comprises be 3k+4+3, formula set up.Comprehensively (1), (2), conclusion known to be proved is set up.

[B3.8.5.8] is about the summary of the number computing formula of this paper method

Conclusion 1.3: distinguishing under the prerequisite of the e on diverse location, the number of the final single file vector that the number of concrete scheme corresponding to any one the scheme mode ρ j in this paper method is generated with scheme mode ρ j is identical, all be designated as Ω [ρ j], have: Ω [ρ j]=(formula is as follows) (definition of σ, referring to definition 1.8) (θ >=2)

$\left\{\begin{array}{c}\left(1\right)---{\mathrm{\&Sigma;}}_{i_1=1}^{\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_1\right)\right]}\{{\mathrm{\&Sigma;}}_{i_2=1}^{\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_2\right)\right]+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{\mathrm{\&theta;}-2}=1}^{\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_{\mathrm{\&theta;}-2}\right)\right]+(i_{\mathrm{\&theta;}-3}-1)}\left\{\mathrm{\&sigma;}\right[\mathrm{\&rho;j}\left(t_{\mathrm{\&theta;}-1}\right)]+(i_{\mathrm{\&theta;}-2}-1)\}\}...\}\\ (\mathrm{\&theta;}\&GreaterEqual;3)\\ \left(2\right)---\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_1\right)\right],(\mathrm{\&theta;}=2)\end{array}\right.$

Conclusion 1.4: might as well define d _θ=σ [ρ j (t _θ)], the formula of conclusion 1.3 is converted into: Ω [ρ j]=(as follows) (definition of σ, referring to definition 1.8) (θ>=2)

$\left\{\begin{array}{c}\left(1\right)---{\mathrm{\&Sigma;}}_{i_1=1}^{d_1}\{{\mathrm{\&Sigma;}}_{i_2=1}^{d_2+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{\mathrm{\&theta;}-1}=1}^{d_{\mathrm{\&theta;}-1}+(i_{\mathrm{\&theta;}-2}-1)}[d_{\mathrm{\&theta;}}+(i_{\mathrm{\&theta;}-1}-1)]\}...\},(\mathrm{\&theta;}\&GreaterEqual;3)\\ \left(2\right)---d_1,(\mathrm{\&theta;}=2)\end{array}\right.$

Conclusion 1.5: according to theorem 1.3, have: Ω [ρ j]=(as follows) (definition of σ, referring to definition 1.8) (θ >=1)

$\left\{\begin{array}{c}\left(1\right)---{\mathrm{\&Sigma;}}_{i_1=1}^{({3u}_1+4)}\{{\mathrm{\&Sigma;}}_{i_2=1}^{({3u}_2+4)+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{\mathrm{\&theta;}-2}=1}^{({3u}_{\mathrm{\&theta;}-2}+4)+(i_{\mathrm{\&theta;}-3}-1)}[({3u}_{\mathrm{\&theta;}-1}+4)+(i_{\mathrm{\&theta;}-2}-1)]\}...\}\\ (\mathrm{\&theta;}\&GreaterEqual;3)\\ \left(2\right)---{3u}_1+4,(\mathrm{\&theta;}=2)\end{array}\right.$

Conclusion 1.6: definition g _θ=3u _θ+ 4, the Ω of conclusion 1.5 [ρ j] formula is converted into: Ω [ρ j]=(as follows) (definition of σ, referring to definition 1.8) (θ>=2)

$\left\{\begin{array}{c}\left(1\right)---{\mathrm{\&Sigma;}}_{i_1=1}^{g_1}\{{\mathrm{\&Sigma;}}_{i_2=1}^{g_2+(i_1-1)}...\{{\mathrm{\&Sigma;}}_{i_{\mathrm{\&theta;}-2}=1}^{g_{\mathrm{\&theta;}-2}+(i_{\mathrm{\&theta;}-3}-1)}[g_{\mathrm{\&theta;}-1}+(i_{\mathrm{\&theta;}-2}-1)]\}...\},(\mathrm{\&theta;}\&GreaterEqual;3)\\ \left(2\right)---g_1,(k=2)\end{array}\right.$

A conclusion 1.7: because altogether produce θ in this paper method implementation process! Individual scheme mode, is distinguishing under the prerequisite of the e on diverse location, and the number of whole concrete schemes that the whole scheme modes in this paper method generate is identical with the number of whole final single file vectors of generation, and number formula is: (θ>=2)

Conclusion 1.8: comprehensive aforesaid each conclusion generates limited concrete scheme and limited final single file vector in this paper method implementation process.The number of the number of concrete scheme and final single file vector determines have definite computing formula and corresponding proof, meets the natural law.

Comprehensive demonstration of [B3.8.5.9] the 1st kind of method

Illustrate: get following possibility matrix solution, and first the syntax vector of finding clear and definite position in this possibility matrix solution is carried out to equivalent substitution.If: possible matrix solution, as follows:

Former possibility matrix solution, is converted into through equivalent substitution:

Set be listed as follows: (set θ=3, θ unequal to 6)

Structure mapping ρ j is as follows: (θ=3, )

$\mathrm{\&rho;j}:\{t_1,t_2,t_3\}\&RightArrow;P_{\mathrm{\&Phi;}}^3,\mathrm{\&rho;j}(t_1,t_2,t_3)\&Element;P_{\mathrm{\&Phi;}}^3,$ j∈N，1≤j≤6

Set π=ρ 1, and ρ 2, and ρ 3, and ρ 4, and ρ 5, and ρ 6} is listed as follows:

The pattern that carries into execution a plan ρ 1, be by recursive algorithm move 2 times.

move as follows: (n ₁∈ N, i ₁∈ N, 1≤i ₁≤ n ₁)

move as follows: (n ₂∈ N, i ₂∈ N, 1≤i ₂≤ n ₂)

According to scheme mode ρ 1, list:

According to scheme mode ρ 1, list:

The pattern that carries into execution a plan ρ 2, be by recursive algorithm move 2 times.

move as follows: (n ₁∈ N, i ₁∈ N, 1≤i ₁≤ n ₁)

move as follows: (n ₂∈ N, i ₂∈ N, 1≤i ₂≤ n ₂)

According to scheme mode ρ 2, list:

According to scheme mode ρ 2, list:

The pattern that carries into execution a plan ρ 2, be by recursive algorithm move 2 times.

move as follows: (n ₁∈ N, i ₁∈ N, 1≤i ₁≤ n ₁)

move as follows: (n ₂∈ N, i ₂∈ N, 1≤i ₂≤ n ₂)

According to scheme mode ρ 3, list:

According to scheme mode ρ 3, list:

The process of the pattern that carries into execution a plan ρ 4--ρ 6, slightly.

By plug hole recursive algorithm important information list be summarized as follows:

Also above-mentioned formula can be converted into: the form representing by the number of predecessor's syntax vector.

The number list of element in predecessor's syntax vector number and Equations of The Second Kind syntax vector:

The important information of expressing plug hole recursive algorithm by the number of predecessor's syntax vector, is listed as follows:

Check whether each final single file vector occurs two sequence valves that position is converse, slightly.

Example comprehensively demonstration is complete.

[B3.8.6] illustrates the 2nd kind of plug hole method

Introduce in detail the whole plug hole method of aforesaid one-sided not order-preserving below.The method can accurately depict each situation of the whole plug hole of limited number of time between the syntax vector that can not find clear and definite position in possibility matrix solution.

[B3.8.6.1] structure plug hole recursive function

The definition of scheme mode, concrete scheme, step, with reference to the 1st kind of method.The difference of the 2nd kind of method and the 1st kind of method below.Construct a plug hole recursive algorithm by this recursive algorithm, just can portray the detailed process of the whole plug hole of aforementioned each time one-sided not order-preserving.Before constitution step recursive algorithm, following 5 definition of given first, as pre-knowledge:

The plug hole recursive algorithm that will construct below k the step of carrying out according to scheme mode ρ j.K wherein represents plug hole recursive algorithm the number of times of operation, carries out the number of times that the whole plug hole of aforesaid one-sided not order-preserving operates.

Definition 2.1: appoint to a syntax vector α, function of a single variable W represents to take out and mark-up syntax vector α.W (α)=α _krepresent to take out syntax vector α, and syntax vector α is labeled as to α _k.

Definition 2.2: appoint to a syntax vector β, function of a single variable Q represents to take out and mark-up syntax vector β.Q (β)=β _krepresent to take out syntax vector β, and syntax vector β is labeled as to β _k.At operation plug hole recursive algorithm in process, be by syntax vector β _kinsert syntax vector α _kin.

Definition 2.3: function of a single variable Z represents distich normal vector α _kmark sequence valve, by syntax vector α _kin from right several the 1st syntax elements mark sequence valve 1, then mark successively from right to left sequence valve 2,3 ..., until by syntax vector α _kin syntax elements all mark is complete.The maximum sequence valve of mark is designated as to n _k.Operation function of a single variable Z obtains:

Definition 2.4: binary function T is illustrated in and uses function of a single variable Z distich normal vector α _kafter mark order, at vectorial α _kon choose right several m _kindividual element, and at m _kunique room is constructed on the right side of individual element, then by syntax vector β _kin the mode of whole plug hole, insert this room.The new vector obtaining after plug hole is designated as: operation binary function T obtains: $T_k(\mathrm{\&alpha;},\mathrm{\&beta;})={\left({\mathrm{\&alpha;}}_k\right)}^{m_k}+<{\mathrm{\&beta;}}_k.$

Definition 2.5: appoint to a syntax vector α, the number of the syntax elements comprising in α is designated as to σ [α].If comprise n syntax elements in syntax vector α, n ∈ N, obviously has: n=σ [α].

Note: in the definition of plug hole recursive algorithm below, can see such equation:

W _k(α)＝T _k-1(α，β)； $T_k(\mathrm{\&alpha;},\mathrm{\&beta;})={\left({\mathrm{\&alpha;}}_k\right)}^{m_k}+<{\mathrm{\&beta;}}_k$

α wherein and β are the implications of independent variable, are abstract marks, can wide in range value.Therefore, above-mentioned notation contradiction not.

Below, according to the aforesaid mapping ρ j of this paper and 4 functions, definition plug hole recursive algorithm as follows: (set )

The number computing formula of [B3.8.6.2] concrete scheme and final single file vector

Lemma 2.1: $\mathrm{\&sigma;}[{\left({\mathrm{\&alpha;}}_k\right)}^{m_k}+<{\mathrm{\&beta;}}_k]=\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_k\right]+\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_k\right]$

Card: according to aforesaid definition, at operation recursive algorithm process in, make blank vector α _kwith plug hole vector β _kin syntax elements all do not have to increase or reduce, for any one syntax elements b:

If 1. b ∈ α _kor b ∈ α _k,

If 2. $b\&NotElement;{\mathrm{\&alpha;}}_k$ And $b\&NotElement;{\mathrm{\&alpha;}}_k,$ ? $b\&NotElement;{\left({\mathrm{\&alpha;}}_k\right)}^{m_k}+<{\mathrm{\&beta;}}_k.$

1. and 2. according to, obviously have: $\mathrm{\&sigma;}[{\left({\mathrm{\&alpha;}}_k\right)}^{m_k}+<{\mathrm{\&beta;}}_k]=\mathrm{\&sigma;}\left[{\mathrm{\&alpha;}}_k\right]+\mathrm{\&sigma;}\left[{\mathrm{\&beta;}}_k\right].$

Card is finished

Lemma 2.2: establish: (k ∈ N, k >=1)

${\mathrm{\&Sigma;\&Psi;}}_k=\left\{\begin{array}{c}\left(1\right)---{\left(({\left(\mathrm{\&rho;j}\left(t_1\right)\right)}^{m_1}+<\mathrm{\&rho;j}{\left(t_2\right))}^{m_2}+<...\right)}^{m_{k-1}}+<\mathrm{\&rho;j}\left(t_k\right),(k\&GreaterEqual;2)\\ \left(2\right)---\mathrm{\&rho;j}\left(t_1\right),(k\&GreaterEqual;1)\end{array}\right.$

Syntax vector ∑ Ψ _kexpression is by ρ j (t ₁), ρ j (t ₂) ..., ρ j (t _k) the syntax vector that obtains through the whole plug hole of one-sided not order-preserving successively.M ₁, m ₂..., m _k-1represent respectively any one room Ser.No. of corresponding vector.Just like the establishment of drawing a conclusion:

$\mathrm{\&sigma;}\left[{\mathrm{\&Sigma;\&Psi;}}_k\right]={\mathrm{\&Sigma;}}_{i=1}^k\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_i\right)\right]$

(note: ${\mathrm{\&Sigma;}}_{i=1}^k\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_i\right)\right]=\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_1\right)\right]+\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_2\right)\right]+...+\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_k\right)\right]$ )

Card: use mathematical induction proof as follows:

(1), if k=1, $\mathrm{\&sigma;}\left[{\mathrm{\&Sigma;\&Psi;}}_1\right]={\mathrm{\&Sigma;}}_{i=1}^1\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_i\right)\right],$ Conclusion is set up.

(2), suppose to set up when k=h, have

When k=h+1, $\mathrm{\&Sigma;}{\mathrm{\&Psi;}}_{h+1}={\left(\mathrm{\&Sigma;}{\mathrm{\&Psi;}}_h\right)}^{m_h}+<\mathrm{\&rho;j}\left(t_{h+1}\right),$ Have:

$\mathrm{\&sigma;}\left[\mathrm{\&Sigma;}{\mathrm{\&Psi;}}_{h+1}\right]={\mathrm{\&sigma;}[\left(\mathrm{\&Sigma;}{\mathrm{\&Psi;}}_h\right)}^{m_h}+<\mathrm{\&rho;j}\left(t_{h+1}\right)]$

According to lemma 2.1, can obtain:

$\mathrm{\&sigma;}[{\left({\mathrm{\&Sigma;\&Psi;}}_h\right)}^{m_h}+<\mathrm{\&rho;j}\left(t_{h+1}\right)]=\mathrm{\&sigma;}\left[{\left({\mathrm{\&Sigma;\&Psi;}}_h\right)}^{m_h}\right]+\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_{h+1}\right)\right]$

According to inductive assumption, can obtain:

$\mathrm{\&sigma;}[{\left({\mathrm{\&Sigma;\&Psi;}}_h\right)}^{m_h}+<\mathrm{\&rho;j}\left(t_{h+1}\right)]=\left({\mathrm{\&Sigma;}}_{i=1}^h\mathrm{\&sigma;}\right[\mathrm{\&rho;j}\left(t_i\right)\left]\right)+\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_{h+1}\right)\right]$

Thereby can obtain: $\mathrm{\&sigma;}[{\left({\mathrm{\&Sigma;\&Psi;}}_h\right)}^{m_h}+<\mathrm{\&rho;j}\left(t_{h+1}\right)]={\mathrm{\&Sigma;}}_{i=1}^{h+1}\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_i\right)\right]$

Can obtain again: ${\mathrm{\&Sigma;\&Psi;}}_{h+1}={\mathrm{\&Sigma;}}_{i=1}^{h+1}\mathrm{\&sigma;}\left[\mathrm{\&rho;j}\left(t_i\right)\right].$

Card is finished

Theorem 2.1: the number of the concrete scheme that any one the scheme mode ρ j in this paper method is generated is designated as Ω [ρ j], has: $\mathrm{\&Omega;}\left[\mathrm{\&rho;j}\right]={\mathrm{\&Pi;}}_{k=1}^{\mathrm{\&theta;}-1}\left({\mathrm{\&Sigma;}}_{i=1}^k\mathrm{\&sigma;}\right[\mathrm{\&rho;j}\left(t_i\right)\left]\right).$ (θ≥2)

Card: according to plug hole recursive algorithm definition, for a scheme mode ρ j arbitrarily, the number of the plug hole situation of k step and k step make blank vector α _kthe number of syntax elements identical.According to aforesaid definition, obviously , according to lemma 2.2, k the step of scheme mode ρ j has individual situation.Because any one scheme mode ρ j has (θ-1) individual step under its command, known according to the multiplicative principle of combinatorics: any one scheme mode ρ j of this paper method is corresponding individual concrete scheme.

Card is finished

Theorem 2.2: the number of the final single file vector that any one the scheme mode ρ j in this paper method is generated is designated as Ω [ρ j], has: $\mathrm{\&Omega;}\left[\mathrm{\&rho;j}\right]={\mathrm{\&Pi;}}_{k=1}^{\mathrm{\&theta;}-1}\left({\mathrm{\&Sigma;}}_{i=1}^k\mathrm{\&sigma;}\right[\mathrm{\&rho;j}\left(t_i\right)\left]\right).$ (θ≥2)

Card: according to the definition of final single file vector sum concrete scheme, the final number of single file vector and the number of concrete scheme equate; According to theorem 2.1, can obtain again.

Card is finished

A conclusion 2.1: this paper method is total θ always! Individual scheme mode, according to theorem 2.1, according to the addition principle of combinatorics, the sum of known concrete scheme is again: ${\mathrm{\&Sigma;}}_{j=1}^{\mathrm{\&theta;}!}\left({\mathrm{\&Pi;}}_{k=1}^{\mathrm{\&theta;}-1}\left({\mathrm{\&Sigma;}}_{i=1}^k\mathrm{\&sigma;}\right[\mathrm{\&rho;j}\left(t_i\right)\left]\right)\right)$

A conclusion 2.2: this paper method is total θ always! Individual scheme mode, according to theorem 2.2, again according to the addition principle of combinatorics, the sum of known final single file vector is: ${\mathrm{\&Sigma;}}_{j=1}^{\mathrm{\&theta;}!}\left({\mathrm{\&Pi;}}_{k=1}^{\mathrm{\&theta;}-1}\left({\mathrm{\&Sigma;}}_{i=1}^k\mathrm{\&sigma;}\right[\mathrm{\&rho;j}\left(t_i\right)\left]\right)\right)$

The example demonstration of [B3.8.6.3] the 2nd kind of method

Illustrate: get following possibility matrix solution, and first the syntax vector of finding clear and definite position in this possibility matrix solution is carried out to equivalent substitution.If: possible matrix solution, as follows:

Former possibility matrix solution, is converted into through equivalent substitution:

Set be listed as follows: set θ=3, θ unequal to 6

Structure mapping ρ j is as follows: (θ=3, )

$\mathrm{\&rho;j}:\{t_1,t_2,t_3\}\&RightArrow;P_{\mathrm{\&Phi;}}^3,\mathrm{\&rho;j}(t_1,t_2,t_3)\&Element;P_{\mathrm{\&Phi;}}^3,j\&Element;N,1\&le;j\&le;6$

Set π=ρ 1, and ρ 2, and ρ 3, and ρ 4, and ρ 5, and ρ 6} is listed as follows:

The pattern that carries into execution a plan ρ 1, be by step recursive algorithm move 2 times.

move as follows:

move as follows:

According to scheme mode ρ 1, operation plug hole function list:

According to scheme mode ρ 1, operation plug hole function list:

The process of the pattern that carries into execution a plan ρ 2--ρ 6, slightly.

By plug hole recursive algorithm important information list be summarized as follows:

Two or more identical final single file vectors are retained to one, and the identical final single file vector of Delete superfluous, finally obtains 210 final single file vectors different between two, fits like a glove with the result of method 1.

Check whether each final single file vector occurs two sequence valves that position is converse, slightly.

Example demonstration is complete.

C certain applications for example

C1 part example 1

Example 1: by pre-service, can remove the impurity in statement, and word element number and type in mark and identification statement.For example, for english statement S=" I can completely understand what what you just said really meant ", it removes the statement S=" I can understand what what you said meant " obtaining after impurity, after it being carried out to the identification of word unit and word cell type mark and numbering, can obtain the data structure of mating with following table.

Statement	Word cell type	Numbering
			I	Noun pronoun unit	1

can?understand	Predicate verb unit	2
			what?A	Subordinate conjunctive word unit	3
what?B	Subordinate conjunctive word unit	4
			you	Noun pronoun unit	5
said	Predicate verb unit	6
			meant	Predicate verb unit	7

The present invention is based on the represented pretreated statement of above data structure is carried out to syntactic analysis, to obtain the composition relation of each word unit in sentence.

Fig. 1 is the process flow diagram of the computer based natural language syntactic structure of the embodiment of the present invention method of resolving.As shown in Figure 1, described method comprises:

Step 110, read pretreated phrase data structure to be resolved, the conjunctive word unit, predicate verb unit and the noun pronoun unit that in described pretreated phrase data structure, only comprise statement, and each word unit is numbered and marks type according to the order in described pretreated statement.

Step 120, to each predicate verb unit, generate corresponding leading question element, subject element, predicate element and object element; The possible value of described leading question element is less than one of the conjunctive word arranged side by side unit of corresponding predicate verb element number or subordinate conjunctive word unit for numbering, or be less than the conjunctive word arranged side by side unit of corresponding predicate verb element number and one by a numbering and be adjacent and number and be less than one of the conjunctive word mix vector that corresponding predicate verb element number and numbering are greater than the subordinate conjunctive word cell formation of this conjunctive word element number arranged side by side, or dummy cell;

The possible value of described subject element is less than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of major term unit is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or at one of syntax vector corresponding to the predicate element of front appearance, or dummy cell;

Described predicate element is corresponding described predicate verb unit;

The possible value numbering of described object element is greater than corresponding predicate verb element number and is less than one of noun pronoun unit of the adjacent predicate verb element number in rear appearance, or the numbering of minimum word unit is greater than corresponding predicate verb element number and is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of the adjacent predicate verb element number in rear appearance, or at one of syntax vector corresponding to the predicate element of rear appearance, or dummy cell.

Particularly, for pretreated statement, establishing its predicate verb unit total quantity is n, and because predicate verb unit only can be as predicate, therefore, all corresponding predicate element in each predicate verb unit, remembers that each predicate verb unit is r _k, k=1 ..., n.

After obtaining predicate element, the Position Number continuing based on each predicate element generates corresponding leading question element, subject element, object element.

I, leading question element

Remember each predicate verb unit r _kcorresponding conjunctive word unit set is:

{x _k}＝Lead _k∪conj _k∪(conj _kοLead _k)∪{e}

Note predicate verb unit r _kcorresponding leading question element is x _k, it may value set be { x _k.Generate predicate verb unit r _kcorresponding leading question element is x _kpossible value set (preferably) comprising:

The possible value of described leading question element is less than one of the conjunctive word arranged side by side unit of corresponding predicate verb element number or subordinate conjunctive word unit for numbering, or be less than the conjunctive word arranged side by side unit of corresponding predicate verb element number and one by a numbering and be adjacent and number and be less than one of the conjunctive word mix vector that corresponding predicate verb element number and numbering are greater than the subordinate conjunctive word cell formation of this conjunctive word element number arranged side by side, or dummy cell.

That is, x _k∈ Lead _k∪ conj _k∪ (conj _kο Lead _k) ∪ { e}.

In above-mentioned formula, set Lead _krepresent that numbering is less than the subordinate conjunctive word unit set of corresponding predicate verb element number; Conj _krepresent that numbering is less than the conjunctive word unit set arranged side by side of corresponding predicate verb element number; (conj _kο Lead _k) representing that by a numbering, being less than the conjunctive word arranged side by side unit of corresponding predicate verb element number and one is adjacent and numbers and be less than the set of conjunctive word mix vector that corresponding predicate verb element number and numbering are greater than the subordinate conjunctive word cell formation of this conjunctive word element number arranged side by side, e represents dummy cell.

For example, for the pretreated statement S=shown in above-mentioned table 1 " I can understand what what you said meant ", have:

R ₁=" can understand ", for r ₁there is { x ₁}={ e}, also, with r ₁the value of corresponding leading question element is dummy cell.

R ₂=" said ", for r ₂there is { x ₂}={ what A, what B, e}, with r ₂corresponding leading question element can value be in first what or second what, that is, and one of " what A " and " what B ", or dummy cell.

R ₃=" meant ", for r ₃there is { x ₃}={ what A, what B, e}, with r ₃corresponding leading question element can value be in first what or second what, that is, and one of " what A " and " what B ", or dummy cell.

II, subject element

Remember each predicate verb unit r _kcorresponding subject noun pronoun unit set is { y _k}=NPI _yk∪ VNP _yk∪ NOMP _k∪ G _k∪ { e} or { y _k}=NPI _yk∪ VNP _yk∪ NOMP _k∪ G _k∪ fy _k∪ { e}.

Note predicate verb unit r _kcorresponding subject element is y _k, it may value set be { y _k.

Generate corresponding subject element y _kpreferably include:

(1) when corresponding predicate verb element number is minimum predicate verb element number, the possible value of described subject element is less than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of its major term unit is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or dummy cell.

Also, when there not being r _k-1time: { y _k}=NPI _yk∪ VNP _yk∪ NOMP _k∪ G _k∪ { e};

Thereby, y _k∈ NPI _yk∪ VNP _yk∪ NOMP _k∪ G _k∪ { e}.

In above-mentioned formula, set NPI _ykrepresent that numbering is less than the pure noun unit set of corresponding predicate verb element number; VNP _ykrepresent that numbering is less than the verb unit set of the noun character of corresponding predicate verb element number; NOMP _knumbering is less than the nominative pronoun unit set of corresponding predicate verb element number; G _kexpression by the numbering of major term unit, be less than corresponding predicate verb element number the group composition of all coordinate noun pronoun mix vectors and set; E represents dummy cell.

(2) when corresponding predicate verb element number is not minimum predicate verb element number, the possible value of described subject element is less than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of major term unit is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or at one of syntax vector corresponding to the predicate verb unit of front appearance, or dummy cell.

Also, when there being r _k-1time: { y _k}=NPI _yk∪ VNP _yk∪ NOMP _k∪ G _k∪ fy _k∪ { e}.

Thereby, y _k∈ NPI _yk∪ VNP _yk∪ NOMP _k∪ G _k∪ fy _k∪ { e}.

In above-mentioned formula, set NPI _ykrepresent that numbering is less than the pure noun unit set of corresponding predicate verb element number; VNP _ykrepresent that numbering is less than the verb unit set of the noun character of corresponding predicate verb element number; NOMP _knumbering is less than the nominative pronoun unit set of corresponding predicate verb element number; G _kexpression by the numbering of major term unit, be less than corresponding predicate verb element number the group composition of all coordinate noun pronoun mix vectors and set; Fy _kbe illustrated in the syntax vector set corresponding to predicate verb unit of front appearance; E represents dummy cell.

For example, for the pretreated statement S=shown in above-mentioned table 1 " I can understand what what you said meant ", have:

R ₁=" can understand ", for r ₁having it is the minimum predicate verb unit of numbering, therefore, and { y ₁}=NOMP ₁∪ { e}={I, e}.

R ₂=" said ", for r ₂there is it not to number minimum predicate verb unit, at r ₁and r ₂between noun pronoun unit only have " you ", and numbering to be less than 2 function be f ₁, therefore, { y ₂}=NOMP ₂∪ fy ₂∪ { e}={I, you} ∪ { f ₁∪ { e}.

R ₃=" meant ", for r ₃it not numbers minimum predicate verb unit, at r ₂and r ₃between there is no noun pronoun unit, and numbering to be less than 3 function be f ₁and f ₂, therefore, have: { y ₃}=NOMP ₃∪ fy ₃∪ { e}={I, you} ∪ { f ₁, f ₂∪ { e}.

III, object element

Remember each predicate verb unit r _kcorresponding object noun pronoun unit set is { z _k}=NPI _zk∪ VNP _zk∪ OBJP _k∪ H _k∪ { e} or { z _k}=NPI _zk∪ VNP _zk∪ OBJP _k∪ H _k∪ fz _k∪ { e}.

Meanwhile, note predicate verb unit r _kcorresponding leading question element is z _k, it may value set be { z _k.

Generate corresponding object element { z _kpreferably include:

(1) when corresponding predicate verb element number is maximum predicate verb element number, the possible value of described object element is greater than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of its minimum word unit is greater than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or dummy cell.

Also, when there not being r _k+1time: { z _k}=NPI _zk∪ VNP _zk∪ OBJP _k∪ H _k∪ { e}.

In above-mentioned formula, set NPI _zkrepresent that numbering is greater than the pure noun unit set of corresponding predicate verb element number; VNP _zkrepresent that numbering is greater than the verb unit set of the noun character of corresponding predicate verb element number; OBJP _krepresent that numbering is greater than the objective case pronoun unit set of the noun character of corresponding predicate verb element number; H _kexpression is greater than the also set of all coordinate noun pronoun mix vectors group composition of corresponding predicate verb element number by the numbering of minimum word unit; E represents dummy cell.

(2) when corresponding predicate verb element number is not maximum predicate verb element number, the possible value of described object element is greater than corresponding predicate verb element number and is less than one of noun pronoun unit of the adjacent predicate verb element number in rear appearance for numbering, or the numbering of its minimum word unit is greater than corresponding predicate verb element number and is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of the adjacent predicate verb element number in rear appearance, or at one of syntax vector corresponding to the predicate verb unit of rear appearance, or dummy cell.

Also, when there being r _k+1time: { z _k}=NPI _zk∪ VNP _zk∪ OBJP _k∪ H _k∪ fz _k∪ { e}.

In above-mentioned formula, set NPI _zkrepresent numbering be greater than corresponding predicate verb element number and be less than the adjacent predicate verb element number in rear appearance pure noun unit set; VNP _zkrepresent to number the verb unit set that is greater than corresponding predicate verb element number and is less than the noun character of the adjacent predicate verb element number in rear appearance; NOMP _krepresent the objective case pronoun unit set that is greater than corresponding predicate verb element number and is less than the adjacent predicate verb element number in rear appearance; H _kexpression by the numbering of minimum word unit, be greater than corresponding predicate verb element number and be less than the adjacent predicate verb element number in rear appearance the group composition of all coordinate noun pronoun mix vectors and set; Fz _kbe illustrated in the syntax vector set corresponding to predicate verb unit of rear appearance; E represents dummy cell.

For example, for the pretreated statement S=shown in above-mentioned table 1 " I can understand what what you said meant ", have:

R ₁=" can understand ", for r ₁it is not the maximum predicate verb unit of numbering, at r ₁and r ₂between there is noun pronoun unit " you ", there is no coordinate noun pronoun mix vector, and numbering to be greater than 1 function be f ₂, f ₃, therefore, { z ₁}=OBJP ₁∪ fz ₁∪ { e}={you} ∪ { f ₂, f ₃∪ { e}.

R ₂=" said ", for r ₂it not numbers maximum predicate verb unit, at r ₁and r ₂between there is no noun pronoun unit, and numbering to be greater than 2 function be f ₃, there is no coordinate noun pronoun mix vector yet, therefore, have: { z ₂}=fz ₂∪ { e}={f ₃∪ { e}.

R ₃=" meant ", for r ₃it is for the maximum predicate verb unit of numbering, at r ₃there is no afterwards noun pronoun unit, also there is no coordinate noun pronoun mix vector, and numbering is greater than 3 function and does not also exist, therefore, { z ₃}={ e}.

Thus, via step 120, process, for above-mentioned example, can generate the value set that obtains each element.

Step 130, according to the possible value of described leading question element, subject element, predicate element, object element, the institute that obtains syntax vector that each predicate verb unit is corresponding is value likely, and described syntax vector comprises leading question element, subject element, predicate element, object element.

As previously mentioned, each subject-predicate matching structure can represent by the mode of syntax vector.According to the operation result of step 120, for the pretreated statement S=shown in above-mentioned table 1 " I can understand what what you said meant ", have:

{r ₁}＝{can?understand}

{x ₁}＝{e}

{y ₁}＝{I，e}

{z ₁}＝{you，f ₂，f ₃，e}

Use the multiplicative principle in combinatorics: f ₁(x ₁, y ₁, r ₁, z ₁)=(seen below list)

Sequence number	Row matrix f 1	Sequence number	Row matrix f 1
				(1-1)	f 1＝(e，I，r 1，you)	(1-5)	f 1＝(e，e，r 1，you)
(1-2)	f 1＝(e，I，r 1，f 2)	(1-6)	f 1＝(e，e，f 1，f 2)
				(1-3)	f 1＝(e，I，r 1，f 3)	(1-7)	f 1＝(e，e，r 1，f 3)
(1-4)	f 1＝(e，I，r 1，e)	(1-8)	f 1＝(e，e，r 1，e)

With sequence valve, replace constant, obtain: f ₁(x ₁, y ₁, r ₁, z ₁)=(seen below list)

Sequence number	Row matrix f 1	Sequence number	Row matrix f 1
				(1-1)	f 1＝(e，1，2，5)	(1-5)	f 1＝(e，e，2，5)
(1-2)	f 1＝(e，1，2，f 2)	(1-6)	f 1＝(e，e，2，f 2)

(1-3)	f 1＝(e，1，2，f 3)	(1-7)	f 1＝(e，e，2，f 3)
				(1-4)	f 1＝(e，1，2，e)	(1-8)	f 1＝(e，e，2，e)

{r ₂}＝{said}

{x ₂}＝{what?A，what?B，e}

{y ₂}＝{I，you，f ₁，e}

{z ₂}＝{f ₃，e}

Use the multiplicative principle in combinatorics: f ₂(x ₂, y ₂, r ₂, z ₂)=(seen below list)

Sequence number	Row matrix f 2	Sequence number	Row matrix f 2
				(2-1)	f 2＝(what?A，I，r 2，f 3)	(2-13)	f 2＝(what?B，f 1，r 2，f 3)
(2-2)	f 2＝(what?A，I，r 2，e)	(2-14)	f 2＝(what?B，f 1，r 2，e)
				(2-3)	f 2＝(what?A，you，r 2，f 3)	(2-15)	f 2＝(what?B，e，r 2，f 3)
(2-4)	f 2＝(what?A，you，r 2，e)	(2-16)	f 2＝(what?B，e，r 2，e)
				(2-5)	f 2＝(what?A，f 1，r 2，f 3)	(2-17)	f 2＝(e，I，r 2，f 3)
(2-6)	f 2＝(what?A，f 1，r 2，e)	(2-18)	f 2＝(e，I，r 2，e)
				(2-7)	f 2＝(what?A，e，r 2，f 3)	(2-19)	f 2＝(e，you，r 2，f 3)
(2-8)	f 2＝(what?A，e，r 2，e)	(2-20)	f 2＝(e，you，r 2，e)
				(2-9)	f 2＝(what?B，I，r 2，f 3)	(2-21)	f 2＝(e，f 1，r 2，f 3)
(2-10)	f 2＝(what?B，I，r 2，e)	(2-22)	f 2＝(e，f 1，r 2，e)
				(2-11)	f 2＝(what?B，you，r 2，f 3)	(2-23)	f 2＝(e，e，r 2，f 3)
(2-12)	f 2＝(what?B，you，r 2，e)	(2-24)	f 2＝(e，e，r 2，e)

With sequence valve, replace constant, obtain: f ₂(x ₂, y ₂, r ₂, z ₂)=(seen below list)

Sequence number	Row matrix f 2	Sequence number	Row matrix f 2
				(2-1)	f 2＝(3，1，6，f 3)	(2-13)	f 2＝(4，f 1，6，f 3)
(2-2)	f 2＝(3，1，6，e)	(2-14)	f 2＝(4，f 1，6，e)
				(2-3)	f 2＝(3，5，6，f 3)	(2-15)	f 2＝(4，e，6，f 3)
(2-4)	f 2＝(3，5，6，e)	(2-16)	f 2＝(4，e，6，e)
				(2-5)	f 2＝(3，f 1，6，f 3)	(2-17)	f 2＝(e，1，6，f 3)
(2-6)	f 2＝(3，f 1，6，e)	(2-18)	f 2＝(e，1，6，e)
				(2-7)	f 2＝(3，e，6，f 3)	(2-19)	f 2＝(e，5，6，f 3)
(2-8)	f 2＝(3，e，6，e)	(2-20)	f 2＝(e，5，6，e)
				(2-9)	f 2＝(4，1，6，f 3)	(2-21)	f 2＝(e，f 1，6，f 3)
(2-10)	f 2＝(4，1，6，e)	(2-22)	f 2＝(e，f 1，6，e)
				(2-11)	f 2＝(4，5，6，f 3)	(2-23)	f 2＝(e，e，6，f 3)
(2-12)	f 2＝(4，5，6，e)	(2-24)	f 2＝(e，e，6，e)

{r ₃}＝{meant}

{x ₃}＝{what?A，what?B，e}

{y ₃}＝{I，you，f ₁，f ₂，e}

{z ₃}＝{e}

Use the multiplicative principle in combinatorics: f ₃(x ₃, y ₃, r ₃, z ₃)=(seen below list)

Sequence number	Row matrix f 3	(3-8)	f 3＝(what?B，f 1，r 3，e)
				(3-1)	f 3＝(what?A，I，r 3，e)	(3-9)	f 3＝(what?B，f 2，r 3，e)
(3-2)	f 3＝(what?A，you，r 3，e)	(3-10)	f 3＝(what?B，e，r 3，e)

(3-3)	f 3＝(what?A，f 1，r 3，e)	(3-11)	f 3＝(e，I，r 3，e)
				(3-4)	f 3＝(what?A，f 2，r 3，e)	(3-12)	f 3＝(e，you，r 3，e)
(3-5)	f 3＝(what?A，e，r 3，e)	(3-13)	f 3＝(e，f 1，r 3，e)
				(3-6)	f 3＝(what?B，I，r 3，e)	(3-14)	f 3＝(e，f 2，r 3，e)
(3-7)	f 3＝(what?B，you，r 3，e)	(3-15)	f 3＝(e，e，r 3，e)

With sequence valve, replace constant, obtain: f ₃(x ₃, y ₃, r ₃, z ₃)=(seen below list)

Sequence number	Row matrix f 3	(3-8)	f 3＝(4，f 1，7，e)
				(3-1)	f 3＝(3，1，7，e)	(3-9)	f 3＝(4，f 2，7，e)
(3-2)	f 3＝(3，5，7，e)	(3-10)	f 3＝(4，e，7，e)
				(3-3)	f 3＝(3，f 1，7，e)	(3-11)	f 3＝(e，1，7，e)
(3-4)	f 3＝(3，f 2，7，e)	(3-12)	f 3＝(e，5，7，e)
				(3-5)	f 3＝(3，e，7，e)	(3-13)	f 3＝(e，f 1，7，e)
(3-6)	f 3＝(4，1，7，e)	(3-14)	f 3＝(e，f 2，7，e)
				(3-7)	f 3＝(4，5，7，e)	(3-15)	f 3＝(e，e，7，e)

Use the multiplicative principle in combinatorics:

|S|＝|f ₁|×|f ₂|×|f ₃|＝8×24×15＝2880

Altogether generating 2880 may matrix solution.

Step 140, according to the institute of all syntax vectors likely value generate at least one syntactic structure may matrix solution, described syntactic structure may matrix solution by forming according to the tactic syntax vector of predicate verb element number.

For the pretreated statement S=shown in above-mentioned table 1 " I can understand what what you said meant " based on f ₁, f ₂and f ₃possible value, can obtain a plurality of may matrix solutions.

Whether the statement that step 150, checking obtain according to syntactic structure possibility matrix solution is identical with described pretreated statement, if identical, by each syntax vector output in this syntactic structure possibility matrix solution, and as one of syntactic structure analysis result.Preferably, utilize word element number to substitute word unit and carry out equivalent substitution, whole plug hole, add operation partially, then whether the statement sequence based on obtaining is that the Serial No. that order increases progressively judges whether identical with pretreated statement.

Step 150 can comprise the steps:

If step 151 exists the sequence valve not occurring in this syntactic structure possibility matrix solution, get rid of this syntactic structure possibility matrix solution; For example, for following possible matrix solution:

Be numbered 4 not appearance of word unit, get rid of.

If step 152 occurs identical sequence valve or occurs identical syntax vector in different syntax vectors, get rid of this syntactic structure possibility matrix solution; For example, for following possible matrix solution:

Be numbered 5 word unit and occurred twice, get rid of.

Step 153, in each may matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there is the intersection contradiction of two syntax vectors after equivalent substitution, get rid of this syntactic structure possibility matrix solution;

For example, for following possible matrix solution:

Above-mentioned matrix is carried out to substitution, f ₂and f ₃the substitution intersection contradiction that has occurred function.Substitution obtains: f ₂=3+ < e+ < 6+ < (4+ < f ₂+ < 7+ < e).There is f in two ends, equation left and right simultaneously ₂, this just occurred () logical contradiction.Get rid of.

Step 154, in each may matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there are two sequence valves that position is converse after equivalent substitution, get rid of this syntactic structure possibility matrix solution;

For example, for following possible matrix solution:

It is carried out to substitution, f ₂=4+ < 5+ < 6+ < 3+ < e+ < 7+ < e, obtains order for (4,5,6,3, e, 7, e), there is the sequence valve that position is converse, get rid of.

Step 155, in any one may matrix solution, if there is no the syntax vector of mutual substitution relation between existence and other syntax vectors, carry out plug hole and operate to obtain the corresponding possibility of all these possibility matrix solutions syntax analytic structure, and whether the statement that checking obtains according to described possibility syntax analytic structure is identical with described pretreated statement, it further comprises:

Step 155.1, first to existing each other the syntax vector of substitution relation to carry out equivalent substitution in this possibility matrix solution, thereby is converted into one group of syntax vector that does not have each other substitution relation by this possibility matrix solution syntax vector in possibility matrix solution is called to first kind syntax vector, the syntax that conversion is obtained vector be called Equations of The Second Kind syntax vector;

Step 155.2, appoints and gets an Equations of The Second Kind syntax vector according to predetermined direction, mark one by one in the sequence valve of each syntax elements; After the sequence valve of mark syntax elements, appoint and get in i syntax elements, only in the first side of this syntax elements, construct unique room; After making sky, appoint and get a syntax vector equations of The Second Kind syntax vector in addition mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, this new syntax vector is designated as and the syntax that whole plug hole is obtained vector, be referred to as the 3rd class syntax vector;

Step 155.3, to the 3rd class syntax vector according to predetermined direction to from vector in first syntax elements of the first side start to vector in the vector that comprises first syntax elements of the second side till each syntax elements, all mark sequence valve; Be positioned at vector in the vector that comprises the element of the first side, does not mark sequence valve; By vector first syntax elements of the second side be designated as will be according to aforementioned manner to vector the syntax vector part of mark, is designated as whipping syntax vector after mark sequence valve, appoint and get a j syntax elements in aforesaid whipping vector, only in this element first side, construct unique room; After making sky, appoint and get an original Equations of The Second Kind syntax vector mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, newly-generated syntax vector is designated as or

The 3rd class syntax vector according to predetermined direction, distich normal vector in each syntax elements all mark sequence valve; After the sequence valve of mark syntax elements, appoint and get one in t syntax elements, in the first side of this syntax elements, construct unique room; After making sky, appoint and get use Equations of The Second Kind syntax vector in the mode of whole plug hole by this vector insert the room of constructing above, and then generate a new vector, this new vector is designated as

Step 155.4, repeated execution of steps 155.3, when the last time makes empty and plug hole step and finishes, the empty and plug hole made carrying out is next time operated, until all Equations of The Second Kind syntaxes are vectorial through last the 3rd class syntax vector of making empty and plug hole step and obtaining all plug hole is complete, finally obtains the 3rd class syntax vector of a single file, and described the 3rd class syntax vector finally obtaining is called to final single file vector;

Step 155.5, if one may syntax all has two sequence valves that position is converse in all described final single file vector corresponding to analytic structure, gets rid of this possibility syntax analytic structure;

Step 155.6, repeated execution of steps 155.2 is to step 155.5 until all may being traversed by syntax analytic structure.

For example, for example as above, a syntactic structure may matrix solution be:

Above-mentioned matrix is converted into linear representation is:

$\left\{\begin{array}{c}{f}_1=e+<1+<2+<5\\ f_2=3+<e+<6+<e\\ f_3=4+<e+<7+<e\end{array}\right.$

Through aforesaid plug hole operation, in each final single file vector, there are two sequence valves that position is converse, get rid of.

For example as above, a syntactic structure may matrix solution be:

Can be linear representation by matrix conversion:

$\left\{\begin{array}{c}{f}_1=e+<1+<2+<f_3\\ f_2=4+<5+<6+<e\\ f_3=3+<f_2+<7+<e\end{array}\right.$

Carry out equivalent substitution operation and obtain statement:

α＝e+＜1+＜2+＜(3+＜(4+＜5+＜6+＜e)+＜7+＜e)

Remove dummy cell e, obtain:

α＝1+＜2+＜(3+＜(4+＜5+＜6)+＜7)

It is identical with pretreated statement, and this nested structure is one of syntactic structure analysis result.

By the above-mentioned matrix of word unit constant substitution, syntactic structure matrix solution can be expressed as:

The linear representation of the S corresponding with this matrix expression is as follows:

$S=\left\{\begin{array}{c}{f}_1(x_1,y_1,r_1,z_1)=I+<\mathrm{can\; unders}\tan d+<f_3\\ f_2(x_2,y_2,r_2,z_2)=\mathrm{what\; B}+<\mathrm{you}+<\mathrm{said}\\ f_3(x_3,y_3,r_3,z_3)=\mathrm{what\; A}+<f_2+<\mathrm{meant}\end{array}\right.$

Accordingly, resolving its syntactic structure of sentence " I can understand what what you said meant " is: I is as the subject of main clause, can understand is as the predicate of main clause, subordinate clause " what what you said meant " is as the object clause of main clause, in this subordinate clause, first what is subordinate clause introducer, and " what you said " is the subject of subordinate clause, meant is the predicate of object clause, and itself does not have object object clause; For " what you said " subordinate clause, it has served as subject clause nested in object clause, and what is introducer, and you is subject, and said is predicate.

Further, described method can also comprise step display, and each syntax vector and corresponding syntax structural relationship in syntactic structure analysis result are shown in human-computer interaction interface with tree structure.

C2 part example 2

Example 2: as another example, the method that the present embodiment is below described is for for example: the resolving of the statement of the labyrinth that " That men who were appointed didn ' t bother the liberals wash ' t remarked upon by the press. " is such.

The word order list of above-mentioned statement after impurity numbering are removed in pre-service is:

Former sentence phrase	Phrase type	Serial number
			That	Subordinate conjunctive word unit	1
men	Noun pronoun unit	2
			who	Subordinate conjunctive word unit	3
were?appointed	Predicate verb unit	4
			didn’t?bother	Predicate verb unit	5
the?liberals	Noun pronoun unit	6

wasn’t?remarked

Predicate verb unit

7

This sentence has three predicate verb unit, is designated as respectively r ₁, r ₂and r ₃.

For r ₁there is { r ₁}={ were appointed}

{ x ₁}={ That, who, e} (e is null character string)

{y ₁}＝{men，e}

{z ₁}＝{f ₂，f ₃，e}

Use the multiplicative principle in combinatorics: f ₁(x ₁, y ₁, r ₁, z ₁)=(seen below list)

Sequence number	Row matrix f 1	Sequence number	Row matrix f 1
				(1-1)	f 1＝(That，men，r 1，f 2)	(1-10)	f 1＝(That，e，r 1，f 3)
(1-2)	f 1＝(who，men，r 1，f 2)	(1-11)	f 1＝(who，e，r 1，f 3)
				(1-3)	f 1＝(e，men，r 1，f 2)	(1-12)	f 1＝(e，e，r 1，f 3)
(1-4)	f 1＝(That，e，r 1，f 2)	(1-13)	f 1＝(That，men，r 1，e)
				(1-5)	f 1＝(who，e，r 1，f 2)	(1-14)	f 1＝(who，men，r 1，e)
(1-6)	f 1＝(e，e，r 1，f 2)	(1-15)	f 1＝(e，men，r 1，e)
				(1-7)	f 1＝(That，men，r 1，f 3)	(1-16)	f 1＝(That，e，r 1，e)
(1-8)	f 1＝(who，men，r 1，f 3)	(1-17)	f 1＝(who，e，r 1，e)
				(1-9)	f 1＝(e，men，r 1，f 3)	(1-18)	f 1＝(e，e，r 1，e)

With sequence valve, replace constant, obtain: f ₁(x ₁, y ₁, r ₁, z ₁)=(seen below list)

Sequence number	Row matrix f 1	Sequence number	Row matrix f 1
				(1-1)	f 1＝(1，2，4，f 2)	(1-10)	f 1＝(1，e，4，f 3)
(1-2)	f 1＝(3，2，4，f 2)	(1-11)	f 1＝(3，e，4，f 3)
				(1-3)	f 1＝(e，2，4，f 2)	(1-12)	f 1＝(e，e，4，f 3)
(1-4)	f 1＝(1，e，4，f 2)	(1-13)	f 1＝(1，2，4，e)
				(1-5)	f 1＝(3，e，4，f 2)	(1-14)	f 1＝(3，2，4，e)
(1-6)	f 1＝(e，e，4，f 2)	(1-15)	f 1＝(e，2，4，e)
				(1-7)	f 1＝(1，2，4，f 3)	(1-16)	f 1＝(1，e，4，e)
(1-8)	f 1＝(3，2，4，f 3)	(1-17)	f 1＝(3，e，4，e)

(1-9)

f 1＝(e，2，4，f 3)

(1-18)

f 1＝(e，e，4，e)

For r ₂there is { r ₂}={ didn ' t bother}

{ x ₂}={ That, who, e} (e is null character string)

{y ₂}＝{men，f ₁，e}

{z ₂}＝{the?liberals，f ₃，e}

Use the multiplicative principle in combinatorics: f ₂(x ₂, y ₂, r ₂, z ₂)=(seen below list)

Sequence number	Row matrix f 2	(2-14)	f 2＝(who，f 1，r 2，f 3)
				(2-1)	f 2＝(That，men，r 2，the?liberals)	(2-15)	f 2＝(e，f 1，r 2，f 3)
(2-2)	f 2＝(who，men，r 2，the?liberals)	(2-16)	f 2＝(That，e，r 2，f 3)
				(2-3)	f 2＝(e，men，r 2，the?liberals)	(2-17)	f 2＝(who，e，r 2，f 3)
(2-4)	f 2＝(That，f 1，r 2，the?liberals)	(2-18)	f 2＝(e，e，r 2，f 3)
				(2-5)	f 2＝(who，f 1，r 2，the?liberals)	(2-19)	f 2＝(That，men，r 2，e)
(2-6)	f 2＝(e，f 1，r 2，the?liberals)	(2-20)	f 2＝(who，men，r 2，e)
				(2-7)	f 2＝(That，e，r 2，the?liberals)	(2-21)	f 2＝(e，men，r 2，e)
(2-8)	f 2＝(who，e，r 2，the?liberals)	(2-22)	f 2＝(That，f 1，r 2，e)
				(2-9)	f 2＝(e，e，r 2，the?liberals)	(2-23)	f 2＝(who，f 1，r 2，e)
(2-10)	f 2＝(That，men，r 2，f 3)	(2-24)	f 2＝(e，f 1，r 2，e)
				(2-11)	f 2＝(who，men，r 2，f 3)	(2-25)	f 2＝(That，e，r 2，e)
(2-12)	f 2＝(e，men，r 2，f 3)	(2-26)	f 2＝(who，e，r 2，e)
				(2-13)	f 2＝(That，f 1，r 2，f 3)	(2-27)	f 2＝(e，e，r 2，e)

With sequence valve, replace constant, obtain: f ₂(x ₂, y ₂, r ₂, z ₂)=(seen below list)

Sequence number	Row matrix f 2	(2-14)	f 2＝(3，f 1，5，f 3)
				(2-1)	f 2＝(1，2，5，6)	(2-15)	f 2＝(e，f 1，5，f 3)
(2-2)	f 2＝(3，2，5，6)	(2-16)	f 2＝(1，e，5，f 3)
				(2-3)	f 2＝(e，2，5，6)	(2-17)	f 2＝(3，e，5，f 3)
(2-4)	f 2＝(1，f 1，5，6)	(2-18)	f 2＝(e，e，5，f 3)
				(2-5)	f 2＝(3，f 1，5，6)	(2-19)	f 2＝(1，2，5，e)

(2-6)	f 2＝(e，f 1，5，6)	(2-20)	f 2＝(3，2，5，e)
				(2-7)	f 2＝(1，e，5，6)	(2-21)	f 2＝(e，2，5，e)
(2-8)	f 2＝(3，e，5，6)	(2-22)	f 2＝(1，f 1，5，e)
				(2-9)	f 2＝(e，e，5，6)	(2-23)	f 2＝(3，f 1，5，e)
(2-10)	f 2＝(1，2，5，f 3)	(2-24)	f 2＝(e，f 1，5，e)
				(2-11)	f 2＝(3，2，5，f 3)	(2-25)	f 2＝(1，e，5，e)
(2-12)	f 2＝(e，2，5，f 3)	(2-26)	f 2＝(3，e，5，e)
				(2-13)	f 2＝(1，f 1，5，f 3)	(2-27)	f 2＝(e，e，5，e)

For r ₃have: { r ₃}={ wasn ' t remarked}

{x ₃}＝{That，who，e}

{y ₃}＝{men，the?liberals，f ₁，f ₂，e}

{z ₃}＝{e}

Use the multiplicative principle in combinatorics: f ₃(x ₃, y ₃, r ₃, z ₃)=(seen below list)

Sequence number	Row matrix f 3	(3-8)	f 3＝(who，f 1，r 3，e)
				(3-1)	f 3＝(That，men，r 3，e)	(3-9)	f 3＝(e，f 1，r 3，e)
(3-2)	f 3＝(who，men，r 3，e)	(3-10)	f 3＝(That，f 2，r 3，e)
				(3-3)	f 3＝(e，men，r 3，e)	(3-11)	f 3＝(who，f 2，r 3，e)
(3-4)	f 3＝(That，the?liberals，r 3，e)	(3-12)	f 3＝(e，f 2，r 3，e)
				(3-5)	f 3＝(who，the?liberals，r 3，e)	(3-13)	f 3＝(That，e，r 3，e)
(3-6)	f 3＝(e，the?liberals，r 3，e)	(3-14)	f 3＝(who，e，r 3，e)
				(3-7)	f 3＝(That，f 1，r 3，e)	(3-15)	f 3＝(e，e，r 3，e)

With sequence valve, replace constant, f ₃(x ₃, y ₃, r ₃, z ₃)=(seen below list)

Sequence number

Row matrix f 3

(3-8)

f 3＝(3，f 1，7，e)

(3-1)	f 3＝(1，2，7，e)	(3-9)	f 3＝(e，f 1，7，e)
				(3-2)	f 3＝(3，2，7，e)	(3-10)	f 3＝(1，f 2，7，e)
(3-3)	f 3＝(e，2，7，e)	(3-11)	f 3＝(3，f 2，7，e)
				(3-4)	f 3＝(1，6，7，e)	(3-12)	f 3＝(e，f 2，7，e)
(3-5)	f 3＝(3，6，7，e)	(3-13)	f 3＝(1，e，7，e)
				(3-6)	f 3＝(e，6，7，e)	(3-14)	f 3＝(3，e，7，e)
(3-7)	f 3＝(1，f 1，7，e)	(3-15)	f 3＝(e，e，7，e)

Use the multiplicative principle in combinatorics:

|S|＝|f ₁|×|f ₂|×|f ₃|＝18×27×15＝7290

Altogether generating 7290 may matrix solution.

To all syntactic structure possibility matrix solutions, operation matrix substitution solver, structural modifications program, can obtain the possible matrix solution as syntactic structure parsing net result:

This example sentence is the typical example sentence that whole plug hole method herein is successfully processed.Through aforesaid whole plug hole processing herein, one of whole plug hole result of above-mentioned possible matrix solution is a following final single file vector: e+ < (1+ < 2+ < (3+ < e+ < 4+ < e)+< 5+ < 6)+< 7+ < e.There is not backward number in this final single file vector, is reasonably final single file vector.This final single file vector, identical with former sentence.This possibility matrix solution is the correct syntactic structure analysis result of this example sentence just also.

The numbering of above-mentioned possibility matrix solution is reduced to word unit, obtains following form:

This matrix is converted into linear representation:

$\left\{\begin{array}{c}{f}_1(x_1,y_1,r_1,z_1)=\mathrm{who}+<e+<\mathrm{were\; appinted}+<e\\ f_2(x_2,y_2,r_2,z_2)=\mathrm{That}+<\mathrm{men}+<{\mathrm{didn}}^{,}\mathrm{t\; bother}+<\\ \mathrm{the\; liberals}\\ f_3(x_3,y_3,r_3,z_3)=e+<f_2+<{\mathrm{wasn}}^{,}\mathrm{t\; remarked}+<e\end{array}\right.$

Removing e obtains:

$\left\{\begin{array}{c}{f}_1(x_1,y_1,r_1,z_1)=\mathrm{who}+<\mathrm{were\; appinted}\\ f_2(x_2,y_2,r_2,z_2)=\mathrm{That}+<\mathrm{men}+<{\mathrm{didn}}^{,}\mathrm{t\; bother}+<\\ \mathrm{the\; liberals}\\ f_3(x_3,y_3,r_3,z_3)=f_2+<{\mathrm{wasn}}^{,}\mathrm{t\; remarked}\end{array}\right.$

Thus, obtain the correct parsing for above-mentioned statement example, that is: f ₃main clause, kernel sentence namely; f ₂f ₃subject, i.e. subject clause; f ₁be attributive clause, modify men.

This example can show the superiority of this method preferably.For above-mentioned statement, current computer industry generally acknowledge two kinds is FA natural language syntactic structure resolver-Berkeley resolver (Berkeley Parser) and Stamford resolver (Stanford Parser) in the world, in submitting to the application, what provide is still wrong analysis result.The result that these two kinds of devices provide is identical.Its result is as follows:

①That?men?didn’t?bother；

②who?were?appointed；

③the?liberals?wasn’t?remarked?upon?by?the?press.

1. be main clause, namely kernel sentence; 3. be object 1., that is, and object clause; 2. be attributive clause, modify men; That is determiner, modifies men.

In the middle of English, if subject clause is positioned at full sentence beginning of the sentence, and guided by that, that cannot omit, even if spoken language is also like this.In the method for the invention; owing to sentence being treated to syntax vector; therefore be just this part of subject clause That men didn ' t bother the liberals; in the process of resolving, reserve sufficient space, protected fully its possibility generating as a complete subordinate sentence.

For the parsing of the subject clause of that guiding this great technical leak of often makeing mistakes, in submitting to the application, above-mentioned two kinds of natural language syntactic structure resolvers advanced in the world still could not make up.

C3 part example 3

Example 3: as another example, the method that the present embodiment is below described is for for example: the resolving of the statement of the labyrinth that " Jack who has a beautiful car is a businessman. " is such.The word order list of above-mentioned statement after impurity numbering are removed in pre-service is:

Former sentence phrase	Phrase type	Serial number
			Jack	Noun pronoun unit	1
who	Subordinate conjunctive word unit	2
			has	Predicate verb unit	3
a?car	Noun pronoun unit	4
			is	Predicate verb unit	5
a?businessman	Noun pronoun unit	6

This sentence has two predicate verb unit, is designated as respectively r ₁and r ₂.

For r ₁there is { r ₁}={ has}

{ x ₁}={ who, e} (e is null character string)

{y ₁}＝{Jack，e}

{z ₁}＝{a?car，f ₂，e}

Use the multiplicative principle in combinatorics: f ₁(x ₁, y ₁, r ₁, z ₁)=(seen below list)

Sequence number	Row matrix f 1	Sequence number	Row matrix f 1
				(1-1)	f 1＝(who，Jack，r 1，a?car)	(1-7)	f 1＝(who，e，r 1，f 2)
(1-2)	f 1＝(e，Jack，r 1，a?car)	(1-8)	f 1＝(e，e，r 1，f 2)
				(1-3)	f 1＝(who，e，r 1，a?car)	(1-9)	f 1＝(who，Jack，r 1，e)

(1-4)	f 1＝(e，e，r 1，a?car)	(1-10)	f 1＝(e，Jack，r 1，e)
				(1-5)	f 1＝(who，Jack，r 1，f 2)	(1-11)	f 1＝(who，e，r 1，e)
(1-6)	f 1＝(e，Jack，r 1，f 2)	(1-12)	f 1＝(e，e，r 1，e)

With sequence valve, replace constant, f ₁(x ₁, y ₁, r ₁, z ₁)=(seen below list)

Sequence number	Row matrix f 1	Sequence number	Row matrix f 1
				(1-1)	f 1＝(2，1，3，4)	(1-7)	f 1＝(2，e，3，f 2)
(1-2)	f 1＝(e，1，3，4)	(1-8)	f 1＝(e，e，3，f 2)
				(1-3)	f 1＝(2，e，3，4)	(1-9)	f 1＝(2，1，3，e)
(1-4)	f 1＝(e，e，3，4)	(1-10)	f 1＝(e，1，3，e)
				(1-5)	f 1＝(2，1，3，f 2)	(1-11)	f 1＝(2，e，3，e)
(1-6)	f 1＝(e，1，3，f 2)	(1-12)	f 1＝(e，e，3，e)

For r ₂there is { r ₂}={ is}

{ x ₂}={ who, e} (e is null character string)

{y ₂}＝{Jack，a?car，f ₁，e}

{z ₂}＝{a?businessman，e}

Use the multiplicative principle in combinatorics: f ₂(x ₂, y ₂, r ₂, z ₂)=(seen below list)

Sequence number	Row matrix f 2	Sequence number	Row matrix f 2
				(2-1)	f 2＝(who，Jack，r 2，a?businessman)	(2-9)	f 2＝(who，Jack，r 2，e)
(2-2)	f 2＝(e，Jack，r 2，a?businessman)	(2-10)	f 2＝(e，Jack，r 2，e)
				(2-3)	f 2＝(who，a?car，r 2，a?businessman)	(2-11)	f 2＝(who，a?car，r 2，e)
(2-4)	f 2＝(e，a?car，r 2，a?businessman)	(2-12)	f 2＝(e，a?car，r 2，e)

(2-5)	f 2＝(who，f 1，r 2，a?businessman)	(2-13)	f 2＝(who，f 1，r 2，e)
				(2-6)	f 2＝(e，f 1，r 2，a?businessman)	(2-14)	f 2＝(e，f 1，r 2，e)
(2-7)	f 2＝(who，e，r 2，a?businessman)	(2-15)	f 2＝(who，e，r 2，e)
				(2-8)	f 2＝(e，e，r 2，a?businessman)	(2-16)	f 2＝(e，e，r 2，e)

With sequence valve, replace constant, f ₂(x ₂, y ₂, r ₂, z ₂)=(seen below list)

Sequence number	Row matrix f 2	Sequence number	Row matrix f 2
				(2-1)	f 2＝(2，1，5，6)	(2-9)	f 2＝(2，1，5，e)
(2-2)	f 2＝(e，1，5，6)	(2-10)	f 2＝(e，1，5，e)
				(2-3)	f 2＝(2，4，5，6)	(2-11)	f 2＝(2，4，5，e)
(2-4)	f 2＝(e，4，5，6)	(2-12)	f 2＝(e，4，5，e)
				(2-5)	f 2＝(2，f 1，5，6)	(2-13)	f 2＝(2，f 1，5，e)
(2-6)	f 2＝(e，f 1，5，6)	(2-14)	f 2＝(e，f 1，5，e)
				(2-7)	f 2＝(2，e，5，6)	(2-15)	f 2＝(2，e，5，e)
(2-8)	f 2＝(e，e，5，6)	(2-16)	f 2＝(e，e，5，e)

Use the multiplicative principle in combinatorics:

|S|＝|f ₁|×|f ₂|＝12×16＝192

Altogether generating 192 may matrix solution.

Can obtain resolving as syntactic structure the possible matrix solution of net result:

This example sentence is the typical example sentence that whole plug hole method herein is successfully processed.Through aforesaid whole plug hole processing herein, above-mentioned possible matrix solution has obtained a unique final single file vector that does not occur backward number: e+ < 1+ < (2+ < e+ < 3+ < 4)+< 5+ < 6.This final single file vector is reasonably final single file vector.The syntax sequence valve numbering of this final single file vector, identical with former sentence.This possibility matrix solution is the correct syntactic structure analysis result of this example sentence just also.

The numbering of above-mentioned possibility matrix solution is reduced to word unit, obtains following form:

This matrix is converted into linear representation:

$\left\{\begin{array}{c}{f}_1(x_1,y_1,r_1,z_1)=\mathrm{who}+<e+<\mathrm{has}+<\mathrm{a\; car}\\ f_2(x_2,y_2,r_2,z_2)=e+<\mathrm{Jack}+<\mathrm{is}+<\mathrm{a\; bu}\sin \mathrm{essman}\end{array}\right.$

Removing e obtains:

$\left\{\begin{array}{c}{f}_1(x_1,y_1,r_1,z_1)=\mathrm{who}+<\mathrm{has}+<\mathrm{a\; car}\\ f_2(x_2,y_2,r_2,z_2)=\mathrm{Jack}+<\mathrm{is}+<\mathrm{a\; bu}\sin \mathrm{essman}\end{array}\right.$

C4 part example 4

Example 4: as another example, the method that the present embodiment is below described is for for example: the resolving of the statement of the parallel construction that " After Jack, Mary and Linda left, I gave my son a new book. " is such.

The word order list of above-mentioned statement after impurity numbering are removed in pre-service is:

Former sentence phrase	Phrase type	Serial number
			After	Subordinate conjunctive word unit	1
Jack	Noun pronoun unit	2
			Mary	Noun pronoun unit	3
and	Conjunctive word arranged side by side unit	4
			Linda	Noun pronoun unit	5
left	Predicate verb unit	6
			I	Noun pronoun unit	7

gave	Predicate verb unit	8
			my?son	Noun pronoun unit	9
a?book	Noun pronoun unit	10

By following steps, comprise and generate coordinate noun pronoun mix vector family:

S2.1 chooses unduplicated two noun pronoun unit:

If A does not have other word unit between these two noun pronoun unit, using these two noun pronoun unit as a coordinate noun pronoun mix vector, and retain this coordinate noun pronoun mix vector;

If there is other word unit in B between these two noun pronoun unit, check between each the word unit between these two noun pronoun unit: if any one the word unit between these two noun pronoun unit, all noun pronoun unit or conjunctive word arranged side by side unit, using selected two noun pronoun unit and all words unit between these two noun pronoun unit as a coordinate noun pronoun mix vector, and retain this coordinate noun pronoun mix vector; Otherwise, do not generate coordinate noun pronoun mix vector;

The capable S2.1 of S2.2 retry, until the array mode of all noun pronoun unit is traversed, generates all coordinate noun pronoun mix vectors that obtain;

If there is coordinate noun pronoun mix vector in this possibility syntax analytic structure of S2.3, all coordinate noun pronoun mix vectors are divided, thereby form several coordinate noun pronoun mix vector families, make: in each coordinate noun pronoun mix vector family, each the coordinate noun pronoun mix vector comprising in this coordinate noun pronoun mix vector family has all comprised two common noun pronoun unit.

S2.4, in each noun pronoun mix vector family, chooses the word unit of the numbering maximum comprising in all noun pronoun mix vectors, and the major term unit as this noun pronoun mix vector family, is used during in order to follow-up generation subject; Choose the word unit of the numbering minimum comprising in all noun pronoun mix vectors, as the minimum word unit of this noun pronoun mix vector family, during in order to follow-up generation object, use.

Generating subject element set { y ₁, { y ₂process in, move subject generating algorithm arranged side by side as follows:

1. A (S) takes out all NPI phrases in former sentence, all VNP phrases, all NOMP phrases, and classifies all NPI phrases in former sentence, all VNP phrases, all NOMP phrases as a set, and this set is designated as to Ψ={ Jack, Mary, Linda, I, my son, a book}={2,3,5,7,9,10}.

2. B (Ψ) represent according to mode get set Ψ={ in 2,3,5,7,9,10}, whole combinations of any two elements, establish set $C_{\mathrm{\&Psi;}}^2=\left\{\right\{\mathrm{2,3}\},\{\mathrm{2,5}\},\{\mathrm{2,7}\},\{\mathrm{2,9}\},\{\mathrm{2,10}\},\{\mathrm{3,5}\},\{\mathrm{3,7}\},\{\mathrm{3,9}\},\{\mathrm{3,10}\},$ $\left\{\mathrm{5,7}\right\},\left\{\mathrm{5,9}\right\},\left\{\mathrm{5,10}\right\},\left\{\mathrm{7,9}\right\},\left\{\mathrm{10,7}\right\},\left\{\mathrm{10,9}\right\}\}.$ B (Ψ)={ { 2,3}, { 2,5}, { 2,7}, { 2,9}, { 2,10}, { 3,5}, { 3,7}, { 3,9}, { 3,10}, { 5,7}, { 5,9}, { 5,10}, { 7,9}, { 10,7}, { 10,9}}.

3. the result of K (α, β) to function of a single variable B (Ψ), to appointing one that gives according to element the arrangement from small to large of the syntax sequence valve in former sentence S.Might as well establish can obtain ordered pair ? the ordered pair generating is:

{<2，3>，<2，5>，<2，7>，<2，9>，<2，10>，<3，5>，<3，7>，<3，9>，<3，10>，<5，7>，<5，9>，<5，10>，<7，9>，<7，10>，<9，10>}；

If set $\mathrm{\&pi;}(<{\mathrm{\&alpha;}}_i^t,{\mathrm{\&alpha;}}_j^t>)=\left\{{\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\right|\mathrm{\&tau;}\left({\mathrm{\&alpha;}}_i^t\right)<\mathrm{\&tau;}\left({\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\right)<\mathrm{\&tau;}\left({\mathrm{\&alpha;}}_j^t\right),{\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\&Element;S\},$ And then set up a continuous word string formula ${\mathrm{\&Phi;}}^t={\mathrm{\&alpha;}}_i^t+<{\mathrm{\&beta;}}_k+<...+{\mathrm{\&beta;}}_l+<{\mathrm{\&alpha;}}_j^t,$ Wherein be in former sentence S from arrive one group of adjacent continuous word string or empty word string, and $\mathrm{\&pi;}(<{\mathrm{\&alpha;}}_i^t,{\mathrm{\&alpha;}}_j^t>)=\{{\mathrm{\&beta;}}_k,...,{\mathrm{\&beta;}}_l\}.$ ? $K(\mathrm{\&alpha;},\mathrm{\&beta;})={\mathrm{\&Phi;}}^t={\mathrm{\&alpha;}}_i^t+<{\mathrm{\&beta;}}_k+<...$ $+<{\mathrm{\&beta;}}_l+<{\mathrm{\&alpha;}}_j^t.$ Φ ¹=2+ < e+ < 3, Φ ²=2+ < 3+ < 4+ < 5, Φ ³=2+ < 3+ < 4+ < 5+ < 6+ < 7, Φ ⁴=2+ < 3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9, Φ ⁵=2+ < 3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9+ < 10, Φ ⁶=3+ < 4+ < 5, Φ ⁷=3+ < 4+ < 5+ < 6+ < 7, Φ ⁸=3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9, Φ ⁹=3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9+ < 10, Φ ¹⁰=5+ < 6+ < 7, Φ ¹¹=5+ < 6+ < 7+ < 8+ < 9, Φ ¹²=5+ < 6+ < 7+ < 8+ < 9+ < 10, Φ ¹³=7+ < 8+ < 9, Φ ¹⁴=7+ < 8+ < 9+ < 10, Φ ¹⁵=9+ < e+ < 10.

4. H (Φ ^t) binary function K (α, β) is generated check: if to appoint to element γ ∈ Φ ^t, and and have: γ=NPI or γ=VNP or γ=NOMP or γ=CONJ or γ=e, by Φ ^tmark change into be called Φ ^tgenerate if set set ${\mathrm{\&Omega;}}_{\mathrm{NP}}=\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6,{\mathrm{\&Phi;}}_{\mathrm{NP}}^{15}\}.$ ? $H(\mathrm{\&alpha;},\mathrm{\&beta;})={\mathrm{\&Omega;}}_{\mathrm{NP}}=\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6,{\mathrm{\&Phi;}}_{\mathrm{NP}}^{15}\}.$

5. M (α, β) represents a set of getting for appointing if set existence correspondence define a set family, this set family is by comprising set all set form, this set family is designated as $I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)=\left\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\right|\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\&SubsetEqual;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;{\mathrm{\&Omega;}}_{\mathrm{NP}}\}.$ ? $I_1\left(\right\{\mathrm{2,3}\left\}\right)=\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2\},$ $I_2\left(\right\{\mathrm{3,5}\left\}\right)=$ $\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6\},$ m (α, β)={ I ₁(2,3}), I ₂(3,5}), I ₃({ 9,10}) }.

6. the result of N (α, β) to binary function M (α, β) for appointing, get set if set there is corresponding set family construct a new set as follows $P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\left\{\mathrm{\&alpha;}\right|\mathrm{\&alpha;}\&Element;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},\mathrm{\&alpha;}\&NotEqual;e,{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\}.$ Can obtain P[I ₁(2,3})]={ 2,3,4,5}, P[I ₂(3,5})]={ 2,3,4,5}, P[I ₃(9,10})]={ 9,10}.

7. the result of U (α) to binary function N (α, β) get if $P^{\max }\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\mathrm{\&delta;},\mathrm{\&delta;}\&Element;P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ To appointing the element γ giving, $\mathrm{\&gamma;}\&Element;P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ There is τ (γ)≤τ (δ).P ^max[I ₁(2,3})]=5, P ^max[I ₂(3,5})]=5, P ^max[I ₃(9,10})]=10.

For r ₁there is { r ₁}={ left}, is numbered 6.The choosing method of corresponding subject is: when there not being r _k-1time: { y _k}=NPI _yk∪ VNP _yk∪ NOMP _k∪ G _k∪ { e};

Wherein:

Wherein:

In above-mentioned formula, G _krepresent maximal value numbering be less than corresponding predicate verb element number all coordinate noun pronoun set family and set.

Have:

Have: $G_1=\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6\}=\{2+<e+<\mathrm{3,2}+<3+<4+<\mathrm{5,3}+<$ $4+<5\}.$ R ₁the set of corresponding subject element is: $\left\{y_1\right\}=\{\mathrm{Jack},\mathrm{Mary},\mathrm{Linda},{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6\}.$

Generating object element set { z ₁, { z ₂process in, move object generating algorithm arranged side by side as follows:

1. A (S) takes out all NPI phrases in former sentence, all VNP phrases, all NOMP phrases, and classifies all NPI phrases in former sentence, all VNP phrases, all NOMP phrases as a set, and this set is designated as to Ψ={ Jack, Mary, Linda, I, my son, a book}={2,3,5,7,9,10}.

2. B (Ψ) represent according to mode get set Ψ={ in 2,3,5,7,9,10}, whole combinations of any two elements, establish set $C_{\mathrm{\&Psi;}}^2=\left\{\right\{\mathrm{2,3}\},\{\mathrm{2,5}\},\{\mathrm{2,7}\},\{\mathrm{2,9}\},\{\mathrm{2,10}\},\{\mathrm{3,5}\},\{\mathrm{3,7}\},\{\mathrm{3,9}\},\{\mathrm{3,10}\},$ $\left\{\mathrm{5,7}\right\},\left\{\mathrm{5,9}\right\},\left\{\mathrm{5,10}\right\},\left\{\mathrm{7,9}\right\},\left\{\mathrm{10,7}\right\},\left\{\mathrm{10,9}\right\}\}.$ B (Ψ)={ { 2,3}, { 2,5}, { 2,7}, { 2,9}, { 2,10}, { 3,5}, { 3,7}, { 3,9}, { 3,10}, { 5,7}, { 5,9}, { 5,10}, { 7,9}, { 10,7}, { 10,9}}.

3. the result of K (α, β) to function of a single variable B (Ψ), to appointing one that gives according to element the arrangement from small to large of the syntax sequence valve in former sentence S.Might as well establish can obtain ordered pair ? the ordered pair generating is:

{<2，3>，<2，5>，<2，7>，<2，9>，<2，10>，<3，5>，<3，7>，<3，9>，<3，10>，<5，7>，<5，9>，<5，10>，<7，9>，<7，10>，<9，10>}；

If set $\mathrm{\&pi;}(<{\mathrm{\&alpha;}}_i^t,{\mathrm{\&alpha;}}_j^t>)=\left\{{\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\right|\mathrm{\&tau;}\left({\mathrm{\&alpha;}}_i^t\right)<\mathrm{\&tau;}\left({\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\right)<\mathrm{\&tau;}\left({\mathrm{\&alpha;}}_j^t\right),{\mathrm{\&beta;}}_{\mathrm{\&epsiv;}}\&Element;S\},$ And then set up a continuous word string formula ${\mathrm{\&Phi;}}^t={\mathrm{\&alpha;}}_i^t+<{\mathrm{\&beta;}}_k+<...+{\mathrm{\&beta;}}_l+<{\mathrm{\&alpha;}}_j^t,$ Wherein be in former sentence S from arrive one group of adjacent continuous word string or empty word string, and $\mathrm{\&pi;}(<{\mathrm{\&alpha;}}_i^t,{\mathrm{\&alpha;}}_j^t>)=\{{\mathrm{\&beta;}}_k,...,{\mathrm{\&beta;}}_l\}.$ ? $K(\mathrm{\&alpha;},\mathrm{\&beta;})={\mathrm{\&Phi;}}^t={\mathrm{\&alpha;}}_i^t+<{\mathrm{\&beta;}}_k+<...$ $+<{\mathrm{\&beta;}}_l+<{\mathrm{\&alpha;}}_j^t.$ Φ ¹=2+ < e+ < 3, Φ ²=2+ < 3+ < 4+ < 5, Φ ³=2+ < 3+ < 4+ < 5+ < 6+ < 7, Φ ⁴=2+ < 3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9, Φ ⁵=2+ < 3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9+ < 10, Φ ⁶=3+ < 4+ < 5, Φ ⁷=3+ < 4+ < 5+ < 6+ < 7, Φ ⁸=3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9, Φ ⁹=3+ < 4+ < 5+ < 6+ < 7+ < 8+ < 9+ < 10, Φ ¹⁰=5+ < 6+ < 7, Φ ¹¹=5+ < 6+ < 7+ < 8+ < 9, Φ ¹²=5+ < 6+ < 7+ < 8+ < 9+ < 10, Φ ¹³=7+ < 8+ < 9, Φ ¹⁴=7+ < 8+ < 9+ < 10, Φ ¹⁵=9+ < e+ < 10.

4. H (Φ ^t) binary function K (α, β) is generated check: if to appoint to element γ ∈ Φ ^t, and and have: γ=NPI or γ=VNP or γ=NOMP or γ=CONJ or γ=e, by Φ ^tmark change into be called Φ ^tgenerate if set set ${\mathrm{\&Omega;}}_{\mathrm{NP}}=\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6,{\mathrm{\&Phi;}}_{\mathrm{NP}}^{15}\}.$ ? $H(\mathrm{\&alpha;},\mathrm{\&beta;})={\mathrm{\&Omega;}}_{\mathrm{NP}}=\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6,{\mathrm{\&Phi;}}_{\mathrm{NP}}^{15}\}.$

5. M (α, β) represents a set of getting for appointing if set existence correspondence define a set family, this set family is by comprising set all set form, this set family is designated as $I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)=\left\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\right|\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\}\&SubsetEqual;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;{\mathrm{\&Omega;}}_{\mathrm{NP}}\}.$ ? $I_1\left(\right\{\mathrm{2,3}\left\}\right)=\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2\},$ $I_2\left(\right\{\mathrm{3,5}\left\}\right)=$ $\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6\},$ m (α, β)={ I ₁(2,3}), I ₂(3,5}), I ₃({ 9,10}) }.

6. the result of N (α, β) to binary function M (α, β) for appointing, get set if set there is corresponding set family construct a new set as follows $P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\left\{\mathrm{\&alpha;}\right|\mathrm{\&alpha;}\&Element;{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}},\mathrm{\&alpha;}\&NotEqual;e,{\mathrm{\&Phi;}}_{\mathrm{NP}}^{\mathrm{\&epsiv;}}\&Element;I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\}.$ Can obtain P[I ₁(2,3})]={ 2,3,4,5}, P[I ₂(3,5})]={ 2,3,4,5}, P[I ₃(9,10})]={ 9,10}.

7. V (β) represents the result to binary function N (α, β) get if $P^{\min }\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right]=\mathrm{\&delta;},\mathrm{\&delta;}\&Element;P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ To appointing the element γ giving, $\mathrm{\&gamma;}\&Element;P\left[I\left(\right\{{\mathrm{\&alpha;}}_i^p,{\mathrm{\&alpha;}}_j^p\left\}\right)\right],$ There is τ (δ)≤τ (γ).P ^min[I ₁(2,3})]=2, P ^min[I ₂(3,5})]=2, P ^min[I ₃(9,10})]=9.

For r ₂there is { r ₂}={ gave}, is numbered 8.The choosing method of corresponding object is: when there not being r _k+1time: { z _k}=NPI _zk∪ VNP _zk∪ OBJP _k∪ H _k∪ { e};

Wherein:

Wherein: when there not being r _k+1time:

In above-mentioned formula, H _krepresent minimum value numbering be greater than corresponding predicate verb unit all coordinate noun pronoun set set family and set.

Have:

Have:

$H_2=\left\{{\mathrm{\&Phi;}}_{\mathrm{NP}}^{15}\right\}=\{9+<e+<10\}.$ R ₂the set of corresponding object element is: $\left\{z_2\right\}=\{\mathrm{my\; son},$ $\mathrm{a\; book},{\mathrm{\&Phi;}}_{\mathrm{NP}}^{15}\}.$

Note: in the process of processing, coordinate noun pronoun mix vector is done as a whole processing; Coordinate noun pronoun combination is to can not be by other syntaxes vector plug holes; When checks sequence value, directly coordinate noun pronoun is combined to comprised syntax sequence valve substitution.

For r ₁there is { r ₁}={ left}

{ x ₁}={ After, and, e} (e is null character string)

$\left\{y_1\right\}=\{\mathrm{Jack},\mathrm{Mary},\mathrm{Linda},{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6,e\}$

{z ₁}＝{f ₂，e}

Use the multiplicative principle in combinatorics: f ₁(x ₁, y ₁, r ₁, z ₁)=(seen below list)

For r ₂there is { r ₂}={ gave}

{ x ₂}={ After, and, e} (e is null character string)

$\left\{y_2\right\}=\{\mathrm{Jack},\mathrm{Mary},\mathrm{Linda},{\mathrm{\&Phi;}}_{\mathrm{NP}}^1,{\mathrm{\&Phi;}}_{\mathrm{NP}}^2,{\mathrm{\&Phi;}}_{\mathrm{NP}}^6,I,f_1,e\}$

$\left\{z_2\right\}=\{\mathrm{my\; son},\mathrm{a\; book},{\mathrm{\&Phi;}}_{\mathrm{NP}}^{15},e\}$

Use the multiplicative principle in combinatorics: f ₂(x ₂, y ₂, r ₂, z ₂)=(seen below list)

With sequence valve, replace constant, slightly.

Use the multiplicative principle in combinatorics:

|S|＝|f ₁|×|f ₂|＝42×108＝4536

Altogether generating 4536 may matrix solution.

Can obtain resolving as syntactic structure the possible matrix solution of net result:

By above-mentioned, may further reduce by matrix solution, obtain following form:

Note: this result is obtained by whole plug hole method.

This matrix is converted into linear representation:

$\left\{\begin{array}{c}{f}_1(x_1,y_1,r_1,z_1)=\mathrm{After}+<\mathrm{Jack\; Marry\; and\; Linda}+<\mathrm{left}+<e\\ f_2(x_2,y_2,r_2,z_2)=e+<I+<\mathrm{gave}+<\mathrm{my\; son\; a\; book}\end{array}\right.$

Removing e obtains:

$\left\{\begin{array}{c}{f}_1(x_1,y_1,r_1,z_1)=\mathrm{After}+<\mathrm{Jack\; Marry\; and\; Linda}+<\mathrm{left}\\ f_2(x_2,y_2,r_2,z_2)=I+<\mathrm{gave}+<\mathrm{my\; son\; a\; book}\end{array}\right.$

C5 part example 5

Example 5: as another example, the method that the present embodiment is below described is for for example: the resolving of the statement of the parallel construction that " Linda was singing, and Mary was dancing. " is such.

The word order list of above-mentioned statement after impurity numbering are removed in pre-service is:

Former sentence phrase	Phrase type	Serial number
			Linda	Noun pronoun unit	1
was?singing	Predicate verb unit	2
			and	Conjunctive word arranged side by side unit	3
Mary	Noun pronoun unit	4
			was?dancing	Predicate verb unit	5

For r ₁there is { r ₁}={ was singing}

{ x ₁}={ e} (e is null character string)

{y ₁}＝{Linda，e}

{z ₁}＝{Mary，f ₂，e}

Use the multiplicative principle in combinatorics: f ₁(x ₁, y ₁, r ₁, z ₁)=(seen below list)

Sequence number	Row matrix f 1
		(1-1)	f 1＝(e，Linda，r 1，Mary)
(1-2)	f 1＝(e，e，r 1，Mary)
		(1-3)	f 1＝(e，Linda，r 1，f 2)
(1-4)	f 1＝(e，e，r 1，f 2)
		(1-5)	f 1＝(e，Linda，r 1，e)
(1-6)	f 1＝(e，e，r 1，e)

For r ₂there is { r ₂}={ was dancing}

{ x ₂}={ and, e} (e is null character string)

{y ₂}＝{Linda，Mary，f ₁，e}

{z ₂}＝{e}

Use the multiplicative principle in combinatorics: f ₂(x ₂, y ₂, r ₂, z ₂)=(seen below list)

Sequence number	Row matrix f 2
		(2-1)	f 2＝(and，Linda，r 2，e)
(2-2)	f 2＝(and，Mary，r 2，e)
		(2-3)	f 2＝(and，f 1，r 2，e)
(2-4)	f 2＝(and，e，r 2，e)
		(2-5)	f 2＝(e，Linda，r 2，e)
(2-6)	f 2＝(e，Mary，r 2，e)
		(2-7)	f 2＝(e，f 1，r 2，e)
(2-8)	f 2＝(e，e，r 2，e)

With sequence valve, replace constant, slightly.

Use the multiplicative principle in combinatorics:

|S|＝|f ₁|×|f ₂|＝6×8＝48

Altogether generating 48 may matrix solution.

The numbering of above-mentioned possibility matrix solution is reduced to word unit, obtains following form:

This matrix is converted into linear representation:

$\left\{\begin{array}{c}{f}_1(x_1,y_1,r_1,z_1)=e+<\mathrm{Linda}+<\mathrm{was}\sin \mathrm{ging}\\ f_2(x_2,y_2,r_2,z_2)=\mathrm{and}+<\mathrm{Mary}+<\mathrm{was\; dancing}\end{array}\right.$

Removing e obtains:

$\left\{\begin{array}{c}{f}_1(x_1,y_1,r_1,z_1)=e+<\mathrm{Linda}+<\mathrm{was}\sin \mathrm{ging}\\ f_2(x_2,y_2,r_2,z_2)=\mathrm{and}+<\mathrm{Mary}+<\mathrm{was\; dancing}\end{array}\right.$

C6 part example 6

Example 6: as another example, the method that the present embodiment is below described is for for example: the resolving of the statement of the parallel construction that " I know that you have a car and that he has a bike. " is such.

The word order list of above-mentioned statement after impurity numbering are removed in pre-service is:

Former sentence phrase	Phrase type	Serial number
			I	Noun pronoun unit	1
know	Predicate verb unit	2
			that?A	Subordinate conjunctive word unit	3
you	Noun pronoun unit	4
			have	Predicate verb unit	5
a?car	Noun pronoun unit	6
			and	Conjunctive word arranged side by side unit	7
that?B	Subordinate conjunctive word unit	8
			he	Noun pronoun unit	9
has	Predicate verb unit	10
			a?bike	Noun pronoun unit	11

Remember each predicate verb unit r _kthe set of corresponding leading question element is:

{x _k}＝Lead _k∪conj _k∪(conj _kοLead _k)∪{e}

Note predicate verb unit r _kcorresponding leading question element is x _k, it may value set be { x _k.Generate predicate verb unit r _kcorresponding leading question element is x _kpossible value set:

{x ₁}＝{Lead ₁}∪{e}＝{that?A，e}；

{x ₂}＝Lead ₂∪conj ₂∪(conj ₂οLead ₂)∪{e}＝{that?A，and，that?B，Ψ，e}。

Above-mentioned two formula derive from the generating algorithm of leading question element: { x _k}=Lead _k∪ conj _k∪ (conj _kο Lead _k) ∪ { e}.Wherein, (conj _kο Lead _k)={ R _k| R _k=conj+ < Lead, conj < _□r _k, Lead < _□r _k, τ (Lead)=τ (conj)+1}; (conj _kο Lead _k) represent that by a numbering, being less than the conjunctive word arranged side by side unit of corresponding predicate verb element number and one is adjacent and numbers and be less than the set of conjunctive word mix vector that corresponding predicate verb element number and numbering are greater than the subordinate conjunctive word cell formation of this conjunctive word element number arranged side by side.

For r ₁there is { r ₁}={ know}

{ x ₁}={ e} (e is null character string)

{y ₁}＝{I，e}

{z ₁}＝{you，f ₂，f ₃，e}

Use the multiplicative principle in combinatorics: f ₁(x ₁, y ₁, r ₁, z ₁)=(seen below list)

Sequence number	Row matrix f 1	Sequence number	Row matrix f 1
				(1-1)	f 1＝(e，I，r 1，you)	(1-5)	f 1＝(e，I，r 1，f 3)
(1-2)	f 1＝(e，e，r 1，you)	(1-6)	f 1＝(e，e，r 1，f 3)
				(1-3)	f 1＝(e，I，r 1，f 2)	(1-7)	f 1＝(e，I，r 1，e)
(1-4)	f 1＝(e，e，r 1，f 2)	(1-8)	f 1＝(e，e，r 1，e)

For r ₂there is { r ₂}={ have}

{ x ₂}={ that A, e} (e is null character string)

{y ₂}＝{you，I，f ₁，e}

{z ₂}＝{a?car，f ₃，e}

Use the multiplicative principle in combinatorics: f ₂(x ₂, y ₂, r ₂, z ₂)=(seen below list)

Sequence number	Row matrix f 2	Sequence number	Row matrix f 2
				(2-1)	f 2＝(that?A，you，r 2，a?car)	(2-13)	f 2＝(that?A，f 1，r 2，f 3)
(2-2)	f 2＝(e，you，r 2，a?car)	(2-14)	f 2＝(e，f 1，r 2，f 3)
				(2-3)	f 2＝(that?A，I，r 2，a?car)	(2-15)	f 2＝(that?A，e，r 2，f 3)
(2-4)	f 2＝(e，I，r 2，a?car)	(2-16)	f 2＝(e，e，r 2，f 3)

(2-5)	f 2＝(that?A，f 1，r 2，a?car)	(2-17)	f 2＝(that?A，you，r 2，e)
				(2-6)	f 2＝(e，f 1，r 2，a?car)	(2-18)	f 2＝(e，you，r 2，e)
(2-7)	f 2＝(that?A，e，r 2，a?car)	(2-19)	f 2＝(that?A，I，r 2，e)
				(2-8)	f 2＝(e，e，r 2，a?car)	(2-20)	f 2＝(e，I，r 2，e)
(2-9)	f 2＝(that?A，you，r 2，f 3)	(2-21)	f 2＝(that?A，f 1，r 2，e)
				(2-10)	f 2＝(e，you，r 2，f 3)	(2-22)	f 2＝(e，f 1，r 2，e)
(2-11)	f 2＝(that?A，I，r 2，f 3)	(2-23)	f 2＝(that?A，e，r 2，e)
				(2-12)	f 2＝(e，I，r 2，f 3)	(2-24)	f 2＝(e，e，r 2，e)

For r ₃there is { r ₃}={ has}

{ x ₃}={ that A, that B, and, Ψ, e} (e is null character string)

{y ₃}＝{you，I，a?car，he，f ₁，f ₂，e}

{z ₃}＝{a?bike，e}

Use the multiplicative principle in combinatorics: f ₃(x ₃, y ₃, r ₃, z ₃)=(seen below list)

Sequence number	Row matrix f 3	Sequence number	Row matrix f 3
				(3-1)	f 3＝(that?A，you，r 3，a?bike)	(3-36)	f 3＝(that?A，you，r 3，e)
(3-2)	f 3＝(that?B，you，r 3，a?bike)	(3-37)	f 3＝(that?B，you，r 3，e)
				(3-3)	f 3＝(and，you，r 3，a?bike)	(3-38)	f 3＝(and，you，r 3，e)
(3-4)	f 3＝(Ψ，you，r 3，a?bike)	(3-39)	f 3＝(Ψ，you，r 3，e)
				(3-5)	f 3＝(e，you，r 3，a?bike)	(3-40)	f 3＝(e，you，r 3，e)
(3-6)	f 3＝(that?A，I，r 3，a?bike)	(3-41)	f 3＝(that?A，I，r 3，e)
				(3-7)	f 3＝(that?B，I，r 3，a?bike)	(3-42)	f 3＝(that?B，I，r 3，e)

(3-8)	f 3＝(and，I，r 3，a?bike)	(3-43)	f 3＝(and，I，r 3，e)
				(3-9)	f 3＝(Ψ，I，r 3，a?bike)	(3-44)	f 3＝(Ψ，I，r 3，e)
(3-10)	f 3＝(e，I，r 3，a?bike)	(3-45)	f 3＝(e，I，r 3，e)
				(3-11)	f 3＝(that?A，a?car，r 3，a?bike)	(3-46)	f 3＝(that?A，a?car，r 3，e)
(3-12)	f 3＝(that?B，a?car，r 3，a?bike)	(3-47)	f 3＝(that?B，a?car，r 3，e)
				(3-13)	f 3＝(and，a?car，r 3，a?bike)	(3-48)	f 3＝(and，a?car，r 3，e)
(3-14)	f 3＝(Ψ，a?car，r 3，a?bike)	(3-49)	f 3＝(Ψ，a?car，r 3，e)
				(3-15)	f 3＝(e，a?car，r 3，a?bike)	(3-50)	f 3＝(e，a?car，r 3，e)
(3-16)	f 3＝(that?A，he，r 3，a?bike)	(3-51)	f 3＝(that?A，he，r 3，e)
				(3-17)	f 3＝(that?B，he，r 3，a?bike)	(3-52)	f 3＝(that?B，he，r 3，e)
(3-18)	f 3＝(and，he，r 3，a?bike)	(3-53)	f 3＝(and，he，r 3，e)
				(3-19)	f 3＝(Ψ，he，r 3，a?bike)	(3-54)	f 3＝(Ψ，he，r 3，e)
(3-20)	f 3＝(e，he，r 3，a?bike)	(3-55)	f 3＝(e，he，r 3，e)
				(3-21)	f 3＝(that?A，f 1，r 3，a?bike)	(3-56)	f 3＝(that?A，f 1，r 3，e)
(3-22)	f 3＝(that?B，f 1，r 3，a?bike)	(3-57)	f 3＝(that?B，f 1，r 3，e)
				(3-23)	f 3＝(and，f 1，r 3，a?bike)	(3-58)	f 3＝(and，f 1，r 3，e)
(3-24)	f 3＝(Ψ，f 1，r 3，a?bike)	(3-59)	f 3＝(Ψ，f 1，r 3，e)
				(3-25)	f 3＝(e，f 1，r 3，a?bike)	(3-60)	f 3＝(e，f 1，r 3，e)
(3-26)	f 3＝(that?A，f 2，r 3，a?bike)	(3-61)	f 3＝(that?A，f 2，r 3，e)
				(3-27)	f 3＝(that?B，f 2，r 3，a?bike)	(3-62)	f 3＝(that?B，f 2，r 3，e)

(3-28)	f 3＝(and，f 2，r 3，a?bike)	(3-63)	f 3＝(and，f 2，r 3，e)
				(3-29)	f 3＝(Ψ，f 2，r 3，a?bike)	(3-64)	f 3＝(Ψ，f 2，r 3，e)
(3-30)	f 3＝(e，f 2，r 3，a?bike)	(3-65)	f 3＝(e，f 2，r 3，e)
				(3-31)	f 3＝(that?A，e，r 3，a?bike)	(3-66)	f 3＝(that?A，e，r 3，e)
(3-32)	f 3＝(that?B，e，r 3，a?bike)	(3-67)	f 3＝(that?B，e，r 3，e)
				(3-33)	f 3＝(and，e，r 3，a?bike)	(3-68)	f 3＝(and，e，r 3，e)
(3-34)	f 3＝(Ψ，e，r 3，a?bike)	(3-69)	f 3＝(Ψ，e，r 3，e)
				(3-35)	f 3＝(e，e，r 3，a?bike)	(3-70)	f 3＝(e，e，r 3，e)

With sequence valve, replace constant, slightly.

Use the multiplicative principle in combinatorics:

|S|＝|f ₁|×|f ₂|×|f ₃|＝8×24×70＝13440

Altogether generating 13440 may matrix solution.

Can obtain resolving as syntactic structure the possible matrix solution of net result:

The linear representation of the S corresponding with this matrix expression is as follows:

$S=\left\{\begin{array}{c}{f}_1(x_1,y_1,r_1,z_1)=e+<I+<\mathrm{know}+<f_2\\ f_2(x_2,y_2,r_2,z_2)=\mathrm{that\; A}+<\mathrm{you}+<\mathrm{have}+<\mathrm{a\; car}\\ f_3(x_3,y_3,r_3,z_3)=\mathrm{and}+<\mathrm{that\; B}+<\mathrm{he}+<\mathrm{has}+<\mathrm{a\; bike}\end{array}\right.$

The correct structure of this example sentence is: I know is as main clause; That Ayou have a car is first object clause that the predicate know of main clause has under its command; And that B he has a bike is second object clause arranged side by side with first object clause; Subordinate conjunctive word unit that A and that B guide respectively two object clauses; Between two object clauses, by conjunctive word arranged side by side unit and, connected; In the process of processing, conjunctive word mix vector Ψ=and that B does as a whole processing; Conjunctive word mix vector Ψ can not be by other syntax vector plug holes; When checks sequence value, two the syntax sequence valves substitution directly conjunctive word mix vector being comprised.Final result, is to regard second object clause as mode with whole plug hole, and plug hole is at the place, end of first object clause.

Fig. 2 is the schematic diagram of the device of a kind of computer based natural language syntactic structure parsing of the present invention, and shown device comprises:

Fetch unit 21, for reading pretreated phrase data structure to be resolved, the conjunctive word arranged side by side unit, subordinate conjunctive word unit, predicate verb unit, the noun pronoun unit that in described pretreated phrase data structure, only comprise statement, and each word unit is numbered according to the order in described pretreated statement, and marks type;

Element generation parts 22, for to each predicate verb unit, generate corresponding leading question element, subject element, predicate element and object element;

Wherein, the possible value of described leading question element is less than one of the conjunctive word arranged side by side unit of corresponding predicate verb element number or subordinate conjunctive word unit for numbering, or be less than the conjunctive word arranged side by side unit of corresponding predicate verb element number and one by a numbering and be adjacent and number and be less than one of the conjunctive word mix vector that corresponding predicate verb element number and numbering are greater than the subordinate conjunctive word cell formation of this conjunctive word element number arranged side by side, or dummy cell;

The possible value of described subject element is less than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of major term unit is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or at one of syntax vector corresponding to the predicate element of front appearance, or dummy cell;

Described predicate element is corresponding described predicate verb unit;

The possible value numbering of described object element is greater than corresponding predicate verb element number and is less than one of noun pronoun unit of the adjacent predicate verb element number in rear appearance, or the numbering of minimum word unit is greater than corresponding predicate verb element number and is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of the adjacent predicate verb element number in rear appearance, or at one of syntax vector corresponding to the predicate element of rear appearance, or dummy cell;

Vector generates parts 23, be used for according to the possible value of described leading question element, subject element, predicate element and object element, the institute that obtains syntax vector that each predicate verb unit is corresponding is value likely, and described syntax vector comprises leading question element, subject element, predicate element and object element;

Matrix generates parts 24, and for according to the institute of all syntax vectors value likely, generating at least one syntactic structure may matrix solution, described syntactic structure may matrix solution by forming according to the tactic syntax vector of predicate verb element number;

Decider 25, whether identical with described pretreated statement for verifying the statement obtaining according to syntactic structure possibility matrix solution, if identical, using each syntax vector in this syntactic structure possibility matrix solution as one of syntactic structure analysis result;

Wherein, described decider 25 is got rid of ineligible syntactic structure feasible solution by following module operation:

First row, except module, if there is the sequence valve not occurring in this syntactic structure possibility matrix solution, is got rid of this syntactic structure possibility matrix solution;

Second row, except module, if occur identical sequence valve or occur identical syntax vector in different syntax vectors, is got rid of this syntactic structure possibility matrix solution;

The 3rd gets rid of module, in each possibility matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there is the intersection contradiction of two syntax vectors after equivalent substitution, getting rid of this syntactic structure may matrix solution;

The 4th gets rid of module, in each possibility matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there are two sequence valves that position is converse after equivalent substitution, getting rid of this syntactic structure may matrix solution;

The 5th gets rid of module, in any one possibility matrix solution, if there is no the syntax vector of mutual substitution relation between existence and other syntax vectors, carry out plug hole and operate to obtain the corresponding possibility of all these possibility matrix solutions syntax analytic structure, and whether the statement that checking obtains according to described possibility syntax analytic structure is identical with described pretreated statement, it further comprises:

The first submodule, first to existing each other the syntax vector of substitution relation to carry out equivalent substitution in this possibility matrix solution, thereby is converted into one group of syntax vector that does not have each other substitution relation by this possibility matrix solution syntax vector in possibility matrix solution is called to first kind syntax vector, the syntax that conversion is obtained vector be called Equations of The Second Kind syntax vector;

The second submodule, appoint and to get an Equations of The Second Kind syntax vector according to predetermined direction, mark one by one in the sequence valve of each syntax elements; After the sequence valve of mark syntax elements, appoint and get in i syntax elements, only in the first side of this syntax elements, construct unique room; After making sky, appoint and get a syntax vector equations of The Second Kind syntax vector in addition mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, this new syntax vector is designated as and the syntax that whole plug hole is obtained vector, be referred to as the 3rd class syntax vector;

The 3rd submodule, to the 3rd class syntax vector according to predetermined direction to from vector in first syntax elements of the first side start to vector in the vector that comprises first syntax elements of the second side till each syntax elements, all mark sequence valve; Be positioned at vector in the vector that comprises the element of the first side, does not mark sequence valve; By vector first syntax elements of the second side be designated as will be according to aforementioned manner to vector the syntax vector part of mark, is designated as whipping syntax vector after mark sequence valve, appoint and get a j syntax elements in aforesaid whipping vector, only in this element first side, construct unique room; After making sky, appoint and get an original Equations of The Second Kind syntax vector mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, newly-generated syntax vector is designated as or

The 3rd class syntax vector according to predetermined direction, distich normal vector in each syntax elements all mark sequence valve; After the sequence valve of mark syntax elements, appoint and get one in t syntax elements, in the first side of this syntax elements, construct unique room; After making sky, appoint and get use Equations of The Second Kind syntax vector in the mode of whole plug hole by this vector insert the room of constructing above, and then generate a new vector, this new vector is designated as

The 4th submodule, repeat the operation of the 3rd submodule, when the last time makes empty and plug hole step and finishes, the empty and plug hole made carrying out is next time operated, until all Equations of The Second Kind syntaxes are vectorial through last the 3rd class syntax vector of making empty and plug hole step and obtaining all plug hole is complete, finally obtains the 3rd class syntax vector of a single file, and described the 3rd class syntax vector finally obtaining is called to final single file vector;

The 5th submodule, if one may syntax all has two sequence valves that position is converse in all described final single file vector corresponding to analytic structure, gets rid of this possibility syntax analytic structure;

The 6th submodule, the operation that repeats to call the second submodule to the five submodules is until all may being traversed by syntax analytic structure.

Further, described device can also comprise:

Result display unit, carries out each syntax vector and corresponding syntax structural relationship in syntactic structure analysis result to show on human-computer interaction interface with tree structure.

The present invention lays particular emphasis on the accurate parsing problem that solves the combined type sentence structure in natural language.Maximum feature of the present invention is: the character that 1. takes full advantage of function of functions; 2. adopt matrix model and linear model to describe syntax formula; 3. use relative theory generator matrix model and the linear model of combinatorics.Use the present invention, can improve the accuracy rate that natural language syntactic structure is resolved.

From the angle of mathematics, natural language is with discreteness feature, and this difficult point in syntactic structure dissection process just.The present invention, by syntax vector and matrix form are carried out to effective combination, does not both destroy the integrality of sentence structure, does not hinder again the inherent composition analyzed among each and the relation between words and phrases.The present invention adopts matrix model and linear model to portray sentence formula, and this had both met the discreteness feature of natural language, has effectively disclosed again the information association on syntactic structure.

Angle from computer technology, the present invention adopts matrix model and linear model, the natural language statement of single file is converted into the linear nested form of layering, thereby the entanglement of having avoided to a great extent computing machine directly the former sentence mark composition of natural language and partition structure to be occurred, so make the program task of computing machine clearer, succinct.Matrix model of the present invention and linear model, be equivalent to draw many parallel runways for the statement of natural language, makes the statement of natural language simultaneously off on many parallel runways, and then therefrom screen correct result; Also being equivalent to provides a plurality of planes for the statement of natural language, processes the statement of natural language simultaneously, and then therefrom screen correct result in a plurality of planes.

In the process of generator matrix, the present invention has used the relative theory of combinatorics to generate all matrix, and then gets rid of one by one, finally obtains at least one possible correct syntactic structure analysis result.In this course, only need to use mathematical principle and information coding, only need to process the numerical value of real number, each step finally implements to checks whether the numerical value of syntax vector is that ascending order is arranged, namely compare the size of real number, and do not relate to the language message of English itself.

Meanwhile, the present invention need to carry out a large amount of mathematical operations, and the computing power of therefore necessary computer, could effectively realize.

To sum up, the present invention is according to mathematical principle and the corresponding computer technology of the subjects such as Abstract Algebra, set theory, combinatorics, computability theory and computational linguistics, use the mathematical thought of function of functions, by setting up matrix model and linear model, structure recursive function carries out the parsing of natural language syntactic structure; The methods such as integrated use mathematical induction meanwhile, prove important conclusion.

The present invention conceive uniqueness, method ingenious, prove full and accurately, take full advantage of the rule of mathematics and Computer Subject, described method accuracy is higher, has certain technical difficulty.

Claims (8)

1. the method that computer based natural language syntactic structure is resolved, comprising:

S1, read pretreated phrase data structure to be resolved, the conjunctive word arranged side by side unit, subordinate conjunctive word unit, predicate verb unit, the noun pronoun unit that in described pretreated phrase data structure, only comprise statement, and each word unit is numbered according to the order in described pretreated statement, and marks type;

S2, to each predicate verb unit, generate corresponding leading question element, subject element, predicate element and object element;

The possible value of described leading question element is less than one of the conjunctive word arranged side by side unit of corresponding predicate verb element number or subordinate conjunctive word unit for numbering, or be less than the conjunctive word arranged side by side unit of corresponding predicate verb element number and one by a numbering and be adjacent and number and be less than one of the conjunctive word mix vector that corresponding predicate verb element number and numbering are greater than the subordinate conjunctive word cell formation of this conjunctive word element number arranged side by side, or dummy cell;

The possible value of described subject element is less than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of major term unit is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or at one of syntax vector corresponding to the predicate element of front appearance, or dummy cell;

Described predicate element is corresponding described predicate verb unit;

The possible value numbering of described object element is greater than corresponding predicate verb element number and is less than one of noun pronoun unit of the adjacent predicate verb element number in rear appearance, or the numbering of minimum word unit is greater than corresponding predicate verb element number and is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of the adjacent predicate verb element number in rear appearance, or at one of syntax vector corresponding to the predicate element of rear appearance, or dummy cell;

S3, according to the possible value of described leading question element, subject element, predicate element and object element, the institute that obtains syntax vector that each predicate verb unit is corresponding is value likely, and described syntax vector comprises leading question element, subject element, predicate element and object element;

S4, according to institute's value likely of all syntax vectors, generating at least one syntactic structure may matrix solution, described syntactic structure may matrix solution by forming according to the tactic syntax vector of predicate verb element number;

Whether the statement that S5, checking obtain according to syntactic structure possibility matrix solution is identical with described pretreated statement, if identical, using each syntax vector in this syntactic structure possibility matrix solution as one of syntactic structure analysis result;

Wherein, S5 comprises in order the following operation of execution successively, gets rid of ineligible syntactic structure feasible solution:

If S5.1 exists the sequence valve not occurring in this syntactic structure possibility matrix solution, get rid of this syntactic structure possibility matrix solution;

If S5.2 occurs identical sequence valve or occurs identical syntax vector in different syntax vectors, get rid of this syntactic structure possibility matrix solution;

S5.3, in each may matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there is the intersection contradiction of two syntax vectors after equivalent substitution, get rid of this syntactic structure possibility matrix solution;

S5.4, in each may matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there are two sequence valves that position is converse after equivalent substitution, get rid of this syntactic structure possibility matrix solution;

S5.5, in any one may matrix solution, if there is no the syntax vector of mutual substitution relation between existence and other syntax vectors, carry out plug hole and operate to obtain the corresponding possibility of all these possibility matrix solutions syntax analytic structure, and whether the statement that checking obtains according to described possibility syntax analytic structure is identical with described pretreated statement, it further comprises:

S5.5.1, first to existing each other the syntax vector of substitution relation to carry out equivalent substitution in this possibility matrix solution, thereby this possibility matrix solution is converted into one group of syntax vector that does not have each other a substitution relation syntax vector in possibility matrix solution is called to first kind syntax vector, the syntax that conversion is obtained vector be called Equations of The Second Kind syntax vector;

S5.5.2, appoint and to get an Equations of The Second Kind syntax vector according to predetermined direction, mark one by one in the sequence valve of each syntax elements; After the sequence valve of mark syntax elements, appoint and get in i syntax elements, only in the first side of this syntax elements, construct unique room; After making sky, appoint and get a syntax vector equations of The Second Kind syntax vector in addition mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, this new syntax vector is designated as and the syntax that whole plug hole is obtained vector, be referred to as the 3rd class syntax vector;

S5.5.3, vectorial to the 3rd class syntax according to predetermined direction to from vector in first syntax elements of the first side start to vector in the vector that comprises first syntax elements of the second side till each syntax elements, all mark sequence valve; Be positioned at vector in the vector that comprises the element of the first side, does not mark sequence valve; By vector first syntax elements of the second side be designated as will be according to aforementioned manner to vector the syntax vector part of mark, is designated as whipping syntax vector after mark sequence valve, appoint and get a j syntax elements in aforesaid whipping vector, only in this element first side, construct unique room; After making sky, appoint and get an original Equations of The Second Kind syntax vector mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, newly-generated syntax vector is designated as or

The 3rd class syntax vector according to predetermined direction, distich normal vector in each syntax elements all mark sequence valve; After the sequence valve of mark syntax elements, appoint and get one in t syntax elements, in the first side of this syntax elements, construct unique room; After making sky, appoint and get use Equations of The Second Kind syntax vector in the mode of whole plug hole by this vector insert the room of constructing above, and then generate a new vector, this new vector is designated as

S5.5.4, repeat S5.5.3, when the last time makes empty and plug hole step and finishes, the empty and plug hole made carrying out is next time operated, until all Equations of The Second Kind syntaxes are vectorial through last the 3rd class syntax vector of making empty and plug hole step and obtaining all plug hole is complete, finally obtains the 3rd class syntax vector of a single file, and described the 3rd class syntax vector finally obtaining is called to final single file vector;

If all there are two sequence valves that position is converse in may the syntax analytic structure corresponding all described final single file vector of one of S5.5.5, get rid of this possibility syntax analytic structure;

S5.5.6, repeat S5.5.2 to S5.5.5 until all may syntax analytic structure be traversed.

2. the method that computer based natural language syntactic structure according to claim 1 is resolved, is characterized in that, S2 comprises generation coordinate noun pronoun mix vector family:

S2.1 chooses unduplicated two noun pronoun unit:

If A does not have other word unit between these two noun pronoun unit, using these two noun pronoun unit as a coordinate noun pronoun mix vector, and retain this coordinate noun pronoun mix vector;

If there is other word unit in B between these two noun pronoun unit, check between each the word unit between these two noun pronoun unit: if any one the word unit between these two noun pronoun unit, all noun pronoun unit or conjunctive word arranged side by side unit, using selected two noun pronoun unit and all words unit between these two noun pronoun unit as a coordinate noun pronoun mix vector, and retain this coordinate noun pronoun mix vector; Otherwise, do not generate coordinate noun pronoun mix vector;

The capable S2.1 of S2.2 retry, until the array mode of all noun pronoun unit is traversed, generates all coordinate noun pronoun mix vectors that obtain;

If there is coordinate noun pronoun mix vector in this possibility syntax analytic structure of S2.3, all coordinate noun pronoun mix vectors are divided, thereby form several coordinate noun pronoun mix vector families, make: in each coordinate noun pronoun mix vector family, each the coordinate noun pronoun mix vector comprising in this coordinate noun pronoun mix vector family has all comprised two common noun pronoun unit.

S2.4, in each noun pronoun mix vector family, chooses the word unit of the numbering maximum comprising in all noun pronoun mix vectors, and the major term unit as this noun pronoun mix vector family, is used during in order to follow-up generation subject; Choose the word unit of the numbering minimum comprising in all noun pronoun mix vectors, as the minimum word unit of this noun pronoun mix vector family, during in order to follow-up generation object, use.

3. computer based natural language syntactic structure analytic method according to claim 1, is characterized in that, generates corresponding subject element and comprises:

When corresponding predicate verb element number is minimum predicate verb element number, the possible value of described subject element is less than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of its major term unit is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or dummy cell.

When corresponding predicate verb element number is not minimum predicate verb element number, the possible value of described subject element is less than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of major term unit is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or at one of syntax vector corresponding to the predicate verb unit of front appearance, or dummy cell.

4. the method that computer based natural language syntactic structure according to claim 1 is resolved, is characterized in that, generates corresponding object element and comprises:

When corresponding predicate verb element number is maximum predicate verb element number, the possible value of described object element is greater than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of its minimum word unit is greater than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or dummy cell.

When corresponding predicate verb element number is not maximum predicate verb element number, the possible value of described object element is greater than corresponding predicate verb element number and is less than one of noun pronoun unit of the adjacent predicate verb element number in rear appearance for numbering, or the numbering of its minimum word unit is greater than corresponding predicate verb element number and is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of the adjacent predicate verb element number in rear appearance, or at one of syntax vector corresponding to the predicate verb unit of rear appearance, or dummy cell.

5. the method for computer based natural language syntactic structure parsing according to claim 1, is characterized in that, in S4 and two steps of S5, utilizes with syntactic structure and may linear representation solution substitute described syntactic structure possibility matrix solution;

Described syntactic structure possibility linear representation solution and described syntactic structure possibility matrix solution are of equal value;

Described syntactic structure may linear representation solution comprise by forming according to the tactic syntax vector expression of predicate verb element number; Described in each, syntax vector expression is the expression formula that leading question element, subject element, predicate element, the object element of corresponding syntax vector added up in order item by item partially.

6. the method that computer based natural language syntactic structure according to claim 1 is resolved, is characterized in that, described method also comprises:

Each syntax vector and corresponding syntax structural relationship in syntactic structure analysis result are shown in human-computer interaction interface with tree structure.

7. the device that computer based natural language syntactic structure is resolved, comprising:

Fetch unit, for reading pretreated phrase data structure to be resolved, the conjunctive word arranged side by side unit, subordinate conjunctive word unit, predicate verb unit, the noun pronoun unit that in described pretreated phrase data structure, only comprise statement, and each word unit is numbered according to the order in described pretreated statement, and marks type;

Element generation parts, for to each predicate verb unit, generate corresponding leading question element, subject element, predicate element and object element;

Wherein, the possible value of described leading question element is less than one of the conjunctive word arranged side by side unit of corresponding predicate verb element number or subordinate conjunctive word unit for numbering, or be less than the conjunctive word arranged side by side unit of corresponding predicate verb element number and one by a numbering and be adjacent and number and be less than one of the conjunctive word mix vector that corresponding predicate verb element number and numbering are greater than the subordinate conjunctive word cell formation of this conjunctive word element number arranged side by side, or dummy cell;

The possible value of described subject element is less than one of noun pronoun unit of corresponding predicate verb element number for numbering, or the numbering of major term unit is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of corresponding predicate verb element number, or at one of syntax vector corresponding to the predicate element of front appearance, or dummy cell;

Described predicate element is corresponding described predicate verb unit;

The possible value numbering of described object element is greater than corresponding predicate verb element number and is less than one of noun pronoun unit of the adjacent predicate verb element number in rear appearance, or the numbering of minimum word unit is greater than corresponding predicate verb element number and is less than one of coordinate noun pronoun mix vector comprising in all coordinate noun pronoun mix vectors family of the adjacent predicate verb element number in rear appearance, or at one of syntax vector corresponding to the predicate element of rear appearance, or dummy cell;

Vector generates parts, be used for according to the possible value of described leading question element, subject element, predicate element and object element, the institute that obtains syntax vector that each predicate verb unit is corresponding is value likely, and described syntax vector comprises leading question element, subject element, predicate element and object element;

Matrix generates parts, and for according to the institute of all syntax vectors value likely, generating at least one syntactic structure may matrix solution, described syntactic structure may matrix solution by forming according to the tactic syntax vector of predicate verb element number;

Decider, whether identical with described pretreated statement for verifying the statement obtaining according to syntactic structure possibility matrix solution, if identical, using each syntax vector in this syntactic structure possibility matrix solution as one of syntactic structure analysis result;

Wherein, described decider is got rid of ineligible syntactic structure feasible solution by following module operation:

First row, except module, if there is the sequence valve not occurring in this syntactic structure possibility matrix solution, is got rid of this syntactic structure possibility matrix solution;

Second row, except module, if occur identical sequence valve or occur identical syntax vector in different syntax vectors, is got rid of this syntactic structure possibility matrix solution;

The 3rd gets rid of module, in each possibility matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there is the intersection contradiction of two syntax vectors after equivalent substitution, getting rid of this syntactic structure may matrix solution;

The 4th gets rid of module, in each possibility matrix solution, by and other syntax vectors between exist the syntax vector of mutual substitution relation all to carry out equivalent substitution, if there are two sequence valves that position is converse after equivalent substitution, getting rid of this syntactic structure may matrix solution;

The 5th gets rid of module, in any one possibility matrix solution, if there is no the syntax vector of mutual substitution relation between existence and other syntax vectors, carry out plug hole and operate to obtain the corresponding possibility of all these possibility matrix solutions syntax analytic structure, and whether the statement that checking obtains according to described possibility syntax analytic structure is identical with described pretreated statement, it further comprises:

The first submodule, first to existing each other the syntax vector of substitution relation to carry out equivalent substitution in this possibility matrix solution, thereby is converted into one group of syntax vector that does not have each other substitution relation by this possibility matrix solution syntax vector in possibility matrix solution is called to first kind syntax vector, the syntax that conversion is obtained vector be called Equations of The Second Kind syntax vector;

The second submodule, appoint and to get an Equations of The Second Kind syntax vector according to predetermined direction, mark one by one in the sequence valve of each syntax elements; After the sequence valve of mark syntax elements, appoint and get in i syntax elements, only in the first side of this syntax elements, construct unique room; After making sky, appoint and get a syntax vector equations of The Second Kind syntax vector in addition mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, this new syntax vector is designated as and the syntax that whole plug hole is obtained vector, be referred to as the 3rd class syntax vector;

The 3rd submodule, to the 3rd class syntax vector according to predetermined direction to from vector in first syntax elements of the first side start to vector in the vector that comprises first syntax elements of the second side till each syntax elements, all mark sequence valve; Be positioned at vector in the vector that comprises the element of the first side, does not mark sequence valve; By vector first syntax elements of the second side be designated as will be according to aforementioned manner to vector the syntax vector part of mark, is designated as whipping syntax vector after mark sequence valve, appoint and get a j syntax elements in aforesaid whipping vector, only in this element first side, construct unique room; After making sky, appoint and get an original Equations of The Second Kind syntax vector mode with whole plug hole is vectorial by syntax insert the room of constructing, and then generate a new syntax vector, newly-generated syntax vector is designated as or

The 3rd class syntax vector according to predetermined direction, distich normal vector in each syntax elements all mark sequence valve; After the sequence valve of mark syntax elements, appoint and get one in t syntax elements, in the first side of this syntax elements, construct unique room; After making sky, appoint and get use Equations of The Second Kind syntax vector in the mode of whole plug hole by this vector insert the room of constructing above, and then generate a new vector, this new vector is designated as

The 4th submodule, repeat the operation of the 3rd submodule, when the last time makes empty and plug hole step and finishes, the empty and plug hole made carrying out is next time operated, until all Equations of The Second Kind syntaxes are vectorial through last the 3rd class syntax vector of making empty and plug hole step and obtaining all plug hole is complete, finally obtains the 3rd class syntax vector of a single file, and described the 3rd class syntax vector finally obtaining is called to final single file vector;

The 5th submodule, if one may syntax all has two sequence valves that position is converse in all described final single file vector corresponding to analytic structure, gets rid of this possibility syntax analytic structure;

The 6th submodule, the operation that repeats to call the second submodule to the five submodules is until all may being traversed by syntax analytic structure.

8. the device that computer based natural language syntactic structure according to claim 7 is resolved, is characterized in that, also comprises:

Result display unit, carries out each syntax vector and corresponding syntax structural relationship in syntactic structure analysis result to show on human-computer interaction interface with tree structure.