FR3028118A3 - COMPRESSION OF COMPUTER DATA - Google Patents

Abstract

The computer data compression method "COMPRESSION-BSP" is a method of coding / compression adaptable to any computer document, without loss and theoretical rate of infinite compression. In base (b), b is a natural integer, it encodes and compresses the original binary code of the document by slicing binary sequences of (2b) bits, indicating the sum of the product of the presence boxes of a so-called "minimum" bit , hence the abbreviation "BSP" for base, sum and product. For b ≥ 9, each bit sequence of (2b) bits is thus encoded / compressed into a bit sequence of (2b-1) bits. These data make it possible to deduce in decoding / decompression the presence boxes of the "Minimum Bit" so by complementarity the boxes of presence of the other bit by concatenations and iterations all the original binary sequence. Given the internal character of the coding (binary sequence transformed into a binary suite), the computer programs associated with the bases (b), (Progc; b (compression) and (Progd; b) (decompression) can be reiterated, hence the infinite theoretical character Compression calculations involve only operations on natural numbers, and special coding / compression packages must be derived from the operation of this method.

Description

The present invention relates to a coding and compression method for computer data. This method may be called "COMPRESSION-BSP", "B" for base, "S" and "P" for sum and product. This method has many advantages over all existing compression methods. Currently there are many computer compression software more or less effective, each specialized in the compression of certain forms of documents (Text, Sound, Film '). The "COMPRESSION-BSP" method adapts to all forms of documents. The "COMPRESSION-BSP" method ensures totally lossless compression, starting at the source of the document and using only arithmetic calculations on natural numbers in its development. The theoretical compression ratio of the "COMPRESSION-BSP" method is infinite, if one takes as definition of the compression ratio the ratio of the computer weight before compression to the computer weight after compression. This means that in practice, after compression, any computer document of any nature whatsoever will be translatable into a series of alphanumeric characters that may vary, depending on the options chosen from one hundred to one thousand characters, ie text document of about fifty kilo-octets. This means that the compression performed by the method "COMPRESSION-BSP" is constant in the sense that it can always be reduced to a minimum physical support part fixed in advance (one or two text pages, very small range of compact -dise, - - - 25 The innovative technical applications are numerous since the original document after compression is reduced to a very low weight and can be coded in ten "sentences" so very easily transmitted by mail, mail, On a conventional CD, hundreds of pieces of music, films and databases can be burned. The manufacture of special coding / compression boxes must be considered, each box being adapted to a specific use (sound, film, transport, space ..) An aviation box can, for example, supplement the classic black boxes of current aircraft, since all information related to the flight of an aircraft can be directly coded and compressed and then transmitted and stored on the ground very quickly and in a minimum of space. This principle coding / compression / transmission can be applied in many fields: space studies (NASA, ....), transport (land, sea, air), archives (storage, ...). In conclusion, the scope of the "COMPRESSION-BSP" method is very wide and varied. 10 35 40 We will now explain in general and in more detail, the compression method "COMPRESSION-BSP". Mathematical properties, particularly natural integers, will be stated and supported by simple examples as the study progresses. We will first study the coding / compression phase and then the decoding / decompression phase.

The first axiom taken into account is the following: any computer (or assimilated) can only interpret binary language, that is to say sequences of 0 and 1. In computer science, 0 and 1 are bit 0 or bit 1.All document, program, software - - -formatics has an original binary code read by the computer. This code is recoverable as a result of the assembly phase by the processor.

The method "COMPRESSION-BSP" acts in coding and compression on this original binary code, so there is no loss at the beginning compared to the original document. We therefore assume a binary sequence of p bits, p integer, that we want to compress. We will define the notion of "base" used in "COMPRESSION-BSP". We say that we code in "Base b", b integer, the cutting of the original sequence of bits into slices of a number of bits equal to (2b). As an example, the base (10) will split into 1024 bits. We will say that we code and compress in "base (10)".

This base (10) will serve as a basis of reference. We will then extend all the results obtained to other bases. We will therefore study the base (10) and demonstrate that it is quite possible to obtain as claimed, lossless compression in a fixed minimum part, which will give an infinite compression ratio. The general principle is to encode / compress a 1024 bit slice into another slice of number bits less than 1024. Concatenations and iterations will compress several 1024 bit slices and so on. We will develop the coding / compression base (10). The necessary mathematical notions will be introduced as and when necessary. All calculations will be performed on natural whole numbers which eliminates the problem of approximations.

The set of natural numbers will be denoted N: N = {0; 1; 2; 3; We denote by N * the set N deprived of 0: N * = 1; 2; 3; 4; We will use first, prime numbers for which we recall some basic concepts.

Definition of a prime number A prime number (abbreviation NP) is an integer that has only two divisors, let itself. Example: 11 is prime and 12 is not prime since 12 = 3x4 = 2x6 1x12.

Properties of the set of prime numbers noted E (NP). E (NP) is an infinite set, ordered and countable. E (NP) = {2; 3; 5; 7; 11; 13 1. Theorem fundamental to prime numbers. Every natural integer is uniquely decomposable into a product of prime powers.

For any integer n, there exists i integer, (NP,) first and (ei) integers such that: n = NP lei X NP 2e 2 x - - - x Npiei Example: 2 100 = 22 x 31 x 52 x 71. Numbering of prime numbers. E (NP) is a countable set so we can classify prime numbers in ascending order and number them on the set N *. Note: NPi = 2; NP2 = 3; NP3 = 5; NP4 = 7; - - - -; NP172 = 1021; In base (10), the 2 == 1024 bits can be represented as occupying 1024 boxes. These boxes can be numbered in two ways.

The boxes are classically numbered in decimal system from box 1 to box 1024. They will be called "decimal boxes" and will be noted (c,) with 1 <i <1 024. We can also number the boxes using decomposition as a product of prime numbers of their decimal numbers. Numbered so they will be called "boxes first" and will be noted (cpi). Example: cPuo = cp (23 31 if). Each bit (0 or 1) occupies a single box following the diagram (Base 10): Boxes 1 2 3 4 120 - - - - 1 024 Decimal Places 1 21 31 22 51 _ _ _ _ 23 x 31 x 5 _ _ _ _ 210 First Values Bits 0 0 1 0 0 - - - - 1 We will call "initial binary sequence" denoted by (SBi) a bit sequence of (2 ") bits, for example, (SBi) will be on 1024 bits in base ( The coded bit sequence obtained after coding / compression will be referred to as the "compressed binary sequence" (SBc) Coding principle: If the positions of a bit are known, the positions of the bit can be easily deduced by difference. another bit therefore all the binary sequence studied Example in base (10): If we know that the bit (0) occupies 300 well-known boxes we can deduce the (1 024 - 300) = 724 boxes occupied by the bit ( 1) Introduces the notion of "Minimum Bit" and its definition.

Definition of the "Minimum Bit". We will call "Minimum Bit" and we will note Bm the bit that is least present in the binary suite studied. In the case of equality of presence between the two bits, the first bit of the sequence will be chosen, ie that which is in box 1. Example in base (10): If one has 300 bits (0) and 724 bits (1) the "Minimum Bit" will be the bit (0), ie Bm = 0.

If we are in the case of equality ie 512 bits (0) and 512 bits (1), we will choose as "Minimum Bit" the first of the sequence (box 1). If SBi = (01110 -01 11) then Bm = 0 If SBi = (11110 -01 11) then Bm 1 We will henceforth study the positions of the "Minimum Bit" (Bm) and transcribe them into another binary sequence as a law of internal composition in the set of binary sequences. The coding sequence (SBc) begins with the value of the "Minimum Bit" (Bm): SBc = {Bm; We then proceed to a study of the box 1 (C1). This will subsequently avoid ambiguities in the calculations. Indeed, 1 = 10 11 20 = 30 = _ _ _ We will thus code the box 1: If the "Minimum Bit" (Bm) is in box 1 (case of presence) then we will code with the bit (1). If the "Minimum Bit" (Bm) is not in box 1 (case of non-presence) then we will code with the bit (0).

Example: Suppose that Bm = 0 If SBi = (0110- - 0- - 11) then (Cl If SBi = (1110- - 0- - 11) then (Cl) is coded in (1) because Bm = 0 is in box 1.) is coded in (0) because Bm = 0 is not in box 1. The beginning of the coding sequence becomes two bits: SBc = {Bm; Case 1; First bit for the value of Bm, second bit for the presence or absence of Bm in box 1.

We will recall a mathematical result essential for the study of coding / compression by "COMPRESSION-BSP". Theorem.

Two integers a and b are perfectly determined if we know their sum S = a + b and their product P = a x b. a and b are the solutions of the following equation of degree 2: x2 - (S xX) + P = 0.

Example. Let = 2 and b = 5, S = 2 + 5 = 7 and P = 2 × 510. We solve the equation X2 - (S xX) + P = 0 <==> X2 - (7xX) + 10 = 0. The discriminant is A = 72 - (4 x 10) = 49 - 40 9 = 32 therefore NiA = = 3.

The solutions are: X, - (S-8) / 2 = (7-3) / 2 = 4/2 = 2 and X2 = (S + 8) / 2 = (7 + 3) / 2 = 10/2 -5. By extension to p nonzero natural integers, different from 1 and different 2 to 2; at, ; - - - -; ap, we deduce that the integers a,; a2; - - - -; ap are perfectly determined by their sum S = a, + a, + - - - - + ap and their product p = a, xa, x- - - - x ap with 2 <a, <a, <- - - - <ap. Example.

Let 3 be integers a = 2; b 3; c 6. We deduce S = 2 + 3 + 6 = 11 and P = 2 X 3 x 6 = 36. S can result from different combinations: S = 11 ---> P = 11 # 36; S 2 + 9 = 11 ---> P = 2x9 = 18 # 36; S = 3 + 8 = 11 ---> P = 3x8 = 24 # 36; 5 = 4 + 7 = 11 ---> P = 4x7 = 28 # 36; S = 5 + 6 = 11 ---> P = 5x6 = 30 # 36; S = 2 + 4 + 5 = 11 ---> P = 2x4x5 = 40 # 36; Only the combination {2; 3; 6} gives a sum S = 11 and a product P = 36. We must make the following remark on the decimal numbering of the boxes. In base (10), the box numbers are the consecutive integers from 1 to 1024 so are non-zero, different two by two and ranked in ascending order. So we will continue coding by coding studies of the sum and the product of the presence boxes of the "Minimum Bit", which will be enough to know perfectly these boxes.

Remember that we always study in base (10) so a slice of -10 2 = 1024 bits. It is also recalled that from now on the value of the "Minimum Bit" Bm and its presence or not in box 1 are known. The study will therefore only cover the 1,023 boxes numbered from 2 to 1024. Boxes {c,; c3; c4; - - - -; c1024} .45 Given the definition of the "Minimum Bit" adopted and the fact that box 1 can be excluded from the study, the "Minimum Bit" Bm will be present in a maximum of 511 (= 512-1) boxes. We will recall a classic formula on the sum of consecutive integers which will serve as a basis for sums calculations. Sum of consecutive integers: The sum of the (n) first integers noted (Sn) is equal to: nx (n + 1) - 2 Example: n = 5; S5 = 1 + 2 + 3 + 4 + 5 = 15 - 5x (5 + 1) = 5x630 2 - = 15 '2 This formula will allow us to deduce the formula giving the value of the maximum sum that can take the sum S decimal places of position of the "Minimum Bit" Bm in any base (b).

In base (b), we study a slice of 2 "bits and the number of squares is Cb = 2b The" Minimum Bit "Bm can occupy at most (2- 1) = 2b-1 1) cases Cb-2 On note therefore Cm this maximum number of boxes, Cm = -Cb -1 - 2 2 Examples: base (5), (b = 5), Cb = 25 = 32, Cm = 15 (- 32-2 - 30 2 2 1 024 -2 1 022 In base (10), (b = 10), Cb = 21 ° = 1024, Cm = 511 (- -) - 2 2 The sum will be obviously maximum when Bm will be positioned in the upper boxes the larger, therefore from the cell (Cm + 3) to the cell (Cb), by calling Sm the maximum sum we get: Sm = {1 + 2 + 3 + - - - - + (Cb - 1 ) + Cb} - {1 + 2 + 3+ - - - + (Cm + 1) + (Cm + 2)} = {Cbx (Cb + 1) / 2} - {(Cb + 2 Cb + 4) / 2} because Cm + 2 - Cb + 2 2 2 2 = {(Cb2 + Cb) / 2} - {Cb2 + (6 x Cb) + 8) / 8} = (4 x cb2 + (4xCb) - Cb2 - (6xCb) - 8) / 8} = (3 x cb 2 - 2xCb - 8) / 8.

Sm = (3 x cb 2 - 2xCb - 8) / 8. We check the formula in bases (4) and (5). Base (4): Cb = 24 = 16; Cm - 16-2 - 7; Cm + 3 = 10; Sm = 10 + 11 + 12 + 13 + 14 + 15 + 16 = 91; (7 upper boxes).

The formula gives: Sm = (3 x Cb2 - 2xCb - 8) / 8 = (3x 162 - 2x16 - 8) / 8 = (3x256-32-8) / 8 = (768 -40) / 8 = 728/8 = 91. Sn = 1 + 2 + 3 + - - - - + n - Base (5): Cb = 2 = 32; CM - 32-2 - 15; Cm + 3 = 18. 2 Sm = 18 + 19 + 20 + 21 + - - - + 30 + 31 + 32 = 375 (15 upper boxes). The formula gives: Sm = (3 x Cb2 - 2xCb - 8) / 8 = (3x322 - 2x32 - 8) / 8 = (3x1024-64-8) / 8- (3072 -72) / 8 = 3000/8 = 375.

The formula can be applied in base (10). Base (10): Cb = = 1024; Cm - 1024-2 - 511; Cm + 3 = 514. 2 Sm = 514+ 515+ 516 + - - - +1 022 + 1,023 + 1,024 (511 upper boxes The formula gives: Sm x cb2 -2 xCb-8) / 8 = (3 x 10242 - 2 x 1024- 8) / 8 (3x1 048 576 - 2 048 - 8) / 8 = (3 145 728 - 2 056) / 8 = 3 143 672 In conclusion the sum S of the decimal boxes of the "Minimum Bit" B m can take 392 959 values from 1 to 392 959 to which must be added the zero sum (S = 0) which represents the particular case for which Bm is in no box (1024 identical bits). We therefore have 392 959 + 1 - 392 960 cases to code. We will recall a property of the binary system.

Property in binary numeration. A number of (n) bits, (n) integer, represents (2) numbers (or cases) in a decimal system. Examples of codings: 1) n = 2; we can code 22 = 4 numbers. 00 - + 0 (case 1); 01 -> 1 (case 2); 10 -> 2 (case 3); 11 -> 3 (case 4). 2) n = 2; we can code 23 = 8 numbers. 000 -> 0 (case 1); 001 -> 1 (case 2); 010 2 (case 3); 011 -> 3 (case 4); 100 -> 4 (case 5); 101 -> 5 (case 6); 110 -> 6 (case 7); 111 -> 7 (case 8). It follows that the sum (S) in base (10) can be encoded on Ns = 19 bits. Indeed we have the following frameworks: -18 2 = 262 144 <392 960 524288 = 2'9.

The coding of the sum, the third step of encoding / compression by the method "COMPRESSION-BSP" is therefore processed. The coding / compression bit sequence (SBc) becomes: SBc = {Bm; Case 1; Sum; In base (10), (SBc) is written in 21 bits: 1 bit for Bm, 1 bit for box 1, 19 bits for sum S. / 8 = 392 959.

We still have to study and code the product of the boxes. We will make some general remarks about the products of numbers. Products of consecutive integers that are written in factorials quickly have very large decimal values. We recall that, n integer, we note and define: "Factorial of n" by: n! = 1 x 2 x 3 x 4 - x (n -1) x n. We will therefore study the product of the boxes without ever calculating its decimal value. One can perfectly express a product by using the decomposition in powers of prime numbers and by using the properties of computation on the powers. Properties of the powers of integers: Definition: Let a and n be two integers: we call base power (a) and exponent (n) that we write a "the product of a by a; n times." = axaxax-- - xa xa (n times). We put a ° = 1. Note: If we know the base (a), the exponent alone perfectly defines the power.

Essential property: The operation "Power" is a group morphism of the multiplicative group on the additive group which results in the relations: Let 4 be integers a, b, m, n: a) am xa "= am + n Example : 22), 23 22 + 3 = 25 b) a (m) n = amxn Example: 2 (2) 3 = 22x 3 = 26 c) a1 1 / a Example: 2-1 = 1 / 2. d) a '---- 1 / am Example: 2-2 = 1 / 22. e) a "' = if / a" Example: 25-2 = 23 = 25/22 - (ab) = if x btm Example: ( 6) 3 = (2 x3) 3 = 23 x 33 A principle of coding is to reason about sums rather than products, as these are very affordable, hence the use of multiplicative group morphisms on an additive group. We will study the product of the boxes by using the decomposition of the number of boxes in powers of prime numbers defined previously.The product obtained will be defined by its decomposition in powers of prime numbers.Since the necessary primes will be known, in finite number, ranked in order nt, the product will be determined by the exponents of these prime numbers, exponents that we will code. We will deepen the study of the coding of the product in base (10). In base (10) there are 172 primes less than 1024.

E (NP) = {2; 3; 5; 7; - - - -; 1,013; 1,019; 1 021}.

These prime numbers are numbered from 1 to 172: NP], = 2; NP2 = 3; NP = 5; - - - -; NP = 1019; - - - -; NP172 = 1021.

We will consider two kinds of exponents for each prime number noted (ei) and (epi) for the i-th prime number. (ei) will be the exponent of (Ni ?,) in the decompositions of the squares into products of powers of prime numbers and (epi) will be the exponent of (Ni ?,) in the product of the first boxes. Example: Ni? ei or NPM for the prime number Ni ?,. We will use the letter M for "Maximum or Maximal": em, and epmi will be maximal exponents in the decomposition of the first boxes and in their product. We will give some examples to clarify these notions. Example 1: Study of the prime number 7 in base (10).

In base 10, we study 1024 boxes numbered from 1 to 1024 in decimal. 7 is the fourth prime number that we note NP4 = 7. In the decomposition of a first cell, NP4 = 7 may not be present: this case will be called the "null case". NP4 = 7 can be exponents 1, 2 or 3. Indeed 74 = 2,401> 1,024. We will say that the maximum exponent of NP4 = 7 is eM4 = 3 in the first case writing. On the other hand {71 = 7; 72 = 49; 73 = 343} <1024 can appear in the decomposition of the first boxes, being less than 1024. The powers of 7 appear in the product of all the multiple boxes of 71, 72 and 73.

In the product of boxes, NP4 = 7 will have a maximal exponent (ep44). NP4 = 7 will have an exponent which can therefore take all the values from 1 to (epm4), this exponent value depending on the number of multiple presence boxes of 7. It will be shown that epm4 = 168. By adding the "null case", we will therefore have all the possibilities for NP4 = 7 in the product of the first boxes, ie 169 in total. It is this number of cases of exponents (+ "null case") that we will code to define NP4 = 7.

Example 2: Study of the prime number 1021 in base (10). NP172 = 1 021. It is obvious that this prime number is either in "zero case", or in exponent 1.

We can not have multiples of 1,021 in base (10) since the smallest possible multiple 2 x 1,021 = 2,042 is greater than 1,024. 10212 is obviously impossible too. We will thus have em172 = epM172 = 1. Finally (NPi72) will have to be coded on two cases (present or not-present) in the product. It can be conjectured that the largest possible exponent in the product will be that of the smallest prime number, NPi = 2, and that the exponents will go decreasing with the growth of prime numbers. We will recall some mathematical notions necessary for the continuation of the study of the product of the first boxes. Whole part of a real number. An integer of a real number x denoted by E (x) is the integer such that: E (x) x <E (x) + 1 Examples: E (3, 14) = 3; E (3) = 3; E (2,9999) = 2. 25 Number of multiples of an integer. Let N and n be two integers with N? n: the number of multiples of n less than or equal to N is equal to E (-n). Examples: 31 N = 31 and n = 2: The multiples of 2 <31 are E () E (15.5) = 15. We have 15 even numbers {2; 4; 6; ---- -; 28; 30} lower than 31. 31 N = 31 and n 5: The multiples of 5 <31 are E (-5-) = E (6,2) = 6. We have 6 multiples of 5 which are { 5; 10; 15; 20; 25; 30 }. 35 Multiple product of an integer. Let N, n, p be three integers such that N> n; nP <N and nP + 1> N then, in the product of all the multiples of (n) less than or equal to N, we have the factor (ne), the exponent (e) 40 being the sum of the integer parts of ratios of N on {n, n2, n3, - - - -, nP produces multiples of n <N = ne; with E (N / n) + E (N / n 2) + E (N / n 3) + --- + E (N / 20 Example 1: N = 31 and n = 5, 52 = 25 <31. the product of all multiples of 5 smaller than or equal to 31 is calculated, the factor 5 will be at the maximum exponent (eP) such that: 51 e nit / 1 E (31/5) + E (31/52) = E (6.02) + E (1.24) = 6 + 1 = 7 We check: 5 x 10 x 15 x 20 x 25 x 30 = (51) 4 21 X 51) X (31 X 51) X (22) X 51) 4 52) X (21 X 31 X = 21 + 2 + 1 x 31 + 1 x 51 + 1 + 1 + 1 + 2 + 1 = 24 x 32 x 57 (5) is at the maximum exponent 7 in products of multiples of (5) less than or equal to 31. Example 2: N = 20 and n = 2, we can calculate the maximal exponent epm of 2 in the product of all multiples of 2, ie all even numbers less than or equal to 21. The maximum exponent of 2 in the multiples of 2 is em = 4 because 24 = 16 <20 and 25 = 32> 20. epm = E (N / n) + E (N / n ') + E (N / / 13) + E (N / n4) = E (20/2) + E (20/22) + E (20/23) + E (20/2) = E (20/2) + E (20/4) + E (20/8) + E (20/16) = E (10) + E (5) + E (2.5) + E (1,25) = 10 + 5 + 2 + 1 = 18. (2) is at the maximum exponent 18 in the products of multiples of (2) less than or equal to 20. We check only on factor 2: 2 is at exponent 1 in 5 multiples of 2 which are { 2, 6, 10, 14, 18}, with exponent 2 in 3 multiples of 2 which are {4; 12; 20 1, with exponent 3 in 1 multiple of 2 which is {8} and at exponent 4 in 1 multiple of 2 which is {16}. In total, we can therefore have the factor 2 in the product of all multiples of 2 less than or equal to 20 with the maximum exponent: ePM = (5 x 1) + (3 x 2) + (1 x 3) + (1 x 4) = 5 + 6 + 3 + 4 = 18.

We study the coding of the product of the basic boxes in base (10) by using these calculation methods to determine the maximum exponent (epmi) in the product of each prime number (NP;) and thus code it in binary. We recall that in base (10), we study a binary sequence of 210 = 1024 bits in 1024 boxes numbered from 1 to 1024 in decimal, that each number of box can be decomposed into product of powers of prime numbers, that the number of prime numbers less than or equal to 1024 is 172, ranked in increasing order of NPi = 2 to NP172 = 1,021. The product of the boxes is decomposed into prime powers and the calculation of its decimal value is not required, which avoids the use of very large numbers. We must study each prime number. We will develop the calculations for the first three prime numbers and then give the necessary results for the following ones. It is recalled that the boxes considered are those of the "Minimum Bit" Bm.45 Study of NP1 = 2 in the product of the first boxes.

NP, = 2 is found in the decomposition of all the even squares (512 squares). Its maximum exponent in the decomposition is em = 10 because 210 = 1024 and -11 2 = 2048> 1024. Note: this result will be found in all bases (b). The maximal exponent of NP, = 2 is equal to (b) in base (b). The maximum exponent (epm,) of NP '= 2 in the product is calculated by the integer parts. E (210/21 + E (210 I 22) + E (210/23 + E (210/24 E (210/25 + E (210/26) + E (210/27 + E (210 / +) E (210/29) + E (210/210) = 210-1 ± 210-2 ± 210-3 ± 210-4 + 210-5 + 210-6 + 210-7 + 210-8 210-9 + 210 -10 = 29 + 28 + 27 + 26 + 25 + 24 + 23 + 22 + 21 + 20 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 1023. Equality epm ' = 1,023 therefore means that NP, = 2 can occur in the product of the prime boxes with an exponent (ep,) of values between 1 and 1,023.

Base (10): epm, = 1023; 1 <ep, <1023. We add the special case ("Case Null") for which NP, = 2 is not present in the product (Bm is only in odd boxes). - In conclusion, in base (10), NPi = 2 has 1,024 (= 1,023 + 1 = 210) possible cases in the product of the first boxes.

The coding of NP = 2 on 10 bits is deduced because 2 '° = 1024. Coding of NP = 2 bits. Note: we can generalize to any base (b) this result obtained in base (10). The prime number NP, = 2 will be encoded in the product of the first boxes on (b) bits. Any other prime number will be coded on (p) bits with p <b. Study of NP2 = 3 in the product of the first boxes.

NP2 = 3 is found in the decomposition of all the multiple boxes of 3 (341 boxes). Its maximum exponent in the decomposition is e2 = 6 because 36 = 729 (729 <1024) and 37 = 2,187 (2,187> 1,024). +40 The maximum exponent (epm2) of NP2 = 3 in the product is calculated by the integer parts. epm2 21 ° /) + E (21 ° / 32) + (216/33) + E (21 ° / 34) + E (210/35) + 210/36) = E (1024/3) + E (1024) / 9) + E (1024/27) + E (1024/81) + E (1024/243) + E (1024/729) = E (341.3) + E (113.7) + E (37, 9) + E (12.6) + E (4.2) + E (1.4) = 341 + 113 + 37 + 12 + 4 + 1-508.

The equality (epm2 = 508) therefore means that NP2 = 3 can occur in the product of the first cells with an exponent ep2 of values between 1 and 508. We add the particular case ("Null Case") for which NP2 = 3 is not present in the product (Bm is in no box multiple of 3). In conclusion, in base (10), NP2 = 3 has 509 (= 508 + 1) possible cases in the product of the first boxes. We deduce the encoding of NP2 3 on 9 bits because 28 = 256 <509 <512 = 29. Coding of NP2 = 3 ---> 9 bits. Study of NP3 = 5 in the product of the first boxes. NP3 = 5 is found in the decomposition of all the multiple boxes of 5 (204 boxes). Its maximum exponent in the decomposition is em3 = 4 because 54 = 625 (625 <1024) and 55 = 3,125 (3,125> 1,024). The maximum exponent epm3 of NP3 = 5 in the product is calculated by whole parts. epm3 _E (210 /) + E (210/52) + 219/53) + E (21 ° / 54) - = E (1024/5) + E (1024/25) + E (1024/125) + E (1024/625) = E (204.8) + E (40.9) + E (8.1) + E (1.6) = 204 + 40 + 8 + 1 = 253. The equality epm3 = 253 therefore means that NP3 = 5 can occur in the product of the first cells with an exponent e3 of values between 1 and 253. We add the particular case ("Null Case") for which NP3 = 5 n is not present in the product (Bm is in no box multiple of 5). In conclusion, in base (10), NP3 = 5 has 254 (253 + 1) possible cases in the product of the first boxes. We deduce the coding of NP3 = 5 on 8 bits because 2 '= 128 <254 <256 = 28 - Coding of NP3 = 5 ---> 8 bits. We will summarize in a table the results obtained for the 172 prime numbers to consider, knowing that the calculations are identical and that the prime numbers will be grouped by the number of coding bits needed.

We will only recall the values of the powers of 2 useful for the binary coding of the exponents of the prime numbers in the product of the first boxes. 21 = 2; 22 = 4; 23 = 8; 24 = 16; 25 = 32; 26 = 64; 27 = 128; 28 = 256 29 = 512; 210 = 1024. Summary table of the product of the basic boxes in base 10 Order of Exponent Values Exponent Number Number of Number First Number of Numbers Maximum Maximum of First Decomposition Bits Product Case Coding Numbers emi i Binary First eple. NPi 2 10 1 023 1 024 10 1 NP2 3 6 508 509 9 1 NP3 5 4 253 254 8 1 NP4 7 3 168 169 8 1 NP5 11 2 101 102 7 1 NP6 13 2 84 85 7 1 Ni-Ni ? ,, 17 -> 31 2 63 -> 34 64 -> 35 6 5 Ni? ' to NP, 8 37 -> 61 1 27 -> 16 28 -> 17 5 7 NP19 to NP31 67 -> 127 1 15 -> 8 16 -> 9 4 13 NP32 to NP54 131 -> 251 1 7 -> 4 8 -> 5 3 23 NP55 to NP97 257 -> 509 1 3 -> 2 4 -> 3 2 43 NP98 to NP172 521 -> 1021 1 1 2 1 75 Note: we have 75 prime numbers which are coded on 1 bit either by (0) or (1). This means that it is perfectly clear if the "Minimum Bit" Bm is present (coding 1) or not present (coding 0) in the 75 boxes numbered by these prime numbers. The number of bits necessary for the coding of the product noted (N p) of the first cells is calculated: p = (lx 10) + (lx 9) + (2 x 8) + (2 x 7) + (5 x 6) + (7x5) + (13x4) + (23x3) + (43x2) + (75x1) = 10 + 9 + 16 + 14 + 30 + 35 + 52 + 69 + 86 + 75 = 396. Coding of the product P -> 396 bits. It is essential to note that in the coding of the product all prime numbers, without exception, are coded by their exponents including the "null case", this making it possible to keep the continuity of the binary coding sequence (SBc).

The coding / compression bit sequence (SBc) becomes: SBc = {Bm; Case 1; Sum; Product}. The coding / compression bit sequence (SBc) is a sequence of 417 bits: Indeed: Bm -> 1 bit; Case 1 -> 1 bit; S- * 19 bits; P -> 396 bits which gives: 1 + 1 + 19 +396 = 417. First conclusion on the method "COMPRESSION-BSP": In base (10), an initial binary sequence (SBi) of 210 = 1024 bits can encode / compress into a 417-bit binary coding bit (SBc) translating the value of the "Minimum Bit" Bm, its presence or not in box 1, the value of the sum of its presence boxes in decimal, the product in decomposition of powers of prime numbers of its first boxes of presence.

From now on, we will call this binary sequence "Binary suite of minimal coding" which will be noted "SBc, min; 10" in base (10). This notion will be applicable to any base (b). Each bit sequence of () bits can be coded as a minimum bit sequence (SBcmin, b).

Subsequent decoding / decompression methods will be further explored. We will only expose the general principle of decoding :( SBcmin, b) gives the values of S and P which make it possible to find the presence boxes of the "Minimum Bit" Bm, as well as its presence or not in box 1, and its value (0 or 1). Since all the presence boxes of Bm are known, the positions of the other bit and hence the totality of the initial binary sequence are complementarily deduced. We will now confirm that the theoretical compression ratio is infinite. At any base (b) we associate a computer program noted (Progc; b), (c) for coding / compression and (b) for base. (Progc; 10) will be the coding / compression computer program associated with the base (10). The ratio of the NBo number of original bits (NBo) to the number of bits after encoding (NBc) will be adopted as a definition of the compression ratio, noted as Tc: Te - NBc It can already be conjectured that (NBc) will be constant, within a limit fixed in advance and that (NBo) can grow indefinitely. 40 45 At first, in base (10), we see that if we code / compress (n) slices of 1024 bits we will obtain (n) slices of 417 bits that we can encode / compress after concatenation, but since 417 is not a power of 2, there will be appearances of remains, complicating the coding phase.

In base (10), we will therefore encode a slice of 210 = 1024 bits into a bit sequence of 2 '° / 2 29 = 512 bits. This final coding binary sequence will be the one we adopt and note (SBc, 10). The 95 bits (512 -417 ----- 95) that are added to the minimum coding bit sequence (SBcmin; 10) will be what will be called the sum precision bits, facilitating the decoding phase. / decompression.

At first we can suppose that we complete by 95 bits (0). Can we generalize this notion to any basis (b) with b? 9. It will be seen that for b <9 the minimal coding does not allow extension. For this reason, the base (9) will be called the threshold base.

Example of approach: Let 22 = 4 slots of 1024 bits or NBo = 22 210 = -12 2 = 4096 bits. The computer program (Progc; 10) encodes each 1024-bit slot into a binary sequence of 29 = 512 bits. By concatenation of these sequences we obtain a new binary sequence of 212/2 = 211 = 2048 bits (2048 = 2 x 1024 bits). (Progc; 10) re-encodes 2 1024-bit slices, each in 512 bits. By concatenation, a last 1024-bit slice is reformed as (Progc; 10) encodes 512 bits. We can no longer act as a base (10). In summary, after 3 passes of (Progc; 10), the coding / compression gives (SBc; 10) in 512 bits. -12 Tc 2/2 = 212-9 = 23 = 8.

To schematize: 22 210 = 212 = 4096 bits - - (Progc; 10) - -> 21 210 = 2 "= 2048 bits - - (Progc; 10) - -> 20 X 21 ° = 211) 1024 bits - - (Progc; 10) - -> 29 = 512 bits.

We can conjecture this result with an original number of original bits NBo 2n 210 = 210 + n bits, with n integer? 0. After (n + 1) coding passes of (Progc; 10) in increments of 1024 bits, the binary coding sequence (SBc) of 2 = 512 bits.45 n> O will be obtained systematically; 2n X 21 ° = 2n + 10 bits - - - (n + 1) passes (Progc; 10) - - -> 29 = 512 bits. Tc - NBo _ 2n + 10 2 = __ 2n + 1 NBc For n tending to infinity, Tc tends to infinity with n. In practice it can be stated that after coding by the "COMPRESSION-BSP" method, any computer document can be compressed into a very finite document (finite computer weight). 10 We will return to this statement after studying other bases and some in-depth studies. Before continuing the study of coding by approaching other bases and the notion of remainder, we will give some simple examples of base coding (10) of a 1024-bit slot, minimal following coding (SBcmin; ). First example of base coding (10) Let the case of the initial binary sequence of 1024 bits equal to 1: (SBi) = {1111 1111}. 20 a) The "Minimum Bit" Bm is obviously 0 because it is present in no box. Coding of Bm-> 0. b) Bm = 0 is not in box 1. Coding of Case 1 - -> 0. c) The sum of the boxes in decimal S is zero. Coding of S - -> (00 --- 00) (19 bits). D) The product of the first boxes P is written with all prime numbers in "zero case" (no NP in the product). We code with 396 bits (0). (SBcmin; 10) = 000- - - 0 - - 000 -> 417 bits (0). To obtain (SBc; 10) it would be enough to complete by 95 bits (0). Second base coding example (10). Either the case of the initial binary sequence of 1024 bits equal to 0: (SBi) = 1 0000 0000 1. a) The "Minimum Bit" Bm is obviously 1 because it is present in no box. Coding of Bm - -> 1. 35 b) Bm = 1 is not in box 1. Coding of Case 1- -> 0. c) The sum of the boxes in decimal S is zero. Coding of S - -> (00 --- 00) (19 bits). d) The product of the first boxes P is written with all prime numbers in "zero case" (no NP in the product). We code with 396 bits (0). 40 (SBcmin; 10) = {100- - 0 - - - 000} - -> bit (1) followed by 416 bits (0). To obtain (SBc; 10) it would be enough to complete by 95 bits (0). Third example of base coding (10). Let there be the case of the initial binary sequence of 1023 bits equal to (0) and a bit (1) in box 210 (C210) * SBi) - {0000-1-0000}. Bit (1) is in box (210) and 210 = 21 X 31 x 51 x 71. a) The "Minimum Bit" Bm is obviously 1 because it is present in a single box. Coding of Bm - -> 1. b) Bm = 1 is not in box 1. Coding of Case 1- -> 0. c) The sum of the boxes in decimal S is 210.

Encoding of S - -> (0000000000011010010) (19 bits). d) The product of the first boxes P is written P = other prime numbers are in "zero case". We code P by 396 bits by encoding the exponent 1 for NP {2; 3; 5; 7 1. 20 Exhibitor 1 for Exhibitor 1 for Exhibitor 1 for Exhibitor 1 for NP = 2 for 10 bits: - -> 0000000001. NP 2 = 3 for 9 bits: - -> 000000001. = 5 for 8 bits: - -> 00000001. = 7 out of 8 bits: - -> 00000001. NP 3 NP 4 21 x31 x51 x 71 so all (SBcmin; 10) = {100000000000011010010000000000100000001 0000000100000- - - 0}. First bit for Bm, second bit for case 1, next 19 bits for sum S, last 396 bits for product P whose first 35 code exponent (1) for NPs {2,3,5,7}, the remaining 168 primes being all coded in "zero cases" therefore by bits (0). To obtain (SBc; 10) it would be enough to complete by 95 bits (0). Fourth example of base coding (10). That is, the case of the initial bit sequence of 1014 bits equal to (1) and 10 bits (0) in boxes (c10, C20, C30, C40, C50, C60, C70, C00, C00, C100). a) The "Minimum Bit" Bm is 0 because it is present in 10 boxes against 1014 for the bit (1). 35 Encoding of Bm - -> 0. b) Bm = 0 is not in box 1. Coding of Case 1 - -> 0. c) The sum of the boxes in decimal S is S = 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 100 = 550. Coding of S - -> (0000000001000111111) (19 bits). d) The product of the first boxes P is written: 40 P = (21 X 51) 4 22 X 51) 4 21 x 31 x 51) x (23 x 51) x (21 x 52) x (22 x 31 x) x (xx 71) x (24 x 51) x (21 x 32 x 51) x (22 x 52) = 21 + 2 + 1 + 3 + 1 + 2 + 1 + 4 + 1 + 2 31 + 1 + 2 51 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 1 + 2 71 45 = 218 x 34 x 512 x 71 We code P by 396 bits by coding the exponents for NP {2; 3; 5; 7}. Exponent 18 for NP = 2 on 10 bits: - -> 0000010010. Exponent 4 for NP 2 = 3 on 9 bits: - -> 000000100. Exponent 12 for NP3 = 5 on 8 bits: - -> 00001100. Exponent 1 for NP4 = 7 on 8 bits: - -> 00000001. (SBcmin; 10) = = {0000000000010001111110000010010000000100000011000000000100 0- - 0}. First bit for Bm, second bit for box 1, next 19 bits for sum S, last 396 bits for product P whose first 35 code exponent for NPs {2,3,5,7}, 168 remaining prime numbers are all coded in "zero cases" therefore by bits (0). To obtain (SBc; 10) it would be enough to complete by 95 bits (0). Note: For the binary coding standard increments have been used. Gray's code could also be used. We will now study other bases, higher and lower than the reference base (10). Only the main results necessary for the coding for each base will be mentioned. Study of bases superior to (10). Study of the base (11): b = 11 We study a slice of 2b = 211 = 2048 bits. We therefore consider 2048 boxes, ie Cb = 211 = 2048. Cb The "Minimum Bit" Bm occupies a maximum (2- 1) of 1,023 boxes. Bm is encoded on one bit. The presence or not in box 1 is coded on a bit. The maximum sum of the boxes is: Sm = (3x cb - 2xC - 8) / 8 = (3 x 20482 - 2 x 2048 - 8) / 8 = (12,582,912 - 4,096 - 8) / 8 - 12,578808 - 1 572 351. 8 Gold 220 _ 1 048 576 <1 572 351 <2 097 152 = 221 therefore (by adding the "Case Null") the sum S will be coded on 21 bits. 40 For the product of the prime boxes we must consider 309 prime numbers numbered NPi = 2 to NP309 = 2039. We know that Ni ?, = 2 will be coded on (b) bits of 11 bits and all the following ones on a number of less than 11 bits. All calculations done, from the integer parts of the multiples ratios, we obtain that the product P must be coded on 705 bits. We note that 137 prime numbers (the largest ones) are coded on 1 bit. The minimum base coding sequence (11) noted (SBcmin; 11) is a bit sequence of 728 bits. Indeed, 1 + 1 + 21 + 705 = 728.

Note that 728 <1024 = 210 = 211/2. (SBcmin; 11) can therefore be supplemented by 296 bits (296 = 1024 - 728) said sum precision to become a bit sequence of 210 = 211/2 = 1024 bits.

Conclusion: (SBc; 11) - - -> 210 = 1024 bits. -11 Any compression of 2 "slices of 2 = 2048 bits will be encoded, after (n + 1) 10 - passes of (Progc; 11) into a binary sequence of 2 = 1024 bits. N? 0; 2n 211 2n +11 bits - - - (n + 1) passes (Progc; 11) - - -> 210 = 1024 bits Note: (SBc; 11) can be encoded / compressed by the action of the base (10) so by (Progc; 10) Study of the base (16): b = 16 We study a slice of 2b = 216 = 65 536 bits, so we consider 65 536 cases, ie Cb = 216 = 65 536. The "Minimum Bit" Bm occupies the maximum (-Cb - 1) or 32 767 boxes 2 Bm is coded on a bit The presence or not in box 1 is coded on one bit The maximum sum of the boxes is: Sm (3x cb - 2xCb - 8) / 8 = (3 x 655362 - 2 x 65 536 - 8) / 8 = 1 610 596 351. Gold 23 ° = 1 073 741 824 <1 610 596 351 <2 147 483 648 = 2 'therefore (adding the "Case Null") the sum S will be coded on 31 bits.For the product of the first boxes we must consider 6,542 primes numbered of 40 NP, = 2 to NP6542 = 65,521. e = 2 will be encoded on (b) 16 bit bits and all subsequent ones on a number of bits less than 16. All calculations made, from the integer parts of the multiples ratios, result in the product P must be coded on 14 251 bits. We notice that 3,030 prime numbers (the largest) are coded on 1 bit.

The base encoding minimum sequence (16) noted (Sbcmin; 16) is a bit sequence of 14,284 bits. Indeed 1+ 1+ 31+ 14 251 = 14,284. We note that 14,284 <32,768 = 215 = 216/2. (Sbcmin; 16) can therefore be supplemented by 18,484 bits (18,484 = 32,768- 14,284) said sum precision to become a bit sequence of 215 = 216/2 = 32,768 bits. Conclusion: n 0; x 216 = 216 + n bits - - - (n + 1) passes (Progc; 16) - -> 215 = 32,768 bits.

Note: (SBc; 16) can be encoded / compressed by the action of the base 15 so (Progc; 15). Conjectures on the bases (20) and (30) Base 20: 220 = 1 048 576 so of the order of one million. Base 30: 23 ° = 1 073 741 824, therefore of the order of one billion These two bases and the associated programs (Progc; 20) and (Progc; 30) encode and compress slices of the order of one million and one billion bits. Each slice will be coded in 220/2 = 219 or 230/2 = 229 bits. It will be admitted that the base (30) is the largest basis that can be studied. The base (30) will be called the maximum base of coding / compression. Study of bases lower than 10. - 1) is 255 cases. Study of the base (9): b = 9 We study a slice of 2 '= 2 = 512 bits. We therefore consider 512 boxes, that is Cb = 2 = 512. Cb The "Minimum Bit" Bm occupies the maximum (2 Bm is coded on a bit) The presence or not in box 1 is coded on one bit.

The maximum sum of the boxes is: Sm = (3 x cb2 -2xCb-8) / 8 = (3 x 5122 -2 x 512-8) / 8 = 98175. Now 216 = 65 536 <98 175 <131 072 = 217 therefore (adding the "Null Case") the sum S will be coded on 17 bits.

For the product of the first boxes, we must consider 97 prime numbers numbered NP, = 2 to NP97 = 509. We know that NP, ---- 2 will be coded on (b) bits or 9 bits and all the following on a number of bits less than 9.

All calculations done, from the integer parts of the multiples ratios, we obtain that the product P must be coded on 233 bits. We note that 43 prime numbers (the largest) are coded on 1 bit.

The minimum base coding sequence (9) noted (Sbcmin; 9) is a 252 bit binary sequence. Indeed, 1 + 1 + 17 + 233 = 252. Note that 252 <256 = 28 = 2 I 2. (Sbcmin; 9) can therefore be supplemented by 4 bits (4 = 256 - 252) called sum precision to become a bit sequence of 28 = 2/2 = 256 bits. Conclusion: (SBc; 9) - - -> 28 = 256 bits.Any compression of 2 "slots of 2512 bits will be coded, after (n + 1) passes of (Progc, 9) in a binary sequence of 28 = 256 bits 2 x 29 = 29 'bits - - - (n + 1) passages (Progc, 9) - - -> 28 = 256 bits Study of the base (8): b = 8 We study a slice of 2 '= 28 = 256 bits - 1) ie 127 boxes.

We therefore consider 256 boxes, ie Cb = 28 = 256. Cb The "Minimum Bit" Bm occupies the maximum (2 Bm is coded on a bit) The presence or not in box 1 is coded on one bit.

The maximum sum of the boxes is: Sm = (3 x cb2 - 2xCb - 8) / 8 = (3 x 2562 - 2 x 256-8) / 8 = 24511. Now 214 = 16,384 <24,511 <32,768 = 215 therefore (adding the "Null Case") the sum S will encode on 15 bits. For the product of the first boxes we must consider 54 prime numbers numbered NP, = 2 to NP54 = 251. It is known that NP, = 2 will be encoded on (b) bits or 8 bits and all the following on a lower number of bits to 8. All calculations done, from the integer parts of the multiples ratios, we obtain that the product P must be encoded on 127 bits. We note that 23 prime numbers (the largest) are coded on 1 bit.

The minimum base coding sequence (8) noted (Sbcmin; 8) is a 144 bit binary sequence. Indeed, 1 + 1 + 15 + 127 = 144. We note that 144> 128 = 27 = 28/2. (Sbcmin; 8) can not be considered as for the other bases studied previously. Conclusion: (SBc; 8) - - -> 144 bits. This base (base 8) can be considered as a finishing base.

Indeed it is a compression base but, if it was used in concatenation of several 256-bit slots would systematically appear remnants of coding. This base can be used on a single passage of (Progc; 8) to bring a more in compression.

We will therefore call the base (9) the threshold base. It is the smallest base that encodes and compresses according to the scheme (SBi) = 2 'bits - - -> 2b-1 = 2b / 2 bits. The base (10) will be called the base of reference. This is the basis on which studies were done. Encoding / compression on 1024 (one thousand) bit slices can yield interesting results. Study of the base (7): b = 7 We study a slice of 2b = 27 = 128 bits.

We therefore consider 128 boxes, ie Cb = 27 = 128. Cb The "Minimum Bit" Bm occupies a maximum (- 1) or 63 boxes. 2 Bm is coded on a bit. The presence or not in box 1 is coded on a bit.

The maximum sum of the boxes is: Sm = (3 x cb - 2xCb - 8) / 8 = (3 x 1282 - 2x 128 - 8) / 8 = 6 111. Or 212 = 4 096 <6 111 <8 192 = 213 therefore (adding the "Null Case") the sum S will be coded on 13 bits.

For the product of the first boxes we must consider 31 prime numbers numbered from NPi = 2 to NP31 = 127. We know that NPi = 2 will be encoded on (b) bits or 7 bits and that all the following ones on a number of bits less than 7.

Any calculations made, from the integer parts of the multiples reports, we obtain that the product P must be coded on 74 bits. We note that 13 prime numbers (the largest) are coded on 1 bit.

The minimum basic coding sequence (7) noted (Sbcmin; 7) is an 89-bit binary sequence. Indeed, 1 + 1 + 13 + 74 = 89. One notices that 89> 64 = 26 = 27/2.

Conclusion: (SBc; 7) - - -> 89 bits. Like the base (8), the base (7) can be considered as a finishing base. Study of the base (6): b = 6 We study a slice of 2 "= 26 = 64 bits so we consider 64 squares, ie Cb = 26 = 64. Cb - 1) or 31 squares The" Minimum Bit "Bm occupies the maximum (2 Bm is coded on a bit.

The presence or not in box 1 is coded on a bit. The maximum sum of the boxes is: Sm = (3 x cb2 - 2xCb - 8) / 8 = (3 x 642 - 2 x 64 - 8) / 8 = 1 519. Gold 210 = 1024 <1519 <2048 = 211 (adding the "Null Case") the sum S will encode on 11 bits. For the product of the first boxes we must consider 18 prime numbers numbered NP, = 2 to NP18 = 61.

It is known that NP, = 2 will be coded on (b) bits or 6 bits and all the following on a number of bits less than 6. All calculations made, from the integer parts of the ratios of multiples, we obtain that the product P must be coded on 42 bits.

We note that 7 prime numbers (the largest) are coded on 1 bit. The minimum basic coding sequence (6) noted (Sbcmin; 6) is a binary sequence of 55 bits. Indeed, 1 + 1 + 11 + 42 = 55. Note that 55> 32 = 25 = 26/2.

Conclusion: (SBc; 6) - - -> 55 bits. Just like the bases (7) and (8), the base (6) can be considered as a base of finishing.40 Study of the base (5): b = 5 We study a slice of 2b = 25 = 32 bits. We therefore consider 32 squares or Cb = 25 = 32.

Cb The "Minimum Bit" Bm occupies at most (2 - 1) 15 boxes. Bm is coded on a bit. The presence or not in box 1 is coded on a bit. The maximum sum of the boxes is: Sm = (3 x cb2 - 2xCb- 8) / 8 = (3 x 322 - 2 x 32 - 8) / 8 = 375. Gold 28 = 256 <375 <512 = 29 therefore (in adding the "Null Case") the sum S will be coded on 9 bits. For the product of the first boxes we must consider 11 prime numbers numbered NP, = 2 to NP '= 31. It is known that NT ', = 2 will be encoded on (b) bits or 5 bits and all the following on a number of bits less than 5. All calculations made, starting from the integer parts of the multiples ratios, we obtain that the product P must be coded on 24 bits. We note that 5 prime numbers (the largest) are coded on 1 bit. The base coding minimal sequence (5) noted (Sbcmin; 5) is a 35 bit bit sequence. Indeed, 1 + 1 + 9 + 24 = 35. We note that 35> 32 = 25 = 28/2. Conclusion: (SBc; 5)> (SBi). The base (5) is not a base of compression. 30 This database can not be used. We can conjecture this result for the bases (b), b <5. In summary: The bases 1, 2, 3, 4, 5 are not bases of compression. The bases 6, 7, 8 are so-called finishing compression bases. The base (9) will be called the threshold base. It is the smallest base that compresses in (SBc) = 2 "/ 2 = 2b-1-bit (no appearance of remains) .All bases (b), with b? 9, compress in (SBc) = 2b 2 = 21.1 bits The base (10) is the reference base and processes 1024-bit slices 40 The base (30) is assumed to be the maximum base and treats slices of the order of one billion bits of bits We will resume the study of coding / compression by the method "COMPRESSION-BSP." We will deepen the conclusions made previously then address the notion of "original rest" to be able to finish the study of the coding phase /compression.

We can consider two cases: 1) In a first case we code with a small number of bases, for example the bases (10), (20), (30) then the base (9) and the bases of finishes. 2) In a second case we code with all the bases of (30) to (6) in decreasing direction. In both cases we will obtain the same results of coding / compression but by different paths. Recall that at any base (b) we associate a computer program coding / compression noted (Progc, b). We will reason on the first case. Let (No) be the original number of bits of the original binary sequence (SBo) to be encoded / compressed. The base (30) therefore (Progc; 30) can act. It is also assumed, first, that (No) is multiple of (239) and of (2 'that is, (No) is of the form () with no integer, no> 0.

Examples: 2 "bits, no = 1, N1 = 2 x bits, n, = 3, N3 = 8 x bits, n, = 0, No = x 239 bits n, = 2, N2 = 4 x Each slice of ( 239) bits being coded by (Progc; 30) in 239/2 = 229 bits and after concatenation of the compressed slices we obtain 20-1 X 23 ° bits. After (n, + 1) passes of (Progc; 30) we will therefore have a compressed binary sequence of 2 '/ 2 = 229 bits We no longer act as a base (30).

We apply the base (20) so (Progc; 20) on a binary sequence of exactly (229) bits (29 X -20 2) bits. After exactly 10 runs of (Progc; 20) we will get a new compressed binary sequence of 229/2 = 219 bits. We no longer act as a base (20).

The base (10) and (Progc; 10) are applied over the sequence of (219) bits, ie (2 X 210) bits. Similarly 10 passes of (Progc; 10) will encode and compress into a binary sequence of 210/2 2 bits or 512 bits.

A base pass (9) of (Progc; 9) encodes and compresses into 256 bits.

Note: one could stop the compression after the action in base (9) or even (10), considering the results of compression obtained. However, we will see what gives the action bases finishing.

The first base of finishing is the base (8). (Progc; 8) encodes and encodes a 256-bit slice into a 144-bit binary sequence. We split this sequence into two sequences: one of 128 bits and one of 16 bits (128 + 16 = 144).

The first 128-bit sequence "base binary sequence" (SBp; 8) and the second 16-bit sequence the "additional binary sequence" in base (8) will be referred to as (SBs; 8). (SBs; 8) is of constant length of 16 bits and will no longer be encoded.

The second base is the base (7). (Progc; 7) encodes and encodes a 128-bit slice into an 89-bit binary sequence. This sequence is broken down into two sequences: one of 64 bits and one of 25 bits (64 + 25 = 89). So we have (SBp; 7) on 64 bits and (SBs; 7) on 25 bits which will no longer be coded.

The third base is the base (6). (Progc; 6) encodes and encodes a 64-bit slice into a 55-bit binary sequence. We can no longer consider another coding / compression. In the end, the compression coding is a bit sequence of 55 bits which must be concatenated with the two additional bit sequences (SBs; 7) and (SBs; 8), the first of 25 bits and the second of 16 bits. The 96 bit binary sequence (96 = 55 + 25 + 16) obtained as a result of the finishing bases will be taken as the final coding / compression bit sequence (SBc).

The coding is coded as "COMPRESSION-BSP" for this first case. No = 2 'x 230 bits with integer no, no> 0.

No - - - - (n, + 1) (Progc; 30) - - - -> 228 - - - - (10) (Progc; 20) - - - -> 218 - - - - (10) (Progc; 10) - - - -> 28 (Progc; 9)> 28 - - - - (Progc; 8) - - - -> 144 = 128 + [16] 40 (Progc; 7) - - - -> 89 = 64 + [25] (Progc; 6) - - - -> 55> SBc <==> 96 bits {55+ [251 bits bits bits bits bits bits bits + 1161 = 96} 45 We will reason about the second case. It is assumed that all bases from (6) to (30) are used so that all associated computer programs (Progc; 6) to (Progc; 30) are written.

The beginning and the end of the coding / compression are identical to the first case: only the intermediate parts different. Let (No) be the original number of bits of the initial binary sequence (SBo) to be encoded / compressed.

Assume No> 23 °, 23 ° = 1 073 741 824. The base (30) therefore (Progc; 30) can act. It is also assumed, in a first step, that (N0) is multiple of (23 °) and of (2 ") that is to say that (N0) is of the form (2 'X 23 ° with no integer, no> 0. Each slot of (23 °) bits being coded by (Progc; 30) in 23 ° / 2 = 229 bits and after concatenation of the compressed slices, one obtains (2'1 X 23 °) bits.

After (no + 1) passes of (Progc; 30), we will have a compressed bit sequence of 23 ° / 2 = 229 bits. We no longer act as a base (30). We apply the base (29) so (Progc; 29) on a binary sequence of exactly (22 °) bits or (2 ° X 229) bits. All that is required is a passage from (Progc; 29) to obtain a new compressed binary sequence of 229/2 = 228 bits. The base (28) and (Progc; 28) are then applied following (2) bits. After a single pass of (Progc; 28), we obtain a bit sequence of (227) bits.

We then apply the base (27) and (Progc; 27) and so on to the base (9) and (Progc; 9). At each passage the binary sequence is divided by 2 and causes the action of the immediately lower base. Following the base (9) and (Progc; 9) is performed as before with the bases and finishing programs (8), (7) and (6) to obtain the same 96-bit compression bit sequence.

SBc <=> 96 bits {55 +125] + 1161 145 We schematize the coding by the method "COMPRESSION-BSP" for this second case. No = 2 'x 230 = 2 "+" bits with integer no, no> 0.

No - - - - (no + 1) (Progc; 30) - - - -> 229 bits. - - - - (Progc; 29) - - - -> 228 bits. - - - - (Progc; 28) - - - -> 227 bits. - - - - (Progc; 27) - - - -> 228 bits. - - - - (Progc; 11) - - - -> 210 bits. - - - - (Progc; 10) - - - -> 2 ° bits. - - - (Progc; 9)> 28 bits - - - - (Progc; 8) - - - -> 144 = 128 + [16] bits. - - - - (Progc; 7) - - - -> 89 = 64+ [25] bits. - - - - (Progc; 6) - - - -> 55 bits. - - - -> SBc <==> 96 bits {55+ [25] + [16]}. We will now study and treat the notion of "original rest" that we will note (Ro). Indeed, we have studied an original number of bits of the form No = 2n ° X 2 "bits with no integer, no> 0 adapted to the first base of compression (base (30) in our case). original rest (Ro) will depend among other things on the first base used.

Let (No) be the number of original bits to encode / compress. Assume No> 23 °. Then there exists necessarily a number, no integer, no> 0 such that No = (2 "x 2") + Ro with 0 <Ro <(x - 1). We can call (x 2 ') the main part of No noted Ppo. It is known to encode and compress the main part at (SBc) = 96 bits.

The rest (Ro) can be treated in the same way. To do this, it suffices to complete (Ro) by a fixed and known bit (for example 0) in a number written (Nro) for It will obviously be necessary to indicate in the definitive encoding the number of bits of the remainder (or the number of (0) ) added). The imperative parameters to be indicated in the final coding will be studied later in order to be able to decode / decompress. With the two methods of coding explained on pages 20 and 21 we will have the same result according to the diagram: No = Ppo + Ro Ppo - - - - (Progc; 30) - - - - (Progc; b) - - - - ( Progc; 9)> SBp = 256 bits. Ro - - - - (Progc; 30) - - - - (Progc; b) - - - - (Progc; 9)> SBr = 256 bits. We concatenate the two sequences (main and remainder) binary (SBp) and (SBr) in a sequence of 512 bits which is coded in base (9) and (Progc; 9) in 256 bits then we apply the bases and programs of fmition.

Note that depending on the number of bits of the rest, it is possible to avoid many coding phases of the rest. Indeed there will always be a base (b) such as the number of bits of the rest Nro, Nro <(2 "x 2"). It will therefore be sufficient to complete the bit (0) to reach this base bit slice (b).

Example: (Ro) is a "rest" of a number of 2 million bits. We have seen that the base (20) processes slices of 220 = 1 048 576 bits. It is therefore preferable to code the "Rest" starting with this base, so with (Progc; 20). 2 millions = 2 x 106 = 2 000 000. Nro = (21 X 220) (2 1U- -6) = 2 097 152 -2 000 000 = 97 152. We thus complete Ro by 97 152 bits (0) to be able to code it as a "Main Party".

This rest complement operation must be indicated in the final encoding. In conclusion, for the coding of the "remainder", if all the bases and programs of (9) to (30) can be used, there will always be a maximum noted base (br) which will initially code this "remainder" after have it completed with a number of bits (0) such as: Nro = (2 "> <Z-bR), n integer If we code with a limited number of bases, it will be sufficient to choose the most appropriate base .

The end coding bit sequence (SBc) is in any case, regardless of the original bit number, a sequence of 96 bits. We note that 96 - 12 x 8 so we can transcribe this sequence in 12 alphanumeric characters coded in octal. Example: SBc <= -> 96 bits <--- => abcdefghijkl} Note: One can imagine to secure the binary coding suite (SBc). Last remark on the coding: Let (No, m) be the initial maximum number of bits that we decide to encode and (No) the original number of bits. We can always decompose (No) according to (NAME) in the form: No = (kx NOM) + (No, r) with k nonzero integer and (No, r) <NAME We will code up to the base of threshold (9) to obtain (k + 1) 256-bit coding sequences which concatenated with possibly adding dummy binary sequences will give after coding a 256-bit binary sequence to which the bases and programs of finishing will be applied to give the binary sequence 96-bit definitive encoding (SBc). Example: We set No, m = 25 x 2 '= 235 = 34 359 738 368, therefore of the order of 34 billion bits. Let No = 110 x 109 110 billion bits. No decomposes into No = (k x No, k,) + (No, r) = (3 x) + (6,920,784,896).

It is assumed that the remainder is completed with bits (0) to obtain a new 235-bit slice. After coding by all the higher bases, we will obtain 4 256-bit slices by the base (9) and (Progc; 9), which will code again in a single 256-bit slice to then finish the coding by the finishing bases in (Sbc) = 96 bits. In the coding it will necessarily be necessary to indicate the number of initial bits to be coded / compressed or parameters which make it possible to find this number: number of processing slots in such a base, number of bits of remainders possibly. This information will obviously be used in the decoding / decompression phase as a decoding limit. It will also be necessary to indicate the origin of the compressed binary sequence (coding in octal, hexadecimal,) and nature of the original document (text, sound, film).

In conclusion, it can be said that any binary sequence can be encoded / compressed into a 96 bit binary sequence. It is necessary to give some additional details and to take into account the physical medium on which will be stored (SBc). An empty text page having a computer weight of the order of 20 KB, we can estimate that any computer document, of whatever nature it is, will be codable and compressible in a document of fixed weight less than or equal to 50 KB, with a compression lossless by the method "COMPRESSION-BSP". We can conclude the coding study as claimed at the beginning. "COMPRESSION-BSP" is a method of lossless compression and theoretical infinite compression rate (Tc = P / Pc, P weight (o) arbitrary, Pc <20KB). We will now study the decoding / decompression phase using the "COMPRESSION-BSP" method. The study will be based on the reference base (10). Explicit methods and results will be exploitable in any base (b).

The decoding / decompression program associated with base (b), d for decoding / decompression will be designated by (Progd; b). Example: (Progd; 10) is the decoding / decompression program associated with the base (10).

We will study the base decoding (10) of a slice of 2 = 1024 bits.

It is recalled that such a slice is coded in a minimum bit sequence (SBCmin; 10) of 417 bits. (SBcmin; 10) = (Bm; Case 1; Sum; Product) This scheme is valid in all bases (b) .This minimal sequence is sufficient to decode / decompress in the initial binary sequence (SBi). decoding is as follows: we know the sum (S) of the decimal boxes and the product (P) of the first boxes of presence of the "Minimum Bit" Bm. Knowing (S) and (P) we find the boxes of the positions of Bm. We form a sum (S °) possible in decimal, decimal boxes (ci, then we form the corresponding product (P °) of the first boxes (cpi) and we compare (P °) to (P): If P ° P then we have not the right combination of boxes and we look for another possible sum If P ° = P then we have the right combination of boxes so we know the positions of Bm completed by its possible presence in box 1. The value of Bm known, we deduce the positions of the other bit so the complete initial binary sequence. shadow of sums to calculate and compare to the product can be very important. Indeed, in minimum coding sequence only (S) is known its decimal value. We saw that we actually coded on 210/2 = 2 = 512 bits to avoid appearances of "leftovers". We will use the 95 (= 512 -417) difference bits to give details on the sum (S) which will facilitate the decoding / decompression phase.

These details on the sum will also depend on the decoding method adopted. The final coding bit sequence (SBc, b) is therefore deduced from the minimum sequence by adding the sum precision bits: (SBc; b) = (Sbcmin; 10); Precisions Sum I (SBc; b) = Bm; Case 1; Sum; Product; Precisions Sum} Example: {(SBc; 10) = {Bm; Case 1; Sum; Product; Precisions Sum = 512 bits.

It is first recalled that the sum considered being those of the box numbers, except box 1, the terms are all different 2 to 2, arranged in ascending order, minus 2 and increased by 1024 in base (10 .45 Accordingly, sums with term 1 will not be used with identical terms, written in decreasing terms. Examples: 1 + 2 + 3; 2 + 5 +5; 5 +4 +3 +2 are sums not to be studied. 2 +3 +4+ 5 is a possible sum. First result on the details of the sums. For any decoding method and for any base (b), b? 10, indicate the number (C) of presence squares of the "Minimum Bit" Bm, which will specify the number of terms of the sum (S). This number (C) will be coded on (b - 1) bits in base (b). Cb -2 Indeed, we saw (page 5, line 7) that Cm - 2 with Cb = 2b. (Cm) is the maximum number of presence boxes of Bm, to which must be added the "Null Case" for which Bm is in no box. We thus have 2-1 possible cases codable on (b - 1) bits. Note: we can not make this precision in base (9) because we have only 4 bits of precision available. Example in base (10): 1024 - 2 1 022 Cm - - 511- - -> 512 cases (addition of the "null case") codable 2 2 on 9 bits because 2 ---- 512. Precisions of sum in base (b) - -> (b - 1) bits for the number of terms (C) of the sum (S).

We recall an equality on integers. Two integers are equal if their prime-power decompositions are equal (same prime numbers and same exponents). We will continue the study of base decoding (10). From now on it is therefore assumed that the number of presence squares (C) of Bm is known: 0 <C <511. There remains 86 (= 95 - 9) sum precision bits to use. Two possible decoding / decompression methods will be explored.40 Incremental decoding method. We will first give two examples of approach to study very simple sums.

Example 1: Let S = 15 be a sum of 2 terms si and s2 S if + s2. We impose the additional condition on s1 and s2: 2 s1 <S2 We can determine all possible sums satisfying these conditions by increments of the terms as follows. We fix s1 = 2; 2 is the smallest possible value for (si) - We deduce s2 = S - 2 = 15 - 2 = 13. We form the sum S1,1 = 2 + 13 = 15. We deduce by incrementing (+1) if si = 2 and by incrementing (-1) of the term s2 = 13 another sum (S1,2) of value 15: S1,2 = 3 + 12 = 15.

By iterations we deduce all possible sums. S1.3 = 4 + 11 = 15; S1.4 = 5 + 10 = 15; S1.5 = 6 + 9 = 15, S16 = 7 + 8 = 15. From (S1,1), we made 5 increments in (+1) and (-1) so 6 are possible.

Note that [E (s2 - si) / 2] +1 = [E (13 - 2) / 2] +1 = E (5,5) + 1 = 5 + 1 = 6. Example 2: We will keep the same value of S. Let S = 15 be a sum of 3 terms: si, S2 and S3; S = S1 + S2 + S3 We impose the additional condition on si, s2 and s3: 2 5. if <s2 <s3. We can determine all the possible sums satisfying these conditions by increments of the terms in the following way: We fix s1,1 = 2 and s1,2 = 3. We deduce if, 3 = S (2 + 3) = 15 - 5 = 10. We form the sum S1,1 = 2 + 3 + 10 = 15. We deduce by incrementing (+1) the term s1,2 = 3 and by incrementing (-1) the term s1,3 = 10 a another sum (S1,2) of value 15: S1,2 = 2 + 4 + 9 = 15.

By iterations, we deduce all possible sums, of first term s1,1 = 2. S1,3 = 2 + 5 + 8 = 15; S1,4 = 2 + 6 + 7 = 15. From (S1,1), we made 3 increments in (+1) and (-1) so 4 sums are possible. S2.2 = 3 + 5 + 7 = 15. We start again by fixing s2,1 = 3 and s2,2 = 4 and we deduce s2,3 = S - (3 + 4) = 15 - 7 = 8. We put S2,1 = 3 + 4 + 8 = 15. We increment in (+1) and (-1) the terms 4 and 7. We obtain: Two sums of first term (3) are possible. We start again by fixing s3,1 = 4 and s3,2 = 5 and we deduce s3,3 = S - (4 + 5) = 15 - 9 = 6. We put S3,1 = 4 + 5 + 6 = 15. No incrementation is possible. We obtain a single sum In total we have obtained 7 (= 4 + 2 + 1) possible sums of value 15 in 3 strictly increasing terms starting from 2. We note that: {[E (s13 - '51, 2 ) / 2 +11+ [E (s23 - s22) / 2 1 +11+ {[E (s33 s3,2) / 2 1 + 1} = {[E (10-3) / 2] +11 + { [E (8 - 4) / 2] 1 + {[E (6-5) 12 1 + 11 = 1E (7/2) + 11+ {E (4/2)} ± {E (1) / 2 +1} E (3.5) +11 + E (2) 1+ {E (0,5) +1} = (3 + 1) + (2) + (0 + 1) = 4 + 2 + 1 = 7. 20 We note therefore that: 1) If the difference between the two terms to be incremented (sk) and (sk_i) (extreme terms) is even we need (i) increments for (i) sums 25 with i = E { (sk - sk_l) / 2} = (sk - sk_i) / 2. 8-2 6 Example: 8 - 2 = 6 pair - -> 2 + 8 - -> 3 + 7- -> 4 + 6 that is 3 - are. 2 Sums (5 + 5, 6 + 4, - - -) are prohibited. 2) If the difference between the two terms to be incremented (sk) and (sk_i 30 (extreme terms) is odd it is necessary (i) increments for [(i) +1] sums with i = E {(sk - Skl) / 2} Example: 8-3 8-3 = 5 odd -> 3 +8 -> 4 + 7 -> 5 + 6 ie3 = E {() + 1 1 35 = E {(-5) + 1} - {E (2,5) +1} = (2 + 1) sums 2 2 We will generalize these mathematical results (for the purpose of computer programming) 40 We will first define the functions of incrementing.

Defmition: One will call "Incrementation of steps (+1)" that one will note (/ +,) the function defined on N by: for all n integer, () (n) = n + 1.

Example: (14.1) (5) = 5 + 1 = 6. Definition: One will call "Incrementation of steps (- 1)" that one will note (Li) the function defined on N by: for all n integer, (L ) (n) = n - 1. Example: Li (5) = 5 - 1 = 4. We will denote by () and () the numbers of increments (/ 44) and (and by (i) the number of simultaneous increments (I +,) and (L) Note that after in increments In applied to (n) the integer (n + LEI) is obtained, Note: (oo - - - - o -) (n) = () (n) = n i-1 - Example: i + 1 = 6: 1 + 1.6 (5) = 5 + 6 = 11. It will be noted that after (L1) increments () applied to (n ) we obtain the integer (n - j1) We write: (0 Li o - - - - - -) (0) = (L11) (n) n - Example: i-1 6 (1-1.6 ) (1 1) = 11 - 6 = 5. We will use the following result: Let S be a sum We assume that S is on (n) terms si; s2; - - - -; sn All terms are integers such that 2 if <s2 <- - - <sn_i <sn 30 We assume that the (n - 1) first integers are consecutive so their sum is (n-1) xn Sn-1 - -1. 2 We deduce the last term sn = S - Sn-1 - By simultaneous increments in () on (sn_i) and on () on (sn), we determine all the sums equal to S, whose (n - 3 ) first terms are consecutive. The number of necessary increments is (i): 40 i = E {[(sn - sn_i) / 2] -1} if (S ,, - sn_i) is even. i = E {[(s' - sn_i) / 2] 1 if (sn - sn_i) is odd. Example 1: Su = 2 + 3 + 4 + 5 + 6 + 18 = 38. (s - sn_i) = 18 - 6 = 12 is even. i = E {[(sn - s) / 2] -11 = E {[(18-6) / 2] -1} = E (4? - -1) 2 = E (6) -1 = 6- 1 = 5. simultaneous increments in (1E1 the 6 sums possible S1,1 = 2 + 3 + 4 + 5 + 6 + 18 = 38. - - -; - on 6 and 18 - -> S1,2 - 1 + 1; - - --I ^; - - .1 + 1;) on (6) and (I -) on 18 give with (= 2 + 3 + 4 + 5 + 7 + 17 = 38. = 2 + 3 + 4 + 5 + 8 + 16 = 38. = 2 + 3 + 4 + 5 + 9 + 15 = 38. = 2 + 3 + 4 + 5 + 10 + 14 = 38. = 2 + 3 + 4 + 5 + 11 + 13 = 38 - on 7 and 17 - -> S13 - on 8 and 16 - -> S1: 1 - on 9 and 15 - -> S1,5 - on 10 and 14- -> S1,6 Note: 2 +3 +4 + 5 + 12 +12 = 38 is a prohibited amount Example2: Si = 2 + 3 + 4 + 5 + 6 + 19 = 39 (s - sn_1) = 19 - 6 = 13 is odd 13 i = E {[(sn - sn_i) / 2] = E {[(19 - 6) / 2]] = E (-2) = E (6.5) = 6. 6 simultaneous increments in (1 + 1) on (6) and (I -) on (19) give with (Si) the 7 possible sums S11 = 2 + 3 + 4 + 5 + 6 +19 --- 39. 1 + 1; - - out of 6 and 19 - -> S1,2 1 + 1; - - on 7 and 18 - -> S1,3 1 + 1; - on 8 and 17 - -> S14 1 + 1; - - on 9 and 16 - -> 5,5 1 + 1; - - on 10 and 15- -> S16; - - on 1 and 14- -> St '= 2 + 3 + 4 + 5 + 7 + 18 = 39. = 2 + 3 + 4 + 5 + 8 +17 = 39. = 2 + 3 + 4 + 5 + 9 + 16 = 39. = 2 + 3 + 4 + 5 + 10 + 15 = 39. = 2 + 3 + 4 + 5 + 11 + 14 = 39. = 2 + 3 + 4 + 5 + 12 + 13 = 39. We can thus determine all the possible sums equal to (S) on (n) terms. It suffices to increment (sn_2) and (sn-1) By () then deduce (sn) and redo simultaneous increments in (14) on (sn_i) and on () on (sr,). We repeat these operations for si = 3 to study the possible sums starting with 3 and 4 until the possible limit.

We have thus described all the possible forms for the sum S, written on (n) integer terms with 2 if <s2 <_ - - <sn_1 <s ,,.

We will only repeat example 1 on page 37, line 6 and continue it. We had: S1,1 = 2 + 3 + 4 + 5 + 6 + 18 = 38. __ (IIi); () -> S1,2 = 2 + 3 +4 +5 +7 +17 = 38. _ (144); () -> S1,3 = 2 + 3 + 4 + 5 + 8 + 16 = 38. _- (1,); () -> S14 = 2 +3 +4 +5 + 9 + 15 = 38. - (); () -> S1.5 = 2 +3 +4 +5 + 10 + 14 = 38. - (144); () -> S1,6 = 2 + 3 + 4 + 5 + 11 + 13 = 38.

Form S2,1 = 2 + 3 + 4 + 6 + 7 + 16 = 38. Onapose () (5) = 6; () () = 7 and () (18) = 16. S2,1 = 2 + 3 +4 +6 + 7 + 16 = 38. _- (); () -> S2,2 = 2 +3 +4 + 6 + 8 + 15 = 38. - (); () -> S2,3 = 2 +3 +4 + 6 + 9 + 14 = 38. - (); (I1) -> S2,4 = 2 +3 +4 + 6 + 10 + 13 = 38. - (1 + 1); () -> S2.5 = 2 +3 +4 + 6 + 11 + 12 = 38. S3,1 -2 +3 +5 + 6 + 7 + 15 = 38 is formed. We put () (5) = 6 and () (16) = 15.

S3,1 = 2 + 3 + 5 + 6 + 7 + 15 = 38. -_ (1 + 1); () -> S3,2 = 2 + 3 + 5 + 6 + 8 + 14 = 38. _- (1 + 1); () -> S3,3 = 2 + 3 + 5 + 6 + 9 + 13 = 38. - (1.1); () -> 53.4 = 2 + 3 + 5 + 6 + 10 + 12 = 38.

S4.1 = 2 +4 +5 + 6 + 7 + 14 = 38. Onaposed (1 + 1) (3) = 4 and () (15) = 14. 54.1 = 2 + 4 + 5 + 6 + 7 + 14 = 38. I + 1) (- -> S4.2 = 2 +4 +5 + 6 + 8 + 13 = 38. I + 1) -> S4,3 = 2 +4 +5 + 6 + 9 + 12 = 38. - - 1-F1; ) -> 54.4 = 2 + 4 + 5 + 6 + 10 + 11 = 38. We form S5,1 = 3 + 4 + 5 + + 7 + 13 = 38. Onapose () (2) = 3 and () (14) = 13.

S5.1 = 3 + 4 + 5 +6 + 7 + 13 = 38. __ (I.f); () -> S5,2 = 3 + 4 + 5 + 6 + 8 + 12 = 38. - (1.41); () -> 55.3 = 3 +4 +5 + 6 + 9 + 11 = 38. S6.1 = 3 +4 + 5 + 7 + 8 + 11 = 38.

Onaposed (1 + 1) (6) = 7; () (7) = 8 and (x) (13) = 11. S6.1 = 3 +4 +5 + 7 + 8 + 11 = 38. - - (I) () - -> S6,2 = 3 +4 +5 + 7 + 9 + 10 = 38.

S7.1 = 3 + 4 + 6 + 7 +8 + 10 = 38. Onaposed (1 + 1) (5) = 6 and (Li) (11) = 10. S7.1 = 3 + 4 + 6 + 7 + 8 + 10 = 38.

Let S8,1 = 3 + 5 + 6 + 7 + 8 + 9 = 38. We put (1 + 1) (4) = 5 and () (10) = 9. Su = 3 + 5 + 6 + 7 + 8 + 9 = 38. No further incrementation is possible. Indeed, the following would give: S9,1 = 4 + 5 + 6 + 7 + 8 + 8 = 38. by posing (LEI) (3) = 4 and () (9) = 8. Given the equality extreme terms (s5 = s6 = 8) the writing of S9,1 is impossible so all the following increments are also.

We got 26 (= 6+ 5 + 4 + 4 + 3 + 2 + 1 + 1) are possible. To know the number of possible sums is not necessary: one will note that this one depends in particular on the differences between two terms (sk) and (that the more the terms are grouped, the less sums will be possible.

Statistically this can translate into a low standard deviation. We will resume the basic decoding study (10) and complete the sequence of "sum accuracies". After having coded the number of boxes of presence of Bm is the number of terms of the sum, it remains 86 bits to use.

By the incremental decoding method, it has been decided to increment the upper extremes first and therefore to fix the first terms starting with the smallest possible.

We know that base (10) there are 1024 boxes so we can code each box on 10 bits by their decimal number. Example: box 10 - - -> 0000010010 starting the coding classically with 0 - - -> 0000000000 on 10 bits.

With 86 (86 = 80 + 6 = 8 x 10 + 6) bits, one can enter 8 Bm presence boxes, including lower boxes, and the remaining 6 bits can fill in the 6 boxes from 2 to 7, box 1 being known. The boxes (c2); (c3); - - - -; (c6); (c7) will be encoded by bit (1) if Bm is present in the box and by bit (0) if Bm is not present in this box.

Example: partial coding (010101), boxes (C2) to (c7), corresponds to the presence of Bm in boxes (c3), (c5) and (c7). The 8 boxes that will be encoded each on 10 bits will therefore be the first 8 boxes of presence of the "Minimum Bit" Bm greater than or equal to 8.

Example: Suppose Bm in boxes {2; 7; 8; 10; 15; 20; 25; 30; 35; 40; 45; 50; 55; 60}. Boxes 2 and 7, (c2) and (c7), will be encoded by bits (1), the boxes {8; 10; 15; 20; 25; 30; 35; 40 will each be encoded on 10 bits. Increments on the sum to determine the boxes {45; 50; 55; 60} will start from the first term (s1) of value 41 (= 40 + 1). This "sum precision" method can be extended to any base (b), b 10. In base (b): let Nr be the number of bits to be used for summation accuracies. Nr = 21) - 1 (Nc, min), with (Nc, min) number of bits of the minimum coding sequence (Sbc, min).

In summary: Details on the sum. Encoding of Sum Precisions in base (b): {Number of Bm boxes of presence encoded on (b - 1) bits; k boxes such that k x 2b <Nr - (b - 1) and (k + 1) x 2> Nr - (b - 1); p boxes from (c2) to (cp + 1) with p = Nr - (b - 1) - (k x 2b) 1. Example in base (10): Precisions of sum b = 10, Nr = 95 = 512 -417, b-1 = 9, k = 8, p = 6. {Number of boxes of presence of Bm; 6 boxes from c2 to c7; 8 boxes from c8 to C1024} - The whole binary coding suite (SBc) is therefore perfectly determined. It is known to define by the incrementing method all the possible sums of value S in (n) integers such that s1 <S <- - - <sn_1 <S ,.

The decoding method is mainly based on the fact, knowing the sum (S) and the product (P) of (n) integers, to define a possible sum by its terms, to break the terms into products of powers of prime numbers, to form the product of all the terms in powers of prime numbers, and finally to compare the product to the given product, knowing that one seeks equality. In our case the terms of the sum are the numbers, in decimal, of the presence squares of the "Minimum Bit" Bm (decimal boxes).

Let Bm be in (C) boxes. Let (S be the sum of the decimal boxes and (P) the product of the first boxes In coding we know the (C) boxes which gives (S) and (P) and in decoding we know the sum (S) and the product (P) which make it possible to find the (C) boxes of presence of Bm.45 Example of simple decoding: Let Bm in the 3 boxes (8; 12; 20) be (c8), (c12) and (C20). coding, we code C = 3, S = 8 + 12 + 20 = 40 and P = (23) x (22 x 31) x (22 x (51) _ 23 + 2 + 2) x (31) x (51) ) 27 x 31 x In decoding it is necessary to find the 3 original spaces knowing C, S, P and that all the terms are strictly increasing starting from 8. One forms 51 = 8 + 9 + 23 = 40 - - -> P1 = 23 x 32 x 231 P. _; - - -> 52 = 8 + 10 + 22 = 40 ---> P2 = 25 x 51 x 111 P. _ 1.1; - -> S3 = 8 + 11 + 21 = 40 ---> P3 = 23 X 31 x 71 x 111 _; - -> S4 = 8 + 12 + 20 = 40 - - -> P4 = 27 x 3 x = P. So the search terms of the sum are (8; 12; 20) so the presence boxes of Bm are (c8), (c12) and (c20). The other bit is in the other 1,021 boxes. Note on comparison of products: for each sum possible, it may be unnecessary to calculate the corresponding product in full. We can calculate the first prime number and its exponent, then compare with P. If this prime number is different from that of P, there is no equality of the products, if there is an identity we compare the second prime number and so right now. We will have equality if all prime numbers are identical in value and exponent. This may make it possible to reduce the number of decoding operations resulting in a decompression speed gain. We will now study and define the notions of sum, product and number of boxes attached. These notions will be applicable to all bases (b) with b> 9. 30 In base (10) we saw that we know perfectly well if the "Minimum Bit" Bm is or not in certain boxes, these boxes are indicated in the coding and there are 91. Box 1 is set apart. Boxes 2 to 7, the first 8 upper boxes or 35 equal to 8 are known, the 75 boxes numbered by the prime numbers NP88 to NP'2 (521-> 1,021), ie a number of 89 (= 6 + 8 + 75). In fact, box 1 is coded in the definition of (SBc), boxes 2 to 7 and the 8 others are boxes coded in the details of the sum, the 75 boxes (NP88 to NP12) are coded in the product, on a single bit specifying the presence or absence of Bm in these boxes. These 89 boxes will be called the annex boxes. We can deduce from it an annexed number of boxes noted Ca, an annexed sum denoted Sa, sum of these boxes, and an annexed product noted Pa, product of these boxes. 45 We have the following frameworks: 0 <Ca <C; 0 <Sa <S; 0 <Pa <P. 15 The part of the decoding study (sums search and product comparison) must focus only on the principal sum denoted Sp with Sp = S-Sa, the main product denoted Pp = -Pa and the number of terms (boxes) principal noted Cp = C - Ca.

Note that Cp <C, Sp <S, Pp <P which partly simplifies the calculations. Example in base (10).

It is assumed that Bm is in 100 boxes (C = 100) whose boxes {2; 7; 10; 20; 30; 40; 50; 60; 70; 80; 541; 839; 971}. We know the sum S and the product P. The 13 boxes {2; 7; 10; 20; 30; 40; 50; 60; 70; 80; 541; 839; 971} are perfectly well known because: 1) 2 and 7 are part of the sum details for boxes 2 to 8. 2) {10; 20; 30; 40; 50; 60; 70; 80} are the first 8 boxes of Bm greater than or equal to 8 and are also encoded in the summation details. 3) {541; 839; 971} are 3 squares equal to prime numbers encoded on a bit in the product. We thus obtain: Ca = 13 = 2 +8 +3. Sa = 2720 = 2 + 7 + 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 541 + 839 + 971. Pa = (21) x (71) x (x 51) x (24 x 51) x (21 x 31 x 51) x (23 x 51 x (21 x 52) x (22 x 3 x 5 ') x ( 21 x 51 x 7 ') x (25 x 5') x (5411) x (8391) x (9711) = 219 x 32 x 59 x 72 x 5411 x 8391 x 9711. So we will put Cp = C - Ca = 100 - 13 = 87, Sp = S - Sa = S - 2 910, Pp = -Pa 35 In decoding we will have to study (Sp) and (Pp) which will give the (Cp) boxes searched then we will add the boxes perfectly well-known appendices, as well as the case of box 1, which will give all the positions of Bm and then by complementarity the boxes of the other bit, hence the original binary sequence 40 We will now study a second dichotomy decoding method. The goal remains the same, ie to find the terms corresponding to a given sum and product.30 In short, the dichotomy method consists in finding by successive approaches the possible values of the first term of the sum, then those of the second. and so on, all the terms of the sum.

We will always do the study on the base (10), the obtained results being able to be extended to any base (b), b? 10. Dichotomy decoding method.

For this method, we will first change the use of the 95 bits of sum precisions We keep the coding of the number of presence boxes of the "Minimum Bit" Bm on 9 bits, ((b - 1) in base ( b)). The number of terms of the sum (S) is thus known. We will use the remaining 86 (= 95 -9) bits to fill in 86 boxes, namely boxes 2 to 87. We will code (1) the boxes in which Bm is present and by (0) the other non-presence boxes of Bm . We will therefore study more than the boxes c88; c88; c1023; c1024 - We consider therefore a sum (S) on (n) integer terms. S = if + s2 + --- + sn_i + sn with 88 <if <s2 - - - <sn-1 <sn <1024. 13 = if x s2 "- - x sn-1 x sn = NP lel X Np 2e2 XX NPiei is also known.

We will seek by minorizations, successive increments and frames the possible values of if. We already know that 88 <if <1024. First result. Let sum (S) be on (n) integer terms, n? 2; if <s2 <- - - - <sn Then necessarily the first term if is such that if <E (-n). The effect sl + sl + --- + if n x if <S so if <E () is if <E (-n) - 1 with E denoting the integer part. Examples: 60 1) S = 60; n = 3; E () = 20. We deduce that if <19. Indeed, if si = 20 is the smallest term, then the minimum sum in 3 consecutive terms 40 is: 20 + 21 + 22 = 63. 63 is greater than 60. By cons 19 + 20 + 21 = 60 is possible. 2) S = 123; n = 5; E (-123) = E (24,6) = 24. We deduce that s1 23. Indeed: 24 + 25 + 26 + 275 + 28 = 130 and 130> 123. For the calculations, results, demonstrations we will note the sum (S) of (n) terms noted s1; s2; - - -; sn with 2 <s1 <s2 <- - - <sn. Then one will apply to the case of the base compression (10), noting (ci) the terms of the sum, (ci) to remember that the terms of the sum are boxes. We thus have a first frame in base (10): 88 <c1 <E (-n) - 1. We will continue the study of the minorizations, additions to obtain finer frameworks One will introduce the notions of "Minimal Sum" ) and "Maximum Sum" associated with a sum.

Definition of the minimum sum. Let (S) be a sum of (n) terms. We always have the condition: 1 <i <n; if whole; 2 <s1 <s2 <- - - <Note (sina) the minimum of (s1). Note (Sin) that we will call the minimal sum associated with (S) the sum of (n) terms defined by: = sno + (sin, 1 + 1) + (+ 2) + - - - + [ sno + (n 1)] + s ,, S with si, = S - {sn ,, i + (sina + 1) + (sina + 2) + - - - + [sina + (n-1) 1 The (n - 1) first terms are the consecutive integers from (sina). Example: S = 50; n = 5; Smi = 2 + 3 + 4 + 5 + [50 - (2 + 3 + 4 + 5)] = 2 + 3 + 4 + 5 + (50 - 14) = 2 + 3 + 4 + 5 + 36 = 50 = S. We will now define the "maximum sum". We will first define the median term of a sum. Median term of a sum Let (S) be a sum of integers on k (terms). We will call the median term of (S) that we will note (smed) the integer defined by: smed = E (); E designating the entire part. 1) Examples: = 10. 2) = 7. 30 S = 30 and k = 3 then SMed = E (-3) = E (10) Note: S = 3 x 10 is a multiple of 3. 30 S = 30 and k = 4 then SMed = E () = E (7,5) Examples of approaches to the maximum sum. Given a given sum S. We will search, starting from the median term, the sum of consecutive integers or with a single difference of 2 of value S. We will separate the study in 2 cases according to the parity of the number of terms ( k). First case. We assume k odd. Example 1: Let S = 30 and k = 3, we see that S is multiple of k (30 = 3 x 10). SMed = E (-s) = E (-3) = E (10) = 10. We will put Seo = 9 + 10 + 11 = 30 = S. (Seo) is in 3 terms centered on smed = 10. (Se0) is the sum of value 30 which is written with the 3 consecutive largest integers possible this which will partly justify the designation of "maximum sum". Indeed 10 + 11 + 12 = 33> 30. Example 2: Let S = 31 and k = 3, we notice that S = (3 x10) + 1. 31 smed = E (-k) = E (-3) = E (10.3) = 10. Sei = 9 + 10 + 12 = 31 = S. (Seo) is in 3 terms. The largest difference is 2, between 10 and 12. (Sei) is obtained from (Seo) by incrementing by (+1) the last term 11. Example 3: Let S = 32 and k = 3, we notice that S = (3 x10) +2. SMed E (-k) = E (-3) = E (10.6) = 10. We will put S m, 2 = 9 + 11 + 12 = 32 = S. (S m, 2) is in 3 terms. The largest difference is 2, between 9 and 11. (Se2) is obtained from (Seo) by incrementing (+1) the last two terms 10 and 11. Example 4: Let S = 33 and k = 3 , we note that S = (3 x11) so S is multiple of 3. 33 SMed E (-s) = E (-3) = E (11) = 11. We will ask Seo = 10 + 11 + 12 = 33 = S. (Seo) is in 3 consecutive terms. We are in the same case as Example 1. We will make a remark on Example 4, on the associated products. Seo = 10 + 11 + 12 = 33 = S - -> Po = 10 x 11 x 12 = 1320. Let S1 be another sum of value S = 33. S1 = 9 + 11 + 13 = 33 = S -> P1 = 9 x 11 x13 = 1287.

It can be seen that P1 <Po. It will be demonstrated later that this type of sum centered on the median term gives the maximum product for S given hence the denomination of "maximum sum". Example 5: Let S = 67 and k = 5. Note that S = (5 x 13) +2. 67 smed = E (--k-) = E (7) = E (13,4) = 13. We will put Seo = 11 + 12 + 13 + 14 + 15 = 65. (Seo) is in 5 consecutive terms of center sied = 13. The rest (r) is 2 so we will make 2 increments of (+1) on the last two terms.

The maximum sum sought will therefore be: Sm = 11 + 12 + 13 + 15 + 16 = 67. Second case. We suppose k par.

Example 1: Let S = 30 and k = 2. Note that S is a multiple of k (30 = 2 x15) 30 smed E (-s) = E (-2) = E (15) = 15. We will put Seo = 14+ 16 = 30 = S. Seo is in 2 terms centered on smed = 15 but in the terms immediately inferior and superior.

Example 2: Let S = 31 and k = 2. Note that S = (2 x15) + 1. 31 smed = E () = E (7) = E (15,5) = 15. We will ask Seo = 14 + 16 and we will deduce Sm = 15 + 16 by incrementing (+1) the term left.

Example 3: Let S = 64 and k = 4. We notice that S = 4x16 = 64. smed = E (-k) = E (-4) = E (16) = 16. 64 We will form (Seo) with the 2 terms immediately below 16 and 2 terms immediately greater than 16. Seo = 14 + 15 + 17 + 18 = 64 = S.

In contrast to the odd k case the median term (smed) does not appear in (Seo) -40 Example 4: Let S = 67 and k = 4. n note that S = (4x16) + 3 = 67. SMed E (- s) = E (-4) = E (16.75) = 16. 67 Seo = 14 + 15 + 17 + 18 = 64. The remainder is r = 3. We will deduce Sm by incrementing by (+1) the 3 (r = 3) terms 14, 15, 18 so we act by circular permutations of left and right.

We obtain: Sm = 15 + 16 + 17 + 19 = 67 = S. We will recall a property on the arithmetic sequences. Arithmetic progression.

A sequence (a) is arithmetic of reason (r) if u ,, 1 = a + r. Note: The set N * of nonzero natural numbers can be considered as an arithmetic sequence of first term 1 and of reason 1. N * = {1; 2 = 1 + 1; 3 = 2 + 1; - - - - 1. The sum of (p) terms of an arithmetic sequence is equal to the number of terms multiplied by the half-sum of the extreme terms. Uk Uk + 1 Uk + 2 up_i = px (uk + up_1) / 2. Example: arithmetic sequence of first-term reason 3 4. 4 + 16 4 + 7 +10 + 13 + 16 = 50 = 5 x (2) = x -20 = 5 x 10 = 50. 2 The formula will be applied to the particular case of consecutive integers. 4 + 8 12 Example: 4 + 5 + 6 + 7 + 8 = 30 = 5 x (2) = 5 x 7 = 5 x 6 = 30. This formula is useful for calculating sums. We will make some mathematical reminders about the notion of congruence for integers. Congruences. Definition. Two integers are congruent modulo (n) if they have the same remainder in the Euclidean division by n. 35 Property. Let n and a be integers; n> 1. Then a is congruent modulo (n) to exactly one of the numbers 1 0; 1; 2; - - -; n - 1 1. We can also write: 40 n> 1; for every integer a, there exists an integer k and integer r called rest such that: a- (kxn) + ravec 0 <r <(n-1). If r = 0 then a is multiple of n. Incrementation (1 + 1) We simply recall that we defined previously the application () By: (/ 44) (a) = a + 1 for any integer (a).

Definition of the maximum sum: Let (S) be a sum of (k) terms. We always have the condition: 1 <i <n; s, integer; 2 <s1 <s2 <- - - <sk We will note (smed) the median term. smed = E (-k) the median term. Note (Sm) that we will call the "Maximum Sum" associated with (S) the sum of (k) terms defined by: First case: k odd.

We denote by k * the integer defined by E (-2) - If S is multiple of k: S = pxk, we will put: (Sm, 0) is centered on (ski &) that is to say that: Sm, 0 = s Med - k *) S Med - 1) + S Med smed +1) + - - - + (smed + k *) = m = S.

We have k consecutive terms left and right of (smed). If S is not multiple of k: S = (p x k) + r with 0 <r <(k - 1). We will deduce Sm from Skd, o by (r) increments (1 + 1) by circular permutations from the last term. Sm = (S Med - k *) + - + (smed - 1) + smed + __ + () (smed + r) + - - + 41,1) (smed + k *) Example: S = 74 en7 terms, S = (7 x10) + 4, k = 7, k * = 3, r = 4. 74 S Med = E (-7) = E (10.5) = 10.

Smx, = 7 + 8 + 9 + 10 + 11 + 12 + 13 = 70. r = 4 then: Sivra = 7 +8 + 9 + () (10) + (.1 + 1) (11) + (1 + 1) (12) + (/,) (13) = 7 + 8 + 9 + 11 + 12 + 13 + 14 = 74 = S.

Second case: k par. We denote by k * the integer defined by E (-2) = -2. If S is multiple of k: S = p xk, we will ask: (Skto) is centered on (smed) that is to say; Sm, c, = (smed - k *) + - - + (smed - 1) + (sm +1) (smed + k *) = Sm.

We have k * consecutive terms left and right of (s), smed not being in Sm, 0.

If S is not multiple of k: S = (p x k) + r with 0 <r <(k - 1). We will deduce (Sm) from (Smm) by (r) increments (1 + 1) by circular permutations from the first term to the left of (smed).

Sm = (smed k *) (smed r) - - - 1 + 1.1) (SMed - 1) ± Sfried +1) ± Sm; k *) Example: S = 74 enterms, S- (6 x12) +2, k = 6, k * = 3, r = 2. 74 smed = E (-6) = E (12.3) = 12. Smm = 9 + 10 + 11 + 13 + 14 + 15 = 72. r = 2 therefore: Sm = 9 + (I + 1) (10) + () (11) +13 + 14 = 9 + 11 + 12 + 13 + 14 = 74 = S. We will now study a property on the produces terms of the "minimum sum" (Sm) and the "maximum sum" (Sm). For that we will demonstrate necessary properties on the products of integers. Properties on whole products.

Property 1: Let be two integers a and b such that: 1 <a <(a + 1) <(b - 1) <b. Thena xb <(a + 1) x (b-1). Indeed: (a + 1) x (b-1) = (a xb) - a + b + 1> (a xb) because (b-a)> and fortiori (b-a) +1> 0. Examples and verifications 1) a = 2 etb = 5donca + 1 = 3 etb - 1 = 4.0nalasuite {2; 3; 4; 5}. = axb = 2x5 = 10 and P2 (a + 1) x (b-1) = 3 x 4 = 12. We have P2> 2) a = 5 etb = 10donca + 1 = 6 etb-1 = 9.0nalasuite { 5; 6; 9; 10}.

P1 = axb = 5 x10 = 50 and P2 - (a + 1) x (b-1) = 6 x 9 = 54.0baby P2> P Property2: We can extend this property to 2> ((k +1) integers such that : 1 <a <(a + 1) <(a + 2) <--- <(a + k) <(bk) <--- <(b-2) <(b-1) <b. obtain P <P1 <P2 <- - - <Pk-1 <Pk with: P = axb; P2 = (a + 1) x (b-1); P2 = (a + 2) x (b-2); ----; Pk-1 = [a + (kl)] x [b- (k-1)]; Pk = (a + k) x (bk).

Examples and verifications. 1) a = 2, b = 5 and k = 2 so a + 1 = 3, a + 2 = 4 and b - 2 = 5, b - 1 = 6, b = 7. Onalasuite {2; 3; 4; 5; 6; 7}. = axb = 2 x 7 = 14; P2 = 3 x 6 = 18; P3 = 4x5 = 20; P3> P2> 2) a = 10, b = 30 and k = 4 so a + 1 = 11, a + 3 = 12, a + 4 = 14 and b - 4 = 26, b - 3 = 27, b - 2 = 28, b 1 = 29. We have the following {10; 11; 12; 13; 14; 26; 27; 28; 29; P1 = 10x30 = 300; P2 = 11x29 = 319; P3 = 12x28 = 336; P4 = 13 x27 = 351; P5 = 14 x26 = 364. We have: P5> P4> P3> P2> P1. The following properties are deduced from the "minimum and maximum sums".

Minimum, Maximum and Product We deduce that for a given sum (S), definitions of the "minimum sum" (Sjp) and the "maximum sum" (Sm) and previous results that the "minimum sum" (S ) will give the product of the minimum terms noted (Pm) and the "maximum sum" (Sm) will give the product of the maximum terms noted (). All intermediate sums (Si) will yield intermediate products (P.). Sm - - - -> Pm = Minimum product. S - - -> Sm - - - -> Pm = Maximum product.

S - - -> Si - - - -> Pi with Pm <Pi <We will use these notions to frame the first term of the sum. Definition of the medium term. the integer denoted (Mo),) defined by Examples: 1) a = 10 and b = 30; Moy Let be two integers a and b such that a <b. We will call "average term" of the pair (a, b) a + b M = E (-2) where E denotes the integer part. a + b 10 + 40 = E (2) -E (2) -E (-2) = E (20) = 20. 2) a = 9 and b = 30; Moy ai-b 9 + 30 39 = E (2) -E (2) = E (-2) = E (19.5) = 19.40 Sum and Minimum and maximum product defined from the average term. Let a, b and X be three integers such that: a <X <b. In our case X will be the first term of a given sum of which we know the value S and of a product whose decomposition in powers of prime numbers is known P. We also know the number of terms of (S) is (n) ). We want to refine the framework of X. We compute the average term of the pair (a, b) that we will note (Moy, 1). The "minimum and maximum sums" are formed from this average term (Moya) as well as the associated products. 1) Minimum sum. Sm = (Aty, 1) + (M0y, 1 + 1) + (M0y, 1 + 2) + - - - + (Moya + k - 2) + sk = S with sk = S {(Mn ,,) + (Moy, 1 + 1) + (M0y, 1 + 2) + - - - + (M0y, 1 + k - 2)} = S - Sm'1 denoting = {(Moy, i) + (M0y, 1 +1) + (M0, +2) + - - - + (Moya + k - 2) = (k-2) x {(M0y, 1) + (Moya + k-2)} 12. 2) Maximum sum. Sm (M0y, 1) + S °. S ° is on (k - 1) terms and it is determined with the median term as seen previously.

The products will be associated with these sums as previously seen: S - - -> Sm Pa, = Minimum product. S - - -> Sm Pm = Maximum product. S - - -> If Pi with Pm <P, <Pm.

Example and verification. Let S = 100; S is on 4 terms, k = 4; S = s, + s2 + s3 + s4 with 10 <sl <20. 10 + 20 The average term is Moya = E (2) - E (-2) = E (15) = 15.

We will form the "minimum and maximum sums" associated with (Moy, 1) - 1) Sm = 15 + 16 + 17 + [100 - (15 + 16 + 17) = 15 + 16 + 17 + 52 = 100 = S We deduce Pm = 15 x16x17x52 = 212 160.40 2) Sm = 15 + S ° = 15 + 85 = 100 = S with S ° = 85 over 3 terms. 85 The median term is (s Med, 1: S Med = E () = E (28,3) = 28; 85 - (3 x 28) +1. We form the sum 27 + 28 +29 = 84 and we gets 85 by incrementing by () the last term 29. S ° = 85 = 27 + 28 + 30 hence Sm = 15 + 27 + 28 + 30 = 100. We deduce Pm = 15 x27x28x30 = 340 200. 3) Either (Si) an intermediate sum. For example Si = 15 + 20 + 25 + 40 = 100.

We deduce Pi = 15 x 20 x 25 x 40 = 300 000. We have 212 160 <300,000 <340 200, ie Pm <Pi <Pm. Application to a coaching.

We will have the following results. (S) and (P) are known: 1) If Pm> P then the first term sought is less than or equal to the average term. 2) If Pm <P then the first term searched is greater than or equal to the average term. 3) If Pm <P <Pm then the average term is a possible value of the first term sought. We then continue by taking the mean terms of the first average term and one of the ends of the first frame and so on which allows to determine lower and larger more and more precise.

The "minimum and maximum sums" associated with the new average terms are reformed as well as the associated products that are compared to the initial product (P). We will support by an example.

Example: Let (S) be the sum of the following 5 terms: S = 20 + 30 + 40 + 50 + 60 = 200. P = 20 x 30 x 40 x 50 x 60 = 72,000,000 = 72 x 106.

Note: We calculate the decimal value of (P) because we are still in "affordable" cases and to support in a relatively simple way the search method of a frame of the first first term. Thereafter and in practice, the products will be compared by their natural logarithms.

We thus try to frame the first term (s,) of (S) knowing that (S) is on k = 5 terms. S = 200, P = 72 x 106, S = s, + s2 + s3 + 54 + s5 with 2 <s, <s2 <53 <54 <s5.

We obtain a first frame by: 200 2 <s1 5_ E (-k-) <-> 2 <s1 <E () <==> 2 s1 5_ E (40) = 40. 2 <s1 <40.

The first average term is calculated: 42 Mc 2 + 40 y, 1 -E (2) = E (7) = E (21) - 21. The "minimum and maximum sums" associated with (Moya) = 21 are determined. Sno = 21 + 22 + 23 + 24 + 110 = 200 which gives Pm, 1 = 21 x 22 x 23 x 24 x 110 = 28 052 640. Salt = 21 + S ° 200 so S ° 179. S ° is on 4 terms . We calculate the median smed term of S °:) - S = 179 Smed = E (-17) = E () = E (44,7) = 44. 179 = (4 x 44) + 3. S ° = ( 1 + 1) (42) + (I +,) (43) + 45 + (1 + 1) (46) = 43 + 44 + 45 + 47 = 179. Sm = 21 + 43 + 44 + 45 + 47 = 200 = S which gives Pm = 21 x 43 x 44 x 45 x 47 = 84 033 180. We see that Pm P 5_ Pm so Moya = 21 is a possible value of (s1 We will therefore calculate the following mean terms on 2 and 21 for try to minimize (s1) and on 21 and 40 to try to increase (s1) Minoration of (s1) 2 + 21 23 We put M0y21 = E (2) = E (7) E (11,5) = 11 We are looking for a reduction so it is useless to calculate a minimal sum and product We form Sm2,1 = 11 ± S ° = 200 so S ° = 189 and S ° is in k = 4 terms S ° 189 We search for the median term smed = () = E (-) = E (47,2) = 47. S ° = 189 = (4 x 47) + 1 and k = 4 is even S ° - = 45+ (1 + 1) ( 46) +48 + 49 = 45 + 47 + 48 + 49 = 189. We deduce Sm = 11 + 45 + 47 + 48 + 49 = 200. We deduce Pm = 11 × 45 × 47 × 48 × 49 = 54,719,280.

We see that Pm <P therefore 11 is a minor of (s1); sl 11.

We can continue by taking as average term: 11 + 21 32 / t4-03,3,1 = E (2) -E (-2) = E (16) = 16. We form S m3,1 = 16 + S ° = 200 so S ° = 184 and S ° is in k = 4 terms. S 184 We look for the median term smed = () = E () = E (46) = 46. S ° ---- 184 = (4 x 46) and k = 4 is even. 44 + 45 + 47 + 48 = 184. We deduce Sm = 16 + 44 + 45 + 47 + 48 = 200. We deduce Pm = 16 × 44 × 45 × 47 × 48 = 71,470,080.

It can be seen that Pm <P therefore 16 is a minor of (s1); s1> 16. We can continue taking as an average term: 16+ 21 37 M0y4,1 = E (2) = E (7) = E (18,5) = 18- Sm4,1 = 18 + S ° is formed = 200 so S ° = 182 and S ° is in k = 4 terms. S 182 We look for the median term smed = E (1-7) = E () = E (45.5) = 45. S ° = 182 = (4 x 45) +2 and k = 4 is even. S = 41 (43) + 1.1 (44) + 46 + 47 = 44 + 45 + 46 + 47 = 182. We deduce Sm = 18 + 44 + 45 + 46 + 47 = 200.

We deduce Pm = 18 x 44 x 45 x 46 x 47 = 77 053 680. It can be seen that Pm> P therefore 18 is a possible value of (s1). We could test with the average term of 16 and 18 or 17 or keep as minor 16. We can either set a maximum of search operations of the minor or a maximum difference between two values of average terms. In the example we will take as the definitive lower bound the value of M0y4,1 = 16. So we have 16 <s1.

Markup of (s1) 21 + 40 61 We set M0y2,2 = E (2) -E () = E (30,5) = 30. We are looking for a markup so it is useless to calculate maximum sum and product. Sm2 2 = 30 + 31 + 32 + 33 + 74 = 200 is formed.

We deduce P: 2.2 = 30 x 31 x 32 x 33 x 74 = 72 673 920. It can be seen that Pm2, 2> P, hence 30 is an upper bound of (sl); sl <30.

We can continue by taking as average term: 51 M0y3,2 = E (21 + 30) = E (-2) = E (25,5) = 25. 2 We form Sni3 2 = 25 + 26 + 27 + 28 + 94 = 200.

We deduce Pm3: 2 = 25 x 26 x 27 x 28 x 94 = 46 191 600. We see that P, 3.2 <P so 25 is a possible value of (si) so if <30. We could test with the average term of 25 and 30 is 27 or keep as an upper bound 30. We will keep 30 as an upper bound. if <30. In conclusion, we have the following (si) frame: 16 <if <30. This method is therefore called a dichotomy because we make cuts by 2 on the extreme terms of the frame to frame the first one. term (if) better. We will continue the study of the example to show that we can now define the possible values of (si). It is known that the product (P) is not given in decimal value but in decomposition of powers of prime numbers. We will then discuss the comparison of products with natural logarithms. In fact P = 2 x 32 x 56 = (72x106 = 72,000,000). We can therefore determine all the possible values of if: 16 <if <30. To continue we fix as first value of (si): s1 = s1,1 = 16. We deduce the sum S1,1 = S - if, / on k - 1 = 4 terms. Sti = 200 - 16 = 184. We deduce the product Pia = P / s11. Pm '13 = 2 x 32 x 56 = (45 x 105 = 4500000). We also search for the possible values of the first term of (S1,1) or the second term (s2) of (S). We do the same for the other three terms. If we find the right combination (S, P) it is finished, otherwise we start again by fixing the second possible value for (si), ie if = s 1,2 = 18. In the example we will find the right combination for if = s1,3 = 20. We will now recall the essential properties of the natural logarithm functions. Natural logarithm. The function "logarithm of Neperian" noted ln (x) is an isomorphism of the multiplicative group on the additive group defined on the set of strictly positive real numbers; x> O. We have the following fundamental properties: 1) a, b real: a> 0 etb> 0 then ln (axb) = ln (a + b). 2) Let a> 0 and b> 0; m and n integers .. ln (am) = mx ln (a). We deduce: ln (am x bn) = mx ln (a) + n x ln (b). 3) The natural logarithm function ln (x) is strictly increasing, ie: a <b <==> ln (a) <ln (b) and a = b <==> ln (a) = ln (b). This function will be used to make comparisons of products P from their decompositions into products of prime factors, P with: P = NPl Np 2e2 2 x _ _ _ _ x Npiel so ei X ln (P) = ln ( Np lel X Np 2e2 x - - - - x Npiel) = (e 1 x ln NPi) + (e2 x ln NP2) + - - - - + (ei x ln NP i). Note 1: Only the natural logarithms of prime numbers will be used. Note2: It can be limited to a small number of decimals. Note3: In the case of logarithmic equality it is obviously necessary to check if the products are equal or if it is an equality due to the approximate values of logarithms. Note4: These logarithms intervene only for comparison studies and in no way in the coding itself which is on natural whole numbers. There is obviously only one method of coding, only the coding of the details of the sum can include different forms specified in advance. Decoding methods are varied. Other decoding / decompression methods based on circular permutations, for example, can be envisaged.

These methods can be cumulated within the same base (b) in particular for large databases (b> 15), those giving a larger number of available bits for sum accuracies. Note: For finishing bases, we can not know the number (C) of spaces and therefore the number of terms of the sum We know only the maximum number (Cm) of spaces. It will therefore be necessary to test all the possible cases with the sum (S) on k terms with 0 k Cm. For each base (b), a decoding / decompression program (Progd, b) is associated. The main function of this program will be the search for possible sums to compare to the products.

The presence boxes of the "Minimum Bit" being determined, it is sufficient to complete by the presence boxes of the second bit to find each initial binary sequence and then by concatenations and iterations the original binary sequence.

The parameters describing the original document make it possible to find the original document that was to be compressed. In conclusion.

The computer coding / compression method "COMPRESSION-BSP" encodes / compresses any computer document by its ORIGINAL binary code, by arithmetic calculations on the natural numbers thus is a compression method WITHOUT LOSS.

The coding / compression by "COMPRESSION-BSP" is summarized in a fixed part, giving the necessary parameters of the document (form, coding, original size) and a binary sequence of very small length of the order of hundred bits. The compression ratio is INFINITE LIMIT.

Let Po and Pc be the computer weights expressed in bytes before and after compression. Po is any. Pc is finite and constant. Method "COMPRESSION-BSP" Document (Doc, o) of weight Po - - - -> Original Binary Suite - - - - (Progc, b) - - - -> Document (Doc, c) of fixed weight Pc, Pc, constant Tc = Po / Pc is of infinite limit. (Doc, c) - - - - (Progd, b) - - - -> (Doc, o) ************************* 45

Claims (5)

CLAIMS1) A method for compressing computer files characterized in that it compresses the source binary code of the digital file to be compressed after splitting into 2b bit sequences, called "bases b", b natural integer. 2) Computer compression method according to claim 1 characterized in that it compresses a binary sequence of 213 bits, b natural number greater than or equal to 9, in another smaller bit sequence is a bit sequence of 2 ' 1 bit. 3) Computer compression method according to any one of the preceding claims characterized in that it compresses all digitized computer files without regard to form, perfectly lossless. 4) Computer compression method according to any one of the preceding claims characterized in that it compresses any computer file of any weight (Kilo-Octects) into a weight file that can be fixed beforehand (minimum weight of the order of 50 KB), which gives a theoretical compression rate of 100% to the computer compression process. 5) Computer compression method according to any one of the preceding claims characterized in that it compresses a binary sequence of 21 bits, b natural number, by studying the sum and the product of the boxes of presence of a bit called "Minimum Bit".