CN110336656A - Binomial APN function and its generation method in a kind of peculiar sign finite field - Google Patents
Binomial APN function and its generation method in a kind of peculiar sign finite field Download PDFInfo
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Abstract
The present invention relates to binomial APN function and its generation methods in a kind of peculiar sign finite field, comprising the following steps: 1, generation finite fieldWherein p is a prime number, p ≡ 3 (mod 4) and p >=7, n are odd numbers;2, construction setAccording to set A construction set B={ u ∈ A | χ (u+1)=- χ (5u+3) }, construction set C=u ∈ A | χ (u+1)=- χ (5u-3) };3, APN function is obtainedWherein u ∈ B ∪ C.The method of construction binomial APN function proposed by the present invention, limitation when not only theoretically breaking through G.J.Ness and T.Helleseth construction binomial APN function to p=3, but also apply also for the fields such as Coding Theory and Combination Design.
Description
Technical field
The present invention relates to field of cryptography, and in particular to binomial APN function and its life in a kind of peculiar sign finite field
At method.
Background technique
Cipher function is component parts particularly important in symmetric cryptosystem and its Cryptographic Properties directly affects entirely
The safety of cryptographic system.Just because of this, cryptanalysis person be often based on cipher function used in cryptographic system and for being
System is attacked.Linear cipher attack and difference cryptographic attack are two kinds of most common and effective cryptographic attack methods, and password
The nonlinearity and the difference uniformity of function are the key that then to measure it to resist linear cryptographic attack and difference cryptographic attack ability
Index.Therefore, the cipher function with good index is constructed with practical significance.
Cipher function difference uniformity index was introduced by Finland professor K.Nyberg in 1993, and then caused people's structure
Make the upsurge with good difference uniformity function.Studies have shown that the function with the best difference uniformity on even property field
APN function, and under certain condition, such as in the even property field of odd dimension, APN function also have be currently known it is best
Nonlinearity.Therefore, APN function can simultaneously effective resist linear cryptographic attack and difference cryptographic attack.Meanwhile APN
Function is in the other fields such as Coding Theory and Combination Design also extensive application.Because of its special property and its in password
Important application in, APN function receive the extensive concern of people, and evoke the great interest that people construct APN function.So
And due to a lack of effective tool, the construction of APN function seems more difficult.The unlimited class of APN function being currently known is still very
It is few, only 6 class APN monomials and 12 class APN multinomials.Meanwhile because on even property field the unlimited class of APN function it is more rare, people
The construction of APN function expanded on general finite domain study.How APN function finite field in is constructed to become
One hot research topic.So far, for any prime number p, finite field FpnUpper known APN function is simultaneously few and with individual event
Based on formula, the method for constructing multinomial APN function is more rare.2007, Norway scholar G.J.Ness and T.Helleseth were mentioned
A kind of finite field F is gone out3nOn binomial APN function building method, accordingly result is published in international information by flagship periodical
It is upper: " A new family of ternary almost perfect nonlinear mappings, IEEE
Transactions on Information Theory, vol.53, no.7, pp.2581-2586, July 2007 are (a kind of new
Ternary almost Nonlinear Mapping, IEEE information theory transactions, volume 53, the 7th phase, the 2581-2586 pages, 2007 years 7
Month) ".But the building method is suitable only for the situation of p=3.
Summary of the invention
Technical problem to be solved by the invention is to provide binomial APN function and its generations in a kind of peculiar sign finite field
Method.
The technical scheme to solve the above technical problems is that
The method that one kind generates a kind of binomial APN function in peculiar sign finite field, comprising the following steps:
Step 1, the value for determining p and n generate finite fieldWherein p is a prime number, p ≡ 3 (mod 4) and p >=7, n
It is odd number;
Step 2, construction setAccording to set A construction set B=u ∈ A | χ
(u+1)=- χ (5u+3) }, construction set C=u ∈ A | χ (u+1)=- χ (5u-3) };Wherein function χ (x) is defined as: if x
=0, then χ (x)=0, if x is square member, χ (x)=1, if x is non-square of member, χ (x)=- 1.
Step 3 obtains APN functionWherein u ∈ B ∪ C.
Binomial APN function in a kind of peculiar sign finite field, the expression formula of the APN function areWherein u ∈ B ∪ C, B={ u ∈ A | χ (u+1)=- χ (5u+3) }, C=u ∈ A | χ (u+1)=-
χ (5u-3) },For finite field, p is a prime number, p ≡ 3 (mod 4) and p >=
7, n be odd number.
The invention has the benefit that any positive odd number n, prime number p ≡ 3 (mod 4) and p >=7, using side of the invention
Method can construct finite fieldOn a kind of binomial APN function.The method of construction binomial APN function proposed by the present invention,
Limitation when not only theoretically breaking through G.J.Ness and T.Helleseth construction binomial APN function to p=3, but also can
Applied to fields such as Coding Theory and Combination Designs.
Detailed description of the invention
Fig. 1 is the algorithm flow chart that APN function is generated in the present invention;
Fig. 2 is the algorithm flow chart that function APN property is verified in the present invention.
Specific embodiment
The principle and features of the present invention will be described below with reference to the accompanying drawings, and the given examples are served only to explain the present invention, and
It is non-to be used to limit the scope of the invention.
As shown in Figure 1, a kind of method for generating a kind of binomial APN function in peculiar sign finite field, comprising the following steps:
(1) value of p and n are taken, wherein p is a prime number, and p ≡ 3 (mod 4) and p >=7, n are an odd numbers;
(2) finite field is generated
(3) when u takes all over finite fieldWhen, construction set
(4) when u takes all over set A, construction set B=u ∈ A | χ (u+1)=- χ (5u+3) };
(5) when u takes all over set A, construction set C=u ∈ A | χ (u+1)=- χ (5u-3) };
(6) APN function is obtainedWherein u ∈ B ∪ C.
Finite field of the present inventionOn secondary multiplicative character χ and the difference uniformity of cipher function define respectively such as
Under:
If f is from finite fieldTo the mapping of itself, enable
Then the difference uniformity of function f is defined as
And abbreviation f is ΔfDifference function, wherein #S indicates to concentrate the number of element in S, and maxS indicates maximum in set S
That element.Particularly, work as ΔfWhen=1, function f is referred to as Perfect Nonlinear Functions;Work as ΔfWhen=2, function f is referred to as almost
Perfect Nonlinear Functions, that is, APN function.
Either in cryptographic system, or in the fields such as Coding Theory or Combination Design, in general, Δf
Value be all require it is the smaller the better.
Verify the existence of function
Below we to verify the APN function constructed in the present invention be it is existing, that is, the u for meeting condition must
It is fixed to exist.Here we illustrate set B non-empty.If setN is the number of element in set B, i.e.,Then
Work as pnWhen > 7,To N >=1.Work as pnWhen=7, it is easily verified that B={ 4 }.Therefore set B non-empty.
The theoretical validation of function APN property
Next we illustrate, are APN functions by the function f (x) that the method in the present invention obtains, i.e., only need to illustrate: right
ArbitrarilyEquation f (x+a)-f (x)=b solution maximum number is 2.
ByCan obtain: Equation f (x+a)-f (x)=b is
1. working asWhen, remember ta=χ (x+a), tx=χ (x).Above-mentioned equation both sides are obtained with multiplied by x (x+a): bx2-
(u(ta-tx)-ab)x+autx+ a=0.
(1) first it is contemplated that the case where b ≠ 0.To taAnd txValue point following four situation consider.Referring to table 1,
It gives the case where equation and its root of every kind of situation.
Table 1
Situation | ta | tx | Equation | x1x2 | Discriminate Δ |
1 | 1 | 1 | bx2+ abx+a (u+1)=0 | b-1a(u+1) | a2b2-4(u+1)ab |
2 | 1 | -1 | bx2(2u-ab) x-a (u-1)=0 | -b-1a(u-1) | a2b2-4ab+4u2 |
3 | -1 | 1 | bx2+ (2u+ab) x+a (u+1)=0 | b-1a(u+1) | a2b2-4ab+4u2 |
4 | -1 | -1 | bx2+ abx-a (u-1)=0 | -b-1a(u-1) | a2b2+4(u-1)ab |
Situation 1:ta=1, tx=1.Above-mentioned equation is are as follows: bx2+ abx+a (u+1)=0.If this situation equation has solution,Because p ≡ 3 (mod 4), n are odd numbers, thus -1 right and wrong square member, i.e. χ (-
1)=- 1.It can thus be concluded that:Due to two satisfactions of equationSo feelings
Shape equation at most only one solution.
Situation 2:ta=1, tx=-1.Above-mentioned equation is are as follows: bx2(2u-ab) x-a (u-1)=0.If equation two are x1,
x2, then we haveWith x1+a,x2+ a is that the equation of root is b (x-a)2-(2u-ab)(x-a)-a
(u-1)=0, abbreviation are as follows: bx2(2u+ab) x+a (u+1)=0.We obtain:By
In χ (u+1)=χ (u-1) and χ (- 1)=- 1 event: χ (x1x2(x1+a)(x2+ a))=χ (x1x2)·χ((x1+a)(x2+ a))=- 1.
So situation equation at most one solution.
To situation 3:ta=-1, tx=1 and situation 4:ta=-1, tx=-1 is respectively adopted similar to situation 2 and situation 1
Discussion method can obtain every kind of situation equation at most only one solution.
Next we illustrate: for given a, b, in situation 1 and situation 4 equation at most only one solve, 2 He of situation
Also at most only one is solved equation in situation 3, to obtain Equation f (x+a)-f (x)=b, at most only there are two solutions.
If situation 1 and situation 4 have solution,With known χ (u+1)
=χ (u-1) contradiction.So situation 1 and 4 equation of situation at most have a solution for given a, b.
Two solutions of equation meet in situation 2If x1Meet χ (x1+ a)=1 and χ (x1)
=-1, thenWith-(x1+ a) ,-(x2+ a) be root equation be represented by b (- x-a)2-(2u-ab)
(- x-a)-a (u-1)=0 is the equation in situation 3 after abbreviation.If two solutions of equation are y1, y2, wherein y1Meet χ (y1+a)
=-1, χ (y1)=1.Then there is y1=-(x2+ a), y2=-(x1+a).So
AndThis and known conditions χ (u+1)=χ (u-1) contradiction.Therefore situation 2 and feelings
Equation also at most has a solution in shape 3.
(2) next it is contemplated that the case where b=0.At this point, Equation f (x+a)-f (x)=b can turn to u (ta-tx) x=
autx+a.To taAnd txValue also we divide four kinds of situations to consider, see Table 2 for details.
Table 2
Situation | ta | tx | Equation | The number of solution of equation |
1 | 1 | 1 | A (u+1)=0 | Equation is without solution (because u ≠ -1) |
2 | 1 | -1 | X=- (2u)-1a(u-1) | At most one solution |
3 | -1 | 1 | X=- (2u)-1a(u+1) | At most one solution |
4 | -1 | -1 | A (u-1)=0 | Equation is without solution (because u ≠ 1) |
Due to χ (u+1)=χ (u-1), therefore also at most only one is solved for equation in a, b, four kinds of given situations.
2. now it is contemplated that the situation of x ∈ {-a, 0 }.Remember b1=a-1(1+u), b2=a-1(1-u).Next we say
It is bright to work as b=b1And b=b2When four kinds of situation equations considering of front to there is more solutions.Work as b=b1When, in situation 1This and ta=1, tx=1 contradiction, therefore equation is without solution.Situation 2,3 and feelings
The discriminate of equation is respectively Δ in shape 42,3=(u-1) (5u+3), Δ4=(u+1) (5u-3).Known χ (u+1)=χ (u-1)
=-χ (5u+3) or χ (u+1)=χ (u-1)=- χ (5u-3), so Δ2,3With Δ4At least one right and wrong square member, therefore four
Kind situation equation at most only one solution.Therefore, Equation f (x+a)-f (x)=b1At most there are two solutions.Work as b=b2When, it is similar can
Obtain four kinds of situation equations at most two solutions.
Since there are z=ab in domain, so that the discriminate of the equation in four kinds of situations is non-square of member, it is fixed to takeThen Equation f (x+a)-f (x)=b in four kinds of situations is without solution, thus in the presence ofSo that there are two solutions by Equation f (x+a)-f (x)=b'.Therefore, verifying explanation is according to the present invention
What method generated is APN function.
The proof of algorithm of function APN property
We provide the algorithm of the APN property of verifying function f (x) below, as shown in Fig. 2, hereWherein u ∈ B ∪ C.
1. times taking
2. set of computationsElement number, be denoted as Na,b;
3. if Na,b> 2, then being concluded that function f (x) not is APN function, and algorithm terminates;If Na,b≤ 2, then it takes another
Group a, the value of b continue to execute step 2;
4. if to allMaximum value be 2, then be concluded that f (x) is APN letter
Number.
Algorithm according to fig. 2, can also verify illustrate that method of the invention generates is APN function.
Embodiment 1
Finite field is given below in embodiment according to the present inventionOn a kind of binomial APN function generation method.
(1) p=7, n=3 are taken.
(2) finite field is generatedWherein α is the primitive element in domain.
(3) when u takes all over finite fieldWhen, construction setBy Magma program
Element can be obtained in set A shaped like αi, wherein i removes the number in table.
(4) when u takes all over set A, construction set B=u ∈ A | χ (u+1)=- χ (5u+3) }.It can must be gathered according to program
Element is shaped like α in Bi, wherein the value of i is located at following table.
9 | 13 | 15 | 19 | 129 | 130 | 133 | 135 | 136 | 139 | 143 | 255 | 257 | 260 | 261 | 268 | 273 |
23 | 26 | 31 | 39 | 40 | 151 | 161 | 165 | 166 | 274 | 278 | 280 | 281 | 284 | 288 | 289 | 290 |
43 | 46 | 48 | 51 | 55 | 63 | 172 | 178 | 182 | 183 | 188 | 295 | 300 | 301 | 304 | 306 | 313 |
75 | 76 | 79 | 86 | 88 | 89 | 90 | 190 | 201 | 202 | 208 | 211 | 214 | 217 | 320 | 322 | 336 |
91 | 99 | 101 | 105 | 109 | 110 | 117 | 167 | 219 | 220 | 226 | 228 | 236 | 247 | 248 | 250 | 317 |
(5) when u takes all over set A, construction set C=u ∈ A | χ (u+1)=- χ (5u-3) }.Utilize journey
Sequence we obtain in set C element shaped like αi, wherein the value of i is given in the table below.
1 | 7 | 11 | 12 | 17 | 129 | 130 | 133 | 135 | 142 | 257 | 259 | 260 | 261 | 262 | 270 | 272 |
30 | 31 | 37 | 40 | 43 | 46 | 48 | 146 | 149 | 151 | 165 | 276 | 280 | 281 | 288 | 300 | 301 |
55 | 57 | 65 | 76 | 180 | 184 | 186 | 190 | 194 | 197 | 202 | 304 | 306 | 307 | 310 | 314 | 322 |
79 | 84 | 86 | 89 | 90 | 97 | 210 | 211 | 214 | 217 | 219 | 222 | 226 | 332 | 336 | 337 | 338 |
19 | 49 | 77 | 102 | 103 | 107 | 109 | 110 | 113 | 117 | 118 | 119 | 124 | 234 | 246 | 247 | 250 |
(6) finite field is obtainedOn APN function
F (x)=ux170+x341,
Wherein the value of u is shaped like αi, i removes the number in table.
1 | 7 | 9 | 11 | 12 | 13 | 15 | 161 | 165 | 166 | 167 | 172 | 314 | 317 | 320 | 322 |
23 | 26 | 30 | 31 | 37 | 178 | 180 | 182 | 183 | 184 | 186 | 188 | 190 | 336 | 337 | 338 |
40 | 43 | 46 | 48 | 49 | 51 | 55 | 197 | 201 | 202 | 208 | 210 | 211 | 214 | 217 | 228 |
57 | 65 | 75 | 76 | 77 | 79 | 84 | 86 | 88 | 89 | 90 | 222 | 226 | 234 | 236 | 246 |
97 | 99 | 101 | 102 | 103 | 105 | 107 | 109 | 110 | 250 | 255 | 257 | 259 | 260 | 261 | 262 |
113 | 117 | 118 | 119 | 124 | 129 | 130 | 270 | 272 | 273 | 274 | 276 | 278 | 280 | 281 | 284 |
133 | 135 | 136 | 139 | 142 | 143 | 146 | 149 | 289 | 290 | 295 | 300 | 301 | 304 | 306 | 307 |
17 | 19 | 39 | 63 | 91 | 151 | 194 | 219 | 220 | 247 | 248 | 268 | 288 | 310 | 313 | 332 |
The foregoing is merely presently preferred embodiments of the present invention, is not intended to limit the invention, it is all in spirit of the invention and
Within principle, any modification, equivalent replacement, improvement and so on be should all be included in the protection scope of the present invention.
Claims (2)
1. the method that one kind generates a kind of binomial APN function in peculiar sign finite field, which comprises the following steps:
Step 1, the value for determining p and n generate finite fieldWherein p is a prime number, and p ≡ 3 (mod 4) and p >=7, n are odd
Number;
Step 2, construction setAccording to set A construction set B=u ∈ A | χ (u+1)
=-χ (5u+3) }, construction set C=u ∈ A | χ (u+1)=- χ (5u-3) };Wherein function χ (x) is defined as: if x=0, then
χ (x)=0, if x is square member, χ (x)=1, if x is non-square of member, χ (x)=- 1;
Step 3 obtains APN functionWherein u ∈ B ∪ C.
2. binomial APN function in a kind of peculiar sign finite field, which is characterized in that the expression formula of the APN function isWherein u ∈ B ∪ C, B={ u ∈ A | χ (u+1)=- χ (5u+3) }, C=u ∈ A | χ (u+1)
=-χ (5u-3) }, For finite field, p is a prime number, p ≡ 3 (mod 4) and p
>=7, n are odd numbers.
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CN107431628A (en) * | 2015-03-27 | 2017-12-01 | 国际商业机器公司 | Broadcast enciphering based on media key block |
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US20120263303A1 (en) * | 2009-12-24 | 2012-10-18 | Shaohua Tang | Group key management approach based on linear geometry |
US20150010117A1 (en) * | 2013-07-08 | 2015-01-08 | Samsung Electronics Co., Ltd. | Apparatus and method for transmitting and receiving signal in broadcasting and communication systems |
CN107431628A (en) * | 2015-03-27 | 2017-12-01 | 国际商业机器公司 | Broadcast enciphering based on media key block |
CN107204841A (en) * | 2017-03-14 | 2017-09-26 | 中国人民武装警察部队工程大学 | A kind of method that many S boxes of the block cipher for resisting differential power attack are realized |
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