A New Approach to Permutation Polynomials over Finite Fields

Transcription

1 A New Approach to Permutation Polynomials over Finite Fields Xiang-dong Hou Department of Mathematics and Statistics University of South Florida Coding, Cryptology and Combinatorial Designs Singapore,

2 outline 1. Introduction 2. Reversed Dickson Polynomials in Characteristic 2 3. Desirable Triples 4. Families of Desirable Triples 5. Open Questions

3 introduction 1. Introduction

4 the polynomial g q,n q = p κ, n 0. There exists a polynomial g n,q F p [x] satisfying (x + a) n = g n,q (x q x). a F q We want to know when g n,q is a PP of F q e. Call the triple (n, e; q) desirable if g n,q is a PP of F q e.

5 Waring s formula g n,q (x) = n q l n q 1 ( n l l n l(q 1) ) x n l(q 1). Not useful for our purpose!

6 recurrence / negative n g 0,q = = g q 2,q = 0, g q 1,q = 1, g n,q = xg n q,q + g n q+1,q, n q. For n < 0, there exists g n,q F p [x, x 1 ] such that (x + a) n = g n,q (x q x). a F q g n,q satisfies the above recurrence relation for all n Z.

7 about g n,q introduced recently ( ) q-ary version of the reversed Dickson polynomial in characteristic 2 q = 2: PPs g 2,n are related to APN; all known desirable triples (n, e; 2) are contained in 4 families. q > 2: several families of desirable triples are found; computer search (q = 3, e 6 and q = 5, e 2) produced many desirable triples that need explanation.

8 reversed Dickson polynomials in char 2 2. Reversed Dickson Polynomials in Characteristic 2

9 p = 2 / reversed Dickson polynomial p = 2, g 2,n F 2 [x] defined by g n,2 (x(1 x)) = x n + (1 x) n. The nth reversed Dickson polynomial D n (1, x) Z[x] is defined by D n (1, x(1 x)) = x n + (1 x) n. g n,2 = D n (1, x) in F 2 [x].

10 g 2,n and APN APN A function f : F q F q is called almost perfect nonlinear (APN) if for each a F q and b F q, the equation f (x + a) f (x) = b has at most two solutions in F q. Power APN A power function x n is an APN function on F q if and only if for each b F q, the equation (x + 1) n x n = b has at most two solutions in F q. g 2,n and power APN x n is an APN on F 2 2e g 2,n is a PP of F 2 e x n is an APN on F 2 e.

11 desirable triples with q = 2 Known desirable triples (n, e; 2) e n ref 2 k + 1, (k, 2e) = 1 Gold 2 2k 2 k + 1, (k, 2e) = 1 Kasami even 2 e + 2 k + 1, k > 0, (k 1, e) = 1 HMSY 5k 2 8k + 2 6k + 2 4k + 2 2k 1 Dobbertin Conjecture. The above table is complete for q = 2 (up to equivalence).

12 desirable triples 3. Desirable Triples

13 equivalence Facts g pn,q = g p n,q. If n 1, n 2 > 0 are integers such that n 1 n 2 (mod q pe 1), then g n1,q g n2,q (mod x qe x). Equivalence. If n 1, n 2 > 0 are in the same p-cyclotomic coset modulo q pe 1, we say that the two triples (n 1, e; q) and (n 2, e; q) are equivalent and we denote this as (n 1, e; q) (n 2, e; q). If (n 1, e; q) (n 2, e; q), then (n 1, e; q) is desirable if and only if (n 2, e; q) is.

14 necessary conditions Assume that (n, e; q) is desirable. gcd(n, q 1) = 1. If q = 2, then gcd(n, 2 2e 1) = 3. If q > 2 or e > 1, then the p-cyclotomic coset of n modulo q pe 1 has cardinality peκ (q = p κ ).

15 power sum Theorem. Let ɛ F q pe such that ɛ qe ɛ = 1. Then x F q e g n,q (x) k = (a,b) F q F q e (aɛ + b) n[ Consequently, (n, e; q) is desirable if and only if (a,b) F q F q e (aɛ+b) n[ c F q (aɛ+b+c) n c F q (aɛ + b + c) n ] k 1. ] k 1 { = 0 if 1 k < q e 1, 0 if k = q e 1.

16 families of desirable triples 4. Families of Desirable Triples

17 easy cases The following triples are desirable. In all these cases g n,q x qe 2 (mod x qe x). (q pe 2, e; q), q > 2. (q 2e q e 1, e; q), q = 3 κ. (3 2e e 2, e; 3).

18 proposition For n = α 0 q α t q t, 0 α i q 1, w q (n) = α α t. Proposition. Let n = α 0 q α t q t, 0 α i q 1. Then 0 if w q (n) < q 1, 1 if w q (n) = q 1, g n,q = α 0 x q0 +(α 0 + α 1 )x q1 + + (α α t 1 )x qt 1 +δ if w q (n) = q, where δ = { 1 if q = 2, 0 if q > 2.

19 the case w q (n) = q Theorem. Let n = α 0 q α t q t, 0 α i q 1, with w q (n) = q. Then (n, e; q) is desirable if and only if gcd ( α 0 + (α 0 + α 1 )x + + (α α t 1 )x t 1, x e 1 ) = 1.

20 a useful lemma Lemma. Let n = α(p 0e + p 1e + + p (p 1)e ) + β, where α, β 0 are integers. Then for x F p e, g n,p (x) = { gαp+β,p (x) if Tr Fpe /F p (x) = 0, x α g β,p (x) if Tr Fp e /F p (x) 0. Note. The lemma does not hold if p is replaced with q. We do not know if there is a q-ary version of the lemma.

21 theorem Theorem. Let p > 2, n = α(p 0e + p 1e + + p (p 1)e ) + β, where α, β 0. Then (n, e; p) is desirable if the following two conditions are satisfied. (i) Both g αp+β,p and x α g β,p are F p -linear on F p e and are 1-1 on Tr 1 F p e /F p (0) = {x F p e : Tr Fp e /F p (x) = 0}. (ii) g β,p (1) 0. Note. There are many instances where (i) and (ii) are satisfied.

22 example Example. Let p = 3, n = 8(1 + 3 e + 3 2e ) + 7. (α = 8, β = 7.) g 8 3+7,3 (x) = g 31,3 (x) = x 30 x 31 x 32 g n,3 (x) = if Tr F3 e /F 3 (x) = 0, x 8 g 7,3 = x 9 if Tr F3 e /F 3 (x) 0. We have g 31,3 (x 3 x) = x + x 3 + x 33. So g 31,3 is 1-1 on Tr 1 F 3 e /F 3 (0) if and only if gcd(1 + x + x 3, x e 1) = x 1. Conclusion: (n, e; 3) is desirable if and only if gcd(1 + x + x 3, x e 1) = x 1.

23 a more interesting family Theorem. Let n = 4( e + 3 2e ) 7. Then (n, e; 3) is desirable. Proof. { g 4 3 7,3 (x) = g 5,3 (x) if Tr g n,3 (x) = F3 e /F 3 (x) = 0, x 4 g 7,3 (x) if Tr F3 e /F 3 (x) 0. We have g 5,3 = x, g 7,3 = x 3 + x 5 x 7. So g n,3 (x) = { x if Tr F3 e /F 3 (x) = 0, x + x 1 x 3 if Tr F3 e /F 3 (x) 0. It is known that x + x 1 x 3 is 1-1 on F 3 e \ Tr 1 F 3 e /F 3 (0). (Hollmann and Xiang 04; Yuan, Ding, Wang, Pieprzyk, 08)

24 theorem For m Z, let m be the integer such that 0 m p e 2 and m m (mod p e 1). Theorem. Let p be a prime. Assume e 0 (mod 2) if p = 2. Let 0 < α, β < p pe 1 such that (i) α p l (mod pe 1 p 1 ) for some 0 l < e; (ii) w p (β) = p 1; (iii) w p ((αp + β) ) = p. Let n = α(1 + p e + + p (p 1)e ) + β and write (αp + β) = a 0 p a t p t, 0 a i p 1. Then (n, e; p) is desirable if and only if gcd(a 0 + (a 0 + a 1 )x + + (a a t 1 )x t 1, x e 1) = 1.

25 proof of the theorem Let x F p e. If Tr Fp e /F p (x) = 0, g n,p (x) = a 0 x p0 + (a 0 + a 1 )x p1 + + (a a t 1 )x pt 1. If Tr Fp e /F p (x) 0, g n,p (x) = x pl N Fp e /F p (x) s, where s is defined by α = p l + s pe 1 p 1. The rest is easy.

26 open questions 5. Open Questions

27 a difficult one Prove that for p = 2, all desirable triples are given in the table. (A similar conjecture for binary power APN has been standing for many years.) Known desirable triples (n, e; 2) e n ref 2 k + 1, (k, 2e) = 1 Gold 2 2k 2 k + 1, (k, 2e) = 1 Kasami even 2 e + 2 k + 1, k > 0, (k 1, e) = 1 HMSY 5k 2 8k + 2 6k + 2 4k + 2 2k 1 Dobbertin

28 another question Recall: Theorem. Let ɛ F q pe such that ɛ qe ɛ = 1. Then (n, e; q) is desirable if and only if { (aɛ+b) n[ ] k 1 (aɛ+b+c) n = 0 if 1 k < q e 1, 0 if k = q e 1. c F q (a,b) F q F q e Question: What can be said about the sum (aɛ + b) n[ (a,b) F q F q e c F q (aɛ + b + c) n ] k 1?

29 a specific questions p = 3, e = 4, n = 20(1 + 3 e + 3 2e ) (α = 20, β = 219.) { (x x 3 x 32 ) 32 if Tr F34/F g n,3 (x) = 3 (x) = 0, [x 20 (x + x 3 ) + x 1 + x] 3 if Tr F3 4/F 3 (x) 0. (n, e; 3) is desirable because of the following curious fact: ( ) x 20 (x + x 3 ) + x 1 + x is a permutation of F 3 4 \ Tr F3 4/F 3 (0). Question: Can ( ) be generalized to F 3 e for a general e?

30 Thank you!