permutation polynomials over finite...

Permutation Polynomials over Finite Fields

Zulfukar Saygı

Department of Mathematics,TOBB University of Economics and Technology,

Ankara, Turkey.

10 April 2015

Zulfukar Saygı Permutation Polynomials over F.F.

Outline

Basic Definitions and Notations

Some Known Results

Motivation

Some New Results and Open Problems

Notations

q be a positive power of a prime,

Fq be a finite field with q elements,

F∗q = Fq \ {0},Fq[x ] be the polynomial ring with the variable x ,

Sn be the symmetric group of order n,

Tr be the trace map from Fqk to Fq,

where Tr(x) = x + xq + · · ·+ xqk−1.

Permutation Polynomials

A polynomial f ∈ Fq[x ] is called a permutation polynomial(PP) of Fq if x → f (x) is a permutation of Fq.

A PP correspond to an element of the symmetric group Sq.There are q! PPs of Fq, all of which are given by the Lagrangeinterpolation

Given a permutation g of Fq, the unique permutationpolynomial Pg (x) of Fq of degree at most q − 1:

Pg (x) =∑a∈Fq

1 − (x − a)q−1)

A PP correspond to an element of the symmetric group Sq.

There are q! PPs of Fq, all of which are given by the Lagrangeinterpolation

Pg (x) =∑a∈Fq

1 − (x − a)q−1)

Pg (x) =∑a∈Fq

1 − (x − a)q−1)

Pg (x) =∑a∈Fq

1 − (x − a)q−1)

A Remark

If f is a PP and a 6= 0, b 6= 0, c ∈ Fq, then f1 = af (bx + c) isalso a PP.

By suitably choosing a, b, c we can arrange to have f1 innormalized form

f1 is monic,f1(0) = 0,when the degree n of f1 is not divisible by char(Fq), thecoefficient of xn−1 is 0.

A Remark

If f is a PP and a 6= 0, b 6= 0, c ∈ Fq, then f1 = af (bx + c) isalso a PP.

By suitably choosing a, b, c we can arrange to have f1 innormalized form

f1 is monic,f1(0) = 0,when the degree n of f1 is not divisible by char(Fq), thecoefficient of xn−1 is 0.

Some well known classes of PPs

Every linear polynomial over Fq is a PP.

Monomials: xn is a PP of Fq iff gcd(n, q − 1) = 1.

Dickson: For a ∈ F∗q, the polynomial

Dn(x , a) =

bn/2c∑i=0

n − i

(n − i

)(−a)ixn−2i

is a PP of Fq iff (n, q2 − 1) = 1.

Linearized: The polynomial L(x) =∑n−1

s=0 asxqs ∈ Fqn [x ] is a

PP of Fqn iff det(aq

)6= 0, 0 ≤ i , j ≤ n − 1.

Dn(x , a) =

bn/2c∑i=0

n − i

(n − i

)(−a)ixn−2i

)6= 0, 0 ≤ i , j ≤ n − 1.

Dn(x , a) =

bn/2c∑i=0

n − i

(n − i

)(−a)ixn−2i

)6= 0, 0 ≤ i , j ≤ n − 1.

More Examples

For odd q, f (x) = x (q+1)/2 + ax is a PP of Fq iff a2 − 1 is anonzero square in Fq.

f (x) = x r (g(xd))(q−1)/d is a PP of Fq if gcd(r , q − 1) = 1,d | q − 1, and g(xd) has no nonzero root in Fq.

Note that if f (x) and g(x) are PPs of Fq then f (g(x)) is aPP of Fq.

More Examples

For odd q, f (x) = x (q+1)/2 + ax is a PP of Fq iff a2 − 1 is anonzero square in Fq.

f (x) = x r (g(xd))(q−1)/d is a PP of Fq if gcd(r , q − 1) = 1,d | q − 1, and g(xd) has no nonzero root in Fq.

Note that if f (x) and g(x) are PPs of Fq then f (g(x)) is aPP of Fq.

Criteria for the PPs

For f ∈ Fq[x ], the following statements are equivalent:

f is a PP of Fq

For each y ∈ Fq, f (x) = y has at least one solution x ∈ Fq

For each y ∈ Fq, f (x) = y has at most one solution x ∈ Fq

For all nontrivial additive characters χ of Fq we have∑a∈Fq

χ(f (a)) = 0

Note that∑

a∈Fqχ(f (a)) =

∑a∈Fq

χ(a) = 0

f is a PP of Fq

χ(f (a)) = 0

Note that∑

a∈Fqχ(f (a)) =

∑a∈Fq

χ(a) = 0

f is a PP of Fq

χ(f (a)) = 0

Note that∑

a∈Fqχ(f (a)) =

∑a∈Fq

χ(a) = 0

Hermite’s criterion

f is a PP of Fq iff

∑x∈Fq

f (x)s =

{0 if 0 ≤ s ≤ q − 2,−1 if s = q − 1.

f is a PP of Fq iff1 f (x) has exactly one root in Fq,2 For each integer t with 1 ≤ t ≤ q − 2 and p 6 | t, the reduction

of (f (x))t mod (xq − x) has degree at most q − 2

where p = char(Fq).

Difficult to apply for a general polynomial.

f is a PP of Fq iff

∑x∈Fq

f (x)s =

{0 if 0 ≤ s ≤ q − 2,−1 if s = q − 1.

where p = char(Fq).

f is a PP of Fq iff

∑x∈Fq

f (x)s =

{0 if 0 ≤ s ≤ q − 2,−1 if s = q − 1.

where p = char(Fq).

Enumeration of PPs

Hermite’s criterion was used by Dickson to obtain allnormalized PPs of degree at most 5.

A list of PPs of degree 6 over finite fields with oddcharacteristic can be found in [D].

[D] L. E. Dickson, The analytic representation of substitutionson a power of a prime number of letters with a discussion ofthe linear group, Ann. of Math. 11 (1896/97), 65–120.

A list of PPs of degree 6 and 7 over finite fields with char=2is presented in [LCX].

[LCX] J. Li, D. B. Chandler, and Q. Xiang, Permutationpolynomials of degree 6 or 7 over finite fields of characteristic2, Finite Fields Appl. 16 (2010) 406–419.

All monic PPs of degree 6 in the normalized form is presentedin [SW].

[SW] C. J. Shallue and I. M. Wanless, Permutationpolynomials and orthomorphism polynomials of degree six,Finite Fields and Their Applications 20 (2013) 84–92

Enumeration of PPs

An Open Problem

Let Nn(q) denote the number of PPs of Fq which have degreen.

Trivial boundary conditions:

N1(q) = q(q − 1),Nn(q) = 0 if n 6= 1 is a divisor of q − 1,∑q−1

n=1 Nn(q) = q!.

Problem: Find Nn(q).R. Lidl and G. L. Mullen, When does a polynomial over a finitefield permute the elements of the field?, II, Amer. Math. Monthly100 (1993) 71–74.

An Open Problem

Let Nn(q) denote the number of PPs of Fq which have degreen.

Trivial boundary conditions:

N1(q) = q(q − 1),Nn(q) = 0 if n 6= 1 is a divisor of q − 1,∑q−1

n=1 Nn(q) = q!.

Problem: Find Nn(q).R. Lidl and G. L. Mullen, When does a polynomial over a finitefield permute the elements of the field?, II, Amer. Math. Monthly100 (1993) 71–74.

Motivation

m a positive integer,

F2 finite field of order 2.

f : Fm2 → Fm

Nf (u, v) := #

{u = x + y ;

v = f (x) + f (y).

u 6= 0 =⇒ Nf (u, v) = 0 or 2

→ f is almost perfect non-linear [APN] .

(Affine Equivalence) Let A, B, C be three affinetransformations of Fm

2 . If A, B are permutations then

f is APN ⇐⇒ A ◦ f ◦ B + C is APN

Motivation

f : Fm2 → Fm

Nf (u, v) := #

{u = x + y ;

v = f (x) + f (y).

u 6= 0 =⇒ Nf (u, v) = 0 or 2

Motivation

f : Fm2 → Fm

Nf (u, v) := #

{u = x + y ;

v = f (x) + f (y).

u 6= 0 =⇒ Nf (u, v) = 0 or 2

Motivation

Flat characterization of APNs{x + y + z + t = 0

all distinct=⇒ f (x) + f (y) + f (z) + f (t) 6= 0

then f is [APN] .

Code Characterization

1 . . . 1 . . . 10 . . . x . . . 1

f (0) . . . f (x) . . . f (1)

if the minimal distance of code(f ) > 4 then f is [APN] .(The code is double-error-correcting (no fewer than 5 cols sum to 0).)

Motivation

then f is [APN] .

1 . . . 1 . . . 10 . . . x . . . 1

f (0) . . . f (x) . . . f (1)

if the minimal distance of code(f ) > 4 then f is [APN] .

(The code is double-error-correcting (no fewer than 5 cols sum to 0).)

Motivation

then f is [APN] .

1 . . . 1 . . . 10 . . . x . . . 1

f (0) . . . f (x) . . . f (1)

if the minimal distance of code(f ) > 4 then f is [APN] .(The code is double-error-correcting (no fewer than 5 cols sum to 0).)

Motivation

Dobbertin constructed several classes of PPs over finite fieldsof even characteristic and used them to prove severalconjectures on APN monomials.

H. Dobbertin, Almost perfect nonlinear power functions onGF (2n): the Niho case, Inform. and Comput. 151 (1999)57–72.H. Dobbertin, Almost perfect nonlinear power functions onGF (2n): the Welch case, IEEE Trans. Inform. Theory 45(1999) 1271–1275.

Motivation

The existence of APN permutations on F22n is a long-termopen problem.

Hou proved that there are no APN permutations over F24 andthere are no APN permutations on F22n with coefficients inF2n .

X.-D. Hou, Affinity of permutations of Fn2, Discrete Appl.

Math. 154 (2006) 313–325.

Recently, Dillon presented the first APN permutation over F26 .

K. A. Browning, J. F. Dillon, M. T. McQuistan, and A. J.Wolfe, An APN permutation in dimension six, In Finite Fields:Theory and Applications, volume 518 of Contemp. Math.,33–42, Amer. Math. Soc., Providence, RI, 2010.

Open Problem Is there any APN permutation on F22n forn ≥ 4.

Motivation

Math. 154 (2006) 313–325.

Motivation

Math. 154 (2006) 313–325.

Motivation

The Kloosterman sum K (a) over F2n is defined for anya ∈ F2n by

K (a) =∑a∈F∗

(−1)Tr(ax+1x )

Shin, Kumar and Helleseth found that the existence of certain3-designs in the Goethals code of length 2n, n odd, over Z4

was equivalent to the identity

1 + a4

)∀a ∈ F2n \ {1}

and they proved this identity for all odd values of n.This relation was extended to the case n even by Helleseth andZinoviev.

Motivation

The Kloosterman sum K (a) over F2n is defined for anya ∈ F2n by

K (a) =∑a∈F∗

(−1)Tr(ax+1x )

Shin, Kumar and Helleseth found that the existence of certain3-designs in the Goethals code of length 2n, n odd, over Z4

was equivalent to the identity

1 + a4

)∀a ∈ F2n \ {1}

and they proved this identity for all odd values of n.This relation was extended to the case n even by Helleseth andZinoviev.

Special PPs

Helleseth and Zinoviev used the PPs(1

x2 + x + δ

+ x over F2n

to derive new identities of Kloosterman sums over F2n ,where δ ∈ F2n with Tr(δ) = 1 and l ∈ {0, 1}.

Recently, PPs with the form

f (x) =(xp

i − x + δ)s

+ L(x)

over the finite field Fq have been extensively studiedwhere , i , s ∈ Z+, δ ∈ Fq, char(Fq) = p and L(x) is alinearized polynomial in Fq[x ].

Special PPs

Helleseth and Zinoviev used the PPs(1

x2 + x + δ

+ x over F2n

to derive new identities of Kloosterman sums over F2n ,where δ ∈ F2n with Tr(δ) = 1 and l ∈ {0, 1}.Recently, PPs with the form

f (x) =(xp

i − x + δ)s

+ L(x)

over the finite field Fq have been extensively studiedwhere , i , s ∈ Z+, δ ∈ Fq, char(Fq) = p and L(x) is alinearized polynomial in Fq[x ].

Some PPs of the form(xp

i − x + δ)s

+ L(x)

T. Helleseth, V. Zinoviev, New Kloosterman sums identitiesover F2m for all m, Finite Fields Appl. 9 (2003) 187-193.

J. Yuan, C. Ding, Four classes of permutation polynomials ofF2m , Finite Fields Appl. 13 (2007) 869-876.

J. Yuan, C. Ding, H. Wang, J. Pieprzyk, Permutationpolynomials of the form (xp − x + δ)s + L(x), Finite FieldsAppl. 14 (2008) 482-493.

X. Zeng, X. Zhu, L. Hu, Two new permutation polynomialswith the form (x2

k+ x + δ)s + x over F2n , Appl. Algebra Eng.

Commun. Comput. 21 (2010) 145-150.

N. Li, T. Helleseth, X. Tang, Further results on a class ofpermutation polynomials over finite fields, Finite Fields Appl.22 (2013) 16-23.

Z. Tu, X. Zeng, C.Li, T. Helleseth, Permutation polynomialsof the form (xp

m − x + δ)s + L(x) over the finite field Fp2m ofodd characteristic, Finite Fields Appl. 34 (2015) 20-35.

Known Explicit PPs

Let p be an odd prime1 For positive integers n and k and δ ∈ Fpn ,(

xpk − x + δ

) pn+12

+ x is a PP of Fpn .

2 For positive integer k and δ ∈ F33k with Tr33k/3k (δ) = 0,(x3

k − x + δ) 33k−1

+ x is a PP of F33k .

3 For positive integers n and k with n|4k and δ ∈ Fpn ,(xp

k − x + δ) pn−1

2 +p2k

± (xpk

+ x) is a PP of Fpn .

4 For a positive integer m and for any δ ∈ F32m ,(x3

m − x + δ)2·3m−1

+ x is a PP of F32m .5 For a positive integer m and δ ∈ F32m , if (Tr32m/3m(δ))2 + 1 = 0

or a square in F3m ,(x3

m − x + δ)3m+2

+ x is a PP of F32m .

New PPs

Theorem

Let n = (t − 1)k , where k is a positive integer, t is an odd integer,gcd(3, t) = 1 and δ ∈ F∗3n .

f (x) = (x3( t−1

− x + δ)s + x and

g(x) = (x3( t−1

− x + δ)s + x3( t−1

+ xare permutation polynomials over F3n with s = 3n−1

t + 1 andTr(δ) = 0.

New PPs

Theorem

Let n = 4k , where k is a positive integer and δ ∈ F∗3n .f (x) = (x3

2k − x + δ)s + x is a permutation polynomial over F3n

for the following cases:

Let k be a positive integer, w be the generator of F3n ,

s = 3(3n−15 ) + 1 and ` = 2

⌊3n/2−1

⌋+ 3n/4.

Then for δ = w ` (mod 2`) ∈ F3n and Tr(δ) = 0, f (x) is a PPover F3n .

Let k = 1 and s = (3n−15 ) + 1. For each δ ∈ F∗3n with

Tr(δ) = 0, then f (x) is a PP over F3n .

Let k = 1 and s = 2(3n−15 ) + 1. Then for each δ ∈ F∗3n f (x) is

a PP over F3n .For this case f (x) + x and f (x) + x3

kare also a permutation

polynomial over F3n .

New PPs

Theorem

Let n = 4k , where k is a positive integer and δ ∈ F∗7n .Let w be the generator of F7n , s = i(7

n−15 ) + 1, where i ∈ {1, 2, 3}

and ` = 2⌊7n/2−1

⌋+ 7n/4.

Then for δ = w ` (mod 2`) ∈ F7n and Tr(δ) = 0,

f (x) = (x72k − x + δ)s + x is a permutation polynomial over F7n .

THANKS ...

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