Semifields, Relative Difference Sets, and Bent Functions Alexander - PowerPoint PPT Presentation

Semifields, Relative Difference Sets, and Bent Functions Alexander Pott Otto-von-Guericke-University Magdeburg December 09, 2013 1 / 34

Outline, or: 2 / 34

Outline, or: Why I am nervous 2 / 34

Outline, or: Why I am nervous ◮ bent functions ... C LAUDE C ARLET , T OR H ELLESETH 2 / 34

Outline, or: Why I am nervous ◮ bent functions ... C LAUDE C ARLET , T OR H ELLESETH ◮ relative difference sets (uninteresting generalization?) 2 / 34

Outline, or: Why I am nervous ◮ bent functions ... C LAUDE C ARLET , T OR H ELLESETH ◮ relative difference sets (uninteresting generalization?) ◮ Z 4 (very old?) 2 / 34

Describe connection between ... ◮ relative difference sets ◮ semifields ◮ projections of relative difference sets ◮ K NUTH operation on semifields ◮ bent functions ◮ Z 4 -valued bent functions 3 / 34

The team ◮ Y UE Z HOU ◮ K AI -U WE S CHMIDT ◮ A LEX . P. 4 / 34

The team ◮ Y UE Z HOU ◮ K AI -U WE S CHMIDT ◮ A LEX . P. ... all this is also related to K ATHY H ORADAM ’s work, but less general and more concrete... 4 / 34

Bent functions, even characteristic ◮ Bent functions f : F 2 n → F 2 such that f ( x + a ) − f ( x ) is balanced for all a � = 0. Example f ( x ) = Trace ( β x 3 ) on F 2 n : Trace ( β ( x 2 a + a 2 x + a 2 )) f ( x + a ) − f ( x ) = Trace ( x 2 β ( a + β a 4 ) + β a 2 ) = hence 1 + a 3 β � = 0 for all a � = 0. Necessary condition n = 2 m is even. 5 / 34

Bent functions, odd characteristic ◮ Bent functions f : F p n → F p such that f ( x + a ) − f ( x ) is balanced for all a � = 0. Example f ( x ) = Trace ( β x 2 ) : f ( x + a ) − f ( x ) = Trace ( 2 x β a ) + β a 2 . Any n . 6 / 34

Vectorial versions, even characteristic ◮ Bent functions F : F n 2 → F k 2 such that F ( x + a ) − F ( x ) is balanced for all a � = 0. Component functions Trace ( β F ( x )) are bent. Hence: Vector space of bent functions. Necessary condition n = 2 m is even and k ≤ m . Example F ( x , y ) = xy ( x , y ∈ F 2 m ) is vectorial bent F 2 m × F 2 m → F 2 m 7 / 34

Vectorial versions, odd characteristic ◮ Bent functions F : F n p → F k p such that F ( x + a ) − F ( x ) is balanced for all a � = 0. Component functions Trace ( β F ( x )) are bent. Hence: Vector space of bent functions. Necessary condition k ≤ n . Example F ( x ) = x 2 : F ( x + a ) − F ( x ) = 2 xa + a 2 . 8 / 34

F : F n p → F k p bent p = 2 p odd n even and k ≤ n k ≤ n 2 k = n : planar functions 9 / 34

Bent functions and relative difference sets If F : F n p → F k p is bent, the set G F := { ( x , F ( x )) : x ∈ F n p } ⊆ F n p × F k p is a relative difference set: 10 / 34

Bent functions and relative difference sets If F : F n p → F k p is bent, the set G F := { ( x , F ( x )) : x ∈ F n p } ⊆ F n p × F k p is a relative difference set: Every element outside { 0 } × F k p has the same number of difference representations g = d − d ′ with d , d ′ ∈ G F : x − y = a , F ( x ) − F ( y ) = b is equivalent to F ( y + a ) − F ( y ) = b 10 / 34

Other groups? ◮ group G ◮ subgroup N (forbidden subgroup) ◮ subset D g ∈ G \ N has constant number of representations g = d − d ′ with d , d ′ ∈ D , no element in N . Example D = { 1 , 2 , 4 } ⊆ Z 8 , forbidden subgroup { 0 , 4 } . In this talk: | D | = | G | | N | , hence from each coset of N exactly one element. 11 / 34

The projection construction If U < N is a normal subgroup of G and D relative difference set, then D / U is a relative difference set in G / U with forbidden subgroup N / U . The size is | D / U | = | D | . One relative difference set produces a chain of relative difference sets. 12 / 34

Planar functions: n = k Definition A function F : F p n → F p n is planar if F ( x + a ) − F ( x ) is a permutation for all a � = 0. We obtain (vectorial) bent functions via projection. p must be odd. 13 / 34

Planar functions: n = k Definition A function F : F p n → F p n is planar if F ( x + a ) − F ( x ) is a permutation for all a � = 0. We obtain (vectorial) bent functions via projection. p must be odd. If p = 2, then generalize to ?? 13 / 34

Two generalizations to characteristic 2 ◮ almost perfect nonlinear functions (APN): F ( x + a ) − F ( x ) = b has at most 2 solutions. ◮ relative difference sets in other groups (not elementary abelian), related to semifields. 14 / 34

Two generalizations to characteristic 2 ◮ almost perfect nonlinear functions (APN): F ( x + a ) − F ( x ) = b has at most 2 solutions. ◮ relative difference sets in other groups (not elementary abelian), related to semifields. Dream : Use the many semifields with p = 2 to construct many APN. 14 / 34

Semifields ( S , + , ⊙ ) is a finite (pre)semifield ( field without associativity ) if ◮ ( S , +) is a finite abelian group. ◮ x ⊙ a = b has a unique solution x if a � = 0. ◮ a ⊙ y = b has a unique solution y if a � = 0. ◮ x ⊙ ( y + z ) = x ⊙ y + x ⊙ z and ( x + y ) ⊙ z = x ⊙ z + y ⊙ z . Example: Finite field! 15 / 34

Semifields ( S , + , ⊙ ) is a finite (pre)semifield ( field without associativity ) if ◮ ( S , +) is a finite abelian group. ◮ x ⊙ a = b has a unique solution x if a � = 0. ◮ a ⊙ y = b has a unique solution y if a � = 0. ◮ x ⊙ ( y + z ) = x ⊙ y + x ⊙ z and ( x + y ) ⊙ z = x ⊙ z + y ⊙ z . Example: Finite field! ◮ T a : S → S with T a ( x ) := x ⊙ a is an isomorphism. ◮ T a + T a ′ = T a + a ′ . Vector space of invertible linear mappings. S is elementary abelian (additive group of a field F p n ), multiplication not always commutative. 15 / 34

Why? Construct projective plane from a semifield: 16 / 34

Why? Construct projective plane from a semifield: ◮ Points: S × S ◮ Lines: { x , m ⊙ x + y : x ∈ S } . 16 / 34

How many? ◮ p odd: quite a few, but not many. L AVRAUW , P OLVERINO (2012). ◮ p = 2: very many commutative K ANTOR (2003). Question Number is not bounded by a polynomial. 17 / 34

Semifields and relative difference sets Any semifield gives rise to a relative difference set in a group of order p 2 n with forbidden subgroup of order p n : 18 / 34

Semifields and relative difference sets Any semifield gives rise to a relative difference set in a group of order p 2 n with forbidden subgroup of order p n : Consider the set S × S with addition ⊕ ( a , b ) ⊕ ( a ′ , b ′ ) = ( a + a ′ , b + b ′ + a ⊙ a ′ ) . 18 / 34

Semifields and relative difference sets Any semifield gives rise to a relative difference set in a group of order p 2 n with forbidden subgroup of order p n : Consider the set S × S with addition ⊕ ( a , b ) ⊕ ( a ′ , b ′ ) = ( a + a ′ , b + b ′ + a ⊙ a ′ ) . Difference set: { ( a , 0 ) : a ∈ S } 18 / 34

What are these strange groups? Let S be commutative: ◮ F 2 n p if p is odd, ◮ Z n 4 if p = 2. Forbidden subgroup: 2 Z n 4 . In the p odd case: Planar function. 19 / 34

The K NUTH cube Basis e i of S � e i ⊙ e j = a i , j , k e k k with a i , j , k ∈ F p . Linear mappings are described by matrices ( a i , k ) . Permuting the indices gives six semifields (K NUTH ). 20 / 34

The K NUTH cube Basis e i of S � e i ⊙ e j = a i , j , k e k k with a i , j , k ∈ F p . Linear mappings are described by matrices ( a i , k ) . Permuting the indices gives six semifields (K NUTH ). If S is commutative, the linear mappings associated with one of the 6 semifields are symmetric: Vector space of symmetric invertible matrices. 20 / 34

What are the projections of RDS, p odd? One semifield in Knuth orbit of a commutative semifield is vector space of symmetric invertible matrices, another one can be described by planar function (RDS). The invertible matrices associated with F x �→ F ( x + a ) − F ( x ) − F ( a ) + F ( 0 ) . are not symmetric. 21 / 34

What are the projections of RDS, p odd? One semifield in Knuth orbit of a commutative semifield is vector space of symmetric invertible matrices, another one can be described by planar function (RDS). The invertible matrices associated with F x �→ F ( x + a ) − F ( x ) − F ( a ) + F ( 0 ) . are not symmetric. p odd: Symmetric invertible matrix ( a i , j ) gives a bent function � f ( x 1 , . . . , x n ) = a i , j x i x j . i , j These are the projections of the planar function F ! 21 / 34

What are the projections of RDS, p odd? One semifield in Knuth orbit of a commutative semifield is vector space of symmetric invertible matrices, another one can be described by planar function (RDS). The invertible matrices associated with F x �→ F ( x + a ) − F ( x ) − F ( a ) + F ( 0 ) . are not symmetric. p odd: Symmetric invertible matrix ( a i , j ) gives a bent function � f ( x 1 , . . . , x n ) = a i , j x i x j . i , j These are the projections of the planar function F ! Nice representation of K NUTH operation in terms of relative difference sets. 21 / 34

p = 2: Difference set in Z n 4 K NUTH gives invertible symmetric matrices ( a i , j ) . Projections are relative difference set in the group Z 4 × Z n − 1 . 2 22 / 34