Duality of bent functions in odd characteristic Alexander Pott - PowerPoint PPT Presentation

Duality of bent functions in odd characteristic Alexander Pott based on work with Ay¸ ca C ¸e¸ smelio˘ glu and Wilfried Meidl July 2017 1 / 31

Outline ◮ p -ary bent functions. ◮ Duality of p -ary regular bent functions. ◮ Duality of p -ary bent function in the non-regular case. ◮ Vectorial dual. ◮ Bent functions and difference sets. 2 / 31

p -ary bent functions Definition A function f : F n p → F p is p -ary bent if � � � � � 2 = p n ζ � a , x � + f ( x ) p x ∈ F n p for all a ∈ F n p . As usual, ζ p = e 2 π i / p is a complex p -th root of unity, and � a , x � is a non-degenerate bilinear form. If p = 2, these are the classical bent functions. 3 / 31

Quadratic examples if p is odd f ( x ) = x T A x where A ∈ F ( n , n ) is a non-singular symmetric matrix. p Without loss of generality f ( x 1 , . . . , x n ) = x 2 1 + x 2 2 + · · · + d · x 2 n where d � = 0. Note: n can be odd. 4 / 31

Proof One can proof this directly, or use Observation f is bent if and only if x �→ f ( x + a ) − f ( x ) is balanced for all a ∈ F n p , a � = 0 : ( x + a ) T A ( x + a ) − x T A x = 2 x T A a + a T A a = b has precisely p n − 1 solutions for all b and all a � = 0 . 5 / 31

p odd: The number theory � ζ � a , x � + f ( x ) � Walsh coefficients f ( a ) = ∈ Z [ ζ p ] p x ∈ F n p If p is even, there are 2 possibilities. If p is odd, there are 2 p possibilities ( Helleseth, Kholosha (2006) ): ◮ p n ≡ 1 mod 4: Walsh coefficients ± ζ j p p n / 2 ◮ p n ≡ 3 mod 4: Walsh coefficients ± i ζ j p p n / 2 Definition ( Kumar, Scholtz, Welch (1985)) f ( a ) = ωζ f ∗ ( a ) p n / 2 for all a for some fixed element ω : Regular : � p Then f ∗ is called the dual of f . Theorem ( Kumar, Scholtz, Welch (1985)) f ∗ is bent if f is regular. 6 / 31

Maiorana-McFarland construction F 2 m → F p p Theorem f ( x , y ) = Trace ( x · π ( y )) + ρ ( y ) is bent if π : F m p → F m p is a permutation and ρ : F m p → F p . Alternative : f ( x , y ) = f y ( x ) + ρ ( y ) where f y : F m p → F p is linear and f y � = f y ′ if y � = y ′ . Alternative : f ( x , y ) = g y ( x ) where g y are affine and the supports of the Walsh spectra are disjoint. 7 / 31

Generalized Maiorana-McFarland construction p → F p be a collection of p s functions such that the Let g y : F m supports of the Walsh spectra are disjoint. If the functions are p s -plateuead, (that is Walsh spectrum takes values 0 and ± p ( m + s ) / 2 ) then f : F m p × F s p → F p with f ( x , y ) = g y ( x ) is p -ary bent. Example If f bent on F k p , then the mappings g y ( x ′ , x ′′ ) = f ( x ′ ) + � y , x ′′ � do the job ( x ′ ∈ F k p , x ′′ , y ∈ F s p ). 8 / 31

Comments Observation The Maiorana-McFarland construction as well as all quadratic examples give regular bent functions. ¸melio˘ Theorem ( Ces glu, McGuire, Meidl (2012)) The (non-regular) generalized Maiorana-McFarland construction gives regular and non-regular bent functions (s � = m). See also Davis, Jedwab (1997) . 9 / 31

Non-regular bent functions and their duals Walsh spectrum has 2 p values ± ωζ i p p n / 2 . We can define a dual even if the spectrum takes 2 p values!! Theorem The duals of the generalized Maiorana-McFarland construction are bent if the g y are regular. 10 / 31

Two more families p = 3 Theorem ( Helleseth, Kholosha (2006), Helleseth, Hollmann, Kholosha, Wang, Xiang (2009)) The following two families of bent functions are regular: ◮ n = 2 k, k odd, α element of order 4(3 k − 1) : � +3 k +1 � 3 n − 1 Trace α x 4 ◮ Coulter, Matthews (1997) � 2 � 3 i +1 Trace ( α x , gcd( i , n ) = 1 , 3 ≤ i ≤ n odd Heavy proofs! 11 / 31

Questions What is special about these functions that is so difficult to prove regularity. Theorem The two families are not in the generalised Maiorana-McFarland family. Note: The Coulter-Matthews example is actually a component of a planar function! 12 / 31

Dual of diagonal quadratic function Theorem If f ( x ) = d 1 x 2 1 + d 2 x 2 2 + . . . + d n x 2 n then the dual is f ∗ ( x ) = − x 2 − x 2 − . . . − x 2 1 2 n 4 d 1 4 d 2 4 d n 13 / 31

Dual of Maiorana-McFarland Theorem Let f ( x , y ) = Trace ( x · π ( y )) then f ∗ ( x , y ) = Trace ( − y · π − 1 ( x )) + ρ ( π − 1 ( y )) where x , y ∈ F p m , π permutation, ρ arbitrary. 14 / 31

The role of the inner product Note that the dual function depends on the choice of the bilinear form. In the Maiorana-McFarland case, we choose Trace( x · x ′ + y · y ′ ) on F p m × F p m . Having a dual does not depend on the choice of the inner product, but self-duality does. 15 / 31

Spreads V = F p m × F p m , U 0 , U 1 , . . . , U p m subspaces of dimension m with pairwise trivial intersection. Let i → γ i be a balanced function from { 1 , 2 , . . . , p m } to { 1 , . . . , p } . Then  1 ≤ i ≤ p m  γ i if z ∈ U i , z � = 0 , f ( z ) =  γ 0 if z ∈ U 0 . is bent. Theorem  z ∈ U ⊥ 1 ≤ i ≤ p m  i , z � = 0 , γ i f ∗ ( z ) =  if z ∈ U ⊥ γ 0 0 . 16 / 31

Self-duality Theorem There are no regular self-dual bent functions if p n ≡ 3 mod 4 . Theorem There are quadratic self-dual bent functions with respect to the classical inner product if there is a symmetrix matrix A such that A 2 = − 1 4 I . Easy if p ≡ 1 mod 4 . We can use these recursively to find self-dual bent functions of large degree. 17 / 31

More examples Theorem Let f be a partial spread bent function for a symplectic spread with respect to an alternating bilinear form � , � . Then f is self-dual with respect to that bilinear form. Theorem ( Helleseth, Kholosha (2006)) Let n = 2 k and t be a positive integer satisfying gcd( t , 3 k + 1) = 1 . Then the function f ( x ) = Trace ( ax t (3 k − 1) ) from F 3 n to F 3 is bent if and only if K k ( a 3 k +1 ) = − 1 . If K k ( a 3 k +1 ) = − 1 then f ( x ) is f ( b ) = 3 k ζ − Trace ( a 3 k b t (3 k − 1) ) . regular bent and � Observation For k = 3 , 5 , 7 there exist self-dual bent functions of that type. with respect to the trace bilinear form. 18 / 31

Further non-regular examples The following functions are not regular: ◮ g 1 : F 3 6 → F 3 with g 1 ( x ) = Trace( ξ 7 x 98 ), where ξ is a primitive element of F 3 6 . ◮ g 2 : F 3 4 → F 3 with g 2 ( x ) = Trace( α 0 x 22 + x 4 ), where α 0 ∈ {± ξ 10 , ± ξ 30 } and ξ is a primitive element of F 3 4 . ◮ g 3 : F 3 3 → F 3 with g 3 ( x ) = Trace( x 22 + x 8 ). ◮ g 4 , g 5 : F 3 6 → F 3 with g 4 ( x ) = Trace( ξ x 20 + ξ 41 x 92 ) , g 5 ( x ) = Trace( ξ 7 x 14 + ξ 35 x 70 ), where ξ is a primitive element of F 3 6 . Helleseth, Kholosha (2006, 2010), Tan, Yang, Zhang (2010). Observation g 3 and g 4 have a bent dual, the others not. g 3 is generalized McFarland, g 1 not. 19 / 31

Infinite family of bent functions without a dual Let 1 , α, β ∈ F p m be linearly independent over F p . If � η (1 + y 1 α + y 2 β ) ǫ − y 1 y 2 | | � = p , p y 1 , y 2 ∈ F p then the function F : F p m × F 2 p → F p F ( x , y 1 , y 2 ) = Trace( x 2 ) + ( y 1 + Trace( α x 2 ))( y 2 + Trace( β x 2 )) is a bent function without a dual. 20 / 31

Recursive Construction Theorem If g is bent and h is bent without a dual, then f ( x , y ) = g ( x ) + h ( y ) has no dual (g : F m p → F p , h : F n p → F p ). We have a second recursive construction. Question Find more functions for which the dual is not bent. 21 / 31

Vectorial bent functions Most constructions of bent functions are “vectorial” constructions. Definition A function F : F n p → F m p is vectorial bent if all component functions x �→ � a , F ( x ) � are bent. If m = n : Planar functions. Example F ( x , y ) = x · y as a mapping F p m × F p m → F p m . Other example: Coulter-Matthews. Question Is there a duality concept for such vectorial functions? Problem Even if the sum of two bent functions is bent, this is not necessarily true for the duals. 22 / 31

Vectorial dual Definition A vectorial bent function has a dual if the set of dual functions form a vector space of bent functions. Example F ( x ) = x 2 on F p m . The dual of Trace( ax 2 ) is Trace( − x 2 4 a ). Hence if we look at the function F p m → F p m , the function has a dual, but there are sub-vectorial functions F p m → F 2 p without a dual: The set 1 { λ a + µ b : λ, µ ∈ F p } do not form a vector space. In general, the duals of planar functions are not planar. 23 / 31

Example of a vectorial dual bent function Theorem The vectorial versions F : F 2 m → F m p of the spread functions have p a vectorial dual which is the same as the original function with respect to the alternating bilinear form (self-dual). Question Are there examples besides x 2 of planar functions whose dual functions also form a planar function? 24 / 31

p = 2: Difference sets Observation f : F n 2 → F 2 is bent if and only if D f := { x ∈ F n 2 : f ( x ) = 1 } is a (non-trivial) difference set in F n 2 : A subset D ⊂ G of a group G is a difference set if any non-zero element g ∈ G has a constant number of representations g = d − d ′ with d , d ′ ∈ D . Non-trivial: 2 ≤ | D | ≤ | G | − 2. 25 / 31