here - Sites personnels de TELECOM ParisTech - Télécom ParisTech

118 Chapter 4. Efficient characterizations for bentnessso that the criterion is equivalent to#H 3 a − #G 3 a = 3 .We could also have used Condition 2b and that its left hand side satisfies⎛⎞∑χ (g a ◦ D 3(x)) = 1 ⎝ ∑∑χ (g a ◦ D 3(x)) − χ ( (Tr ) m 1 x−1+ g a ◦ D ) 3(x) ⎠2x∈F ∗ 2 m x∈F ∗ 2 mx∈F ∗ 2 m ,Trm 1 (x −1 )=1= 1 (( ) ( −2 m − 1 + #G 3 a − −2 m + #Ha))32= 1 ( #G3a − #Ha 3 − 1 ) ;2and decuce the same reformulation.If b = 1, using the previous calculations, the first term in Condition 3 of Theorem 4.1.2 satisfies∑2χ (g a ◦ D 3 (x)) = #G 3 a − #Ha 3 − 1 ;x∈F ∗ 2 m ,Tr m 1 (x−1 )=1and the second term satisfies∑3whence the reformulation.x∈F ∗ 2 m ,Tr m 1 (x−1 )=1χ (g a (x)) = 3 2 (#G a − #H a − 1) ;Here all the curves are also Artin–Schreier curves. So, for a fixed subset of indices R, we alsoget a test in polynomial time and space in m. However, the complexity of the point countingalgorithms also depends on the genera of the curves, and so on the degrees of the polynomialsdefining them. Denoting by r max the maximal index as above, the genus of H 3 a (respectivelyG 3 a) is (3r max + 1)/2 (respectively (3r max − 1)/2), so approximately three times that of H a(respectively G a ). Therefore, the associated test will be much slower than for Boolean functionsof the family of Charpin and Gong for a given subset R: we have to compute the cardinalities oftwo curves of genera (3r max + 1)/2 and (3r max − 1)/2 if b is primitive, or four curves of genera(3r max + 1)/2, (3r max − 1)/2, (r max + 1)/2 and (r max − 1)/2 if b = 1, instead of two curves ofgenera (r max + 1)/2 and (r max − 1)/2. Hence, we propose another reformulation of the Mesnagercriterion involving slightly less computations.Theorem 4.2.8 (Reformulation of the second Mesnager criterion). The notation is as in Theorem4.2.7. If b is a primitive element of F 4 , then f a,b is hyper-bent if and only if#G 3 a − 1 2 (#G a + #H a ) = − 3 2 .If b = 1, then f a,1 is hyper-bent if and only if2#G 3 a − 5 2 #G a + 1 2 #H a = 3 2 .Proof. We use the fact that m is odd, so that the function x ↦→ D 3 (x) = x 3 + x is a permutationof the set { x ∈ F ∗ 2 | ( )m Trm 1 x−1= 0 } (see the papers of Berlekamp, Rumsey and Solomon [13,Theorem 2] and Charpin, Helleseth and Zinoviev [47] for the case of D 3 , or more generally thearticle of Dillon and Dobbertin [71]), and similar arguments as previously.