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4.2. Reformulation in terms of cardinalities of curves 1171. it ensures that the smooth projective models of the curves H a and G a are imaginaryhyperelliptic curves and such curves are way easier to manipulate than real hyperellipticcurves;2. for efficiency reasons r max should be as small as possible, so the natural choice for the theindices in a cyclotomic coset will be the coset leaders which are odd integers.In fact, the curves H a and G a are even Artin–Schreier curves. As was the case for ellipticcurves, Theorem 3.2.18 states that there exist efficient algorithms to compute the cardinalities ofsuch curves. Thus, Lisoněk obtained an efficient test for hyper-bentness of Boolean functionsin the class described by Charpin and Gong. The polynomial defining H a (respectively G a ) isindeed of degree r max + 2 (respectively r max ), so the curve is of genus (r max + 1)/2 (respectively(r max − 1)/2). The complexity for testing a Boolean function in this family is then dominated bythe computation of the cardinality of a curve of genus (r max + 1)/2, which is polynomial in m fora fixed r max (and so fixed genera for the curves H a and G a ).We now show that a similar reformulation can be applied to the different versions of thesecond criterion of Mesnager for Boolean functions with multiple trace terms.Theorem 4.2.7 (Reformulation of the second Mesnager criterion). The notation is as in Theorem4.1.2. Moreover, let H a and G a be the (affine) curves defined over F 2 m byG a : y 2 + y = ∑ r∈Ra r D r (x) ,H a : y 2 + xy = x + x 2 ∑ r∈Ra r D r (x) ;and let H 3 a and G 3 a be the (affine) curves defined over F 2 mbyG 3 a : y 2 + y = ∑ r∈Ra r D r (D 3 (x)) ,H 3 a : y 2 + xy = x + x 2 ∑ r∈Ra r D r (D 3 (x)) .If b is a primitive element of F 4 , then f a,b is hyper-bent if and only if#H 3 a − #G 3 a = 3 .If b = 1, then f a,1 is hyper-bent if and only if(#G3a − #H 3 a)−32 (#G a − #H a ) = 3 2 .Proof. If b is a primitive element of F 4 , according to Proposition 4.2.5 the left hand side ofCondition 2c in Theorem 4.1.2 satisfies∑χ ( Tr m (1 x−1 ) + g a (D 3 (x)) ) = −2 m + #Ha 3 ,x∈F ∗ 2 mand according to Proposition 4.2.4 the right hand side of Condition 2c in Theorem 4.1.2 satisfies2 m − 2 w H (g a ◦ D 3 ) + 3 = −2 m + 3 + #G 3 a ,