Research article Special Issues

On a class of bent, near-bent, and 2-plateaued functions over finite fields of odd characteristic

  • The main purpose of this paper is to study a class of the p-ary functions fλ,u,v(x)=Trk1(λxpk+1)+Trn1(ux)Trn1(vx) for any odd prime p and n=2k,λGF(pk),u,vGF(pn). With the help of Fourier transforms, we are able to subdivide the class of all fλ,u,v into sublcasses of bent, near-bent and 2-plateaued functions. It is shown that the choice of λ,u and v, ensuring that f is bent, 2-plateaued or near-bent, is directly related to finding the subset AGF(p)3. The efficient method for defining the set AGF(p)3 is described in detail.

    Citation: Samed Bajrić. On a class of bent, near-bent, and 2-plateaued functions over finite fields of odd characteristic[J]. AIMS Mathematics, 2022, 7(2): 1971-1981. doi: 10.3934/math.2022113

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  • The main purpose of this paper is to study a class of the p-ary functions fλ,u,v(x)=Trk1(λxpk+1)+Trn1(ux)Trn1(vx) for any odd prime p and n=2k,λGF(pk),u,vGF(pn). With the help of Fourier transforms, we are able to subdivide the class of all fλ,u,v into sublcasses of bent, near-bent and 2-plateaued functions. It is shown that the choice of λ,u and v, ensuring that f is bent, 2-plateaued or near-bent, is directly related to finding the subset AGF(p)3. The efficient method for defining the set AGF(p)3 is described in detail.



    Bent functions are extreme combinatorial objects with several areas of application, such as coding theory, maximum length sequences, cryptography, and the theory of difference sets to name a few. Boolean bent functions were introduced by Rothaus [6], who also considered two classes of bent functions. Among other equivalent characterizations of bent functions, the one that is most often used is a characterization of bent functions as a class of Boolean functions having so-called flat Walsh spectrum. It means that for any bent function over GF(2)n, its Hamming distance to any affine function in n variables is constant, including the distance to the all-zero function (or all-one function). The bent property of f is commonly specified in terms of its flat Walsh spectrum, that is requiring that Wf(λ)=±2n/2 for any λGF(2n), where,

    Wf(λ)=xGF(2n)(1)f(x)+Trn1(λx). (1.1)

    Here, Trn1:GF(2n)GF(2) denotes the absolute trace function defined by Trn1(x)=x+x2++x2n1.

    On the other hand, a generalization to finite fields of odd characteristic was first suggested in [3] and a function f:GF(pn)GF(p) is bent if and only if its Fourier transform defined by

    ˆf(a)=xGF(pn)ωf(x)Trn1(ax), (1.2)

    is flat, that is, |ˆf(a)|=pn/2 for any aGF(pn). Here, ω=e2πıp denotes the primitive root of unity where ı=1. Some surveys of known results on p-ary bent functions can be found in [1,3,5] and the references therein.

    The classes of plateaued functions introduced by Zheng and Zhang [10] are good candidates for designing cryptographic functions since they possess various desirable cryptographic characteristics. They can be balanced for both odd and even number of variables, and they do not possess nonzero linear structures. In particular, the functions whose Fourier spectrum belong to {0,±pn+12} are known as near-bent functions and, they play a significant role in certain cryptographic primitives.

    In 2016 and 2017, G. Xu et. al. [8,9] characterised the Fourier transform of the p-ary functions of the form

    Trk1(λxpk+1)+Trn1(ux)Trn1(vx), (1.3)

    for the special case p=3 and n=2k,λGF(pk),u,vGF(pn). The authors proved that some of these p-ary functions are also bent under certain conditions. Furthermore, a construction of quadratic ternary bent, near-bent and 2-plateaued functions from some known ternary bent functions are presented.

    In this work we generalise the method for computing Fourier transform originally given in [8] for multinomial trace functions f=fλ,u,v:GF(pn)GF(p) of the form Trk1(λxpk+1)+Trn1(ux)Trn1(vx) for any odd prime p, where n=2k,λGF(pk),u,vGF(pn). It is shown that the choice of λ,u and v, ensuring that f is bent, 2-plateaued or near-bent is directly related to finding the subset AGF(p)3 (cf. Section 4). Moreover, we give an efficient method for defining the set A that corresponds for any odd prime p. This is not the case in the previous works, where the set A is only explicitly defined just for the case p=3.

    The rest of this article is organized as follows. In Section 2 some basic definitions and notions are given. The analysis of the Fourier transform of the function f is presented in Section 3. Some sufficient conditions for a given function to be bent, 2-plateaued and near-bent are derived in Section 4. Some concluding remarks are found in Section 5.

    In the sequel, let Fpn denote the finite Galois field GF(pn) consisting of pn elements. The group of units of Fpn, denoted by Fpn, is a cyclic group consisting of pn1 elements. An element αFpn is said to be a primitive element if it is a generator of the multiplicative group Fpn. Fpn is an n-dimensional vector space over Fp. After fixing an Fp-basis (γ0,,γn1) of Fpn, the mapping

    Fnp(α0,,αn1)α0γ0++αn1γn1Fpn

    defines an isomorphism of vector spaces over Fp.

    A bent function f(x) is called regular if for every aFpn the normalized Fourier coefficient pn2ˆf(a) equals to complex p-th root of unity, that is, pn2ˆf(a)=ωf(a) where the function f(a) is called the dual of f(x). A binary bent function is always regular. For odd p, a p-ary bent function f(x) may not be regular, but its Fourier transform coefficients [3] satisfy

    ˆf(a)={±pn2ωf(a),ifp1(mod4)±ıpn2ωf(a),ifp3(mod4)and n is odd. (2.1)

    Such a function is called weakly regular, if for all aFpn, ı is fixed.

    If for all aF(pn),ˆf(a){0,±pn+s2}, where 0sn, then a p-ary function f(x) is called s-plateaued. In the case of s=1,f is called near-bent.

    The trace function Trnk:FpnFpk, a mapping to the subfield Fpk, where kn, is defined as

    Trnk(x)=x+xpk+xp2k++xp(n/k1)k, for all xFpn. (2.2)

    Let fλ,u,v(x)=Trk1(λxpk+1)+Trn1(ux)Trn1(vx), where n=2k,λFpk,u,vFpn. In this section we compute the Fourier transform of the function f(x) for any odd prime p. This result will help us to divide the class of functions fλ,u,v into subclasses of bent, near-bent and 2-plateaued functions.

    Before we prove the main theorem we need the following preparatory result.

    Lemma 1. [2,4] Let n=2k and λFpk. For any odd prime p, the p-ary monomial g(x)=Trk1(λxpk+1) is a weakly regular bent function. Moreover, for aFpn the corresponding Fourier transform coefficient of g(x) is equal to

    ˆg(a)=pkωTrk1(λ1apk+1). (3.1)

    The authors in [8] computed the Fourier transform of the function fλ,u,v for p=3. In what follows, we continue the work of [8,9] and present the general form of the Fourier transform of fλ,u,v for any odd prime p.

    Theorem 1. Let n=2k be a positive integer and u,vFpn. Define the function f(x) by

    f(x)=g(x)+Trn1(ux)Trn1(vx), (3.2)

    where g(x)=Trk1(λxpk+1) is a p-ary function defined on Fpn. Then for any aFpn, the corresponding Fourier transform coefficient of f=fλ,u,v is equal to

    ˆf(a)=1p0i,j<pωijˆg(ajv+iu). (3.3)

    Proof. For i,j=0,,p1 and u,vFpn define the sets Ti={xFpnTrn1(ux)=i} and denote

    Si(a)=xTiωg(x)Trn1(ax),Ri(ajv)=xTiωg(x)Trn1((ajv)x).

    The Fourier transform coefficient of f(x) at a is equal to

    ˆf(a)=xFpnωf(x)Trn1(ax)=xFpnωg(x)+Trn1(ux)Trn1(vx)Trn1(ax)=xT0ωg(x)Trn1(ax)+xT1ωg(x)Trn1((av)x)+xT2ωg(x)Trn1((a2v)x)++xTp1ωg(x)Trn1((a(p1)v)x)=S0(a)+R1(av)+R2(a2v)++Rp1(a(p1)v)=S0(a)+p1j=1Rj(ajv)

    Let us first compute S0(a). By definition of Ti we have

    xTiωg(x)Trn1((a+u)x)=xTiωg(x)Trn1(ax)i=ωiSi(a).

    Therefore, for any bFp the Fourier transform of the function g at the point a+bu can be computed as

    ˆg(a+bu)=p1i=0ωbiSi(a). (3.4)

    Since ωC is a primitive p-root of unity, 1+ω+ω2++ωp1=0. Hence, adding p equations of (3.4) gives

    S0(a)=1pp1i=0ˆg(a+iu). (3.5)

    Next we calculate Rj(ajv). Similar to the above, using the definition of Ti, we have

    ˆg(ajv+bu)=p1i=0ωbiRi(ajv)

    The last expression can be written in the matrix form as

    [ˆg(ajv)ˆg(ajv+u)ˆg(ajv+2u)ˆg(ajv+(p1)u)]=[111111ωp1ωp2ω2ω1ωp2ω2(p2)ω(p2)(p2)ω(p1)(p2)1ωω2ωp2ωp1][R0(ajv)R1(ajv)R2(ajv)Rp1(ajv)].

    where the coefficient matrix represents Discrete Fourier Transform (DFT) matrix. It is well-known that the inverse DFT matrix is

    1p[111111ωω2ωp2ωp11ωp2ω2(p2)ω(p2)(p2)ω(p1)(p2)1ωp1ωp2ω2ω].

    Therefore, we have

    Rj(ajv)=1pp1i=0ωijˆg(ajv+iu). (3.6)

    Then the desired conclusion follows from (3.5) and (3.6).

    In what follows, we give explicit formulas for computing S0(a) and Rj(ajv) given by (3.5) and (3.6), respectively.

    Let g(x)=Trk1(λxpk+1). According to (3.1) and using the fact Trn1(x)=Trk1(Trnk(x)), we have

    ˆg(a+iu)=pkωTrk1(λ1apk+1)iTrn1(λ1apku)ipk+1Trk1(λ1upk+1),i=0,,p1.

    Denote c1=Trn1(λ1apku) and t1=Trk1(λ1upk+1). By (3.5) it follows

    S0(a)=1ppkωTrk1(λ1apk+1)[1+ωc1t1+ω2c12pk+1t1++ω(p1)c1(p1)pk+1t1].

    Since ω is a primitive p-th root of unity it holds ωpk=1, i.e., ωpk+1=ω. Therefore

    S0(a)=1ppkωTrk1(λ1apk+1)[1+ωc1t1+ω2c122t1++ω2c122t1+ωc1t1]. (3.7)

    On the other side,

    ˆg(ajv+iu)=pkωTrk1(λ1(av)pk+1)ijTrn1(λ1apku)+ijTrn1(λ1vpku)(ij)pk+1Trk1(λ1upk+1).

    Denote c2=Trn1(λ1apkv),t2=Trk1(λ1vpk+1) and t0=Trn1(λ1vpku). By (3.6) it follows

    Rj(ajv)=1ppkωTrk1(λ1apk+1)[ωjc2jjt2+ωj+jc2jjt2c1+jt0t1+ω2j+jc2jjt22c1+2jt022t1+ω(p1)j+jc2jjt2(p1)c1+j(p1)t0(p1)(p1)t1]. (3.8)

    In the next section we apply the above computation to construct some (near)-bent and 2-plateaued functions in the form of (3.2).

    To make it easier construction of bent, near-bent and 2-plateaued functions, we are going to construct a subset A of F3p,p>3, which will help us to separate bent from non-bent functions. Its construction is described in the following way:

    A={(0,1,14modp),(0,2,a02),,(0,p1,a0(p1)),(1,1,1),(1,2,a12),(1,3,a13),,(1,p1,a1(p1)),(2,1,a21),(2,2,a22),(2,3,a23),,(2,p1,a2(p1)),(3,1,4),(3,2,a32),(3,3,a33),,(3,p1,a3(p1)),(p32,1,a(p32)1),(p32,2,a(p32)2),,(p32,p1,a(p32)(p1)),(p12,1,a(p32)1),(p12,2,a(p32)2),,(p12,p1,a(p32)(p1)),(p3,1,1),(p3,2,a12),(p3,3,a13),,(p3,p1,p1),(p2,1,14modp),(p2,2,a02),(p2,3,a13),,(p2,p1,a0(p1)),(p1,0,0),,(p1,0,p1),(p1,1,0),,(p1,p1,0)} (4.1)

    where aij{0,1,,p1} and the boxed triples are fixed for any prime number p. The remaining coefficients aij must satisfy the following conditions:

    1. In each row, the product of the second and third numbers of a triple satisfies

    1ai12ai23ai3(p1)ai(p1)modp

    2. The first p12 rows are symmetric to the next p12 rows in such a way that first and (p2) row, second and (p3), third and (p4), and so on, have the same second and third number of each triple.

    3. In each column, the sum of the third numbers of a triple satisfies

    a0j+a1j+a2j++a(p32)j0modp

    Remark 1. Note that we have p1 rows with p1 triples, and the last row has 2p1 triples, which means that the cardinality of the subset A is

    #A=(p1)(p1)+(2p1)=p2.

    Remark 2. The equation in third condition has more than one solution for any odd prime p>7. In total, there is (p72)!(p72)! possible solutions taking into account the first two conditions.

    Example 1. Let us compute the set A for p=11. At the beginning we have

    A={(0,1,14mod11),(0,2,a02),,(0,9,a09),(0,10,a010),(1,1,1),(1,2,a12),,(1,9,a19),(1,10,a110),(2,1,a21),(2,2,a22),,(2,9,a29),(2,10,a210),(3,1,4),(3,2,a32),,(3,9,a39),(3,10,a310),(4,1,a41),(4,2,a42),,(4,9,a49),(4,10,a410),(5,1,a41),(5,2,a42),,(5,9,a49),(5,10,a410),(6,1,4),(6,2,a32),,(6,9,a39),(6,10,a310),(7,1,a21),(7,2,a22),,(7,9,a29),(7,10,a210),(8,1,1),(8,2,a12),,(8,9,a19),(8,10,a110),(9,1,14modp),(9,2,a02),,(9,9,a09),(9,10,a010),(10,0,0),(10,0,1),,(10,9,0),(10,10,0)}

    With the help of boxed triples and given conditions the remaining coefficients in the rows can be computed. For instance, in the first row, since 143mod11, we are solving the following equations

    132a029a0910a0103mod11

    It is easy to see that a02=7,a03=1,a04=9,a05=5,a06=6,a07=2,a08=10,a09=4,a010=8. After other computations we get

    A={(0,1,3),(0,2,7),(0,3,1),(0,4,9),(0,5,5),(0,6,6),(0,7,2),(0,8,10),(0,9,4),(0,10,8)(1,1,1),(1,2,6),(1,3,4),(1,4,3),(1,5,9),(1,6,2),(1,7,8),(1,8,7),(1,9,5),(1,10,10)(2,1,a21),(2,2,a22),(2,3,a23),(2,4,a24),(2,5,a25),(2,6,a26),(2,7,a27),(2,8,a28),(2,9,a29),(2,10,a210),(3,1,4),(3,2,2),(3,3,5),(3,4,1),(3,5,3),(3,6,8),(3,7,10),(3,8,6),(3,9,9),(3,10,7),(4,1,a41),(4,2,a42),(4,3,a43),(4,4,a44),(4,5,a45),(4,6,a46),(4,7,a47),(4,8,a48),(4,9,a49),(4,10,a410),(5,1,a41),(5,2,a42),(5,3,a43),(5,4,a44),(5,5,a45),(5,6,a46),(5,7,a47),(5,8,a48),(5,9,a49),(5,10,a410),(6,1,4),(6,2,2),(6,3,5),(6,4,1),(6,5,3),(6,6,8),(6,7,10),(6,8,6),(6,9,9),(6,10,7),(7,1,a21),(7,2,a22),(7,3,a23),(7,4,a24),(7,5,a25),(7,6,a26),(7,7,a27),(7,8,a28),(7,9,a29),(7,10,a210),(8,1,1),(8,2,6),(8,3,4),(8,4,3),(8,5,9),(8,6,2),(8,7,8),(8,8,7),(8,9,5),(8,10,10),(9,1,3),(9,2,7),(9,3,1),(9,4,9),(9,5,5),(9,6,6),(9,7,2),(9,8,10),(9,9,4),(9,10,8),(10,0,0),(10,0,1),(10,0,2),(10,0,3),(10,0,4),(10,0,5),(10,0,6),(10,0,7),(10,0,8),(10,0,9),(10,0,10),(10,1,0),(10,2,0),(10,3,0),(10,4,0),(10,5,0),(10,6,0),(10,7,0),(10,8,0),(10,9,0),(10,10,0)}

    With the help of third condition we can calculate the rest of the coefficients. In fact, we only need to solve the following equation

    3+1+a21+4+a410mod11a21+a413mod11

    The possible solutions are a21=5,a41=9, or a21=9,a41=5, or a21=8,a41=6 or a21=6,a41=8. We can choose whatever solution we want. Therefore, for a21=9 and a41=5 we have

    A={(0,1,3),(0,2,7),(0,3,1),(0,4,9),(0,5,5),(0,6,6),(0,7,2),(0,8,10),(0,9,4),(0,10,8)(1,1,1),(1,2,6),(1,3,4),(1,4,3),(1,5,9),(1,6,2),(1,7,8),(1,8,7),(1,9,5),(1,10,10)(2,1,9),(2,2,10),(2,3,3),(2,4,5),(2,5,4),(2,6,7),(2,7,6),(2,8,8),(2,9,1),(2,10,2)(3,1,4),(3,2,2),(3,3,5),(3,4,1),(3,5,3),(3,6,8),(3,7,10),(3,8,6),(3,9,9),(3,10,7)(4,1,5),(4,2,8),(4,3,9),(4,4,4),(4,5,1),(4,6,10),(4,7,7),(4,8,2),(4,9,3),(4,10,6)(5,1,5),(5,2,8),(5,3,9),(5,4,4),(5,5,1),(5,6,10),(5,7,7),(5,8,2),(5,9,3),(5,10,6)(6,1,4),(6,2,2),(6,3,5),(6,4,1),(6,5,3),(6,6,8),(6,7,10),(6,8,6),(6,9,9),(6,10,7)(7,1,9),(7,2,10),(7,3,3),(7,4,5),(7,5,4),(7,6,7),(7,7,6),(7,8,8),(7,9,1),(7,10,2)(8,1,1),(8,2,6),(8,3,4),(8,4,3),(8,5,9),(8,6,2),(8,7,8),(8,8,7),(8,9,5),(8,10,10)(9,1,3),(9,2,7),(9,3,1),(9,4,9),(9,5,5),(9,6,6),(9,7,2),(9,8,10),(9,9,4),(9,10,8)(10,0,0),(10,0,1),(10,0,2),(10,0,3),(10,0,4),(10,0,5),(10,0,6),(10,0,7),(10,0,8),(10,0,9),(10,0,10),(10,1,0),(10,2,0),(10,3,0),(10,4,0),(10,5,0),(10,6,0),(10,7,0),(10,8,0),(10,9,0),(10,10,0)}

    The following conjecture is the main result of this section and shows how the set A defined above can be used in the construction of quadratic bent, near-bent and 2-plateaued functions. Unfortunately, we are unable to give a complete proof, for the following reasons: first, not all triples of the set A are described in a general way, so we are not able to verify all triples, and second, the triples of F3p cannot be described in a way that is suitable for our proof technique. Therefore, we give a partial proof, and complete proof remains as an interesting problem for all mathematics enthusiasts.

    Conjecture 1. Let n=2k with k>1 and λFpk,u,vFpn. Let the subset AF3p be defined as in (4.1) and fλ,u,v(x) be a p-ary function defined as f=fλ,u,v(x)=Trk1(λxpk+1)+Trn1(ux)Trn1(vx). Denote a triple T=(Trn1(λ1vpku),Trk1(λ1upk+1),Trk1(λ1vpk+1)). Then

    1. If TF3pA, then f is bent.

    2. If TA{(p1,0,0)}, then f is near-bent.

    3. If T=(p1,0,0), then f is a 2-plateaued function.

    Idea of Proof. Let us show the proof technique for the third statement. According to (3.7) and (3.8) we have

    S0(a)=1ppkωTrk1(λ1apk+1)[1+ωc1+ω2c1++ω2c1+ωc1]Rj(ajv)=1ppkωTrk1(λ1apk+1)[ωjc2+ωj+jc2c1++ωj(p1)+jc2(p1)c1].

    Then, using equation (3.3) we get

    ˆf(a)=1ppkωTrk1(λ1apk+1)[p1i=0(ωic2+ωi(c2+p)c1++ωi(c2+(p1)p)(p1)c1)]

    If c2=0, then

    ˆf(a)=1ppkωTrk1(λ1apk+1)[p+ωc1p+ω2c1p++ω(p1)c1p]=pkωTrk1(λ1apk+1)[1+ωc1+ω2c1++ω(p1)c1].

    If c1=0, then ˆf(a)=pk+1ωTrk1(λ1apk+1), otherwise is 0.

    If c20, then ˆf(a)=0.

    Therefore, |ˆf(a)|{0,pk+1} for all aFpn. Then f is 2-plateaued.

    This technique can also be used to prove either of the first two statements, but we are unable to check all possible triples.

    Example 2. Let p=7 and define the function f(x)=Trk1(λx7k+1)+Trn1(ux)Trn1(vx). By (4.1) we can define the set A as

    A={(0,1,2),(0,2,1),(0,3,3),(0,4,4),(0,5,6),(0,6,5),(1,1,1),(1,2,4,(1,3,5),(1,4,2),(1,5,3),(1,6,6),(2,1,4),(2,2,2),(2,3,6),(2,4,1),(2,5,5),(2,6,3),(3,1,4),(3,2,2),(3,3,6),(3,4,1),(3,5,5),(3,6,3),(4,1,1),(4,2,4),(4,3,5),(4,4,2),(4,5,3),(4,6,6),(5,1,2),(5,2,1),(5,3,3),(5,4,4),(5,5,6),(5,6,5),(6,0,0),(6,0,1),(6,0,2),(6,0,3),(6,0,4),(6,0,5),(6,0,6),(6,1,0),(6,2,0),(6,3,0),(6,4,0),(6,5,0),(6,6,0)}

    In what follows, we give some triples (t0,t1,t2) such that the function f is bent, near-bent and 2-plateaued, respectively. It can be easily verified that for any TF37A the function f is bent. For instance, if T=(2,3,4), then

    ˆf(a)={7kωTrk1(λ1a7k+1),if (c1,c2)={(0,0)}7kωTrk1(λ1a7k+1)+1,if (c1,c2)={(1,0),(1,1),(2,3),(2,6),(5,1),(5,4),(6,0),(6,6)}7kωTrk1(λ1a7k+1)+2,if (c1,c2)={(1,3),(1,5),(3,0),(3,3),(4,0),(4,4),(6,2),(6,4)}7kωTrk1(λ1a7k+1)+3,if (c1,c2)={(0,2),(0,5),(1,4),(2,4),(2,5),(5,2),(5,3),(6,3)}7kωTrk1(λ1a7k+1)+4,if (c1,c2)={(2,0),(2,2),(3,1),(3,2),(4,5),(4,6),(5,0),(5,5)}7kωTrk1(λ1a7k+1)+5,if (c1,c2)={(0,3),(0,4),(2,1),(3,4),(3,6),(4,1),(4,3),(5,6)}7kωTrk1(λ1a7k+1)+6,if (c1,c2)={(0,1),(0,6),(1,2),(1,6),(3,5),(4,2),(6,1),(6,5)}.

    Similarly, for any TA(6,0,0) the function f is near-bent. For instance, if T=(1,1,1), then

    ˆf(a)={±7k+12ωTrk1(λ1a7k+1),if(c1,c2)={(0,0),(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}0,otherwise.

    At the end, if T=(6,0,0) then f is 2-plateaued, i.e.,

    ˆf(a)={7k+1ωTrk1(λ1a7k+1),if(c1,c2)={(0,0)}0,otherwise.

    The exact computation of the Fourier transform for a given p-ary function of the form fλ,u,v has been established. Also, certain conditions such that a given p-ary function is bent, near-bent or 2-plateaued has been presented. It will be interesting to completely determine the distribution of the Fourier spectrum of these functions and try to determine the Fourier transform of the other p-ary functions using the presented method.

    The research presented in this paper was financially supported by the Slovenian Research Agency under research program No. P2-0037 – Future Internet technologies: Concepts, architectures, services and socio-economic issues.

    The author declares that he has no conflict of interest.



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