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Research article

The nonlinearity and Hamming weights of rotation symmetric Boolean functions of small degree

  • Received: 14 January 2020 Accepted: 14 May 2020 Published: 22 May 2020
  • MSC : Primary 06E30, 94C10

  • Let e, l and n be integers such that 1e<n and 3ln. Let i denote the least nonnegative residue of imodn. In this paper, we investigate the following Boolean function Fnl,e(xn)=n1i=0xixi+exi+2e...xi+(l1)e, which plays an important role in cryptography and coding theory. We introduce some new sub-functions and provide some recursive formulas for the Fourier transform. Using these recursive formulas, we show that the nonlinearity of Fnl,e(xn) is the same as its weight for 5l7. Our result confirms partially a conjecture of Yang, Wu and Hong raised in 2013. It also gives a partial answer to a conjecture of Castro, Medina and Stănică proposed in 2018. Our result extends the result of Zhang, Guo, Feng and Li for the case l=3 and that of Yang, Wu and Hong for the case l=4.

    Citation: Liping Yang, Shaofang Hong, Yongchao Xu. The nonlinearity and Hamming weights of rotation symmetric Boolean functions of small degree[J]. AIMS Mathematics, 2020, 5(5): 4581-4595. doi: 10.3934/math.2020294

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  • Let e, l and n be integers such that 1e<n and 3ln. Let i denote the least nonnegative residue of imodn. In this paper, we investigate the following Boolean function Fnl,e(xn)=n1i=0xixi+exi+2e...xi+(l1)e, which plays an important role in cryptography and coding theory. We introduce some new sub-functions and provide some recursive formulas for the Fourier transform. Using these recursive formulas, we show that the nonlinearity of Fnl,e(xn) is the same as its weight for 5l7. Our result confirms partially a conjecture of Yang, Wu and Hong raised in 2013. It also gives a partial answer to a conjecture of Castro, Medina and Stănică proposed in 2018. Our result extends the result of Zhang, Guo, Feng and Li for the case l=3 and that of Yang, Wu and Hong for the case l=4.


    Let Fn2 be the vector space of dimension n over the two-element field F2={0,1}. A Boolean function fn(x0,,xn1) in n variables is a map from Fn2 to F2. Boolean functions have wide applications to different scientific areas, like information theory, electrical engineering, game theory, cryptography and coding theory. In 1999, Piepryzyk and Qu [8] introduced a kind of special Boolean functions such that their evaluations on every cyclic inputs are the same, which is called rotation symmetric Boolean functions (abbr. RSBFs). Further, Piepryzyk and Qu [8] showed that RSBFs are useful in the design of fast hashing algorithms with strong cryptographic properties. Since then, RSBFs have attracted much attention for their wide applications in cryptography and coding theory ([2,5,9]).

    For brevity, let the vectors (x0,x1,...,xn1) and (c0,c1,...,cn1) in Fn2 be denoted by xn and cn respectively. The Hamming weight of a function fn(xn) is the number of xnFn2 satisfying fn(xn)=1 and denoted by wt(fn). For any two n variables Boolean functions fn(xn) and gn(xn), the Hamming distance between fn(xn) and gn(xn) is defined as wt(fn+gn), and denoted by d(fn,gn). We define the linear function Lcn by Lcn(xn):=cnxn, where is the vector dot product. The nonlinearity of fn is defined as Nfn:=min{d(fn,Lcn)cnFn2}. Since hashing algorithm employing RSBFs with degree 2 as components cannot resist the linear and differential attacks [7], we need to use higher-degree RSBFs with high nonlinearity to protect from differential attack. As early as 1998, Filiol and Fontaine [4] studied the nonlinearity of RSBFs up to 9 variables. In this paper, we study the relation between Hamming weights and nonlinearity of RSBFs with small degree.

    By i we denote the least nonnegative residue of imodn. Let e and l be integers such that 1e<n and 2ln. Then the l-th rotation symmetric Boolean function Fnl,e in n variables generated by the monomial x0xex2e...x(l1)e is defined as

    Fnl,e=Fnl,e(x0,...,xn1):=n1i=0xixi+exi+2e...xi+(l1)e.

    In 2009, Kim, Park and Hahn [6] explored the nonlinearity of Fn2,e, and proved that if ngcd(n,e) is even, then NFn2,e=wt(Fn2,e), otherwise, NFn2,ewt(Fn2,e). In 2010, Ciungu [3] showed that the linearity of Fn3,1 is the same as its weight for the case l=3 and 3|n. Zhang, Guo, Feng and Li [11] totally proved the equality of the linearity of Fn3,1 and its weight. Later on, Yang, Wu and Hong [10] investigated the case l=4 and proved that wt(Fn4,e)=NFn4,e. Furthermore, Yang, Wu and Hong [10] proposed the following conjecture.

    Conjecture 1.1. [10] Let e1 and l5 be any given integer. Then the nonlinearity of Fnl,e is equal to its weight.

    Recently, Castro, Medina and Stănică [2] showed that the Walsh transforms of symmetric and rotation symmetric Boolean functions satisfy the linear recurrence with integer coefficients, and suggested the following conjecture:

    Conjecture 1.2. [2] Let l>1 be a fixed integer. The sequence of {NFil}il satisfies the linear recurrence whose characteristic polynomial is given by

    xl2(xl2+xl2++x+1)=0.

    In this paper, we dedicate to prove that Conjecture 1.1 is true for some small degree cases, which also confirms Conjecture 1.2 for l=5,6 and 7. Particularly, we prove the following result.

    Theorem 1.3. Let l{5,6,7} and e and n be integers such that 1e<n and n2l1. Then NFnl,e=wt(Fnl,e).

    This paper is organized as follows. In Section 2, we study Fourier transform of Boolean functions and obtain some recursive formulas. In Section 3, we give the proof of Theorem 1.3. The final section is devoted to some remarks.

    We define the Fourier transform of the Boolean function fn(xn) at cnFn2 to be

    ^fn(cn):=xnFn2(1)fn(xn)+cnxn.

    Obviously, we have ^fn(cn)=2n2wt(fn+Lcn). In particular, one has ^fn(0)=2n2wt(fn).

    First of all, we introduce some notations. We let Fnl:=Fnl,1. Let i,j and k be nonnegative integers such that 1kji+1. Let m and t be nonnegative integers such that tm. Let X(i,j,0):=0, Y(m,0):=0,

    X(i,j,k):=k1r=0js=i+rxs and Y(m,t):=tr=1mrs=0xs.

    Let n be an integer with nl. Then we let

    tn:=nls=0xsxs+1...xs+l1.

    For any integers i,j with 0i,jl1, we let

    fni,j(xn):=tn+X(n(l1),n1,i)+Y(l1,j).

    Now we let n>l. It is easy to check that if xn1=0, then tn=tn1 and X(n(l1),n1,i)=0 for any 0il1. This implies that if xn1=0, then

    tn+X(n(l1),n1,i)=tn1 (2.1)

    with 0il1. If xn1=1, then we derive that

    tn+X(n(l1),n1,i)={tn1+X(nl,n2,i+1),if i=0,1,...,l2,tn1+X(nl,n2,i)+1,if i=l1. (2.2)

    Lemma 2.1. Let l and n be integers such that n>l5. Let i and j be integers such that 0i,jl1. If 0il2, then

    ^fni,j(cn)=^fn10,j(cn1)+(1)cn1^fn1i+1,j(cn1).

    If i=l1, then

    ^fni,j(cn)=^fn10,j(cn1)(1)cn1^fn1i,j(cn1).

    Proof. We only prove the relation ^fnl1,0(cn)=^fn10,0(cn1)(1)cn1^fn1l1,0(cn1). The remaining relations can be handled similarly. It follows from (2.1) and (2.2) that

    ^fnl1,0(cn)=(xn:xn1=0+xn:xn1=1)(1)tn+X(n(l1),n1,l1)+cnxn=xn1(1)tn1+cn1xn1+xn1(1)tn1+X(nl,n2,l1)+1+cn1+cn1xn1=^fn10,0(cn1)(1)cn1^fn1l1,0(cn1)

    as desired. Thus Lemma 2.1 is proved.

    Let (cnl+1,cnl+2,,cn1)Fl12. If n2l1, then Fnl can be written as

    Fnl=tnl+1+l2i=1llj=1xnl+j+i.

    Now we let (xnl+1,,xn1) take all values in Fl12. If l=5, then

    ^Fn5(cn)=3s=2(1+(1)cns)^fn40,0(cn4)+(1)cn1(1+(1)cn3)^fn40,1(cn4)+(1)cn4(1+(1)cn2)^fn41,0(cn4)+2s=1(1)cns^fn40,2(cn4)+(1)cn42s=1(1)cns^fn41,2(cn4)+(1)cn4+cn1^fn41,1(cn4)+3s=1(1)cns^fn40,3(cn4)+2s=1(1)cn(5s)^fn42,0(cn4)+2s=1(1)cn(5s)(1)cn1^fn42,1(cn4)+3s=1(1)cn(5s)^fn43,0(cn4)+4s=1(1)cns^fn44,4(cn4).

    If l6, then ^Fnl(cn) can be decomposed as follows:

    ^Fnl(cn)=l2s=2(1+(1)cns)^fn(l1)0,0(cn(l1))+l4j=1jt=1(1)cntl2s=j+2(1+(1)cns)^fn(l1)0,j(cn(l1))+l4i=1it=1(1)cn(lt)li2s=2(1+(1)cns)^fn(l1)i,0(cn(l1))+l4i=1j+il4j=1it1=1(1)cn(lt1)jt2=1(1)cnt2li2s=j+2(1+(1)cns)^fn(l1)i,j(cn(l1))+l3t=1(1)cnt^fn(l1)0,l3(cn(l1))+(1)cn(l1)l3t=1(1)cnt^fn(l1)1,l3(cn(l1))+l4i=2l2ij=l3iit1=1(1)cn(lt1)jt2=1(1)cnt2^fn(l1)i,j(cn(l1))+(1)cn(l1)l4t=1(1)cnt^fn(l1)1,l4(cn(l1))+l2t=1(1)cnt^fn(l1)0,l2(cn(l1))+l3t=1(1)cn(lt)^fn(l1)l3,0(cn(l1))+l3t=1(1)cn(lt)(1)cn1^fn(l1)l3,1(cn(l1))+l2t=1(1)cn(lt)^fn(l1)l2,0(cn(l1))+l1t=1(1)cnt^fn(l1)l1,l1(cn(l1)). (2.3)

    Now we let cn=(0,,0). Clearly, we have fni,j(x0,...,xn1)=fnj,i(xn1,...,x0). Hence ^fni,j(0)=^fnj,i(0). Thus we conclude that if l=5, then

    ^Fn5(0)=4^fn40,0(0)+4^fn40,1(0)+2^fn40,2(0)+2^fn41,2(0)+^fn41,1(0)+2^fn40,3(0)+^fn44,4(0).

    And if l6, then

    ^Fnl(0)=l4i=0j+il4j=02l3ij^fn(l1)i,j(0)+^fn(l1)1,l4(0)+2^fn(l1)0,l2(0)+l4i=2l2ij=l3i^fn(l1)i,j(0)+21i=0^fn(l1)i,l3(0)+^fn(l1)l1,l1(0). (2.4)

    Let [x] denote the largest integer that is less than or equal to the real number x. In what follows, we give some recursive relations about ^fni,j(cn) for any integer i and j with 0i,jl1.

    Lemma 2.2. Let n,l,i,j and k be integers such that l5, n2l, 0i,jl1 and 1kn. Let cn=(c0,...,cn1)Fn2 such that cn1=0. Let ck be the vector consisting of the first k bits of cn. If i=l1, then

    ^fni,j(cn)={2l22s=12st=2(1)cnt^fn(2s+1)0,j(cn(2s+1))+2lt=2(1)cnt^fnll1,j(cnl),if i is odd,2[l22]s=12st=2(1)cnt^fn(2s+1)0,j(cn(2s+1))+2l1t=2(1)cnt^fnl0,j(cnl),if i is even.

    If i=l2, then

    ^fni,j(cn)={2l12s=12st=2(1)cnt^fn(2s+1)0,j(cn(2s+1))+2lt=2(1)cnt^fnll1,j(cnl),if i is odd,2[l12]s=12st=2(1)cnt^fn(2s+1)0,j(cn(2s+1))+2l1t=2(1)cnt^fnl0,j(cnl),if i is even.

    If 0il3, then

    ^fni,j(cn)={2l2s=1st=1(1)cnt^fn(s+1)0,j(cn(s+1))+2l1t=2(1)cnt^fnl0,j(cnl),    if i=0,2l3s=1st=1(1)cnt^fn(s+1)0,j(cn(s+1))+2l2t=2(1)cnt^fn(l1)0,j(cn(l1))+2lt=2(1)cnt^fnll1,j(cnl),                                                if i=1,2li2s=1st=1(1)cnt^fn(s+1)0,j(cn(s+1))+2l1it=2(1)cnt(^fn(li)0,j(cn(li))+[i2]s=12s1k=0(1)cn(li+k)^fn(li+2s)0,j(cn(li+2s))+ik=0(1)cn(li+k)^fnll1,j(cnl)),                                                                             if i2 is odd,2li2s=1st=1(1)cnt^fn(s+1)0,j(cn(s+1))+2l1it=2(1)cnt(^fn(li)0,j(cn(li))+i2s=12s1k=0(1)cn(li+k)^fn(li+2s)0,j(cn(li+2s))),               if i2 is even.

    Proof. Note that cn1=0. We divide the proof into the following three cases.

    Case 1. i=l1. For any integer j with 0jl1, by Lemma 2.1 we conclude that

    ^fnl1,j(cn)=^fn10,j(cn1)+(1)1+cn1^fn1l1,j(cn1)=^fn20,j(cn2)+(1)cn2^fn21,j(cn2)(^fn20,j(cn2)+(1)1+cn2^fn2l1,j(cn2))=2(1)cn2^fn30,j(cn3)+(1)cn2+cn3(^fn32,j(cn3)+(1)3^fn3l1,j(cn3))==2[l22]s=12st=2(1)cnt^fn(2s+1)0,j(cn(2s+1))+l1t=2(1)cnt(^fnl+1l2,j(cnl+1)+(1)l1^fnl+1l1,j(cnl+1)).

    If l is even, then by Lemma 2.1 one derives that

    ^fnl1,j(cn)=2l22s=12st=2(1)cnt^fn(2s+1)0,j(cn(2s+1))+2lt=2(1)cnt^fnll1,j(cnl).

    If l is odd, it then follows from Lemma 2.1 that

    ^fnl1,j(cn)=2[l22]s=12st=2(1)cnt^fn(2s+1)0,j(cn(2s+1))+2l1t=2(1)cnt^fnl0,j(cnl).

    This finishes the proof of Lemma 2.2 in this case.

    Case 2. i=l2. Then by Lemma 2.1, we have

    ^fnl2,j(cn)=^fn10,j(cn1)+^fn1l1,j(cn1)=2^fn20,j(cn2)+(1)cn2^fn21,j(cn2)+(1)1+cn2^fn2l1,j(cn2)=2^fn20,j(cn2)+(1)cn2+cn3(^fn32,j(cn3)+(1)2^fn3l1,j(cn3))==2[l12]s=12s1t=1(1)cnt^fn2s0,j(cn2s)+l1t=2(1)cnt(^fnl+1l2,j(cnl+1)+(1)l2^fnl+1l1,j(cnl+1)).

    If l is even, then one derives that

    ^fnl2,j(cn)=2[l12]s=12s1t=1(1)cnt^fn2s0,j(cn2s)+2l1t=2(1)cnt^fnl0,j(cnl).

    If l is odd, then

    ^fnl2,j(cn)=2l12s=12s1t=1(1)cnt^fn2s0,j(cn2s)+2lt=2(1)cnt^fnll1,j(cnl).

    Thus Lemma 2.2 is proved in this case.

    Case 3. 0il3. Then by Lemma 2.1, we have

    ^fni,j(cn)=^fn10,j(cn1)+(1)cn1^fn1i+1,j(cn1)==2li2s=1st=1(1)cnt^fn(s+1)0,j(cn(s+1))+l1it=2(1)cnt(^fn(li1)li2,j(cn(li1))+^fn(li1)l1,j(cn(li1))):=2li2s=1st=1(1)cnt^fn(s+1)0,j(cn(s+1))+l1it=2(1)cntTn(li1). (2.5)

    Note that 0il3. From Lemma 2.1 we derive that

    Tn(li1)=2^fn(li)0,j(cn(li))+(1)cn(li)(^fn(li)li1,j(cn(li))^fn(li)l1,j(cn(li))). (2.6)

    If i=0, then it is easy to see that Tn(l1)=2^fnl0,j(cnl). It follows from (2.5) that

    ^fn0,j(cn)=2l2s=1st=1(1)cnt^fn(s+1)0,j(cn(s+1))+2l1t=2(1)cnt^fnl0,j(cnl).

    If i=1, then by (2.6) and Lemma 2.1, we have

    Tn(li1)=2^fn(l1)0,j(cn(l1))+2(1)cn(l1)+cnl^fnll1,j(cnl).

    Thus from (2.5), we obtain that

    ^fn1,j(cn)=2l3s=1st=1(1)cnt^fn(s+1)0,j(cn(s+1))+2l2t=2(1)cnt^fn(l1)0,j(cn(l1))+2lt=2(1)cnt^fnll1,j(cnl).

    If 2il3, then by (2.6) and Lemma 2.1, we have

    Tn(li1)=2^fn(li)0,j(cn(li))+2[i2]s=12s1k=0(1)cn(li+k)^fn(li+2s)0,j(cn(li+2s))+ik=0(1)cn(li+k)(1+(1)i+1)^fnll1,j(cnl).

    Hence we obtain that if i is odd, then

    Tn(li1)=2^fn(li)0,j(cn(li))+2[i2]s=12s1k=0(1)cn(li+k)^fn(li+2s)0,j(cn(li+2s))+2ik=0(1)cn(li+k)^fnll1,j(cnl).

    By (2.5), one deduces that

    ^fni,j(cn)=2li2s=1st=1(1)cnt^fn(s+1)0,j(cn(s+1))+2l1it=2(1)cnt(^fn(li)0,j(cn(li))+[i2]s=12s1k=0(1)cn(li+k)^fn(li+2s)0,j(cn(li+2s))+ik=0(1)cn(li+k)^fnll1,j(cnl)).

    If i is even, then

    Tn(li1)=2^fn(li)0,j(cn(li))+2i2s=12s1k=0(1)cn(li+k)^fn(li+2s)0,j(cn(li+2s)).

    From (2.5) we derive that

    ^fni,j(cn)=2li2s=1st=1(1)cnt^fn(s+1)0,j(cn(s+1))+2l1it=2(1)cnt(^fn(li)0,j(cn(li))+i2s=12s1k=0(1)cn(li+k)^fn(li+2s)0,j(cn(li+2s))).

    Hence Lemma 2.2 holds in this case.

    This completes the proof of Lemma 2.2.

    Lemma 2.3. Let n,l,i,j and k be integers such that l5, n2l, 0i,jl1 and 1kn. Let cn=(c0,...,cn1)Fn2 such that cn1=1. Let ck denote the first k bits of cn. Then

    ^fni,j(cn)={(1)cn2^fn21,j(cn2)+(1)1+cn2^fn2i+2,j(cn2),          if 0il3,(1)cn2^fn21,j(cn2)+(1)cn2^fn2l1,j(cn2),             if i=l2,2^fn20,j(cn2)+2[l12]s=22s1t=2(1)cnt^fn2s0,j(cn2s)+2l1t=2(1)cnt^fnl0,j(cnl),                                                                    if i=l1 is odd,2^fn20,j(cn2)+2l12s=22s1t=2(1)cnt^fn2s0,j(cn2s)+2lt=2(1)cnt^fnll1,j(cnl),                                                                    if i=l1 is even.

    Proof. We divide the proof into the following three cases.

    Case 1. 0il3. Then by Lemma 2.1 we have

    ^fni,j(cn)=^fn10,j(cn1)^fn1i+1,j(cn1)=^fn20,j(cn2)+(1)cn2^fn21,j(cn2)^fn20,j(cn2)(1)cn2^fn2i+2,j(cn2)=(1)cn2^fn21,j(cn2)+(1)1+cn2^fn2i+2,j(cn2).

    Thus Lemma 2.3 is true in this case.

    Case 2. i=l2. From Lemma 2.1 one deduces that

    ^fnl2,j(cn)=^fn10,j(cn1)^fn1l1,j(cn1)=^fn20,j(cn2)+(1)cn2^fn21,j(cn2)^fn20,j(cn2)+(1)2+cn2^fn2l1,j(cn2)=(1)cn2^fn21,j(cn2)+(1)cn2^fn2l1,j(cn2).

    Hence Lemma 2.3 holds for this case.

    Case 3. i=l1. It follows from Lemma 2.1 that

    ^fnl1,j(cn)=^fn10,j(cn1)(1)cn1^fn1l1,j(cn1)=2^fn20,j(cn2)+(1)cn2(^fn21,0(cn2)^fn2l1,j(cn2))==2^fn20,j(cn2)+2[l12]s=22s1t=2(1)cnt^fn2s0,j(cn2s)+l1t=2(1)cnt(^fnl+1l2,j(cnl+1)+(1)l2^fnl+1l1,j(cnl+1)).

    If l is even, then one derives that

    ^fnl1,j(cn)=2^fn20,j(cn2)+2[l12]s=22s1t=2(1)cnt^fn2s0,j(cn2s)+2l1t=2(1)cnt^fnl0,j(cnl)

    as desired.

    If l is odd, then

    ^fnl1,j(cn)=2^fn20,j(cn2)+2l12s=22s1t=2(1)cnt^fn2s0,j(cn2s)+2lt=2(1)cnt^fnll1,j(cnl)

    as required. Hence Lemma 2.3 is proved in this case.

    This finishes the proof of Lemma 2.3.

    Lemma 2.4. Let l,n,j be integers such that l5, n2l and 0jl1. Then for integer i with 0il1, we have

    ^fni,j(0)=2ls=2^fnsi,j(0). (2.7)

    Proof. It follows from Lemma 2.2 that for any integer j with 0jl1, we have

    ^fn0,j(0)=2ls=2^fns0,j(0).

    Thus (2.7) is true when i=0 and 0jl1. Note that ^fni,j(0)=^fnj,i(0). Then for 0il1, one has

    ^fni,0(0)=2ls=2^fnsi,0(0).

    Then by the definition of ^fni,j(0) and the assumption n2l1, we can conclude that (2.7) holds for any integer 0i,jl1. Hence Lemma 2.4 is proved.

    In the following, we show that the weight of Fnl satisfies a linear recurrence, which is also proved by Castro, Chapman, Medina and Sepˊulveda [1].

    Proposition 2.5. Let l and n be integers such that l5 and n2l. Then

    ^Fnl(0)=2ls=2^Fnsl(0).

    Proof. We only prove the case l6 since the case l=5 can be done similarly. Let l6. It then follows from (2.4) and Lemma 2.4 that

    ^Fnl(0)=2l4i=0j+il4j=02l3ijls=2^fnl+1si,j(0)+2ls=2^fnl+1s1,l4(0)+41i=0ls=2^fnl+1si,l3(0)+2l4i=2l2ij=l3ils=2^fnl+1si,j(0)+4ls=2^fnl+1s0,l2(0)+2ls=2^fnl+1sl1,l1(0).

    Then by (2.4), we deduce that

    ^Fnl(0)=2ls=2(l4i=0j+il4j=02l3ij^fnl+1si,j(0)+^fnl+1s1,l4(0)+1i=02^fnl+1si,l3(0)+l4i=2l2ij=l3i^fnl+1si,j(0)+2^fnl+1s0,l2(0)+^fnl+1sl1,l1(0))=2ls=2^Fnsl(0).

    The proof of Proposition 2.5 is complete.

    Lemma 2.6 (Proposition 3.1, [10]). Let fn(xn) be a Boolean function with n variables. If ^fn(0)=max{|^fn(cn)|:cnFn2}, then Nfn=wt(fn).

    In this section, we show Theorem 1.3. For this purpose, we need the following lemma.

    Lemma 3.1. Let n and l be nonnegative integers such that 5l7. Let cn=(c0,,cn1)Fn2. If cn such that c1=1, then for all 0i,jl1, we have

    |^fni,j(cn)|<12l1^Fn+l1l(0). (3.1)

    Proof. We prove Lemma 3.1 by induction on n. When ln2l1 and c10, using Maple 17 we can check that (3.1) holds for all 0i,jl1. For example, the case l=n=5 is given in Table 1, we can check that ^fnij(c5)124^F2l1l(0)=42024. In what follows, we let n2l. Assume that Lemma 3.1 holds for any positive integer k less than n1. Now we prove that Lemma 3.1 is true for the case k=n. Since c1=1, we divide the proof into the following two cases.

    Table 1.  The evaluations of ^f5i,j(c5), 0i,j4, where c5=(c0,...c4)F52 with c10 is denoted by 0i4ci2i.
    c 2 3 6 7 10 11 14 15 18 19 22 23 26 27 30 31
    ^f50,0 2 -2 -2 2 -2 2 2 -2 -2 2 2 -2 2 -2 -2 2
    ^f50,1 2 -2 -2 2 -2 2 2 -2 2 -2 -2 2 -2 2 2 -2
    ^f50,2 6 -6 -6 6 2 -2 -2 2 -2 2 2 -2 2 -2 -2 2
    ^f50,3 10 -10 6 -6 -2 2 2 -2 2 -2 -2 2 -2 2 2 -2
    ^f50,4 -10 10 -6 6 2 -2 -2 2 -2 2 2 -2 2 -2 -2 2
    ^f51,0 2 2 -2 -2 -2 -2 2 2 -2 -2 2 2 2 2 -2 -2
    ^f51,1 6 -2 -6 2 -6 2 6 -2 -2 -2 2 2 2 2 -2 -2
    ^f51,2 6 -2 -6 2 2 -6 -2 6 -2 -2 2 2 2 2 -2 -2
    ^f51,3 14 -10 2 -6 -6 2 6 -2 -2 -2 2 2 2 2 -2 -2
    ^f51,4 -10 14 -6 2 2 -6 -2 6 -2 -2 2 2 2 2 -2 -2
    ^f52,0 -2 -2 2 2 2 2 -2 -2 2 2 -2 -2 -2 -2 2 2
    ^f52,1 -2 -2 2 2 2 2 -2 -2 6 -2 -6 2 -6 2 6 -2
    ^f52,2 2 -6 -2 6 6 -2 -6 2 2 2 -2 -2 -2 -2 2 2
    ^f52,3 6 -10 -6 10 10 -6 -10 6 2 2 -2 -2 -2 -2 2 2
    ^f52,4 -10 6 -6 10 2 2 -2 -2 -2 6 2 -6 2 -6 -2 6
    ^f53,0 2 2 -2 -2 -2 -2 2 2 -2 -2 2 2 2 2 -2 -2
    ^f53,1 6 -2 -6 2 -6 2 6 -2 -2 -2 2 2 2 2 -2 -2
    ^f53,2 6 -2 -6 2 2 -6 -2 6 -2 -2 2 2 2 2 -2 -2
    ^f53,3 10 -6 -2 -2 -2 -2 10 -6 2 -6 6 -2 -2 6 -6 2
    ^f53,4 -6 10 -2 -2 -2 -2 -6 10 -6 2 -2 6 6 -2 2 -6
    ^f54,0 -2 -2 2 2 2 2 -2 -2 2 2 -2 -2 -2 -2 2 2
    ^f54,1 -2 -2 2 2 2 2 -2 -2 6 -2 -6 2 -6 2 6 -2
    ^f54,2 -2 -2 2 2 2 2 -2 -2 6 -2 -6 2 2 -6 -2 6
    ^f54,3 2 -6 6 -2 -2 6 -6 2 10 -6 -2 -2 -2 -2 10 -6
    ^f54,4 -6 2 -2 6 6 -2 2 -6 -6 10 -2 -2 -2 -2 -6 10

     | Show Table
    DownLoad: CSV

    Case 1. cn1=0. From Lemma 2.2 and Theorem 2.5, then

    |^fni,j(cn)|<22l1(^Fn2+l1l(0)+^Fn3+l1l(0)++^Fnl+l1l(0))=12l1^Fn+l1l(0).

    Hence Lemma 3.1 is true in this case.

    Case 2. cn1=1. Then by Lemma 2.3 and Theorem 2.5 we have

    |^fni,j(cn)|<22l1(^Fn2+l1l(0)+^Fn3+l1l(0)++^Fnl+l1l(0))=12l1^Fn+l1l(0).

    Thus Lemma 3.1 holds for this case.

    This finishes the proof of Lemma 3.1.

    Now we present the proof of Theorem 1.3.

    Proof of Theorem 1.3. From Lemma 2.6 we conclude that to prove the validity of Theorem 1.3 for e=1, it suffices to prove that

    ^Fnl(0)=max{|^Fnl(cn)|:cnFn2}.

    From the definition of Fnl, we conclude that for any integer 0jn1,

    ^Fnl(c0,c1,,cn1)=^Fnl(cj,cj+1,,cn+j1).

    Without loss of generality, we suppose that c10. From Lemma 3.1, it follows that

    ^fn(l1)i,j(cn(l1))<12l1^Fnl(0).

    It follows from (2.3) that ^Fnl(cn) is the sum of ^fn(l1)i,j(cn(l1)). Hence ^Fnl(cn)<^Fnl(0). Namely, Theorem 1.3 is true for the case e=1.

    Now let e>1. Let s=gcd(n,e) and t=n/s. Then

    Fnl,e(xn)=n1i=1xixi+exi+2exi+(l1)e=s1k=0t1j=0xk+jexk+je+exk+je+2exk+je+(l1)e:=s1k=0gtk(xk,xe+k,...,x(t1)e+k). (3.2)

    For 0ks1,0jt1, substituting xk+je by y(k)j, one gets that

    gtk(xk,xe+k,...,x(t1)e+k)=t1j=0y(k)jy(k)j+1y(k)j+2y(k)j+(l1).

    Let ctk=(ck,ce+k,...,c(t1)e+k) and xtk=(xk,xe+k,...,x(t1)e+k). For any ctk0, we have showed that ^gtk(ctk)<^gtk(0). Then for all cn=(c0,...,cn1)0, it follows from the definition of Fourier transform and (3.2) that

    |^Fnl1,e(cn)|=|n1i=0(1)xixi+exi+2exi+(l1)e+cnxn|=|s1k=0t1i=0(1)xk+jexk+je+exi+je+(l1)e+ctkxtk|=|s1k=0^gtk(ctk)|<s1k=0^gtk(0)=^Fnl1,e(0).

    Thus Theorem 1.3 is true for the case e>1.

    This concludes the proof of Theorem 1.3.

    In Section 2, we present some recursive formulas for the Fourier transform of fni,j(xn) and Fnl(xn) for all positive integers i,j,l and n such that l5, 0i,jl1 and n2l. So to prove the truth of Conjectures 1.1 and 1.2, it is enough to show that for any integer n such that ln2l1, the inequality

    |^fni,j(cn)|<12l1^Fn+l1l(0) (4.1)

    is true when c1=1. When l is an integer greater than 7 but not too large, by some direct calculations one can derive that (4.1) holds. But for the general case when l>7, one meets some obstructions when one tries to prove (4.1). Hence we propose the following conjecture.

    Conjecture 4.1. Let i,j,l,n be integers such that l>7, ln2l1 and 0i,jl1 and let cl=(c0,,cl1)Fl2 with c1=1. Then

    |^fni,j(cn)|<12l1^Fn+l1l(0).

    Let q=pr with p being a prime and r>1. One can form the exponential sum of a function F:FnqFq as follows:

    SFq(F)=xFnqe2πipTrFq/Fp(F(x)),

    where TrFq/Fp represents the field trace function from Fq to Fp. Castro, Chapman, Medina and Sepˊulveda [1] showed that the sequence of {SFq(Fnl)}nl satisfies the linear recurrence whose characteristic polynomial is given by

    Xlql2k=0(q1)kXl2k=0.

    On the other hand, by Lemma 2.4, we know that for any integer i and j with 0i,jl1, the sequence {SF2(fni,j)}nl satisfies the linear recurrence whose characteristic polynomial is given by

    Xl2l2k=0Xl2k=0.

    We believe that the same holds for the general finite field. That is, we suggest the following conjecture as the conclusion of this paper.

    Conjecture 4.2. Let i,j and l be integers with l3 and 0i,jl1. The sequence {SFq(fni,j)}nl satisfies the linear recurrence whose characteristic polynomial is given by

    Xlql2k=0(q1)kXl2k=0.

    The authors would like to thank the anonymous referees for careful reading of the manuscript and helpful comments and corrections that improved the presentation of this paper. Hong was supported partially by the Fundamental Research Funds for the Central Universities.

    We declare that we have no conflict of interest.



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    [2] F. N. Castro, L. A. Medina and P. Stănică, Generalized Walsh transforms of symmetric and rotation symmetric Boolean functions are linear recurrent, Appl. Algebra Eng. Comm., 29 (2018), 433-453. doi: 10.1007/s00200-018-0351-5
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    [7] S. Mariai, T. Shimoyama and T. Kaneko, Higher order differential attack using chosen higher order differences, International Workshop on Selected Areas in Cryptography, 1556 (1998), 106-117, Springer-Verlag, Berlin.
    [8] J. Pieprzyk and C. X. Qu, Fast hashing and rotation-symmetric functions, J. Univers. Comput. Sci., 5 (1999), 20-31.
    [9] P. Stănică and S. Maitra, Rotation symmetric Boolean functions count and cryptographic properties, Discrete Appl. Math., 156 (2008), 1567-1580. doi: 10.1016/j.dam.2007.04.029
    [10] L. P. Yang, R. J. Wu and S. F. Hong, Nonlinearity of quartic rotation symmetric Boolean functions, Southeast Asian Bull. Math., 37 (2013), 951-961.
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