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АЛГЕБРО-ЛОГИЧЕСКИЕ МЕТОДЫ В ИНФОРМАТИКЕ И ИСКУССТВЕННЫЙ ИНТЕЛЛЕКТ
ALGEBRAIC AND LOGICAL METHODS IN COMPUTER SCIENCE AND ARTIFICIAL INTELLIGENCE
Серия «Математика» 2019. Т. 30. С. 125-140
Онлайн-доступ к журналу: http: / / mathizv.isu.ru
ИЗВЕСТИЯ
Иркутского государственного ■университета
УДК 519.714.4 MSG 68Q17
DOI https://doi.org/10.26516/1997-7670.2019.30.125
Complexity Lower Bound for Boolean Functions in the Class of Extended Operator Forms*
A. S. Baliuk
Irkutsk State University, Irkutsk, Russian Federation
Abstract. Starting with the fundamental work of D.E.Muller in 1954, the polynomial representations of Boolean functions are widely investigated in connection with the theory of coding and for the synthesis of circuits of digital devices. The operator approach to polynomial representations, proposed in the works of S. F. Vinokurov, made it possible, on the one hand, to uniformly describe all known types of polynomial forms of Boolean functions, and, on the other hand, to generalize them to the case of expansions by the operator images of arbitrary odd function, not only conjunction.
In the study of polynomial and, in the general case, operator forms, one of the main questions is obtaining lower and upper bounds of the complexity of the representation of Boolean functions in various classes of forms. The upper bounds of complexity are actually algorithms for minimizing Boolean functions in a particular class of forms.
The lower bounds of complexity can be divided into two types: combinatorial and effective. Combinatorial lower bounds make it possible to prove the existence of Boolean functions, having high complexity, without finding the explicit form of these functions. Effective lower bounds are based on explicit constructing Boolean functions that have high complexity in a particular class of forms.
In this paper, using an algebraic extension of a finite field of order 2, we obtain a lower bound for the complexity of Boolean functions in the class of extended operator forms. This lower bound strengthens the previously known lower bounds for this class
* This work was supported by Russian Foundation for Basic Research, grant N 1901-00200.
of operator forms and is becoming asymptotically optimal if the sequence of Mersenne primes is infinite.
Keywords: Boolean function, lower bound, extension of finite field, Mersenne prime.
1. Introduction
In the initial work [9] Muller introduced several polynomial forms of Boolean functions. Since that, these and many other polynomial forms were widely investigated.
The uniform approach to polynomial forms of Boolean functions were proposed in [12], using the notion of operators and their bundles. In section 2 of the current paper we suggest another way to represent operators and bundles, using vectors and matrices. Such a way could be naturally generalized to multivalued functions, including functions over finite fields [3].
One of the problems in the area of Boolean functions polynomial representation is obtaining lower bounds of complexity in particular classes of polynomial, and more general, of operator forms. This paper is devoted to obtaining lower bound for the class of extended operator forms. To achieve this result we developed a method for counting zeros in vectors over arbitrary finite field, which is described in section 3.
An extended list of references on the complexity for polynomial forms of Boolean functions and multivalued functions can be found in [11].
2. Matrix representations of bundles of operators
Definition 1. A word an...a\ over the alphabet {t),e,p} will be called n-ary operator.
Let us construct the map v from the set of operators to Boolean vectors recursively as follows: v(an ... ai) = v(an) <g> v(an-1... ai) for n ^ 2 and v(Q) = (11), v(t) = (01), v(p) = (10) for n = 1. The symbol <g> denotes the tensor product of vectors. For the sake of convenience let us introduce the vector (1), which corresponds to the 0-ary operator 0, i.e. to empty word. As tensor product <g> is an associative operation we can simply write v(an ... ai) = v(cin) 0 ■■■ 0 v(ai).
Let N = 2n and cr1,...,(TJV be all pairwise different binary n-tuples ordered lexicographically such that j = 1 + <r^2ra_1 + • • • + a322l + a[2° where denotes ith component of the tuple cr-7.
For every tuple S = (gi,... ,gN) of n-ary Boolean functions let us define a matrix Ms in the following way:
/ Mil ... MiN \
ms = ( • ••. • ) (2.1)
\mn2 ... Mnn J,
where Mjk = gj(a^,...,a1 )• For every n-ary Boolean function g let us define the tuple Sg = (gi,... ,gN), assuming that for all 1 ^ j ^ N
gj(xn,..., xi) = g(xra ® aN-j+i,..., xi ® aN-+i). (2.2)
Proposition 1. g(xn,...,xi) — xn ' ... ' x i iff Msg is an identity matrix.
Proof. Let g(xn,... ,xi) — xn ' ... ' x i • By definition, Sg = (g i,...,gN), where gj (xn,..., x i) = (xn ® aN-j+i) ■ ... ■ (x i ® a N-j+i )• If Msg has the form (2.1), then Mjk = ^ ® aN-j+i) ■ ... ■ (af ® aN-j+i)• The binary tuples a i,...,aN are ordered lexicographically. Thus, a i = (0,..., 0) and aN = (1,..., 1). Further, a1 is the k-th tuple from the beginning, and aN-j+i is the j-th tuple from the end. For a1, there is exactly one tuple which differs from a1 in each component. It is aN-1+i. Consequently, Mjk = (an ® aN-j+i) ■ ... ■ (af ® a N-j+i) = 1 if and only if j = k. Otherwise, Mji = 0. This means that the matrix Msg is the identity matrix.
Conversely, let M be the identity NxN matrix of the form (2.1), and let n-ary Boolean functions g i,... ,gN are given by gj (an,..., a 1) = Mjk, where 1 ^ j, k ^ N. Then gj (an, ...,a 1) = 1 if and only if Mjk = 1, i.e. k = j• Further, gj(xn,..., xi) = 1 only if xn = a3n,..., xi = aj• This means that gj(xn,..., x i) = (xn ® ah) ■ ... ■ (xi ® aj)• Since ( ah,..., aj) is different from aj in each element, we have (an,..., aj) = aN-j+i, and therefore gj(xn,..., xi) = (xn ® aN-j+i) ■ ... ■ (x i ® a N-j+i). This means that (g i,... ,gN) = Sg, whereg(xn,... ,xi) — xn'. . . *x i, and M = Msg ■ □
Following [5], let us define the action of an operator an ... a i on an n-ary Boolean function g as follows: an... a ig = fn, where for all 1 ^ m ^ n
{fm— i (xn, . . . , xm,, . . . x i) ® /m— i (xn, . . . , x i), if am = d; fm—i(xn,...,xi), if am = e; (2.3)
fm— i (xn, . . . , xm+ i , xam, xm— i, . . . x i ), if am — p;
and fo(xn,... ,x i) = g(xn,..., xi).
For every n-ary Boolean function f let us introduce the binary vector Vf, assuming Vf = (V,..., VN) where V1 = f (an,..., a1) for all 1 ^ k ^ N.
Proposition 2. For every n-ary Boolean function g and every n-ary operator an ... ai if f = an ... a ig, then Vf = v(a„ ... a i)MSg.
Proof. Let On... ai be an n-ary operator. Recall through J0 the set of indices m for which am = d, and through Jp the set of indices rri for which am = p. Denote by P(Jo) the set of all subsets of J0, including the empty set. First of all, note that since j = 1 + <r^2ra_1 + • • • + a\2° then
a-
holds for all 1 ^ j ^ 2m 1 ^ N. Define sets of integers
Im={ 1 + E + E | 5 G
seJp s&s
ssim s4m
Obviously, j ^ 2m for all j € Im. Also define the vectors Vm = v(am ... ai).
By induction, we will show that if fm is defined in the same way as in (2.3), then Vp = 1 if and only if 2m - j + 1 € Imi as well as
fm{(Tn, ...,o\)= E g(an
j&Im
By the basis, we have I0 = {1}, V° = (1), a1 = (0,..., 0). Thus, Vf = 1 if and only if 2° — j + 1 G To, and
E © < • • •, © <4) = 9(*n, = ■■■,
j€l0
By the step of induction, we take am. 1) If am = д, then Im = ViU^, where Гт_х = {j + 2m~l \ j e Im-1}-By the induction hypothesis, /m_i(<r^,..., erf) = Yj9{an®an, • • •, o\®a{).
j&Im-1
Thus, =
j^m-l j&Im-1
= ^^ g(akn ® ..., cr^j ® a3m,..., a\ ф a\) = /m_i(cr^,..., ..., erf) ie/m-i
/m(o"ra, • • • , = /m-l(o"ra, • • • , ) + /m-l(o"n, • • • , • • • , CF\ ) = E g(ekn® ек1®е{)+ E J^®^-^®^
= E
j&Im
Also, we have = (11) <g> Vm~l = (V^1,..., V£l\, ..., V£z\).
Hence, if j ^ 2m~1, then V™ = 1 if and only if 2m~l -j + le /m_b and if 2m-1 < j ^ 2"», then V™ = 1 if and only if 2"1"1 - (j - 2"1"1) + 1 e Im-1-
In the first case, we have 2m — (j + 2m_1) +1 € Im-1, in the second case we have 2m-j + l e Im-1 and, therefore, V™ = 1 if and only if 2m-j + l € Im-
2) If am = e, then Im = Im_u
fm((Tn, ■ ■ ■ , of) = /m- • • • ) ) =
= E © ■ ■ ■ , © ffi) = E © < • • • , ©
j€lm-1 j&Im
Since Vm = (01) eg) Vm~l = (0,..., 0, V^'1,..., we have V™ = 1
if and only if 2m~l - (j - 2"1"1) + 1 € Im-i- The latter means that 2m-j + 1 € Im.
3) If am = p, then Im = {j + 2m~1 \ j e Im-1} and
j&I-m j&Im-l
= /m-l(Cra) • • • j • • • > al) = fm{&ni • • • i al)-
In this case, Vm = (10) eg) Vm~l = (V™-1,..., 0,..., 0) and, there-
fore, Vp = 1 if and only if 2m~l -j + 1 € Im-1 or 2m - j + 1 € Im, which is the same.
At this point, we have f(ak,..., erf) = ^ $(<7^ © ah,..., erf © <rj) for
j&In
f = On ... a\g. Now consider fc-th element of the product v(an ... a\)Msg-According to (2.2) it is equal to
1V
E © <~j+\ ■ ■ •, o\ © = E 9« ®al...,a\® a[).
j=i jein
This means that Vf = v(an ... a\)Msg and completes the proof. □
Definition 2. A bundle of n-ary operators is a set 21, which contains of N pairwise different n-ary operators.
Definition 3. A bundle is called generated by a pair or just pair-generated if it can be represented as 21 = {a^ ... a}, a^ ... aj,..., ... af} where aj = aj tf aj = ® and aj=aj if aj = 1 • In ^is case, the operators a^ ... a} and ... af are called generators or generating operators for the bundle 21.
An NxN Boolean matrix M represents a bundle of n-ary operators 21 = {a^... a\ | 1 ^ k ^ N} if the elements of fc-th row of the matrix M are pairwise equal to the corresponding elements of the vector v(akn ... a^). As operators in a bundle can be ordered in various ways, a matrix, representing the bundle, is not uniquely determined. But all such matrices can be reduced to each other by permutation of their rows.
For the sake of convenience, let us introduce the following notation. Let V = (Vi,..., Vm) be a Boolean vector. Then, the number of zero elements of the vector V is denoted by Z(V), i.e. Z(V) = #{г | Vi = 0, 1 ^ г ^ m}.
Definition 4. Let 21 = {akn ...a\ | 1 ^ к ^ N} be a bundle of n-ary operators. If every n-ary Boolean function f can be represented as
f{xi, ...,xn) = Ciah ... a\{xn-...-xi)®-... af {xn-.. .-xi) (2.5)
where С = (C\,... ,CN) is a Boolean vector, then the bundle 21 is called base and the value L<%(f) = N — Z{C) is called the complexity of the representation of Boolean function f by images of operators from the bundle 21.
Proposition 3. If 21 is a base bundle of n-ary operators, and is a matrix, representing the bundle 21, then is поп-degenerate. Moreover, for arbitrary n-ary Boolean functions f it holds that L^(f) = N — Z(VfM^1).
Proof. Let 21 = {ak ... | 1 ^ к ^ N} be a base bundle of n-ary operators. For each j, 1 ^ j ^ N, take a Boolean vector C-7 = (C^,..., C3N) and a Boolean function fj(xn,..., x\) = (xn ф a■ ... ■ (x\ ® crf--^1) such that fj (xn,..., = C{ fl^ ... cij (xn •...•Ж1)ф'''ф C3Na% ... fl^ (xn ■... • Ж1). By Definition 4, such a vector C-7 exists for every 1 ^ j ^ N.
Let the function g(xn,..., x\) = xn-.. .-x\. By Proposition 1, the matrix Msg is the identity NxN matrix. Thus, from Proposition 2 it follows that
vfi = C(v(ai ... а}) Ф • • • Ф CjNv{a% ... af) or Vfj = CjMя in vector form. Consider a matrix whose rows are vectors Vfx,..., VfN. This is exactly the matrix Msg since fj satisfies (2.2). Let M be a matrix whose rows are vectors C1,..., CN. Then we have the matrix equality Msg = MMa. Since Msg is the identity matrix, both matrices M and Мд are necessarily non-degenerate.
Let / be an arbitrary n-ary Boolean function and (2.5) hold. Then ¿si(/) = N — Z(C). As shown above, (2.5) can be represented in vector form as Vf = CMSince Мд is non-degenerate, the inverse matrix M^1 exists. So С = VfM~l and L^/) = N- Z{VfM~l). □
Definition 5. The complexity of an n-ary Boolean function f in the set К of base bundles of n-ary operators is the value L^(/), which is defined as LkU) = тт{Ы/) I 21 € K).
Proposition 4. For arbitrary n-ary Boolean function f and every set К of base bundles of n-ary operators the value Lx(f) can be calculated by the expression LK(f) = N - max{Z(VfM~l) \ 21 € K}.
Proof. Let the matrices Мд and M^ represent the same base bundle 21 of n-ary operators, and let / be arbitrary n-ary Boolean function. According to Proposition 3, the expression (2.5) in vector form can be represented as
Vf = CMy, or Vf = C'M.depending on the choice of the representing matrix. Since the matrices M<% and M^ differ from each other only by the permutation of the rows, the vectors C and C' also differ in the same permutation of their elements. Thus, Z(C) = Z(C') and, consequently, L<n(f) does not depend on the choice of the representing matrix. The rest of the proof follows directly from Definition 5 and Proposition 3. □
Definition 6. For a given bundle 21 = {a^... a\ | 1 ^ k ^ N}, generated by pair, its extension E% is a set of bundles E% = {21} U | 1 ^ j ^ N}, where iBJ = {a^ ... a\ | 1 ^ k ^ N, k / j} U {bn ... bi} and bn ... b\ is an
1V
operator such that v(bn ... bi) = ^ ... a\).
k=1
By Proposition 3.10 of [6], the operator bn ... b\ from Definition 6 always exists and is uniquely determined by a pair-generated bundle 21. By Theorem 3.17, in [6] all bundles in including 21 itself, are the base bundles. It is also true for n = 0, since 21 = {0} and E^ = {21} for this case.
Definition 7. The set of all pair-generated bundles ofn-ary operators will be called the class of pair-generated bundles ofn-ary operators and will be denoted as H(ra). The set ExH^ = (J E^ will be called the extended class
sieH<n)
of pair-generated bundles ofn-ary operators. Proposition 5. For arbitrary n-ary Boolean function f
LExH(n)(/) = mm {N - Z{VfM~l), 1 + Z{VfM~1)}.
sieH<n)
Proof. It is known (see Expression (3) in [5]) that for every n-ary Boolean function / it holds that ¿ea(/) = mm{L^(f), N + 1 — L^(f)}. By Proposition 3, LEa(/) = min {N - Z(VfM~l), 1 + Z(VfM~1)}. This leads to the desired expression. □
Let S be a set of 2x2 Boolean matrices. The set S®n is defined as S®n = {Mn(g>■ ■ -<g)Mi | Mi € S}, where <g> is Kronecker product of matrices. The set consists of exactly one lxl matrix which only element is equal to 1. The set of all non-degenerate 2x2 Boolean matrices will be denoted as KRO2.
Proposition 6. Let 21 € H^ be a pair-genera,ted bundle of n-ary operators. Then there exists a matrix M € KRofra such that M represents 21. And vice versa, for every matrix M € KRofra there exists 21 € H^ such that M represents 21.
Proof. Let 21 be a pair-generated bundle and bn.. .b\ and cn ... ci be its generators. By induction on m, let us show that if 2lTO is the bundle
generated by the pair bm...b\ and cTO...ci, then there exists a matrix Mam € KRofm representing 2lm.
The basis of induction is obvious, since 2lo = {0}, v(0) = (1), M<%0 = (1), and KRof0 = {(1)}.
Let m > 0 and 2lTO = {a^ ... a\ | 1 ^ k ^ 2m} be generated by the pair bm ... b\ and cm ... ci, such that ak = bj if ak = 0 and ak = cj if ak = 1. {■v(akm ■■■ af) | 1 < k < 2m~1} = {u(bm) eg) ^a^... af) | 1 < k < 2™"1}, since cr^j = 0 whenever 1 ^ k ^ 2m_1. This means that the 2m_1x2"7' matrix Mo, whose rows are the vectors ... a}),..., v(a... afm ), can be expressed as Mo = v(bm) ®M2lm_1 if the vector v(bm) is considered as 1x2 matrix. Similarly, the 2m~lx2m matrix M\ whose rows are exactly the vectors v(am +1 • • • aj™ +1),..., v(a2m ■ ■ ■ af™)) can be expressed as M\ = v(cm) <g> M2im_1. Thus, the 2mx2m matrix, consisting of the rows of the matrices Mo and Mi, represents the bundle 2lTO and can be denoted by M^. Moreover, = M*®M2tm_1, where M* is the 2x2 matrix, whose rows are v(bm) and v(cm). Since bm and cm must be different (otherwise the set 21 m contains less than 2m elements), the vectors v(bm) and v(cm) are also different and non-zero. This means that M* is non-degenerate and, thus, belongs to Krc>2. Hence, by the induction hypothesis, Mam € KRof"1.
Since 21 = 2lra, we have a matrix M € KRof"-, which represents 21.
Conversely, let M = Mn <g> ■ ■ ■ <g> Mi, where Mj € Kro2 and
Mj=(Mj[0,0] Mj[0,1]
Mj[ 1,0] Mj[ 1,1]
By the definition of Kronecker product, the k-th row in M can be written as the vector (Mn[ak, 0] Mn[akn, 1]) eg) • • • eg) (Mi[a\, 0] Mi[a\, 0]). Since the rows of each matrix Mj are non-zero and are not equal to each other, there are unary operators bj and cj such that the first row in Mj is represented by the vector v(bj) and the second one by the vector v(cj). Moreover, bj / cj. Thus, the k-th row of the matrix M can be represented as • • -®v(ai)
where ak = bj if ak = 0 and ak = cj if ak = 1. By Definition 3 the bundle 21 = {ak ... a^ | 1 ^ k ^ N} is generated by the pair of the operators bn ... bi, cn ... ci, and the matrix M represents 21. □
Corollary 1. For every n-ary Boolean function
LExH(n)(/)= min {N-Z{ViM),l + Z{ViM)}.
Me KRof11
Proof By Proposition 5 LExH(„)(/) = min {N-Z(VfM~l), 1+Z(VfM~1)}.
sieH<n)
By Proposition 6 LExH(„)(/) = min {N — Z(VfM~l), 1 + Z(VfM~1)}.
Me KRof11
Since the set KRO2 consists of all non-degenerate 2x2 matrices, a matrix
M belongs to KRofra together with the matrix M_1. It follows that LExH(n)(/)= min {N-Z(yfM),l + Z(VfM)}. □
MeKRof"
3. Counting zeros in vectors over finite fields
In this section several notions of theory of finite field will be used. Non familiar reader can obtain missing information in [7].
Let FgS be a finite field of order qs, and let ( be its primitive element. Let £ be a linear map from finite field FgS onto its subfield Fg such that £(af3 + 5) = o£(J3) + i{5) for every aeF, and f3,5 e FgS.
Proposition 7. #{i I 1(C) = 0, o < t < qs - 2} = qs~l - 1.
Proof. For each a € ¥q, denote by Sa the set {(3 € FgS | ¿((3) = a}. Since £ is onto, every Sa is non-empty. Let us fix some 5 € Si. For each a, consider the set S'a = {a<5 + (3 \ (3 € So}. Since £(aS + (3) = a for all (3 € So, S'a C Sa for every a € Fg. As aS + (3i / aS + /?2 whenever (3i / /?2, we get S'a = Sa and #Sa = #S0. The sets Sa are pairwise distinct and together contain all elements from ¥qS. Thus, #Sa = #Fgs/^Fg = qs~l. Therefore, #{i | £{?) = 0, 0 < t < q° - 2} = #(50 \ {0}) = qs~l - 1. □
For each vector V = (Vi,..., Vn) which components belongs to the field FgS put £(V) = (£(Vi),... ,£(Vn)).
For integers t and j let us define series of maps from FgS to complex numbers as follows: Xj{C') = e~2m3t!r, where r = It is easy to see
that the map %. is a multiplicative character of finite field FgS.
Let p be a prime integer such that q = pk for some integer k. An absolute
Q fc_
trace for finite field Fg is defined by Trg(a) = ap H-----hap for all a € Fg.
It is known that for every a € ¥q the value Trg(a) belongs to Zp. Let us define a map ipe from FgS to complex numbers, which maps each element ¡3 € FgS to ipe((3) = _ it eaSy to see that the map ipe is an
additive character of finite field FgS.
Definition 8. A Gauss sum for multiplicative character %. and additive
qs —2
character tpe of finite field F qs is defined by G(xj,ipl,) = Y1 Xj( O^AO-
t=o
It is known (see theorem 5.11 in [7]) that if % . and ipe are both non trivial, then IG^x-,^)! = qS//2. It is easy to see that is non trivial for all integers j ^ 0 (mod r), and tjje is also non trivial for above defined £.
Lemma 1. Let a vector V = ((dl,..., (dN) for some integers d\,... ,d
N>
r = UJ = e2"Ki!r. Then Z(£(V)) = y-N + R(V), where R(V) is
r— 1 N
given by R(V) = ±Z G(X],A) £ ujdk-j=l k=l
Proof. The proof technique is taken from Chapter 12 of [4].
First of all, note that ¥q = {0} U {(mr | 0 ^ m ^ q - 2}, since (r is a generator of the multiplicative group of the subfield Fg. As (mr € ¥q and £ is linear, we have £((t+mr) = (mr£((*). Thus, if £((d) = 0, then there is a unique integer t such that O^t^r — 1, d = t (mod r), and £((*) = 0. Let us apply this observation to Z(£(V)) as follows.
1V r—1 1V
w)) = E1 = E E1 (3-1)
k= 1 t=0 k= 1 £(Cdk)=0 i(d=0 4=t(mod r)
The following well-known equation can be easily proved if we consider it as a geometric progression.
= Sr 'lid = t (modr) f32)
[0 if d^t (mod r). ' J
Applying this equation to (3.1), we get
r—1 N r—1 r—1 r—1 1V
mv)) = \ E E E = \ E (E <»-*) E -3dk ^) t=0 k=l j=0 j=0 t=0 k=1
Introduce the value E* as follows and, using similar transformations as in (3.1) and observing that (jj~3d = uj~3t whenever d = t (mod r), we get
qs-2 r— 1 qs-2 r-1 qs - 2 r-1
^; = Ew"Ji = E E= E^E^-dE-^ t=0 t=0 d=0 t=0 d=0 i=0 i(Ct)=0 ¿(0=0 d=t (mod r) ¿(O=0 d=t (mod r) ¿(O=0
By Proposition 7 Eq = qs~l — 1. Using the equality uf = 1, we get
qs-2 q-2qs-2 qs - 2 q-2qs-2
E) = Ew~Ji - E Ew~Ji = Ew~Ji -EE
ш-з{г-тг)
t=0 m=0 i=0 i=0 m=0 i=0
¿(0=Cmr i(Omr)=1
If 0 < j < r, the first sum is zero, as indicated in (3.2). So we have
q-2qs-2 qs-2
я; = -££*-* = a-
m=0 t=0 i=0
¿(0=1 ¿(0=1
Let v = e2m!v. Since Trg(a) takes each value from Zp k times when a runs through all values from ¥q, it follows that
E v^mr) = + £ = -1. (3.4)
m=0 a€ Fq
Split the Gauss sum G{Xj^i>) by zero and non-zero images of t.
qs-2 qs — 2 q-2 qs-2
= ExAC'WC) = ExACWC) + E E^WC4)
i=0 i=0 m=0 i=0
¿(C4)=o e(ct)=Cmr
Consider the first part of the previous equation.
eWwc4) = '¿U^ = = % t=0 t=0 t=0
¿(C4)^ ^(Ct)=o ^(Ct)=o
Now consider the second part, applying (3.4) just before the end.
q-2qs-2 q-2qs-2 q-2 qs-2
E E^ccucc4) = E = E Ew"J(i_mr)
m=0 t=0 m=0 t=0 m=0 t=0
e(Ct)=Cmr e(ct)=Cmr e(ct~mr)=i
q-2 qs — 2 qs-2 q-2 qs-2
= y^ Ew_Ji = E E = - Ew"Ji = ^
m=0 t=0 t=0 m=0 t=0 ^
i(Ct)=i i(Ct)=i ^(Ct)=!
Putting it all together, we have G{Xj^i>) = ^rjE* and
r—1
t=o y y
e(C*)=o
Recall that this is true only for 0 < j < r. From (3.3) it follows that
r—1 1V r— 1 r—1 JV
zw»o) =; (E!) E1 + 7 E(E •"-*) E =
t=0 fc=l j = l t=0 fc=l i(Ct)=o ¿(0=0
/V F* 1 r-1 E* N as~l 1 1 r-1 N
—^y = ^—— n+- y G (x; A) y uJdk
to — 1 r^q-l^ qs - 1 or ^ VAj' ^
y i=i y fc=i y y j=l k=1
This completes the proof. □
Lemma 2. Let V = ((dnl,... ,(d"q)<g>- • -^(C^11, • • • ,(dlq) for some integers
(hi,---, (hq, -■-, dnl, -■-, dnq, r = UJ = e2m/r. Then
r— 1 n
where R(V) = ^ £ GiXp^t) II {ujdtl H-----hwJ(it«) . Moreover, if r is
j=i t=l
prime and for every t, 1 ^ t ^ n, among the numbers dti, ■ ■ ■, dtq there are incomparable modulo r, then R(V) = O ((<? — 2 + 2 cos f)n) = o(qn).
Proof. The number of elements in the vector V is equal to qn. Let Vk denote the fc-th element in V. Each integer k in the range 1 ^ k ^ qn can be
uniquely represented as k = l+(kn—l)qn~1+(kn-i — l)qn~2-\-----\-(ki — l)q°,
where 1 ^ kj ^ q, 1 ^ j ^ n. By the definition of tensor product <S>, Vk = (dnkn ■ ■■■■ (dlki, i.e. Vk = (d«kn+-+dik1 _ Qn the other hand,
Y\ [ujjdtl + • • • + ujdt* j = £ UJDk t=1 k=1
where Dk = uj3dnkn ■... ■ uj3dlki = w^'1™^^ referring to the previous
representation of k. After this observation, the first part of Lemma 2 is essentially Lemma 1, slightly reformulated.
Now consider the case when r is a prime integer, and evaluate the value of |E(V)|. By Theorem 5.11 in [7], \G(Xj^n)\ = y/Wi since ipe and Xj are nontrivial characters if 0 < j < r — 1.
Consider the value of (w-7^1 H-----h w-"^91. Without loss of generality, let
dti and dt2 be incomparable modulo r. Thus, denoting d* = dt2 — dti,
\u)jdtl+--- + u)jdt«\ = \u)jdtl\-\l + u)jd* +---+U)j^~dtl) \ ^ \l + u)jd*\+q-2
As u)jd* = e27Tijd*/r, we have
|l+c^*| = ^(l+cos^)2 + sin2^l = ^2 + 2cos^l=2|cos2fl|
Since j and d* are relatively prime with r, we have |cos^-| < 1 and, moreover, |cos^-| ^ cos^, which completes the proof. □
complexity lower bound of extended operator forms 137 4. Lower bound in the class of extended operator forms
Theorem 1. Let p = 2s — 1 be a Mersenne prime, £ be a linear map from finite field F2s onto F2, ( be a primitive element of ¥2°, and n-ary Boolean function f is represented by a vector Vf = £ ((1, C)®"-). Then
(4-1)
Proof Let M € KRof™. Then M = Mn <g> • • • <g> Mi where Mj € Kro2.
VfM = £ ((1, C fjn) M = £ ((1, C fjn(Mn eg) • • • eg) Mi)) = ¿(((l,C)Mra)<g>...<g>((l,C)Mi))
Since Kro2^{(; ?),(oO,(?o),(lo),(?í)}
(1,0m, e {(1,C),(1 + c,c),(1,1 + c),(c,1),(c,1 + c),(1 + c,1)}- As c is a
generator of the multiplicative group of the finite field F2s there exists an integer t such that ! + ( = (* and 1 < t < p. Recall also that 1 = Since 0, 1, and t are incomparable modulo p, we can apply Lemma 2, which gives us the following: Z{VfM) = + o{2n) = - 2ra + o{2n) for
every M € KRof™.
By Corollary 1, LExH(n)(/) = min {2n - Z(VfM), 1 + Z(VfM)}.
Me KRof11
Thus, LExhW (/) = min { (I + 2» - o(2n), (l - ¿ W + 1 + 0(2»)} Me KRofnl-v J v J J
which leads us to expression (4.1). □
Note that the largest currently known Mersenne prime is 28258"33 — 1 [1]. From Theorem 1 it follows that there exists an n-ary Boolean function
/, such that LExH(„)(/) > - 282B89!,34_2) ~ 0This is asymptotically stronger than the lower bound of the form LExH(„)(/) > — y^) 2n, previously obtained in [5].
Corollary 2. If the sequence of Mersenne primes is infinite then for every e > 0 there exist an n-ary Boolean function f, such that
Wo(/)> Q-e) 2ra — o(2n).
Proof Given e > 0 take a Mersenne prime p such that p > Since the sequence of Mersenne primes is infinite, such p exists. Thus, ^ < e, and using Theorem 1, we obtain the desired result. □
5. Conclusion
In this paper we have proposed a general approach to obtain lower bounds of complexity in a certain class of polynomial forms of Boolean functions. Lemma 6 and lemma 8 in [2] can be considered as a special case of lemma 1 and lemma 2 of this work. As showed in [2] (see theorems 1 and 2) lower bounds in [8; 10] can be also obtained as a consequences of lemma 1 of this work.
References
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Aleksandr Baliuk, Candidate of Sciences (Physics and Mathematics), Irkutsk State University, 1, K. Marx st., Irkutsk, 664003, Russian Federation tel.: (3952)242210, e-mail: [email protected].
Received 09.09.19
Нижняя оценка сложности булевых функций в классе расширенных операторных форм
А. С. Балюк
Иркутский государственный университет, Иркутск, Российская Федерация
Аннотация. Полиномиальные представления булевых функций активно исследуются в связи с применением в теории кодирования и для синтеза схем цифровых устройств, начиная с основопологающей работы Мюллера. Операторный подход к полиномиальным представлениям предложенный в работах Винокурова позволил, с одной стороны, единообразно описать все известные виды полиномиальных форм булевых функций, с другой стороны, обобщить их на случай разложений по образом нечетных функций, отличных от конъюнкции.
При исследовании полиномиальных и, в общем случае, операторных форм один из главных вопросов — это получение оценок сложности представления булевых функций в различных классах форм. Верхние оценки сложности фактически представляют собой алгоритмы минимизации булевых функций в том или ином классе форм.
Нижние оценки сложности можно разделить на два вида: комбинаторные и эффективные. Комбинаторные оценки позволяют доказать существование булевых функций, имеющих высокую сложность, без нахождения явного вида этих функций. Эффективные же нижние оценки основаны на конструировании в явном виде булевых функций, имеющих высокую сложность в том или ином классе форм.
В настоящей работе с использованием алгебраического расширения конечного поля порядка 2 получена нижняя оценка сложности булевых функций в классе расширенных операторных форм. Данная оценка усиливает ранее известные оценки для данного класса операторных форм и будет являться асимптотически оптимальной в случае, если последовательность простых чисел Мерсенна бесконечна.
Ключевые слова: булевы функции, нижние оценки сложности, расширение конечного поля, простые числа Мерсенна.
Список литературы
1. А000043 - OEIS // The On-Line Encyclopedia of Integer Sequences. URL: https://oeis.org/A000043 (дата обращения: 24.08.2019)
2. Балюк А. С., Зинченко А. С. Нижние оценки сложности поляризованных полиномов над конечными полями // Сибирский математический журнал. 2019. Т. 60, № 1, С. 3-13. https://doi.org/10.33048/smzh.2019.60.101
3. Балюк А. С., Янушковский Г. В. Операторные полиномиальные формы функций над конечными полями // Труды IX Международной конференции «Дискретные модели в теории управляющих систем». М. : МАКС Пресс, 2015. С. 28-30.
4. Berndt В. С., Evans R. J., Williams К. S. Gauss and Jacobi sums. Toronto : John Wiley & Sons Inc., 1998. 600 p.
5. Францева А. С. Сложность представлений булевых функций в классах расширенных двупорожденных операторных форм // Сибирские электронные математические известия. 2019. Т. 16. С. 523-541. https://doi.org/10.33048/semi.2019.16.034
6. Избранные вопросы теории булевых функций / под ред. С. Ф. Винокурова, Н. А. Перязева. М. : Физматлит, 2001. 192 с.
7. Лидл Р., Нидеррайтер Г. Конечные поля : пер. с англ. М.: Мир, 1988. Т. 1. 430 с.
8. Маркелов Н. К. Нижняя оценка сложности функций трехзначной логики в классе поляризованных полиномов // Вестник Московского университета. Сер. 15, Вычислительная математика и кибернетика. 2012. Вып. 3, С. 40-45.
9. Muller D. Е. Application of Boolean algebra to switching circuit design and to error detection // IRE Trans. Electron. Comput. 1954. Vol. EC—3, Issue 3. P. 6-12. https://doi.org/10.1109/IREPGELC.1954.6499441
10. Перязев H. А. Сложность булевых функций в классе полиномиальных поляризованных форм // Алгебра и логика. 1995. Т. 34, № 3. С. 323-326.
11. Селезнева С. Н. Верхняя оценка длины функций над конечным полем в классе псевдополиномов // Журнал вычислительной математики и математической физики. 2017. Т. 57, № 5. С. 899-904. https://doi.org/10.7868/S0044466917050118
12. Винокуров С. Ф. Смешанные операторы в булевых функциях и их свойства. Иркутск : Иркутский университет, 2000. 36 с. (Дискретная математика и информатика ; вып. 12).
Александр Сергеевич Балюк, кандидат физико-математических наук, доцент, Институт математики, экономики и информатики, Иркутский государственный университет, Российская Федерация, 664003, г. Иркутск, ул. К. Маркса, 1, тел.: (3952)242210, e-mail: [email protected]
Поступила в редакцию 09.09.19
