Thin circulant matrices and lower bounds on the complexity of some Boolean operators

0.17 MB·

by  M. I. Grinchuk

Thin circulant matrices and lower bounds on the
complexity of some Boolean operators
∗
M. I. Grinchuk, I. S. Sergeev
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1
0
2

n
Abstract
a
J
0 We prove a lower bound Ω kk2+l2lN2−k+kll+2 on the maximal possible
3 weight of a (k,l)-free (that is, fr(cid:16)ee of all-ones (cid:17)k l submatrices) Boolean
×
circulant N N matrix. The bound is close to the known bound for the
]
×
C class of all (k,l)-free matrices. As a consequence, we obtain new boundsfor
C several complexity measures of Boolean sums’ systems and a lower bound
s. Ω(N2log−6N) on the monotone complexity of the Boolean convolution of
c
order N.
[

1
Keywords: complexity, circulant matrix, thin matrix, Zarankiewicz problem, mono-
v
7 tone circuit, rectifier circuit, Boolean sum, Boolean convolution.
5
5
8
1 Introduction
0
.
1
0 Hereafter, a Boolean matrix is called (k,l)-free (or thin) if it does not contain an
7
all-ones k l submatrix. In the case k = l we write simply k-free. Further, assume
1 ×
: 2 k l.
v ≤ ≤
i An N N matrix (ci,j) is circulant (or cyclic), if either ci,j = c0,(i+j)modN for
X ×
all i,j, or c = c for all i,j.
i,j 0,(i−j)modN
r
a In [2] the first author proved the existence of k-free Boolean circulant N N
×
matrices of weight1 Ω k−4N2−√3/k and obtained corollaries for the complexity2
ofBooleansums’syste(cid:16)ms3 withcircu(cid:17)lant matrices, withrespect toimplementation
∗Original text published in Russian in Diskretnyi Analiz i Issledovanie Operatsii (Discrete
analysis and operations research). 2011. 18(5), 38–53.
1Weight of a (Boolean) matrix is the number of non-zero entries in it.
2The reader can find the notions of complexity, depth, rectifier circuit, circuit of functional
elements e.g. in [4, 5].
3Boolean sum is a function of the form x1 ... xn. A system of Boolean sums with an
N N matrix (c ) is a mapping with compon∨ents ∨N c x , 1 i N.
× i,j j=1 i,j j ≤ ≤
W
1
via rectifier circuits of depth 2 or unbounded depth. Precisely, the bound for the
first measure is Ω(N2log−10N), and for the second it is Ω(N2log−12N).
In fact, the method has a potential for improvement of the above bounds,
which is of interest due to connection to the Zarankiewicz problem (the problem
is discussed in details e.g. in [6]). This potential is in application of a more
accurate bound on the cardinality of the sum of two sets in a Euclidean space
following from [8, 10].
Below, we show the existence of (k,l)-free circulant N N matrices of weight
×
Ω kk2+l2lN2−k+kll+2 . For comparison, the classic Erd¨os—Spencer result [6] states
jus(cid:16)t a slightly be(cid:17)tter bound Ωk,l N2−kk+l−l−12 in the class of all (k,l)-free matrices.
Hence, for a system of Boolea(cid:16)n sums wi(cid:17)th an appropriate circulant matrix the
following complexity bounds hold:
— Ω(N2log−6N) with respect to implementation via circuits of functional
elements4 over the basis , ;
{∨ ∧}
— Ω(N2log−5N) with respect to implementation via circuits over the basis
, or via rectifier circuits;
{∨}
— Ω(N2log−4N) with respect to implementation via depth-2 rectifier circuits.
The paper [1] considers the ratio λ(N) = max L∨(A), where L (A) is the
A L⊕(A) ∨
circuit complexity of the Boolean sums’ system with matrix A over the basis ,
{∨}
L (A) is the circuit complexity of the linear operator with matrix A over the basis
⊕
, and the maximum is taken over all Boolean N N matrices. The result of
{⊕} ×
the present paper leads to a bound λ(N) = Ω N , which in a sense
(logN)6loglogN
close to an upper bound λ(N) = O N . (cid:16) (cid:17)
logN
As another corollary, we obtain(cid:16)that(cid:17)the circuit complexity of the Boolean
convolution of order N over the basis , is Ω(N2log−6N). Specifically, this
{∨ ∧}
bound holds for the number of disjunctors (that is, -gates) in any monotone
∨
circuit computing the convolution. Some recent papers (e.g. [3, 7]) mention the
bound Ω(N3/2) as a record, though a stronger bound follows from [2] directly5.
The obtained lower bound is close to the trivial upper bound O(N2).
4
Further, we simply call them circuits.
5The bound Ω(N3/2) corresponds to the number of disjunctors in a monotone circuit (the
survey [3] is inaccurate at this point). However, the recent paper [7] declares the same bound
for the number of conjunctors ( -gates; proof is omitted there).
∧
2
2 Some properties of “rectangles”
Now, wepresent themainresult followingtheproofstrategyfrom[2]. Letk,l N,
∈
2 k l. Denote Z = N 0 . We define rectangle as an element of the set
+
≤ ≤ ∪{ }
R = (x ,...,x ,y ,...,y ) Zk+l (x = x ), (y = y ) .
k,l { 1 k 1 l ∈ + | ∀i6=j i 6 j ∀i6=j i 6 j }
Let E = (a ,...,a ,b ,...,b ) be a rectangle. Let m(E) = a +b 1 i
1 k 1 l i j
|{ | ≤ ≤
k, 1 j l denote the number of points in the rectangle E.
≤ ≤ }|
Consider the system S(E) of linear equations
x +y = x +y a +b = a +b , 1 r, u k, 1 s, v l
r s u v r s u v
{ | ≤ ≤ ≤ ≤ }
over the field R. The set of solutions constitutes a linear subspace T in
E
Rk+l. Let n(E) be its dimension. Let C(E) denote the set of rectangles
(x ,...,x ,y ,...,y ) satisfying S(E) and failing to satisfy any other equation
1 k 1 l
{ }
x +y = x +y (in [2], C(E) is called equivalence class).
r s u v
We have to estimate the number of rectangles with bounded (by a number N)
coordinates and fixed number of points. An implicit relation between the number
of rectangles and the number of points will be further established with the help
of intermediate parameter n(E). First, we will count the number of rectangles
E with a given value of n(E). Next, we will derive relations between n(E) and
m(E).
To roughly estimate the number of rectangles with bounded coordinates 0
≤
x ,...,y < N in C(E) we use the following lemma.
1 l
Lemma 1. Let N N. Then C(E) 0,...,N 1 k+l Nn(E).
∈ | ∩{ − } | ≤
Proof. The coordinates x ,...,x ,y ,...,y of a vector from T are defined by
1 k 1 l E
values of n(E) free variables. There are at most Nn(E) ways to arrange such
values, given that the vector is from C(E) 0,...,N 1 k+l.
∩{ − }
The second lemma estimates the number of classes with a given value of n(E).
(We use notation Ck for binomial coefficients.)
n
Lemma 2. Let n N. Then C(E) n(E) = n Ck+l−n.
∈ |{ | }| ≤ k2l2
Proof. The class C(E) is uniquely defined by the system S(E), which in its turn
is uniquely defined by a linearly independent subsystem of k + l n equations.
−
The number of such subsystems is bounded from above by the number of ways to
choose k +l n equations from k2l2 ones.
−
Now, we manage to obtain relations between n(E) and m(E). This piece of
proof differs from [2].
Let ξ denote the unit vector in the space Rk+l with i-th coordinate being 1
i
and other coordinates being 0.
3
Let n = n(E). For unification, let us introduce notation x = y , 1 i l.
i+k i
≤ ≤
Set
x¯ = (1,...,1,0,...,0), y¯= (0,...,0,1,...,1).
k l k l
Notice that x¯,y¯ T (regardless of E). Let T′ be the space of solutions of the
∈ E| {z } | {z } E| {z } | {z }
system
S′(E) = S(E) x = y = 0 .
1 1
∪{ }
Then dimT′ = n 2 and T = T′ + αx¯ + βy¯ α,β R (A + B hereafter
E − E E { | ∈ }
denotes the element-wise sum (Minkowski sum) of sets A and B). Write
n−2
T′ = (x ,...,x ) x = α x , 1 i k +l ,
E 1 k+l i i,j ij ≤ ≤
( (cid:12) )
(cid:12) Xj=1
(cid:12)
(cid:12)
where x ,...,x is a set of free(cid:12)variables of the system S′(E), and α are real
i1 in−2 i,j
constants. Then setting i = 1 and i = k +1 we conclude that x ,...,x is
n−1 n i1 in
a set of free variables of the system S(E), and
n
T = (x ,...,x ) x = α x , 1 i k +l ,
E ( 1 k+l (cid:12) i i,j ij ≤ ≤ )
(cid:12) Xj=1
(cid:12)
(cid:12)
where αi,n−1 = αk+j,n = 1, and αi,n(cid:12) = αk+j,n−1 = 0 for 1 i k, 1 j l.
≤ ≤ ≤ ≤
Consider a linear mapping ψ from Rk+l to the space Rn−2 with Euclidean
E
metrics and orthonormal basis e ,...,e defined by ψ : ξ n−2α e for
{ 1 n−2} E i → j=1 i,j j
any i. In particular, ψ (ξ ) = ψ (ξ ) = 0 (0 hereafter stands for the zero vector
E 1 E k+1
P
of a space if it does not lead to a misunderstanding).
Set A = ψ ( ξ ,...,ξ ), B = ψ ( ξ ,...,ξ ).
E E 1 k E E k+1 k+l
{ } { }
RecallthatthedimensiondimAofasetAinaEuclideanspaceistheminimum
of dimensions of affine subspaces containing A.
Lemma 3. A +B = m(E), dim(A +B ) = n 2.
E E E E
| | −
Proof. The first equality holds due to the following chain of equivalent transfor-
mations:
a +b = a +b
r s u v
⇐⇒
( (x ,...,x ,y ,...,y ) T = x +y = x +y )
1 k 1 l E r s u v
∈ ⇒ ⇐⇒
n n
(x ,...,x ,y ,...,y ) T = (α +α )x = (α +α )x
1 k 1 l ∈ E ⇒ r,j k+s,j ij u,j k+v,j ij
!
j=1 j=1
X X
(α +α = α +α )
j,1≤j≤n r,j k+s,j u,j k+v,j
⇐⇒ ∀ ⇐⇒
4
(α +α = α +α )
j,1≤j≤n−2 r,j k+s,j u,j k+v,j
∀ ⇐⇒
n−2 n−2
ψ (ξ +ξ ) = (α +α )e = (α +α )e = ψ (ξ +ξ ).
E r k+s r,j k+s,j j u,j k+v,j j E u k+v
j=1 j=1
X X
Thesecond equality is straightforward, since 0 A B and e ,...,e
E E 1 n−2
∈ ∩ { } ⊂
A B .
E E
∪
In the next section, we will estimate m(E).
3 The cardinality of the sum of two sets in a
Euclidean space
The following result is due to I. Ruzsa [10].
Theorem 1 (Ruzsa [10]). Let A and B be finite sets in the Euclidean space Rn
satisfying A B and dim(A+B) = n. Then
| | ≤ | |
n(n+1)
A+B n A + B .
| | ≥ | | | |− 2
Ruzsa also provided a more accurate bound
|A|−1
A+B B + min n, B i .
| | ≥ | | { | |− }
i=1
X
W.l.o.g. we can assume 0 A B throughout this section.
∈ ∩
Already, these bounds are sufficient to principally achieve results announced
in the introduction. However, the bounds are not asymptotically tight for large n.
On the contrary, the bound A+B n2/4 established in [2] is rough for small
| | ≥ ⌊ ⌋
n (though its advantage is the simplicity of the proof). The method [8] allows to
exhibit tight bounds.
Let e ,...,e beanorthonormalbasisofaEuclideanspaceEn. Following[8],
1 n
{ }
we define long simplex as a set F of the form
me m = 0,..., F k e ,...,e , (1)
{ 1| | |− }∪{ i1 ik−1}
with numbers 1,i ,...,i being pairwise different, k 0.
1 k−1
≥
The next lemma is a reformulation of the Corollary 3.8 [8].
Lemma 4. Under conditions of Theorem 1 the minimum of A + B is either
| |
A B (in this case dimA + dimB = n), or it is witnessed by a pair of long
| || |
simplices.
5
The proof can be found in [8]. It is crucial to observe that the sets A and
B delivering the minimum in the lemma satisfy the definition (1) with the same
basis and, in particular, with the same vector e .
1
Tight bounds (for any values of parameters) were not determined in [8].
Though, they can be easily derived from the lemma above.
Theorem 2. Let A,B Rn, K = A B = L, and dim(A + B) = n. We
⊂ | | ≤ | |
have:
(i) if n = K +L 2, then A+B = KL;
− | |
(ii) if n L K, then A+B L+n(K 1);
≤ − | | ≥ −
(iii) if L K n L, then
− ≤ ≤
(n L+K)(n L+K +1)
A+B (n+1)K − − ;
| | ≥ − 2
(iv) if L n K +L 3, then
≤ ≤ −
(K +L n)(K +L n 1)
A+B KL − − − .
| | ≥ − 2
Proof. In the case dimA+dimB = n, the set A+ B has the maximal possible
cardinality KL, thus, (i) follows. Therefore, in the case n < K +L 2, we may
−
assume that dimA+dimB > n.
So, by Lemma 4, it suffices to consider sets A, B being long simplices (1).
Assume w.l.o.g.
A = C D D , B = C D D ,
A A B B
∪ ∪ ∪ ∪
where
C = me m = 0,...,K s s 1 ,
A 1 A
{ | − − − }
C = me m = 0,...,L s s 1 ,
B 1 B
{ | − − − }
D = e ,...,e , D = e ,...,e , D = e ,...,e ,
{ 2 s+1} A { s+2 s+sA+1} B { s+sA+2 n}
s = D , s = D , s = D , s+s +s = n 1. Hence,
A A B B A B
| | | | | | −
A+B = C +C + (C C )+D + C +D +
A B A B A B
| | | | | ∪ | | |
C +D + D +D + D +(D D ) + D +D .
B A A B A B
| | | | | ∪ | | |
It can be verified directly that
C +C = K +L s n, (C C )+D = s(max L s , K s s),
A B A B B A
| | − − | ∪ | { − − }−
C +D = (K s s )s , C +D = (L s s )s ,
A B A B B A B A
| | − − | | − −
6
s(s+1)
D+D = , D+(D D ) = s(s +s ), D +D = s s .
A B A B A B A B
| | 2 | ∪ | | |
Summing all, we obtain
s(s+1)
A+B = (s +1)K+(s +1)L+s max L s , K s n s s . (2)
A B B A A B
| | · { − − }− − 2 −
Thus, the problem reduced to finding the minimum of the expression (2). Let s∗,
s∗, s∗ denote the values of parameters s, s , s delivering this minimum. Let us
A B A B
list restrictions on the parameters:
s+s +s = n 1, s+s K 1, s+s L 1. ( )
A B A B
− ≤ − ≤ − ∗
Consider (ii). Suppose n L K. Then
≤ −
L s K +(n s ) K K s .
B B A
− ≥ − ≥ ≥ −
Thus, minimization of (2) (with eliminated constant terms) is equivalent to max-
imization of the expression
s(s+1)
s (n+L K 1 s )+ . (3)
B B
− − − 2
For a fixed s the value of (3) grows when s decreases (and s increases ac-
A B
cordingly), since 2s < n + L K 1 and due to the fact that the function
B
− −
x(a x) monotonically grows in the interval [0, a/2]. Yet, the conditions ( ) are
− ∗
not violated. Hence, s∗ = 0.
A
Set s = n 1 s. Then, afterelimination of constant terms theexpression (3)
B
− −
reduces to
s(s+1)
((L K) n)s.
− 2 − − −
Consequently, s∗ = 0. By the assignment s = n 1 and s = s = 0 in (2), we
B A
−
derive the inequality (ii).
Let us prove (iii). Assume L K n L. Consider two cases.
− ≤ ≤
Case A. Suppose L s K s . As above, the problem reduces to maxi-
B A
− ≥ −
mization of (3). Note that for a fixed s the value of (3) grows with decreasing of
B
s (and corresponding increasing of s), and the conditions ( ) are not violated.
A
∗
Therefore, either s∗ L K and s∗ = 0, or s∗ = L K +s∗.
B ≤ − A B − A
In the former subcase, assign s = n 1 s . Then, after elimination of
B
− −
constant terms the expression (3) reduces to
s (2(L K) 1 s ),
B B
− − −
hence, s∗ L K 1, L K .
B ∈ { − − − }
7
In the latter subcase, assign s = L K +s and s = n 1 L+K 2s .
B A A
− − − −
Then, the expression (3) has the form
s (s +L K n).
A A
− −
The second factor is s n, and so it is negative. Consequently, s∗ = 0, and
B − A
s∗ = L K follows as well, as in the previous subcase.
B −
Case B. Suppose L s K s . Then,
B A
− ≤ −
s+s = n 1 s L 1 s K 1 s K 1.
A B B A
− − ≤ − − ≤ − − ≤ −
So, only the first of conditions ( ) is essential. Here, minimization of (2) is equiv-
∗
alent to maximization of the expression
s(s+1)
s (n L+K 1 s )+ . (4)
A A
− − − 2
For a fixed s the value of (4) grows, when s increases and s accordingly de-
A B
creases, thus, s∗ = L K +s∗. That is the very situation already discussed in
B − A
the second subcase of the case A.
Via assignment s = 0, s = L K, s = n 1 L+K in (2), we obtain the
A B
− − −
inequality (iii) (the assignment is in a sense correct also in the case L K = n).
−
Now, turn to (iv). Assume L n K +L 3. Again, consider two cases.
≤ ≤ −
Case A. Suppose L s K s . In this case, the latter of conditions ( )
B A
− ≥ − ∗
follows from the second:
s+s s+s +L K L 1.
B A
≤ − ≤ −
Again, the problem is to maximize the expression (3). Observe that for a fixed
s the value of (3) grows when s decreases (and s correspondingly increases),
B A
and conditions ( ) are not violated. Hence, s∗ = s∗ L+K (it is the minimal
∗ A B −
possible value of s for a fixed s ).
A B
Under the assignment s = n+L K 1 2s and elimination of constant
B
− − −
terms, the expression (3) reduces to
s (s n L+K).
B B
− −
Since 2s < (L K +s )+(n s ) = n+L K, the second factor is negative
B A A
− − −
and greater than the first factor by absolute value. Consequently, the maximum
is achieved on the minimal possible value of s under the conditions ( ). Hence,
B
∗
we deduce that s∗ = n K.
B −
Case B. Suppose L s K s . In this case, the second condition in ( )
B A
− ≤ − ∗
is inessential:
s+s s+s L+K K 1.
A B
≤ − ≤ −
8
We have to maximize (4). Observe that it grows when s is fixed, s increases and
A
s decreases, and conditions ( ) are fulfilled. Thus, s∗ = L K +s∗. So, we are
B ∗ B − A
under the conditions of the already investigated case A.
Under assignment s = n L, s = n K, s = L + K 1 n in (2), we
A B
− − − −
exhibit the inequality (iv).
As follows from the proof, the bounds of the theorem are achievable.
Under the conditions of Theorem 2, define the function
n+2
ρ(K,L) = max . (5)
1≤n≤K+L−2 A+B
| |
Lemma 5. ρ(2,L) = L+2. If K 3, then
2L ≥
K +L K +L 1 2(L+2) K +L+2
ρ(K,L) = max , − , < .
KL KL 3 K(2L K +1) KL
(cid:26) − − (cid:27)
In particular, ρ(K,K) = 2(K+2).
K(K+1)
Proof. Define additionally ρ(K,L,n) = n+2 . By the definition, ρ(K,L) =
minA,B|A+B|
max ρ(K,L,n).
n
First, we need to verify that the function ρ(K,L,n) achieves its maximum at
the endpoints of intervals defined in pp. (ii)–(iv) of Theorem 2.
In the case 1 n L K, the function
≤ ≤ −
L+n(K 1) L 2K +2
ρ−1(K,L,n) = − = K 1+ −
n+2 − n+2
is evidently monotone (hereafter, we consider ρ(K,L,n) as a function of vari-
able n).
In the case L K n L, denote n′ = n (L K). Then
− ≤ ≤ − −
n′(n′ +1)+2K
ρ−1(K,L,n) = K .
− 2(n′ +L K +2)
−
The subtrahend function is convex downward for n′ 0, since it has the form
cn′(n′+1)+a with a,b,c 0. Therefore, with respect to≥the interval [0, K] it takes
n′+1+b ≥
its maximal value in the endpoints (it holds for K 3; for K = 2 the argument of
≥
the maximum lies in the interval [0, 1]). Consequently, there takes its maximum
the function ρ(K,L,n).
In the case L n K +L 3, denote n′ = n L. Then
≤ ≤ − −
2(n′ +2)K +(K n′)(K n′ 1)
ρ−1(K,L,n) = K − − − =
− 2(n′ +L+2)
n′(n′ +1)+K(K +3)
= K .
− 2(n′ +L+2)
9
We treat this case the same way as the previous one.
Thus, for K 3 we have
≥
arg max ρ(K,L,n) 1, L K, L, K +L 3, K +L 2 ,
1≤n≤K+L−2 ∈ { − − − }
arg max ρ(2,L,n) 1, L 2, L 1, L .
1≤n≤K+L−2 ∈ { − − }
Let us check that ρ(K,L,1) ρ(K,L,K +L 2). Indeed,
≤ −
3 4 1 1 K +L
ρ(K,L,1) = + = = ρ(K,L,K +L 2),
K +L 1 ≤ K +L ≤ K L KL −
−
due to the well-known inequality a2 + c2 (a+c)2, where b,d > 0.
b d ≥ b+d
Notice further that
1 1 1 K
ρ(K,L,L K) = 1+ 1+ = ρ(K,L,K +L 2).
− K L (K 1) ≤ K L −
(cid:18) − − (cid:19) (cid:18) (cid:19)
Yet,
L+1 L+2
ρ(2,L,L 1) = = ρ(2,L,L).
− 2L 1 ≤ 2L
−
Therefore, it is proved that ρ(2,L) = ρ(2,L,L) = L+2 and
2L
ρ(K,L) = max ρ(K,L,K +L 2), ρ(K,L,K +L 3), ρ(K,L,L) =
{ − − }
K +L K +L 1 2(L+2)
max , − , .
KL KL 3 K(2L K +1)
(cid:26) − − (cid:27)
Applying the simple estimation
2(L+2) L+(K +3) L L+(K +3) K K +L+2
= 2L−K+1 K+1 < ,
K(2L K +1) KL ≤ KL KL
−
the inequality ρ(K,L) < K+L+2 can be easily checked. The last statement of the
KL
lemma concerning ρ(K,K) is easy to verify.
4 Weight of thin circulant matrices
A circulant matrix is entirely defined by its one row, say, the first row. Let
c = c , 0 j N 1, denote the entries of the row, where N is the size of the
j 0,j
≤ ≤ −
matrix. For convenience, assume that the other entries satisfy c = c
i,j (i+j)modN
(that is, 1-uniform diagonals of the matrix are parallel to the secondary diagonal).
10