UDC 519.95

MATHEMATICS

G. S. PLESNEVICH

ON LARGE INTERVALS OF A RANDOM BOOLEAN FUNCTION

(Presented by Academician S. L. Sobolev, January 4, 1968)

As is well known, a Boolean function (f) of (n) variables may be regarded as the characteristic function of the corresponding subset (N_f) of the set (E_n) of vertices of the (n)-dimensional unit cube ((^1)). A (k)-dimensional face (a \subset E_n) of the (n)-dimensional cube (otherwise, an interval of length (k), or a (k)-interval) is called an interval of the Boolean function (f) if (N_f \supset a).

With each vertex (x \in E_n) we associate a random variable (\omega_x), taking the values 0 and 1 with equal probabilities, i.e.
[
P(\omega_x=0)=P(\omega_x=1)=\frac12 .
]
Suppose that the random variables (\omega_x), (x \in E_n), are mutually independent. We shall call a realization of the collection ({\omega_x}), (x \in E_n), of these random variables a random Boolean function (of (n) variables). Thus, the probability that a random Boolean function coincides with an arbitrary fixed Boolean function is equal to (2^{-2^n}).

Introduce the random variables (\xi_{n,k}) and (\zeta_n), expressing respectively the number of (k)-intervals and the maximum of the lengths of intervals of a random Boolean function depending on (n) variables.

In the present note it is shown that, under certain conditions, the distribution of the random variable (\xi_{n,k}) is approximated by a Poisson distribution. As a consequence of this fact we obtain an asymptotic formula for the mean value of the random variable (\zeta_n) (in other words, for the mean dimension of a Boolean function depending on (n) variables).

Theorem 1. Let (n \to \infty) and (k \to \infty) in such a way that the quantity
[
\lambda=\lambda(n,k)=\binom{n}{k}\cdot 2^{\,n-k-2^k}
]
remains bounded. Then, for any fixed (r),
[
P(\xi_{n,k}=r)=\frac{\lambda^r}{r!}e^{-\lambda}+o(1).
]

Theorem 2. Let (\chi(n)=\min{k:\lambda(n,k)\leq 1}). Then
[
M\zeta_n=\chi(n)-e^{-\lambda(n,\chi(n)-1)}-e^{-\lambda(n,\chi(n))}+o(1),\qquad n\to\infty .
]

Remark 1. It is easy to show that (\chi(n)) is equal to ([\log_2 n]) or ([\log_2 n]+1) (depending on (n)).

Remark 2. If (1\leq \xi_{n,k}\leq 2k+1), then (\zeta_n=k). Consequently, in this case (\xi_{n,k}) is in fact equal to the number of largest intervals of the random Boolean function. (Indeed, if (\zeta_n>k), then the random Boolean function has at least one interval of length (k+1), and consequently at least (2(k+1)) intervals of length (k).)

It is known that the (n)-dimensional cube (E_n) contains exactly
[
m=\binom{n}{k}\cdot 2^{\,n-k}
]
distinct (k)-intervals. With each interval (a) associate the random variable (X_a), equal to one when (a) is an interval of the random Boolean function, and equal to zero otherwise. Then it is obvious that
[
\xi_{n,k}=\sum X_a,
\tag{1}
]
where the sum is taken over all (k)-intervals (a) in (E_n).

Since (MX_\alpha = 2^{-2k}), by (1) we have (M\xi_{n,k}=m\cdot 2^{-2k}=\lambda(n,k)). In proving Theorem 1 we shall use the following formula, valid for any random variable (Y) taking nonnegative integer values:

[
P(Y=r)=\frac{1}{r!}\sum_{s=0}^{\infty}(-1)^s\frac{M^{(s+r)}(Y)}{s!}.
\tag{2}
]

In this formula (M^{(t)}(Y)) denotes the factorial moment of order (t) of the random variable (Y), i.e.
[
M^{(t)}(Y)=M[Y(Y-1)\ldots(Y-t+1)],
]
if (t\geq 1), and (M^{(0)}(Y)=1); here we put (M^{(t)}(Y)=0) if all possible values of (Y) are less than (t). The validity of formula (2) is easily established. Expanding in a Taylor series at (z=1) the (r)-th derivative (f^{(r)}(z)) of the generating function (f(z)) of the random variable (Y), we obtain

[
f^{(r)}(z)=\sum_{s=0}^{\infty}\frac{f^{(r+s)}(1)}{s!}(z-1)^s
=\sum_{s=0}^{\infty}\frac{M^{(s+r)}(Y)}{s!}(z-1)^s .
\tag{3}
]

Moreover,

[
\left|\sum_{s=t}^{\infty}\frac{M^{(s+r)}(Y)}{s!}(z-1)^s\right|
\leq
\frac{f^{(t+r)}(\theta |z-1|)}{t!}|z-1|^t
\quad (0\leq \theta\leq 1).
\tag{4}
]

Substituting (z=0) in (3) and taking into account that (f^{(r)}(0)=r!P(Y=r)), we obtain (2). From (4), also by substituting (z=0), we obtain the estimate

[
\left|\sum_{s=t}^{\infty}\frac{M^{(s+r)}(Y)}{s!}(-1)^s\right|
\leq
\frac{M^{(t+r)}(Y)}{t!}.
\tag{5}
]

In proving Theorem 1, we may assume that (k<\log_2 n+1). Indeed,

[
P(\xi_{n,k}=0)=P\bigl(\pi{X_\alpha=0}\bigr)
=1-P\bigl(U{X_\alpha=1}\bigr)>
]

[

1-\sum P(X_\alpha=1)
=1-m\cdot 2^{-2k}
=1-\lambda(n,k).
\tag{6}
]

Therefore, if (k\geq \log_2 n+1), then (P(\xi_{n,k}=0)\to 1), (e^{-\lambda}\to 1), since obviously (\lambda\to 0). Consequently, for any (r\geq 1)

[
P(\xi_{n,k}=r)\to 0,\qquad \frac{\lambda^r}{r!}e^{-\lambda}\to 0.
]

It is clear that, in view of (2) and (5), to prove Theorem 1 it is sufficient to show that for every (s=0,1,2,\ldots)

[
M^{(s)}(\xi_{n,k})=\lambda^s+o(1),
\tag{7}
]

if (n) and (k) increase as indicated in the hypothesis of the theorem.

It is easy to verify that if (0\leq s\leq m), then

[
\sum X_\alpha\left(\sum X_\alpha-1\right)\ldots\left(\sum X_\alpha-s+1\right)
=
\sum X_{\alpha_1}X_{\alpha_2}\ldots X_{\alpha_s},
]

where the sum on the right is taken over all ordered sets of distinct (k)-intervals. Hence

[
M^{(s)}(\xi_{n,k})=\sum M\left(X_{\alpha_1}X_{\alpha_2}\ldots X_{\alpha_s}\right)
\quad (0\leq s\leq m).
\tag{8}
]

Before proving (7), let us introduce a definition. We shall call a set of intervals in (E_n) connected if, for any two intervals (\alpha) and (\beta) from this set, there exists a sequence (\alpha_1,\alpha_2,\ldots,\alpha_p) of intervals, also belonging to this set, such that (\alpha_1=\alpha), (\alpha_p=\beta), and (\alpha_{i-1}\cap\alpha_i\neq\varnothing) ((1<i\leq p)). It is not difficult to show that for a given connected (t)-element set of (k)-intervals ({\alpha_1,\alpha_2,\ldots,\alpha_t}) there су-

there exists a suitable interval (\beta) of length (kl) such that (\beta \supset \alpha_i) ((1 \leq i \leq t)). Hence it follows that there exist no more than
[
\binom{kt}{t}^{\,t}\binom{n}{kt}2^{\,n-kt}\quad (< n^{2kt}\cdot 2^n)
]
connected sets consisting of (t) intervals of length (k). An arbitrary set (a) of (k)-intervals in (E_n) decomposes into connected components (i.e., into maximal connected subsets). Denote by ([a]i) the number of all connected components of the set (a) that consist of (i) elements. If (\langle t_1,t_2,\ldots,t_s\rangle) is an arbitrary set of nonnegative integers such that
[
\sum i t_i=s,}^{s
\tag{9}
]
then, obviously, there exist no more than
[
m^{t_1} n^{2k(t_2+\cdots+t_s)}\cdot 2^{n(t_2+\cdots+t_s)}
= m^{t_1}\cdot 2^{(n+2k\log_2 n)(t_2+\cdots+t_s)}
\tag{10}
]
distinct (s)-element sets (a) of (k)-intervals in which ([a]i=t_i) ((1\leq i\leq s)). We split the sum in (8) into two partial sums, assigning to the first partial sum ((\Sigma_1)) all terms of (8) that correspond to (s)-tuples of pairwise nonintersecting (k)-intervals, and to the second partial sum ((\Sigma_2)) all the remaining terms of (8). For a given (k)-interval there exist exactly
[
\sum2^i\quad ( < (2nk)^k)}^{k-1}\binom{n-k}{i}\binom{n}{i
]
distinct (k)-intervals intersecting it. Therefore the number of terms in the sum (\Sigma_1) (coinciding with the number of all ordered (s)-tuples of pairwise nonintersecting (k)-intervals) is less than (m^s), but greater than
[
m{m-(2nk)^k}\cdots{m-(s-1)(2nk)^k}
\quad (> m^s-s^2(2nk)^k m^{s-1}).
]
Hence, and from the fact that
[
M(X_{\alpha_1}X_{\alpha_2}\cdots X_{\alpha_s})
=MX_{\alpha_1}MX_{\alpha_2}\cdots MX_{\alpha_s}=2^{-s\cdot 2^k}
]
for every term of (\Sigma_1), the following estimate for the sum (\Sigma_1) follows:
[
\lambda^s-s^2(2nk)^k\cdot 2^{-2^k}\lambda^{s-1}<\Sigma_1<\lambda^s .
\tag{11}
]

We shall now show that the sum (\Sigma_2) is infinitely small when (n) and (k) increase as indicated in the condition of Theorem 1. Let (a={\alpha_1,\alpha_2,\ldots,\alpha_s}) be an arbitrary set of (k)-intervals. Then
[
M(X_{\alpha_1}X_{\alpha_2}\cdots X_{\alpha_s})
\leq 2^{-t_1\cdot 2^k-3\cdot 2^{k-1}(t_2+\cdots+t_s)},
\tag{12}
]
where (t_i=[a]i). Indeed, if the intervals (\alpha) and (\beta) belong to different connected components of the set (a), then they have no common vertices and, consequently, (M(X\alpha X_\beta)=MX_\alpha\cdot MX_\beta). On the other hand, for any (k)-intervals (\alpha) and (\beta) one always has
[
M(X_\alpha X_\beta)\leq 2^{-3\cdot 2^{k-1}},
]
since the union of these intervals contains at least
[
2^k+2^k-2^{k-1}=3\cdot 2^{k-1}
]
vertices. Therefore the mathematical expectation of the product of random variables (X_\alpha) corresponding to distinct (k)-intervals (\alpha) from any connected set does not exceed (2^{-3\cdot 2^{k-1}}), provided only that this product contains at least two factors. Hence (12) follows.

From (10) and (12) it follows that
[
\sum_{[a]i=t_i} M(X)}X_{\alpha_2}\cdots X_{\alpha_s
\leq
\lambda^{t_1}\cdot
2^{(n+k\log_2 n-3\cdot 2^{k-1})(t_2+\cdots+t_s)} .
\tag{13}
]
(Here the sum is taken over all ordered (s)-tuples (\langle\alpha_1,\alpha_2,\ldots,\alpha_s\rangle) of distinct intervals of length (k), and such that ([a]_i=t_i), where (a={\alpha_1,\alpha_2,\ldots,\alpha_s}).) We show that the sum (13) tends to zero if the set (a) contains at least one pair of intersecting intervals. Indeed, in this case (t_1\leq s-2) and (t_2+\cdots+t_s>0), and it is sufficient

but show that (n+2k\log_2 n-3\cdot 2^{k-1}\to-\infty). Thanks to the assumption that (k<\log_2 n+1), and to the fact that (n-k-2^k<\log_2\lambda), we have

[
n+2k\log_2 n-3\cdot 2^{k-1}
= n+k(\log_2 n+3/2)+{}^3/2(-k-2^k) <
]
[
< n+(\log_2 n+1)(2\log_2 n+{}^3/2)-{}^3/2 n+{}^3/2\log_2\lambda\to-\infty.
\tag{14}
]

Obviously,

[
\Sigma_2=\sum_{[a]j=t_j}\sum M(X),}X_{a_2}\ldots X_{a_s
]

where the outer summation on the right-hand side is taken over all solutions (\langle t_1,t_2,\ldots,t_s\rangle) of equation (9) in nonnegative integers, moreover such that (t_1\le s-2). For fixed (s) there are, obviously, only finitely many such solutions. (More precisely, their number is not greater than (s^s/s!).) Hence it follows that (\Sigma_2) tends to zero. (More precisely, by (13) and (14) we have, starting from some (n):

[
\Sigma_2<\frac{s^s}{s!}\lambda^{s-2}\cdot 2^{-(1/2-\varepsilon)n},
]

where (\varepsilon) is any positive number.) Together with (11) this gives (7). Theorem 1 is proved.

Remark 3. From the proof of Theorem 1 there essentially follows the following estimate of the order of approximation of the distribution of the random variable (\xi_{n,k}) by the Poisson distribution:

[
P(\xi_{n,k}=r)-\frac{\lambda^r}{r!}e^{-\lambda}
=O\left(2^{-(1/2-\varepsilon)n}\right),
]

where (\varepsilon) is any fixed positive number.

Passing to the proof of Theorem 2, note that

[
M\xi_n=\sum_{k=1}^{n} P(\xi_n\ge k)=
]

[
=\sum_{k=1}^{\chi(n)-2} P(\xi_n\ge k)
+P(\xi_n\ge \chi(n)-1)+P(\xi_n\ge \chi(n))
+\sum_{k=\chi(n)+1}^{n} P(\xi_n\ge k).
]

Since, obviously, (\xi_n