Depth-Independent Lower bounds on the Communication Complexity of Read-Once Boolean Formulas

📝 Abstract

We show lower bounds of $\Omega(\sqrt{n})$ and $\Omega(n^{1/4})$ on the randomized and quantum communication complexity, respectively, of all $n $-variable read-once Boolean formulas. Our results complement the recent lower bound of $\Omega(n/8^d)$ by Leonardos and Saks and $\Omega(n/2^{\Omega(d\log d)})$ by Jayram, Kopparty and Raghavendra for randomized communication complexity of read-once Boolean formulas with depth $d $. We obtain our result by “embedding” either the Disjointness problem or its complement in any given read-once Boolean formula.

💡 Analysis

We show lower bounds of $\Omega(\sqrt{n})$ and $\Omega(n^{1/4})$ on the randomized and quantum communication complexity, respectively, of all $n $-variable read-once Boolean formulas. Our results complement the recent lower bound of $\Omega(n/8^d)$ by Leonardos and Saks and $\Omega(n/2^{\Omega(d\log d)})$ by Jayram, Kopparty and Raghavendra for randomized communication complexity of read-once Boolean formulas with depth $d $. We obtain our result by “embedding” either the Disjointness problem or its complement in any given read-once Boolean formula.

📄 Content

A read-once Boolean formula f : {0, 1} n → {0, 1} is a function which can be represented by a Boolean formula involving AND and OR such that each variable appears, possibly negated, at most once in the formula. An alternating AND-OR tree is a layered tree in which each internal node is labeled either AND or OR and the leaves are labeled by variables; each path from the root to the any leaf alternates between AND and OR labeled nodes. It is well known (see eg. [HW91]) that given a read-once Boolean formula f : {0, 1} n → {0, 1} there exists a unique alternating AND-OR tree, denoted T f , with n leaves labeled by input Boolean variables z 1 , . . . , z n , such that the output at the root, when the tree is evaluated according to the labels of the internal nodes, is equal to f (z 1 . . . z n ). Given an alternating AND-OR tree T , let f T denote the corresponding read-once Boolean formula evaluated by T .

Let x, y ∈ {0, 1} n and let x ∧ y, x ∨ y represent the bit-wise AND, OR of the strings x and y respectively. For f : {0, 1} n → {0, 1}, let f ∧ : {0, 1} n × {0, 1} n → {0, 1} be given by f ∧ (x, y) = f (x ∧ y). Similarly let f ∨ : {0, 1} n × {0, 1} n → {0, 1} be given by f ∨ (x, y) = f (x ∨ y). Recently Leonardos and Saks [LS09], investigated the two-party randomized communication complexity, denoted R(•), of f ∧ , f ∨ and showed the following. (Please refer to [KN97] for familiarity with basic definitions in communication complexity.)

In the theorem, the depth of a tree is the number of edges on a longest path from the root to a leaf. Independently, Jayram, Kopparty and Raghavendra [JKR09] proved randomized lower bounds of Ω(n/2 Ω(d log d) ) for general read-once Boolean formulas and Ω(n/4 d ) for a special class of “balanced” formulas. It follows from results of Snir [Sni85] and Saks and Wigderson [SW86] (via a generic simulation of trees by communication protocols [BCW98]) that for the read-once Boolean formula with their canonical tree being a complete binary alternating AND-OR trees, the randomized communication complexity is O(n 0.753… ), the best known so far. However in this situation, the results of [LS09,JKR09] do not provide any lower bound since d = log 2 n for the complete binary tree. We complement their result by giving universal lower bounds that do not depend on the depth. Below Q(•) represents the two-party quantum communication complexity.

Theorem 2 Let f : {0, 1} n → {0, 1} be a read-once Boolean formula. Then,

Remark:

1. Note that the maximum in Thoerem 1 and 2 is necessary since for example if f is the AND of the n input bits then it is easily seen that R(f ∧ ) is 1.

2. This fact is easy to observe for balanced trees, as is also remarked in [LS09].

In this section we show the proof of Theorem 2. We start with the following definition.

Definition 1 (Embedding) We say that a function g 1 : {0, 1} r × {0, 1} r → {0, 1} can be embedded into a function g 2 : {0, 1} t ×{0, 1} t → {0, 1}, if there exist maps h a : {0, 1} r → {0, 1} t and h b : {0, 1} r → {0, 1} t such that ∀x, y ∈ {0, 1} r , g 1 (x, y) = g 2 (h a (x), h b (y)).

It is easily seen that if g 1 can be embedded into g 2 then the communication complexity of g 2 is at least as large as that of g 1 . Let us define the Disjointness problem DISJ n : {0, 1} n × {0, 1} n → {0, 1} as DISJ n (x, y) = i=1,…,n (x i ∨ y i ) (where the usual negation of the variables is left out for notational simplicity). Similarly the Non-Disjointness problem NDISJ n : {0, 1} n × {0, 1} n → {0, 1} is defined as NDISJ n (x, y) = i=1,…,n (x i ∧ y i ). We shall also use the following well-known lower bounds.

Recall that for the given read-once Boolean formula f : {0, 1} n → {0, 1} its the canonical tree is denoted T f . We have the following lemma which we prove in Section 2.1.

1. Let T f have its last layer consisting only of AND gates. Let m 0 be the largest integer such that DISJ m0 can be embedded into f ∨ and m 1 be the largest integer such that NDISJ m1 can be embedded into f ∨ . Then m 0 m 1 ≥ n.

2. Let T f have its last layer consisting only of OR gates. Let m 0 be the largest integer such that DISJ m0 can be embedded into f ∧ and m 1 be the largest integer such that NDISJ m1 can be embedded into f ∧ . Then m 0 m 1 ≥ n.

With this lemma, we can prove the lower bounds on max{R(f ∧ ), R(f ∨ )} and max{Q(f ∧ ), Q(f ∨ )} as follows. For an arbitrary read-once formula f with n variables, consider the sets of leaves L odd = {leaves in T f on odd levels}, L even = {leaves in T f on even levels} At least one of the two sets has size at least n/2; without loss of generality, let us assume that it is L odd . Depending on whether the root is AND or OR, this set consisting only of AND gates or OR gates, corresponding to case 1 or 2 in Lemma 3. Then by the lemma, either DISJ √ n/2 or NDISJ √ n/2 can be embedded in f (by setting the leaves in L even to 0’s). By Fact 1 and 2, we get the lower bounds in Theorem 2.

We shall prove the first statement; th

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