Adaptivity Gaps for the Stochastic Boolean Function Evaluation Problem

Abstract

We consider the Stochastic Boolean Function Evaluation (SBFE) problem where the task is to efficiently evaluate a known Boolean function f on an unknown bit string x of length n. We determine f(x) by sequentially testing the variables of x, each of which is associated with a cost of testing and an independent probability of being true. If a strategy for solving the problem is adaptive in the sense that its next test can depend on the outcomes of previous tests, it has lower expected cost but may take up to exponential space to store. In contrast, a non-adaptive strategy may have higher expected cost but can be stored in linear space and benefit from parallel resources. The adaptivity gap, the ratio between the expected cost of the optimal non-adaptive and adaptive strategies, is a measure of the benefit of adaptivity. We present lower bounds on the adaptivity gap for the SBFE problem for popular classes of Boolean functions, including read-once DNF formulas, read-once formulas, and general DNFs. Our bounds range from $\varOmega (\log n)$ to $\varOmega (n/\log n)$, contrasting with recent O(1) gaps shown for symmetric functions and linear threshold functions.

A Additional Proofs

Proof

(Proof of Theorem 1). Suppose f is a read-once DNF formula. We will prove that for unit costs and the uniform distribution, there is a non-adaptive strategy S such that $\text {cost}(f,S) \le O(\log n) \cdot \textsf{OPT}_{\mathcal {A}}(f)$.

Let m be the number of terms in f. Because each variable $x_i$ appears in at most one term, we have that $m \le n$. As a warm-up, we begin by proving adaptivity gaps for two special cases of f.

Case 1: All Terms have at Most $2\log n$ Variables. Under the uniform distribution and with unit costs, the $p_i$ are all equal, and the $c_i$ are all equal. Thus in this case, the optimal adaptive strategy described previously tests terms in increasing order of length. The adaptive strategy skips in the sense that if it finds a variable in a term that is false, it moves to the next term without testing the remaining variables in the term. Suppose we eliminate skip** from the optimal adaptive strategy, making the strategy non-adaptive. Since all terms have at most $2\log n$ variables, this increases the testing cost for any given x by a factor of at most $2\log n$. Thus the cost of evaluating f(x) for a fixed x increases by a factor of at most $2 \log n$ from an optimal adaptive strategy to a non-adaptive strategy, leading to an adaptivity gap of at most $2 \log n$.

Case 2: All Terms have More than $2\log n$ Variables Consider the following non-adaptive strategy that operates in two phases. In Phase 1, the strategy tests a fixed subset of $2 \log n$ variables from each term, where the terms are taken in increasing length order. In Phase 2, it tests the remaining untested variables in fixed arbitrary order. Since each term has more than $2\log n$ variables, the value f can only be determined in Phase 1 if a false variable is found in each term during that phase.

Say that an assignment x is bad if the value of f cannot be determined in Phase 1, meaning that a false variable is not found in every term during the phase. The probability that a random x satisfies all the tested $2 \log n$ variables of a particular term is $1/n^2$. Then, by the union bound, the probability that x is bad is at most $m/n^2 \le n/n^2 = 1/n$.

Now let us focus on the good (not bad) assignments x. For each good x, our strategy must find a false variable in each term of f, which requires at least one test per term for any adaptive or non-adaptive strategy. The cost incurred by our non-adaptive strategy on a good x is at most $2m\log n$, since the strategy certifies that $f(x)=0$ by the end of Phase 1. Therefore, the expected cost incurred by our non-adaptive strategy S is

$$\begin{aligned} \text {cost}(f, S)&\le \Pr (x \text{ good}) \cdot \mathbb {E}[\text {cost}(f,x,S) | x \text{ good}] \\&+\Pr (x \text{ bad}) \cdot \mathbb {E}[\text {cost}(f,x,S) | x \text{ bad}] \\&\le 1 \cdot 2m\log n + \frac{1}{n} \cdot n \le 3m \log n \end{aligned}$$

using the fact that $\mathbb {E}[\text {cost}(f,x,S) | x \text{ bad}] \le n$, since there are only n tests, with unit costs.

The expected cost of any strategy, including the optimal adaptive strategy, is at least

$$\begin{aligned} \textsf{OPT}_\mathcal {A}(f)&\ge \min _{S \in \mathcal {A}} \Pr (f(x)=0) \cdot \mathbb {E}[\text {cost}(f,x,S)|f(x)=0] \ge P(f(x)=0) \cdot m \\&= (1-\Pr (f(x)=1)) \cdot m \ge (1-\Pr (x \text{ bad})) \cdot m \ge \left( 1- \frac{1}{n} \right) \cdot m \ge \frac{m}{2} \end{aligned}$$

for $n \ge 2$. It follows that the adaptivity gap is at most $6 \log n$.

Case 3: Everything Else. We now generalize the ideas in the above two cases. Let f be a read-once DNF that does not fall into Case 1 or Case 2. We can break this DNF into two smaller DNFs, $f = f_1 \vee f_2$ where $f_1$ contains the terms of f of length at most $2\log n$ and $f_2$ contains the terms of f of length greater than $2 \log n$.

Let S be the non-adaptive strategy that first applies the strategy in Case 1 to $f_1$ and then, if $f_1(x)=0$, the strategy in Case 2 to $f_2$. Since S cannot stop testing until it determines the value of f, in the case that $f_1(x)=0$, it will test all variables in $f_1$ and then proceed to test variables $f_2$.

Let $S^*$ be the optimal adaptive strategy for evaluating read-once DNFs, described above. We know $S^*$ will test terms in non-decreasing order of length since all tests are equivalent. So, like S, $S^*$ tests $f_1$ first and then, if $f_1(x)=0$, it continues to $f_2$. It follows that we can write the expected cost of S on f as

$$\begin{aligned} \mathbb {E}[\text {cost}(f,x,S)] = \mathbb {E}[\text {cost}(f_1,x,S_1)] + \Pr (f_1(x)=0) \cdot \mathbb {E}[\text {cost}(f_2, x, S_2)|f_1(x)=0] \end{aligned}$$

where $S_1$ is the first stage of S, where $f_1$ is evaluated, and $S_2$ is the second stage of S, where $f_2$ is evaluated. Notice that, by the independence of variables, $\mathbb {E}[\text {cost}(f_2, x, S_2)|f_1(x)=0] = \mathbb {E}[\text {cost}(f_2, x, S_2)]$. We can similarly write the expected cost of $S^*$ on f. Then the adaptivity gap is

$$\begin{aligned} \frac{\textsf{OPT}_\mathcal {N}(f)}{\textsf{OPT}_\mathcal {A}(f)} \le \frac{\mathbb {E}[\text {cost}(f_1,x,S_1)] + \Pr (f_1(x)=0) \cdot \mathbb {E}[\text {cost}(f_2, x, S_2)]}{\mathbb {E}[\text {cost}(f_1,x,S^*_1)] + \Pr (f_1(x)=0) \cdot \mathbb {E}[\text {cost}(f_2, x, S^*_2)]} \end{aligned}$$

(6)

where $S^*_1$ is $S^*$ applied to $f_1$ and $S^*_2$ is $S^*$ beginning from the point when it starts evaluating $f_2$.

Using the observation that $(a+b)/(c+d) \le \max \{a/c, b/d\}$ for positive real numbers a, b, c, d, we know that

$$\begin{aligned} (6) \le \max \left\{ \frac{\mathbb {E}[\text {cost}(f_1,x,S_1)]}{\mathbb {E}[\text {cost}(f_1,x,S^*_1)]}, \frac{\mathbb {E}[\text {cost}(f_2,x,S_2)]}{\mathbb {E}[\text {cost}(f_2,x,S^*_2)]} \right\} = O(\log n) \end{aligned}$$

where the upper bound follows from the analysis of Cases 1 and 2.

Proof

(Proof of Theorem 2). Suppose f is a read-once DNF formula. For unit costs and the uniform distribution, we will show that $\textsf{OPT}_\mathcal {N}(f) \ge \varOmega (\log n) \cdot \textsf{OPT}_\mathcal {A}(f)$.

For ease of notation, assume $\sqrt{n}$ is an integer. Consider a read-once DNF f with $\sqrt{n}$ terms where each term has $\sqrt{n}$ variables. By examining the number of tests in each term, we can write the optimal adaptive cost as

$$\begin{aligned} \textsf{OPT}_\mathcal {A}(f) \le \sqrt{n} \sum _{i=1}^{\sqrt{n}} \frac{i}{2^i} \le \sqrt{n} \sum _{i=1}^{\infty } \frac{i}{2^i} = 2 \sqrt{n}. \end{aligned}$$

The key observation is that, within a term, the adaptive strategy queries variables in any order since each variable is equivalent to any other. Then the probability that the strategy queries exactly $i \le \sqrt{n}$ variables is $1/2^i$.

Next, we will lower bound the expected cost of the optimal non-adaptive strategy

$$\begin{aligned} \textsf{OPT}_\mathcal {N}(f)&= \min _{S \in \mathcal {N}} \mathbb {E}_{x \sim \{0,1\}^n}[\text {cost}(f,x,S)] \\&\ge \min _{S \in \mathcal {N}} \Pr (f(x)=0) \mathbb {E}[\text {cost}(f,x,S)|f(x)=0] \end{aligned}$$

where $x \sim \{0,1\}^n$ indicates x is drawn from the uniform distribution. First, we know $\Pr (f(x)=0) \ge .5$. To see this, consider a random input $x \sim \{0,1\}^n$. The probability that a particular term is true is $1/2^{\sqrt{n}}$ so the probability that all terms are false (i.e., $f(x)=0$) is

$$\begin{aligned} \left( 1-\frac{1}{2^{\sqrt{n}}} \right) ^{\sqrt{n}} = \left( \left( 1-\frac{1}{2^{\sqrt{n}}} \right) ^{2^{\sqrt{n}}} \right) ^{\sqrt{n}/2^{\sqrt{n}}} \ge \left( \frac{1}{2e}\right) ^{\sqrt{n}/2^{\sqrt{n}}} \ge .5 \end{aligned}$$

where the first inequality follows from the loose lower bound that $(1-1/x)^x \ge 1/(2e)$ when $x \ge 2$ and the second inequality follows when $n \ge 8$. Second, we know

$$\begin{aligned}&\mathbb {E}[\text {cost}(f,x,S)|f(x)=0] \\&\ge \Pr (\text {one term needs} \varOmega (\log n) \text {tests}|f(x)=0) \cdot \frac{\log _4 n}{2} \cdot \frac{\sqrt{n}}{2} \end{aligned}$$

where we used the symmetry of the terms to conclude that if any term needs $\varOmega (\log n)$ tests to evaluate it then any non-adaptive strategy will have to spend $\varOmega (\log n)$ on half the terms in expectation.

All that remains is to lower bound the probability one term requires $\varOmega (\log n)$ tests given $f(x)=0$. Observe that this probability is

$$\begin{aligned}&1-(1-\Pr (\text {a particular term needs} \varOmega (\log n) \text {tests} |f(x)=0))^{\sqrt{n}} \\&\ge 1-\left( 1-\frac{1}{\sqrt{n}}\right) ^{\sqrt{n}} \ge 1- \frac{1}{e} \ge .63 \end{aligned}$$

where we will now show the first inequality. We can write the probability that a particular term needs $\log _4(n)/2$ tests given $f(x)=0$ as

$$\begin{aligned} \Pr&\left( x_1=1|f(x)=0 \right) \cdots \Pr \left( x_{\log _4(n)/2}=1|f(x)=0, x_1=\cdots =x_{\log _4(n)/2-1}=1 \right) \\&= \frac{2^{\sqrt{n}-1}-1}{2^{\sqrt{n}}-1} \cdots \frac{2^{\sqrt{n}-1-\log _4(n)/2}-1}{2^{\sqrt{n}-\log _4(n)/2}-1} \ge \left( \frac{2^{\sqrt{n}-1-\log _4(n)/2}-1}{2^{\sqrt{n}-\log _4(n)/2}-1}\right) ^{\log _4(n)/2} \\&\ge \left( \frac{1}{4} \right) ^{\log _4(n)/2} = \frac{1}{\sqrt{n}}. \end{aligned}$$

For the first equality, we use the observation that conditioning on $f(x)=0$ eliminates the possibility every variable is true so the probability of observing a true variable is slightly smaller. For the first inequality, notice that $(2^{i-1}-1)/(2^{i}-1)$ is monotone increasing in i. For the second, observe that $i \ge \sqrt{n} - \log _4(n)/2$ for our purposes and so $(2^{i-1}-1)/(2^{i}-1) \ge 1/4$ when $n \ge 16$.

Proof

(Proof of Theorem 4). Suppose f is a read-once DNF. For unit costs and arbitrary probabilities, we prove $\textsf{OPT}_\mathcal {N}(f) \ge \varOmega (\sqrt{n}) \cdot \textsf{OPT}_\mathcal {A}(f)$.

Consider the read-once DNF with $m=2\sqrt{n}$ identical terms where each term has $\ell =\sqrt{n}/2$ variables. In each term, let one variable have $1/\ell$ probability of being true and the remaining variables have a $(\ell /m)^{1/(\ell -1)}$ probability of being true. Within a term, the optimal adaptive strategy will test the variable with the lowest probability of being true first. Using this observation, we can write

$$\begin{aligned} \textsf{OPT}_\mathcal {A}(f) \le \left[ \Pr (x_1=0) \cdot 1 + \Pr (x_1=1) \cdot \ell \right] \cdot m \\ \le \left[ (1-1/\ell ) \cdot 1 + (1/\ell ) \cdot \ell \right] \cdot m \le 4 \sqrt{n} \end{aligned}$$

where $x_1$ is the first variable tested in each term. The first inequality follows by charging the optimal adaptive strategy for all $\ell$ tests in the term if the first one is true. The second inequality follows since the variable with probability $1/\ell$ of being true is tested first for $n \ge 18$ (i.e., $1/\ell < (\ell /m)^{1/(\ell -1)}$ for such n).

In order to lower bound the cost of the optimal non-adaptive strategy, we will argue that there is a constant probability of an event where the non-adaptive strategy has to test $\varOmega (n)$ variables. In particular,

$$\begin{aligned} \textsf{OPT}_\mathcal {N}(f) \ge&\min _{S \in \mathcal {N}} \Pr (\text{ exactly } \text{ one } \text{ term } \text{ is } \text{ true}) \\ \cdot&\mathbb {E}[\text {cost}(f,x,S)| \text{ exactly } \text{ one } \text{ term } \text{ is } \text{ true}]. \end{aligned}$$

By the symmetry of the terms, observe that

$$\mathbb {E}[\text {cost}(f,x,S)| \text{ exactly } \text{ one } \text{ term } \text{ is } \text{ true}] \ge \sqrt{n}/2 \cdot \sqrt{n} = n/2.$$

That is, the optimal non-adaptive strategy has to search blindly for the single true term among all $2\sqrt{n}$ terms, making $\sqrt{n}/2$ tests each for half the terms in expectation.

All that remains is to show there is a constant probability exactly one term is true. The probability a particular term is true is $(1/\ell )((\ell /m)^{1/(\ell -1)})^{(\ell -1)} = 1/m$. Since all variables are independent, the probability that exactly one of the m terms is true is

$$\begin{aligned}&m \cdot \Pr (\text {a term is true}) \cdot \Pr (\text {a term is false})^{m-1} \\ =&m \cdot \frac{1}{m} \cdot \left( 1-\frac{1}{m} \right) ^{m-1} \ge \frac{1}{2e}^{(m-1)/m} \ge \frac{1}{2e}. \end{aligned}$$

It follows that $\textsf{OPT}_\mathcal {N}(f) \ge \frac{1}{2e} \cdot \frac{n}{2} = \varOmega (n)$ so the adaptivity gap is $\varOmega (\sqrt{n})$.

Proof

(Proof of Theorem 5). Suppose f is a read-once formula. For arbitrary costs and the uniform distribution, $\textsf{OPT}_\mathcal {N}(f) \ge \varOmega (n^{1-\epsilon }/\log n) \cdot \textsf{OPT}_\mathcal {A}(f)$.

Define $W(w) := w^{1-\epsilon } \log _2(w^{1-\epsilon })$ for positive real numbers w.³ We will choose $n_\epsilon$ in terms of the function W so that $W(n) < n$ for $n \ge n_\epsilon$. First, consider the first and second derivatives of W:

$$\begin{aligned} W'(w)&= \frac{1-\epsilon }{w^\epsilon } \left( \log _2(w^{1-\epsilon }) + \frac{1}{\log 2} \right) \\ W''(w)&= \frac{1-\epsilon }{w^{1+\epsilon }} \left[ -\epsilon \left( \log _2(w^{1-\epsilon }) + \frac{1}{\log 2}\right) + \frac{1-\epsilon }{\log 2} \right] . \end{aligned}$$

For fixed $\epsilon > 0$, observe that as w goes to infinity, $W(w) < w$, $W'(w) < 1$, and $W''(w) < 0$. Therefore there is some point $n_\epsilon$ so that for all $n \ge n_\epsilon$, the slope of W is decreasing, the slope of W is less than the slope of n, and W(n) is less than n. Equivalently, $n \ge W(n) = n^{1-\epsilon }\log _2(n^{1-\epsilon })$. We will use this inequality when lower bounding the asymptotic behavior of the adaptivity gap.

For $n \ge n_\epsilon$ we construct the n-variable read-once DNF formula f as follows. First, let $r_n$ be a real number such that $n = n^{1-r_n}\log _2(n^{1-r_n})$. We know that $r_n$ exists for all $n \ge 4$ by continuity since $n^{1-0} \log _2(n^{1-0}) \ge n \ge n^{1-1} \log _2(n^{1-1})$. Let f be the read-once DNF formula with m terms of length $\ell$, where $\ell = \log _2(n^{1-r_n})$ and $m=2^{\ell }$. Thus the total number of variables in f is $m\ell = n^{1-r_n}\log _2(n^{1-r_n})=n$ as desired. We assume for simplicity that $\ell$ is an integer. The bound holds by a similar proof without this assumption.

To obtain our lower bound on evaluating this formula, we consider expected evaluation cost with respect to the uniform distribution and the following cost assignment: in each term, choose an arbitrary ordering of the variables and set the cost of testing the ith variable in the term to be $2^{i-1}$.

Consider a particular term. Recall the optimal adaptive strategy for evaluating a read-once DNF formula presented at the start of Sect. 2. Within a term, this optimal strategy tests the variables in non-decreasing cost order, since each variable has the same probability of being true. Since it performs tests within a term until finding a false variable or certifying the term is true, we can upper bound the expected cost of this optimal adaptive strategy in evaluating f as follows:

$$\begin{aligned} \textsf{OPT}_\mathcal {A}(f) \le m \cdot \left[ \frac{1}{2}\cdot (1) + \frac{1}{4}\cdot (1+2) + \ldots + \frac{1}{2^\ell }\cdot (1+\ldots +2^{\ell -1}) \right] \le m \cdot \ell . \end{aligned}$$

In contrast, the optimal non-adaptive strategy does not have the advantage of stop** tests in a term when it finds a false variable. We will lower bound the expected cost of the optimal non-adaptive strategy in the case that exactly one term is true. By symmetry, any non-adaptive strategy will have to randomly search for the term and so pay $2^{\ell }$ for half the terms in expectation.

All that remains is to show there is a constant probability exactly one term is true. The probability that a particular term is true is $1/2^{\ell }$ and so the probability that exactly one term is true is

$$\begin{aligned} m \cdot \frac{1}{2^\ell } \cdot \left( 1-\frac{1}{2^\ell }\right) ^{m-1} \ge \frac{m}{2^\ell } \cdot \left( \frac{1}{2e} \right) ^{(m-1)/2^\ell } \ge \frac{1}{2e} \end{aligned}$$

where the last inequality follows since $m=2^\ell$. Then the expected cost $\textsf{OPT}_\mathcal {N}(f)$ of the optimal non-adaptive strategy is at least

$$\begin{aligned} \Pr (\text{ exactly } \text{ one } \text{ term } \text{ is } \text{ true}) \cdot 2^\ell \cdot \frac{m}{2} = \varOmega (m \cdot 2^\ell ) = \varOmega (m \cdot n^{1-r_n}) \ge \varOmega (m \cdot n^{1-\epsilon }) \end{aligned}$$

where we used that $2^\ell = n^{1-r_n}$ and $n^{1-r_n} \log _2(n^{1-r_n}) = n \ge n^{1-\epsilon } \log _2(n^{1-\epsilon })$ since $n \ge n_\epsilon$. It follows that the adaptivity gap is $\varOmega (n^{1-\epsilon }/\log n)$.