Sensitivity Versus Certificate Complexity of Boolean Functions

Abstract

Sensitivity, block sensitivity and certificate complexity are basic complexity measures of Boolean functions. The famous sensitivity conjecture claims that sensitivity is polynomially related to block sensitivity. However, it has been notoriously hard to obtain even exponential bounds. Since block sensitivity is known to be polynomially related to certificate complexity, an equivalent of proving this conjecture would be showing that the certificate complexity is polynomially related to sensitivity. Previously, it has been shown that $bs(f) \le C(f) \le 2^{s(f)-1} s(f) - (s(f)-1)$. In this work, we give a better upper bound of $bs(f) \le C(f) \le \max \left( 2^{s(f)-1}\left( s(f)-\frac{1}{3}\right) , s(f)\right)$ using a recent theorem limiting the structure of function graphs. We also examine relations between these measures for functions with 1-sensitivity $s_1(f)=2$ and arbitrary 0-sensitivity $s_0(f)$.

This work was supported by the European Union Seventh Framework Programme (FP7/2007-2013) under projects RAQUEL (Grant Agreement No. 323970) and ERC Advanced Grant MQC, and the Latvian State Research Programme NeXIT project No. 1.

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

1 Introduction

Sensitivity and block sensitivity are two well-known combinatorial complexity measures of Boolean functions. The sensitivity of a Boolean function, s(f), is just the maximum number of variables $x_i$ in an input assignment $x=(x_1, \ldots , x_n)$ with the property that changing $x_i$ changes the value of f. Block sensitivity, bs(f), is a generalization of sensitivity to the case when we are allowed to change disjoint blocks of variables.

Sensitivity and block sensitivity are related to the complexity of computing f in several different computational models, from parallel random access machines or PRAMs [7] to decision tree complexity, where block sensitivity has been useful for showing the complexities of deterministic, probabilistic and quantum decision trees are all polynomially related [5, 6, 13].

A very well-known open problem is the sensitivity vs. block sensitivity conjecture which claims that the two quantities are polynomially related. This problem is very simple to formulate (so simple that it can be assigned as an undergraduate research project). At the same time, the conjecture appears quite difficult to solve. It has been known for over 25 years and the best upper and lower bounds are still very far apart. We know that block sensitivity can be quadratically larger than sensitivity [3, 14, 16] but the best upper bounds on block sensitivity in terms of sensitivity are still exponential [1, 11, 15].

Block sensitivity is polynomially related to a number of other complexity measures of Boolean functions: certificate complexity, polynomial degree and the number of queries to compute f either deterministically, probabilistically or quantumly [6]. This gives a number of equivalent formulations for the sensitivity vs. block sensitivity conjecture: it is equivalent to asking whether sensitivity is polynomially related to any one of these complexity measures.

Among the many equivalent forms of the conjecture, relating sensitivity to certificate complexity C(f) might be the combinatorially simplest one. Certificate complexity being at least c simply means that there is an input $x=(x_1, \ldots , x_n)$ that is not contained in an $(n-(c-1))$-dimensional subcube of the Boolean hypercube on which f is constant. Therefore, in this paper we focus on the “sensitivity vs. certificate complexity” form of the conjecture.

1.1 Related Work

New Approaches to the Sensitivity Conjecture. Recently, there have been multiple developments in various approaches to the sensitivity conjecture. Gilmer et. al. interpret the problem through the cost of a novel communication game [8]. Gopalan et. al. investigate the properties of Boolean functions with low sensitivity [9]. Lin and Zhang give a bound on block sensitivity in terms of sensitivity and the alternating number of the function [12].

Upper Bounds on bs(f) and C(f) in Terms of s(f). There has been a substantial amount of work on reducing the gap between sensitivity and block sensitivity measures. The first non-trivial upper bound is due to Simon [15]:

$$\begin{aligned} bs(f) \le 4^{s(f)} s(f). \end{aligned}$$

(1)

Kenyon and Kutin [11] improved the bound to

$$\begin{aligned} bs(f) \le \frac{e}{\sqrt{2\pi }} e^{s(f)} \sqrt{s(f)}. \end{aligned}$$

(2)

Recently, Ambainis et. al. [1] showed an even better estimate:

$$\begin{aligned} bs(f) \le 2^{s(f)-1} s(f) - (s(f)-1). \end{aligned}$$

(3)

The essense of this result lies in the following relation between certificate complexity and sensitivity:

$$\begin{aligned} C_0(f) \le 2^{s_1(f)-1} s_0(f) - (s_1(f)-1). \end{aligned}$$

(4)

Note that any bound for $C_0(f)$ also holds for $C_1(f)$ symmetrically (in this case, $C_1(f) \le 2^{s_0(f)-1} s_1(f) - (s_0(f)-1)$).¹

1.2 Our Results

In this work, we give improved upper bounds for the “sensitivity vs. certificate complexity” problem. Our main technical result is

Theorem 1

Let f be a Boolean function which is not constant. If $s_1(f) = 1$, then $C_0(f) = s_0(f)$. If $s_1(f) > 1$, then

$$\begin{aligned} C_0(f) \le 2^{s_1(f)-1}\left( s_0(f)-\frac{1}{3}\right) . \end{aligned}$$

(5)

A similar bound for $C_1(f)$ follows by symmetry. This implies a new upper bound on block sensitivity and certificate complexity in terms of sensitivity:

Corollary 1

Let f be a Boolean function. Then

$$\begin{aligned} bs(f) \le C(f) \le \max \left( 2^{s(f)-1}\left( s(f)-\frac{1}{3}\right) , s(f)\right) . \end{aligned}$$

(6)

On the other hand, the function of Ambainis and Sun [3] gives the separation of

$$\begin{aligned} C_0(f) = \left( \frac{2}{3} + o(1)\right) s_0(f)s_1(f) \end{aligned}$$

(7)

for arbitrary values of $s_0(f)$ and $s_1(f)$. For $s_1(f) = 2$, we show an example of f that achieves

$$\begin{aligned} C_0(f) = \left\lfloor \frac{3}{2} s_0(f) \right\rfloor = \left\lfloor \frac{3}{4} s_0(f)s_1(f) \right\rfloor . \end{aligned}$$

(8)

We also study the relation between $C_0(f)$ and $s_0(f)$ for functions with low $s_1(f)$, as we think these cases may provide insights into the more general case.

If $s_1(f)=1$, then $C_0(f)=s_0(f)$ follows from (4). So, the easiest non-trivial case is $s_1(f)=2$, for which (4) becomes $C_0(f) \le 2 s_0(f) - 1$.

For $s_1(f) = 2$, we prove a slightly better upper bound of $C_0(f) \le \frac{9}{5} s_0(f)$. We also show that $C_0(f) \le \frac{3}{2} s_0(f)$ for $s_1(f) = 2$ and $s_0(f) \le 6$ and thus our example (8) is optimal in this case. We conjecture that $C_0(f) \le \frac{3}{2} s_0(f)$ is a tight upper bound for $s_1(f) = 2$.

Our results rely on a recent “gap theorem” by Ambainis and Vihrovs [4] which says that any sensitivity-s induced subgraph G of the Boolean hypercube must be either of size $2^{n-s}$ or of size at least $\frac{3}{2} 2^{n-s}$ and, in the first case, G can only be a subcube obtained by fixing s variables. Using this theorem allows refining earlier results which used Simon’s lemma [15] – any sensitivity-s induced subgraph G must be of size at least $2^{n-s}$ – but did not use any more detailed information about the structure of such G.

We think that further research in this direction may uncover more interesting facts about the structure of low-sensitivity subsets of the Boolean hypercube, with implications for the “sensitivity vs. certificate complexity” conjecture.

2 Preliminaries

Let $f: \{0,1\}^n \rightarrow \{0,1\}$ be a Boolean function on n variables. The i-th variable of an input x is denoted by $x_i$. For an index set $P \subseteq [n]$, let $x^P$ be the input obtained from an input x by flipping every bit $x_i$, $i \in P$.

We briefly define the notions of sensitivity, block sensitivity and certificate complexity. For more information on them and their relations to other complexity measures (such as deterministic, probabilistic and quantum decision tree complexities), we refer the reader to the surveys by Buhrman and de Wolf [6] and Hatami et al. [10].

Definition 1

The sensitivity complexity s(f, x) of f on an input x is defined as

$$\begin{aligned} s(f,x) = \left| \left\{ i \,\Big |\, f(x) \ne f\left( x^{\{i\}}\right) \right\} \right| . \end{aligned}$$

(9)

The b -sensitivity $s_b(f)$ of f, where $b \in \{0,1 \}$, is defined as $\max (s(f,x) \mid x \in \{0,1\}^n, f(x)=b)$. The sensitivity s(f) of f is defined as $\max (s_0(f), s_1(f))$.

We say that a vertex x has full sensitivity if $s(f,x) = s_{f(x)}(f)$.

Definition 2

The block sensitivity bs(f, x) of f on an input x is defined as the maximum number t such that there are t pairwise disjoint subsets $B_1, \ldots , B_t$ of [n] for which $f(x) \ne f\left( x^{B_i}\right)$. We call each $B_i$ a block. The b -block sensitivity $bs_b(f)$ of f, where $b \in \{0,1 \}$, is defined as $\max (bs(f,x) \mid x \in \{0,1\}^n, f(x)=b)$. The block sensitivity bs(f) of f is defined as $\max (bs_0(f), bs_1(f))$.

Definition 3

A certificate c of f on an input x is defined as a partial assignment $c: P \rightarrow \{0,1\}, P \subseteq [n]$ of x such that f is constant on this restriction. We call |P| the length of c. If f is always 0 on this restriction, the certificate is a 0-certificate. If f is always 1, the certificate is a 1-certificate.

Definition 4

The certificate complexity C(f, x) of f on an input x is defined as the minimum length of a certificate that x satisfies. The b -certificate complexity $C_b(f)$ of f, where $b \in \{0,1 \}$, is defined as $\max (C(f,x) \mid x \in \{0,1\}^n, f(x)=b)$. The certificate complexity C(f) of f is defined as $\max (C_0(f), C_1(f))$.

In this work we look at $\{0, 1\}^n$ as a set of vertices for a graph $Q_n$ (called the n -dimensional Boolean cube or hypercube) in which we have an edge (x, y) whenever $x=(x_1, \ldots , x_n)$ and $y=(y_1, \ldots , y_n)$ differ in exactly one position. We look at subsets $S \subseteq \{0,1\}^n$ as subgraphs (induced by the subset of vertices S) in this graph.

Definition 5

Let c be a partial assignment $c: P \rightarrow \{0,1\}, P \subseteq [n]$. An $(n-|P|)$ -dimensional subcube of $Q_n$ is a subgraph G induced on a vertex set $\{ x \mid \forall i \in P \, (x_i = c(i)) \}$. It is isomorphic to $Q_{n-|P|}$. We call the value $\dim (G) = n-|P|$ the dimension and the value |P| the co-dimension of G.

For example, a subgraph induced on the set $\{x \mid x_1 = 0, x_2 = 1\}$ is a $(n-2)$-dimensional subcube. Note that each certificate of length l corresponds to a subcube of $Q_n$ with co-dimension l.

Definition 6

Let G be a subcube defined by a partial assignment $c : P \rightarrow \{0,1\}, P \subseteq [n]$. Let $c' : P \rightarrow \{0,1\}$ where $c'(i) \ne c(i)$ for exactly one $i \in P$. Then we call the subcube defined by $c'$ a neighbour subcube of G.

For example, the sets $\{x \mid x_1 = 0, x_2 = 0\}$ and $\{x \mid x_1 = 0, x_2 = 1\}$ induce two neighbouring subcubes, since their union is a subcube induced on the set $\{x \mid x_1=0\}$.

We also extend the notion of Hamming distance to the subcubes of $Q_n$:

Definition 7

Let G and H be two subcubes of $Q_n$. Then the Hamming distance between G and H is defined as $d(G, H) = \min _{\begin{array}{c} x \in G \\ y \in H \end{array}} d(x, y)$, where d(x, y) is the Hamming distance between x and y.

Definition 8

Let G and H be induced subgraphs of $Q_n$. By $G \cap H$ denote the intersection of G and H that is the graph induced on $V(G) \cap V(H)$. By $G \cup H$ denote the union of G and H that is the graph induced on $V(G) \cup V(H)$. By $G \setminus H$ denote the complement of G in H that is the graph induced by $V(G) \setminus V(H)$.

Definition 9

Let G and H be induced subgraphs of $Q_n$. By R(G, H) denote the relative size of G in H:

$$\begin{aligned} R(G, H) = \frac{|V(G \cap H)|}{|V(H)|}. \end{aligned}$$

(10)

We extend the notion of sensitivity to the induced subgraphs of $Q_n$:

Definition 10

Let G be a non-empty induced subgraph of $Q_n$. The sensitivity $s(G,Q_n,x)$ of a vertex $x \in Q_n$ is defined as ${\left| \left\{ i \,\Big |\, x^{\{i\}} \notin G\right\} \right| }$, if $x \in G$, and ${\left| \left\{ i \,\Big |\, x^{\{i\}} \in G\right\} \right| }$, if $x \notin G$. Then the sensitivity of G is defined as $s(G,Q_n) = \max (s(G,Q_n,x) \mid x \in G)$.

Our results rely on the following generalization of Simon’s lemma [15], proved by Ambainis and Vihrovs [4]:

Theorem 2

Let G be a non-empty induced subgraph of $Q_n$ with sensitivity at most s. Then either $R(G, Q_n) = \frac{1}{2^s}$ and G is an $(n-s)$-dimensional subcube or $R(G, Q_n) \ge \frac{3}{2}\cdot \frac{1}{2^s}$.

3 Upper Bound on Certificate Complexity in Terms of Sensitivity

In this section we prove Corollary 1. In fact, we prove a slightly more specific result.

Theorem 1

Let f be a Boolean function which is not constant. If $s_1(f) = 1$, then $C_0(f) = s_0(f)$. If $s_1(f) > 1$, then

$$\begin{aligned} C_0(f) \le 2^{s_1(f)-1}\left( s_0(f)-\frac{1}{3}\right) . \end{aligned}$$

(11)

Note that a similar bound for $C_1(f)$ follows by symmetry. For the proof, we require the following lemma.

Lemma 1

Let $H_1$, $H_2$, $\ldots$, $H_k$ be distinct subcubes of $Q_n$ such that the Hamming distance between any two of them is at least 2. Take

$$\begin{aligned} {T = \bigcup _{i=1}^k H_i, \quad \quad \quad T' = \left\{ x \,\Big |\, \exists i \,\left( x^{\{i\}} \in T\right) \right\} \setminus T.} \end{aligned}$$

(12)

If $T \ne Q_n$, then $|T'| \ge |T|$.

Fig. 1.

A schematic representation of the 0-certificate $S_0$ and its neighbour cubes $S_1, S_2, \ldots , S_m$. The shaded parts represent the vertices in the subcubes for which the value of f is 1.

Proof

If $k = 1$, then the co-dimension of $H_1$ is at least 1. Hence $H_1$ has a neighbour cube, so $|T'| \ge |T| = |H_1|$.

Assume $k \ge 2$. Then $n \ge 2$, since there must be at least 2 bit positions for cubes to differ in. We use an induction on n.

Base case. $n = 2$. Then we must have that $H_1$ and $H_2$ are two opposite vertices. Then the other two vertices are in $T'$, hence $|T'| = |T| = 2$.

Inductive step. Divide $Q_n$ into two adjacent $(n-1)$-dimensional subcubes $Q_n^0$ and $Q_n^1$ by the value of $x_1$. We will prove that the conditions of the lemma hold for each $T \cap Q_n^b$, $b \in \{0,1\}$. Let $H_u^b = H_u \cap Q_n^b$. Assume $H_u^b \ne \varnothing$ for some $u \in [k]$. Then either $x_1 = b$ or $x_1$ is not fixed in $H_u$. Thus, if there are two non-empty subcubes $H_u^b$ and $H_v^b$, they differ in the same bit positions as $H_u$ and $H_v$. Thus the Hamming distance between $H_u^b$ and $H_v^b$ is also at least 2. On the other hand, $Q_n^b \not \subseteq T$, since then k would be at most 1.

Let $T_b = T \cap Q_n^b$ and ${T_b' = \left\{ x \,\Big |\, x \in Q_n^b, \exists i \, \left( x^{\{i\}} \in T_b\right) \right\} \setminus T_b}$. Then by induction we have that $|T_b'| \ge |T_b|$. On the other hand, $T_0 \cup T_1 = T$ and $T_0' \cup T_1' \subseteq T'$. Thus

$$\begin{aligned} |T'| \ge |T_0'| + |T_1'| \ge |T_0| + |T_1| = |T|. \end{aligned}$$

(13)

   $\square$

Proof of Theorem 1. Let z be a vertex such that $f(z) = 0$ and $C(f, z) = C_0(f)$. Pick a 0-certificate $S_0$ of length $C_0(f)$ and $z \in S_0$. It has $m=C_0(f)$ neighbour subcubes which we denote by $S_1, S_2, \ldots , S_m$ (Fig. 1).

We work with the graph G induced on the vertex set $\{x \mid f(x) = 1\}$. Since $S_0$ is a minimum certificate for z, $S_i \cap G \ne \varnothing$ for $i \in [m]$.

As $S_0$ is a 0-certificate, it gives 1 sensitive bit to each vertex in $G \cap S_i$. Then $s(G \cap S_i, S_i) \le s_1(f)-1$.

Suppose $s_1(f) = 1$, then for each $i \in [m]$ we must have that $G \cap S_i$ equals to the whole $S_i$. But then each vertex in $S_0$ is sensitive to its neighbour in $G \cap S_i$, so $m \le s_0(f)$. Hence $C_0(f) = s_0(f)$.

Otherwise $s_1(f) \ge 2$. By Theorem 2, either $R(G,S_i) =\frac{1}{2^{s_1(f)-1}}$ or $R(G,S_i) \ge \frac{3}{2^{s_1(f)}}$ for each $i \in [m]$. We call the cube $S_i$ either light or heavy respectively. We denote the number of light cubes by l, then the number of heavy cubes is $m-l$. We can assume that the light cubes are $S_1, \ldots , S_l$.

Let the average sensitivity of the inputs in $S_0$ be $as(S_0) = \frac{1}{|S_0|} \sum _{x \in S_0} s_0(x)$. Since each vertex of G in any $S_i$ gives sensitivity 1 to some vertex in $S_0$, $\sum _{i=1}^m R(G, S_i) \le as(S_0)$. Clearly $as(S_0) \le s_0(f)$. We have that

$$\begin{aligned} l \frac{1}{2^{s_1(f)-1}} + (m-l) \frac{3}{2^{s_1(f)}}&\le as(S_0) \le s_0(f) \end{aligned}$$

(14)

$$\begin{aligned} m \frac{3}{2^{s_1(f)}} - l \frac{1}{2^{s_1(f)}}&\le as(S_0) \le s_0(f). \end{aligned}$$

(15)

Then we examine two possible cases.

Case 1. $l \le (s_0(f)-1)2^{s_1(f)-1}$. Then we have

$$\begin{aligned} m \frac{3}{2^{s_1(f)}} -(s_0(f)-1) \frac{2^{s_1(f)-1}}{2^{s_1(f)}}&\le as(S_0) \le s_0(f) \end{aligned}$$

(16)

$$\begin{aligned} m \frac{3}{2^{s_1(f)}}&\le s_0(f) + \frac{1}{2} (s_0(f)-1) \end{aligned}$$

(17)

$$\begin{aligned} m \frac{3}{2^{s_1(f)}}&\le \frac{3}{2} s_0(f) - \frac{1}{2} \end{aligned}$$

(18)

$$\begin{aligned} m&\le 2^{s_1(f)-1}\left( s_0(f)-\frac{1}{3}\right) . \end{aligned}$$

(19)

Case 2. $l = (s_0(f)-1)2^{s_1(f)-1}+\delta$ for some positive integer $\delta$. Since $s_1(f) \ge 2$, the number of light cubes is at least $2(s_0(f)-1) + \delta$, which in turn is at least $s_0(f)$.

Let $\mathcal F = \{ F \mid F \subseteq [l], |F| = s_0(f)\}$. Denote its elements by $F_1, F_2, \ldots , F_{|\mathcal F|}$. We examine $H_1, H_2, \ldots , H_{|\mathcal F|}$ – subgraphs of $S_0$, where $H_i$ is the set of vertices whose neighbours in $S_j$ are in G for each $j \in F_i$. By Theorem 2, $G \cap S_i$ are subcubes for $i \le l$. Then so are the intersections of their neighbours in $S_0$, including each $H_i$.

Let $N_{i,j}$ be the common neighbour cube of $S_i$ and $S_j$ that is not $S_0$. Suppose $v \in S_0$. Then by $v_i$ denote the neighbour of v in $S_i$. Let $v_{i,j}$ be the common neighbour of $v_i$ and $v_j$ that is in $N_{i,j}$.

Next we will show the following:

Proposition 1

The Hamming distance between any two subcubes $H_i$ and $H_j$, $i \ne j$ is at least 2.

Proof

Assume there is an edge (u, v) such that $u \in H_i$ and $v \in H_j$. Then $u_k \in G$ for each $k \in F_i$. Since $i \ne j$, there is an index $t \in F_j$ such that $t \notin F_i$. The vertex u is sensitive to $S_k$ for each $k \in F_i$ and, since $|F_i| = s_0(f)$, has full sensitivity. Thus $u_t \notin G$. On the other hand, since each $S_k$ is light, $u_k$ has full 1-sensitivity, hence $u_{k,t} \in G$ for all $k \in F_i$. This gives full 0-sensitivity to $u_t$. Hence $v_t \notin G$, a contradiction, since $v \in H_j$ and $t \in F_j$.

Thus there are no such edges and the Hamming distance between $H_i$ and $H_j$ is not equal to 1. That leaves two possibilities: either the Hamming distance between $H_i$ and $H_j$ is at least 2 (in which case we are done), or both $H_i$ and $H_j$ are equal to a single vertex v, which is not possible, as then v would have a 0-sensitivity of at least $s_0(f)+1$.

Let $T = \bigcup _{i=1}^{|\mathcal F|} H_i$. We will prove that $T \ne S_0$. If each of $H_i$ is empty, then $T = \varnothing$ and $T \ne S_0$. Otherwise there is a non-empty $H_j$. As $s_1(f) \ge 2$, by Theorem 2 it follows that $\dim (G \cap S_k) = \dim (S_k) - s_1(f)+1 \le \dim (S_0) - 1$ for each $k \in [l]$. Thus $\dim (H_j) \le \dim (S_0) - 1$, and $H_j \ne S_0$. Then it has a neighbour subcube $H_j'$ in $S_0$. But since the Hamming distance between $H_j$ and any other $H_i$ is at least 2, we have that $H_j' \cap H_i = \varnothing$, thus T is not equal to $S_0$.

Therefore, $H_1, H_2, \ldots , H_{|\mathcal F|}$ satisfy all the conditions of Lemma 1. Let $T'$ be the set of vertices in $S_0 \setminus T$ with a neighbour in T. Then, by Lemma 1, $|T'| \ge |T|$ or, equivalently, $R(T', S_0) \ge R(T, S_0)$.

Then note that $R(T', S_0) \ge R(T, S_0) \ge \frac{\delta }{2^{s_1(f)-1}}$, since $R(G,S_i)= \frac{1}{2^{s_1(f)-1}}$ for all $i \in [l]$, there are a total of $(s_0(f)-1)2^{s_1(f)-1}+\delta$ light cubes and each vertex in $S_0$ can have at most $s_0(f)$ neighbours in G.

Let $S_h$ be a heavy cube, and $i \in [|\mathcal F|]$. The neighbours of $H_i$ in $S_h$ must not be in G, or the corresponding vertex in $H_i$ would have sensitivity $s_0(f)+1$.

Let $k \in F_i$. As $S_k$ is light, all the vertices in $G \cap S_k$ are fully sensitive, therefore all their neighbours in $N_{k,h}$ are in G. Therefore all the neighbours of $H_i$ in $S_h$ already have full 0-sensitivity. Then all their neighbours must also not be in G.

This means that vertices in $T'$ can only have neighbours in G in light cubes. But they can have at most $s_0(f)-1$ such neighbours each, otherwise they would be in T, not in $T'$. As $R(T', S_0) \ge \frac{\delta }{2^{s_1(f)-1}}$, the average sensitivity of vertices in $S_0$ is at most

$$\begin{aligned} as(S_0)&\le s_0(f) R(S_0 \setminus T', S_0) + (s_0(f)-1)R(T', S_0) \end{aligned}$$

(20)

$$\begin{aligned}&\le s_0(f) \left( 1-\frac{\delta }{2^{s_1(f)-1}}\right) +(s_0(f)-1)\frac{\delta }{2^{s_1(f)-1}} \end{aligned}$$

(21)

$$\begin{aligned}&= s_0(f)-\frac{\delta }{2^{s_1(f)-1}}. \end{aligned}$$

(22)

Then by inequality (15) we have

$$\begin{aligned} m \frac{3}{2^{s_1(f)}} - \left( (s_0(f)-1)2^{s_1(f)-1}+\delta \right) \frac{1}{2^{s_1(f)}} \le s_0(f)-\frac{\delta }{2^{s_1(f)-1}}. \end{aligned}$$

(23)

Rearranging the terms, we get

$$\begin{aligned} m \frac{3}{2^{s_1(f)}}&\le \left( (s_0(f)-1)2^{s_1(f)-1}+\delta \right) \frac{1}{2^{s_1(f)}} + s_0(f)-\frac{\delta }{2^{s_1(f)-1}} \end{aligned}$$

(24)

$$\begin{aligned} m \frac{3}{2^{s_1(f)}}&\le s_0(f) + \frac{1}{2}(s_0(f)-1) - \frac{\delta }{2^{s_1(f)}} \end{aligned}$$

(25)

$$\begin{aligned} m \frac{3}{2^{s_1(f)}}&\le \frac{3}{2} s_0(f) - \frac{1}{2} - \frac{\delta }{2^{s_1(f)}} \end{aligned}$$

(26)

$$\begin{aligned} m&\le 2^{s_1(f)-1}\left( s_0(f) - \frac{1}{3}\right) - \frac{\delta }{3}. \end{aligned}$$

(27)

   $\square$

Theorem 1 immediately implies Corollary 1:

Proof of Corollary 1. If f is constant, then $C(f) = s(f) = 0$ and the statement is true. Otherwise by Theorem 1

$$\begin{aligned} C(f)&= \max (C_0(f), C_1(f)) \end{aligned}$$

(28)

$$\begin{aligned}&\le \max _{b \in \{0,1\}}\left( \max \left( 2^{s_{1-b}(f)-1}\left( s_b(f)-\frac{1}{3}\right) , s_b(f)\right) \right) \end{aligned}$$

(29)

$$\begin{aligned}&\le \max \left( 2^{s(f)-1}\left( s(f)-\frac{1}{3}\right) , s(f)\right) \end{aligned}$$

(30)

On the other hand, $bs(f) \le C(f)$ is a well-known fact.    $\square$

4 Relation Between $C_0(f)$ and $s_0(f)$ for $s_1(f) = 2$

Ambainis and Sun exhibited a class of functions that achieves the best known separation between sensitivity and block sensitivity, which is quadratic in terms of s(f) [3]. This function also produces the best known separation between 0-certificate complexity and 0/1-sensitivity:

Theorem 3

For arbitrary $s_0(f)$ and $s_1(f)$, there exists a function f such that

$$\begin{aligned} C_0(f) = \left( \frac{2}{3} + o(1)\right) s_0(f)s_1(f). \end{aligned}$$

(31)

Thus it is possible to achieve a quadratic gap between the two measures. As $bs_0(f) \le C_0(f)$, it would be tempting to conjecture that quadratic separation is the largest possible. Therefore we are interested both in improved upper bounds and in functions that achieve quadratic separation with a larger constant factor.

In this section, we examine how $C_0(f)$ and $s_0(f)$ relate to each other for small $s_1(f)$. If $s_1(f) = 1$, it follows by Theorem 1 that $C_0(f) = s_0(f)$. Therefore we consider the case $s_1(f) = 2$.

Here we are able to construct a separation that is better than (31) by a constant factor.

Theorem 4

There is a function f with $s_1(f) = 2$ and arbitrary $s_0(f)$ such that

$$\begin{aligned} C_0(f) = \left\lfloor \frac{3}{4} s_0(f)s_1(f)\right\rfloor = \left\lfloor \frac{3}{2} s_0(f)\right\rfloor . \end{aligned}$$

(32)

Proof

Consider the function that takes value 1 iff its 4 input bits are in either ascending or descending sorted order. Formally,

$$\begin{aligned} \textsc {Sort}_4(x) = 1 \Leftrightarrow (x_1 \le x_2 \le x_3 \le x_4) \vee (x_1 \ge x_2 \ge x_3 \ge x_4). \end{aligned}$$

(33)

One easily sees that $C_0(\textsc {Sort}_4) = 3$, $s_0(\textsc {Sort}_4) = 2$ and $s_1(\textsc {Sort}_4) = 2$.

Denote the 2-bit logical AND function by $\textsc {And}_2$. We have $C_0(\textsc {And}_2) = s_0(\textsc {And}_2) = 1$ and $s_1(\textsc {And}_2) = 2$.

To construct the examples for larger $s_0(f)$ values, we use the following fact (it is easy to show, and a similar lemma was proved in [3]):

Fact 1

Let f and g be Boolean functions. By composing them with OR to $f \vee g$ we get

$$\begin{aligned} C_0(f \vee g)&= C_0(f) + C_0(g), \end{aligned}$$

(34)

$$\begin{aligned} s_0(f \vee g)&= s_0(f) + s_0(g), \end{aligned}$$

(35)

$$\begin{aligned} s_1(f \vee g)&= \max (s_1(f), s_1(g)). \end{aligned}$$

(36)

Suppose we need a function with $k = s_0(f)$. Assume k is even. Then by Fact 1 for $g = \bigvee _{i=1}^{\frac{k}{2}} \textsc {Sort}_4$ we have $C_0(g) = \frac{3}{2} k$. If k is odd, consider the function $g = \left( \bigvee _{i=1}^{\frac{k-1}{2}} \textsc {Sort}_4\right) \vee \textsc {And}_2$. Then by Fact 1 we have $C_0(g) = 3 \cdot \frac{k-1}{2} + 1 = \left\lfloor \frac{3}{2} k \right\rfloor$.    $\square$

A curious fact is that both examples of (31) and Theorem 4 are obtained by composing some primitives using OR. The same fact holds for the best examples of separation between bs(f) and s(f) that preceded the [3] construction [14, 16].

We are also able to prove a slightly better upper bound in case $s_1(f) = 2$.

Theorem 5

Let f be a Boolean function with $s_1(f) = 2$. Then

$$\begin{aligned} C_0(f) \le \frac{9}{5} s_0(f). \end{aligned}$$

(37)

Proof

Let z be a vertex such that $f(z) = 0$ and $C(f,z) = C_0(f)$. Pick a 0-certificate $S_0$ of length $m=C_0(f)$ and $z \in S_0$. It has m neighbour subcubes which we denote by $S_1$, $S_2$, $\ldots$, $S_{m}$. Let $n'=n-m = \dim (S_i)$ for each $S_i$.

We work with a graph G induced on a vertex set $\{x \mid f(x) = 1\}$. Let $G_i = G \cap S_i$. As $S_0$ is a minimal certificate for z, we have $G_i \ne \varnothing$ for each $i \in [m]$. Since any $v \in G_i$ is sensitive to $S_0$, we have $s(G_i, S_i) \le 1$. Thus by Theorem 2 either $G_i$ is an $(n'-1)$-subcube of $S_i$ with $R(G_i : S_i) = \frac{1}{2}$ or $R(G_i : S_i) \ge \frac{3}{4}$. We call $S_i$ light or heavy, respectively.

Let $N_{i,j}$ be the common neighbour cube of $S_i$, $S_j$ that is not $S_0$. Let $G_{i,j} = G \cap N_{i,j}$. Suppose $v \in S_0$. Let $v_i$ be the neighbour of v in $S_i$. Let $v_{i,j}$ be the neighbour of $v_i$ and $v_j$ in $N_{i,j}$.

Let $S_i, S_j$ be light. By $G_i^0, G_j^0$ denote the neighbour cubes of $G_i, G_j$ in $S_0$. We call $\{S_i, S_j\}$ a pair, iff $G_i^0 \cup G_j^0 = S_0$. In other words, a pair is defined by a single dimension. Also we have either $z_i \notin G$ or $z_j \notin G$: we call the corresponding cube the representative of this pair.

Proposition 2

Let $\mathcal P$ be a set of mutually disjoint pairs of the neighbour cubes of $S_0$. Then there exists a 0-certificate $S_0'$ such that $z \in S_0'$, $\dim (S_0') = \dim (S_0)$ and $S_0'$ has at least $|\mathcal P|$ heavy neighbour cubes.

Proof

Let $\mathcal R$ be a set of mutually disjoint pairs of the neighbour cubes of $S_0$. W.l.o.g. let $S_1, \ldots , S_{|\mathcal R|}$ be the representatives of $\mathcal R$. Let $F_i$ be the neighbour cube of $S_i \setminus G$ in $S_0$. Let $B_{\mathcal R} = \bigcap _{i=1}^{|\mathcal R|} F_i$. Suppose $S_0+x$ is a coset of $S_0$ and $x_t = 0$ if the t-th dimension is not fixed in $S_0$: let $B_{\mathcal R}(S_0+x)$ be $B_{\mathcal R}+x$.

Pick $\mathcal R \subseteq \mathcal P$ with the largest size, such that for each two representatives $S_i$, $S_j$ of $\mathcal R$, $B_{\mathcal R}(N_{i,j})$ is a 0-certificate.

Next we prove that the subcube $S_0'$ spanned by $B_{\mathcal R}, B_{\mathcal R}(S_1), \ldots , B_{\mathcal R}\left( S_{|\mathcal R|}\right)$ is a 0-certificate. It corresponds to an $|\mathcal R|$-dimensional hypercube $Q_{|\mathcal R|}$ where $B_{\mathcal R}(S_0+x)$ corresponds to a single vertex for each coset $S_0+x$ of $S_0$.

Let $T \subseteq Q_{|\mathcal R|}$ be the graph induced on the set $\{v \mid v \text { corresponds to } B_{\mathcal R}(S_0+x), B_{\mathcal R}(S_0+x) \text { is} \text { not a 0-certificate}\}$. Then we have $s(T, Q_{|\mathcal R|}) \le 2$. Suppose $B_{\mathcal R}$ corresponds to $0^{|\mathcal R|}$. Let $L_d$ be the set of $Q_{|\mathcal R|}$ vertices that are at distance d from $0^{|\mathcal R|}$. We prove by induction that $L_d \cap T = \varnothing$ for each d.

Proof

Base case. $d \le 2$. The required holds since all $B_{\mathcal R}, B_{\mathcal R}(S_i), B_{\mathcal R}(N_{i, j})$ are 0-certificates.

Inductive step. $d \ge 3$. Examine $v \in L_d$. As v has d neighbours in $L_{d-1}$, $L_{d-1} \cap T = \varnothing$ and $s(T, Q_{|\mathcal R|}) \le 2$, we have that $v \notin T$.

Let k be the number of distinct dimensions that define the pairs of $\mathcal R$, then $k \le |\mathcal R|$. Hence $\dim (S_0') = |\mathcal R| + \dim (B_{\mathcal R}) = |\mathcal R| + (\dim (S_0)-k) \ge \dim (S_0)$. But $S_0$ is a minimal 0-certificate for z, therefore $\dim (S_0') = \dim (S_0)$.

Note that a light neighbour $S_i$ of $S_0$ is separated into a 0-certificate and a 1-certificate by a single dimension, hence we have $s(G,S_i,v) = 1$ for every $v \in S_i$. As $S_i$ neighbours $S_0$, every vertex in its 1-certificate is fully sensitive. The same holds for any light neighbour $S_i'$ of $S_0'$.

Now we will prove that each pair in $\mathcal P$ provides a heavy neighbour for $S_0'$. Let $\{S_a, S_b\} \in \mathcal P$, where $S_a$ is the representative. We distinguish two cases:

• $B_{\mathcal R}(S_b)$ is a 1-certificate. Since $S_b$ is light, it has full 1-sensitivity. Therefore, $v \in G$ for all $v\in B_R(N_{i, b})$, for each $i \in [|\mathcal R|]$. Let $S_b'$ be the neighbour of $S_0'$ that contains $B_{\mathcal R}(S_b)$ as a subcube. Then for each $v \in B_{\mathcal R}(S_b)$ we have $s(G,S_b',v) = 0$. Hence $S_b'$ is heavy.

• Otherwise, $\{S_a, S_b\}$ is defined by a different dimension than any of the pairs in $\mathcal R$. Let $\mathcal R' = \mathcal R \cup \{S_a, S_b\}$. Examine the subcube $B_{\mathcal R'}$. By definition of $\mathcal R$, there is a representative $S_i$ of $\mathcal R$ such that $B_{\mathcal R'}(N_{i,a})$ is not a 0-certificate. Let $S'_a$ be the neighbour of $S_0'$ that contains $B_{\mathcal R}(S_a)$ as a subcube. Then there is a vertex $v \in B_{\mathcal R'}(S_a)$ such that $s(G, S_a', v) \ge 2$. Hence $S_a'$ is heavy.   $\square$

Let $\mathcal P$ be the largest set of mutually disjoint pairs of the neighbour cubes of $S_0$. Let l and $h = m- l$ be the number of light and heavy neighbours of $S_0$, respectively. Each pair in $\mathcal P$ gives one neighbour in G to each vertex in $S_0$. Now examine the remaining $l-2 | \mathcal P|$ light cubes. As they are not in $\mathcal P$, no two of them form a pair. Hence there is a vertex $v \in S_0$ that is sensitive to each of them. Then $s_0(f) \ge s_0(f,v) \ge |\mathcal P| + (l-2 | \mathcal P|) = l - |\mathcal P|$. Therefore $|\mathcal P| \ge l - s_0(f)$.

Let q be such that $m=q s_0(f)$. Then there are $q s_0(f)-l$ heavy neighbours of $S_0$. On the other hand, by Proposition 2, there exists a minimal certificate $S_0'$ of z with at least $l-s_0(f)$ heavy neighbours. Then z has a minimal certificate with at least $\frac{(q s_0(f)-l)+(l-s_0(f))}{2} = \frac{q-1}{2}\cdot s_0(f)$ heavy neighbour cubes.

W.l.o.g. let $S_0$ be this certificate. Then $l = qs_0(f) - h \le (q - \frac{q-1}{2}) s_0(f) = \frac{q+1}{2}\cdot s_0(f)$. As each $v \in G_i$ for $i \in [m]$ gives sensitivity 1 to its neighbour in $S_0$,

$$\begin{aligned} l \frac{1}{2} + h\frac{3}{4}&\le s_0(f). \end{aligned}$$

(38)

Since the constant factor at l is less than at h, we have

$$\begin{aligned} \frac{q+1}{2} \cdot s_0(f) \cdot \frac{1}{2} + \frac{q-1}{2} \cdot s_0(f) \cdot \frac{3}{4} \le s_0(f) \end{aligned}$$

(39)

By dividing both sides by $s_0(f)$ and simplifying terms, we get $q \le \frac{9}{5}$.    $\square$

This result shows that the bound of Corollary 1 can be improved. However, it is still not tight. For some special cases, through extensive casework we can also prove the following results:

Theorem 6

Let f be a Boolean function with $s_1(f) = 2$ and $s_0(f) \ge 3$. Then

$$\begin{aligned} C_0(f) \le 2s_0(f) - 2. \end{aligned}$$

(40)

Theorem 7

Let f be a Boolean function with $s_1(f) = 2$ and $s_0(f) \ge 5$. Then

$$\begin{aligned} C_0(f) \le 2s_0(f) - 3. \end{aligned}$$

(41)

The proofs of these theorems are available online in the full version of the paper [2].

These theorems imply that for $s_1(f) = 2$, $s_0(f) \le 6$ we have $C_0(f) \le \frac{3}{2} s_0(f)$, which is the same separation as achieved by the example of Theorem 4. This leads us to the following conjecture:

Conjecture 1

Let f be a Boolean function with $s_1(f) = 2$. Then

$$\begin{aligned} C_0(f) \le \frac{3}{2} s_0(f). \end{aligned}$$

(42)

We consider $s_1(f) = 2$ to be the simplest case where we don’t know the actual tight upper bound on $C_0(f)$ in terms of $s_0(f), s_1(f)$. Proving Conjecture 1 may provide insights into relations between C(f) and s(f) for the general case.