# Monotonicity Testing Lower Bounds via Communication Complexity

Prateek Kumar

Nov 9, 2018

CS5030: Communication Complexity
Class Presentation

## Property Testing

Let $D$, $R$ be two sets, then preperty $\mathcal{P}$ is a set of functions from $D$ to $R$.

Examples: Linearity, Monotonicity

Property Testing: Given a black box which outputs $f(x)$ when provided with query $x$, goal is to decide whether $f \in \mathcal{P}$ using as few queries to the black box as possible.

Exact property testing requires $\Theta(|D|)$ queries. So, our focus will be on randomized algorithms.

## $\epsilon$-Property Testing

Decide whether $f \in \mathcal{P}$ or $f$ is $\epsilon$-far from $\mathcal{P}$.

$f$ is $\epsilon$-far from $\mathcal{P}$ iff $\forall g \in \mathcal{P}$, $f$ and $g$ differ at $\geq \epsilon|D|$ values.

Query Complexity: $\min_{T \in \text{testers}} \max_{f : D \to R} Q(T, f)$

Testers can be adaptive or non-adaptive, have one-sided or two-sided errors.

## Monotonicity testing upper bounds

Boolean case: $D = \{0, 1\}^n, R = \{0, 1\}$
$f : \{0, 1\}^n \to \{0, 1\}$ is monotone iff $f(x_0) \leq f(x_1)$ for all $x_0, x_1$ such that $x_0$ and $x_1$ differ at $i$th bit only and $x_0$ has 0 as $i$th bit and $x_1$ has 1 as the $i$th bit.

Theorem: For $\epsilon$-monotone testing (randomized), the upper bound on number of queries is $O(\frac{n}{\epsilon})$. For larger ranges, it is $O(\frac{n\log_2(|R|)}{\epsilon})$ (Proof at the end, if time available)

## Monotonicity testing lower bounds

First we will see bounds for larger ranges($|R| = \Omega(n)$).

Theorem: For large enough ranges, $R$ and $\epsilon = \frac{1}{8}$ every adaptive monotonicty tester with 2-sided error uses $\Omega(n)$ queries.

### Proof of the theorem

We will reduce $\text{DISJ}$ to $\frac{1}{8}$-monotonicity testing.

Let $T$ be a randomized tester to distinguish monotone and $\frac{1}{8}$-far from monotone.

We will develop a public-coin randomized protocol for $\text{DISJ}$.

Let $A, B \subseteq [n]$ and let Alice and Bob have $A$ and $B$ respectively.

Lemma: Let $A, B \subseteq [n]$, function $h_{AB} : 2^{[n]} \to \mathbb{Z}$ defined as

$$h_{AB}(S) = 2|S| + (-1)^{|S \cap A|} + (-1)^{|S \cap B|}$$

then,

1. $A \cap B = \phi$ implies $h$ is monotone.
2. $A \cap B \neq \phi$ implies $h$ is $\frac{1}{8}$-far from monotone.

Let us assume the lemma is true (proof, we will see in sometime)

### Protocol

Alice and Bob will agree on random bits and feed those bits to their own copies of the tester $T$.

The tester has to decide whether $h_{AB}$ is monotone or $\frac{1}{8}$-far from monotone.

The tester will query for $h_{AB}(S)$ for some $S \in \{0, 1\}^n$.

Alice and Bob will together decide the value of $h_{AB}(S)$ and feed their testers.

1. Until $T$ halts:
• Let $S$ be the query asked by $T$.
• Alice sends $(-1)^{|S \cap A|}$ to Bob. (1-bit)
• Bob sends $(-1)^{|S \cap B|}$ to Alice. (1-bit)
• Both compute $h_{AB}(S)$ and feed their testers.
2. If $T$ accepts $h_{AB}$ then Alice/Bob declare $\text{DISJOINT}$, else they declare $\text{NOT DISJOINT}$

Communication complexity of the protocol $= 2(\text{No. of queries asked by }T)$

Since $\text{DISJ}$ has $\Omega(n)$ public coin randomized complexity, the query complexity of monotone is $\Omega(n)$.

## Proof of the lemma

Case (i) $A \cap B = \phi$

Let $S \subseteq [n] - \{i\}$.

Since $A$ and $B$ are disjoint, $i \notin$ at least one of $A$ and $B$.

$h_{AB}(S \cup \{i\}) - h_{AB}(S) \geq 0$
$\forall i \in [n], S \subseteq [n] - \{i\}$

$\therefore$ $h_{AB}$ is monotone.

Case (ii) $A \cap B \neq \phi$

Let $i \in A \cap B$.

Consider $S \subseteq [n] - \{{i\}}$ such that $(-1)^{|S \cap A|} = 1$ and $(-1)^{|S \cap B|} = 1$.

In this case,
$h_{AB}(S \cup \{i\}) - h_{AB}(S) < 0$.

There are at least $2^{n-1}/4 = 2^n/8$ possible values for $S$ for a given pair $A, B$. (Reason, next slide)

Consider sets $A’ = A - B, B’ = B - A$.

$Pr_{S}[|S \cap C| \mod 2 = 0] = \frac{1}{2}$ for any non-empty set $C \subseteq [n]$

Assuming $A’$ and $B’$ to be non-empty.

Since $A’, A \cap B, B’$ are all mutually disjoint, we have that the events $(|S \cap A’| \mod 2 = 0)$, $(|S \cap (A \cap B)| \mod 2 = 0)$, $(|S \cap B’| \mod 2 = 0)$ are all mutually independent.

$\therefore Pr_{S}((-1)^{|S \cap A|} = 1 \text{ and } (-1)^{|S \cap B|} = 1) = Pr_{S}(|S \cap A’| \mod 2 = 0$ and $|S \cap (A \cap B)| \mod 2 = 0$ and $|S \cap B’| \mod 2 = 0) + Pr_{S}(|S \cap A’| \mod 2 = 1$ and $|S \cap (A \cap B)| \mod 2 = 1$ and $|S \cap B’| \mod 2 = 1) = \frac{1}{2}*\frac{1}{2}*\frac{1}{2} + \frac{1}{2}*\frac{1}{2}*\frac{1}{2} = \frac{1}{4}$

Note: When $A’$ is empty, then both $A \cap B$ and $B’$ must have even number of elements and same probability of 14 will be obtained.

For each of the possible pair $(S, S \cup \{i\})$, we have to change one of the values of the function $h_{AB}$ (i.e. either $h_{AB}(S)$ or $h_{AB}(S \cup \{i\})$) to make it monotone.

Total no. of such modifications $\geq \frac{2^n}{8}$
$\therefore$ $h_{AB}$ is $\frac{1}{8}$-far from monotone.

### Query complexity when $|R| = \Omega(\sqrt{n})$

Trick is to truncate $h_{AB}$ to obtain $h’_{AB}$ as follows:

1. $h_{AB}(S)$ $< n - c\sqrt{n} \implies h’_{AB} = n - c\sqrt{n}$.
2. $h_{AB}(S)$ $> n + c\sqrt{n} \implies h’_{AB} = n + c\sqrt{n}$.
3. $h’_{AB}(S) = h_{AB}(S)$ in all other cases.

Case (i) When $A, B$ are disjoint, truncating gives another monotone function.

Case (ii) When $A, B$ are not disjoint, $h’_{AB}$ turns out to be $\frac{1}{16}$-far from monotone.
First, see that $h_{AB}$ and $h’_{AB}$ differ in at most $\frac{1}{16}$ entries (using Chebyshev’s inequlaity).
Hamming distance obeys triangle inequality. So $h’_{AB}$ is $\frac{1}{16}$-far from montone.

### Query complexity when $|R| = o(\sqrt{n})$

Let $m = |R|^2$

Consider a function $g : [m] \to R$. In this case, we define $h : [n] \to R$ as $h(S) = g(S \cap [m])$

$g$ is monotone $\implies$ $h$ is monotone (Clear from def.).

Next, we will show that
$g$ is $\epsilon$-far from monotone $\implies$ $h$ is $\epsilon$-far from monotone.

Suppose $h$ is not $\epsilon$-far from monotone $\implies$ $\exists$ monotone $h’$ such that $Pr_{X \subseteq [m], Y \subseteq [n]-[m]}[h(X \cup Y) \neq h’(X \cup Y)] \leq \epsilon$.

Using averaging argument,
$\exists Y_0$ such that $Pr_{X \subseteq [m]}[h(X \cup Y_0) \neq h’(X \cup Y_0)] \leq \epsilon$

Define $g’$ as $g’(X) = h’(X \cup Y_0)$.
$\therefore Pr_{X \subseteq [m]}[g(X) \neq g’(X)] \leq \epsilon$
$\implies$ $g$ is also not $\epsilon$-far from monotone.

To test $g$, we can test $h$ and return the result.

Since $g$ requires $\Omega(m)$ queries, $h$ also requires $\Omega(m) = \Omega(|R|^2)$ queries.

## Monotonicity testing upper bounds

Theorem: For $\epsilon$-monotone testing (randomized) and boolean range, the upper bound on number of queries is $O(\frac{n}{\epsilon})$. For larger ranges, it is $O(\frac{n\log_2(|R|)}{\epsilon})$.

(Proof on the board)

1. Communication Complexity (for Algorithm Designers), Tim Roughgardden (Chapter 8).
2. Property Testing Lower Bounds Via Communication Complexity, Eric Blais, Joshua Brody, Kevin Matulef (Section 4).