# Lecture 19

## P vs NP: State of the Art

Conjecture $$P\neq NP$$. For any $$NP-complete$$ $$L$$ and any $$c>0$$, there is no algorithm that computes $$L$$ in time $$O(n^c)$$.

Conjecture (Strong Exponential-time Hypothesis) There is a constant $$\delta > 0$$ such that every algorithm that computes $$SAT$$ has time complexity $$\Omega(2^{\delta n})$$.

Theorem 1 There exists a language such that any algorithm in the RAM model that computes $$L$$ has running time $$\Omega(n)$$.

Theorem 2 There is a language $$L$$ such that any multi-tape TM that decides $$L$$ has running time $$\Omega(n(\log^*n)^{1/4})$$

Theorem 3 There is a language $$L$$ such that any single-tape TM that decides $$L$$ has running time $$\Omega(n^2)$$

## Communication Complexity

Definition A communication protocol $$\pi$$ for functions mapping $$\mathcal{X}\times\mathcal{Y}$$ to $$\mathcal{Z}$$ is a binary tree where

1. every internal node $$v$$ is labelled with a function $$a_v:\mathcal{X}\rightarrow\{0,1\}$$ or $$b_v:\mathcal{Y}\rightarrow\{0,1\}$$
2. every leaf is labelled with an element $$z\in\mathcal{Z}$$

Definition The protocal $$\pi$$ computes $$f:\mathcal{X}\times\mathcal{Y}\rightarrow\mathcal{Z}$$ if for every $$x\in\mathcal{X}$$ and $$y\in\mathcal{Y}$$, the leaf in $$\pi$$ reached by $$(x,y)$$ is labelled $$f(x,y)$$.

Definition The cost of $$\pi$$ is the (maximum) depth of the tree, (= max number of bits that Alice and Bob exchange to compute $$f(x,y)$$ over all possible $$(x,y)\in\mathcal{X}\times\mathcal{Y}$$

Definition $$D(f)$$ The (deterministic) communication complexity of $$f:\mathcal{X}\times\mathcal{Y}\rightarrow\mathcal{Z}$$ is the maximum cost of a protocol $$\pi$$ that computes $$f$$.

Proposition 1 Every function $$f:\{0,1\}^n\times\{0,1\}^n\rightarrow\{0,1\}$$ has communication complexity $$D(f)\leq n+1$$.

Proposition 2 The function $$f(x,y)=\oplus_{i\in [n]}(x_i\oplus y_i)$$ has communication complexity $$D(f)\leq 2$$.

Proof $$\oplus_{i\in [n]}(x_i\oplus y_i)=(\oplus_{i\in [n]}x_i)\oplus (\oplus_{i\in [n]}y_i)$$

## Equality

Definition $$EQ:\{0,1\}^n\times\{0,1\}^n\rightarrow\{0,1\}$$ is $$EQ(x,y)=1$$ if $$x=y$$, $$EQ(x,y)=0$$ otherwise.

Theorem $$D(EQ)=\Omega(n)$$

Proof Let $$\pi$$ be any protocol that computes $$EQ$$.

Consider every leaf labelled $$1$$ in $$\pi$$. Let's tag such a leaf with $$x\in\{0,1\}^n$$ if $$(x,x)$$ leads to that leaf in $$\pi$$

Claim 1 No leaf of $$\pi$$ can be tagged with $$2$$ distinct $$x,x'\in\{0,1\}^n$$

Proof By contradiction. Assume $$(x,x)$$ and $$(x', x')$$ both lead to the same leaf. But then $$(x,x')$$ also defines the same path in $$\pi$$ to a leaf that output $$1$$.

Claim 1 implies that $$\pi$$ mush have $$\geq 2^n$$ labeeled $$1$$ (Pigeonhole principle)

If $$\pi$$ has depth $$d$$, it has $$\leq 2^d$$ leaves, so it must have depth $$\geq n$$.