6.841 Advanced Complexity Theory Feb 18, 2009 Lecture 5 Lecturer: Madhu Sudan Scribe: Yang Cai

1 Overview

• P ARIT Y∈ / AC0.

• Random Restriction • Switching Lemma DNF → CNF

2 Introduction

We first introduce two sets of classes. • ACk: Class of functions computable by polynomial size and O((log n)k) depth circuits over {∞ − AND, ∞ − OR,NOT } gates.

• NCk: Class of functions computable by polynomial size and O((log n)k) depth circuits over {2 − AND, 2 − OR,NOT } gates.

k k+1 k+1 S k S k We know that for any k, AC ⊆ NC ⊆ AC , so k AC = k NC . Through this lecture, we assume circuits in ACk are organized to have alternating levels of AND and OR gates. Because all NOT gates can be moved to the first level, and since the AND and OR gates have infinite fan-in, we can combine any consecutive AND or OR levels. So we can consider the depth as the number of AND and OR levels. The goal of this lecture is to prove the following theorem P ARIT Y∈ / AC0. By P ARIT Y we mean X P ARIT Y (x1, x2, . . . , xn) = xi (mod 2). i

Theorem 1 [Furst, Saxe, Sipser; Ajtai; Yao; H¨astad;Razborov; Razborov-Smolensky] P ARIT Y∈ / AC0.

3 Random Restriction

If x1, x2, . . . , xn are variables, a random restriction on them is to randomly set values for most of the variables. Formally, we say a random restriction ρ with parameter p, if for every i, we leave xi unset with probability 1−p p, restrict it as xi = 0 or xi = 1 with equal probability 2 .  0 with prob. 1−p  2 xi = xi with prob. p  1−p 1 with prob. 2

If a function f has n variables, after this restriction, we get the function f|ρ with about pn variables. Then

P ARIT Y |ρ (x1, x2, . . . , xn) = P ARIT Y (xi1 , xi2 , . . . , xit )(⊕1) (t ≈ pn)

5-1 4 Switching Lemma

Lemma 2 (Switching Lemma) Let F be a DNF formula on n variables with size S ≤ nc1 . Let ρ be a random √1 restriction with p = n , 1 Pr[F |ρ depends on ≥ C variables] ≤ n2c1

C is a constant decided by c1.

We will first use Switching Lemma to prove P ARIT Y∈ / AC0, then prove the Switching Lemma. Notice that we can always set the circuit’s bottom 2 levels to be DNF’s, because if it’s a CNF, we can just negate the circuit.

Theorem 1 P ARIT Y∈ / AC0 Proof We prove this by induction. The base of induction is not hard to see that, if a circuit of depth 2 computes P ARIT Y then its size is O(2n). If for every c and depth d, no depth d circuit of size S = computes P ARIT Y . Now we prove this is also true for d + 1. c1 Say G is a circuit with depth d + 1 and size n that computes P ARIT Y (x1, x2, . . . , xn). √1 Hit G with random restriction ρ with parameter p = n . We know the following:(a) According to √ √ − n pn n Chernoff bound, we know with probability (1 − 2 ), there are at least 2 = 2 variables are unset. (b) According to Switching Lemma, for every DNF formula, it depends on only C variables with probability 1 1− S2 . (c) Since there are at most S DNF formula at the bottom, all depth 2 gates depend on ≤ C variables with probabilty 1 − 1 . When all depth 2 gates depend on ≤ C variables, we can replace the bottom levels S √ by CNF of size 2C , then G becomes G. G computes PARITY of n unrestricted variables in a circuit with √ e e 2 c1 C n depth d and size n 2 = poly( 2 ). Contradiction to the induction hypothesis.

5 Proof of Switching Lemma

Let the DNF F = T1 ∨ T2 ∨ ... ∨ Tm. We restrict variables in two stages Stage 1 √ Restrict variable with probability p, i.e.  √ 0 with prob. 1− p  √2 x = x with prob. p i i √  1− p  1 with prob. 2

f −→ f|ρ1 . Stage 2 √ Restrict variables in f|ρ1 with probability p, i.e.  √ 0 with prob. 1− p  √2 x = x with prob. p i i √  1− p  1 with prob. 2

f|ρ1 −→ f|ρ1∪ρ2 . Stage 1

• Case 1: Terms with fan-in ≥ 4 log S √ 1 + p 2 1 Pr[Any T erm with fan − in ≥ 4 log S 6= 0] ≤ ( )4 log S ≤ ( )4 log S ≤ 2 3 S3

5-2 Therefore, 1 Pr[∃ T erm with fan − in ≥ 4 log S doesn0t become 0] ≤ S2 • Case 2: Terms with fan-in≤ 4 log S √ 1 Pr[T depends on c variables] ≤ (4 log S)c0 ( p)c0 ≤ i 0 S3 Therefore, 1 Pr[∃ T erm with fan − in ≤ 4 log S, depends on ≥ c variables] ≤ 0 S2

. c0 is a constant depend on c1, so now the DNF is a c0-DNF.

Stage 2

c0 • Case 1: ∃ many disjoint Term T1,T2,...,Tl, ≥ 3 4 log S. 1 Pr[T = 1] ≥ ( )c0 i 3 1 Pr[T 6= 1] ≤ (1 − )c0 i 3

1 c0 1 Pr[∃ T = 1] ≥ 1 − ((1 − )c0 )l = 1 − 2−l/3 ≥ 1 − i 3 S2

• Case 2: The max number of disjoint Ti’s ≤ 4 log S

c0 Claim: ∃ set H with 4c03 log S variables, s.t. ∀i H ∩ Ti 6= ∅. Proof Select disjoint Ti’s greedily. When stop, variables in H hit all Ti’s

– Part a: First restrict variables of H; As in Case 2 of Stage 1, we can find a constant b, s.t. the number of unset variables in H ≤ b.

– Part b: Now we restrict variables not in H. And we use induction on c0. 0 Induction hypothesis: Any (c0 − 1)-DNF under random restriction depends on only C variables with high probability, C0 is a constant.

Now we want to prove that c0-DNF under random restriction also only depends on constant many variables with high probability. b 0 Claim: With high probability c0-DNF only depends on C ≤ b + 2 C variables. Proof After restricting variables in H, enumerate all possible assignments of the b unset vari- ables. After assignning values to these variables, the c0-DNF becomes a (c0 − 1)-DNF. According to the induction hypothesis, we know this depends on C0 variables under restriction. So the b 0 c0-DNF depends on at most b + 2 C variables.

5-3