CSC2429 / MAT1304: Circuit Complexity January 31, 2019 Lecture 4: Monotone Formulas for Majority; AC0 Circuits Instructor: Benjamin Rossman Scribe: Dmitry Paramonov Overview Section 1 Valiant's Monotone Formulas for Majority Section 2 AC0 Circuits 1 Valiant's Monotone Formulas for Majority Suppose you want to find whether the majority of inputs to a given circuit are on. This is a simple function, but with a surprisingly complex size. n n P Definition 1. The Hamming weight of some vector x 2 f0; 1g is given by jxj = xi = jfxi 2 i=1 [n]: xi = 1gj. ( n 1; jxj > 2 The majority function is given by the function MAJn(x1; : : : ; xn) = n . 0; jxj ≤ 2 n For convenience, we will let n be odd. This means that the latter case reduces to jxj < 2 . Now, suppose that we choose some i 2 [n] uniformly at random. What is the probability that xi is 1, conditioned on the value of MAJn(x)? 1 1 Claim 2. If MAJn(x) = 0, then Pri2[n](xi = 1) ≤ 2 − 2n . 1 1 If MAJn(x) = 1, then Pri2[n](xi = 1) ≥ 2 + 2n n n−1 Proof. Suppose that MAJn(x) = 0. Then, jxj < 2 . Because n is odd, jxj ≤ 2 . jfi 2 [n]: xi = 1gj Pr (xi = 1) = i2[n] n jxj = n n−1 ≤ 2 n 1 1 = − 2 2n The proof for MAJn(x) = 1 follows analogously. 1 Definition 3. Consider any m > n.A random projection from y to x is a function π : fy1; : : : ; ymg ! fx1; : : : ; xng, where each yi is mapped to an xj chosen uniformly at random, inde- pendently of the other yi. For a monotone formula F (y), let Fπ(x) be the monotone formula obtained from F by replacing each variable yi with π(yi). Note that for any fixed x, all the π(yi) are independent Bernoulli variables, each with probability jxj 1 1 n . This is a value that is at least 2n away from 2 , subject to the above inequalities. m Definition 4. For any f : f0; 1g ! f0; 1g define the output probability µf : [0; 1] ! [0; 1] as the probability of f(y1; : : : ; ym) being 1 given that each yi is an independent Bernoulli random variable with probability p. µf (p) = Pr (f(y1; : : : ; yn) = 1) y1;:::;ym2Bern(p) 3 2 For example, µMAJ3 (p) = p + 3p (1 − p). Note also that when f is monotone and non-constant, then µf is increasing, with µf (0) = 0 and µf (1) = 1. Claim 5. µf⊗g(p) = µf (µg(p)) Proof. Suppose that f ⊗ g : f0; 1gm×k ! f0; 1g, with g : f0; 1gk ! f0; 1g and f : f0; 1gm ! f0; 1g. Note that if y1;1; : : : ; ym;k are all chosen independently and uniformly at random with probability p, then the output of each copy of g used by f ⊗g is independent of the outputs of the others. Note that for any copy of g, the probability of its output being 1 is then µg(p). Thus, the copy of f in f ⊗g perceives the copies of g as being m independent random variables, each chosen to be 1 with probability µg(p). Thus, the output of f is 1 with probability µf (µg(p)). Corollary 6. (k) µf ⊗k (p) = µ (p) Now, we wish to show some properties of composing MAJ3 with itself. (c log n) 1 1 1 Lemma 7. There is a constant c such that µ (x) < n for 0 ≤ x ≤ − . Likewise, MAJ3 2 2 2n (c log n) 1 1 1 µ (x) > 1 − n for 0 ≤ x ≥ − . MAJ3 2 2 2n Proof. Note that because MAJ3 is monotone and non-constant, so µMAJ3 is increasing. Likewise, so is µ(k) for all k ≥ 1. Thus, it suffices to prove the desired relations for x = 1 − 1 and for MAJ3 2 2n 1 1 x = 2 + 2n . 1 1 Thus, suppose that p = 2 − δ, where δ ≤ 4 . 3 µ (p) = ( − 2δ2)δ MAJ3 2 1 ≤ − 1:25δ 2 2 1 1 Thus, as long as p is of the form p = 2 − δ for δ ≤ 4 , each application of µMAJ3 to p multiplies the 5 1 δ value by 4 . Thus, within O(log n) applications of µMAJ3 , the probability falls to be below 4 . 1 3 2 2 3 Now, suppose that p ≤ 4 . Then p + 3p (1 − p) ≤ 3p ≤ 4 p. So after an additional O(log n) 1 applications of µMAJ3 , the value of p falls below 2n . O(log n) 1 1 1 Thus, we see that µ ( − ) < n . MAJ3 2 2n 2 Furthermore, because µ is rotationally symmetric about ( 1 ; 1 ), we see that µO(log n)( 1 + 1 ) > MAJ3 2 2 MAJ3 2 2n 1 1 − 2n . This property can then be used to bound the size of the MAJn function. Theorem 8 (Valiant 1984). MAJn has polynomial size monotone formulas. Proof. Let t = c log n. We will construct a complete tree of MAJ3 functions, all composed with ⊗t each other, of depth t. In other words, we will construct a circuit for MAJ3 . ⊗t Note that because MAJ3 can be implemented by composing many instances of circuits solving ⊗t MAJ3, and because there is a monotone formula of size 5 for MAJ3, we see that Lmon(MAJ3 ) ≤ t t Lmon(MAJ3) ≤ 5 . ⊗t Fix a monotone formula F of this size for MAJ3 . Suppose that F is defined over variables y1; : : : ; y3t . Let π : fy1; : : : ; y3t g ! fx1; : : : ; xng be a random projection. Then, consider Fπ. n Consider any fixed x 2 f0; 1g where MAJn(x) = 0. As we noted, π(y1); : : : ; π(y3t ) are independent jxj 1 1 Bernoulli random variables with expectation n ≤ 2 − 2n . Thus, consider the probability of getting 1 returned by Fπ, over all choices of π. Pr(Fπ(x) 6= MAJn(x)) = Pr(Fπ(x) = 1) π π = Pr (F (y1; : : : ; y3t ) = 1) jxj y1;:::;y3t 2Bern( n ) jxj = µ ( ) F n 1 1 ≤ µ ( − ) F 2 2n 1 < 2n 1 Similarly, if x is fixed such that MAJn(x) = 1, we still get that Prπ(Fπ(x) 6= MAJn(x)) < 2n . 3 Thus, let us take a union bound, to see the probability that Fπ differs from MAJn on some input. n X Pr(9x 2 f0; 1g such that Fπ(x) 6= MAJn(x)) ≤ Pr(Fπ(x) 6= MAJn(x)) π π x2f0;1gn X 1 < 2n x2f0;1gn 1 = 2n 2n = 1 Thus, because the probability is less than 1, there must exist some π such that Fπ(x) = MAJn(x) n for all x 2 f0; 1g . Thus, for that π, Fπ is a monotone formula for MAJn. t t c log n c log 5 Thus, we see that Lmon(MAJn) ≤ leafsize(Fπ) ≤ 5 . By our choice of t, 5 = 5 = n . For 7 c < 3 (which can be shown by a more thorough argument), this shows that Lmon(MAJn) ≤ n . Thus, MAJn has polynomial monotone formula size. Curiously, Valiant's original proof used a different function, V (a; b; c; d) = (a ^ b) _ (c ^ d). This 1 had µV (p) be different, and not passing through 2 , but it could be modified to work. He also projected onto the set fx1; : : : ; xn; 0g, in order for it to work. However, his approach proved that 5:27 Lmon(MAJn) = O(n ), the best known upper bound. 2 At the same time, it is known that L(MAJn) = Ω(n ). However, the above upper bound is not explicit. You can't construct formulas using that proof. In O(1) 1983, Ajtai, Komlos and Szemeredi provided a set of explicit n monotone formulas for MAJn. However, their formulas were of size n1073 , which is not very practical. The best-known explicit formulas nowadays are n6000, which is still very large, which shows a difference between explicit and non-explicit proofs. 1.1 Slice Functions Definition 9. A boolean function f : f0; 1gn ! f0; 1g is a slice function if there is some k 2 f0; : : : ; ng such that if jxj < k then f(x) = 0 and if jxj > k then f(x) = 1. n A treshold function is a function THRk;n : f0; 1g ! f0; 1g such that THRk;n(x) is 1 if and only if jxj ≥ k. This is a generalization of MAJn. Note that threshold functions are slice functions, and note that all slice functions are always mono- tone. Theorem 10 (Berkowitz 1982). If f is a slice function then Lmon(f) ≤ L(f) · poly(n), and Cmon(f) ≤ C(f) + poly(n). Proof. We know that MAJn has polynomial-size monotone formulas. Thus, we can also compute any threshold function THRk;n using a polynomial-size monotone formula, by padding the input of THRk;n with 0's and 1's, and then taking the majority of the result. 4 Note that you will never need more than n input bits of padding, so at worst, you need to have to solve MAJ2n. Let F be a De Morgan formula computing some k-slice function f. Express this in negation normal form. 0 0 Now, consider a monotone formula THRk;n ^ F , where F is equivalent to F , but all instances of xi are replaced by THRk;n(x1; : : : ; xi−1; 0; xi+1; : : : ; xn). If there are less than k inputs which are 1, then THRk;n is 0, so this function evaluates to 0. If more than k inputs are 1, then 0 THRk;n(x1; : : : ; xi−1; 0; xi+1; : : : ; xn) will always be 1, so F will be 1, as all its inputs are 1.