Computational Complexity Theory, Fall 2010 September 29 Lecture 11: Valiant-Vazirani and Toda's Theorem Lecturer: Kristoffer Arnsfelt Hansen Scribe: Andreas Hummelshøj ] P class of functions Definition 1. P ]P ]P P = class of languages computed in polynomial time with oracle access to L]SAT = fhφ, kij]SAT (φ) = kg. Clearly being able to count the number of satisfying assignment to a Boolean formula, allows one to decide if such an assignment exists. Lemma 2. NP ⊆ P ]P , also P ]P ⊆ P SP ACE. Our goal will be to prove the following theorem, that shows that counting is in fact much more powerful than it immediately seems. Theorem 3. Toda's Theorem PH ⊆ P ]P n Definition 4 (Pair-wise independent hash functions). Let Hn;k be a finite set of functions f0; 1g ! k 0 f0; 1g . We say that Hn;k is a family of pair-wise independent hash functions if for x; x 2 f0; 1gn ; x 6= x0 and for all y; y0 2 f0; 1gk: 1 P r h(x) = y ^ h(x0) = y0 = : 22k Example Hn;k: Let A be a k × n matrix over GF(2) (=G[eneral]F[ield] (2)). Let b be a dim-k vector over n k GF(2). Let Hn;k : f0; 1g ! f0; 1g be given by: Hn;k(x) = Ax + b: Proposition 5. Hn;k above is a pair-wise independent family of hash functions. Proof. Pick x 6= x0, y 6= y0. P r h(x) = y ^ h(x0) = y0 =P r Ax + b = y ^ Ax0 + b = y0 0 0 0 0 =P r Ax + b = y Ax + b = y P r Ax + b = y =2−k · 2−k 1 Lemma 6. (Valiant-Vazirani) Let Hn;k be a pair-wise independent family of hash functions and let S ⊆ f0; 1gn be such that: 2k 2k ≤ jSj ≤ : 4 2 k 1 Pick h 2 Hn;k uniformly at random. Then P r There is a unique x 2 S : h(x) = 0 ≥ 8 . Proof. h i P r There is a unique x 2 S : h(x) = 0k h i =P r 9x 2 S : h(x) = 0k ^ 8x0 2 Sn fxg : h(x0) 6= 0k X h i = P r h(x) = 0k ^ 8x0 2 Sn fxg : h(x0) 6= 0k (= since the events for different x are disjoint) x2S X h i h i = P r h(x) = 0k − P r h(x) = 0k ^ 9x0 2 Sn fxg : h(x0) = 0k x2S 0 1 X h ki X h k 0 ki ≥ @P r h(x) = 0 − P r h(x) = 0 ^ h(x ) = 0 A (only ≥ since events are not disjoint) x2S x02Snfxg 0 1 X −k X −2k = @2 − 2 A x2S x06=x =jSj 2−k − (jSj − 1) 2−2k jSj jSj ≥ 1 − 2k 2k 2k=4 2k=2 1 ≥ 1 − = 2k 2k 8 Theorem 7. Valian-Vazirani There is a probabilistic polynomial time algorithm that, given a boolean formula φ produces another boolean formula τ sucht that: 1 If φ 2 SAT then P r[φ ^ τ has a unique satisfying assignment] ≥ 8n , If φ 3 SAT then P r[φ ^ τ 2 SAT ] = 0. Proof. We shall use the hash family given in the example above. Let S ⊆ f0; 1gn be the set of assingments satisfying φ. Pick k 2 f2; 3; : : : ; n + 1g uniformly at random. Pick A; b uniformly at random. k Output φ ^ "hA;b(x) = 0 ". 1 k k With probability n , we have 2 =4 ≤ jSj ≤ 2 =2. The result follows from the lemma. 2 k "hA;b(x) = 0 " has a polynomial size formula: Ax + b = 0 k 0 n 1 ^ X ≡ @ aijxj + bj = 0A i=1 j=1 k 0 n 1 ^ M ≡ @ aij ^ xj ⊕ bj = 0A i=1 j=1 L V is done by a binary tree of depth log n, by one of depth log k.(aij ^ xj ⊕ bj = 0) is a simple formula. Summing up, we get a formula of depth O(log n), giving a total size of nO(1), as promised. Definition 8. Notation: For a boolean formula φ, let ]φ denote the number of satisfying assign- ments. L L Definition 9 ( quantifier). Let φ(x) be a boolean formula. We say x2f0;1gn φ(x) is true if ]φ is odd. Definition 10 (L P ). L P = class of languages L where there is a polynomial p and R 2 P such that: n o p(jxj) x 2 L () y 2 f0; 1g hx; yi 2 R is odd: ( ) M M SAT = φ φ ≡ φ(x) and φ is true : x Proposition 11. L SAT is complete for ]P . Proof. Cook's theorem preserves the number of witnesses, hence also parity. Remark: L P is the least significant bit of a ]P function. Lemma 12. (First part of Toda's Theorem) For all k, there is a probabilistic algorithm A that, on input hΨ; mi, where is a quantified formula with k blocks of quantifiers, runs in polynomial time and outputs a formula φ = A( ; 1m) such that: h M i −m 2 ΣkSAT )P r φ 2 SAT ≥ 1 − 2 h M i −m 2= ΣkSAT )P r φ 2 SAT ≤ 2 Lemma 13. There are polynomial time algorithms that, given formulas φ in n variables and in m variables, outputs formulas φ · in m + n variables and φ + in max(m; n) + 1 variables, such that: ](φ · ) =(]φ) · (] ) ](φ + ) =(]φ) + (] ) 3 Proof. Define (φ · )(x; y):≡ (φ(x)) ^ ( (y)). (Assume wlog m ≤ n): (φ + )(x; z):≡ (z ^ φ(x1; : : : ; xn)) _ (z ^ (x1; : : : ; xm) ^ xm+1 ^ · · · ^ xn). Boolean expressions of L-quantified formulas: " # " # M M M φ(x) ^ (y) () (φ · )(x; y) x y x;y " # M M : φ(x) () (φ + 1)(x; z1) x x " # " # " # " #! M M M M φ(x) _ (y) () : : φ(x) ^ : (y) x y x y Note this extends to taking Boolean AND and OR of polynomially many L-quantified formulas (in fact, we can just form a binary tree), and these can be computed efficiently. Proof of Lemma 12. Proof by induction. k = 1 ≡ 9x 2 f0; 1gn 0(x). (Remark: one-sided error makes success-amplification very easy.) Run the algorithm from Valian-Vazirani's Theorem for K trials to get τ1; : : : ; τK . Set K _ M 0 φ := ( (x) ^ τi(x)): i=1 x (Here we can use W by the previous remarks, and get a L-quantified formulas equivalent to φ, which is in fact the formula we output.) Analysis: Suppose is false. Then P r [φ 2 L SAT ] = 0. Suppose is true. Then 0 L 1 L 1 K P r [ (x) ^ τi(x) 2 SAT ] ≥ 8n . Thus P r [φ 2 SAT ] ≥ 1− 1 − 8n . Pick K = O(n·m) 1 K −m such that (1 − 8n ) ≤ 2 . Observation: The above also works even if has free variables. Suppose we have l free variables y. The above will immediately give that for any choice of y, with probability at least 1 − 2−m we have (y) 2 SAT , φ(y) 2 L SAT . We shall in fact want to have the stronger statement that P r[8y : (y) 2 SAT , φ(y) 2 L SAT ] > 1 − 2−m. But this statement follows by a union bound over all y from the even stronger statement, 8y : P r[ (y) 2 SAT , φ(y) 2 L SAT ] > 1−2−ml, so we can just do as above with K = O(lmn). n 0 0 k − 1 ) k Let :≡ 9x1 2 f0; 1g (x1), where is a Πk−1SAT formula. By induction hypothesis (note: here we use that L P is closed under complements) there is a probabilistic 4 0 L algorithm that on input (x1) produces a ⊕SAT formula, β(x1) ≡ z ρ(x1; z), with free variables x1, such that 0 −(m+1) P r 8x1 : (x1) 2 Πk−1SAT , β(x1) 2 ⊕SAT ≥ 1 − 2 Run again the Valiant-Vazirani algorithm for K independent trials, and define K _ M φ := (β(x1) ^ τi(x1)): i=1 x1 K 0 _ M 0 φ := (x1) ^ τi(x1) : i=1 x1 Here φ is the L SAT formula that we output. We use φ0 for the analysis below. Since φ and φ0 are equivalent with probability at least 1 − 2−(m+1), we can just work with φ0 and then at the end add 2−(m+1) to the error probability. The calculation is now as in the base case. Suppose is false. Then P r []φ0 is odd] = 0. 0 1 Suppose is true. Then P r []( (x1) ^ τi(x1)) is odd] ≥ 8n . Thus P r []φ is odd] ≥ 1 − 1 K 1 K −(m+1) 1 − 8n . Pick K = O(m · n) such that (1 − 8n ) ≤ 2 . (We don't use ⊕SAT here, 0 since is a quantified formula. One could of course have defined ⊕Πk−1SAT). Now, the total error is ≤ 2−(m+1) + 2−(m+1) = 2−m. Again, similarly to the base case, the argument also works if φ has free variables. 5.