Lecture 4: Unique-SAT, Parity-SAT, and Approximate Counting

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1 Advaced Complexity Theory Sprig 206 Lecture 4: Uique-SAT, Parity-SAT, ad Approximate Coutig Prof. Daa Moshkovitz Scribe: Aoymous Studet Scribe Date: Fall 202 Overview I this lecture we begi talkig about coutig. A example of a coutig problem would be to fid the umber of satisfyig assigmets a formula has. First, we will discuss U iquesat ad the Valiat-Vazirai Theorem, ad the SAT. The we will show that approximate coutig is i BP P NP. Next time we will cover Toda s Theorem which says that P H P #P, which shows that the power to cout exactly is eough to solve ay problem i the polyomial hierarchy. 2 Mai Sectio 2. Uique-SAT Defiitio: U iquesat is a promise problem, meaig that we are give a guaratee o the iput, which distiguishes betwee the followig two classes: U Y ES = {φ φ has satisfyig assigmet } U NO = {φ φ has 0 satisfyig assigmets } The UiqueSAT problem is: Give φ U Y ES U NO, distiguish betwee φ U Y ES ad φ U NO. If φ has more tha oe satisfyig assigmet, there is o guaratee o the output. 2.2 Valiat-Vazirai Theorem This theorem was proved by Valiat ad Vazirai i 986[]. The theorem states that there is a radomized reductio from SAT to U iquesat. The reductio is as follows: φ SAT = R(φ) U Y ES with probability p(), where p() is a polyomial i φ / SAT = R(φ) U NO always Notice that this does ot seem very good sice p() is very small, but this result is very useful for may other results i complexity. Ay improvemet o this result would cause a improvemet i may other results as well.

2 2.2. Proof of Valiat-Vazirai Theorem The idea of the proof is to costruct a reductio i such a way that the ew formula is the same as the old but icludes a extra term which uiquely idetifies oe satisfyig assigmet. I order to uiquely idetify oe satisfyig assigmet, we will hash the assigmets. So we get R(φ) = φ (h(x) = 0), where h : {0, } {0, } m is a radomly selected hash fuctio. However, for this to work, we eed the hash fuctio to ot have collisios, so that h(x) = 0 for oly oe satisfyig assigmet. We will pick h(x) radomly from a family H, to be discussed ext. Crucial Poit: Choosig the correct size of the set to be hashed to, 2 m, is very importat. If we choose m to be too large, othig will be mapped to 0. If we choose a m that is too small, there will be collisios so multiple thigs might get mapped to 0. We pick a value for m at radom from 0,,,. This is where the p() the probability that we chose the correct value for m is. comes from because Defiitio. A pairwise idepedet hash family H = {h : {0, } {0, } m } has the property that x y {0, }, a, b {0, } m, Pr(h(x) = a h(y) = b) = ( ) 2 2 m. Examples of pairwise idepedet hash families:. H = all fuctios 2. H = {Ax + b A {0, } m, b {0, } m }, where the matrix-vector multiplicatio is performed over F 2. Now suppose we chose m such that 2 i # satisfyig assigmets < 2 i+, where m = i + 2, which will happe with probability. For all x { 0, }, defie the evet U x to be that x is the uique satisfyig assigmet that h maps to 0 (h may map other strigs to 0, but those caot be satisfyig assigmets). The Pr(U x ) Pr(h(x) = 0) Pr(h(x) = 0 h(y) = 0) 2 m N 2 2m, 2 m+ y x, is sat. assg. where N is the umber of satisfyig assigmets. We wat to fid Pr( x U x ). Sice these are disjoit evets for each x, Pr( xu x ) = N Pr(Ux ) 2 m+ 8 x is sat. assg.. So with probability 8, φ SAT = R(φ) U Y ES. If φ / SAT, R(φ) U NO because there are o satisfyig assigmets. No oe kows how to get rid of the factor, but if we talk about SAT istead of UiqueSAT we ca do much better. 2.3 Parity-SAT Defiitio 2. SAT = {φ The umber of satisfyig assigmets to φ is eve }. 2

3 2.3. Valiat-Vazirai for SAT We ca modify the Valiat-Vazirai reductio above to achieve the followig for SAT : φ SAT = R (φ) SAT with probability Ω( ) φ / SAT = R (φ) / SAT always Proof of Valiat-Vazirai for SAT The reductio is as follows: give a formula φ, we first perform the U iquesat reductio φ R(φ), which with probability Ω( ) will give a formula that s uiquely satisfiable (if φ was satisfiable to begi with). The, R (φ) will be the formula (y R(φ)(x)) (y x = 0). Note that the satisfyig assigmets for R (φ) are: y =, x satisfyig for R(φ). y = 0, x = 0 If R(φ) has oe satisfyig assigmet, the R (φ) has two satisfyig assigmets, ad so R (φ) SAT. If R(φ) has o satisfyig assigmets, the R (φ) has oe satisfyig assigmet, ad so R (φ) / SAT. Sice these two are the oly possibilities with probability Ω( ), the theorem holds Amplified Valiat-Vazirai We ca improve the Ω( ) probability for the SAT reductio sigificatly: φ SAT = R (φ) SAT with probability exp( ) φ / SAT = R (φ) / SAT always Proof of Amplified Valiat-Vazirai Let k = 2. Ivoke the reductio R k times, each with idepedet radomess, ad relabel the variables so that every output formula is usig a ew set of variables. The we get φ,...φ k o disjoit sets of variables. Eve though R has a low probability of succeedig, we oly eed it to succeed oce, so ivokig it may times will give us the result we wat. k Defie φ = i= φ i. The the umber of satisfyig assigmets for φ is k (# satisfyig assigmets to φ i ). Notice that if eve oe term i the above product is eve, the product will be eve. So if the umber of satisfyig assigmets for φ i is odd for all i, the φ / SAT. If for some i the umber of satisfyig assigmets to φ i is eve, the φ SAT. The probability that R (φ) SAT give that φ SAT is ( Ω( 2 )) e, because each reductio attempt is idepedet of oe aother ad ( ) e. i 3

4 2.4 Approximate Coutig We ll show that give some boolea formula φ, distiguishig betwee the followig two cases: The umber of satisfyig assigmets for φ is 2 k+ The umber of satisfyig assigmets for φ is 2 k is i BP P NP. This meas that we ca 2-approximate the umber of satisfyig assigmets to a boolea formula i BP P NP, because we ca just try each possible value for k ad the check if the umber of satisfyig assigmets is less tha or equal to 2 k util we fid the oe which gives us a 2-approximatio. Note: Ay problem where you eed to approximate the sum of oegative umbers will work this way. Oce the umbers are allowed to be egative, it becomes much harder Algorithm for Approximate Coutig The idea of the algorithm is to hash the set of all assigmets to a set of size less tha 2 k. The N P oracle ca give us satisfyig assigmets. The algorithm works as follows: Pick h : {0, } {0, } m at radom from a pairwise idepedet hash family, with m = k 5. I the case that k < 5, just use brute force. Ask the prover for > 48 assigmets x that satisfy φ ad h( x) = 0. Notice that this algorithm is i BP P NP. Now we just eed to prove that the algorithm is correct Correctess of Algorithm Lemma 3. For ay S {0, }, S 4 2 m, ɛ 2 ( X S 2 m ɛ S ) where X = {x S h(x) = 0}. Pr h H Proof. The idea is to boud the variace of X. Write X = x S I h(x)=. 0 We ca boud the variace of this sum because H is a pairwise idepedet hash family, ad the variace of a sum of pairwise idepedet variables is simply the sum of the variaces. We get that ( ) Var(X) = S Pr(I ) Pr(I ) 2 h(x)=0 h(x)= 0. 2 m, 4 The we ca use Chebyshev s Iequality, Pr( X E[X] ɛe[x]) stated lemma result. Var(X), ad we get the ɛ 2 E[X] 2 4

5 Now there are two cases:. S 2 k+, take ɛ = 4 The S 4 2 m = 64 2 m ɛ 2 With probability 3 4, the umber of satisfyig assigmets hashed to 0 is 64( ± 4) > 48 If this is larger tha 48 there are > 48 satisfyig assigmets ad the prover ca give them to us. 2. S < 2 k, take ɛ = 2 The there is some S such that S S ad S = 2 k = 32 2 m With probability 3 4, the umber of satisfyig assigmets hashed to 0 is 32( ± 2) 48 So i this case there are ot more tha 48 satisfyig assigmets, so the prover caot provide more tha 48 satisfyig assigmets. We ca the amplify the 3 probability by repetitio to get a arbitrarily large probability of success Coclusio I this lecture we proved the Valiat-Vazirai Theorem for U iquesat ad SAT, ad the saw a algorithm i BP P NP for approximate coutig. Next time, we will go from approximate coutig to exact coutig. Refereces [] Valiat, L.; Vazirai, V. NP is as easy as detectig uique solutios, Theoretical Computer Sciece 47:

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