How to Cope with NP-Completeness

Transcription

1 How to Cope wth NP-Completeness Kurt Mehlhorn June 12, 2013 NP-complete problems arse frequently n practce. In today s lecture, we dscuss varous approaches for copng wth NP-complete problems. When NP-completeness was dscovered, algorthm desgners could take t as an excuse. It s true, that I have not found an effcent algorthm. However, nobody wll ever fnd an effcent algorthm (unless P = NP). Nowadays, NP-completeness s consdered a challenge. One searchers for exact algorthms wth runnng tme c n wth small c, one searches for approxmaton algorthms wth a good approxmaton rato, one tres to prove that the problem s hard to approxmate unless P = NP, one searches for specal cases that are easy to solve or easy to approxmate, and one ntroduces parameters besdes nput sze to descrbe the problem and to analyze the runnng tme n. Contents 1 Exact Algorthms Enumeraton Branch-and-Bound-and-Cut Approxmaton Algorthms The Knapsack Problem: A PTAS va Scalng and Dynamc Programmng Set Cover: A Greedy Algorthm Maxmum Satsfablty: A Randomzed Algorthm Vertex Cover Hardness of Approxmaton The PCP-Theorem (Probablstcally Checkable Proofs) Max-3-SAT s hard to approxmate From Max-3-SAT to Vertex Cover Approxmaton Algs for Problems n P General Search Heurstcs 13 4 Parameterzed Complexty Parameterzed Complexty for Problems n P

2 1 Exact Algorthms 1.1 Enumeraton The smplest exact algorthm s complete enumeraton. For example, n order to fnd out whether a boolean formula over varables x 1 to x n s satsfable, we could smply terate over all 2 n possble assgnments. Many short-cuts are possble. For example, f the formula contans a unt-clause,.e., a clause consstng of a sngle lteral, ths lteral must be set to true and hence only half of the assgnments have to be tred. Note that unt-clauses arse automatcally when we systematcally try assgnments, e.g., f x 1 x 2 x 3 s a clause and we have set x 1 and x 2 to false, then x 3 must be set to true. There are many rules of ths knd that can be used to speed-up the search for a satsfyng assgnment. We refer the reader to [NOT06] for an n-depth dscusson. There has been sgnfcant progress on speedng-up SAT-solvers n recent years and good open source mplementatons are avalable. Local Search There s a smple randomzed algorthm due to Schönng (1999) that runs n tme (4/3) n where n s the number of clauses and decdes 3-SAT wth hgh probablty [Sch99]. The algorthm works as follows (adapted from [Sch01]). We assume that all clauses have at most k lterals. ) n do T = 2 ln 1 ε tmes (ths choce guarantees an error probablty of ε ( 2(k 1) k Choose an assgnment x {0,1} n unformly at random; do C k n tmes (C k s a (small) constant; see below) f x s satsfyng, stop and accept (ths answer s always correct) Let C be a clause that s not satsfed choose a lteral n C at random and flp the value of the lteral n x. declare the formula unsatsfable (ths s ncorrect wth probablty at most ε) The algorthm s surprsngly smple to analyze. The analyss determnes the value of T as a functon of the desred error probablty. Suppose that the nput formula F s satsfable and let x be some satsfyng assgnment. We splt the set of all assgnments nto blocks. Block B contans all assgnments wth Hammng dstance exactly to x. Let x be our current assgnment and let C be a clause that s not satsfed by x. Then all lterals of C are set to false by x and l 1 lterals of C are set to true by x. If we flp the value of one of these l lterals, the Hammng dstance of x and x decreases by one. If we flp the value of one of the other at most k l lterals of C, the Hammng dstance wll ncrease by one. Thus, f x belongs to B, the next x belongs to B 1 wth probalty at least 1/k and belongs to B +1 wth probablty at most 1 1/k. The algorthm accepts at the latest when x B 0 ; t may accept earler when x equals some other satsfyng assgnment. >= 1/k >= 1/k +1 1 <= 1 1/k <= 1 1/k 2

3 So n order to analyze the algorthm, t suffces to analyze the random walk wth states 0, 1, 2,... and transton probablty 1/k from state to state 1 and probablty 1 1/k from state to state + 1. We want to know the probablty of reachng state 0 when startng n state and the expected length of a walk leadng us from state to state 0. What should we expect? The walk has a strng drft away from zero and hence the probablty of ever reachng state 0 should decrease exponentally n the ndex of the startng state. For the same reason, most walks wll never reach state 0. However, walks that reach state 0 should be relatvely short. Ths s because long walks are more lkely to dverge. The ntal assgnment x has some Hammng dstance to x. Clearly, ths dstance s bnomally dstrbuted,.e., ( ) n prob(hammng dstance = ) = 2 n. For ths (nfnte) Markov chan t s known (see, for example, [GS92]) assumng k 3: The probablty 1 of reachng 0 from. exponentally n. More precsely, 1 prob(absorbng state 0 s reached process started n state ) = ( k 1 ). Walks startng n state and reachng state 0 are relatvely short 2. More precsely, E[number of steps untl state 0 s reached process started n state and the absorbng state 0 s reached k ]) =C, where C k = k2 2k+2 k(k 2) artcle. Schoenng states C k = k(k 2).. WARNING: ths s NOT the constant that s gven n Schoenng s WARNING k 2 k = 1 + k 2 2 wthout proof. My calculaton yelds C k = Combnng these facts wth the ntal probablty for state, we obtan 1 Let p be the probablty of reachng state 0 from state 1. Such a walk has the followng form: the last transton s from 1 to 0 (probablty 1/k). Before that t performs t 0 loops,..e, t goes from 1 to 2 and then returns wth probablty p back to 1 (probablty ((k 1)/k)p) t. Thus p = 1 k ( k 1 p) t = 1 1 t 0 k k 1 k 1 k p ) = 1 k (k 1)p. Ths equaton has solutons p = 1 and p = k 1 1. We exclude the former on semantc reasons. Thus the probablty of reachng state 0 from state 1 s 1/(k 1) and the probablty of reachng t from state s the -th power of ths probablty. 2 Let L be the expected length of a walk from 1 to 0. Such a walk has the followng form: the last transton s from 1 to 0 (length 1). Before that t performs exactly t 0 loops,..e, t goes from 1 to 2 and then returns back to 1 (probablty ((k 1)/k)p) t (1/k), where p s as n the prevous tem) and length t(l + 1). Thus L = 1 + t 0 ( k 1 k = 1 (L + 1) d dk p) t 1 k t(1 + L) = 1 + (L + 1) tk t 1 = 1 + (L + 1) t 1 t 1 d dk ( k t ) 1/k 1 1/k = 1 (L + 1) d dk (k 1) 1 = 1 + (L + 1)(k 1) 2. Ths solves to L = k2 2k+2 k(k 2). The expected length of a walk startng and and reachng 0 s tmes ths number. 3

4 for the probablty of reachng state 0: prob(the absorbng state 0 s reached) = 0 j n ( ) n 2 n (1/(k 1)) ) n ( = = 2 n ( k 1 and for the expected number of steps of a walk reachng state 0: ( n E[# of steps untl 0 s reached 0 s reached] = C k 0 ) 2 n = C k n 2 n 1 k 2(k 1) ) n ( ) n = C kn 1 2. By Markov s nequalty the probablty that a non-negatve random varables s less than twce ts expectaton s at least 1/2. Thus prob(0 s reached n at most C k n steps 0 s reached) 1/2. Combnng ths wth the probablty that 0 s reached, we obtan prob(after at most C k n steps the state 0 s reached) (1/2)(k/(2(k 1))) n. Let p = (1/2)(k/(2(k 1))) n. The probablty that x s reached from a random x s at least p. If we repeat the experment T tmes, the probablty that we never reach x s bounded by (1 p) T = exp(t ln(1 p)) exp( T p) (recall that ln(1 p) p for 0 p < 1). In order to have error probablty at most ε, t therefore suffces to choose T such that exp( T p) ε,.e., T 1 p ln 1 ( ) 2(k 1) n ε = 2 ln 1 k ε. ( ) Theorem 1 (Schoenng) In tme O(2 2(k 1) n k ln 1 ε poly(n,m)) one can decde satsfablty of formulas n varables, m clauses, and at most k lterals per clause wth (one-sded) error ε. A derandomzaton wth essentally the same tme bound was obtaned recently by Moser and Schreder []. 1.2 Branch-and-Bound-and-Cut For optmzaton problems, the branch-and-bound and branch-and-bound-and-cut paradgm are very useful. We llustrate the latter for the Travellng Salesman Problem. Recall the subtour elmnaton LP for the travelng salesman problem n a graph G = (V,E). Let c e be the cost of edge e. We have a varable x e for each edge. subject to e δ(v) mn c e x e e x e = 2 x e 2 e δ(s) x e 0 for each vertex v for each set S V, /0 S V 4

5 We solve the LP above. If we are lucky, the soluton s ntegral and we found the optmal tour. In general, we wll not be lucky. We select a fractonal varable, say x e, and branch on t,.e., we generate the subproblems x e = 0 and x e = 1. We solve the LP for both subproblems. Ths gves us lower bounds for the cost of optmal tours for both cases. We contnue to work on the problem for whch the lower bound s weaker (= lower). We may also want to ntroduce addtonal cuts. There are addtonal constrants known for the TSP. There are also generc methods for ntroducng addtonal cuts. We refer the reader to [ABCC06] for a detaled computatonal study of the TSP. 2 Approxmaton Algorthms Approxmaton construct provably good solutons and run n polynomal tme. Algorthms that run n polynomal tme but do not come wth a guarantee on the qualty of the soluton found are called heurstcs. Approxmaton algorthms may start out as heurstcs. Improved analyss turns the heurstc nto an approxmaton algorthm. Problems vary wdely n ther approxmablty. 2.1 The Knapsack Problem: A PTAS va Scalng and Dynamc Programmng The Knapsack Problem s easly stated. We are gven n tems,, each wth a value and a weght. Let v and w be the value and the weght of the -th tem. We are also gven a weght bound W. The goal s to determne a subset S [n] of the tems, maxmzng the value S v and obeyng the weght constrant S w W. We may assume w W for all. The Greedy Heurstc: We order the tems by value per unt weght,.e., by v/w. We start wth the empty set S of tems and then terate over the tems n decreasng order of value-perunt-weght-rato. If an tem fts,.e., the current weght plus the weght of the tem does not exceed the weght bound, we add t, otherwse we dscard t. Let V g be the value computed. The greedy heurstc s smple, but no good. Consder the followng example. We have two tems: the frst one has value 1 and weght 1 and the second one has value 99 and weght 100. The weght bound s 100. The greedy heurstc consders the frst tem frst (snce 1/1 > 99/100) and adds t to the knapsack. Havng added the frst tem, the second tem wll not ft. Thus the heurstc produces a value of 1. However, the optmum s 99. A small change turns the greedy heurstc nto an approxmaton algorthm that guarantees half of the optmum value. Return the maxmum of V g and max v. Lemma 1 The modfed greedy algorthm acheves a value of V opt /2 and runs n tme O(nlogn). Proof: Order the tems by decreasng rato of value to weght,.e., v 1 /w 1 v 2 /w 2... v n /w n. Let k be mnmal such that w w k+1 > W. Then w w k W. The greedy algorthm wll pack the frst k tems and maybe some more. Thus V g v v k. 5

6 The value of the optmal soluton s certanly bounded by v v k + v k+1. Thus V opt v v k + v k+1 V g + maxv 2max(V g,maxv ). The tme bound s obvous. We sort the tem accordng to decreasng rato of value to weght and then terate over the tems. Dynamc Programmng: We assume that the values are ntegers. We show how to compute an optmal soluton n tme O(n 2 v max ). Clearly, the maxmum value of the knapsack s bounded by nv max. We fll a table B[0,nv max ] such that, at the end, B[s] s the mnmum weght of a knapsack of value s for 0 s nv max. We fll the table n phases 1 to n and mantan the nvarant that after the -th phase: { } B[s] = mn w j x j ; v j x j = s and x j s {0,1}. 1 j 1 j The mnmum of the empty set s. The entres of the table are readly determned. Wth no tems, we can only obtan value 0, and we do so at a weght of zero. Consder now the -th tem. We can obtan a value of s n one of two ways: ether by obtanng value s wth the frst 1 tems or by obtanng value s v wth the frst 1 tems. We choose the better of the two alternatves. B[0] = 0 and B[s] = for s 1. for from 1 to n do for s = nv max step 1 downto v do B[s] = mn(b[s],b[s v ] + w ) let s be maxmal such that B[s] W; return s. Lemma 2 The dynamc programmng algorthm solves the knapsack problem n tme O(n 2 v max ). Proof: Obvous. A small trck mproves the runnng tme to O(nV opt ). We mantan a value K whch s the maxmum value seen so far. We ntalze K to zero and replace the nner loop by K = K + v for s = K step 1 downto v do B[s] = mn(b[s],b[s v ] + w ) Observe that ths runnng tme s NOT polynomal as v max may be exponental n the sze of the nstance. Scalng: We now mprove the runnng tme by scalng at the cost of gvng up optmalty. We wll see that we can stay arbtrarly close to optmalty. Let S be an nteger. We wll fx t later. Consder the modfed problem, where the scale the values by S,.e., we set ˆv = v /S. We can compute the optmal soluton to the scaled problem n tme O(n 2 v max /S). We wll next show that an optmal soluton to the scaled problem s an excellent soluton to the orgnal problem. 6

7 Let x = (x 1,...,x n ) be an optmal soluton of the orgnal nstance and let y = (y 1,...,y n ) be an optmal soluton of the scaled nstance,.e., { } V opt = v x = max v z ; w z W and z s {0,1},and 1 n 1 n { } ˆv y = max ˆv z ; 1 n w z W and z s {0,1} 1 n Let V approx = v y be the value of the knapsack when flled accordng to the optmal soluton of the scaled problem. Then Thus V approx = v y = S S S v S y y S v x S v S ( v S 1)x v x S x snce y s an optmal soluton of scaled nstance v x ns = V opt ns. snce x s an optmal soluton of orgnal nstance V approx (1 ns V opt )V opt. It remans to choose S? Let ε > 0 be arbtrary. Set S = max(1, εv opt /n ). Then V approx (1 ε)v opt, snce V approx = V opt f S = 1. The runnng tme becomes O(mn(nV opt,n 2 /ε)). Ths s nce, but the defnton of S nvolves a quantty that we do not know. The modfed greedy algorthm comes to rescue. It determnes V opt up to a factor of two. We may use the approxmaton nstead of the true value n the defnton of S. We obtan. Theorem 2 Let ε > 0. In tme O(n 2 /ε) one can compute a soluton to the knapsack problem wth V approx (1 ε)v opt. 2.2 Set Cover: A Greedy Algorthm We are gven subsets S 1 to S n of some ground set U. Each set come wth a cost c(s ) 0. The goal s to the cheapest way to cover U,.e., mn c(s ) subject to I S = U. I The greedy strategy apples naturally to the set cover problem: we always pck the most effectve set,.e., the set that covers uncovered elements at the smallest cost per newly covered element. 7.

8 More formally, let C be the set of covered elements before an teraton. Intally, C s the empty set. Defne the cost-effectveness of S as c(s )/ S \C,.e., as the cost per uncovered elements covered by S. C = /0 whle C U do Let mnmze c(s )/ S \C. C = C S set the prce of every e S \C to prce(e) = c(s )/ S \C. output the sets pcked. Let OPT be the cost of the optmum soluton. For the analyss, number the elements n the ground set n the order n whch they covered by the algorthm above. Let e 1 to e n be the elements n ths numberng. Lemma 3 For all k, prce(e k ) OPT/(n k + 1). Proof: Consder the teraton n whch e k s covered. Clearly, the sets n the optmal soluton cover the remanng elements for a cost of at most OPT,.e., at a cost of OPT/( U \ C OPT/(n k + 1) per element. Thus there must be a set whch covers elements at ths cost or less. The algorthm pcks the most cost-effectve set and hence prce(e k ) OPT/(n k + 1). Theorem 3 The greedy algorthm for the set cover problem produces a soluton of cost at most H n OPT, where H n = 1 + 1/2 + 1/ /n s the n-th Harmonc number. Proof: By the precedng Lemma, the cost of the soluton produced by the greedy algorthm s bounded by 1 OPT 1 k n n k + 1 = OPTH n. No better approxmaton rato can be proved for ths algorthm as the followng example shows. The ground set has sze n. We have the followng sets: a sngleton set coverng element at cost 1/ and a set coverng all elements at cost 1 + ε. The optmal cover has a cost of 1 + ε. However, the greedy algorthm chooses the n sngletons. Hardness of Approxmaton: the greedy one. It s unlkely that there s a better algorthm for set-cover than Theorem 4 If there s an approxmaton algorthm for set cover wth approxmaton rato o(logn) then NP ZTIME(n O(loglogn) ). ZTIME(T (n)) s the set of problems whch have a Las Vegas algorthm wth runnng tme T (n). 8

9 2.3 Maxmum Satsfablty: A Randomzed Algorthm We consder formulae n 3-CNF. Max-3-SAT s the problem of satsfyng a maxmum number of clauses n a gven formula Φ. Lemma 4 Let Φ be a formula n whch every clause has exactly dstnct three lterals. There s a randomzed algorthm that satsfes an expected number of 7/8 of the clauses. Proof: Consder a random assgnment x. For any clause C, the probablty that x satsfes C s 7/8 because exactly one out of the eght possble assgnments to the varables n C falsfes C. Thus the expected number of satsfed clauses s 7/8 tmes the number of clauses. It s crucal for the argument above that every clause has exactly three lterals and that ths lterals are dstnct. Clauses of length 1 are only satsfed wth probablty 1/2 and clauses of length 2 only wth probablty 3/4. So the randomzed algorthm does not so well when there are short clauses. However, there s an algorthm that does well on short clauses. See the book by Vazran or by Motwan and Raghavan. A student ponted out to me that the algorthm above s easly derandomzed. Let {C 1,...,C m } be a set of clauses (not necessarly of length 3). Let I be an ndcator varable for the -th clause. It s one f C s satsfed and zero otherwse. Then E[# of satsfed clauses] = E[I I m ] = E[I 1 ] E[I m ] = A clause wth k dstnct lterals s true wth probablty 1 2 k. We can now use the method of condtonal expectons. Note that prob(c s true). 1 m E[# of satsfed clauses] = (E[# of satsfed clauses x 1 = 0]+E[# of satsfed clauses x 1 = 1])/2. And hence one of the two expectons on the rght s at least the expecton on the left. Snce we can compute the expectons, we can select the better choce between x 1 = 0 and x 1 = 1. In ths way, we fnd determnstcally an assgnment that has the same guarantee as the randomzed algorthm. What can one acheve determnstcally? Consder the followng two assgnments. Set all varables to TRUE or set all varables to FALSE. Snce any clause s satsfed by all but one of the assgnments to ts lterals, every clause s satsfed by one of the assgnments. Hence one assgnment satsfes at least half of the clauses. Karloff-Zwck have shown how to satsfy at 7/8 of the clauses n any satsfable formula. Ths s not the same as a 7/8-approxmaton algorthm. 2.4 Vertex Cover Consder the vertex cover problem. We are gven a graph G = (V,E). The goal s to fnd the smallest subset of the vertces that covers all the edges. Lemma 5 A factor 2-approxmaton to the optmum vertex cover can be computed n polynomal tme. 9

10 Proof: Compute a maxmal matchng M and output the endponts of the edges n the matchng. A maxmal matchng can be computed greedly. We terate over the edges n arbtrary order. If an edge s uncovered we add t to the matchng. Clearly, the sze of the cover s 2 M. An optmal cover must contan at least one endpont of every edge n M. 2.5 Hardness of Approxmaton Ths secton follows the book by Vazran [Vaz03]. How does one show that certan problems do not have good approxmaton algorthms unless P = NP. The key are gap-ntroducng and gap-preservng reductons. Assume now that we have a polynomal tme reducton from SAT to MAX-3-SAT wth the followng property. It maps an nstance Φ of SAT to a nstance Ψ of MAX-3-SAT such that f Φ s satsfable, Ψ s satsfable. f Φ s not satsfable, every assgnment to the varables n Ψ satsfes a fracton less than (1 α) of the clauses of Ψ. where α > 0 s a constant. Theorem 5 In the stuaton above, there s no approxmaton algorthm for Max-3-SAT that acheves an approxmaton rato of 1 α, unless P = NP. Proof: Assume otherwse. Let Φ be any formula and let Ψ be constructed by the reducton. Construct a maxmzng assgnment x for Ψ usng the approxmaton algorthm and let m be the number of clauses n Ψ. If x satsfes at least (1 α)m clauses of Ψ, declare Φ satsfable. Otherwse, declare Φ unsatsfable. The correctness s almost mmedate. If Φ s satsfable, there s a satsfyng assgnment for Ψ and hence the assgnment produced by the approxmaton algorthm must satsfy at least (1 α)m clauses of Ψ. Thus the algorthm declares Φ satsfable. On the other hand, f Φ s not satsfable, any assgnment for the varables n Ψ satsfes less than (1 α)m clauses of Ψ. In partcular, the assgnment returned by the approxmaton algorthm cannot do better. Thus Ψ s declared unsatsfable The PCP-Theorem (Probablstcally Checkable Proofs) We are all famlar wth the characterzaton of NP va wtnesses (proofs) and polynomal tme verfers. A language L belongs to NP f and only f there s a determnstc polynomal tme Turng machne V (called the verfer) wth the followng property. On nput x, V s provded wth an addtonal nput y, called the proof, such that: If x L, there s a y such that V accepts. If x L, there s no y such that V accepts. 10

11 In probabstcally checkable proofs, we make the verfer a probablstc machne (ths makes the verfer stronger), but allow t to read only part of the proof n any partcular run. A language L belongs to PCP(logn,1) f and only f there s polynomal tme Turng machne V (called the verfer) and constants c and q wth the followng propertes. On nput x of length n, V s provded wth x, a proof y, and a random strng 3 r of length clogn. It performs a computaton queryng at most q bts of y. The q bts probed depend on x and r, but not 4 on y. If x L, then there s a proof y that makes V accept wth probablty 1. If x L, then for every proof y, V accepts wth probablty less than 1/2. The probabltes are computed wth respect to the random choces r. Theorem 6 (Arora, Lund, Motwan, Sudan, and Szegedy) NP = PCP(logn,1). The drecton PCP(logn,1) NP s easy (Guess y and then run V for all random strngs of length clogn. Accept f V accepts for all random strngs.) The other drecton s a deep theorem. The crux s brngng the error probablty below 1/2. Wth a larger error bound, the statement s easy. Consder the followng algorthm for SAT. The verfer nterprets y as an assgnment and the random strng as the selector of a clause C. It checks whether the assgnment satsfes the clause. Clearly, f the nput formula s satsfable, the verfer wll always accept (f provded wth a satsfyng assgnment). If the nput formula s not satsfable, there s always at least one clause that s not satsfed. Therefore the verfer wll accept wth probablty at most 1 1/m, where m s the number of clauses. The PCP-theorem gves rse to an optmzaton problem whch does not have a 1/2-approxmaton algorthm unless P = NP. Maxmze Accept Probablty: Let V be a PCP(logn,1)-verfer for SAT. On nput Φ, fnd a proof y that maxmzes the acceptance probablty of V. Theorem 7 Maxmze Accept Probablty has no factor 1/2 approxmaton algorthm unless P = NP. Proof: Assume otherwse. Let Φ be any formula and let y be the proof returned by approxmaton alg. We run V wth y and all random strngs and compute the acceptance probablty p. If p s less than 1/2 we declare Φ non-satsfable, If p 1/2, we declare Φ satsfable. Why s ths correct? If Φ s satsfable, there s proof wth acceptance probablty 1. Hence the approxmaton alg must return a proof wth acceptance probablty at least 1/2. If Φ s not satsfable, there s no proof wth acceptance probablty at least 1/2. In partcular, for the proof returned by the approxmaton alg, we must have p < 1/2. 3 It s convenent to make V a determnstc Turng machne and to provde the randomness explctly. 4 Ths requrement s not essental. Assume we would allow the bts read to depend on y. The frst bt nspected s ndependent of y, the second bt may depend on the value of the frst bt, and so on. So there are only q 1 potental bts that can be read. 11

12 2.7 Max-3-SAT s hard to approxmate We gve a gap-ntroducng reducton from SAT to MAX-3-SAT. Let V be a PCP(logn,1)-verfer for SAT. On an nput Φ of length n t uses a random strng r of length clogn and nspects (at most) q postons of the proof. Thus there are at most qn c bts of the proof that are nspected for some random strng. Let B = B 1...B qn c be a bt-strng that encodes these places of the proof. The other places of the proof are rrelevant. We wll construct a formula Ψ over varables B 1 to B qn c wth the desred propertes. If Φ s satsfable, Ψ s satsfable. If Φ s not satsfable, then any assgnment does not satsfy a constant fracton of the clauses n Ψ. For any random strng r, the verfer nspect a partcular set of q bts, say the bts ndexed by (r,1) to (r,q). Let f r (B (r,1),...,b (r,q) ) be a boolean functon of q varables that s one f and only f the verfer accepts wth random strng r and the q bts of the proof as gven by B (r,1) to B (r,q). Let us pause and see what we have acheved. (A) If Φ s satsfable, there s a proof y that makes V accept wth probablty one. Set B accordng to y. Then all functons f r, r {0,1} clogn, evaluate to true. (B) If Φ s not satsfable, every proof makes V accept wth probablty less than 1/2. Hence every B falsfes more than half of the functons f r. Each f r s a functon of q varables and hence has a conjunctve normal form Ψ r wth at most 2 q clauses. Each clause s a dsjuncton of q lterals. An assgnment that falsfes f r falsfes at least one clause n Ψ r. Let Ψ be the conjuncton of the Ψ r,.e., Ψ = r Ψ r. We fnally turn Ψ nto a formula Ψ wth exactly three lterals per clause usng the standard trck. Consder a clause C = x 1 x 2... x q. Introduce new varables y 1 to y q 2 and consder C = (x 1 x 2 y q ) ( y 1 x 3 y 2 )... ( y q 2 x q 1 x q ). An assgnment satsfyng C can be extended to an assgnment satsfyng C. Conversely, an assgnment falsfyng C falsfes at least one clause of C. What have we acheved? Ψ conssts of no more than n c 2 q (q 2) clauses. Each clause has 3 lterals. (A) If Φ s satsfable, there s a proof y that makes V accept wth probablty one. Set B accordng to y. Then all functons f r, r {0,1} clogn, evaluate to true and hence Ψ s satsfed. (B) If Φ s not satsfable, every proof makes V accept wth probablty less than 1/2. Hence every B falsfes more than half of the functons f r. Hence B falsfes more than n c /2 clauses n Ψ. Let α = (n c /2)/(n c 2 q (q 2)) = 1/((q 2)2 q+1 ). Theorem 8 The above s a gap ntroducng reducton from SAT to MAX-3-SAT wth α = 1/((q 2)2 q+1 ). The theorem can be sharpened consderably. Hastad showed that the theorem holds for any α = 1/8 ε for any ε > 0. 12

13 2.8 From Max-3-SAT to Vertex Cover Step 1: One shows that Max-3-SAT stays hard to approxmate f one ntroduces the addtonal constrant that each varable appears at most 30 tmes. Step 2: From Max-3-SAT(30) to vertex cover. One uses the standard reducton. 2.9 Approxmaton Algs for Problems n P 3 General Search Heurstcs 4 Parameterzed Complexty 4.1 Parameterzed Complexty for Problems n P References [ABCC06] D. Applegate, B. Bxby, V. Chvatal, and W. Cook. The Travelng Salesman Problem: A Computatonal Study. Prnceton Unversty Press, [GS92] [NOT06] [Sch99] [Sch01] G. R. Grmmett and D. R. Strzacker. Probablty and Random Processes. Oxford Unversty Press, R. Neuwenhus, A. Olveras, and C. Tnell. Solvng SAT and SAT modulo theores: From an abstract Davs-Putnam-Logemann-Loveland procedure to DPLL(T). Journal of the ACM, 53(6): , U. Schönng. A probablstc algorthm for k-sat and constrant satsfacton problems. FOCS, Uwe Schönng. New algorthms for k-sat based on the local search prncple. In Mathematcal Foundatons of Computer Scence 2001, volume 2136 of Lecture Notes n Computer Scence, pages Sprnger Berln Hedelberg, [Vaz03] V. Vazran. Approxmaton Algorthms. Sprnger,