Consensus Patterns (Probably) Has no EPTAS

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1 Consensus Patterns (Probably) Has no EPTAS Chrstna Boucher Chrstne Lo Danel Lokshantov June 23, 2015 Abstract Gven n length-l strngs S = {s 1,..., s n } over a constant sze alphabet Σ together wth an nteger l, where l L, the objectve of Consensus Patterns s to fnd a length-l strng s, a substrng t of each s n S such that d(t, s) s mnmzed. Here d(x, y) denotes the Hammng dstance between the two strngs x and y. Consensus Patterns admts a PTAS [L et al., JCSS 2002] s fxed parameter tractable when parameterzed by the objectve functon value [Marx, SICOMP 2008], and although t s a well-studed problem, mprovement of the PTAS to an EPTAS seemed elusve. We prove that Consensus Patterns does not admt an EPTAS unless FPT=W[1], answerng an open problem from [Fellows et al., STACS 2002, Combnatorca 2006]. To the best of our knowledge, Consensus Patterns s the frst problem that admts a PTAS, and s fxed parameter tractable when parameterzed by the value of the objectve functon but does not admt an EPTAS under plausble complexty assumptons. The proof of our hardness of approxmaton result combnes parameterzed reductons and gap preservng reductons n a novel manner. 1 Introducton Lanctot et al. [15] ntated the study of dstngushng strng selecton problems n bonformatcs, where we seek a representatve strng satsfyng some dstance constrants from each of the nput strngs. The Consensus Patterns problem falls wthn ths broad class of strngology problems. Gven n length-l strngs S = {s 1,..., s n } over a constant sze alphabet Σ together wth an nteger l, where l L, the objectve of Consensus Patterns s to fnd a length-l strng s, a length-l substrng t of each s n S such that d(t, s) s mnmzed. Here d(x, y) denotes the Hammng dstance between the two strngs x and y. One specfc applcaton of Consensus Patterns n bonformatcs s the problem of fndng transcrpton factor bndng stes [15, 22]. Transcrpton factors are protens that bnd to promoter regons n the genome and have the effect of regulatng the expresson of one or more genes. Hence, the regon where a transcrpton factor bnds s very well-conserved, and the problem of detectng such regons can be extrapolated to the problem of fndng the substrngs {t 1,..., t n }. Consensus Patterns s NP-hard even when the alphabet s bnary [16], so we do not expect a polynomal-tme algorthm for the problem. On the other hand, the problem admts a polynomal tme approxmaton scheme (PTAS), whch fnds a soluton that s at most a factor (1 + ɛ) worse than the optmum [16] n n O( 1 ɛ 2 log 1 ɛ ) -tme. Whle a superpolynomal dependence of the runnng tme on 1 ɛ s mpled by the NP-hardness of Consensus Patterns, there s stll room for faster approxmaton schemes for the problem and so a sgnfcant effort has been nvested n attemptng on provng tghter bounds on the runnng tme of the PTAS [4, 5]. If the exponent of the polynomal n the runnng tme of a PTAS s ndependent of ɛ then the PTAS s called Department of Computer Scence and Engneerng, Unversty of Calforna, San Dego 1

2 an effcent PTAS (EPTAS). An nterestng queston, posed by Fellows et al. [9] s whether Consensus Patterns admts an EPTAS. The dfference n runnng tme of a PTAS and an EPTAS can be qute dramatc. For nstance, runnng a O(2 1/ɛ n)-tme algorthm s reasonable for ɛ = 1 10 and n = 1000, whereas runnng a O(n 1/ɛ )-tme algorthm s nfeasble on ths same nput. Hence, consderable effort has been devoted to mprovng PTASs to EPTASs, and showng that such an mprovement s unlkely for some problems. For example, Arora [2] gave a n O(1/ɛ) -tme PTAS for Eucldean TSP, whch was then mproved to a O(2 O(1/ɛ2) n 2 )-tme algorthm n the journal verson of the paper [3]. On the other hand Independent Set admts a PTAS on unt dsk graphs [14] but Marx [18] showed that t does not admt an EPTAS assumng FPT W[1] a wdely beleved assumpton from parameterzed complexty. Many more examples of PTASs that have been mproved to EPTASs, and problems for whch there exsts a PTAS but the exstence of an EPTAS has been ruled out under the assumpton that FPT W[1] can be found n the survey of Marx [19]. In ths paper we show that assumng FPT W[1], Consensus Patterns does not admt an EPTAS, resolvng the open problem of Fellows et al. [9]. Snce Consensus Patterns has a PTAS and s FPT, standard methods for rulng out an EPTAS cannot be appled. We dscuss ths n more detals n Secton 1.1. Our proof avods ths obstacle by combnng gap presevng reductons and parameterzed reductons n a novel manner. 1.1 Methods Our lower bounds are proved under the assumpton FPT W[1], a standard assumpton n parameterzed complexty that we wll brefly dscuss here. In a parameterzed problem every nstance I comes wth a parameter k. A parameterzed problem s sad to be fxed parameter tractable (FPT) f there s an algorthm solvng nstances of the problem n tme f(k) I O(1) for some functon f dependng only on k and not on I. The class of all fxed parameter tractable problems s denoted by FPT. The class W[1] of parameterzed problems s the basc class for fxed parameter ntractablty, FPT W[1] and the contanment s beleved to be proper. A parameterzed problem Π wth the property that an FPT algorthm for Π would mply that FPT=W[1] s called W[1]-hard. Thus demonstratng W[1]-hardness of a parameterzed problem mples that t s unlkely that the problem s FPT. We refer the reader to the textbooks [6, 8, 11, 21] for a more thorough dscusson of parameterzed complexty. W[1]-hardness s frequently used to rule out EPTAS s for optmzaton problems, snce an EPTAS for an optmzaton problem automatcally yelds a FPT algorthm for the correspondng decson problem parameterzed by the value of the objectve functon [19]. More specfcally, f we set ɛ = 1 2α, where α s the value of the objectve functon, then a (1 + ɛ)-approxmaton algorthm would dstngush between yes and no nstances of the problem. Hence, an EPTAS could be used to solve the problem n O(f(ɛ)n O(1) ) = O(g(α)n O(1) )-tme. Hence, f a problem s W[1]-hard when parameterzed by the value of the objectve functon then the correspondng optmzaton problem does not admt an EPTAS unless FPT=W[1]. To the best of our knowledge, all known results rulng out EPTASs for problems for whch a PTAS s known use ths approach. However, ths approach cannot be used to rule out an EPTAS for Consensus Patterns because Consensus Patterns parameterzed by d has been shown to be FPT by Marx [17]. Thus, dfferent methods are requred to rule out an EPTAS for Consensus Patterns. In hs survey, Marx [19] ntroduces a hybrd of FPT reductons and gap preservng reductons and argues that t s concevable that such a reducton could be used to prove that a problem that has a PTAS and s FPT parameterzed by the value of the objectve functon does not admt an EPTAS unless FPT=W[1]. We show that Consensus Patterns does not admt an EPTAS unless FPT=W[1], gvng the frst example of ths phenomenon. 2

3 Prelmnares A PTAS for a mnmzaton problem fnds a (1 + ɛ)-approxmate soluton n tme I f(1/ɛ) for some functon f. An approxmaton scheme where the exponent of I n the runnng tme s ndependent of ɛ s called an effcent polynomal tme approxmaton scheme (EPTAS). Formally, an EPTAS s a PTAS whose runnng tme s f(1/ɛ) O(1) I O(1). Let L, L N be two parameterzed problems. We say that L fpt-reduces to L f there are functons f, g : N N, and an algorthm that gven an nstance (I, k) runs n tme f(k) I f(k) and outputs an nstance (I, k ) such that k g(k) and (I, k) L (I, k ) L. These reductons work as expected; f L fpt-reduces to L and L s FPT then so s L. Furthermore, f L fpt-reduces to L and L s W[1]-hard then so s L. Let s be a strng over the alphabet Σ. We denote the length of s as s, and the jth character of s as s[j]. Hence, s = s[1]s[2]... s[ s ]. For a set S of strngs of the same length we denote by S[] as {s[] : s S}. Thus, f the same character appears at poston n several strngs t s counted several tmes n S[]. For an nterval P = {, + 1,..., j 1, j} of ntegers, defne s[p ] to be the substrng s[]s[ + 1]... s[j] of s. For a set S of strngs and nterval P defne S[P ] to be the (mult)set {s[p ] : s S}. For a set S of length-l strngs we defne the consensus strng of S, denoted as c(s), as the sequence where c(s)[] s the most-frequent character n S[] for all l. Tes are broken by selectng the lexcographcally frst such character, however, we note that the te-breakng wll not affect our arguments. We denote the sum Hammng dstance between a strng, s, and a set of strngs, S, as d(s, s). Observe that the consensus strng c(s) mnmzes d(s, c(s)) mplyng that no other strng x s closer to S than c(s). However, some x c(s) could acheve d(s, x) = d(s, c(s)) and we refer to such strngs as majorty strngs because they are obtaned by pckng a most-frequent character at every poston wth tes broken arbtrarly. We wll use standard concentraton bounds for sums of ndependent random varables. In partcular, the followng varant of the Hoeffdng s bound [13] gven by Grmmett and Strzaker [12, p. 476] wll be needed. Proposton 1. (Hoeffdng s bound) Let X 1, X 2,...X n be ndependent random varables such that a X b for all. Let X = Σ X and the expected value of X be E[X] then t follows that: ( ) 2t 2 Pr[X E[X] t] exp Σ n =1 (b a ) 2. 2 Hardness of Approxmatng Colored Consensus Strng wth Outlers To show that Consensus Patterns does not admt an EPTAS we wll frst demonstrate hardness the followng problem, that we call Colored Gap-Consensus Strng wth Outlers. When defnng parameterzed gap problems, we follow the notaton of Marx [19]. 3

4 Colored Gap-Consensus Strng wth Outlers (CCWSO) Input: A (mult)set of n length-l strngs S = {s 1,..., s n } over a fnte alphabet Σ, an nteger n n, a parttonng of S nto n sets S = S 1 S 2... S n, a ratonal ɛ and two ntegers D yes and D no wth D no D yes (1+ɛ) such that ether (a) there exsts a set S such that S S = 1 for every and d(s, c(s )) D yes or (b) for every S such that S S = 1 for every we have d(s, c(s )) D no. Parameter: 1/ɛ Queston: Is there an S such that d(s, c(s )) D yes? The am of ths secton s to prove the followng lemma. Lemma 1. Gap-Colored Consensus Strng wth Outlers s W[1]-hard. The proof of Lemma 1 s by reducton from the MultColored Clque (MCC) problem. Here nput s a graph G, an nteger k and a partton of V (G) nto V 1 V 2... V k such that for each, G[V ] s an ndependent set. The task s to determne whether G contans a clque C of sze k. Observe that such a clque must contan exactly one vertex from each V, snce for each we have C V 1. It s well-known that MCC s W[1]-hard [10]. Gven an nstance (G, k) of MCC we produce n f(k)n O(1) -tme an nstance (S 1, S 2,... S n ) of Colored Gap-Consensus Strng wth Outlers. We wll say that a subset S of S such that S S = 1 for every n s a potental soluton to the CCWSO nstance. Our constructed nstance wll have the followng property. If G has a k-clque then there exsts a potental soluton S such that d(s, c(s )) D yes. On the other hand, f no k-clque exsts n G then for each potental soluton S we have d(s, c(s )) D no. The values of D yes and D no wll be chosen later ( n the) proof, however, we note that the crucal pont of the constructon s that D no h(k) D yes. Hence, a f(ɛ)(nl) O(1) -tme algorthm for Gap-Consensus Strng wth Outlers could be used to solve to solve the MCC problem n tme g(k)n O(1) by settng ɛ = 1 2h(k). Thus, the reducton s a parameterzed, gap-creatng reducton where the sze of the gap decreases as k ncreases but the decrease s a functon of k only. Constructon. We descrbe how the nstance (S 1, S 2,... S n ) s constructed from (G, k). Our constructon s randomzed, and wll succeed wth probablty 2 3. To prove Lemma 1 we have to change the constructon to make t determnstc but for now let us not worry about that. We start by consderng the nstance (G, k) and let E(G) = {e 1, e 2,... e m }. In the reducton we ( wll create one strng s for every edge e E(G). We partton the edge set E(G) nto sets k ) 2 sets E{p,q} where 1 p, q k as follows; e E p,q f e has one endpont n V p and the other n V q. The edge e E p,q has two endponts, one n V p and the other n V q. The strng s s nserted nto the set S {p,q} and the set S of strngs n the nstance of Gap-Colored Consensus Strng wth Outlers wll be exactly S = S {p,q}. p,q p q We set n = ( ) k 2, and use exactly the partton of S nto the sets S{p,q} as the partton nto n sets n the nstance. Thus, pckng a potental soluton S corresponds to pckng a set of edges wth exactly one edge from each of the sets E {p,q}. 4

5 There are K = k (k 1) (k 2) ordered trples of ntegers from 1 to k. Consder the lexcographc orderng of such trples. As an example, f k = 3 ths orderng s (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). For each from 1 to K, let σ() be the th trple n ths orderng. Thus, for k = 3, we have that σ(4) = (2, 3, 1). The functons σ 1, σ 2 and σ 3 return the frst, second and thrd entry of the trple returned by σ. Contnung our example for the case that k = 3, we have σ 1 (4) = 2, σ 2 (4) = 3 and σ 3 (4) = 1. Based on G and k, we select an nteger l. The exact value of l wll be dscussed later n the proof, for now the reader may thnk of l as some functon of k tme log n. We construct a set Z = z 1, z 2,... z m of strngs, Z wll act as a pool of random bts n our constructon. For each edge e E(G) we make a strng z as follows. z = a σ(1) a σ(2)... a σ(k) For every m and p K, the strngs ã, ã and aσ(p) are random bnary strngs of length l. For each p K and vertex u V σ1 (p) we make an dentfcaton strng d p (u) of length l. Let be the smallest nteger such that the edge e s ncdent to u. We set d p (u) = a σ(p). Notce that the other endpont of e a vertex not n V p. Thus, for any other vertex v V p dstnct from u we have that d p (v) = a σ(p) j for some nteger j. We now make the set S of strngs n our nstance. For each edge e E(G) we make a strng s as follows. s = a σ(1) a σ(2)... a σ(k) For each x K we defne a x usng the followng rules. Let e = uv wth u V p and v V q. If σ 1 (x) = p and σ 2 (x) = q or σ 1 (x) = p and σ 3 (x) = q, we set a σ(x) = d x (u). If σ 1 (x) = q and σ 2 (x) = p or σ 1 (x) = q and σ 3 (x) = p, we set a σ(x) = d x (v). Otherwse we set a σ(x) = a σ(x). For 1 p K we defne B p = {(p 1)l + 1, (p 1)l + 2,... (p 1)l + l}, and wll refer to B p as the p th block of the nstance. Notce that for every m and p K we have s [B p ] = a σ(p). We set L = K l and N = S = m, ths concludes the constructon. Recall that n s the sze of the soluton S sought for and observe that L s the length of the constructed strngs n S. Analyss. We consder the constructed strngs s as random varables, and for every j the character s [j] s also a random varable whch takes value 1 wth probablty 1/2 and 0 wth probablty 1/2. Observe that for any two postons j and j such that j j and any and the random varables s [j] and s [j ] are ndependent. On the other hand s [j] and s [j] could be dependent. However, f s [j] and s [j] are dependent then, by constructon s [j] = s [j]. Let S S be a potental soluton. Here we consder S as a set of random strng varables, rather than a set of strngs. We are nterested n studyng d(s, c(s )) for dfferent choces of the set S. We can wrte out d(s, c(s )) as and d(s, c(s )) = K d(s [B p ], c(s )[B p ]) (1) p=1 d(s [B p ], c(s )[B p ]) = j B p d(s [j], c(s )[j]). Thus, for each p K we have that d(s [B p ], c(s )[B p ]) s a sum of l ndependent random varables, each takng values from 0 to n. Hence, when l s large enough d(s [B p ], c(s )[B p ]) 5

6 s sharply concentrated around ts expected value. Usng a unon bound (over the choces of p) we can show that d(s, c(s )) s sharply concentrated around ts expectaton as well. We turn our attenton to E[d(S, c(s ))] for dfferent choces of S. The two man cases that we dstngush between s whether S corresponds to the set of edges of a clque n G or not. Note that a potental soluton S corresponds to a set of E edges wth exactly one edge e {p,q} E {p,q} for every (unordered) par p,q. In the remander of ths secton S s a potental soluton and E s the edge set correspondng to S. For each par p,q of ntegers, e {p,q} s the unque edge n E E {p,q}. We wll determne whether E s the set of edges of a clque usng the followng observaton, whose proof s obvous and hence omtted. Observaton 1. E s the edge set of a clque n G f and only f for every ordered trple (a, b, c) of dstnct ntegers between 1 and k the edge e {a,b} and the edge e {a,c} are ncdent to the same vertex n V a. In the constructed nstance the block B p such that σ(p) = (a, b, c) s responsble for performng the check for the trple (a, b, c). Before proceedng we need some addtonal defntons regardng random walks on the ntegers. Let v be a vector of postve ntegers. We defne the random varable X v = W v where W s a random vector wth same dmenson as v, such that each coordnate of W s drawn from { 1, 1} unformly at random. The varable X v s nterpreted as follows: start a one-dmensonal random walk at 0, n each step of the walk we go left or rght wth probablty 1/2. However, the length of the dfferent steps vares, n step the walk jumps v[] to the left or rght. The value of X v s the offset from the orgn at the end of the walk. The total length of the random walk s v[] whereas the number of steps of the walk s the dmenson of v. We defne the random varable Xr,t = + X v where v s a vector wth r 2t entres that are 1 and t entres that are 2. Intutvely Xr,t s the offset from 0 of a random walk startng at of length r, wth t steps of length 2 and the remanng steps of length 1. We set x r,t = E[ Xr,t ]. The next lemma characterzes the expectaton of d(s [B p ], c(s )[B p ]), subject to the case dstncton on whether the soluton S passes or fals the test of Observaton 1 for the trple σ(p). Lemma 2. Let p K and let σ(p) = (a, b, c). If e {a,b} and e {a,c} are ncdent to the same vertex n V a, then E[d(S [B p ], c(s )[B p ])] = l (n /2 x 0 n,1). If e {a,b} and e {a,c} are not ncdent to the same vertex n V a, then E[d(S [B p ], c(s )[B p ])] = l (n /2 x 0 n,0). Proof. Every strng s S corresponds to an edge e E(G). If e / E a,b E a,c then s [B p ] = a σ(p) = a σ(p). On the other hand, f e = E a,b E a,c then s [B p ] = a σ(p) = d p (u), where u s the vertex of V a ncdent to e. Let j be a poston n B p. Consder the case that the two edges e {a,b} and e {a,c} n E are not ncdent to the same vertex n V a. Then d(s [j], c(s [j])) s dstrbuted as n /2 X v where v s a n -dmensonal vector of 1s. Specfcally for all s S the s [j]s are ndependent so c(s [j]) s the majorty character out of n characters ndependently drawn from {0, 1}, and d(c(s [j], S [j])) s the number of occurrences of the mnorty character. Ths s dstrbuted as n /2 X v. Consder now the case that the two edges e {a,b} and e {a,c} n E are ncdent to the same vertex n V a. From the constructon of the strngs a σ(p) t follows that all of the characters n S [j] are ndependent wth the excepton of the two characters n the strngs correspondng to the two edges e {a,b} and e {a,c} ; these two characters are equal. Therefore d(s [j], c(s [j])) s 6

7 dstrbuted as n /2 X v where v s a n 1 dmensonal vector wth 1 entry of value 2 and n 1 entres wth value 1. Lnearty of expectaton mples the lemma. We now defne E yes as follows. E yes = K l (n /2 x 0 n,1) Observe that Equaton 1, Lemma 2 and lnearty of expectaton mmedeately mples that f E s the set of edges of a clque then E[d(S, c(s ))] = E yes. Furthermore, By Lemma 2 each trple (a, b, c) of dstnct ntegers from 1 to k such that the edge e {a,b} and e {a,c} are not ncdent to the same vertex n V a wll contrbute exactly l (n /2 x 0 n,0 ) nstead of l (n /2 x 0 n,1 ) to the expectaton E[d(S, c(s ))]. Ths proves the followng lemma. Lemma 3. Let t be the number of ordered trples (a, b, c) of dstnct ntegers from 1 to k such that the edge e {a,b} and e {a,c} are not ncdent to the same vertex n V a. Then E[d(S, c(s ))] = E yes + t l (x 0 n,1 x 0 n,0) To conclude the analyss we need to show that as the number of trples t that fal the test of Observaton 1 ncreases, so does the expected value of d(s, c(s )). To that end, all we need to prove s that x 0 n,1 x0 n,0 > 0. We wll prove ths by dfferentatng x0 n,t wth respect to t. Clam 1. x 0 n,0 < x0 n,1. Furthermore we can compute x0 n,0 and x0 n,1 n tme polynomal n n. Proof. Recall that x r,t = E[ Xr,t ] where Xr,t s a random varable denotng the fnal poston of a random walk of length r, wth t double steps, startng at. Here s an nteger and mght be negatve. Condtonal expectaton yelds the followng recurrence for x r,t, r 2t 0. f r = 0, x r,t = (x +1 r 1,t + x 1 r 1,t )/2 f r > 2t, (x +2 r 2,t 1 + x 2 r 2,t 1 )/2 f t 1. It s easy to see that one of the three cases must apply when r 2t 0 - and x r,t s only defned for these values. Observe that f r > 2t and t 1 then both the second and the thrd case apply. The recurrence above also yelds a polynomal tme algorthm to compute x r,t. Now, for ntegers, r, t such that r 1 and t 2 defne δx r,t = x r,t x r,t 1. The recurrence above together wth defnton of δx r,t yelds the followng recurrence for δx r,t, for r 2t and t 1. 0 f r = 2, 2, 1/2 f r = 2, = 1, δx r,t = 1 f r = 2, = 0, (δx +1 r 1,t + δx 1 r 1,t )/2 f r > 2t, (δx +2 r 2,t 1 + δx 2 r 2,t 1 )/2 f t 2. A straghtforward nducton usng ths recurrence shows that δx 0 r,1 > 0 for all r 1, provng that x 0 n,0 < x0 n,1, as clamed. We now defne as follows, = x 0 n,1 x 0 n,0, and note that Clam 1 mples that > 0. Furthermore, note that depends only on n = ( k 2), so s a computable functon of k. Defne E no = E yes + l (2) 7

8 Observe that E no /E yes K n, and that therefore E no /E yes 1 + 1/h(k) for a functon h dependng only on k. Lemma 3, Clam 1 and the defnton of E no mples the followng lemma, whch summarzes the analyss up untl now. Lemma 4. If E s the edge set of a clque n G, then E[d(S [B p ], c(s )[B p ])] = E yes. Otherwse E[d(S [B p ], c(s )[B p ])] E no. From the defntons of E yes and E no t follows that there exst constants κ yes and κ no dependng only on k such that E yes = κ yes l and E no = κ no l. Furthermore, κ yes < κ no and the value of κ yes and κ no can be computed n tme f(k) for some functon f. Set κ yes = (2κ yes + κ no )/3 and κ no = (κ yes + 2κ no )/3. Then κ yes < κ yes < κ no < κ no. We set D yes = κ yesl and D no = κ nol. Notce that κ yes κ yes = κ no κ no. A randomzed analogue of Lemma 1. Before provng Lemma 1 we argue that the randomzed constructon works. Specfcally, we show that f Gap-Consensus Strng Wth Outlers s W[1]-hard under randomzed FPT-reductons. The results proved n ths secton are not used n the proof of Lemma 1, but they provde useful nsghts on how the determnstc constructon works. Lemma 5. For any potental soluton S, any p K and any real x > 0 we have the followng nequalty. [ ] ( ( x ) ) 2 P d(s [B p ], c(s )[B p ]) E[d(S [B p ], c(s )[B p ])] > x l 2 exp 2 l n. Proof. We have that d(s [B p ], c(s )[B p ]) = j B p d(s [j], c(s )[j]). The d(s [j], c(s )[j]) s are ndependent of each other, and therefore d(s [B p ], c(s )[B p ]) s the sum of l ndependent random varables takng values from 0 to n. The statement of the lemma follows from Hoeffdng s nequalty (Proposton 1). We now defne l. Ths value for l s only vald for the randomzed constructon, and a dfferent value for l s used n the proof of Lemma 1. ( Kn ) 2 ( l = κ ln 20Km n ). (3) yes κ yes Recall that m s the number of edges n the graph G, so m n 2 and hence l 1 f log n for some f dependng only on k. Lemma 6. If G has a k-clque C, let S be the set of strngs correspondng to edges of C. Then P [d(s, c(s 1 )) > D yes ] 10(m) n. If G does not contan a k-clque, then the probablty that S contans a potental soluton S such that d(s, c(s )) < D no s at most 1/10. Proof. If G has a k-clque C, let S be the set of strngs correspondng to edge endponts of edges n C. By Lemma 5 t follows that for any p K we have [ P d(s [B p ], c(s )[B p ]) E[d(S [B p ], c(s )[B p ])] > κ yes κ yes K ( ( κ ) yes κ 2 yes 2 exp 2 l) Kn ] l 1 10Km n. 8

9 The unon bound over all choces of p K, together wth Equaton 1 yelds [ ] P d(s, c(s )) E[d(S, c(s ))] > (κ yes κ yes ) l 1 10m n. Snce D yes E yes = (κ yes κ yes )l, t follows that P [d(s, c(s )) > D yes ] 1 10m. n On the other hand, consder a set S of sze n that does not correspond to the edge endponts of a clque. An argument dentcal to the one above (but usng that κ yes κ yes = κ no κ no) shows that P [d(s, c(s )) < D no ] 1 10mn. Snce there are at most mn choces for potental solutons S, the unon bound mples the second statement of the lemma. We now prove a randomzed analogue of Lemma 1. Lemma 7. If Colored Gap-Consensus Strng Wth Outlers s FPT then W[1] randomzed FPT. Proof. Assmng that Colored Gap-Consensus Strng Wth Outlers has an algorthm wth runnng tme f(ɛ)(nl) O(1) we gve a randomzed fxed parameter tractable algorthm for MCC wth two sded error. We construct the nstance to Colored Gap-Consensus Strng Wth Outlers as descrbed, and set ɛ = Dno D yes 1 = κ no κ 1. If the algorthm for Colored Gap-Consensus yes Strng Wth Outlers concludes that there potental soluton S such that d(s, c(s )) D yes the algorthm returns that the nput graph G contans a k-clque, otherwse we return that G has no k-clque. The constructon takes tme O(f(k)n O(1) ) for some functon f, and ɛ depends only on k. Hence the total runnng tme s g(k)n c for some functon g. Thus the algorthm termnates n FPT tme. If G contans a k-clque, then by Lemma 6, wth probablty at least (m) n 1 1 n k there s a set S of sze n such that d(s, c(s )) D yes. If ths event occurs, the algorthm for Colored Gap-Consensus Strng Wth Outlers wll correctly fnd such a set and correctly return yes. Hence the probablty of false negatves s at most 1 n k. If G does not contan a k-clque, then by Lemma 6, wth probablty at least 9/10 for every set S of sze n we have d(s, c(s )) > D no. If ths event occurs the algorthm correctly returns no and hence the probablty f false postves s at most 1/10. Ths mples that there s a randomzed fxed parameter tractable algorthm for MCC, whch n turn shows that W[1] randomzed FPT. A Determnstc Constructon and Proof of Lemma 1. In order to prove Lemma 1 we need to make the constructon determnstc. We only used randomness to construct the set Z, all other steps are determnstc. We now show how Z can be computed determnstcally nstead of beng selected at random, preservng the propertes of the reducton. For ths, we need the concept of near p-wse ndependence defned by Naor and Naor [20]. The orgnal defnton of near p-wse ndependence s n terms of sample spaces, we defne near p-wse ndependence n terms of collectons of bnary strngs. Ths s only a notatonal dfference, and one may freely translate between the two varants. Defnton 1 ([20]). A set C = {c 1, c 2,... c t } of length l bnary strngs s (ɛ, p)-ndependent f for any subset C of C of sze p, f a poston t s selected unformly at random, then α {0,1} p P [C [] = α] 2 p ɛ. 9

10 Naor and Naor [20] and Alon et al. [1] gve determnstc constructons of small nearly k-wse ndependent sample spaces. Reformulated n our termnology, Alon et al. [1] prove a slghtly stronger verson of the followng theorem. Theorem 1 ([1]). For every t, p, and ɛ there s a (ɛ, p)-ndependent set C = {c 1, c 2,... c t } of bnary strngs of length l, where l = O( 2k k log t ɛ ). Furthermore, C can be computed n tme O( C O(1) ). We use Theorem 1 to construct the set Z. We set ɛ = κ yes κ yes K n and construct an (ɛ, n )-ndependent set C of 2m strngs. These strngs have length l = f log(n) for some f dependng only on k, and C can be constructed n tme O(gn O(1) ) for some g dependng only on k. we set z = c c... c, where we used K copes of c such that z s a strng of length L. That s, n the constructon of z we set a σ(p) = c for all p K. The remanng part of the constructon,.e the constructon of S from Z remans unchanged. To dstngush between the determnstcally constructed S and the randomzed constructon, we refer to the determnstcally constructed S as S det. We now prove that for every potental soluton Sdet S det, f S s the set of strngs n the randomzed constructon that corresponds to the same edges as Sdet, then d(s det, c(s det )) s almost equal to E[d(S, c(s ))]. When consderng E[d(S, c(s ))] we consder the randomzed constructon, but wth the same choce of l as n the constructon of S det, so that the strngs n S and S det have the same length. For a subset I of {1, 2,..., m} defne S (I) = {s S : I} and Sdet (I) = {s S det : I}. The constructon of S det (and S) from Z mples that for every x K, there exsts a functon f x : N N such that for any m, s [B x ] = z f() [B x ]. For any I {1, 2,..., m} and x K we defne Z (I, x) to be an arbtrarly chosen subset of Z of sze n such that {z fx() : I} Z (I, x). The reason we dd not defne Z (I, x) as exactly {z fx() : I} s that the functon f x s not njectve, and we want to ensure Z (I, x) = n. The defnton of Z (I, x) ensures that for every I {1, 2,..., m} of sze n, the strng sets S (I)[B x ] and Sdet (I)[B x] are functons of Z (I, x)[b x ]. Even stronger, for every j B x we have that the strngs S (I)[j] and Sdet (I)[j] are functons of Z (I, x)[j]. Strctly speakng S (I)[j], Sdet (I)[j] and Z (I, x)[j] are mult-sets of characters, but we can thnk of them as strngs by, for example, readng the characters n S (I)[j] as s [j] for all I n ncreasng order. Snce the determnstc and randomzed constructons are dentcal (except for the constructon of Z) the strngs S (I)[j] and Sdet (I)[j] depend on Z (I, x)[j] n exactly the same way. An mmedate mplcaton of the fact that S (I)[B x ] and Sdet (I)[B x] are functons of Z (I, x)[b x ], s that the dstances d(s (I)[j], c(s (I)[j])) and d(sdet (I)[j], c(s det (I)[j])) are also functons of Z (I, x)[j]. We now gve these functons a name. For every set I {1, 2,..., m} of sze n and nteger x < K defne d I x : {0, 1} n {0, 1,..., n } to be a functon such that for any j B x, f Z (I)[j] = α then d(s (I)[j], c(s (I)[j])) = d I x(α) and d(sdet (I)[j], c(s det (I)[j])) = di x(α). For every set I {1, 2,..., m} of sze n and nteger x K we have the followng expresson for d(s (I) det [B x ], c(s (I) det [B x ])). d (S det (I)[B x], c(s det (I)[B x])) = l α {0,1} n P [Z (I)[j] = α] d I j (α) (4) 10

11 Here the probablty P [Z (I)[j] = α] s taken over random selectons of j from B x. For the randomzed constructon we have that P [Z (I)[j] = α] = 1 2n, whch yelds the followng expresson. E [d (S (I)[B x ], c(s 1 (I)[B x ]))] = l d I 2 n j (α) (5) α {0,1} n Combnng Equatons 4 and 5 yelds the followng bound. d (Sdet (I)[B x], c(sdet (I)[B x])) E[d(S (I)[B x ], c(s (I)[B x ]))] ( = l P [Z (I)[p] = α] 1 ) d I 2 n j (α) α {0,1} n l ɛ n Summng Equaton 6 over 1 x K yelds the desred bound for every I {1, 2,..., 2m} of sze n. d (S det (I), c(s det (I))) E[d(S (I), c(s (I)))] l K ɛ n l (κ yes κ yes ) (7) Equaton 7 allows us to fnsh the proof of Lemma 1. For any potental soluton S that corresponds to a clque n G, we have that E[d(S (I), c(s (I)))] = E yes = lκ yes, and so by Equaton 7, d (Sdet (I), c(s det (I))) lκ yes = D yes. For any potental soluton S of sze n that does not correspond to a clque n G, we have that E[d(S (I), c(s (I)))] E no = lκ no, and so by Equaton 7, d (S det (I), c(s det (I))) lκ no = D no. Snce Dno D yes 1 + δ for some δ dependng only on k, the constructon s an fpt-reducton from MCC to Gap-Consensus Strng Wth Outlers, completng the proof of Lemma 1. 3 Hardness of Approxmatng Consensus Patterns To show that Consensus Patterns does not have an EPTAS unless F P T = W [1] we ntroduce the followng gap varant of the problem. Gap-Consensus Patterns Input: A set S = {s 1,..., s n } of length-l strngs over a constant sze alphabet Σ together wth an nteger l, where l L, a ratonal ɛ and ntgers D yes and D no wth D no D yes (1 + ɛ) such that the followng holds. Ether there s a length-l substrng t of each s n S such that d(t, s) D yes or for every collecton t 1,... t n such that t s a length-l substrng s we have d(t, s) D no. Parameter: 1/ɛ Queston: Is there a length-l substrng t of each s n S such that d(t, s) D yes? We wll now gve a fpt-reducton from Gap-Colored Consensus Strng wth Outlers to gap- Consensus Patterns. The man ngredent n our reducton s a gadget strng w. The strng w has length L 1 (to be determned later), and for every 1, w[] = 1 f = j 2 for an nteger j and w[] = 0 otherwse. We wll say that an nteger s a square f = j 2 for some nteger j. Thus w[] s 1 f and only f s a square. 11 (6)

12 Lemma 8. For postve ntegers x, y and z such that z L 1 4, x < y and y + z L 1 we have d(w[{x, x + 1,..., x + z}], w[{y, y + 1,..., y + z}]) L1 16 Proof. To lower bound d(w[{x, x + 1,..., x + z}], w[{y, y + 1,..., y + z}]) t s suffcent to fnd the number of values for between 0 and z such that w[x + ] = 1 but w[y + ] = 0, that s x + s a square but y + s not. Let 1, 2,... t be all the values for such that x + s square, n ncreasng order. We prove that f y + j s square then y + j+1 s not. In partcular, suppose y + j s square. Let r x and r y be the ntegers such that x + j = r 2 x and y + j = r 2 y. Snce x < y we have r x < r y. Furthermore, x + j+1 = (r x + 1) 2. Hence y + j+1 = r 2 y + j+1 j = r 2 y + ((r x + 1) 2 r 2 x) < r 2 y + ((r y + 1) 2 r 2 y) = (r y + 1) 2. But then y + j+1 can t be square. It follows that there are at least t 2 values for such that x + s a square but y + s not. It remans to lower bound t. The gap between a square number and the next square number s less than 2 2 L 1. Thus the number of square numbers n {x, x + 1,..., x + z} s at least L d(w[{x, x + 1,..., x + z}], w[{y, y + 1,..., y + z}]) L L 1 L1 8. Hence Gven an nstance n, S = S 1 S 2... S n of Gap-Colored Consensus Strng wth Outlers we construct an nstance of Gap-Cosensus Patterns as follows. Frst we ensure that all of the (mult) sets S contan the same number of strngs; f S < S j for some, j we can make duplcates of strngs n S untl equalty s obtaned. Ths does not affect any other aspects of the nstance, snce a soluton S has to pck one strng from each S. Let l be the length of all the strngs n S. We choose L 1 such that L1 16 > n l and construct a gadget strng w of length L 1. For every n we make a strng ŝ from the set S. Let S = s 1, s2,..., st. We defne ŝ = w s 1 w s 2 w... w s t. and set L = L 1 + l. We keep the values of D yes and D no. Ths concludes the constructon. Lemma 9. For every S = {s 1,..., s n } S such that s S for all there s a collecton T = t 1,... t n such that t s a length L substrng of ŝ and d(c(t ), T ) d(c(s ), S ). Proof. For every, set t 1 = w s. Snce s S we have that t 1 s a length L substrng of ŝ. Set c = w c(s ), we have that d(c(t ), T ) d(c, T ) d(c(s ), S ). Lemma 10. For every collecton T = t 1,... t n such that t s a length L substrng of ŝ and d(c(t ), T ) n l there s a subset S = {s 1,..., s n } S such that s S for all and d(c(s ), S ) d(c(t ), T ). Proof. For every we can decompose t nto t = w[{a + 1,..., L}] s w[{1,..., a }] for a non-negatve nteger a L, where s S. If a = 0 then t = w s whle a = L gves t = s w. Set S = {s 1,..., s n }. We need to show that d(c(s ), S ) d(c(t ), T ). It s suffcent to show that for every, j we have a = a j because then all the s s algn n the decomposton of the t s and so d(c(s ), S ) d(c(t ), T ) holds. We prove that f a a j for some, j then d(c(t ), T ) d(t, t j ) > d(t, t j ) > n l, contradctng the assumpton of the lemma. If a a j, wthout loss of generalty a < a j. Then we can decompose t = w 1 z w 2 s w3 and t j = w1 j s j wj 2 s j w3 j such that the followng propertes hold. The lengths of w 1, w2 and w 3 equals the lengths of wj 1, w2 j and wj 3 respectvely, z and z j both have length l, and w 1, w2, w3, w1 j, w2 j, w3 j are all substrngs of w. Snce l L 1 4 we have that one of w 1, w2, w3 have length at least L 1 4. Wthout loss of 12

13 generalty ths s w 1. We have that d(t, t j ) d(w1, w1 j ). Furthermore, snce a a j we have w 1 w 1 j and hence by Lemma 8 t follows that d(w1, w1 j ) L1 16 > n l. But ths mples that d(t, t j ) > n l. By the trangle nequalty we have d(c(t ), T ) d(t, t j ) > n l yeldng the desred contradcton. The constructon, together wth Lemmata 1, 9 and 10 yeld the followng result. Lemma 11. Gap-Consensus Patterns s W[1]-hard. Snce an EPTAS for Consensus Patterns could be used to solve Gap-Consensus Patterns n tme f(ɛ)(nl) O(1), Lemma 11 mples our man result. Theorem 2. Consensus Patterns does not have an EPTAS unless FPT=W[1]. 4 Conclusons and Future Work We have shown that Consensus Patterns does not admt an EPTAS unless FPT=W[1]. Our result rules out the possblty of a (1+ɛ) approxmaton algorthms wth runnng tme f(1/ɛ)n O(1), whle the best PTAS for Consensus Patterns has runnng tme n O(1/ɛ4). Hence there s stll a sgnfcant gap between the known upper and lower bounds, and obtanng tghter bounds warrants further nvestgaton. References [1] N. Alon, O. Goldrech, J. Håstad and R. Peralta, Smple Constructon of Almost k-wse Independent Random Varables. Random Struct. Algor., 3(3): , [2] S. Arora. Polynomal Tme Approxmaton Schemes for Eucldean TSP and Other Geometrc Problems. Proc of 37th FOCS, pages 2-11, [3] S. Arora, Polynomal Tme Approxmaton Schemes for Eucldean Travelng Salesman and other Geometrc Problems. J. ACM, 45, 5: , [4] B. Brejová, D.G. Brown, I.M. Harrower, and T. Vnar. New Bounds for Motf Fndng n Strong Instances. Proc. of 17th CPM, pages , [5] B. Brejová, D.G. Brown, I.M. Harrower, A. López-Ortz and T. Vnar. Sharper Upper and Lower Bounds for an Approxmaton Scheme for Consensus-Pattern. Proc. of 16th CPM, pages 1 10, [6] M. Cygan, F. V. Fomn, D. Lokshtanov, L. Kowalk, D. Marx, M. Plpczuk, M. Plpczuk, and S. Saurabh. Parameterzed Algorthms. Sprnger, In press. [7] C. Lo, B. Kakaradov, D. Lokshtanov, and C. Boucher. SeeSte: Effcently Fndng Co-occurrng Splce Stes and Exon Splcng Enhancers. arxv: v1. [8] R. G. Downey and M. R. Fellows. Fundamentals of Parameterzed Complexty. Texts n Computer Scence. Sprnger, [9] M.R. Fellows, J. Gramm, and R. Nedermeer. On the parameterzed ntractablty of motf search problems. Combnatorca, 26: , [10] M.R. Fellows, D. Hermeln, F.A. Rosamond, and S. Valette. On the parameterzed complexty of multple-nterval graph problems. Theor. Comput. Sc., 410(1):53 61, 2009 [11] J. Flum, and M. Grohe. Parameterzed Complexty Theory. Sprnger-Verlag, [12] F. Grmmett, and D. Strzaker. Probablty and random processes. Oxford Unversty Press, 3 edton, [13] W. Hoeffdng. Probablty Inequaltes for Sums of Bounded Random Varables. J. Amer. Statstcal Assoc., 58(301): 13 30,

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