Fast FPTalgorithms for cleaning grids


 Shavonne Andrews
 6 years ago
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1 Fast FPTalgoithms fo cleaning gids Josep Diaz Dimitios M. Thilikos Abstact We conside the poblem that given a gaph G and a paamete k asks whethe the edit distance of G and a ectangula gid is at most k. Weexaminethegenealcase whee the edit opeation ae vetex/edge emovals and additions. If the dimensions of the gid ae given in advance, then we give a paameteized algoithm that uns in 2 O(log k k) + O(n 3 )steps.inthecasewheethedimensionsofthegidaenotgiven we give a paameteized algoithm that uns in 2 O(log k k) + O(k log k n 3 )steps. We insist on paameteized algoithms with unning times whee the elation between the polynomial and the nonpolynomial pat is additive. Fo this, ou algoithm design is based on the technique of kenelization. In paticula we pove that fo each vesion of the above poblem thee exists a kenel of size O(k 4 ). 1 Intoduction An inteesting poblem with many applications in Compute Science is the poblem of measuing the degee of similaity of two gaphs G and H, wheebydegeeofsimilaity is conside to be the minimum numbe of edit opeations (inset, delete, and modify) needed to tansfom one gaph into the othe. This measue of similaity between gaphs is also denoted as the edit distance by analogy with the elated poblem on stings. A paticulaly impotant application has been the ecognition of objects and shapes, whee the objects ae epesented by special type of labelled tees and use the opeations, deletion, insetion and elabelling of nodes. The poblem is NPComplete in geneal gaphs and can be computed by dynamic pogamming in polynomial time (see fo ex. [10]). Futhe wok has been done of computing the edit distance of moe involve weighted tees, called shock tees, using and extended sets of opeations: slice of edges, contact of edges, defom of edges, (see fo ex. [8]) In this pape, we conside decision poblems associated with theeditingdistance between G and H when H is a gid of size p q (we call such gid (p, q)gid and we denote it as H p,q ). An edit opeation on a gaph G can be eithe the emoval o the insetion of eithe an edge o a vetex. We epesent these opeations by the membes Depatament de Llenguatges i Sistemes Infomàtics, Univesitat Politècnica de Catalunya, Campus Nod Edifici Ω, c/jodi Giona Salgado 13, E08034, Bacelona, Spain. This eseach was suppoted by the spanish CICYT poject TIN (GRAMMARS). The fist autho was patially suppoted by the Distinció pe a la Pomoció de la Receca de la GC,
2 of the set U = {eout, ein, vout, vin}. Given E U,wedenoteEdist(G, H) asthe edit distance of G and H with espect to E, whichistheminimumnumbeofopeations in the set E that when applied to G can tansfom it to H. Ou study adopts the point of view of paameteized complexity intoduced by Downey and Fellows (see [2]). We conside paameteizations ofhadpoblems,i.e. poblems whose input contains some (in geneal small) paamete k and a some main pat. A paameteized poblem belongs in the class FPT if it can be solved by an algoithm of time complexity g(k) n O(1) whee g is a supepolynomial function of k and n is the size of the poblem (we call such an algoithm FPTalgoithm). A popula technique on the design of paameteized algoithms is kenelization. Biefly,thistechnique, consists in finding a polynomial time eduction of a paameteized poblem to itself in awaythatthesizesoftheinstancesofthenewpoblem,wecallitkenel depend only on the paamete k. The function that bounds the size of the main pat of the educed poblem detemines the size of the kenel and is usually of polynomial (on k) size. Clealy, a paameteized poblem admits a eduction to a poblem kenel, then it is in FPT because any bute foce algoithm can solve the educed poblem in time that does not depend on the main pat of the oiginal poblem. Notice also that this technique povides FPTalgoithms of time complexity g(k)+n O(1) whee the nonpolynomial pat g(k) isjustadditivetotheoveallcomplexity. The fist esult of this pape is the classification in FPT of the following poblem by giving a kenel of size O(k 4 )oo(k 3 )dependingonthesizeofthegidweaelooking fo (see Section 3) kalmost gid Input: AgaphG, twopositiveintegesp, q, anonnegativeintegek, andasete U. Paamete: Anonnegativeintegek. Question: Can G be tansfomed to a (p, q)gid afte at most k edit opeations fom E? Notice that the nonpaameteized vesion of kalmost Gid is NPcomplete, as the poblem with q = 1andwithE consisting only in the easing a vetex opeation, is equivalent to the Longest Path poblem, i.e. the poblem of given a gaph G and aconstantk 0decideifG contains a simple path of length at least k (we just set k V (G) k). Theefoe, ou paameteization also includes the dual paameteization of the Longest Path, inthesensethatnowthepaameteisnotthelength of the path but the numbe of vetices in G that ae absent in such a path. The pimal paameteization of the Longest Path poblem was known to be in FTP 1 [6, 1]. 1 In an abuse of notation, we indistinctly efe to FPT poblem o to poblem in the FPT class 2
3 Howeve, no esult was known, so fa, fo he dual paameteiztion of Longest Path. In Section 4 we conside the following moe geneal poblem: kalmost some Gid Input: AgaphG and a set E U. Paamete: Anonnegativeintegek. Question: Decide if thee exist some pai p, q such that Gdist(G, H p,q ) k. Clealy, the above poblem can be solved applying the algoithm fo kalmost Gid fo all possible values of p and q. This implies an algoithm of time complexity O(log n(g(k) +n O(1) )). In Section 4, we explain how to avoid this O(log n) ovehead and pove that thee exists also a time O(g(k) +n O(1) )algoithmfothekalmost some Gid poblem. That way, both of ou algoithms can be seen as pepocessing algoithms that educe the size of the poblem input to a function depending only on the paamete k and not on the main pat of the poblem. Adiffeentbutsomehowelatedsetsofpoblems,whichhave eceived plenty of attention is the following: Given a gaph G and a popety Π, decide what is the minimum numbe of edges (nodes) that must be emoved to obtain asubgaphofg with popety Π. In geneal all these poblems ae NPcomplete [9, 4, 7]. Some of those poblems have been studied fom the paameteized point of view [5], howeve, all these poblems have the chaacteistic that the popety Π must be an heeditay popety. We stess that the kalmost Gid and kalmost any Gid poblems completely diffeent natue, as the popety of containing a gid is not heeditay. 2 Definitions and basic esults All gaphs we conside ae undiected, loopless and without multipleedges. Define the neighbouhood of a vetex v V (G) asn G (v) ={u V (G) (u, v) E(G)}. Let (G) =max{ N G (v) v V (G)} (i.e. (G) isthemaximumdegeeofavetexina gaph). We call (p, q)gid any gaph H p,q that is isomophic to a gaph with vetex set {0,...,p 1} {0,...,q 1} and edge set {{(i, j), (k, l)} (i, j), (k, l) V (H) and i k + j l =1}. The bode of H s, is defined as B () (H s, )=((i, 0), (i, 1) i =1,..., 1). We call a vetex (edge) of H s, intenal if it is not in (none of its endpoints is in) B () (H s, ). 3
4 Figue 1: A gaph G whee dist(g, H 15,15 ) 13 with a (6, 15)band J in it and the esult of a 3contaction of J in G. All the algoithms and esults in this pape will be fo the geneal case whee E = U. The othe cases ae staightfowad simplifications of ou esults 2.Subsequently,wewill also dop the E pat in the Edist(G, H) notation.ifdist(g, H) k we will denote as V dead /E dead the vetices/edge that should be emoved and V add /E add the vetices/edges that should be added towads tansfoming G to H. Without loss of geneality, we assume that the tansfomation pocedue gives pioity fist to vetex emovals then to edge emovals, then to vetex insetions, and finally to edge insetions. We also use the notation k 1 = V dead, k 2 = E dead, K 3 = V add and k 4 = E add. Finally, we obseve that k 1 + k 2 + k 3 + k 4 k. We say that a gaph G contains an band J of width s if the following thee conditions ae satisfied: a) G contains H s, as induced subgaph. b) Thee is no edge {x, y} E(G) suchthatx V (G) V (H) andy V (H) B () (J). If G is a gaph and H p,q is a (p, q)band in G whee p 3andq 1, the esult of the ccontaction of H p,q in G whee 0 c p 3, is the gaph obtained if we contact in G all the edges of H p,q that ae in the set {{(i, m), (i +1,m)} 2 i c +1, 1 m q}. To denote the esult of the opeation we just descibed, we use the notation contact(g, H p,q,c). Notice that outine contact(g, H p,q,c)unso(cq) insteps.foan example of a ccontaction, see Figue 1. Lemma 1 Let G be a gaph containing a (3 + c, q)band H 3+c,q fo some c> k q,andlet G = contact(g, H p,q,c). Thendist(G, H p,q ) k iff dist(g,h p c ) k. Poof: 2 Ou esults also hold fo moe geneal sets of opeations given that they locally change the stuctue of the input gaph. 4
5 We call an edge of a gaph qexteme if q>1andbothitsendpointshavedegee3 o p =1andbothitsendpointshavedegee2. Fothesetofqexteme edges of a gaph G we use the notation E (q) ext (G). findedgemaxband(g, q, e) Input: AgaphG, apositiveintegeq and a qexteme edge e of G. Output: Retun a maximal length (3 + c, q)band H 3+c,q whee c 0ande is also an exteme edge of H. If such a(3 + c, q)band does not exists, then etun NO. 1 let e = {y 1,0,z 1,0 } and set L {y 1,0 }, R {z 1,0 }, i 2 2 while i q and thee ae vetices y i,0 and z i,0 such that {y i,0,y i,0 }, {z i,0,z i,0 }, {y i,0,z i,0 } E(G) (d(y i,,0 )=d(z i,0 )=4andi<q)o(d(y i,0 )=d(z i,0 )=3andi = q) o (d(y i,0 )=d(z i,0 )=1andq =1), and N(y i,0 ) (L R {y i,0 } = N(z i,0 ) (L R {z i,0 } =, do set L L {y i,0 }, R R {z i,0 },andi i +1, 3 if i = q +1,thendo begin set 1, fo i =1,...,q,let{z i, } = N(z i, 1 ) R while N(z i, ) (L R) ={z i, 1 }, 1 i<j q {z i,,z j, } E(G) j = i +1, and 1 i q (d(z i, )=4and1<i<q)o((d(z i, )=3andi {1,q}) o ((d(z i, )=2andq =1),do R R {z 1,,...,z q, }; +1;foi =1,...,q, {z i, } N(z i, 1 ) R if N(z i, ) R = {z i, 1 } and 1 i<j q {z i,,z j,z } E(G) j = i +1,then R R {z 1,,...,z q, },othewise 1 set l 1, fo i =1,...,q,let{y i,l } = N(y i,l 1 ) L while N(y i,l ) (L R) ={y i,l 1 }, 1 i<j q {y i,l,y j,l } E(G) j = i +1, and 1 i q (d(y i,l )=4and1<i<q)o((d(y i,l )=3andi {1,q}), do L L {y 1,l,...,y q,l }; l l +1;foi =1,...,q, {y i,l } N(y i,l 1 ) L if N(y i,l ) L = {y i,l 1 } and 1 i<j q {y i,l,y j,l } E(G) j = i +1,then L L {y 1,l,...,y q,l },othewisel l 1 if + l>0thenetung[r L] end 4 etun NO. Lemma 2 The algoithm findedgemaxband(g, q, e) is coect. If the answe is NO the algoithm finishes in O(q) steps. If the answe is a vetex set with (3 + c)q vetices, then the algoithm finishes in O(qc) steps. 5
6 Poof: Lemma 3 If G is a gaph whee (G) d and dist(g, H p,q ) k whee p q, theng contains at most (p 2) + (q 2) + 2dk qextemal edges. Poof: Suppose that G contains moe than (p 2) + (q 2) + 2dk qextemal edges. Notice fist that the emoval of the vetices in V dead can decease the size of this set by at most 2dk 1.ThentheemovaloftheedgesofE dead can futhe decease it by at most 5k 2. The addition of V add does not decease this numbe and finally, the addition of E add can decease it by 2k 4.Ask 1 + k 2 + k 4 k, wehavethath will have moe than (p 2) + (q +2) extemal edges, acontadiction as H contains exactly (p 2) + (q 2) qextemal edges. findmaxband(g, q, c) Input: AgaphG, andtwopositiveintegesq,c Output: Retun a maximal length (3 + c,q)band H 3+c,q whee c c. (3 + c,q)band does not exists, then etun NO. If such a 1 Let E E (q) ext (G) 2 fo while E do begin let e be an edge in E. if findedgemaxband(g, q, e) =H 3+c,q then end 3 etun NO if c c, thenetunh 3+c,q, othewise, set E E E whee E ae the qintenal exteme edges ofh 3+c,q, othewise set E E {e} Lemma 4 Algoithm findedgemaxband(g, q) is coect. In case of negative answe it uns in O(q E q ext (G) ) steps. In case of a positive answe it uns in O(qc Eq ext ) steps. Poof: Suppose that G can be tansfomed to H afte a sequence of edit opeations. We call a vetex v G safe if is not emoved by a vetex emoval opeation in this sequence. If a vetex of G is not live then we call it dead. We also say that an safe vetex v is a dity vetex if some of the edit opeations altes the set of edges incident to v in G (i.e. eithe adds o emoves some edge incident to v). If a safe vetex is not dity, when we 6
7 call it clean. Finally,wecallavetexofH that is not a vetex of G (i.e. was intoduced duing some vetex insetion opeation) new. Clealy, if dist(g, H) k, thenthee ae at most k new vetices in H and at most k dead vetices in G and this implies the following lemma. Lemma 5 Let G and H be two gaphs whee dist(g, H) k and such that the tansfomation of G to H involves k 1 vetex emovals and k 2 vetex additions. Then, V (H) k 2 V (G) V (H) + k 1. Poof: Lemma 6 Let G be a gaph such that dist(g, H p,q ) k. Supose also that G does not contain any (3 + c, q)band whee q c>kand H p,q contains d dity vetices. Then p ( k q +3)(d + k +1). Poof: Assume that H p,q contains d dity vetices and at most k 3 = V in new vetices. Let l = k q.noticethath p,q contains p (l +3) 1distinct(butpossiblyovelapping) (l + 3,q)bands. We denote this collection as B. A new vetex can be in columns at most l +3 membes of B and a dity vetex can be in columns of at most l +1 membes of B. If H contains a membe of B that does not contain columns with new vetices o inteio columns with dity vetices, then p (l +3)+1 k 3 (l +3) d(l +1)> 0. Theefoe p l +2+k 3 (l +3)+d(l +1) l +2+k(l +3)+d(l +1). Lemma 7 Let G be a gaph with a vetex v of degee moe than k +4 in G and let H p,q be a gid. Then any way to tansfom G to H p,q should involve the opeation of the emoval of v. Inpaticula,dist(G, H p,q ) k iff dist(g[v (G) v],h) k 1. Poof: Lemma 8 Let G be a gaph whee (G) b and let H be some gid. If dist(g, H) k then H will contain at most max{b, 2} k dity vetices. Poof: If v is a dity vetex of H this means that it is eithe adjacent to a vetex in V dead o is incident to an edge in E add E dead. Notice that each vetex in X dead is can be adjacent to at most b dity vetices of H. AlsoanedgeinE dead E dead can be incident to at most two dity vetices of H. Theefoe, each edit opeation ceates at most max{2,b} dity vetices and the esult follows. 3 Looking fo a given gid H p,q In this section, we show that the kalmost gid poblem is in FPT. The next lemma intoduces the function f to bound the size of the kenel. 7
8 Lemma 9 Let G be a gaph whee dist(g, H p,q ) k. supposealsothat (G) b and that G does not contain any (3 + c, q)band whee c>k/qo any (3 + c, p)band whee c>k/p.then V (G) f(k, b, p, q) whee f(k, b, p, q) = k + k 2 if p k and q k k +5k(b k + k +1) if min{p, q} k<max{p, q} k +16(b k + k +1) 2 if p>kand q>k Poof: Fom Lemma 5 we get that V (G) p q +k. FomLemma8,thedityvetices of H will be at most b k. Fom Lemma 6 we get that p ( k q +3)(b k + k +1) and q ( k q +3)(b k + k +1). If p k and q k then clealy V (G) k + k2. If q k then p ( k q +3)(b k + k +1) ( k q +4)(b k + k +1) and theefoe V (G) k+p q k+q( k q +4)(b k+k+1) k+5k(b k+k+1). The symmetic analysis woks when p k. Ifnowp, q > k then k q = k p =1andtheefoep, q 4(b k+k+1). We conclude that V (G) k +16(b k + k +1) 2. KenelfoCheckGid(G, p, q, k) Input: AgaphG, twopositiveintegesp, q, andanonnegativeintegek Output: Eithe etuns NO, meaning that that dist(g, H p,q ) >k, o etuns a gaph G and a tiple p,q,k such that dist(g, H p,q ) k iff dist(g,h p,q ) k. 0. Setk = k, G = G, p = p and q = q. 1. AslongasG has a vetex of degee k +4emove it fom G and set k k 1. If afte the end of this poccess k < 0thenetun NO. 2. Applyanyofthefollowingpoceduesaslongasthisispossible: As long as findmaxband(g,q, k q +1) = H 3+c,q,thenG = contact(g,h 3+c,q,c) and set p p c. As long as findmaxband(g,p, k p +1) = H 3+c,p,thenG = contact(g,h 3+c,p,c) and set q q c. 3 If V (G ) >f(k,k+4,p,q )thenetun NO. 4. etung,p,q,k. Theoem 1 Algoithm KenelfoCheckGid(G, p, q, k) is coect and if it outputs the gaph G then V (G ) f(k, k +4,p,q). Moeove,itunsinO(pq(p + q + k 2 )) steps. Poof: Step 1 is justified by Lemma 7. Notice that befoe the algoithm entes Step 2, (G) k + 4. Step 2 ceates equivalent instances of the poblem because of Lemma 1. Fom Lemma 4, Each time a contaction of step 2 is applied, the coesponding q band (pband) equies O(q c(p + q + k 2 )) (O(p c(p + q + k 2 ))) steps. As the the sum of the lengths of the bands cannot exceed q(p) weobtainthatstep2 equies, in total, O(pq(p+q +k 2 )) steps. Moeove, befoe the check of Step 3, alltheconditionsof 8
9 Lemma 9 ae satisfied and theefoe, if the algoithm etuns G,p,q,k,then V (G ) f(k,k+4,p,q ) f(k, k +4,p,q). CheckGid(G, p, q, k) Input: AgaphG, twopositiveintegesp, q, andanonnegativeintegek Output: Eithe etuns NO, meaning that that dist(g, H p,q ) >k,o etuns a sequence of at most k opeations that tansfoms G to H p,q. 1. foanysetv dead V (G) ofatmostk vetices do if E(G[V (G) V dead ]) 2pq p q + k then fo any set E dead of at most k V dead edges of E(G[V dead ]) do fo any i =0,...,k V dead E dead do begin let G be the gaph obtained fom G afte the emoval of the vetices in V dead and the edges in E dead and the addition of a set V add of i new vetices. let S be the set of vetices in G with degee less than 4. if S 2p +2q 4+2k then do fo any E add S S whee E add k V dead E dead V add do if (V (G ),E(G ) E add )isisomophictoh p,q then etun asequencecontainingthesequenceofopeations 2. etun NO end defined by the sets V dead, E dead, V add,ande add Theoem 2 Algoithm CheckGid(G, p, q, k) is coect and uns in O(16 k (2n +3k) 2k ) steps. Poof: we ecall the notation k 1 = V dead, k 2 = E dead, K 3 = V add and k 4 = E add. The fist fo of the algoithm guesses the set of vetices V dead that should be emoved while tansfoming G to H p,q.clealy,theeae ( n k) = O(n k 1 )waystomakethisguess. Fom Lemma 5 these guesses ae at most (pq + k 1 ) k 1. If this is a coect guess, then the emaining gaph will be a subgaph of H p,q with at most k edges moe. As H p,q has (p 1)q +(q 1)p edges, then G[V (G) V dead ]willhaveatmost2pq p q + k 2 edges and the filte of the algoithm befoe the second fo is coect. Thesecond fo guesses the set E dead of edges that should be emoved and thee ae at most ( 2pq p q+k2 k 2 ) (2pq + k2 ) k 2 ways to make such a guess. Then comes the tun to guess 9
10 the set of vetices V add that should be added towads constucting H p,q.aswecannot add moe than k 3 vetices, the thid fo makes at most k 3 calls one fo each possible quantity of vetices in V add.noticethatifalltheguessesdonesofaaecoect,g is a subgaph of H p,q with k 4 edges less than H p,q.ash p,q has 2(p 1)+2(q 1) vetices of degee less than 4 V (G )shouldcontainasubsets of at most 2p+2q 4+2k 4 vetices of degee less than 4 (an missing edge can ham the degee of at most 2 vetices). Clealy, the edges to add in G towads constucting H p,q will have endpoints in S and theefoe each choice of an edge to add will be done by a set of ( 2p+2q 4+2k 4 ) 2 4(p + q + k4 ) 2 candidate edges. As thee ae at most k 4 guesses of edges to be done in the fouth fo we finally have at most ( 4(p+q+k 4 ) ) 2 k 4 (4(p + q + k4 )) 2k 4 edge sets to guess. Afte than, we can veify whethe the esulting gaph is isomophic to H p,q,whichcanbedonein O(pq) steps. Resuming,thetotalunningtimeisO((pq + k 1 ) k1 (2pq + k 2 ) k2 k 3 (4(p + q + k 4 )) 2k4 pq) =O(16 k (p + q + k) 2k ). As p + q 2pq, usinglemma5weobseve that (p + q + k) 2pq + k 2(pq k)+3k 2n +3k and we have the equied time bound. Almostgid(G, p, q, k) Input: AgaphG, twopositiveintegesp, q, andanonnegativeintegek Output: Eithe etuns NO, meaning that that dist(g, H p,q ) >k,o etuns a sequence of at most k opeations that tansfoms G to H p,q. 1. IfKenelfoCheckGid(G, p, q, k) =(G,p,q,k )thenetuncheckgid(g, p, q, k) 2. etun NO Theoem 3 Algotihm Almostgid(G, p, q, k) solves the kalmostgid poblem in 2 O(k log k) + O(n 3 ) steps. Poof: The algoithm we fist calls the algoithm KenelfoCheckGid(G, p, q, k) that gives some answe in O(pq(p + q + k 2 )) = O(n 3 )steps. etuns that dist(g, H p,q ) >k. Iftheansweis NO then If the algoithm etuns (G,p,q,k )thensolvek AlmostGid with input G,p,q,k using bute foce and etun the coesponding answe. Fom Theoem 1 we have that V (G ) k +16(k 2 +5k +1) 2 = O(k 4 ). Fom Theoem 2, CheckGid(G, p, q, k) equies(o(k)) 8k steps. 4 Looking fo any gid Fo α, β, γ N, wedefinepods(α, β, γ) ={(p, q) N 2 p α and γ β p q γ + β}. 10
11 Clealy, to solve the kalmost some Gid poblem it is enough to apply Check Gid(G, p, q, k) foall(p, q) A(n, k) = n k i n+k pods(i, n, k). Clealy, A(n, k) 2k 2 log(k + n). Theefoe, kalmost some Gid can be solved afte O(k 2 log(n + k)) calls of Almostgid(G, p, q, k) whichgivesaunningtimeofo(log n)(2 O(k log k) )+ O(k 2 n 3 log n) steps. Wecallthisalgoithmcheckallcases(G, k). We stess that thee ae cases whee A(n, k) 1 2 log n and theefoe, that way, we may not avoid the log n ovehead. In what emains we will explain an altenative appoach that gives unning times of the same type as the case of kalmost Gid. We call band collection C of a gaph G acollectionof C bands in G of widths s 1,...,s C whee no pai of them have common inteio vetices. The width of such a collection is equal to i=1,..., C (s i 2). findmaxbandcollection(g,, w) Input: AgaphG and two positive integes and w Output: Retun YES ifg contains an band collection of width w; othewise, etun NO. 1 Let E ext be the set of extemal edges of G, t 0 2 while E ext and t<wdo begin t 0 Let e be an edge in E ext set E E ext while findedgemaxband(g,, e) =H 3+c, do begin set E E E whee E ae the exteme edges of H 3+c, set t t + c +1 let e be an edge in E end set E ext E ext {e} end 3 if t w then etun YES, othewise etun NO. Lemma 10 The algoithm findmaxbandcollection(g,, w) is coect and uns in O(w E ext (G) + n) steps. Poof: 11
12 We fist need some conditions on when it is possible to make an estimation of the dimensions of a gid. Lemma 11 Let G be a gaph whee (G) b and let H p,q be a gid whee dist(g, H) k. Then If G does not contain any qband collection of width > k then V (H) (b k +2) 2. Poof: H p,q contains a single element qband collection C H of width p 2. We apply back the k opeations that tansfomed G to H. IfweemoveanedgeinY in,inthe wost case, its endpoints may belong in neighboing columns of C H and this can emove at most 2 units fom the width of C H.IfweaddbackanedgeinY out,inthewostcase, its endpoints may belong in diffeent columns of C H and this can emove at most 4 units fom the width of C H.IffinallyweaddbackavetexinX out,inthewostcase,allits neighbous will belong into diffeent columns and this may educe the width of C H by (G) b. Resuming, we have that G contains a q band collection of width at least p 2 e in e out b x out p 2 b k which implies that p b k +2. Woking symmetically, we also conclude that q b k +2andtheesultfollows. Lemma 12 Let G be a gaph whee (G) b and let H be a gid whee dist(g, H) k. Then if G contains an band collection of width > 3k then H = H s, fo some s [ n k, n+k ]. Moeove at most two such band collections exist and fo each, s will be unique when >2k. Poof: Let C be the inteio columns in such a collection of bands. We will apply the edit opeations that tansfom G to H. The emoval of a vetex in X out can educe the width of C by at most 3. of C by at most 3. The emoval of an edge in Y out can educe the width The addition of an edge in Y in can educe the width of C by at most 2. Theefoe, in the wost case, H still contains a collection of bands of width 3k +1 3k >1. So, thee exist at least one band of width at least 1 in H and this means that H = H s, fo some s 1. Suppose now that thee ae thee integes 1 < 2 < 3 and that G contains an i band collection of width > 3k, foi =1, 2, 3. Let also I i, =1, 2, 3betheinteio columns of each of these band collections. Clealy, fo any i<j,i i I j. Accoding to the pevious analysis, afte the edit opeations, thee, diffeent, size i bands of width al least 1 will suvive which is impossible to exist in a ectangula gid. Notice now that s = n x out + x in. Let δ = x out x in. We obtain that s [n δ, n + δ] whichimpliesthats [ n k ]containstwointeges,then n+k [ n k, n+k, n+k ], as δ k. If now the inteval n k 1whichimpliesthat 2k. 12
13 ChecksomeGid(G, k) Input: A gaph Gand a nonnegative intege k Output: The answe to the kalmostsomegid poblem with instance G and paamete k. 1. As long as G has a vetex of degee k +4emove itfomg and 2. R set k k 1. If afte the end of this poccess k<0thenetun NO. 3. fo i =1,...,n,iffindmaxbandcollection(G, i, 3k +1)= YES, then set R R {i} 4. If R =0thenif V (G) >k+((k +4)k +2) 2 then etun NO, othewise, goto Step 7 5. Fo any R, If >2k, then begin If N [ n k, n+k ]=, then etun NO, othewise, etun AlmostGid(G,, s, k) whee{s} = N [ n k end othewise, fo (i, j) pods(2k,n,k), begin, n+k ]. If AlmostGid(G, i, j, k) =YES, then etun AlmostGid(G, i, j, k). end 6. Retun NO. 7. checkallcases(g, k). Theoem 4 Algoithm ChecksomeGid(G, k) is coect and the kalmostanygid poblem is in FPT and can be solved in time 2 O(k log k) + O(k log k n 3 ). Poof: Step 1 is justified by Lemma 7 and we may assume that befoe the algoithm entes Step 2, (G) k +4. Steps 2 4 ae based on Lemma 11. Step 3 involves O(n) callsoffindmaxbandcollection(g, i, 3k +1)whichequiesO(kn 2 )stepsbecause of Lemma 10. Theoefoe, Step 3 equies O(kn 3 )steps. FinallytheloopinStep5 is coect because of Lemma 12. Note that the fist loop in step 5 is applied at most 2 times and that pods(2k, n, k) = O(k log k). Theefoe, step 5 uns in 2 O(k log k) + O(k log k n 3 )steps. Aswenoticedinthebeginingofthissectioncheckallcases(G, k) equies O(log n )(2 O(k log k) + O(log n n 3 k 2 )) steps whee n is the numbe of vetices of G in step 7 whee n O(k 4 )andthismeansthatstep7 equies 2 O(k log k) steps. Refeences [1] N, Alon, R. Yuste and U. Zwick. ColoCoding. Jounal of the ACM, 42(4), ,
14 [2] G. Downey and M. Fellows. Paameteized Complexity. SpingeVelag,1999. [3] H. Kaplan, R. Shami and R. Tajan.. Tactability of paameteized completion poblems on chodal, stongly chodal and pope inteval gaphs. SIAM Jounal of Computing, 28, ,1999. [4] J. Lewis and M. Yannakakis. The nodedeletion poblem fo heeditaypopeties is NPComplete. Jounal Comput. and Systems Sci. 20(2), , [5] Leizhen Cai. FixedPaamete Tactability of Gaph Modification Poblems fo Heeditay Popeties. Infomation Pocessing Lettes, 58(4) , 1996 [6] B. Monien. How to find paths efficiently. Annals of Discete Mathematics, 25, , [7] A. Natanzon, R. Shami and R. Shaan. Complexity classification of some edge modification poblems. Discete Applied Mathematics, 113(1), ,2001. [8] T. Sebastian, P. Klein, and B. Kimia. Recognition of shapes by editing thei shock gaphs. IEEE Tansactions on Patten Matching and Machine Intelligence, 26, , [9] M. Yannakakis. Node and Edge Deletion NPcomplete poblems. ACM Symposium on Theoy of Computing (STOC), ,1978. [10] K. Zhang and D. Sasha. Simple fast algoithms fo the editing distance between tees and elated poblems. SIAM Jounal on Computing 18, ,
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