Unique perfect phylogeny is intractable

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1 Unque perfect phylogeny s ntractable Mchel Habb a, Juraj Stacho b, a LIAFA CNRS and Unversté Pars Dderot Pars VII, Case 7014, Pars Cedex 13, France b DIMAP and Mathematcs Insttute, Unversty of Warwck, Coventry CV4 7AL, Unted Kngdom Abstract A phylogeny s a tree capturng evoluton and ancestral relatonshps of a set of taxa (e.g., speces). Reconstructng phylogenes from molecular data plays an mportant role n many areas of contemporary bologcal research. A phylogeny s perfect f (n rough terms) t correctly captures all nput data. Determnng f a perfect phylogeny exsts was shown to be ntractable n 1992 by Mke Steel [32] and ndependently by Bodlaender et al. [4]. In lght of ths, a related problem was proposed n [32]: gven a perfect phylogeny, determne f t s the unque perfect phylogeny for the gven dataset, where the dataset s provded as a set of quartet (4-leaf) trees. It was suggested that ths problem may be more tractable [32], and determnng ts complexty became known as the Quartet Challenge [33]. In ths paper, we resolve ths queston by showng that the problem s CoNP-complete. We prove ths by relatng perfect phylogenes to satsfyng assgnments of Boolean formulas. To ths end, we cast the queston as a chordal sandwch problem. As a partcular consequence of our method, we show that the unque mnmal chordal sandwch problem s CoNP-complete, and countng mnmal chordal sandwches s #P-complete. Key words: perfect phylogeny, chordal graph, trangulaton, chordal sandwch, ntractablty, unque soluton 1. Introducton One of the major efforts n molecular bology has been the computaton of phylogenetc trees, or phylogenes, whch descrbe the evoluton of a set of speces from a common ancestor. A phylogenetc tree for a set of speces s a tree n whch the leaves represent the speces from the set and the nternal nodes represent the (hypothetcal) ancestral speces. One standard model for descrbng the speces s n terms of characters, where a character s an equvalence relaton on the speces set, parttonng t nto dfferent character states. In ths model, we also assgn character states to the (hypothetcal) ancestral speces. The desred property s that for each state of each character, the set of nodes n the tree havng that character state forms a connected subgraph. When a phylogeny has ths property, we say t s perfect. The Perfect Phylogeny problem [20] then asks for a gven set of characters defnng a speces set, does there exst a perfect phylogeny? Note that we allow that states of some characters are unknown for some speces; we call such characters partal, otherwse we speak of full characters. Ths approach to constructng phylogenes has been studed snce the 1960s [8, 25, 26, 27, 35] and was gven a precse mathematcal formulaton n the 1970s [12, 13, 14, 15]. In partcular, Buneman [7] showed that the Perfect Phylogeny problem reduces to a specfc graphtheoretc problem, the problem of fndng a chordal completon of a graph that respects a prescrbed colourng. In fact, the two problems are polynomally equvalent [23]. Thus, usng ths formulaton, t has been proved that the Perfect Phylogeny problem s NP-hard n [4] and ndependently n [32]. These two results rely on the fact that the nput may contan partal characters. In fact, the characters n these constructons only have two states. If we nsst on full characters, the stuaton s dfferent as for any fxed number r of character states, the problem can be solved n tme polynomal [1] n the sze of the nput (and exponental n r). In partcular, for r = 2 (or r = 3), the soluton exsts f and only f t exsts for every par (or trple) of characters [15, 24]. Also, when the number of characters s k (even f there are partal characters), the complexty [28] s polynomal n the number of speces (and exponental n k). Correspondng author Emal addresses: habb@lafa.unv-pars-dderot.fr (Mchel Habb), j.stacho@warwck.ac.uk (Juraj Stacho) Preprnt submtted to Elsever January 9, 2013

2 Another common formulaton of ths problem s the problem of a consensus tree [10, 19, 32], where a collecton of subtrees wth labelled leaves s gven (for nstance, the leaves correspond to speces of a partal character). Here, we ask for a (phylogenetc) tree such that each of the nput subtrees can be obtaned by contractng edges of the tree (we say that the tree dsplays the subtree). The problem does not change [31] f we only allow partcular nput subtrees, the so-called quartet trees, whch have exactly sx vertces and four leaves. Ths follows from the fact that every ternary phylogenetc tree (all nternal nodes have degree 3) can be unquely descrbed by a collecton of quartet trees [31]. However, a collecton of quartet trees does not necessarly unquely descrbe a ternary phylogenetc tree. (Note that some authors use the term bnary tree [5, 31] or subcubc tree for what we call here a ternary tree as defned n [30].) Ths leads to a natural queston (frst posed n [32]): What s the complexty of decdng whether or not a collecton of quartet trees unquely descrbes a (ternary) phylogenetc tree? Intally, t was suggested [32] that ths problem may be more tractable. Indeed, a pror t s possble that unque solutons only exst for specal collectons of quartet trees and thus have specal structure whch could be easy to test. However, as the problem was open for a number of years, and perhaps from experence wth real datasets, t became more clear that ths probably s not the case. Ths was reflected n the problem beng conjectured to be ntractable by Mke Steel who named t Quartet Challenge and lsted t on hs personal webpage [33] alongsde wth other challengng research problems from the area of phylogenetcs. In partcular, to emphasze the mportance of the problem, a prce of $100 was offered for the frst proof of ntractablty. In ths paper, we resolve the problem by showng that t s ndeed ntractable. Namely, we show the followng. Theorem 1. It s CoNP-complete to determne, gven a ternary phylogenetc X-tree T and a collectonqof quartet subtrees on X, whether or nott s the only phylogenetc tree that dsplaysq. To prove ths theorem, we nvestgate the graph-theoretcal formulaton of the problem [7] and vew t through the noton of chordal sandwch [17]. In contrast, an alternatve proof of the theorem, whch recently appeared as [5], s based on the betweenness property, extendng the hardness result of [32]; our proof extends the hardness from [4]. In lght of ths, we note that there are specal cases of the problem that are known to be solvable n polynomal tme. For nstance, ths s so f the collecton Q contans a subcollecton Q wth the same set L of labels of leaves and wth Q = L 3. However, fndng such a subcollecton s known to be NP-complete. For these and smlar results, we refer the reader to [3]. We prove Theorem 1 by descrbng a polynomal-tme reducton from the unqueness problem for ONE-IN-THREE- 3SAT, whch s CoNP-complete by [22]. Theorem 2. [22] It s CoNP-complete to decde, gven an nstance I of ONE-IN-THREE-3SAT, and a truth assgnment σ that satsfes I, whether or not σ s the unque satsfyng truth assgnment for I. We extract ths from [22] by encodng the problem as the ternary relaton{(0, 0, 1),(0, 1, 0),(1, 0, 0). We check that ths relaton s not: 0-vald, 1-vald, Horn, ant-horn, affne, 2SAT, or complementve. Thus the unqueness of the satsfablty problem correspondng to ths relaton s CoNP-complete by [22]. Our constructon n the reducton s essentally a modfcaton of the constructon of [4] whch proves NP-hardness of the Perfect Phylogeny problem. Recall that the constructon of [4] produces nstances Q that have a perfect phylogeny f and only f a partcular boolean formula Φ s satsfable. Whle studyng ths constructon, we mmedately observed that these nstances Q have, n addton, the property that Φ has a unque satsfyng assgnment f and only f there s a unque mnmal restrcted chordal completon of the partal partton ntersecton graph of Q (for defntons see 2). Ths s precsely one of the two necessary condtons for unqueness of perfect phylogeny as proved by Semple and Steel n [30] (see Theorem 5). Thus by modfyng the constructon of [4] to also satsfy the other condton of unqueness of [30], we obtaned the constructon that we present n ths paper. Note that, however, unlke [4] whch uses 3SAT, we had to use a dfferent problem n order for the constructon to work correctly. Also, to prove that the constructon s correct, we employ a varant of the characterzaton of [30] that uses the more general chordal sandwch problem [17] nstead of the restrcted chordal completon problem (see Theorem 8). In fact, by way of Theorems 6 and 7, we establsh a drect connecton between the problem of perfect phylogeny and the chordal sandwch problem, whch apparently has not been yet observed. (Note that the connecton to the (restrcted) chordal completon problem of coloured graphs as mentoned above [7, 23] s a specal case of ths.) Fnally, as a corollary, we obtan the followng result whch s very nterestng by tself. 2

3 Corollary 3. The unque mnmal chordal sandwch problem s CoNP-complete. The problem of countng the number of mnmal chordal sandwches s #P-complete. The frst part follows drectly from Theorems 2 and 9, whle the second part follows from Theorem 9 and [9]. The paper s structured as follows. In 2, we descrbe some prelmnary defntons and results needed for the constructon n our reducton. In partcular, we descrbe, based on [30], necessary and suffcent condtons for the exstence of a unque perfect phylogeny n terms of the mnmal chordal sandwch problem (cf. [16, 17]). The proof of ths characterzaton s postponed untl 5. In 3 and 4, we present our hardness reducton, frst nformally and then formally. We state the two unqueness condtons (Theorems 9 and 10) relatng satsfyng assgnments of an nstance I of ONE-IN-THREE-3SAT to mnmal chordal sandwches and phylogenetc trees unquely determned by these assgnments. The proofs are presented later n 6 and 7. In 8, we put these results together to prove Theorem 1. We conclude n 9 wth some other consequences and open questons related to ths work. 2. Prelmnares We mostly follow the termnology of [30, 31] and the graph-theoretcal notons of [34]. In ths paper, a graph s always smple, undrected, wth no loops or parallel edges. For a graph G = (V, E), we wrte V(G) to denote ts vertex set and E(G) to denote ts edge set. We wrte uv for the edge (u, v) E(G), and say that u, v are neghbours or adjacent n G. For a vertex v V(G), we denote by N G (v) the neghbourhood of v n G,.e, the set of neghbours of n G. We wrte N G [v] for N G (v) {v. When approprate, we drop the ndex G and smply wrte N(v) and N[v]. For a set X V(G), we denote by G[X] the subgraph of G nduced by X,.e., the graph wth vertex set X and edges uv such that u, v X and uv E(G). We wrte G X for the graph G[V(G)\ X]. Smlarly, for a set of edges F E(G), we wrte G F for the graph wth vertex set V(G) and edge set E(G)\ F. We wrte G x as a shorthand for G {x. We say that X s a clque of G f G[X] s a complete graph (.e., has all possble edges). A vertex v V(G) s a smplcal vertex of G f all ts neghbours are parwse adjacent. A graph s a chordal graph f t does not contans an nduced cycle of length four or more. A perfect elmnaton orderng of a graph G s an orderng v 1, v 2,..., v n of the vertces of G such that for every {1... n, the vertex s a smplcal vertex of G [ {v 1,..., ],.e., all ts neghbours among {v 1..., 1 are parwse adjacent. It s well-known [11] that a graph s chordal f and only f t admts a perfect elmnaton orderng. Let X be a non-empty set. An X-tree s a par (T, φ) where T s tree and φ : X V(T) s a mappng such that φ 1 (v) = for all vertces v V(T) of degree at most two. An X-tree(T, φ) s ternary f all nternal vertces of T have degree three. Two X-trees(T 1, φ 1 ),(T 2, φ 2 ) are somorphc f there exsts an somorphsm ψ : V(T 1 ) V(T 2 ) between T 1 and T 2 that satsfes φ 2 = ψ φ 1. An X-tree (T, φ) s a phylogenetc X-tree (or a free X-tree n [30]) f φ s a bjecton between X and the set of leaves of T. A partal partton of X s a partton of a non-empty subset of X nto at least two sets. If A 1, A 2,..., A t are these sets, we call them cells of ths partton, and denote the partton A 1 A 2... A t. If t = 2, we call the partton a partal splt. A partal splt A 1 A 2 s trval f A 1 = 1 or A 2 = 1. A quartet tree s a ternary phylogenetc tree wth a label set of sze four, that s, a ternary tree T wth 6 vertces, 4 leaves labelled a, b, c, d, and wth only one non-trval partal splt {a, b {c, d that t dsplays. Note that such a tree s unambguously defned by ths partal splt. Thus, n the subsequent text, we dentfy the quartet tree T wth the partal splt{a, b {c, d, that s, we say that {a, b {c, d s both a quartet tree and a partal splt. Let T = (T, φ) be an X-tree, and let π = A 1 A 2... A t be a partal partton of X. Let F E(T) be a set of edges of T. We say that F dsplays π n T f for all dstnct, j {1... t, the sets φ(a ) and φ(a j ) are subsets of the vertex sets of dfferent connected components of T F. We say that T dsplays π f there s a set of edges that dsplays π n T. Further, an edge e of T s dstngushed by π f every set of edges that dsplays π n T contans e. Let Q be a collecton of partal parttons of X. An X-tree T dsplays Q f t dsplays every partal partton n Q. An X-tree T = (T, φ) s dstngushed by Q f every nternal edge of T s dstngushed by some partal partton n Q; we also say that Q dstngushes T. The set Q defnes T f T dsplays Q, and all other X-trees that dsplay Q are somorphc to T. Note that f Q defnes T, then T s necessarly a ternary phylogenetc X-tree, snce otherwse 3

4 {1, 2 {2, {1, 4 {5, {3, 4 {3, 5 a) b) c) d) Fgure 1: a) quartet trees Q, b), c) two X-trees dsplayng Q and dstngushed by Q, d) nt (Q); dotted lnes represent the edges n forb(q). resolvng any vertex ether of degree four or more, or wth multple labels results n a non-somorphc X-tree that also dsplaysq(also, see [30, Proposton 2.6]). See Fg. 1 for an llustraton of these concepts. The partal partton ntersecton graph ofq, denoted by nt(q), s a graph whose vertex set s {(A, π) where A s a cell of π Q and two vertces(a, π),(a, π ) are adjacent just f the ntersecton of A and A s non-empty. A chordal completon of a graph G = (V, E) s a chordal graph G = (V, E ) wth E E. A restrcted chordal completon of nt(q) s a chordal completon G of nt(q) wth the property that f A 1,A 2 are cells of π Q, then (A 1, π) s not adjacent to (A 2, π) n G. A restrcted chordal completon G of nt(q) s mnmal f no proper subgraph of G s a restrcted chordal completon of nt(q). The problem of perfect phylogeny s equvalent to the problem of determnng the exstence of an X-tree that dsplays the gven collecton Q of partal parttons. In [7], t was gven the followng graph-theoretcal characterzaton. Theorem 4. [7, 31, 32] Let Q be a set of partal parttons of a set X. Then there exsts an X-tree that dsplays Q f and only f there exsts a restrcted chordal completon of nt(q). Of course, the X-tree n the above theorem mght not be unque. For the problem of unqueness, Semple and Steel [30, 31] descrbe necessary and suffcent condtons for when a collecton of partal parttons defnes an X-tree. Theorem 5. [30] Let Q be a collecton of partal parttons of a set X. Let T be a ternary phylogenetc X-tree. Then Q defnest f and only f: () T dsplaysq and s dstngushed byq, and () there s a unque mnmal restrcted chordal completon of nt(q). In order to smplfy our proof of Theorem 1, we now descrbe a varant of the above theorem that, nstead, deals wth the noton of chordal sandwch [17]. Let G = (V, E) and H = (V, F) be two graphs on the same set of vertces wth E F =. A chordal sandwch of (G,H) s a chordal graph G = (V, E ) wth E E and E F =. We say that E are the forced edges and F are the forbdden edges. (For other possble formulatons of ths noton, see [17].) A chordal sandwch G of (G,H) s mnmal f no proper subgraph of G s a chordal sandwch of(g,h). The cell ntersecton graph of Q, denoted by nt (Q), s the graph whose vertex set s {A where A s a cell of π Q and two vertces A, A are adjacent just f the ntersecton of A and A s non-empty. Let forb(q) denote the graph whose vertex set s that of nt (Q) n whch there s an edge between A and A just f A,A are cells of some π Q. See Fg. 1d for an example. The relatonshp between the noton of partal partton ntersecton graph and the cell ntersecton graph s captured by the followng theorem. Theorem 6. Let Q be a collecton of partal parttons of a set X. Then there exsts a bjectve mappng between the mnmal restrcted chordal completons of nt(q) and the mnmal chordal sandwches of(nt (Q), forb(q)). (The proof of ths theorem s rather techncal and t s presented as 5.) Ths combned wth Theorem 4 yelds that there exsts a phylogenetc X-tree that dsplays Q f and only f there exsts a chordal sandwch of(nt (Q), forb(q)). Conversely, we can express every nstance of the chordal sandwch problem as a correspondng nstance of the problem of perfect phylogeny as follows. 4

5 Theorem 7. Let (G, H) be an nstance of the chordal sandwch problem. Then there exsts a collectonqof partal splts such that there s a bjectve mappng between the mnmal chordal sandwches of (G, H) and the mnmal restrcted chordal completons of nt(q). In partcular, there exsts a chordal sandwch for(g, H) f and only f there exsts a phylogenetc tree that dsplaysq. PROOF. Consder the nstance(g, H) where G = (V, E) and H = (V, F) are two graphs wth E F =. Wthout loss of generalty, we may assume that each connected component of G has at least three vertces. (We can safely remove any component wth two or fewer vertces wthout changng the number of mnmal chordal completons, snce every such component s already chordal.) We defne the collecton Q of partal splts (of the set E) as follows: for every edge xy F, we construct the partal splt D x D y, where D x are the edges of E ncdent to x, and D y are the edges of E ncdent to y. By defnton, the vertex set of the graph nt (Q) s precsely {D v v V. Further, t can be easly seen that the mappng ψ that, for each v V, maps v to D s an somorphsm between G and nt (Q). (Here, one only needs to verfy that D u = D mples u = v; for ths we use that each component of G has at least three vertces.) Moreover, forb(q) s precsely {ψ(x)ψ(y) xy F by defnton. Therefore, by Theorem 6, there s a one-to-one correspondence between the mnmal chordal sandwches of (G, H) and the mnmal restrcted chordal completons of nt(q). Ths proves the frst part of the clam; the second part follows drectly from Theorem 4. As an mmedate corollary, we obtan the followng desred characterzaton. Theorem 8. Let Q be a collecton of partal parttons of a set X. Let T be a ternary phylogenetc X-tree. Then Q defnest f and only f: () T dsplaysq and s dstngushed byq, and () there s a unque mnmal chordal sandwch of ( ) nt (Q), forb(q). We remark that the man techncal advantage of ths theorem over Theorem 5 s that t s less restrctve; t allows us to construct nstances wth arbtrary sets of forbdden edges rather than just wth forbdden edges between vertces of the same colour. Ths makes our proof of Theorem 1 much smpler and more manageable. 3. Overvew of the proof Consder an nstance I of ONE-IN-THREE-3SAT. The nstance I conssts of n varables v 1,..., v n and m clauses C 1,...,C m each of whch s a dsjuncton of exactly three lterals (.e., varables or ther negatons ). To smplfy the presentaton, we shall denote lterals by captal letters X, Y, etc., and ndcate ther negatons by X, Y, etc. (For nstance, f X = then X =, and f X = then X =.) A truth assgnment for the nstance I s a mappng σ : {v 1,..., v n {0, 1 where 0 and 1 represent false and true, respectvely. To smplfy the notaton, we wrte = 0 and = 1 n place of σ( ) = 0 and σ( ) = 1, respectvely, and extend ths notaton to lterals X,Y, etc.,.e., wrte X = 0 and X = 1 n place of σ(x) = 0 and σ(x) = 1, respectvely. A truth assgnment σ s a satsfyng assgnment for I f n each clause C j exactly one of the three lterals evalues to true. That s, for each clausec j = X Y Z, ether X = 1, Y = 0, Z = 0, or X = 0, Y = 1, Z = 0, or X = 0, Y = 0, Z = 1. By standard arguments, we may assume that no varable appears twce n the same clause, snce otherwse we can replace the nstance I by an equvalent nstance wth ths property. In partcular, we can replace each clause of the form v j by clauses x v j and x v j where x s a new varable, and replace each clause of the form v j by clauses v j x, v j x, and v j x where x s agan a new varable. Note that these two transformatons preserve the number of satsfyng assgnments, snce n the former the new varable x has always the truth value of whle n the latter x s always false n any satsfyng assgnment of ths modfed nstance. In what follows, we dscuss the followng objects arsng from the nstance I: the set of labels X I, 5

6 the collectonq I of quartet trees whose leaves are labelled by elements ofx I, the ternary tree T I, and the labellng φ σ : X I V(T I ) of the leaves of T I, where σ s a satsfyng assgment for I, whch together yeld the phylogenetcx I -treet σ = (T I, φ σ ). The formal defntons of these objects s gven as 4. We then prove that the satsfyng assgnments to I are n bjecton wth the mnmal chordal sandwches of nt (Q I ), the cell ntersecton graph of Q I, and forb(q I ). Further, we show that every satsfyng assgnment σ for I defnes a perfect phylogeny for Q I, namely the tree T σ = (T I, φ σ ), that s dstngushed by Q I. These together wll mply Theorem 1, the man result of ths paper. We summarze ths as the followng two theorems. Theorem 9. There s a bjectve mappng between the satsfyng assgnments of the nstance I and the mnmal chordal sandwches of(nt (Q I ), forb(q I )). Theorem 10. If σ s a satsfyng assgnment for I, thent σ = (T I, φ σ ) s a ternary phylogenetcx I -tree that dsplays Q I and s dstngushed by Q I. We present the proofs of these theorems as 6 and 7, respectvely. In the rest of ths secton, we nformally dscuss the constructons nvolved to prepare the reader for the techncal nature of the proofs that wll follow. Before descrbng the collectonq I, let us brefly revew the constructon from [4] that proves NP-hardness of the Perfect Phylogeny problem. For convenence, we descrbe t n terms of the chordal sandwch problem whose nput s a graph wth (forced) edges and forbdden edges. In [4], one smlarly consders a collectonc 1,...,C m of 3-lteral clauses, and treats t as an nstance I of 3-SATISFIABILITY. From ths nstance, one constructs a graph where each varable corresponds to two shoulders S v and S v, and where each lteral W n a clause C j corresponds to a par of knees K j W and Kj. In addton, there are two specal vertces the head H and the foot F. All shoulders are adjacent to W the head whle all knees are adjacent to the foot. Further, f occurs n the clause C j (postvely or negatvely), then the vertces H, S v, K j, F, K j, S v form an nduced 6-cycle (see Fg. 2a). Also, fc j = X Y Z, then the vertces K j X, Kj Y, Kj Z nduce a trangle wth pendant edges Kj X Kj Y, Kj Y Kj Z, and Kj Z Kj (the clause gadget, see Fg. 2b). X Fnally, the edge between H and F s forbdden n the desred chordal sandwch, and so s the edge between S v and S v, and between K j and K j (the dotted edges n Fg. 2) for all ndces and j for whch these vertces exst. The man dea of ths constructon s that each of the 6-cycles allows only two possble chordal sandwches: ether the path H, K j, S v, F s added, or the path H, K j, S v, F s added (the authors of [4] call ths path the Mark of Zorro ). These two choces correspond to assgnng the value true or false, respectvely, and the constructon ensures that ths choce s consstent over all clauses. Ths only produces satsfyng assgnments to 3-SATISFIABILITY, snce we notce that no chordal sandwch adds a trangle on K j X, Kj Y, Kj Z. One can try to use ths constructon to prove Theorem 1 (we explan later why ths fals). Indeed, t can be observed that the truth assgnments satsfyng the clausesc 1,...,C m are n one-to-one correspondence wth the mnmal chordal sandwches of the above graph G. To see ths, one descrbes all edges that we are forced to have n the sandwch after the marks of Zorro are added accordng to a satsfyng assgnment. It turns out that these edges yeld a chordal sandwch, and thus a mnmal chordal sandwch. From G, usng Theorems 6 and 7, one can further construct a collecton Q of partal splts (phylogenetc trees) such that the satsfyng assgnments of the clauses C 1,...,C m are n a one-to-one correspondence wth the mnmal chordal sandwches of (nt (Q), forb(q)). In partcular, ths collecton Q satsfes the condton () of Theorem 8 f and only f the clauses C 1,...,C m have a unque satsfyng assgnment. Snce ths s CoNP-complete to determne [22], t would seem lke we almost have a proof of Theorem 1. Unfortunately, we are mssng a crucal pece whch s the phylogenetc tree T satsfyng the condton () of Theorem 8 for the collecton Q. A straghtforward constructon of such a tree based on [30] yelds a phylogenetc tree that s dstngushed byq, but whose nternal nodes may have 6

7 H H K j Z S v S v S v S v a) K j K j K j Mark of Zorro K j b) K j X K j Y K j Z F F K j Y K j X Fgure 2: Confguratons from [4]. A L j Y S j W S j W K j Z a) K j W H W B H W L j W b) K j Y K j Y F j K j W L j X K j X K j Z L j Z D j p D j p+1 K j X Fgure 3: Confguratons from our constructon (note that, on the left, W s a lteral, ether or, and s the p-th lteral of the clause C j ) degree hgher than three. If we try to fx ths (by resolvng the hgh-degree nodes n order to get a ternary tree), the resultng tree may no longer be dstngushed by Q. Moreover, the collecton Q may not consst of quartet trees only. For all these reasons, we need to modfy the constructon of G. Frst, we dscuss how to modfy G so that t corresponds to a collecton of quartet trees. To do ths, we must ensure that the neghbourhood of each vertex conssts of two clques (wth possbly edges between them). We construct a new graph G I by modfyng G as follows. Instead of one head H, we ntroduce, for each varable, two heads H v, H v, and an auxlary head A. For a lteral W n the clause C j, we ntroduce two shoulders S j W and S j W, and, as before, two knees Kj W and Kj W, but also an addtonal auxlary knee Lj W. Further, for each clause C j, we ntroduce a foot F j and three auxlary feet D j 1, Dj 2, and Dj 3. Fnally, we add one addtonal vertex B known as the backbone. The resultng modfcatons to the 6-cycles and the clause gadgets can be seen n Fg. 3a and 3b. (The forbdden edges are agan ndcated by dotted lnes.) Note that, unlke n the case of G, ths s not a complete descrpton of G I as we need to add some addtonal (forced) edges and forbdden edges not shown n these dagrams n order to make the reducton work. Ths s rather techncal and we omt ths for brevty. From the constructon, we conclude that, just lke n G, the 6-cycles of G I (Fg. 3a) admt only two possble knds of sandwches, and ths s consstent over dfferent clauses. However, unlke n G, the chordal sandwches of G I no longer correspond to satsfyng assgnments of 3-SATISFIABILITY but rather to satsfyng assgnments of ONE-IN-THREE-3-SAT. Fortunately, the unqueness varant of ths problem s CoNP-complete (see Theorem 2). Now, from G I, we construct a collecton Q I of quartet trees. To do ths, we cannot smply use Theorem 7 as before, snce ths may create partal parttons that do not correspond to quartet trees. Moreover, even f we use [31] to replace these parttons by an equvalent collecton of quartet trees, ths process may not preserve the number of solutons. We need a more careful constructon. We recall that each vertex v of G I belongs to two clques that completely cover ts neghbourhood; we assgn greek letters to these two clques (to dstngush them from vertces), and assocate them wth v. In partcular, we use the followng symbols: α W, β j W, γj 1, γj 2, γj 3, λj, δ, µ where W s a lteral and j {1... m. They defne specfc clques of G I as follows. The letter α W defnes the clque of G I consstng of all heads and 7

8 shoulders of W. The letter β j W corresponds to the clque formed by the shoulder Sj W and the knees Kj W, Lj (f exsts). W Further, λ j s the clque on F j, D j 1, Dj 2, Dj 3, Kj X, Kj Y, Kj Z where C j = X Y Z, whle the clque for γ j p where p {1, 2, 3 s formed by D j p, K j W, Lj U where W and U are the p-th and(p 1)-th (modulo 3) lterals ofc j. Fnally, δ corresponds to the clque contanng B and all heads H W, whle µ corresponds to the clque wth B and all feet F j. From ths, we construct the collecton Q I by consderng every forbdden edge uv of G I and by constructng a partal partton wth two cells n whch one cell s the set of clques assgned to u and the other s the set of clques assgned to v. Snce we assgn to each vertex of G I exactly two clques, ths yelds parttons correspondng to quartet trees. For nstance, n Fg. 3b, we have a forbdden edge K j X Kj X where Kj X s assgned clques βj X, λj, and K j X s assgned β j X, γj 1. Ths yelds a quartet tree {βj X, λj {β j X, γj 1. Fnally, snce by constructon every vertex of G I s ncdent to at least one forbdden edge, we conclude that G I = nt (Q I ). Ths completes the overvew of the constructon. From ths, the proof of Theorem 9 follows, essentally along the same lnes as the unqueness property we dscussed for G. That s, we descrbe the edges that are forced n the sandwch by a satsfyng assgnment for I, treated as an nstance of ONE-IN-THREE-3SAT, and prove that ths yelds a chordal sandwch,.e., a mnmal chordal sandwch. To complete the result, we need to explan how to construct a phylogenetc tree correspondng to a satsfyng assgnment σ for I, namely the tree T σ = (T I, φ σ ), and show that t dsplays and s dstngushed by the trees n Q I, as stated n Theorem 10. As ths s rather techncal, we nstead dscuss a small example here. The example nstance I + conssts of four varables v 1, v 2, v 3, v 4 and three clauses C 1 = v 1 v 2 v 3, C 2 = v 1 v 2 v 4, andc 3 = v 3 v 2 v 4. The unque satsfyng assgnment assgns true to v 1, v 4 and false to v 2, v 3. The correspondng phylogenetc tree T = (T, φ) s shown n Fg. 4. α v2 γ 1 1 β 1 v 1 γ 2 3 β 2 v 4 γ 3 2 β 3 v 2 α v1 β 1 v 2 α v3 α v4 λ 1 λ 2 λ 3 β 1 v 1 β 2 v 2 β 1 v 3 β 2 v 4 γ 1 3 γ 2 2 γ 3 1 β 2 v 1 β 3 v 2 β 3 v 3 β 3 v 4 β 1 v 3 β 2 v 2 β 3 v 3 α v1 α v2 α v3 α v4 γ 1 2 β 1 v 2 γ 2 1 β 2 v 1 γ 3 3 β 3 v 4 δ µ Fgure 4: The phylogenetc tree for the example nstance I +. For nstance, one of the quartet trees n Q I + s π = {α v1, β 1 v 1 {α v1, β 1 v 1 representng the forbdden edge of G I + between S 1 v 1 and S 1 v 1. It s easy to verfyt dsplays π. Another example fromq I + s π = {β 1 v 1, λ 1 {β 1 v 1, γ 1 1 representng the forbdden edge K 1 v 1 K 1 v 1. Agan, t s dsplayed byt, but ths tme one nternal edge of T s contaned n every set of edges of T that dsplays π n T ; hence, ths edge s dstngushed by π. Ths way we can verfy all other quartet trees nq I + and conclude that they are dsplayed byt and they dstngusht. Now, wth the help of Theorem 8, ths allows us to prove that gven an nstance I of ONE-IN-THREE-3SAT and a satsfyng assgnment σ for I, one can n polynomal tme construct a phylogenetc tree T and a collecton of quartet treesqsuch thatt s the unque tree defned byq f and only f σ s the unque satsfyng assgnment for I. Combned wth Theorem 2, ths proves Theorem 1. That concludes ths secton. 4. Formal Constructon Let I be an nstance of ONE-IN-THREE-3SAT consstng of n varables v 1,..., v n and m clausesc 1,...,C m each of whch s a dsjuncton of exactly three lterals. Assume that no varable appears twce n the same clause. 8

9 For each {1... n, we let denote all ndces j such that or appears n the clause C j. In the followng, we defne the set X I, ntroduce notaton for some of ts 2-element subsets, and usng these defne the collectonq I Defnton ofx I The set X I conssts of the followng elements: α v, α v for each {1... n, β j, β j for each {1... n and j, γ j 1, γj 2, γj 3, λj for each j {1... m, δ and µ Selected subsets ofx I { B = µ, δ For each {1,..., n: H v = {α v, δ, H v = {α v, δ, A = {α v, α v, Sv j = {α v, β j, S j = {α v, β j for all j v For each j {1... m where C j = X Y Z: { F j = λ j, µ, { { K {β j = j X X, γj 1, K jy = β j Y, γj 2, K jz = β j Z, γj 3, { { K j X = β j X, λj, K j Y = β j Y, λj, K j Z {β = j Z, λj, { L j X {β = j X, γj 2, L j Y = β j Y, γj 3,, L j Z = { β j Z, γj 1 D j 1 = { γ j 1, λj, D j 2 = { γ j 2, λj, D j 3 = { γ j 3, λj 4.3. Defnton ofq I The collectonq I of quartet trees s defned as the unon of the followng sets: { A B {1...n j {1...m {1...n j,j 1 < n j {D j 1 B, Dj 2 B, Dj 3 B {S j S j j {1...m where C j =X Y Z {H v S j, H v S j, H v S j v, H v S j v {1...n j,j and j<j {1...n j and j<j m {1...n j {1...m { j K X Kj X, Kj Y Kj Y, Kj Z Kj Z, Kj X Lj X, Kj Y Lj Y, Kj Z Lj Z S j Y Kj X, Sj Z Kj Y, Sj X Kj Z, Sj Z Lj X, Sj X Lj Y, Sj Y Lj Z {S j K j, S j K j {K j v F j, K j F j {H v F j, H v F j Note that n each clause C j = X Y Z there s a partcular type of symmetry between the lterals X, Y, and Z. In partcular, f we replace, n the above, the ndces X, Y, Z and 1, 2, 3 as follows: X Y Z X and , we obtan precsely the same defnton of Q I as the above. We shall refer to ths as the rotatonal symmetry between X, Y, Z. Now, we formally defne the tree T I correspondng to the nstance I. For satsfyng assgnments σ, we also defne the labellng φ σ of the leaves of T I by the elements of X I. Ths (as we prove later n Theorem 10) wll consttute a perfect phylogeny, anx I -treet σ = (T I, φ σ ), for the collectonq I. 9

10 A 1 A 2 A n B 1 B 2 B m y 0 y 1 y 2 y n u 1 u 2 u m u 0 Fgure 5: The tree T I Defnton of the tree T I { { { V(T I ) = y 0, y 1, y 1,..., y n, y n a 1, a 1,..., a n, a n c j, zj {1... n and j {u 0, u 1,..., u m {x j 1, xj 2, xj 3, xj 4, xj 5, xj 6, bj 1, bj 2, bj 3, gj 1, gj 2, gj 3,lj j {1... m { { { E(T I ) = y 1 y 1, y 2y 2,..., y ny n a 1 y 1, a 2y 2,... a ny n c j n zj j =1 {y 0 y 1, y 1 y 2, y 2 y 3,..., y n 1 y n {y n u 1, u 1 u 2, u 2 u 3,..., u m 1 u m, u m u 0 { u j x j 1, xj 1 xj 2, xj 2 xj 3, xj 2 xj 4, xj 4 xj 5, xj 4 xj 6, bj 1 xj 6, bj 2 xj 3, bj 3 xj 5, gj 1 xj 6, gj 2 xj 1, gj 3 xj 3,lj x j 5 j {1... m { a zj 1, z j 1 z j 2,..., z j t 1 z j t, z j t y {1... n and j 1 < j 2 <... < j t are elements of 4.5. Defnton of the labellng φ σ Let σ be a satsfyng assgnment for the nstance I. The mappng φ σ : X I V(T I ) s defned as follows: φ σ (δ) = y 0 and φ σ (µ) = u 0, for each {1... n: f = 1, then φ σ (α v ) = a, φ σ (α v ) = a, and φ σ(β j ) = c j for all j, f = 0, then φ σ (α v ) = a, φ σ (α v ) = a, and φ σ(β j ) = c j for all j, for each j {1... m wherec j = X Y Z: f X = 1, then φ σ (β j X ) = bj 1, φ σ(β j Y ) = bj 2, φ σ(β j Z ) = bj 3, φ σ (γ j 1 ) = gj 1, φ σ(γ j 2 ) = gj 2, φ σ(γ j 3 ) = gj 3, φ σ(λ j ) = l j, f Y = 1, then φ σ (β j Y ) = bj 1, φ σ(β j Z ) = bj 2, φ σ(β j X ) = bj 3, φ σ (γ j 2 ) = gj 1, φ σ(γ j 3 ) = gj 2, φ σ(γ j 1 ) = gj 3, φ σ(λ j ) = l j, f Z = 1, then φ σ (β j Z ) = bj 1, φ σ(β j X ) = bj 2, φ σ(β j Y ) = bj 3, φ σ (γ j 3 ) = gj 1, φ σ(γ j 1 ) = gj 2, φ σ(γ j 2 ) = gj 3, φ σ(λ j ) = l j, For llustraton of the constructon of T I and φ σ, see Fg. 5 and 6. 10

11 A a g j 1 b j 1 B j γ j 1 β j X c j 1 c j 2 c j t a. z j 1 z j 2 z j t y y l j b j 3 x j 5 g j 2 x j 2 x j 6 x j 4 x j 1 x j 3 g j 3 b j u j λ j β j Z x j 5 γ j 2 x j 6 x j 4 x j 2 x j 3 x j 1 γ j 3 β j Y u j Fgure 6: a) subtree A for the varable, b) subtree B j for the clause C j, c) labellng of leaves of B j when σ(x) = 1, σ(y) = σ(z) = Perfect Phylogenes and Mnmal Chordal Sandwches In ths secton, we prove Theorem 6. As a partcular consequence of ths theorem, we obtan Theorem 8, whch allows us to cast the problem of unqueness of perfect phylogenes as a mnmal chordal sandwch problem. We need to ntroduce some addtonal tools. The followng s a standard property of mnmal chordal completons. Lemma 11. Let G be a chordal completon of G. Then G s a mnmal chordal completon of G f and only f for all uv E(G )\E(G), the vertces u, v have at least two non-adjacent common neghbours n G. PROOF. Suppose that G s a mnmal chordal completon. Let uv E(G )\E(G), and let G = G uv. Snce G s a mnmal chordal completon and uv E(G), we conclude that G s not chordal. Thus, there exsts a set C V(G ) that nduces a cycle n G. Snce G s chordal, C does not nduce a cycle n G. Ths mples u, v C, and hence, us the unque chord of G [C]. So, we conclude C = 4, because otherwse G [C] contans an nduced cycle. Let x, y be the two vertces of C\{u, v. Clearly, xy E(G ) and both x and y are common neghbours of u and n G, as requred. Conversely, suppose that G s not a mnmal chordal completon. Then by [29], there exsts an edge uv E(G )\E(G) such that G us a chordal graph. If the vertces u, v have non-adjacent common neghbours x, y n G, then{u, x, v, y nduces a 4-cycle n G uv. Ths s mpossble as we assume that G us chordal. That concludes the proof. Usng ths tool, we prove the followng two mportant lemmas. Lemma 12. Let G be a graph and G be a mnmal chordal completon of G. N G (u) N G (v), then also N G (u) N G (v). If G contans vertces u, v wth PROOF. Let u, v be vertces of G wth N G (u) N G (v). Let B = N G (u)\ N G (v) and A = N G (u) N G (v). Assume for contradcton that B =, and let A 1 denote the vertces of A wth at least one neghbour n B. Look at the graph G 1 = G [A 1 B {v]. By the defnton of A 1 and B, the vertex s adjacent to each vertex n A 1 and non-adjacent to each vertex n B. Hence, no vertex n A 1 s a smplcal vertex of G 1, snce t s adjacent to v and at least one vertex n B. Now, consder w B. By the defnton of B, we have that w s adjacent n G to u but not v. Thus, uw s not an edge of G, snce N G (u) N G (v) and N G (v) N G (v). So, by Lemma 11, the vertces u, w have non-adjacent common neghbours x, y n G. Snce x, y are adjacent to u, we have x, y A B. In fact, snce w has no neghbours n A\ A 1, we conclude x, y A 1 B. Thus, w s not a smplcal vertex of G 1. Ths proves that no vertex of G 1, except possbly for v, s smplcal n G 1. Also, G 1 s not a complete graph, snce B =, and v has no neghbour n B. Recall that G 1 s chordal because G s. Thus, by the result of Drac [11], t follows that G 1 must contan at least two non-adjacent smplcal vertces, but ths s clearly mpossble. Hence, we must conclude B =. In other words, N G (u) N G (v) as promsed. 11

12 Lemma 13. Let G be a graph, and let H be a graph obtaned from G by substtutng complete graphs for the vertces of G. Then there s a one-to-one correspondence between mnmal chordal completons of G and H. PROOF. Let v 1,v 2,...,v n be the vertces of G. Snce H s obtaned from G by substtutng complete graphs, there s a partton C 1... C n of V(H) where each C nduces a complete graph n H, and for all dstnct, j {1... n: ( ) each x C, y C j satsfy v j E(G) f and only f xy E(H). We defne the followng mappng Ψ: f G s a graph wth vertex set V(G), then H = Ψ(G ) denotes be the graph constructed from G by consderng each {1... n, substtutng the set C for the vertex, and makng all vertces n C parwse adjacent. Thus, for all dstnct, j {1... n: ( ) each x C, y C j satsfy v j E(G ) f and only f xy E(H ). We prove that Ψ s a bjecton between the mnmal chordal completons of G and H whch wll yeld the lemma. Let G be a mnmal chordal completon of G, and let H = Ψ(G ). Clearly, H s chordal, snce G s chordal, and chordal graphs are closed under the operaton of substtutng a complete graph for a vertex. Also, observe that V(H) = V(H ). If xy E(H) where x, y C for some {1... n, then also xy E(H ), snce C nduces a complete graph n H. If xy E(H) and x C, y C j for dstnct, j {1... n, then v j E(G) by ( ), mplyng v j E(G ), snce E(G) E(G ). Hence, xy E(H ) by ( ). Ths proves that E(H) E(H ), and thus, H s a chordal completon of H. To prove that H s a mnmal chordal completon of H, t suffces, by Lemma 11, to show that for all xy E(H )\ E(H), the vertces x, y have at least two non-adjacent common neghbours n H. Consder xy E(H )\E(H), and let, j {1... n be such that x C and y C j. Snce xy E(H) and C nduces a complete graph n H, we conclude = j. Thus, by ( ), we have v j E(G ), and so, v j E(G )\E(G) by ( ). Now, recall that G s a mnmal chordal completon of G. Thus, by Lemma 11, the vertces, v j have non-adjacent common neghbours v k, v l n G. So, we let w C k and z C l. By ( ), we conclude wz E(H ), snce v k v l E(G ). Moreover, ( ) also mples that z, w are common neghbours of x, y, snce v k, v l are common neghbours of, v j. Ths proves that x, y have non-adjacent common neghbours, and thus shows that H s a mnmal chordal completon of H. Conversely, let H be a mnmal chordal completon of H. Let G be the graph wth V(G ) = V(G) such that v j E(G ) f and only f there exsts x C, y C j wth xy E(H ). Let {1... n and consder vertces x, y C n the graph H. Recall that C nduces a complete graph n H. Ths mples that xy E(H) and both x and y are adjacent n H to every z C \{x, y. Further, by ( ), f z C j where j =, then x, y are both adjacent to z f v j E(G), and x, y are both non-adjacent to z f v j E(G). Ths shows that N H (x) = N H (y), and hence, N H (x) = N H (y) by Lemma 12 and the fact that H s a mnmal chordal completon of H. Ths proves that H = Ψ(G ), and hence, G s chordal. In fact, E(G) E(G ) by ( ) and ( ). Thus G s a chordal completon of G. It remans to show that G s a mnmal chordal completon of G. Agan, t suffces to show that for each v j E(G )\E(G), the vertces, v j have non-adjacent common neghbours n G. Consder v j E(G )\E(G), and let x C and y C j. So, = j and xy E(H ) by ( ). Further, xy E(H )\E(H) by ( ) and the fact that v j E(G). So, the vertces x, y have non-adjacent common neghbours w, z n H by Lemma 12 and the fact that H s a mnmal chordal completon of H. Let k,l {1... n be such that w C k and z C l. Snce xz E(H ) but wx E(H ), we conclude by ( ) that = k. By symmetry, also = l, j = k, and j = l. Further, k = l, snce wx E(H ) and C k nduces a complete graph n H. Thus, ( ) mples that v k, v l are non-adjacent common neghbours of, v j n G, snce w, z are non-adjacent common neghbours of x, y n H. Ths proves that G s ndeed a mnmal chordal completon of G. That concludes the proof. Now, we are fnally ready to prove Theorem 6. PROOF OF THEOREM 6. We observe that the graph nt(q) can be obtaned by substtutng complete graphs for the vertces of nt (Q). Namely, for each vertex A of nt (Q), we substtute A by the complete graph on vertces C A = {(A, π) π Q and A s a cell of π. Thus, by Lemma 13, there s a bjecton Ψ between the mnmal chordal completons of nt(q) and nt (Q). By translatng the condton ( ) from the proof of Lemma 13, we 12

13 conclude that f G s a mnmal chordal completon of nt (Q), then H = Ψ(G ) s the graph whose vertex set s that of nt(q) and n whch for all A, A V(G ): ( ) all meanngful π, π Q satsfy AA V(G ) (A, π)(a, π ) V(H ). We show that Ψ s a bjecton between the mnmal restrcted chordal completons of nt(q) and the mnmal chordal sandwches of(nt (Q), forb(q)). Frst, let H be a mnmal restrcted chordal completon of nt(q). Then G = Ψ 1 (H ) s a mnmal chordal completon of nt (Q). Consder two cells A 1, A 2 of π Q. Snce H s a restrcted chordal completon of nt(q), we have that (A 1, π) s not adjacent to (A 2, π) n H. Thus, A 1 A 2 E(G ) by ( ). Ths shows that G contans no edge from forb(q). Thus G s a mnmal chordal sandwch of (nt (Q), forb(q)), snce t s also a mnmal chordal completon of nt (Q). Conversely, let G be a mnmal chordal sandwch of (nt (Q), forb(q)). Then H = Ψ(G ) s a mnmal chordal completon of nt(q). Consder two cells A 1, A 2 of π Q. Snce A 1 A 2 s an edge of forb(q), and G s a mnmal chordal sandwch of (nt (Q),forb(Q)), we have A 1 A 2 E(G ). Thus, (A 1, π)(a 2, π) E(H ) by ( ). Ths shows that H s a mnmal restrcted chordal completon of nt(q). That concludes the proof. 6. Mnmal Chordal Sandwches and Boolean Satsfablty In ths secton, we prove Theorem 9. We consder an nstance I of ONE-IN-THREE-3SAT, and carefully analyze chordal sandwches of (nt (Q I ), forb(q I )). For a truth assgnment σ for the nstance I, we construct graphs G σ, G σ, and Gσ, startng from nt (Q I ). We show that f σ s a satsfyng assgnment for I, then Gσ s a mnmal chordal sandwch of (nt (Q I ), forb(q I )). Conversely, for every mnmal chordal sandwch G of (nt (Q I ), forb(q I )), we descrbe a satsfyng assgnment σ for I such that G = Gσ. From ths the theorem wll follow. For later, we need the followng smple propertes. The proofs are straghtforward and left to the reader. Lemma 14. Let G be a chordal graph, and let a, b be non-adjacent vertces of G. Then every two common neghbours of a and b are adjacent. Lemma 15. Let G be a chordal graph, and C = {a, b, c, d, e be a 5-cycle n G wth edges ab, bc, cd, de, ae. (a) If bd, ce E(G), then ac, ad E(G), and (b) f bd, be E(G), then ac E(G). Lemma 16. Let G be a chordal graph, and C = {a, b, c, d, e, f be a 6-cycle n G wth edges ab, bc, cd, de, e f, a f. (a) If bd, ce, d f E(G), then ac, ad, ae E(G), (b) f bd, ce, c f E(G), then ac, ad E(G), and (c) f be, b f, ce, c f E(G), then ad E(G). To assst the reader n followng the subsequent arguments, we now lst here the clques of nt (Q I ) accordng to the elements from whch they arse: δ: B, H v1,..., H vn, H v1,..., H vn µ: B, F 1,..., F m For each {1... n where j 1, j 2,..., j k are the elements of : α v : H v, A, S j 1 v, S j 2 v,..., S j t, α v : H v, A, S j 1 v, S j 2 v,..., S j t, For each j {1... m wherec j = X Y Z: λ j : K j X, Kj Y, Kj Z, Dj 1, Dj 2, Dj 3, Fj 13

14 γ j 1 : Kj X, Lj Z, Dj 1 γ j 2 : Kj Y, Lj X, Dj 2 γ j 3 : Kj Z, Lj Y, Dj 3 β j X : Sj X, Kj X β j Y : Sj Y, Kj Y β j Z : Sj Z, Kj Z β j X : Sj X, Kj X, Lj X β j Y : Sj Y, Kj Y, Lj Y β j Z : Sj Z, Kj Z, Lj Z We start wth a useful lemma descrbng an mportant property of nt (Q I ). Lemma 17. Let G be a chordal sandwch of(nt (Q I ), forb(q I )), and let {1... n. Then (a) there exsts W {, such that for all j, the vertex K j W s adjacent to B, and (b) for each j, and each W {,, f K j W s adjacent to B, then the vertces Sj W, Kj W, Lj W adjacent to B, A, H W, H W, F j. (See Fg. 7a) (f exsts) are PROOF. Let {1... n. Frst, we observe the followng. ( ) for each j, each W {,, at least one of S j W, Kj W s adjacent to B. We may assume that S j W s not adjacent to B, otherwse we are done. Observe that Sj W s adjacent to Kj W, snce β j W Kj W Sj W. Moreover, there exsts p {1, 2, 3 such that Kj W Dj p contans λ j or γ j p, mplyng that K j W s adjacent to D j p. Also, F j s adjacent to D j p and B, snce λ j D j p F j and µ B F j, respectvely. Further, H W s adjacent to S j W and B, snce α W H W Sj W and δ H W B. Fnally, H W s not adjacent to Fj, and B s not adjacent to D j p, snce H W F j and D j p B are n Q I. So, by Lemma 16 appled to the cycle {K j W, Sj W, H W, B, Fj, Dp, j we conclude that K j W s adjacent to B. Ths proves ( ). Now, to prove (a), suppose for contradcton that there are j, j such that both K j v and K j are not adjacent to B. Then by ( ), both Sv j and S j v are adjacent to B. Note also that A s adjacent to both Sv j, S j v, snce α v A Sv j and α v A S j v. Further, note that A B and Sv j S j v are not edges of G, snce A B and Sv j S j v are n Q I. But then G contans an nduced 4-cycle on{s j, A, S j, B, whch s mpossble, snce G s chordal. Ths proves (a). For (b), suppose that K j W s adjacent to B for j and W {,. Frst observe that K j W s adjacent to Sj W, and the vertex K j W s adjacent to Sj W, snce βj W Kj W Sj W and βj W Kj W Sj W. Moreover, there exsts p {1, 2, 3 such that K j W Dj p and K j W Dj p contan γ j p and λ j, respectvely, or λ j and γ j p, respectvely. Ths mples that K j W and K j W are adjacent to Dj p. Also, A s adjacent to S j W and Sj W, snce α W A S j W and α W A S j W. Further, A L j Y K j Z S j W S j W a) H W B HW b) K j Y K j Y K j W F j K j W L j X K j X K j Z L j Z D j p K j X Fgure 7: Chordal completon edges for a) W = 1, b) X = 1, Y = 0, Z = 0. 14

15 note that D j pb, A B, K j W Kj W, and Sj W Sj W are not edges of G, snce D j p B, A B, K j W Kj W, and Sj W Sj W are n Q I. Ths mples that K j W s not adjacent to B, snce otherwse G contans an nduced 4-cycle on {K j W, B, Kj W, Dj p. So, by ( ), we have that S j W s adjacent to B. Thus, Lemma 15 appled to {Kj W, Sj W, A, S j W, B yelds that Kj W s adjacent to A and S j W. So, by Lemma 14 appled to {Sj W, Kj W, Dj p, K j W, we have that Sj W s adjacent to Dj p. Now, observe that H W, H W are adjacent to both A and B, snce α W H W A, α W H W A, and δ B H W H W. Thus, by Lemma 14 appled to {u, A, u, B where u {S j W, Kj W and u {H W, H W, we conclude that S j W and Kj W are adjacent to both H W and H W. Smlarly, we observe that F j s adjacent to B and D j p, snce µ F j B and λ j Dp j F j. Thus, Lemma 14 appled to {u, B, F j, Dp j yelds that S j W and Kj W are also adjacent to F j. Lastly, suppose that L j W exsts. Then there s q {1, 2, 3 such that γj q Dq L j j W, mplyng that Lj W s adjacent to Dq. j Moreover, F j s adjacent to Dq j and B, snce λ j Dq j F j and µ F j B. Also, H W s adjacent to B, S j W, and the vertex S j W s adjacent to Lj W, snce δ B H W, α W H W Sj W, and βj W Sj W Lj W. Further, H W Fj and D j qb are not edges of G, snce H W F j and D j q B are n Q I. Also, S j W B s not an edge of G, snce otherwse G contans an nduced 4-cycle on {S j W, B, Sj W, A. Thus, by Lemma 15 appled to {L j W, Sj W, H W, B, Fj, D j q, we conclude that L j W s adjacent to H W, B, and Fj. Moreover, by Lemma 15 appled to {L j W, B, Sj W, A, S j, we W conclude that L j W s adjacent to A. Fnally, recall that H W s adjacent to both A and B. Thus, Lemma 14 appled to {L j W, A, H W, B yelds that L j W s also adjacent to H W. That concludes the proof. Now, let σ be a truth assgnment for the nstance I. Recall that, for smplcty, we wrte X = 0 and X = 1 n place of σ(x) = 0 and σ(x) = 1, respectvely. To facltate the arguments n the subsequent proofs, we ntroduce a namng conventon for the vertces n nt (Q I ) smlar to that of [4], as we already ndcated n 3. The vertces S j W for all meanngful choces of j and W are called shoulders. For a fxed j, we call them shoulders of the clausec j, and for a fxed W, we call them shoulders of the lteral W. A shoulder s a a true shoulder f W = 1. Otherwse, t s a false shoulder. The vertces K j W, Lj W for all meanngful choces of j and W are called knees. For a fxed j, we call them knees of the clause C j, and for a fxed W, we call them knees of the lteral W. A knee s a true knee f W = 1. Otherwse, t s a false knee. The vertces A, D j p, H W, F j for all meanngful choces of ndces are called A-vertces, D-vertces, H-vertces, and F-vertces, respectvely. Based on σ, we defne the followng three graphs: G σ, G σ, and G σ Defnton of G σ The graph G σ s constructed from nt (Q I ) by performng the followng steps: () make B adjacent to all true knees and true shoulders 15