Documents de Travail du Centre d Economie de la Sorbonne

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1 Documents de Traval du Centre d Econome de la Sorbonne Detectng nduced subgraphs Benamn LEVEQUE, Davd Y. LIN, Frédérc MAFFRAY, Ncolas TROTIGNON Mason des Scences Économques, boulevard de L'Hôptal, Pars Cedex 13 ISSN : X

2 Detectng nduced subgraphs Benamn Lévêque, Davd Y. Ln, Frédérc Maffray, Ncolas Trotgnon October 9, 27 Abstract An s-graph s a graph wth two nds of edges: subdvsble edges and real edges. A realsaton of an s-graph B s any graph obtaned by subdvdng subdvsble edges of B nto paths of arbtrary length (at least one). Gven an s-graph B, we study the decson problem Π B whose nstance s a graph G and queston s Does G contan a realsaton of B as an nduced subgraph?. For several B s, the complexty of Π B s nown and here we gve the complexty for several more. Our NP-completeness proofs for Π B s rely on the NP-completeness proof of the followng problem. Let S be a set of graphs and d be an nteger. Let Γ d S be the problem whose nstance s (G, x, y) where G s a graph whose maxmum degree s at most d, wth no nduced subgraph n S and x, y V (G) are two non-adacent vertces of degree 2. The queston s Does G contan an nduced cycle passng through x, y?. Among several results, we prove that Γ 3 s NPcomplete. We gve a smple crteron on a connected graph H to decde whether Γ + {H} s polynomal or NP-complete. On the other hand, the polynomal cases rely on the algorthm three-n-a-tree, due to Chudnovsy and Seymour. AMS Mathematcs Subect Classfcaton: 5C85, 68R1, 68W5, 9C35 Key words: detectng, nduced, subgraphs 1 Introducton In ths paper graphs are smple and fnte. A subdvsble graph (s-graph for short) s a trple B = (V, D, F) such that (V, D F) s a graph and D F =. The edges n D are sad to be real edges of B whle the edges n F are sad to be subdvsble edges of B. A realsaton of B s a graph obtaned from B by subdvsng edges of F nto paths of arbtrary length (at least one). The problem Laboratore G-SCOP, 46 Avenue Félx Vallet, 3831 Grenoble Cedex, France. (benamn.leveque@g-scop.npg.fr, Frederc.Maffray@g-scop.npg.fr) Prnceton Unversty, Prnceton, NJ, dyln@prnceton.edu Unversté Pars I, Centre d Économe de la Sorbonne, boulevard de l Hôptal, Pars cedex 13, France. ncolas.trotgnon@unv-pars1.fr Ths wor has been partally supported by ADONET networ, a Mare Cure tranng networ of the European Communty. 1

3 T H Fgure 1: s-graphs yeldng trvally polynomal problems B 1 B 2 B 3 Fgure 2: Pyramds, prsms and thetas Π B s the decson problem whose nput s a graph G and whose queston s Does G contan a realsaton of B as an nduced subgraph?. On fgures, we depct real edges of an s-graph wth straght lnes, and subdvsble edges wth dashed lnes. Several nterestng nstance of Π B are studed n the ltterature. For some of them, the exstence of a polynomal tme algorthm s trval, but efforts are devoted toward optmzed algorthms. For example, Alon, Yuster and Zwc [2] solve Π T n tme O(m 1.41 ) (nstead of the obvous O(n 3 ) algorthm), where T s the s-graph depcted on Fgure 1. Ths problem s nown as trangle detecton. Taran and Yannaas [1] solve Π H n tme O(n + m) where H s the s-graph depcted on Fgure 1. For some Π B s, the exstence of a polynomal tme algorthm s non-trval. A pyramd (resp. prsm, theta) s any realsaton of the s-graph B 1 (resp. B 2, B 3 ) depcted on fgure 2. Chudnovsy and Seymour [5] gave an O(n 9 )-tme algorthm for Π B1 (or equvalently, for detectng a pyramd). As far as we now, that s the frst example of a soluton to a Π B whose complexty s non-trval to settle. In contrast, Maffray and Trotgnon [8] proved that Π B2 (or detectng a prsm) s NP-complete. Chudnovsy and Seymour [4] gave an O(n 11 )-tme algorthm for P B3 (or detectng a theta). Ther algorthm reles on the soluton of a problem called three-n-a-tree, that we wll defne precsely and use n Secton 2. The three-n-tree algorthm s qute general snce t can be used to solve a lot of Π B problems, ncludng the detecton of pyramds. These facts are a motvaton for a systematc study of Π B. A further motvaton s that very smlar s-graphs can lead to a drastcally dfferent complexty. The followng example may be more strng than pyramd/prsm/theta : Π B4, Π B6 are polynomal and Π B5, Π B7 are NP-complete, where B 4,...,B 7 are the s-graphs depcted on fgure 3. Ths wll be proved n secton 3.1. Notaton and remars By C ( 3) we denote the cycle on vertces, by K l (l 1) the clque on l vertces. We denote by I l (l 1) the tree on l + 5 vertces obtaned by 2

4 B 4 B 5 B 6 B 7 Fgure 3: Some s-graphs wth pendng edges Fgure 4: I 1 tang a path of length l wth ends a, b, and addng four vertces, two of them adacent to a, the other two to b; see Fgure 4. When a graph G contans a graph somorphc to H as an nduced subgraph, we wll often say G contan an H. Let (V, D, F) be an s-graph. Suppose that (V, D F) has a vertex of degree one ncdent to an edge e. Then Π (V,D {e},f \{e}) and Π (V,D\{e},F {e}) have the same complexty, because a graph G contans a realsaton of (V, D {e}, F \ {e}) f and only f t contans a realsaton of (V, D \ {e}, F {e}). For the same reason, f (V, D F) has a vertex of degree two ncdent to the edges e f then Π (V,D\{e} {f},f \{f} {e}), Π (V,D\{f} {e},f \{e} {f}) and Π (V,D\{e,f},F {e,f}) have the same complexty. If F 1 then Π (V,D,F) s clearly polynomal. Thus n the rest of the paper, we wll consder only s-graphs (V, D, F) such that: F 2; no vertex of degree one s ncdent to an edge of F; every nduced path of (V, D F) wth all nteror vertces of degree 2 and whose ends have degree 2 has at most one edge n F. Moreover, ths edge s ncdent to an end of the path; every nduced cycle wth at most one vertex v of degree at least 3 n (V, D F) has at most one edge n F and ths edge s ncdent to v f v exsts (f t does not then the cycle s a component of (V, D F)). 2 Detecton of holes wth prescrbed vertces Let (G) be the maxmum degree of G. Let S be a set of graphs and d be an nteger. Let Γ d S be the problem whose nstance s (G, x, y) where G s a graph such that (G) d, wth no nduced subgraph n S and x, y V (G) are two non-adacent vertces of degree 2. The queston s Does G contan a hole passng through x, y?. For smplcty, we wrte Γ S nstead of Γ + S (so, the graph n the nstance of Γ S has unbounded degree). Also we wrte Γ d nstead 3

5 of Γ d (so the graph n the nstance of Γd has no restrcton on ts nduced subgraphs). Benstoc [3] proved that Γ = Γ s NP-complete. For S = {K 3 } and S = {K 1,4 }, Γ S can be shown to be NP-complete, and a consequence s the NP-completeness of several problems of nterest: see [8] and [9]. In ths secton, we try to settle Γ d S for as many S s and d s as we can. In partcular, we gve the complexty of Γ S when S contans only one connected graph and of Γ d for all d. We also settle Γ d S for some cases when S s a set of cycles. The polynomal cases are ether trval, or are a drect consequence of an algorthm of Chudnovsy and Seymour. The NP-complete cases follow from several extensons of Benstoc s constructon. Polynomal cases Chudnovsy and Seymour [4] proved that the problem whose nstance s a graph and three vertces a, b, c, whose queston s Does the graph contans a tree passng through a, b, c as an nduced subgraph? can be solved n O(n 4 ). We call ths algorthm three-n-a-tree. Three-n-a-tree can be used drectly to solve Γ S for several S s. Let us call subdvded claw any tree wth one vertex u of degree 3, three vertces v 1, v 2, v 3 of degree 1 and all the other vertces of degree 2. Theorem 2.1 Let H be a graph on vertces that s ether a path or a subdvded claw. There s an O(n )-tme algorthm for Γ {H}. proof Here s an algorthm for Γ {H}. Let (G, x, y) be an nstance of Γ H. If H s a path on vertces then every hole n G s on at most vertces. Hence, by a brute-force search on every -tuple, we wll fnd a hole through x, y f there s any. Now we suppose that H s a subdvded claw. So 4. For convenence, we put x 1 = x, y 1 = y. Let x, x 2 (resp. y, y 2 ) be the two neghbors of x 1 (resp. y 1 ). Frst chec whether there s n G a hole C through x 1, y 1 such that the dstance between x 1 and y 1 n C s at most 2. If = 4 or = 5 then the vertex-set of any such hole must be ncluded n {x, x 1, x 2, y, y 1, y 2 }, so t can be found n constant tme. Now suppose 6. For every l-tuple (x 3,...,x l+2 ) of vertces of G, wth l 5, test whether P = x x 1 x l+2 y 2 y 1 y s an nduced path, and f so delete the nteror vertces of P and ther neghbors except x, y, and loo for a shortest path from x to y. Ths wll fnd the desred hole f there s one, after possbly swappng x, x 2 and dong the wor agan. Ths taes tme O(n 3 ). Now we assume that n every hole through x 1, y 1, the dstance between x 1, y 1 s at least 1. Let be the length of the unque path of H from u to v, = 1, 2, 3. Note that = Let us chec every ( 4)-tuple z = (x 3,...,x 1+1, y 3,..., y 2+ 3 ) of vertces of G. For such a ( 4)-tuple, test whether x x 1 x 1+1 and P = y y 1 y 2+ 3 are nduced paths of G wth no edge between them except possbly x 1+1y If not, go to the next ( 4)-tuple, but f yes, delete the nteror vertces of P and ther neghbors except y, y Also delete the neghbors of x 2,...,x 1, except x 1, x 2,..., x 1, x 1+1. Call G z the resultng graph and run three-n-a-tree n 4

6 G z for the vertces x 1, y 2+ 3, y. We clam that the answer to three-n-a-tree s YES for some ( 4)-tuple f and only f G contans a hole through x 1, y 1 (after possbly swappng x, x 2 and dong the wor agan). If G contans a hole C through x 1, y 1 then up to a symmetry ths hole vsts x, x 1, x 2, y 2, y 1, y n ths order. Let us name x 3,...,x 1+1 the vertces of C that follow after x 1, x 2, and let us name y 3,..., y 2+ 3 those that follow after y 1, y 2. Note that all these vertces exst and are parwse dstnct snce n every hole through x 1, y 1 the dstance between x 1, y 1 s at least 1. So the path from y to y 2+ 3 n C \ y 1 s a tree of G z passng through x 1, y 2+ 3, y, where z s the ( 4)-tuple (x 3,..., x 1+1, y 3,..., y 2+ 3 ). Conversly, suppose that G z contans a tree T passng through x 1, y 2+ 3, y, for some ( 4)-tuple z. We suppose that T s vertex-ncluson-wse mnmal. If T s a path vstng y, x 1, y 2+ 3 n ths order, then we obtan the desred hole of G by addng y 1, y 2,..., y to T. If T s a path vstng x 1, y, y 2+ 3 n ths order, then we denote by y the neghbor of y 2+ 3 along T. Note that T contans ether x or x 2. If T contans x, then there are three paths n G: y T x x 1 x 1, y T y y 3+2 and y y 1 y 3. These three paths form a subdvded claw centered at y that s long enough to contan an nduced subgraph somorphc to H, a contradcton. If T contans x 2 then the proof wors smlarly wth y T x 1+1 x 1 x 1 nstead of y T x x 1 x 1. If T s a path vstng x 1, y 2+ 3, y n ths order, the proof s smlar, except that we fnd a subdvded claw centered at y If T s not a path, then t s a subdvded claw centered at a vertex u of G. We obtan agan an nduced subgraph of G somorphc to H by addng to T suffcently many vertces of {x,...x 1+1, y,..., y 2+ 3 }. NP-complete cases (unbounded degree) Many NP-completeness results can be proved by adaptng Benstoc s constructon. We gve here several polynomal reductons from the problem 3- Satsfablty of Boolean functons. These results are gven n a framewor that nvolves a few parameters, so that our result can possbly be used for other problems of the same type. Recall that a Boolean functon wth n varables s a mappng f from {, 1} n to {, 1}. A Boolean vector ξ {, 1} n s a truth assgnment for f f f(ξ) = 1. For any Boolean varable z on {, 1}, we wrte z := 1 z, and each of z, z s called a lteral. An nstance of 3-Satsfablty s a Boolean functon f gven as a product of clauses, each clause beng the Boolean sum of three lterals; the queston s whether f admts a truth assgnment. The NP-completeness of 3-Satsfablty s a fundamental result n complexty theory, see [6]. Let f be an nstance of 3-Satsfablty, consstng of m clauses C 1,..., C m on n varables z 1,..., z n. For every nteger 3 and paramaters α {1, 2}, β {, 1}, γ {, 1}, δ {, 1, 2, 3}, ε {, 1}, ζ {, 1} such that f α = 2 then ε = β = γ, let us buld a graph G f (, α, β, γ, δ, ε, ζ) wth two specalzed vertces x, y of degree 2. There wll be a hole contanng x and y n G f (, α, β, γ, δ, ε, ζ) f and only f there exsts a truth assgnment for f. In G f (, α, β, γ, δ, ε, ζ) (we wll sometmes wrte G f for short), there wll be two 5

7 nds of edges: blue and red. The reason for ths dstncton wll appear later. For each varable z ( = 1,...,n), prepare a graph G(z ) wth 4 vertces a,r, b,r, a,r, b,r, r {1,..., } and 4(m + 2)2 vertces t,2p+r, f,2p+r, t,2p+r, f,2p+r, p {,..., m + 1}, r {,..., 2 1}. Add blue edges so that the four sets {a,1,... a,, t,,..., t,2(m+2) 1, b,1,...,b, }, {a,1,...a,, f,,...,f,2(m+2) 1, b,1,...,b, }, {a,1,... a,, t,,..., t,2(m+2) 1, b,1,...,b, }, {a,1,...a,, f,,..., f,2(m+2) 1, b,1,..., b, } all nduce paths (and the vertces appear n ths order along these paths). See Fgure 5. Add red edges accordng to the value of α, β, γ, as follows. If α = 1 then, for every p = 1,..., m + 1, add all edges between {t,2p, t,2p+β } and {f,2p, f,2p+γ }, between {f,2p, f,2p+γ } and {t,2p, t,2p+β }, between {t,2p, t,2p+β } and {f,2p, f,2p+γ }, between {f,2p, f,2p+γ } and {t,2p, t,2p+β }. If α = 2 then, for every p = 1,..., m, add all edges between {t,2p+ 1, t,2p+ 1+β } and {f,2p+ 1, f,2p+ 1+γ } ; for every p = 1,...,m + 1, add all edges between {f,2p+ 1, f,2p+ 1+γ } and {t,2p, t,2p+β }, between {t,2p, t,2p+β } and {f,2p, f,2p+γ }, between {f,2p, f,2p+γ } and {t,2(p 1)+ 1, t,2(p 1)+ 1+β }. See Fgures 6, 7. For each clause C ( = 1,...,m), wth C = y 1 y2 y3, where each y q (q = 1, 2, 3) s a lteral from {z 1,..., z n, z 1,..., z n }, prepare a graph G(C ) wth 2 vertces c,p, d,p, p {1,..., } and 6 vertces u q,p, q {1, 2, 3}, p {1,...,2}. Add blue edges so that the three sets {c,1,..., c,, u q,1,..., uq,2, d,1,...d, }, q {1, 2, 3} all nduce paths (and the vertces appear n ths order along these paths). Add red edges accordng to the value of δ. If δ =, add no edge. If δ = 1, add u 1,1 u2,1, u1,2 u2,2. If δ = 2, add u 1,1 u2,1, u1,2 u2,2, u1,1 u3,1, u1,2 u3,2. If δ = 3, add u1,1 u2,1, u 1,2 u2,2, u1,1 u3,1, u1,2 u3,2, u2,1 u3,1, u2,2 u3,2. See Fgure 8. The graph G f (, α, β, γ, δ, ε, ζ) s obtaned from the dsont unon of the G(z ) s and the G(C ) s as follows. For = 1,...,n 1, add blue edges b, a +1,1 and b, a +1,1. Add a blue edge b n, c 1,1. For = 1,...,m 1, add a blue edge d, c +1,1. Introduce the two specal vertces x, y and add blue edges xa 1,1, xa 1,1 and yd m,, yb n,. See Fgure 9. Add red edges accordng to f, ε, ζ. For q = 1, 2, 3, f y q = z, then add all possble edges between {f,2+ 1, f,2+ 1+ε } and {u q,, uq,+ζ } and between {f,2+ 1, f,2+ 1+ε } and {uq,, uq,+ζ }; whle f yq = z then add all possble edges between {t,2+ 1, t,2+ 1+ε } and {u q,, uq,+ζ } and between {t,2+ 1, t,2+ 1+ε } and {uq,, uq,+ζ }. See Fgure 1. Clearly the sze of G f (, α, β, γ, δ, ε, ζ) s polynomal (actually quadratc) n the sze n + m of f, and x, y are non-adacent and both have degree two. Lemma 2.2 f admts a truth assgnment f and only f G f (, α, β, γ, δ, ε, ζ) contans a hole passng through x, y. proof Recall that f α = 2 then ε = β = γ. We wll prove the lemma for β =, γ =, ε =, ζ = because the proof s essentally the same for the other possble values. 6

8 p 1 2 m 1 m m + 1 r t,2p+r a,1 a, b, f,2p+r b,1 t,2p+r a,1 b, f,2p+r a, b,1 Fgure 5: The graph G(z ) (only blue edges are depcted) p 1 2 m 1 m m + 1 r t,2p+r a,1 a, b, f,2p+r b,1 t,2p+r a,1 b, f,2p+r a, b,1 Fgure 6: The graph G(z ) when α = 1, β =, γ = p 1 2 m 1 m m + 1 r t,2p+r a,1 a, b, f,2p+r b,1 t,2p+r a,1 b, f,2p+r a, b,1 Fgure 7: The graph G(z ) when α = 2, β =, γ = 7

9 1p 2 1 u 1,p u 2,p c,1 c, d,1 d, u 3,p Fgure 8: The graph G(c ) when δ = 3 x y Fgure 9: The whole graph G f Fgure 1: Red edges between G(z ) and G(c ) when ε = ζ = 8

10 Suppose that f admts a truth assgnment ξ {, 1} n. We can buld a hole n G by selectng vertces as follows. Select x, y. For = 1,..., n, select a,p, b,p, a,p, b,p for all p {1,...,}. For = 1,...,n, select c,p, d,p for all p {1,...,}. If ξ = 1 select t,p, t,p for all p {,..., 2(m + 2) 1}. If ξ = select f,p, f,p for all p {,...,2(m + 2) 1}. For = 1,...,m, snce ξ s a truth assgnment for f, at least one of the three lterals of C s equal to 1, say y q = 1 for some q {1, 2, 3}. Then select uq,p for all p {1,..., 2}. Now t s a routne matter to chec that the selected vertces nduce a cycle Z that contans x, y, and that Z s chordless, so t s a hole. The man pont s that there s no chord n Z between some subgraph G(C ) and some subgraph G(z ), for that would be ether an edge t,p u q,r wth yq = z and ξ = 1, or, symmetrcally, an edge f,p u q,r wth yq = z and ξ =, and n ether case ths would contradct the way the vertces of Z were selected. Conversely, suppose that G f (, α, β, γ, δ, ε, ζ) admts a hole Z that contans x, y. (1) For = 1,..., n, Z contans at least 4 +4(m+2) vertces of G(z ): 4 of these are a,p, a,p, b,p, b,p where p {1,..., }, and the others are ether the t,p, t,p s or the f,p, f,p s where p {,...,2(m + 2) 1}. Let us frst deal wth the case = 1. Snce x Z has degree 2, Z contans a 1,1,...,a 1, and a 1,1,..., a 1,. Hence exactly one of t 1,, f 1, s n Z. Lewse exactly one of t 1,, f 1, s n Z. If t 1,, f 1, are both n Z then there s a contradcton: ndeed, f α = 1 then, t 1,,..., t 1,2 and f 1,,...,f 1,2 must all be n Z, and snce t 1,2 sees f 1,2, Z cannot go through y; and f α = 2 the proof s smlar. Smlarly, t 1,, f 1, cannot both be n Z. So, there exsts a largest nteger p 2(m + 2) 1 such that ether t 1,,..., t 1,p and t 1,,...,t 1,p are all n Z or f 1,,..., f 1,p and f 1,,..., f 1,p are all n Z. We clam that p = 2(m+2) 1. For otherwse, some vertex w n {t 1,p, t 1,p, f 1,p, f 1,p } s ncdent to a red edge e of Z. If α = 1 then, up to a symmetry, we assume that t 1,,...,t 1,p and t 1,,..., t 1,p are all n Z. Let w be the vertex of e that s not w. Then w (whch s ether an f 1,, an f 1, or a u, ) s a neghbor of both t 1,p, t 1,p. Hence, Z cannot go through y, a contradcton. Ths proves our clam when α = 1. If α = 2, we dstngush between the followng sx cases. Case 1: p = 1. Then e = t 1, 1 f 1,2. Clearly t 1,,..., t 1, 1 must all be n Z. If t 1,,...,t 1,2 are n Z, there s a contradcton because of t 1,2 f 1,2, and f f 1,,...,f 1,2 are n Z, there s a contradcton because of e. Case 2: p = 2l where 1 l m + 1 and w = t 1,2l. Then e IS t 1,2l f 1,2l+ 1 or t 1,2l f 1,2l. In ether case, t 1,2l,..., t 1,2l+ 1 are all n Z, and there s a contradcton because of the red edge f 1,2l+ 1 t 1,2l+ 1 or t 1,2(l 1)+ 1 f 1,2l, or when l = m + 1 because of b 1,1. Case 3: p = 2l where 1 l m + 1 and w = f 1,2l. Then e s f 1,2l t 1,2(l 1)+ 1 or t 1,2l f 1,2l. In ether case, f 1,2l,..., f 1,2l+ 1 are all n Z, and there s a contradcton because of the red edge t 1,2(l 1)+ 1 f 1,2(l 1)+ 1 or t 1,2l f 1,2l+ 1, or when l = 1 because of a 1,. Case 4: p = 2l + 1 where 1 l m and w = t 1,2l+ 1. Then e s t 1,2l+ 1 f 1,2l+ 1, t 1,2l+ 1 f 1,2(l+1), or t 1,2l+ 1u q, for some, q. In the last case, there s a contradcton snce t 1,2l+ 1 Z also sees uq,. For the same reason, t 1,2l+ 1 uq, s not an edge of Z and t 1,2l+ 1,..., t 1,2(l+1) are 9

11 all n Z. So there s a contradcton because of the red edge t 1,2l f 1,2l+ 1 or t 1,2(l+1) f 1,2(l+1). Case 5: p = 2l + 1 where 2 l m and w = f 1,2l+ 1. Then e s f 1,2l+ 1 t 1,2l+ 1, f 1,2l+ 1 t 1,2l, or f 1,2l+ 1u q, for some, q. In the last case, there s a contradcton snce f 1,2l+ 1 Z also sees uq,. For the same reason, f 1,2l+ 1 uq, s not an edge of Z and f 1,2l+ 1,..., f 1,2(l+1) are all n Z. So there s a contradcton because of the red edge t 1,2l f 1,2l or t 1,2l+ 1 f 1,2(l+1). Case 6: p = 2(m + 1) + 1 and w = f 1,2(m+1)+ 1. Then there s a contradcton because of the red edge t 1,2(m+1) f 1,2(m+1). Ths proves our clam. Snce p = 2(m + 2) 1, b 1,1 s n Z. We clam that b 1,2 s n Z. For otherwse, the two neghbors of b 1,1 n Z are t 1,2(m+2) 1 and f 1,2(m+2) 1. Ths s a contradcton because of the red edges t 1,2m+ 1 f 1,2(m+1), t 1,2(m+1) f 1,2(m+1)+ 1 (f α = 2) or t 1,2(m+1) f 1,2(m+1), t 1,2(m+1) f 1,2(m+1) (f α = 1). Smlarly, b 1,1, b 1,2 are n Z. So b 1,1,..., b 1, and b 1,1,...,b 1, are all n Z. Ths proves (1) for = 1. The proof for = 2 s essentally the same as for = 1, and by nducton the clam holds up to = n. Ths proves (1). (2) For = 1,...,m, Z contans c,1,..., c,, d,1,..., d, and exactly one of {u 1,1,..., u1,2 }, {u2,1,...,u2,2 }, {u3,1,...,u3,2 }. Let us frst deal wth the case = 1. By (1), b n, s n Z and so c 1,1,..., c 1, are all n Z. Consequently exactly one of u 1 1,1, u 2 1,1, u 3 1,1 s n Z, say u 1 1,1 up to a symmetry. Note that the neghbour of u 1 1 n Z \ c 1 cannot be a vertex among u 2 1,1, u3 1,1 for ths would mply that Z contans a trangle. Hence u1 1,2,...,u1 1, are all n Z. The neghbour of u 1 1, n Z \ u1 1, 1 cannot be n some G(z ) (1 n). Else, up to a symmetry we assume that ths neghbor s t 1,p, p {,...,2(m + 2) 1}. If t 1,p Z, there s a contradcton because then t 1,p s also n Z by (1) and t 1,p would be a thrd neghbour of u 1 1, n Z. If t 1,p / Z, there s a contradcton because then the neghbor of t 1,p n Z \ u 1 1, must be t 1,p+1 (or symmetrcally t 1,p 1 ) for otherwse Z contans a trangle. So, t 1,p+1, t 1,p+2,... must be n Z, tll reachng a vertex havng a neghbor f 1,p or f 1,p n Z (whatever α). Thus the neghbour of u1 1, n Z \ u1 1, 1 s u1 1,+1. Smlarly, we prove that u 1,+2,...,u 1,2 are n Z, that d 1,1,...,d 1, are n Z, and so the clam holds for = 1. The proof of the clam for = 2 s essentally the same as for = 1, and by nducton the clam holds up to = m. Ths proves (2). Together wth x, y, the vertces of Z found n (1) and (2) actually nduce a cycle. So, snce Z s a hole, they are the members of Z and we can replace at least by exactly n (1). We can now mae a Boolean vector ξ as follows. For = 1,...,n, f Z contans t,, t, set ξ = 1; f Z contans f,, f, set ξ =. By (1) ths s consstent. Consder any clause C (1 m). By (2) and up to symmetry we may assume that u 1, s n Z. If y1 = z for some {1,.., n}, then the constructon of G mples that f,2+ 1, f,2+ 1 are not n Z, so t,2+ 1, t,2+ 1 are n Z, so ξ = 1, so clause C s satsfed by x. If y 1 = z for some {1,...,n}, then the constructon of G f mples that 1

12 t,2+ 1, t,2+ 1 are not n Z, so f,2+ 1, f,2+ 1 are n Z, so ξ =, so clause C s satsfed by z. Thus ξ s a truth assgnment for f. Theorem 2.3 Let 5 be an nteger. Then Γ {C3,...,C,K 1,6} and Γ {I1,...,I,C 5,...,C,K 1,4} are NP-complete. proof It s a routne matter to chec that the graph G f (, 2,,,,, ) contans no C l (3 l ) and no K 1,6 (n fact t has no vertex of degree at least 6). So Lemma 2.2 mples that Γ {C3,...,C,K 1,6} s NP-complete. It s a routne matter to chec that the graph G f (, 1, 1, 1, 3, 1, 1) contans no K 1,4, no C l (5 l ) and no I l (3 l ). So Lemma 2.2 mples that Γ {K1,4,C 5,...,C,I 3,...,I } s NP-complete. Complexty of Γ {H} when H s a connected graph Theorem 2.4 Let H be a connected graph. Then ether : H s a path or a subdvded claw and Γ {H} s polynomal. H contans one of K 1,4, I for some 1, or C l for some l 3 as an nduced subgraph and Γ {H} s NP-complete. proof If H contans one of K 1,4, I for some 1, or C l for some l 3 as an nduced subgraph then Γ {H} s NP-complete by Theorem 2.3. Else, H s a tree snce t contans no C l, l 3. If H has no vertex of degree at least 3, then H s a path and Γ {H} s polynomal by Theorem 2.1. If H has a sngle vertex of degree at least 3, then ths vertex has degree 3 because H contans no K 1,4. So, H s a subdvded claw and Γ {H} s polynomal by Theorem 2.1. If H has at least two vertces of degree at least 3 then H contans an I l, where l s the length of the unque path of H onnng these two vertces. Ths s a contradcton. Interestngly, a smlar theorem was proved by Aleseev: Theorem 2.5 (Aleseev, [1]) Let H be a connected graph that s not a path nor a subdvded claw. Then the problem of fndng a maxmum stable set n H-free graphs s NP-hard. But the complexty of the maxmum stable set problem s not nown n general for H-free graphs when H s a path or a subdvded claw. See [7] for a survey. NP-complete cases (bounded degree) Here, we wll show that Γ d s NP-complete when d 3 and polynomal when d = 2. If S s any fnte lst of cycles C 1, C 2,..., C m, then we wll also show that Γ 3 S s NP-complete as long as C 6 / S. Let f be an nstance of 3-Satsfablty, consstng of m clauses C 1,..., C m on n varables z 1,..., z n. For each clause C ( = 1,...,m), wth C = y 3 2 y 3 1 y 3, then y ( = 1,...,3m) s a lteral from {z 1,..., z n, z 1,..., z n }. 11

13 α 1+ α 2+ α 3+ α 4+ α α 1 α α 2 α 3 α 4 β β 1+ β 2+ β β 3+ β 4+ β 1 β 2 β 3 β 4 Fgure 11: The graph G(y ) Let us buld a graph G f wth two specalzed vertces x and y of degree 2 such that (G f ) = 3. There wll be a hole contanng x and y n G f f and only f there exsts a truth assgnment for f. For each lteral y ( = 1,...,3m), prepare a graph G(y ) on 2 vertces α, α, α 1+,..., α 4+, α 1,..., α 4, β, β, β 1+,..., β 4+, β 1,..., β 4. (We drop the subscrpt n the labels of the vertces for clarty). For = 1, 2, 3 add the edges α + α (+1)+, β + β (+1)+, α α (+1), β β (+1). Also add the edges α 1+ β 1, α 1 β 1+, α 4+ β 4, α 4 β 4+, αα 1+, αα 1, α 4+ α, α 4 α, ββ 1+, ββ 1, β 4+ β, β 4 β. For each clause C ( = 1,...,m), prepare a graph G(C ) wth 1 vertces c 1+, c 2+, c 3+, c 1, c 2, c 3, c +, c 12+, c, c 12. (We drop the subscrpt n the labels of the vertces for clarty). Add the edges c 12+ c 1+, c 12+ c 2+, c 12 c 1, c 12 c 2, c + c 12+, c + c 3+, c c 12, c c 3. c 1+ c 1 c 12+ c 12 c + c 2+ c 2 c c 3+ c 3 Fgure 12: The graph G(C ) 12

14 For each varable z ( = 1,...,n), prepare a graph G(z ) wth 2z + 2z + vertces, where z s the number of tmes z appears n clauses C 1,..., C m and z + s the number of tmes z appears n clauses C 1,..., C m. Let G(z ) consst of two nternally dsont paths P + and P wth common endponts d + and d and lengths 1 + 2z and 1 + 2z + respectvely. Label the vertces of P + as d +, p+,1,..., p+,2f, d and label the vertces of P as d +, p,1,..., p,2g, d. The fnal graph G f (see fgure 2) wll be constructed from the dsont unon of all the graphs G(y ), G(C ), and G(x ) wth the followng modfcatons: For = 1,...,3m 1, add the edges α α +1 and β β +1. Ths creates one connected chan of the graphs G(y ). For = 1,...,m 1, add the edge c c Ths creates one connected chan of the graphs G(C ). For = 1,..., n 1, add the edge d d+ +1. Ths creates one connected chan of the graphs G(x ). For = 1,..., n, let y n1,..., y nz be the occurrences of x over all lterals. For = 1,...,z, delete the edge p+,2 1 p+,2 and add the four edges p +,2 1 α2+ n, p +,2 1 β2+ n, p +,2 α3+ n, p +,2 β3+ n. For = 1,...,n, let y n1,..., y nz + be the occurrences of x over all lterals. For = 1, 2,...,z +, delete the edge p,2 1 p,2 and add the four edges p,2 1 α2+ n, p,2 1 β2+ n, p,2 α3+ n, p,2 β3+ n. For = 1,...,m and = 1, 2, 3, add the edges α 2 β 2 3( 1)+ c+, β 3 3( 1)+ c. Add the edges α 3md + 1 and β 3mc + 1 Add the vertex x and add the edges xα 1 and xβ 1. Add the vertex y and add the edges yc m and yd n. 3( 1)+ c+, α 3 3( 1)+ c, It s easy to verfy that (G f ) = 3, that G f s polynomal (actually lnear) n the sze n + m of f, and that x, y are non-adacent and both have degree two. Lemma 2.6 f admts a truth assgnment f and only f G f contans a hole passng through x and y. proof Frst assume that f admts a truth assgnment ξ {, 1} n. We wll pc a set of vertces that nduce a hole contanng x and y. P + d + P d Fgure 13: The graph G(z ) 13

15 x y Fgure 14: The fnal graph G f 1. Pc vertces x and y. 2. For = 1,...,3m, pc the vertces α, α, β, β. 3. For = 1,...,3m, f y s satsfed by ξ, then pc the vertces α 1+, α 2+, α 3+, α 4+, β 1+, β 2+, β 3+, and β 4+. Otherwse, pc the vertces α 1, α 2, α 3, α 4, β 1, β 2, β 3, and β For = 1,...,n, f ξ = 1, then pc all the vertces of the path P + and all the neghbors of the vertces n P + of the form α 2+ or α 3+ for any. 5. For = 1,..., n, f ξ =, then pc all the vertces of the path P and all the neghbors of the vertces n P of the form α 2+ or α 3+ for any. 6. For = 1,...,m, pc the vertces c + and c. Choose any {3 2, 3 1, 3} such that ξ satsfes y. Pc vertces α 2, and α 3. If = 3 2, then pc the vertces c 12+, c 1+, c 1, c 12. If = 3 1, then pc the vertces c 12+, c 2+, c 2, c 12. If = 3, then pc the vertces c 3+ and c 3. It suffces to show that the chosen vertces nduce a hole contanng x and y. The only potental problem s that for some, one of the vertces α 2+, α3+, α 2, or α3 was chosen multple tmes. If α 2+ and α 3+ were pced n Step 3, then y s satsfed by ξ. Therefore, α 2+ and α 3+ were not chosen n Step 4 or Step 5. Smlarly, f α 2 and α 3 were pced n Step 6, then y s satsfed by ξ and α 2 and α 3 were not pced n Step 3. Thus, the chosen vertces nduce a hole n G contanng vertces x and y. Now assume G f contans a hole H passng through x and y. The hole H must contan α 1 and β 1 snce they are the only two neghbors of x. Next, ether both α 1+ 1 and β1 1+ are n H, or both α 1 1 and β1 1 are n H. Wthout loss of generalty, let α 1+ 1 and β1 1+ be n H (the same reasonng that follows wll hold true for the other case). Snce β1 1 and α 1 1 are both neghbors of two members n H, they cannot be n H. Thus, α 2+ 1 and β1 2+ must be n H. Snce α 2+ 1 and β1 2+ have the same neghbor outsde G(y 1 ), t follows that H must contan α 3+ 1 and β1 3+. Also, H must contan α4+ 1 and β1 4+. Suppose that α 4 1 and β1 4 are n H. Because α 1 has the same neghbor as β1 outsde G(y 1 ) for = 2, 3, t follows that H must contan α 3 1, α2 1, and α 1 1. But then H s not a hole contanng b, a contradcton. Therefore, α4 1 and β1 4 cannot both be n H, so H must contan α 1, β 1, α 2, and β 2. By nducton, we see for = 1, 2,...,3m that H must contan α, α, β, β. Also, for each, ether H contans α 1+, α 2+, α 3+, α 4+, β 1+, β 2+, β 3+, β 4+ or H contans α 1, α 2, α 3, α 4, β 1, β 2, β 3, β 4. As a result, H must also contan d + 1 and c+ 1. By symmetry, we may assume H contans p + 1,1 and α2+ for some. Snce α 1+ s adacent to two vertces 14

16 n H, H must contan α 3+. Smlarly, H cannot contan α4+, so H contans p + 1,2 and p+ 1,3. By nducton, we see that H contans p+ 1, for = 1, 2,...,z+ and d 1. If H contans p, then H must contan p 1,z 1, for = z,..., 1, a contradcton. Thus, H must contan d + 2. By nducton, for = 1, 2,...,n, we see that H contans all the vertces of the path P + or P and by symmetry, we may assume H contans all the neghbors of the vertces n P + or P of the form α 2+ or α 3+ for any. Smlarly, for = 1, 2,..., m, t follows that H must contan c + and c. Also, H contans one of the followng: and ether α 2 and α 3 or β 2 and β 3 (where α 2 c 12+, c 1+, c 1, c 12 s adacent to c 1+ ). c 12+, c 2+, c 2, c 12 and ether α 2 and α 3 or β 2 and β 3 (where α 2 s adacent to c 2+ ). c 3+ and c 3 and ether α 2 and α 3 or β 2 and β 3 (where α 2 s adacent to c 3+ ). We can recover the satsfyng assgnment ξ as follows. For = 1, 2,..., n, set ξ = 1 f the vertces of P + are n H and set ξ = f the vertces of P are n H. By constructon, t s easy to verfy that at least one lteral n every clause s satsfed, so ξ s ndeed a satsfyng assgnment. Theorem 2.7 The followng statements hold: For any d Z wth d 2, the problem Γ d s NP-complete when d 3 and polynomal when d = 2. If H s any fnte lst of cycles C 1, C 2,...,C m such that C 6 / H, then Γ 3 H s NP-complete. proof In the above reducton, (G f ) = 3 so Γ d s NP-complete for d 3. When d = 2, there s a smple O(n) algorthm. Any hole contanng x and y must be a component of G so pc the vertex x and consder the component C of G that contans x. It taes O(n) tme to verfy whether C s a hole contanng x and y or not. To show the second statement, let K be the length of the longest cycle n H. In the above reducton, do the followng modfcatons. For = 1, 2, 3 and = 1, 2,...,3m, replace the edges α + α(+1)+, α α(+1), β + β (+1)+, and β β (+1) by paths of length K. For = 1, 2,..., 3m 1, replace the edges α α +1 and β β +1 by paths of length K. Replace the edges xα 1 and xβ 1 by paths of length K. Ths new reducton s polynomal n n and m and s H-free. The proof of Lemma 2.6 stll holds for ths new reducton so therefore, Γ 3 H s NP-complete. 15

17 3 Π B for some specal s-graphs 3.1 Holes wth pendng edges and trees Here, we study Π B4,..., Π B7 where B 4,...,B 7 are the s-graphs depcted on Fgure 3. Our motvaton s smply to gve a strng example and to pont out that surprsngly, pendng edges of s-graphs matter and that even an s-graph wth no cycle can lead to NP-complete problems. Theorem 3.1 There s an O(n 13 )-tme algorthm for Π B4 but Π B5 s NPcomplete. proof A realsaton of B 4 has exactly one vertex of degree 3 and one vertex of degree 4. Let us say that the realsaton H s short f the dstance between these two vertces n H s at most 3. Detectng short realsatons of B 4 can be done n tme n 9 as follows: for every 6-tuple F = (a, b, x 1, x 2, x 3, x 4 ) such that G[F] has edge-set {x 1 a, ax 2, x 2 b, bx 3, bx 4 } and for every 7-tuple F = (a, b, x 1, x 2, x 3, x 4, x 5 ) such that G[F] has edge-set {x 1 a, ax 2, x 2 x 3, x 3 b, bx 4, bx 5 }, delete x 1,...,x 5 and ther neghbors except a, b. In the resultng graph, chec whether a and b are n the same component. The answer s YES for at least one 7-or-6-tuple f and only f G contans at least one short realsaton of B 4. Here s an algorthm for Π B4, assumng that the entry graph G has no short realsaton of B 4. For every 9-tuple F = (a, b, c, x 1,..., x 6 ) such that G[F] has edge-set {x 1 a, bx 2, x 2 x 3, x 3 x 4, cx 5, x 5 x 3, x 3 x 6 } delete x 1,..., x 6 and ther neghbors except a, b, c. In the resultng graph, run three-n-a-tree for a, b, c. It s easly checed that the answer s YES for some 7-tuple f and only f G contans a realsaton of B 4. Let us prove that Π B5 s NP-complete by a reducton of Γ 3 to Π B5. Snce by Theorem 2.7, Γ 3 s NP-complete, ths wll complete the proof. Let (G, x, y) an nstance of Γ 3. Prepare a new graph G : add four vertces x, x, y, y to G and add four edges xx, xx, yy, yy. Snce (G) 3, t s easly seen that G contan a hole passng through x, y f and only f G contans a realsaton of B 5. The proof of the theorem below s omtted snce t s smlar to the proof of Theorem 3.1. Theorem 3.2 There s an O(n 14 )-tme algorthm for Π B6 but Π B7 s NPcomplete. 3.2 Induced subdvsons of K 5 Here, we study the problem of decdng whether a graph contan an nduced subdvson of K 5. More precsely, we put : sk 5 = ({a, b, c, d, e},, ( ) {a,b,c,d,e} 2 ). Theorem 3.3 Π sk5 s NP-complete. proof We consder an nstance (G, x, y) of Γ 3. Let us denote by x, x the two neghbors of x and by y, y the two neghbors of y. 16

18 a b a b e e x x x x x x G G y y y y y y c d c d Fgure 15: Graphs G and G Let us buld a graph G by addng fve vertces a, b, c, d, e. We add the edges ab, bd, dc, ca, ea, eb, ec, ed, ax, bx, cy, dy. We delete the edges xx, xx, yy, yy. We defne a very smlar graph G, the only change beng that we do not add edges cy, dy but edges cy, dy nstead. See fgure 15. Now n G (and smlarly G ) every vertex has degree at most 3, except for a, b, c, d, e. We clam that G contans a hole gong through x and y f and only f at least one of G, G contans an nduced subdvson of K 5. Indeed, f G contans a hole passng through x, x, y, y, y, x n that order then G obvously contans an nduced subdvson of K 5, and the hole passes n order through x, x, y, y, y, x then G contans such a subgraph. Conversely, f G (or symmetrcally G ) contans an nduced subdvson of K 5 then a, b, c, d, e must be the vertces of the underlyng K 5, because they are the only vertces wth degree at least 4. Hence there s a path from x to y n G \ {x, y} and a path from x to y n G \ {x, y}, and consequently a hole gong through x, y n G. 3.3 Π B for small B s Here, we survey the complexty Π B when B has at most four vertces. By the remars n the ntroducton, f V 3 then Π (V,D,F) s polynomal. Up to symmetres, we are left wth twelve s-graphs on four vertces. 17

19 For the followng two s-graphs, there s a polynomal algorthm usng threen-a-tree: The next two s-graphs yeld an NP-complete problem: (by Π {C4}) (by Π {K3}) For the remanng eght ones, we do not now the answer: As a concluson, we would le to pont out that every detecton problem assocated to an s-graph for whch a polynomal tme algorthm s nown can be solved by usng three-n-a-tree or by some easy brute-force enumeraton. References [1] V.E. Aleseev. On the local restrctons effect on the complexty of fndng the graph ndependence number. Combnatoral-algebrac methods n appled mathematcs, 132:3 13, Gory Unversty Press, Gory, n Russan. [2] N. Alon, R. Yuster, and U. Zwc. Fndng and countng gven length cycles. In Proceedngs of the 2nd European Symposum on Algorthms. Utrecht, The Netherlands, pages , [3] D. Benstoc. On the complexty of testng for odd holes and nduced odd paths. Dscrete Math., 9:85 92, See also Corrgendum by B. Reed, Dscrete Math., 12, (1992), p. 19. [4] M. Chudnovsy and P. Seymour. The three-n-a-tree problem. Manuscrpt. [5] M. Chudnovsy, G. Cornuéols, X. Lu, P. Seymour, and K. Vušovć. Recognzng Berge graphs. Combnatorca, 25: , 25. [6] M.R. Garey and D.S. Johnson. Computer and Intractablty : A Gude to the Theory of NP-completeness. W.H. Freeman, San Franssco, [7] A. Hertz and V. V. Lozn. The maxmum ndependent set problem and augmentng graphs. Manuscrpt. [8] F. Maffray and N. Trotgnon. Algorthms for perfectly contractle graphs. SIAM Jour. of Dscrete Math., 19(3): , 25. [9] F. Maffray, N. Trotgnon, and K. Vušovć. Algorthms for square- 3P C(, )-free Berge graphs. SIAM Journal on Dscrete Mathematcs. To appear. [1] R.E. Taran and M. Yannaas. Smple lnear-tme algorthms to test chordalty of graphs, test acyclcty of hypergraphs, and selectvely reduce acyclc hypergraphs. SIAM Journal on Computng, 13: ,