Cycle transversals in bounded degree graphs


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1 Electronic Notes in Discrete Mathematics 35 (2009) Cycle transversals in bounded degree graphs M. Groshaus a,2,3 P. Hell b,3 S. Klein c,1,3 L. T. Nogueira d,1,3 F. Protti c,d,1,3 a Universidad de Buenos Aires, FCEyN, Dpto de Computación,CONICET,Argentina b Simon Fraser University, Canada c Universidade Federal do Rio de Janeiro (UFRJ), Brazil. d Universidade Federal Fluminense (UFF), Brazil. Abstract In this work we consider the problem of finding a minimum C k transversal (a subset of vertices hitting all the induced chordless cycles with k vertices) in a graph with bounded maximum degree. In particular, we seek for dichotomy results as follows: for a fixed value of k, finding a minimum C k transversal is polynomialtime solvable if k p, andnphard otherwise. Keywords: transversal, Htransversal, Hsubgraph, Hfree graph 1 Introduction The graphs considered in this work are simple, connected and finite. Let H be a fixed family of graphs. An Hsubgraph of a graph G is an induced subgraph of G isomorphic to a member of H. A graph is Hfree if it contains no Hsubgraph. An Htransversal of a graph G is a subset T V (G) such that T intersects all the Hsubgraphs of G. Clearly, if T is an Htransversal of G then G T is Hfree. Moreover, if T is small (minimum) then G T is a large (maximum) induced Hfree subgraph of G. For a fixed family H, the general decision problem named Htransversal can be formulated as follows: given a graph G andanintegerl, decide whether G contains an Htransversal T such that T l. Yannakakis proved that this problem is NPcomplete [6]. Many problems in graphs can be considered in the context of transversals. For example, if H = {C 2k+1 k 0}, then Htransversal corresponds to the maximum induced bipartite subgraph problem. The table below shows known examples. Some references for this table are [1,2,3,4,5,6]. In the table, T denotes an Htransversal of G. 1 Partially supported by Brazilian agencies CNPq and FAPERJ. 2 Partially supported by UBACyT Grants X456, Cod X143 PICT ANCyT Grant Eaddresses: /$ see front matter 2009 Elsevier B.V. All rights reserved. doi: /j.endm
2 190 M. Groshaus et al. / Electronic Notes in Discrete Mathematics 35 (2009) G H G T general odd cycles bipartite general {K 2} stable set general {K 3} trianglefree general {P 3} disjoint union of cliques general {P 4} cograph chordal {K 3} forest interval {K 1,3} indifference bipartite {P 4} disjoint union of bicliques chordal bipartite {C 4} forest perfect {K l } (l 1)colorable Let k denote a fixed integer, k 3. In this work we investigate the case H = {C k }. (C k denotes a chordless cycle with k vertices.) We consider the following problem, named C k  transversal: given a graph G (with bounded maximum degree Δ) and an integer l, doesg contain a C k transversal of size at most l? In particular, we seek for dichotomy results as follows: for a fixed value of Δ, C k transversal is polynomialtime solvable if k p, and NPcomplete otherwise. Alternatively, we can fix k and determine p such that C k transversal is polynomialtime solvable if Δ p, and NPcomplete otherwise. The table below summarizes the complexity results dealt with in this work. Δ=2 Δ=3 Δ 4 k =3 P P NPc k =4 P P NPc k 5 P NPc NPc If Δ = 2, minimum C k transversals are trivially obtained in polynomial time for any k, since in this case the input graph is a disjoint union of paths and cycles. In Section 2, we show that C k transversal for maximum degree three graphs is polynomialtime solvable for k 4 and NPcomplete otherwise. For maximum degree four graphs, such a dichotomy is not possible: we show in Section 3 that C k transversal is NPcomplete for any fixed k 3. This NPcompleteness result trivially extends to Δ 5. In view of the hardness of finding minimum C 3 transversals (or triangletransversals) for Δ = 4, polynomial cases and ideas for an approximation algorithm are presented in Section 3, where we describe a decomposition theorem for maximum degree four graphs and reduction rules. 2 Maximum degree three graphs An edge e E(G) is called a kfree edge if e is contained in no induced C k of G. Theorem 2.1 C 3 transversal is polynomial time solvable for maximum degree three graphs. Proof. Let G with Δ = 3. Clearly, if e is a 3free edge of G then T is a triangletransversal
3 M. Groshaus et al. / Electronic Notes in Discrete Mathematics 35 (2009) of G if and only if T is a triangletransversal of G e. Thus, to find a minimum triangletransversal of G, first remove 3free edges; next, observe that each connected component of the remaining graph can be a triangle, a K 4 or a diamond (K 4 minus one edge). Hence a minimum triangletransversal consists of one vertex per component. A4bracelet is a cubic graph with vertex set V = {a 1,...,a j,b 1,...,b j } and edge set E = {a i b i 1 i j} {a i a i+1,b i b i+1 1 i j 1} {a 1 a j,b 1 b j }.Atwisted 4bracelet is defined similarly, with edges a 1 b j,b 1 a j instead of a 1 a j,b 1 b j. Theorem 2.2 C 4 transversal is polynomial time solvable for maximum degree three graphs. Proof. The result follows from the fact that a graph G with V (G) 6, Δ = 3 and containing no 4free edges is a subgraph (not necessarily induced) of a bracelet or twisted bracelet. The next result completes the dichotomy for Δ = 3: Theorem 2.3 C k transversal is NPcomplete for maximum degree three graphs, for any fixed k 5. 3 Maximum degree four graphs For graphs with maximum degree four, we have: Theorem 3.1 C k transversal is NPcomplete for maximum degree four graphs, for any fixed k 3. Anaïve kapproximation algorithm is possible for finding C k transversals in general graphs, for a fixed k 3. Given a graph G, initially set T = and C :=. At each step: (i) locate an induced C k,sayc (which can be found in polynomial time, since k is fixed); (ii) set T := T V (C) andc := C {C}; (iii) remove the vertices in V (C) fromg. Repeat (i) (iii) until there are no more C k s. Observe that the collection C is a C k packing, thatis, a collection of vertexdisjoint C k s. Also, T is clearly a C k transversal. If T is a minimum C k transversal, we have T C. Since T = k C, it follows that T / T k. The above naïve algorithm produces triangletransversals with size at most three times the optimum. Nonetheless, better behaviors might be achieved after applying some reductions on a maximum degree four input graph. We need the following definitions. A tie is a graph formed by five vertices a, b, c, d, z where d(z) =4anda, b, c, d induce 2K 2. The vertex z is called a bond. A piece is a maximum degree four connected graph containing no 3free edges and no bonds. The following theorem characterizes pieces. Theorem 3.2 Let G be a piece. Then G is one of the graphs in Figure 1. The proof of Theorem 3.2 is a consequence of the following two lemmas. A piece G is said to be minimal if G z is not a piece for any z V (G).
4 192 M. Groshaus et al. / Electronic Notes in Discrete Mathematics 35 (2009) H (n 3) n H (n 7) n G G G G G G G G G G G G Fig. 1. Pieces. The graph H n (n 3) is formed by two paths u 1 u 2...u n/2 and v 1 v 2...v n/2,plus the following edges: u i v i and u i v i+1,1 i n/2 1; u n/2 v n/2 ; and, if n is odd, u n/2 v n/2. The graph H n (n 7) is formed by a copy of H n plus the following edges: v 1 u n/2 ; v 1 v n/2 ;and u 1 v n/2 (if n is even) or u 1 v n/2 (if n is odd). Lemma 3.3 If G is a minimal piece then either G = H 3 or G = H n, for n 7. Lemma 3.4 If G is a nonminimal piece then G is one of the graphs H n (n 4),G 4, G 5i (1 i 5), G 6j (1 j 5), G 7. A direct consequence of Theorem 3.2 is: Corollary 3.5 Let G be a maximum degree four graph containing no bonds. Then a minimum triangletransversal of G can be obtained in polynomial time. Proof. After removing the 3free edges of G, each of its connected components is a piece, for which a minimum triangletransversal is easily obtained. We analyze now maximum degree four graphs that may contain bonds. We can restrict our analysis to connected graphs without 3free edges. The following definition describes a decomposition for such graphs: Definition 3.6 Let G be a maximum degree four connected graph without 3free edges. The piece decomposition of G is the collection of pieces obtained by splitting each bond of
5 M. Groshaus et al. / Electronic Notes in Discrete Mathematics 35 (2009) G into two vertices, each having two adjacent neighbors, as shown in Figure 2. Each piece of the collection is also said to be a piece of G. Fig. 2. Piece decomposition. A piece decomposition of G can be obtained in polynomial time by locating its bonds. Definition 3.7 Let G be a piece and v V (G). If d(v) = 2 then v is called a connector, otherwise an inner vertex. The template of G is a sequence (t 0,t 1,...,t k ) such that: (i) k is the number of connectors of G; (ii) if G is a piece of a graph H with minimum triangletransversal T,andi is the number of connectors of G belonging to T, then t i is the number of inner vertices of G belonging to T (observe that, for every piece, t i depends only on i, i.e., this value is independent of which group of i connectors lies in T ). For example, the templates of G 55 and G 61 are, respectively, (1, 1, 1, 0) and (2, 1, 1, 1). The template of H n,forn 4, depends on the value of n: if n =3j then it is ( n, n 3, n 3 ); if n =3j + 1 then it is ( n 1, n 1, n 4 n 2 ); and if n =3j + 2 then it is (, n 2, n 2) The piece H 3 is special, since all of its vertices are connectors and the case i = 0 cannot occur for it. To be coherent with Definition 3.7, the template of H 3 is (1, 0, 0, 0). Templates will be helpful to describe reduction rules that eliminate almost all types of pieces of a maximum degree four input graph G. Reduction rules. Let G be a maximum degree four connected graph without 3free edges. Perform the piece decomposition of G. LetT be a minimum triangletransversal of G to be computed, initially empty. 1. If G contains only one bond and only one piece then G is a 3bracelet, a graph obtained from H n (for n 8) by collapsing its degreetwo vertices. In this case, T is easily obtained. (In the cases below, G is not a 3bracelet, therefore every piece of G is an induced subgraph of G.) 2. If G contains G 4 as a piece then G = G 4, since G 4 contains no connectors; hence, T can be trivially obtained. The same argument applies to G 51,G 53,G 54,G 63,G 64 and H n,for
6 194 M. Groshaus et al. / Electronic Notes in Discrete Mathematics 35 (2009) n 7. Thus we can exclude these pieces from consideration. 3. For each piece H of G isomorphic to G 52, choose two inner vertices v, w V (H), one of them with degree four. Include v, w in T, and remove from G all the inner vertices of H. An analogous procedure can be applied to any piece isomorphic to G 65, provided that v, w are not adjacent to a same connector of H. 4. For each piece H of G isomorphic to G 62,letv be the connector of H and w the inner vertex of H whose neighbors induce C 4. Include v, w in T, and remove all the vertices of H from G (including v). Update the piece decomposition of G by removing v from another piece of G containing it. 5. The templates of G 7 and H 7 are identical. Thus, transform every piece H of G isomorphic to G 7 into another piece isomorphic to H 7, as follows: if v and w are the connectors of H, and xy is an edge of H such that x is adjacent to v and y is adjacent to w, then remove xy from G. 6. The template of G 61 is (2, 1, 1, 1). Note that it can be obtained by adding one to each t i in the template of H 3.Thus,ifH is a piece of G isomorphic to G 61 where v, w, x are its inner vertices, construct a graph G by removing v, w, x and adding the edges vw, vx, wx. This corresponds to replacing H by a copy of H 3. It is easy to see that there exists a triangletransversal of G with size q if and only if there exists a triangletransversal of G with size q +1. Thus,G 61 can be excluded from consideration. 7. For n =3j +2 (j 1), the template of H n saysthatthenumberofinnerverticesto be included in T is always the same. Thus, for each piece H isomorphic to H 3j+2 for some j 1, include in T a suitable subset of j inner vertices of H, and remove from G all the inner vertices of H. (In the case of H 5, for instance, the degreefour inner vertex must be included in T.) 8. For n =3j +1(j 2) the template of H n can be obtained by adding j 1toeacht i in thetemplateofh 4.Thus,ifH is a piece of G isomorphic to H 3j+1 for some j 2, construct agraphg by first removing all the inner vertices of H except the two neighbors v, w of some connector of H, and next linking v, w to the other connector of H. This corresponds to replacing H by a copy of H 4. Again, it is easy to see that there exists a triangletransversal of G with size q if and only if there exists a triangletransversal of G with size q + j 1. Thus, H 3j+1,forj 2, can also be excluded from consideration. 9. For n =3j (j 2) the template of H n can be obtained by adding j 1toeacht i in the template of H 3 (for i 2). Thus, if H is a piece of G isomorphic to H 3j for some j 2, construct a graph G by first removing all the inner vertices of H, and next creating a triangle using the connectors of H together with a new vertex x. This corresponds to replacing H by a copy of H 3. Since x is a degreetwo vertex, we can assume that x/ T. Hence, there exists a triangletransversal of G with size q if and only if there exists a triangletransversal of G with size q + j 1. Thus, H 3j,forj 2, can also be excluded from consideration. 10. At this point, the original graph G may have been converted into a disconnected graph; then we deal with each connected component separately. Hence, let G still stand for a connected graph. The only possible pieces of G are now H 3, H 4 and G 55. In fact, we can
7 M. Groshaus et al. / Electronic Notes in Discrete Mathematics 35 (2009) eliminate H 4 by adding, for each piece isomorphic to H 4 with inner vertices v and w, new vertices x, y, z and new edges xv, xw, xy, xz, yz. That is, the only pieces of G are now H 3 and G 55 (called crown). At this point, let us call G reduced graph. The application of the rules is completed. We remark that the maximum degree four graph constructed in the reduction of Theorem 3.1 contains only triangles and crowns as pieces. Hence, C 3 transversal remains NPcomplete for maximum degree four graphs containing only such pieces. By excluding the crowns, we have the following result: Theorem 3.8 C 3 transversal is polynomial time solvable for a maximum degree four graph G when its reduced graph contains no piece isomorphic to a crown. Proof. Let G be the reduced graph of G. Construct an intersection graph P(G ) as follows: the pieces of G (triangles) are the vertices of P(G ), and two vertices of P(G )areadjacent if they share a bond of G. Take a maximum matching M of P(G ). Let S be the subset of Munsaturated vertices of G. An optimal triangletransversal T of G is formed as follows: for each edge e M, include in T the corresponding bond of G, and for each vertex of S include in T any vertex of the corresponding piece of G. We are currently analyzing the performance of the following approximation algorithm: given the reduced graph G, for each crown C whose degree four vertices are u C and v C, include u C in T and remove u C,v C from G. Apply to the resulting graph the method described in Theorem 3.8. References [1] D. Cornaz and A. R. Mahjoub. The maximum induced bipartite subgraph problem with edge weights. Submitted manuscript. [2] P. C. Fishburn. Interval Orders and Interval Graphs. Wiley, New York, [3] M. R. Garey and D. S. Johnson. Computers and Intractability: A Guide to the Theory of NPCompleteness. W. H. Freeman, New York, [4]M.C.Golumbic.Algorithmic Graph Theory and Perfect Graphs. Wiley, New York, [5] G. Manic and Y. Wakabayashi. Packing triangles in lowdegree graphs and indifference graphs. Proc. European Conference on Combinatorics, Graph Theory and Applications (EuroComb 05), Berlin, Germany, Discrete Mathematics and Theoretical Computer Science (DMTCS), Vol. AE 2005, pp [6] M. Yannakakis. Node and edgedeletion NPcomplete problems. Proc. of the Tenth Annual ACM Symposium on Theory of Computing STOC 78, pp , 1978, ACM Press.
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