Cycle transversals in bounded degree graphs

Cycle transversals in bounded degree graphs

M. Groshaus ^a

P. Hell ^b

S. Klein ^c

L. T. Nogueira ^d

F. Protti ^c

a

Universidad de Buenos Aires, FCEyN, Dpto de Computaci´ on, 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

-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

-transversal is polynomial-time solvable if k ≤ p, and NP -hard otherwise.

Keywords: transversal, H-transversal, H-subgraph, H-free graph

1 Introduction

The graphs considered in this work are simple, connected and finite. Let H be a fixed family of graphs. An H-subgraph of a graph G is an induced subgraph of G isomorphic to a member of H. A graph is H-free if it contains no H-subgraph. An H-transversal of a graph G is a subset T ⊆ V (G) such that T intersects all the H-subgraphs of G. Clearly, if T is an H-transversal of G then G − T is H-free. Moreover, if T is small (minimum) then G − T is a large (maximum) induced H-free subgraph of G.

For a fixed family H, the general decision problem named H -transversal can be formulated as follows: given a graph G and an integer , decide whether G contains an H-transversal T such that |T | ≤ . Yannakakis proved that this problem is NP-complete [ 6].

Many problems in graphs can be considered in the context of transversals. For example, if H = {C

| k ≥ 0}, then H -transversal 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 H-transversal of G.

1

Partially supported by Brazilian agencies CNPq and FAPERJ.

2

Partially supported by UBACyT Grants X456, Cod X143 PICT ANCyT Grant 1562.

3 E-addresses: [email protected], [email protected], [email protected],{loana,fabio}@ic.uff.br

1571-0653/$ – see front matter © 2009 Elsevier B.V. All rights reserved.

www.elsevier.com/locate/endm

doi:10.1016/j.endm.2009.11.032

G H G − T general odd cycles bipartite

general {K2} stable set

general {K3} triangle-free

general {P3} disjoint union of cliques

general {P4} cograph

chordal {K3} forest

interval {K1,3} indifference

bipartite {P4} disjoint union of bicliques chordal bipartite {C4} forest

perfect {K} ( − 1)-colorable

Let k denote a fixed integer, k ≥ 3. In this work we investigate the case H = {C

}. (C

denotes a chordless cycle with k vertices.) We consider the following problem, named C

- transversal : given a graph G (with bounded maximum degree Δ) and an integer , does G contain a C

-transversal of size at most ?

In particular, we seek for dichotomy results as follows: for a fixed value of Δ, C

-transversal is polynomial-time solvable if k ≤ p, and NP-complete otherwise. Alternatively, we can fix k and determine p such that C

-transversal is polynomial-time solvable if Δ ≤ p, and NP- complete otherwise. The table below summarizes the complexity results dealt with in this work.

Δ = 2 Δ = 3 Δ≥ 4

k = 3 P P NP-c

k = 4 P P NP-c

k ≥ 5 P NP-c NP-c

If Δ = 2, minimum C

-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

-transversal for maximum degree three graphs is polynomial-time solvable for k ≤ 4 and NP-complete otherwise. For maximum degree four graphs, such a dichotomy is not possible: we show in Section 3 that C

-transversal is NP-complete for any fixed k ≥ 3.

This NP-completeness result trivially extends to Δ ≥ 5.

In view of the hardness of finding minimum C

-transversals (or triangle-transversals) 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 k-free edge if e is contained in no induced C

of G.

Theorem 2.1 C

-transversal is polynomial time solvable for maximum degree three graphs.

Proof. Let G with Δ = 3. Clearly, if e is a 3-free edge of G then T is a triangle-transversal

of G if and only if T is a triangle-transversal of G − e. Thus, to find a minimum triangle- transversal of G, first remove 3-free edges; next, observe that each connected component of the remaining graph can be a triangle, a K

or a diamond ( K

minus one edge). Hence a minimum triangle-transversal consists of one vertex per component. 2 A 4-bracelet is a cubic graph with vertex set V = {a

, . . . , a

, b

, . . . , b

} and edge set E = {a

b

|1 ≤ i ≤ j} ∪ {a

a

, b

b

|1 ≤ i ≤ j − 1} ∪ {a

a

, b

b

}. A twisted 4-bracelet is defined similarly, with edges a

b

, b

a

instead of a

a

, b

b

.

Theorem 2.2 C

-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 4-free edges is a subgraph (not necessarily induced) of a bracelet or twisted

bracelet. 2

The next result completes the dichotomy for Δ = 3:

Theorem 2.3 C

-transversal is NP-complete 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

-transversal is NP-complete for maximum degree four graphs, for any fixed k ≥ 3.

A na¨ıve k-approximation algorithm is possible for finding C

-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

, say C (which can be found in polynomial time, since k is fixed); (ii) set T := T ∪ V (C) and C := C ∪ {C}; (iii) remove the vertices in V (C) from G. Repeat (i)–(iii) until there are no more C

’s. Observe that the collection C is a C

-packing, that is, a collection of vertex-disjoint C

’s. Also, T is clearly a C

-transversal. If T

is a minimum C

-transversal, we have |T

| ≥ |C |. Since |T | = k|C |, it follows that |T |/|T

| ≤ k.

The above na¨ıve algorithm produces triangle-transversals 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) = 4 and a, b, c, d induce 2K

. The vertex z is called a bond. A piece is a maximum degree four connected graph containing no 3-free 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).

H’ (n n ≥⁷⁾

G G G G G G

4 51 52 53 54 55

H (n n ≥³⁾

G G G G G G

61 62 63 64 65 7

Fig. 1. Pieces. The graph H

( n ≥ 3) is formed by two paths u

u

. . . u

and v

v

. . . v

, plus the following edges: u

v

and u

v

, 1 ≤ i ≤ n/2 − 1; u

v

; and, if n is odd, u

v

. The graph H

( n ≥ 7) is formed by a copy of H

plus the following edges: v

u

; v

v

; and u

v

(if n is even) or u

v

(if n is odd).

Lemma 3.3 If G is a minimal piece then either G = H

or G = H

, for n ≥ 7.

Lemma 3.4 If G is a non-minimal piece then G is one of the graphs H

(n ≥ 4), G

, G

(1 ≤ i ≤ 5), G

(1 ≤ j ≤ 5), G

.

A direct consequence of Theorem 3.2 is:

Corollary 3.5 Let G be a maximum degree four graph containing no bonds. Then a mini- mum triangle-transversal of G can be obtained in polynomial time.

Proof. After removing the 3-free edges of G, each of its connected components is a piece, for which a minimum triangle-transversal is easily obtained. 2

We analyze now maximum degree four graphs that may contain bonds. We can restrict our analysis to connected graphs without 3-free edges. The following definition describes a decomposition for such graphs:

Definition 3.6 Let G be a maximum degree four connected graph without 3-free edges.

The piece decomposition of G is the collection of pieces obtained by splitting each bond of

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

, t

, . . . , t

) such that: (i) k is the number of connectors of G; (ii) if G is a piece of a graph H with minimum triangle- transversal T , and i is the number of connectors of G belonging to T , then t

is the number of inner vertices of G belonging to T (observe that, for every piece, t

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

and G

are, respectively, (1 , 1, 1, 0) and (2, 1, 1, 1). The template of H

, for n ≥ 4, depends on the value of n: if n = 3j then it is (

,

,

); if n = 3j + 1 then it is (

,

,

); and if n = 3j + 2 then it is (

,

,

).

The piece H

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

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 3-free edges.

Perform the piece decomposition of G. Let T be a minimum triangle-transversal of G to be computed, initially empty.

1. If G contains only one bond and only one piece then G is a 3-bracelet, a graph obtained from H

(for n ≥ 8) by collapsing its degree-two vertices. In this case, T is easily obtained.

(In the cases below, G is not a 3-bracelet, therefore every piece of G is an induced subgraph of G.)

2. If G contains G

as a piece then G = G

, since G

contains no connectors; hence, T

can be trivially obtained. The same argument applies to G

, G

, G

, G

, G

and H

, for

n ≥ 7. Thus we can exclude these pieces from consideration.

3. For each piece H of G isomorphic to G

, 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

, provided that v, w are not adjacent to a same connector of H.

4. For each piece H of G isomorphic to G

, let v be the connector of H and w the inner vertex of H whose neighbors induce C

. 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

and H

are identical. Thus, transform every piece H of G isomorphic to G

into another piece isomorphic to H

, 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

is (2 , 1, 1, 1). Note that it can be obtained by adding one to each t

in the template of H

. Thus, if H is a piece of G isomorphic to G

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

. It is easy to see that there exists a triangle-transversal of G

with size q if and only if there exists a triangle-transversal of G with size q + 1. Thus, G

can be excluded from consideration.

7. For n = 3j + 2 (j ≥ 1), the template of H

says that the number of inner vertices to be included in T is always the same. Thus, for each piece H isomorphic to H

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

, for instance, the degree-four inner vertex must be included in T .)

8. For n = 3j + 1 (j ≥ 2) the template of H

can be obtained by adding j − 1 to each t

in the template of H

. Thus, if H is a piece of G isomorphic to H

for some j ≥ 2, construct a graph G

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

. Again, it is easy to see that there exists a triangle-transversal of G

with size q if and only if there exists a triangle-transversal of G with size q + j − 1.

Thus, H

, for j ≥ 2, can also be excluded from consideration.

9. For n = 3j (j ≥ 2) the template of H

can be obtained by adding j − 1 to each t

in the template of H

(for i ≤ 2). Thus, if H is a piece of G isomorphic to H

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

. Since x is a degree-two vertex, we can assume that x / ∈ T . Hence, there exists a triangle-transversal of G

with size q if and only if there exists a triangle-transversal of G with size q + j − 1. Thus, H

, for j ≥ 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

, H

and G

. In fact, we can

eliminate H

by adding, for each piece isomorphic to H

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

and G

(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

-transversal remains N P -complete for maximum degree four graphs containing only such pieces. By excluding the crowns, we have the following result:

Theorem 3.8 C

-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

) are adjacent if they share a bond of G

. Take a maximum matching M of P(G

). Let S be the subset of M -unsaturated vertices of G

. An optimal triangle-transversal 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

. 2 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

and v

, include u

in T and remove u

, v

from G

. Apply to the resulting graph the method described in Theorem 3.8.

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