Cycle transversals in bounded degree graphs

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


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  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 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 k -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 3 -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.

Maximum degree three graphs
An 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 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 triangletransversal of G, first remove 3-free 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 triangle-transversal consists of one vertex per component.2 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 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: is NP-complete for maximum degree three graphs, for any fixed k ≥ 5.

Maximum degree four graphs
For graphs with maximum degree four, we have: The above naïve 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 2 .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).
Author's personal copy 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 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 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.
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 Author's personal copy 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.This corresponds to "replacing" H by a copy of H 3 .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 61 can be excluded from consideration.
7. For n = 3j + 2 (j ≥ 1), the template of H n says that the number of inner vertices to 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 degree-four 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 − 1 to each t i in the template of H 4 .Thus, if H is a piece of G isomorphic to H 3j+1 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 4 .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 3j+1 , for j ≥ 2, can also be excluded from consideration.9.For n = 3j (j ≥ 2) the template of H n can be obtained by adding j − 1 to each t 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 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 3j , 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 3 , H 4 and G 55 .In fact, we can for maximum degree four graphs, for any fixed k ≥ 3. A naïve k-approximation 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 , 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 k 's.Observe that the collection then H-transversal corresponds to the maximum induced bipartite 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 Alternatively, we can fix k and determine p such that C k -transversal is polynomial-time solvable if Δ ≤ p, and NPcomplete otherwise.The table below summarizes the complexity results dealt with in this work.
ktransversal: given a graph G (with bounded maximum degree Δ) and an integer , does G contain a C k -transversal of size at most ?In particular, we seek for dichotomy results as follows: for a fixed value of Δ, C k -transversal is polynomial-time solvable if k ≤ p, and NP-complete otherwise.
4.For each piece H of G isomorphic to G 62 , let v 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, if H 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.