On Variants of Induced Matchings Andreas Brandstädt University of Rostock, Germany (joint work with Raffaele Mosca and Ragnar Nevries)

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Presentation transcript:

On Variants of Induced Matchings Andreas Brandstädt University of Rostock, Germany (joint work with Raffaele Mosca and Ragnar Nevries)

Distance-k Matchings Let G = (V,E) be a undirected finite simple graph. An edge set M  E is a matching in G if the edges in M are mutually vertex-disjoint. An edge set M is an induced matching in G [Kathie Cameron 1989] (also called strong matching [Golumbic, Laskar 1993]) if the mutual distance of edges in M is  2.

Distance-k Matchings An edge set M is a distance-k matching in G if the mutual distance of edges in M is  k (called  -separated matching [Stockmeyer, Vazirani 1982]).

Matchings e1e1 e2e2 e3e3 e4e4 e5e5 e6e6

e1e1 e2e2 e3e3 e4e4 e5e5 e6e6

Induced Matchings e1e1 e2e2 e3e3 e4e4 e5e5 e6e6

Line Graph For graph G = (V, E), let L(G ) = (E, E ) with edges xy  E  x  y  denote the line graph of G.

Line Graph e1e1 e2e2 e3e3 e1e1 e2e2 e3e3 G L(G)L(G)

e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 G L(G)L(G)

e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 e1e1 e2e2 e3e3 e4e4 e5e5 e6e6

Graph Powers For graph G = (V, E), let G k = (V, E k ) with xy  E k  dist G (x,y)  k denote the k-th power of G.

Induced Matchings L(G) 2 is the square of the line graph of G, i.e., the vertex set of L(G) 2 is E, and two edges of G are adjacent in L(G) 2 if they share a vertex or are connected by an edge in G.

Induced Matchings L(G) 2 is the square of the line graph of G, i.e., the vertex set of L(G) 2 is E, and two edges of G are adjacent in L(G) 2 if they share a vertex or are connected by an edge in G. Fact. Induced matchings in G  independent vertex sets in L(G) 2.

Induced Matchings e1e1 e2e2 e3e3 e4e4 e5e5 e6e6

e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 L(G)L(G)

e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 L(G)2L(G)2

Maximum Induced Matchings Maximum Matching Problem: Find a maximum matching of largest size. Maximum Induced Matching (MIM) Problem: Find a max. induced matching of largest size. NP-complete [Stockmeyer, Vazirani 1982, Kathie Cameron 1989]

Maximum Induced Matchings The Maximum Induced Matching Problem remains NP-complete for very restricted bipartite graphs [Ko, Shepherd 2003, Lozin 2002] and for line graphs (and thus also for claw-free graphs) [Kobler, Rotics 2003].

Maximum Induced Matchings G chordal  L(G) 2 chordal [Cameron 1989]. G circular-arc graph  L(G) 2 circular-arc graph [Golumbic, Laskar 1993] G cocomparability graph  L(G) 2 cocomparability graph [Golumbic, Lewenstein 2000] G weakly chordal  L(G) 2 weakly chordal [Cameron, Sritharan, Tang 2003] stronger result for AT-free graphs [J.-M. Chang 2004]

Maximum Induced Matchings Hence: MIM in polynomial time for chordal graphs [Cameron 1989] circular-arc graphs [Golumbic, Laskar 1993] cocomparability and interval dimension k graphs [Golumbic, Lewenstein 2000] AT-free graphs [J.-M. Chang 2004] weakly chordal graphs [Cameron, Sritharan, Tang 2003]

Distance-k Matchings L(G) k is the k-th power of the line graph of G, i.e., the vertex set of L(G) k is E, and two edges of G are adjacent in L(G) k if their distance in L(G) is at most k. Fact. Distance-k matchings in G  independent vertex sets in L(G) k.

Chordal Graphs Graph G is chordal if it contains no chordless cycles of length at least four.

Chordal Graphs Graph G is chordal if it contains no chordless cycles of length at least four. Chordal graphs have many facets: - clique separators - clique tree - simplicial elimination orderings - intersection graphs of subtrees of a tree...

Graph Powers [Duchet, 1984]: Odd powers of chordal graphs are chordal. But: Even powers of chordal graphs are in general not chordal.

Powers of Chordal Graphs [K. Cameron, 1989]: G chordal  L(G) 2 chordal.

Powers of Chordal Graphs [K. Cameron, 1989]: G chordal  L(G) 2 chordal  MIM problem in polynomial time on chordal graphs.

Powers of Chordal Graphs [K. Cameron, 1989]: G chordal  L(G) 2 chordal  MIM problem in polynomial time on chordal graphs. even in linear time! [B., Hoang 2005].

Powers of Chordal Graphs [K. Cameron, 1989]: G chordal  L(G) 2 chordal  MIM problem in polynomial time on chordal graphs But: If G is chordal then in general, L(G) 3 is not chordal.

Maximum Distance-3 Matchings Theorem. The Maximum Distance-3 Matching Problem for chordal graphs is NP-complete.

Maximum Distance-3 Matchings Theorem. The Maximum Distance-3 Matching Problem for chordal graphs is NP-complete. (Reduction from Maximum Independent Set for any graph [GT20])

Powers of strongly chordal graphs A graph is strongly chordal if it is chordal and sun-free. Theorem. [Lubiw 1982; Dahlhaus, Duchet 1987; Raychaudhuri 1992] For every k  2: G strongly chordal  G k strongly chordal.

Powers of strongly chordal graphs Recall: G chordal  L(G) 2 chordal  MIM problem in polynomial time on strongly chordal graphs. But: If G is strongly chordal then in general, L(G) 3 is not strongly chordal.

L(G) 3 not strongly chordal e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 e1e1 e2e2 e3e3 e4e4 e5e5 e6e6

Maximum Distance-3 Matchings Theorem. If G is strongly chordal then L(G) 3 is chordal.

Maximum Distance-3 Matchings Theorem. If G is strongly chordal then L(G) 3 is chordal. Corollary. The Maximum Distance-k Matching Problem for strongly chordal graphs is solvable in polynomial time for every k  1.

Dominating Induced Matchings An induced matching M is a dominating induced matching (d.i.m.) in G if M intersects every edge in G. In other words: M is dominating in L(G) and M is independent in L(G) 2.

Dominating Induced Matchings Example: e1e1 e4e4 e2e2 e3e3 e5e5 e6e6

Dominating Induced Matchings Note that there are graphs (even trees) without such an edge set: e1e1 e4e4 e2e2 e3e3 e5e5

Dominating Induced Matchings The Dominating Induced Matching (DIM) Problem is: Given a graph, does it have a dominating induced matching? Also called the Efficient Edge Domination (EED) Problem.

Dominating Induced Matchings Theorem [Grinstead, Slater, Sherwani, Holmes 1993] The EED problem is NP-complete in general and efficiently solvable for series-parallel graphs.

Dominating Induced Matchings Theorem [Lu, Tang 1998] The EED problem is NP-complete for bipartite graphs, and is efficiently solvable for bipartite permutation graphs. Theorem [Cardozo, Lozin 2008] The EED problem is NP-complete for (very special) bipartite graphs, and is efficiently solvable for claw- free graphs.

Dominating Induced Matchings Theorem [B., Nevries 2009] The EED problem is solvable in linear time for chordal bipartite graphs; polynomial time for hole-free graphs.

Dominating Induced Matchings Proposition. Let M be a d.i.m. Then: (i)M contains exactly one edge of every triangle.

Dominating Induced Matchings Proposition. Let M be a d.i.m. Then: (i)M contains exactly one edge of every triangle. (ii)M contains no edge of any C 4.

Dominating Induced Matchings Proposition. Let M be a d.i.m. Then: (i)M contains exactly one edge of every triangle. (ii)M contains no edge of any C 4. (iii)Every mid-edge of a diamond is in M. (iv)The peripheral edges of any butterfly are in M.

Dominating Induced Matchings Proposition. Let M be a dim. Then: (i)M contains exactly one edge of every triangle. (ii)M contains no edge of any C 4. (iii)Every mid-edge of a diamond is in M. (iv)The peripheral edges of any butterfly are in M. (v)Graphs with dim are (K 4,gem,long-antihole)-free.

Dominating Induced Matchings Lemma. Let M be a d.i.m. in a chordal bipartite graph G, and let Q = X  Y be a 2-connected component in G. Then either M dominates all vertices in X and none in Y or vice versa.  reduction to the EED problem in a (K 4 -free) block graph G via the following gadget:

Dominating Induced Matchings Lemma. A chordal bipartite graph G has a d.i.m. M  the (K 4 -free) block graph G has a d.i.m. M, and the weights of M and M coincide.

Dominating Induced Matchings Note that hole-free graphs with d.i.m. are weakly chordal (no long antiholes). Theorem. For hole-free graphs, the minimum weight dominating induced matching problem can be solved in polynomial time.

Thank you for your attention!