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Problems on Paired Graphs. Stephen Hedetniemi, Alice McRae, Dee Parks, Kelly Wise and come join in…. Very General Idea Input: Graph G = (V,E) and set S = {S 1 , S 2 , …S k } of disjoint subsets of V.
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Problems on Paired Graphs Stephen Hedetniemi, Alice McRae, Dee Parks, Kelly Wise and come join in…
Very General Idea Input: Graph G = (V,E) and set S = {S1, S2, …Sk} of disjoint subsets of V. Consider problems on graphs where we are making decisions about the vertices in S. The idea is to add a constraint that if two vertices u,v are in the same set in S, then we “treat them the same way” in the problem solution.
Examples • Graph Coloring: if vertex u and v are in the same set, they are colored the same. • Dominating Set, Independent Set, Packing, Vertex Cover, etc: if vertex u and v are in the same set, then they are either both chosen for the dominating set, or both excluded… or perhaps they must both have the same number of neighbors in the solution set, etc… • If the problems involve assigning weights to the vertices, then two vertices in the same set, must be assigned the same weight… or the sum of the weights of their neighborhoods must be the same. • … one could do edges as well as vertices • … perhaps infinite possibilities…
Some Combinations Don’t Make Sense Example: • Graph coloring: if “treating the same” meant giving the same color, we wouldn’t want two adjacent vertices in the same set. • We could require that each set S = {S1, S2, …, Sk} be independent.
Cost of the Solution Is the cost/profit of a set the same as the sum of the cost/profit of the individual vertices? No Easier to Find Problem Solutions If S = {}, or |Si| = 1 for each Si S, then we are back to the original problem.
Applications? Package deals? Politics? Map coloring is an obvious one…
A Graph that is not 4-colorable? D E C B A F A is a colony of E. They should be colored the same color.
Graph Homomorphisms • Homomorphism of a graph G=(V,E) into a graph G’=(V’,E’): f: V V’ such that uv E f(u)f(v) E’. • Homomorphic image graph: f(G) = (f(V),f(E)), f(E) = {f(u)f(v) | uv E} • Note f(u) = f(v) implies uv E. • All the vertices in a given “empire” are identified. By becoming one vertex, they are forced to receive the same color.
D Graph G P = { {A,E} } A B C E F Identify {A,E}. The resulting graph is K5. D A B C E D F B E which has chromatic number 5. C F
M-pire graphs • Heawood (1890). Empires are limited to m regions (homomorphic images of planar graphs where the pre-image of a vertex is at most cardinality m) m-pires (m 2) can be colored with 6m colors. • Jackson and Ringel (1983). Show how to find m-pires whose homomorphic images are K6m.
Earth/moon maps: Special type of 2-pire graphs. Each earth vertex can have a colony on the moon. • Joan Hutchinson (1993) applications of moon graphs to testing of printed circuit boards. • Stewart(2008) Open problem whether there are earth/moon graphs that require 12 colors. Sulanke found an earth/moon graph requiring 9 colors.
Pairing of a Graph • A pairing P of a graph G is a collection of disjoint pairs of vertices. • Let P(G) be the graph obtained from G by identifying each pair of vertices in P. • If the pairs in P are independent, then P(G) is a homomorphic image of G.
Some problems What is the best way to pair the graph, … or the worst, … with respect to chromatic number? Pr(G) : the minimum chromatic number of a homomorphic image of G. Pr(G): the maximum chromatic number of a homomorphic image of G. Pr(G) = (G) kPr(G) (G) // there are k pairs
Finding (P(G)) • In general, this is NP-complete. • What if G is a path? • a cycle? • a tree? • graphs where (G) is easy to compute?
Paired Cycles are 4-colorable(with one exception) (Brooks, 1941) If G is a connected graph other than a clique or an odd cycle, then (G) (G). D B E C F (C10) = 5
Paired Trees and Paths • For any tree T, (T) 4 • Complexity of finding (P(G)): Instance: P(G), where G is a path Question: Is P(G) 3-colorable? Thm. Determining if P(G) is 3-colorable is NP-complete even when G is restricted to a path (or a cycle). Proof. Chromatic number is NP-complete for 4-regular graphs (proof for cycles, just find an Eulerian circuit through a 4-regular graph G). Chromatic number is NP-complete for graphs with degree sequence 4,4,4,4,4,4,…4,3,3 (same deal for an Eulerian trail)
Beyond chromatic numberIndependent Sets Given: Graph G and a pairing P. • Find: A largest independent set S, such that {u,v} P, u S v S. • Find: A smallest maximal independent set S, such that {u,v} P, u S v S. • An independent set in P(G) will be an independent set in G, and an independent set in G with the property {u,v} P, u S v S will be an independent set in P(G), but the sizes of the sets may be different.
Use the size of the independent set in G… or the sum of the cardinalities of the pre-images of the vertices in an independent set in a homomorphic image of G… Or a weighted independence in the homomorphic image of G. Given graph G, pairing P, a set S V(G) respects the pairing P if {u,v} P, u S v S.
i(G,P) = the cardinality of a smallest maximal independent set in a graph G that respects independent pairing P. (G,P) = the cardinality of a maximum independent set in a graph G that respects indep. pairing P. i-Pr(G) = the smallest i(G,P) over all possible indep. pairings P. i+Pr(G) = the largest i(G,P) over all possible indep. pairings of P -Pr(G) = the smallest (G,P) over all possible indep, pairings of P. +Pr(G) = the largest (G,P) over all possible indep. pairings of P.
-Pr(G) +Pr(G) = (G) • i-Pr(G) i(G) i+Pr(G) P1 = {{A,I},{B,H},{C,E},{D,I}} P2 = {{F,A}} S1 = {F} S2 = {A,B,C,F} J A H B D E F I C
Independent Set: Instance: Graph G, integer k Question: Does G have an independent set of cardinality k? Known NP-complete for cubic graphs (Garey and Johnson). From this, easy construction to show NP-complete for 4-regular graphs and also for graphs with degree sequence 4,4,4,…,4,3,3.
If all vertices are paired, then we can use the paired homomorphic image of G because every vertex has weight 2. i(G,P) is NP-complete even for paths and cycles… (if i(G) is NP-complete for 4-regular graphs and graphs with degree sequence 4,4,4,4…4,3,3) (G,P) is NP-complete even for paths and cycles.
Independent Dominating Sets Instance: Graph G, Independent pairing P Question: Does G have an independent dominating set that respects P? This is NP-complete.
Dominating Sets PAIRINGS DO NOT HAVE TO BE INDEPENDENT Problem Type 1: (G,P). Given: Graph G, pairing P Find a dominating set that respects P of minimal cardinality. Problem Type 2: +Pr(G,P). Given a graph G Find a pairing P that maximizes (G,P).
Relationships Theorem: -Pr(G) = (G) +Pr(G) 2(G) Interesting one: +Pr(G) 2(G). Let S be a dominating set for G (without respect to any P). Find a dominating set S’ that respects a pairing P by taking any vertex in a pair with a vertex in S.
When is +Pr(G) 2(G) • For all paths Pn, (Pn) = n/3 +Pr(Pn) 2(Pn) n mod 6 1. +Pr(Pn) 2(Pn) - 1, n mod 6 = 1 . • For all cycles, Cn, , (Cn) = n/3 +Pr(Cn) 2(Cn), n mod 6 = 0. +Pr(Cn) 2(Cn) - 1, n mod 6 = 3. +Pr(Cn) 2(Cn) – 2, otherwise
Achieving 2* for paths/cycles Might think the best idea is to simply pair consecutive vertices, but then as n gets big, you will take every other pair, and achieve about n/2 instead of 2n/3. One way is to group by 6. A B B C C A
Lots of Questions Need Answering • Independent Set problems for paths and cycles • We’ve looked at very few graphs – paths and cycles • Different set properties • Different ways of defining vertices being “treated” the same. • Earth/moon problem: 9,10,11, or 12
