Conjectures and Questions in Graph Reconfiguration

Transcription

1 Conjectures and Questions in Graph Reconfiguration Ruth Haas, Smith College Joint Mathematics Meetings January 2014

2 The Reconfiguration Problem Can one feasible solution to a problem can be transformed into another by some allowable set of moves, while maintaining feasibility at all steps? E.g., Power Supply Problem

3 Reconfiguration: Can you get from solution A to solution B? *The reconfiguration problem.*

4 Reconfiguration: Can you get from solution A to solution B? *The reconfiguration problem.* How many steps might it take to go between the two solutions?

5 Reconfiguration: Can you get from solution A to solution B? *The reconfiguration problem.* How many steps might it take to go between the two solutions? How many ways can you get from one solution to another?

6 Reconfiguration: Can you get from solution A to solution B? *The reconfiguration problem.* How many steps might it take to go between the two solutions? How many ways can you get from one solution to another? Can you get from a particular solution to any other?

7 Reconfiguration: Can you get from solution A to solution B? *The reconfiguration problem.* How many steps might it take to go between the two solutions? How many ways can you get from one solution to another? Can you get from a particular solution to any other? How hard is it to tell?

8 Conjecture Reconfiguration is hard.

9 Conjecture Reconfiguration is hard. More specifically: Conjecture The reconfiguration problem is hard when the problem is hard.

10 Theorem (Ito, Demaine, Harvey, Papadimitriou, Siderie, Ueharaf, Uno, 2011) The following problems are PSPACE-complete: independent set reconfiguration, clique reconfiguration, vertex cover reconfiguration, set cover reconfiguration, integer programming reconfiguration, power supply reconfiguration. T. Ito et al. / Theoretical Computer Science 412 (2011) a b c Fig. 1. A sequence of feasible solutions for the power supply problem. lution of I); and fix a polynomially testable symmetric adjacency relation A on the set of feasible solutions, that is, lynomial-time algorithm such that, given an instance I of S and two feasible solutions y 0 and y 00 of I, it determines wheth and y 00 are adjacent. (In almost all problems discussed in this paper, the feasible solutions can be considered as sets

11 The reconfiguration graph. Vertices are feasible solutions. Edge between two represents one application of the reconfiguration rule. So the reconfiguration problem asks whether the two solutions are in the same connected component.

12 Reconfiguration in graph coloring The k-coloring graph of G, C k pgq, vertices are the proper k-colorings of G two k-colorings joined by an edge if they differ in color on just one vertex of G. C 3 pk 2 q

13 Approximating number of colorings of G with k colors. Think of coloring graph as a Markov system. 1/2 1/4 1/4 1/4 1/4 1/4 1/4 Run the Markov process to convergence enough times. Use the number of states found to approx. total number of colorings. Requires: 1) C k pgq to be connected ( mixing ) 2) Convergence must be fast enough. rapid mixing

14 Application to theoretical physics: The Glauber dynamics of an anti-ferromagnetic Potts model at zero temperature. Systems like: magnetism lattice gases spin glasses Discrete collection of atoms for which certain pairs must have different spin values Individual atoms can change spin, one at a time.

15 Reconfiguration of colorings is hard. Theorem (Cereceda,van den Heuvel, Johnson, 2009) The decision problem: Is C 3 pgq connected? is conp-complete. But in some cases it is easy...

16 Theorem (Jerrum, 1995) For all G, if k ą 2 then C k pgq is connected and the chain converges in optimal time Opn log nq.

17 Theorem (Jerrum, 1995) For all G, if k ą 2 then C k pgq is connected and the chain converges in optimal time Opn log nq. Theorem (Vigoda) k ě 11{6 then C k pgq is connected and the chain converges in polynomial time.

18 Theorem (Jerrum, 1995) For all G, if k ą 2 then C k pgq is connected and the chain converges in optimal time Opn log nq. Theorem (Vigoda) k ě 11{6 then C k pgq is connected and the chain converges in polynomial time. There are graphs G for which C p `1q pgq is disconnected.

19 Theorem (Jerrum, 1995) For all G, if k ą 2 then C k pgq is connected and the chain converges in optimal time Opn log nq. Theorem (Vigoda) k ě 11{6 then C k pgq is connected and the chain converges in polynomial time. There are graphs G for which C p `1q pgq is disconnected. Question For what value of k (as a function of, χ, n??) can we insure C k connected?

20 Proposition (Cereceda, van den Heuvel, and Johnson, 2008) There is no function F pχq, so that for all graphs G and integers More interesting extremal graphs k ą F pχpgqq, C k pgq is connected. Eg. graph K m,m L- m (matching), : K m,m minus χpgq a perfect 2 and m matching coloring ( musing 3) all m colors on each part is an isolated vertex in C m pgq (L m )+1 = D(L m )+1 = m m m has frozen m -colourings hence L m is not m -mixing

21 Recall colpgq = 1` least max back degree (under all vert. orderings) Theorem (Dyer, Flaxman, Frieze, Vigoda, 2006) For any graph G and integer k ě colpgq ` 1, C k pgq is connected.

22 Recall colpgq = 1` least max back degree (under all vert. orderings) Theorem (Dyer, Flaxman, Frieze, Vigoda, 2006) For any graph G and integer k ě colpgq ` 1, C k pgq is connected. In fact:

23 Recall colpgq = 1` least max back degree (under all vert. orderings) Theorem (Dyer, Flaxman, Frieze, Vigoda, 2006) For any graph G and integer k ě colpgq ` 1, C k pgq is connected. In fact: Theorem (Choo and MacGillivray, 2011) If k ě colpgq ` 2, then C k pgq is Hamiltonian.

24 Recall colpgq = 1` least max back degree (under all vert. orderings) Theorem (Dyer, Flaxman, Frieze, Vigoda, 2006) For any graph G and integer k ě colpgq ` 1, C k pgq is connected. In fact: Theorem (Choo and MacGillivray, 2011) If k ě colpgq ` 2, then C k pgq is Hamiltonian. In special cases reconfiguration has nice structure.

25 Non-isomorphic colorings. Consider 3-colorings of P , 2123, , 2321, , 2121, , 2323, , 2132, , 2312, , 2131, , 2313, 3212

26 Non-isomorphic colorings. Consider 3-colorings of P , 2121, , 2323, , 2123, , 2321, , 2132, , 2312, , 2131, , 2313, 3212

27 Non-isomorphic colorings. Consider 3-colorings of P , 2121, , 2323, , 2123, , 2321, , 2131, , 2313, , 2132, , 2312, 3213

28 Non-isomorphic colorings. Consider 3-colorings of P , 2121, , 2323, , 2123, , 2321, , 2131, , 2313, 3212 I 3 pp 4 q 1231, 2132, , 2312, 3213

29 Non-isomorphic colorings. Consider just the canonical representatives 1212, 2121, , 2323, , 2123, , 2321, , 2131, , 2313, , 2132, , 2312, 3213

30 Non-isomorphic colorings. Consider just the canonical representatives 1212, 2121, , 2323, , 2123, , 2321, , 2131, , 2313, , 2132, , 2312, 3213 Can k pp 4 q in blue and additional edges of I k pp 4 q in red.

31 Comparing the coloring graphs C k pgq all colorings I k pgq isomorphic classes Can k pgq canonical representatives Note that I k pgq is obtained from C k pgq by contracting pairs of isomorphic colorings. Thus if C k pgq is connected, then I k pgq is connected.

32 Comparing the coloring graphs C k pgq all colorings I k pgq isomorphic classes Can k pgq canonical representatives Note that I k pgq is obtained from C k pgq by contracting pairs of isomorphic colorings. Thus if C k pgq is connected, then I k pgq is connected. The converse is false.

33 Comparing the coloring graphs The 5 cycle:

34 Comparing the coloring graphs The 5 cycle: Can 3 pc 5 q is disconnected.

35 Comparing the coloring graphs The 5 cycle: Can 3 pc 5 q is disconnected. I 3 pc 5 q is C 5.

36 Comparing the coloring graphs The 5 cycle: Can 3 pc 5 q is disconnected. I 3 pc 5 q is C 5. C 3 pc 5 q is two disjoint 15 cycles.

37 Comparing the coloring graphs The 5 cycle: Can 3 pc 5 q is disconnected. I 3 pc 5 q is C 5. C 3 pc 5 q is two disjoint 15 cycles. So I k pgq can be connected while C k pgq disconnected.

38 Different orders give different Canonical graphs.

39 Different orders give different Canonical graphs

40 Different orders give different Canonical graphs

41 Different orders give different Canonical graphs Two orders for CanpP 4 q

42 Connectivity Theorem Every connected graph n ą 2 that is not complete has an order π such that Cank π pgq is disconnected for sufficiently large k. G must have an induced P 3. t112 u t123 u

43 Question Under what conditions is Cank π pgq connected?

44 Question Under what conditions is Cank π pgq connected? In general the decision problem is hard.

45 Question Under what conditions is Cank π pgq connected? In general the decision problem is hard. Conjecture If k ą f pcolpgqq then Cank π pgq is connected... for some order π.

46 Hamiltonian Canonical colorings Theorem (H, 2012) For k ě 3 colors, any tree, T, there exists an order π such that Cank π pt q has a Hamilton cycle.

47 Hamiltonian Canonical colorings Theorem (H, 2012) For k ě 3 colors, any tree, T, there exists an order π such that Cank π pt q has a Hamilton cycle. Theorem (H, MacGillivray, 2012) If G is a split graph then there exists an order π such that Can π t pgq has a Hamilton path, for t ě χpgq ` 1. Further if there exists a partition into simplicial vertices, C, and a clique, D, such that the clique is size ωpgq χpgq and C ě 2 then Can π t pgq has a Hamilton cycle.

48 Hamiltonian Canonical colorings But Can π 6 pk 2,2,2q Proposition Can π k pk n 1,n 2,...,n r q has no Hamilton path, for any order π of the vertices, when n 1, n 2, n 3 ě 2.

49 Connected Canonical colorings Theorem (H, MacGillivray) If G is perfect and χpgq ωpgq ColpGq then there exists an order π such that Cank π pgq is connected for all k ě χpgq ` 1.

50 Connected Canonical colorings Theorem (H, MacGillivray) If G is perfect and χpgq ωpgq ColpGq then there exists an order π such that Cank π pgq is connected for all k ě χpgq ` 1. Theorem (H, MacGillivray, 2013) Let G be a bipartite graph on n vertices, then there exists an ordering π of the vertices such that Cank π pgq is connected for k ě n{2 ` 2.

51 Conjecture If k ą f pcolpgqq then there exists an order π such that Can π k pgq is connected.

52 Conjecture If k ą f pcolpgqq then there exists an order π such that Cank π pgq is connected. Conjecture If k V pgq then there exists an order π such that Can π k pgq is connected.