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Combinatorial algorithms. Torsten Mütze. The Hamilton cycle problem. Problem: Given a graph , does it have a Hamilton cycle ?. fundamental problem with many applications ( special case of travelling salesman problem ). computational point of view :
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Combinatorial algorithms Torsten Mütze
The Hamilton cycleproblem • Problem:Given a graph, doesithave a Hamilton cycle? • fundamentalproblemwithmanyapplications(specialcase of travellingsalesmanproblem) • computational point of view: • no efficientalgorithmknown (NP-complete[Karp 72]), i.e. brute-forceapproachessentially best possible
The Hamilton cycleproblem • named after Sir William Rowan Hamilton(1805-1865) and his `Icosian Game’ • Most advanced version of this puzzle: Conjecture[Lovász 70]: Everyconnectedvertex-transitivegraph has a Hamilton cycle (apart fromfiveexceptions). • vertex-transitive = graph `looks the same’ from every vertex • various related/opposing conjectures, e.g. [Babai 95] • This thesis: solutions of several long-standing special cases
Partial results • Known cases: • and prime • [Turner 67], [Alspach 79], [Marušič 86, 87], [Chen 98], [Kutnar, Marušič 08] • Every connected and vertex-transitive graph has a cycle of length at least [Babai 79] • Many results known for Cayley graphs: • abelian groups [Marušič 83] • finite p-groups [Witte 86] • every finite group has gen. set of size[Pak, Radoičič 09]
Combinatorialalgorithms • Goal:generate all objects of a combinatorialclassefficiently • Examples: • binarytrees • triangulations • permutations • words • partitions • linear extensions • spanningtrees • matchings • … aabc 1234 ccba 1243 ccca 1423
Flip graphs • Goal:generate all objects of a combinatorialclassefficiently • fundamental taskwithmanyapplications • ultimately: eachnewobject in constanttime • consecutiveobjectsmaydifferonly`a littlebit‘ • defines a highlysymmetricgraph, oftenevenvertex-transitive 111 321 101 011 110 231 312 213 132 001 100 010 123 • associahedron • permutahedron • hypercube 000 • Questions: Existence of a Hamilton cycle? Efficientalgorithm?
Applications Song Duration Problem: Partition songsintosubsets A and B of sizeorwithminimum play lengthdifference. • 2:432 3:203 2:18…7 1:58 Partition is hard! 1234567 1234567 1000011 1000011 0110110 1000111 Question: Isthere a cyclicsingle-fliplisting of all bitstrings of lengthwithormany 1-bits forevery ?
The middlelevelsconjecture Define a graph : 10110 10101 01101 01011 00111 11100 11010 11001 10011 01110 10001 01100 00110 00101 00011 11000 10100 10010 01010 01001 • is a subgraph of the -dimensional hypercube • isvertex-transitive Question: Isthere a cyclicsingle-fliplisting of all bitstrings of lengthwithormany 1-bits forevery ? Middlelevelsconjecture: has a Hamilton cyclefor all .
The middlelevelsconjecture • notoriouslydifficultproblemraised in the 1980s • hardestopenproblem in [Knuth 11] withdifficulty 49/50 • mentioned in popularbooks[Diaconis, Graham 12], [Winkler 04], and in survey[Gowers 17] • ≈25 previouspaperswith partial results ([Kierstead, Trotter 88], [Savage, Winkler 95], [Felsner, Trotter 95], [Johnson 04], …) • has severalfar-rangingimplications • specialcase of Lovász‘ conjecture Middlelevelsconjecture: has a Hamilton cyclefor all .
Ourresults Theorem 1 [M. 16; Proc. LMS]: The middlelevelsgraph has a Hamilton cycleforevery . Theorem 2 [M. 16; Proc. LMS]: The middlelevelsgraph has distinct Hamilton cycles. Remarks: number of automorphismsisonly ,so Theorem 2 isnot an immediate consequence of Theorem 1
Ourresults Theorem 1 [M. 16; Proc. LMS]: The middlelevelsgraph has a Hamilton cycleforevery . Theorem 2 [M. 16; Proc. LMS]: The middlelevelsgraph has distinct Hamilton cycles. Remarks: number of Hamilton cyclesis at most ,so Theorem 2 is best possible
Ourresults Theorem 1 [M. 16; Proc. LMS]: The middlelevelsgraph has a Hamilton cycleforevery . Theorem 2 [M. 16; Proc. LMS]: The middlelevelsgraph has distinct Hamilton cycles. • Remarks: • Werecentlyfound a short and moreaccessibleproofforthosetheorems[Gregor, M., Nummenpalo 18; Discrete Analysis] • 40 pages 9 pages • 27 lemmas 2 lemmas • 88 formulas 9 formulas
Proofideas Step 1:Build a 2-factor in thegraph 2-factor
Proofideas Step 1:Build a 2-factor in thegraph Step 2: Connect the cycles in the 2-factor to a singlecycle 2-factor flippable pair
Proofideas Step 1:Build a 2-factor in thegraph Step 2: Connect the cycles in the 2-factor to a singlecycle 2-factor flippable pair
Proofideas Step 1:Build a 2-factor in thegraph Step 2: Connect the cycles in the 2-factor to a singlecycle The middlelevelsgraphdoes not have 4-cycles, so weuse 6-cycles 2-factor flippable pair
Proofideas Step 1:Build a 2-factor in thegraph Step 2: Connect the cycles in the 2-factor to a singlecycle The middlelevelsgraphdoes not have 4-cycles, so weuse6-cycles 2-factor flippable pair
Proofideas Step 1:Build a 2-factor in thegraph Step 2: Connect the cycles in the 2-factor to a singlecycle 2-factor Auxiliarygraph 1 1 4 4 3 3 2 2 8 5 8 6 5 6 flippable pairs (disjoint) 7 7
Proofideas Lemma 1:Ifisconnected, then has a Hamilton cycle. Lemma 2:If has distinctspanningtrees, then has distinct Hamilton cycles. 2-factor Auxiliarygraph 1 1 4 4 3 3 2 2 8 5 8 6 5 6 flippable pairs (disjoint) 7 7
The crucialreduction Provethat auxiliarygraph isconnected (has manyspanningtrees) Provethat middlelevelsgraph has a Hamilton cycle (many Hamilton cycles)
Algorithmicresults Theorem 3 [M., Nummenpalo 17; SODA]: Thereis an algorithmwhichfor a givenvertex of the middlelevelsgraphcomputes the nextone on a Hamilton cycle in time . • initialization time is and requiredspaceis • Remarks • C++ codeavailable on ourwebsite
BipartiteKnesergraphs • integer parameters and • vertices = all -element and -element subsets of • edges = • iff is the middle • Examples levelsgraph {2,3,4} {1,3,4} {1,2,4} {1,2,3} {1,2} {1,3} {2,3} • 1 {1} {2} {3} {4} {1} {2} {3}
IsHamiltonian? • Conjecture: • For all and the graph has Hamilton cycle. • raisedby[Simpson 91], and Roth (see[Gould 91], [Hurlbert 94]) • anotherinstance of Lovász‘ conjecture • middlelevelsconjecture
IsHamiltonian? • Knownresults: • has a Hamilton cycleif • [Shields, Savage 94] • [Chen 03] (followingearlierworkby[Simpson 94], [Hurlbert 94], [Chen 00]) • ???
Ourresults • Theorem 4 [M., Su 17; Combinatorica]: • For all and the graph has Hamilton cycle. Remark: simple inductionproof, assuming the validity of the middlelevelsconjecture
Generalized MLC 11...1 Conjecture[Savage 93], [Gregor, Škrekovski 10]:For any and ,the middlelevels ofhave a Hamilton cycle. levels • Known results: 00...0 [Gray 53] [El-Hashash, Hassan 01], [Locke, Stong 03] [Gregor, Škrekovski 10] ? [Gregor, Jäger, M.; ICALP 18] [M. 16] Middlelevelsconjecture
Generalized MLC 11...1 Conjecture[Savage 93], [Gregor, Škrekovski 10]:For any and ,the middlelevels ofhave a Hamilton cycle. levels 00...0 • Theorem 5 [Gregor, M. 17; STACS+TCS]: • For any and anyinterval not part of thisconjecture,the subgraph of withlevels in thisinterval has an `almost‘ Hamilton cycle, and wehavecorrespondingconstant-time generation algorithms.
Knesergraphs Knesergraphs • integer parameters and • vertices = all -element subsets of • edges = • iff Petersen graph Completegraph {1,2} {1} {2} {3,5} {3,4} {4,5} {2,3} {1,5} {4} {3} {2,4} {1,4} {2,5} {1,3}
Knesergraphs Knesergraphs • introducedby[Lovász 78] to proveKneser‘sconjecture • have long been conjectured to have Hamilton cycle, with one notable exception, the Petersen graph • sparsest and therefore hardest case is when • anotherinstance of Lovász‘ conjecture • odd graphs • degree , which is logarithmic in number of vertices Observation:Hamiltonicity of impliesHamiltonicity of .
Knesergraphs Knesergraphs • conjecture that is Hamiltonian for raised in the 70s[Meredith, Lloyd 72], [Biggs 79] • , [Balaban 72] • , [Meredith, Lloyd 72+73] • [Mather 76] • [Shields, Savage 04] has a Hamilton cycle if • [Heinrich, Wallis 78] • [B. Chen, Lih 87] • [Y. Chen 00], [Y. Chen, Füredi 02] • [Y. Chen 03] several other partial results [Johnson 04], [Johnson, Kierstead 04], etc.
Ourresults • Theorem 6 [M., Nummenpalo, Walczak 18; STOC]: • For all , the oddgraph has aHamilton cycle. • Theorem 7 [M., Nummenpalo, Walczak 18; STOC]: • For all and , the graph has aHamilton cycle. • Theorem 8 [M., Nummenpalo, Walczak 18; STOC]: • For all , the oddgraph has at least distinctHamilton cycles.
Proofideas Step 1:Build a 2-factor in thegraph 2-factor • based on Dyck words of length[M., Standke, Wiechert 17; EurJC] • all cycles have the same length • number of cycles = th Catalan number
Proofideas Step 1:Build a 2-factor in thegraph Step 2: Connect the cycles in the 2-factor to a singlecycle 2-factor flippable triple flipping 6-cycle flippable triples
Proofideas Step 1:Build a 2-factor in thegraph Step 2: Connect the cycles in the 2-factor to a singlecycle 2-factor auxiliaryhypergraph 2 2 1 1 5 5 4 4 3 3 7 7 6 6 flippable triples
Proofideas • translates problem into proving that has a loosespanning tree • connectivity is not enough; construct spanning tree directly 2-factor auxiliaryhypergraph 2 2 1 1 5 5 4 4 3 3 7 7 6 6 flippable triples
Resultssummary • Middle levels conjecture • [M., Weber 2012; Journal of CombinatorialTheorySeries A] • [M. 2016; Proceedings of the London Mathematical Society] • [Gregor, M., Nummenpalo 2018; Discrete Analysis] • [M., Nummenpalo 2015; ESA] • [M., Nummenpalo 2017; SODA] • [Gregor, M. 2017; STACS], [Gregor, M. 2017; Theoretical Computer Science] • (Bipartite) Kneser graphs • [M., Su 2017; Combinatorica] • [M., Standke, Wiechert 2018; European Journal of Combinatorics] • [M., Nummenpalo, Walczak 2018; STOC]