Proof of the middle levels conjecture Torsten Mütze The middle layer graph Consider the cube 11...1 level 3 111 level 2 110 101 011 level 1 100 010 001 level 0 000 00...0 Middle layer graph The middle layer graph Middle layer of 11100 11010 11001 10110 10101 10011 01110 01101 01011 00111 11000 10100 10010 10001 01100 01010 01001 00110 00101 00011 • bipartite, connected • number of vertices: • degree: The middle layer graph Middle layer of • • • • 11100 11010 11001 10110 10101 10011 01110 01101 01011 00111 11000 10100 10010 10001 01100 01010 01001 00110 00101 00011 bipartite, connected number of vertices: degree: automorphisms: bit permutation + inversion, The middle layer graph Middle layer of • • • • 11100 11010 11001 10110 10101 10011 01110 01101 01011 00111 11000 10100 10010 10001 01100 01010 01001 00110 00101 00011 bipartite, connected number of vertices: degree: automorphisms: bit permutation + inversion, The middle layer graph Middle layer of • • • • 11100 11010 11001 10110 10101 10011 01110 01101 01011 00111 11000 10100 10010 10001 01100 01010 01001 00110 00101 00011 bipartite, connected number of vertices: degree: automorphisms: bit permutation + inversion, • vertex-transitive The middle levels conjecture Conjecture: The middle layer graph contains a Hamilton cycle for every • probably first mentioned in [Havel 83], [Buck, Wiedemann 84] • also attributed to Dejter, ErdÅ‘s, Trotter [Kierstead, Trotter 88] and various others • exercise (!!!) in [Knuth 05] . The middle levels conjecture Conjecture: The middle layer graph contains a Hamilton cycle for every Motivation: • Gray codes • Conjecture [Lovász 70]: Every connected vertex-transitive graph contains a Hamilton path. . History of the conjecture Numerical evidence: The conjecture holds for all [Moews, Reid 99], [Shields, Savage 99], [Shields, Shields, Savage 09], [Shimada, Amano 11] History of the conjecture Asymptotic results: The middle layer graph contains a cycle of length • [Savage 93] • [Felsner, Trotter 95] • [Shields, Winkler 95] • [Johnson 04] History of the conjecture Other relaxations and partial results: [Kierstead, Trotter 88] [Duffus, Sands, Woodrow 88] [Dejter, Cordova, Quintana 88] [Duffus, Kierstead, Snevily 94] [Hurlbert 94] [Horák, Kaiser, Rosenfeld, Ryjácek 05] [Gregor, Škrekovski 10] … Our results Theorem 1: The middle layer graph contains a Hamilton cycle for every Theorem 2: The middle layer graph contains . different Hamilton cycles. Remarks: number of automorphisms is only , so Theorem 2 is not an immediate consequence of Theorem 1 Our results Theorem 1: The middle layer graph contains a Hamilton cycle for every Theorem 2: The middle layer graph contains Remarks: number of Hamilton cycles is at most so Theorem 2 is best possible . different Hamilton cycles. , Proof ideas Step 1: Build a 2-factor in the middle layer graph Step 2: Connect the cycles in the 2-factor to a single cycle Structure of the middle layer graph Structure of the middle layer graph Structure of the middle layer graph A Hamilton cycle Catalan numbers Structure of the middle layer graph A Hamilton cycle Structure of the middle layer graph A Hamilton cycle Structure of the middle layer graph A Hamilton cycle Structure of the middle layer graph A Hamilton cycle Step 1: Build a 2-factor Construction from [M., Weber 12] ??? isomorphism (bit permutation + inversion) Step 1: Build a 2-factor Construction from [M., Weber 12] 2-factor isomorphism (bit permutation + inversion) Step 1: Build a 2-factor Construction from [M., Weber 12] • parametrizing yields different 2-factors • essentially only one can be analyzed: = plane trees with edges 2-factor Fundamental problem: varying changes globally Step 2: Connect the cycles New ingredient: Flippable pairs 2-factor is a flippable pair, if there is a flipped pair such that Step 2: Connect the cycles New ingredient: Flippable pairs 2-factor is a flippable pair, if there is a flipped pair such that Step 2: Connect the cycles New ingredient: Flippable pairs 2-factor flippable pairs yield different 2-factors + very precise local control …we can construct many flippable pairs Step 2: Connect the cycles New ingredient: Flippable pairs 2-factor Auxiliary graph Step 2: Connect the cycles Lemma 1: If is connected, then the middle layer graph contains a Hamilton cycle. Lemma 2: If contains the middle layer graph contains 2-factor different spanning trees, then different Hamilton cycles. Auxiliary graph The crucial reduction Prove that middle layer graph contains a Hamilton cycle (many Hamilton cycles) Prove that is connected (has many spanning trees) Analysis of 2 leaves 6 leaves 5 leaves 4 leaves = plane trees with 3 leaves edges Analysis of 2 leaves 6 leaves 5 leaves 4 leaves = plane trees with 3 leaves edges Thank you!