Grin and berg it Mar. 24th Grinberg's Thm. (1976) Let G be a plane graph with n vertices and a hamiltonian cycle C. Let ai be the number of faces of G in the interior of C whose boundary contains exactly i edges. Let bi be the number of faces of G in the exterior of C whose boundary contains i edges. Then: ∑ (i − 2)(ai − bi) = 0 i=3 An example: An example: A hamiltonian cycle An example: 5 4 4 5 4 4 4 An example: 5 4 4 5 4 4 4 (3-2)(a3 - b3) + (4-2)(a4 - b4)+(5-2)(a5 - b5) = (3-2)(0-0) + (4-2)(4-1)+(5-2)(0-2) = 0 + 6 + -6 = 0 Diagonals A diagonal of a cycle is an edge of the graph joining to non-adjacent vertices of the cycle Proof: Let d denote the number of diagonals of G in the interior of C. The removal of any diagonal results in one less interior region. The removal of d diagonals results in one region and so there are d-1 interior faces. Hence: d + 1 = ∑ ai i=3 d = ∑ ai − 1 ( i=3 ) Proof continued: ∑ iai = 2d + n i=3 The sum of the number of ia = 2 a − 2 + n edges bounding each ∑ i ∑ i i=3 ( i=3 ) region will count each diagonal twice. ∑ iai = ∑ 2ai − 2 + n i=3 ( i=3 ) ∑ iai − ∑ 2ai = n − 2 i=3 i=3 ∑ (i − 2)ai = n − 2 = ∑ (i − 2)bi i=3 i=3 ∑ (i − 2)ai − ∑ (i − 2)bi = 0 i=3 i=3 ∑ (i − 2)(ai − bi) = 0 i=3 Herschel Graph Planar? Hamiltonian? Eulerian? Bipartite? Regular? Connected? Eight vertices of Three vertices of degree three degree four Herschel Graph Planar, Connected, not Regular, and not Eulerian Herschel Graph Bipartite and not Hamiltonian (11 vertices and then hw problem) 9 regions Each region has four sides (4-2)*(a-b) != 0 Herschel Graph Not Hamiltonian (Grinberg) The Tutte Graph 2 8 3 9 4 10 1 5 6 7 The Tutte Graph 8 9 10 If there is a hamiltonian cycle assume this edge is not in it. 1 2 7 3 6 There is a hamiltonian path 4 5 starting at vertex 5 and ending at vertex 7 Tutte's Fragment Add an edge to 5 7 make the path a cycle Tutte's Fragment One region of degree eight (infinite face) One region of degree three Two regions of degree four Five regions of degree five 5 7 Tutte's Fragment (3-2)(1-0) + (8-2)(0-1) + (4-2)(a4-b4) + (5-2)(a5-b5) = 0 1 - 6 + (4-2)(a4-b4) + (5-2)(a5-b5) = 0 2(a4-b4) + 3(a5-b5) = 5 2(0-2) + 3(a5-b5) = 5 2(2-0) + 3(a5-b5) = 5 2(1-1) + 3(a5-b5) = 5 3(a5-b5) = 9 3(a5-b5) = 1 3(a5-b5) = 5 a5-b5 = 3 and a5+b5 = 5 1 is not divisible by 3 5 is not divisible by 3 a5 = 4 Both regions with four sides are exterior! Tutte's Fragment Both edges must Each edge of cycle must be on cycle be on both an interior and exterior region So at least one of the four sided regions must be on the interior, a contradiction. 5 7 Must be opposite of infinite face so interior.