notes-100406 - MATH 682 Notes Combinatorics and Graph...

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Unformatted text preview: MATH 682 Notes Combinatorics and Graph Theory II 1 Hamiltonian properties 1.1 Hamiltonian Cycles Last time we saw this generalization of Diracs result, which we shall prove now. Proposition 1 (Ore 60) . For a graph G with nonadjacent vertices u and v such that d ( u )+ d ( v ) | G | , it follows that G is Hamiltonian if and only if G + e is Hamiltonian, for e = { u,v } . Proof. In one direction this implication is trivial: if G is Hamiltonian, then clearly G + e is Hamil- tonian. To prove the other direction, let G + e contain a Hamiltonian cycle C . If the Hamiltonian cycle doesnt use e , then C lies in G , so that G is Hamiltonian as well. If, however, e is in C , we do not immediately see a Hamiltonian cycle in G : instead, we have a Hamiltonian path C- e in G . Let us denote the vertices of this path C- e by u = v v 1 v k = v in G . Since this path visits every vertex of G , we know that every neighbor of u and every neighbor of v is in the path, and since u 6 v , we know every neighbor of u and every neighbor of v is in { v 1 ,...,v k- 1 } . Let us consider the set A of neighbors of u , which clearly is a subset of { v 1 ,...,v k- 1 } of size d ( u ). Now consider the set B of successors of v s neighbors; that is to say, points v i such that v i- 1 v ; clearly this will be a subset of { v 1 ,...,v k } of size d ( v ). We now have two subsets of { v 1 ,...,v k } of sizes d ( u ) and d ( v ); since d ( u ) + d ( v ) | G | > k , these two sets must overlap, so there is some i such that both u v i and v i- 1 v . Using this value, we can produce a Hamiltonian cycle in G : u v 1 v i- 1 v v k- 1 v k- 2 v i v In consequence of this, we may add edges between any two non-adjacent vertices of total degree | G | without affecting Hamiltonicity. Definition 1. The Hamiltonian closure C ( G ) of a graph G is the graph resulting from adding edges between nonadjacent vertices of degree sum | G | until it is impossible to do so any further. A graph is closed if for every pair u and v of vertices, either d ( u ) + d ( v ) < | G | or u and v are adjacent. The definition above contains a hidden assumption: namely, that the procedure described has a unique result regardless of choice of vertices. Proposition 2. The result of the Hamiltonian-closure procedure is the same regardless of order of edge-addition choice. Proof. Consider a closure which adds specific edges e 1 ,...,e k . We will prove by induction that any order of edge-addition must add the edge e i for each i . For the base case, note that if e i = { u i ,v i } , then d G ( u 1 )+ d G ( v 1 ) | G | ; since adding edges can only increase degree, this inequality will remain true at every stage of edge-addition; thus it will become impossible to add more edges only after e 1 is added, so every edge-addition order will add e 1 eventually....
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notes-100406 - MATH 682 Notes Combinatorics and Graph...

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