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Introduction to Graph Theory Lecture 11: Eulerian and Hamiltonian Graphs

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Hamiltonicity As opposed to Eulerian graphs, we now want to make sure we visit each vertex just once. The famous application for such a graph is the traveling salesman problem Starting from the home city, the salesman should visit each city to show his merchandise.

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Terminology A graph with a spanning path is called traceable. A graph with a spanning cycle is called hamiltonian. The spanning cycle is called hamiltonian cycle. A traceable non-hamiltonian graph is called semi-hamiltonian Having only hamiltonian path, not cycle

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Observations Unlike eulerian graphs, it is often hard to determine if a graph is hamiltonian. Some necessary conditions for a graph to be hamiltonian If graph G is bipartite, but not equitable, then not hamiltonian. The graph must be 2-connected (no cut vertex) Can we see why?

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Hamiltonicity of Hypercubes Hamiltonicity of a constructed graph can be derived from the component graphs. : consisting of 2 copies of G (G and G’). If G is traceable, then is hamiltonian. Taking the spanning path P in G with endpoint x and y Taking its mirror image path P’ in G’ with endpoint x’ and y’ Adding xx’ and yy’ to the path is hamiltonian, so are

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Sufficient Conditions There exist many such conditions that guarantee that a given graph is hamiltonian. We just study one of them (popular and easier) Theorem 5.2: If G is a graph of order such that, then G is hamiltonian.

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Proof of Theorem 5.2 If n=3, then implies that, which is hamiltonian. Now suppose n>3 and let be a path of maximum length in G Observation1: Since, P must contain at least vertices. Reason: The only possible additional adjacencies of v 1 and v k lie within the set Observation2: There must be some vertex such that and are adjacent while and are adjacent.

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(cont) Now we know that G contains a cycle Suppose If there is a vertex u not in C. Using observation1, u must be adjacent to at least one vertex on C => a longer path than P. Therefore impossible. Therefore C contains all vertices, and thus a hamiltonian cycle.

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Some Remarks The converse of Theorem is not true. If G is hamiltonian, it needs not have the property Take for as an example Can you tell the difference between necessary and sufficient conditions?

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Homogeneous Connected Given any vertex, there exists a hamiltonian path starting at v. Hamiltonian graph is homogeneous connected. Simply by deleting an edge of the spanning cycle incident with v. Please note that the reverse is not always true. E.g Peterson graph

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Hamiltonian Connected Given any pair of vertices x and y, there is an x-y halmiltonian path for G For example, If G is hamiltonian connected, it is homogenous connected If order is at least 3, a hamiltonian-connected => hamiltonian Proof?

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Hypercubes We know that are hamiltonian for But hypercubes are not hamiltonian connected Picking any two vertices x and y of the same color. in the bipartite hypercube. has even order, thus the hamiltonian path is even order with end vertices of opposite colors Therefore, there is no x-y hamiltonian path. But are hamilton laceable.

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Hamilton Laceble An equitable bipartite graph G is hamilton laceble if there exists an x-y spanning path for G for any two oppositely colored vertices x and y. Theorem 5.4: is hamilton laceable Proof by induction (some typos in the textbook)

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Proof of Theorem 5.4 Basic case: true for n=1 or 2. Hypothesis: Assume true for n=k (Q k ) Prove that the theorem is true for n=k+1 (Q k+1 ) Let r and b be vertices of Q k+1, s.t. r is red and b is blue. Split Q k+1 into two copies of Q k, R and B, such that We know that R is hamiltonian and bipartite, so there exists a cycle spanning C containing r and its neighbor x. Let x’ be a neighbor of x in B. Since x is blue, x’ must be red. By the hypothesis, we can find an b-x’ spanning path L in B. The required r-b spanning path for Q k+1 is M, xx’,L, where M is the spanning path from r to x on C.

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Application: The Traveling Salesman We want to find the hamiltonian cycle of minimum weight --- least cost b a f c e d

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Some Remarks In applications the problems are much more overwhelming There is no efficient method for solving the traveling salesman problem There are approximation techniques for this problem The problem is intrinsically difficult to solve.

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