## Graph Theory Lecture Notes12

### Hamiltonian Chains and Paths

Def: Hamiltonian Chain, Hamiltonian Path, Hamiltonian Circuit, Hamiltonian Cycle

It is an NP-Complete problem to determine if a graph has a Hamiltonian chain or circuit.

Ex: Following the edges of a Dodecahedron.

This was an example due to Hamilton. He tried to market it as a puzzle. Each vertex of the dodecahedron was labeled with the name of a city, and one was to find a circuit using the edges of the dodecahedron which visited each city once and only once.

Ex: The Travelling Salesman Problem

This NP-complete problem requires finding a Hamiltonian circuit in a graph whose edge weights have a minimum sum.

### Sufficient Conditions for the Existence of a Hamiltonian Circuit

Theorem (Ore, 1960): Let G be a graph with n >= 3 vertices. If for each pair, x, y of non-adjacent vertices, deg(x) + deg(y) >= n, then G has a Hamiltonian circuit.

Ore's theorem gives a sufficient, but not necessary condition. Consider any circuit graph with 5 or more vertices.

Corollary (Dirac): If the degree of each vertex is >= n/2, then G has a Hamiltonian circuit.

Corollary: If G has a pair of non-adjacent vertices x and y, such that deg(x) + deg(y) >= n, then G is Hamiltonian if and only if G U {x,y} is Hamiltonian.

Def: Closure of G, c(G).

Corollary: G is Hamiltonian iff c(G) is Hamiltonian.

Theorem: Let G be a graph with at least 3 vertices. If c(G) is a complete graph then G is Hamiltonian.

There are corresponding theorems for the existence of a Hamiltonian cycle in digraphs.

### Tournaments

Def: Tournament

Theorem: (Rédei, 1934) Every tournament has a Hamiltonian path.

Pf: By induction on the number of vertices.

If there is more than one Hamiltonian path in a tournament, the vertices do not have a unique ranking.

Theorem: A tournament has a unique Hamiltonian path if and only if the tournament is transitive.

Pf: Suppose that D is a transitive tournament having two Hamiltonian paths, H1 and H2. Let u and v be the vertices on these paths where the two paths first differ. If H1 first goes to u, then v must follow it on this path, and in H2 , u must follow v. By transitivity in H1, there is an arc from u to v, while by transitivity in H2, there is an arc from v to u. This is a contradiction since D is a tournament.

Now suppose that a tournament D has a unique Hamiltonian path. Let this path be a1, a2, ..., an . Let am be the first vertex in this path that has an arc coming to it from a higher numbered vertex. Let ak be the largest numbered vertex that has an arc going to am. If m is not 1 and k is not n, then there is an arc from am-1 to am+1 (otherwise the minimality of m is contradicted), and an arc from am to ak+1 (otherwise the maximality of k is contradicted). Now the path, a1, ..., am-1, am+1, ..., ak, am, ak+1, ..., an is a second Hamiltonian path (contradiction). In a similar manner, we see that if m = 1 and k is not n, we will obtain a second path a2, ...., ak, a1, ak+1, ..., an. If m is not 1 and k = n, we get a second path, a1, ..., am-1, am+1, ..., an, am. And if, m = 1 and k = n, we get a2, ..., an, a1. So, in all cases, we obtain a contradiction. Therefore, there is no vertex am, i.e., there is an arc from ai to aj iff i < j. Now if there are arcs from ai to aj and from aj to ak, then i < j and j < k, so i < k and there is an arc from ai to ak, i.e., D is a transitive tournament.

### Topological Sorting

Def: Topological order, Acyclic digraph

Lemma: Every acyclic digraph has a vertex with no incoming arcs.

Pf: If D is an acyclic digraph, let D' be the digraph with the same vertices as D but whose arcs are all reversed. D' is acyclic since D was. Take a path of maximal length in D'. Consider the last vertex, v, in this path. There is no arc from v to any other vertex in the path since D' is acyclic. There is no arc from v to any other vertex since the path was of maximal length. Thus, v has out degree 0. In D, vertex v has no incoming arcs.

Theorem: A digraph D has a topological order if and only if D is acyclic.

Pf: A digraph with a topological order can clearly have no cycles since all arcs are oriented from smaller labeled vertices to larger labeled ones. On the other hand, suppose that D is acyclic. By the lemma, there is a vertex with no incoming arcs. Label it 1 and remove it and its arcs from D. The new digraph is still acyclic, so there is a vertex in it with no incoming arcs. Label it 2, remove it and its arcs. Continue in this way until all the vertices of D have been labeled. This labeling is a topological order.