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# Tag Info

## Hot answers tagged hamiltonian-path

2

Yes it exists and the existence of a linear ordering not only shows the existence of the topological oredering over the vertices but also that this ordering is unique. It is not hard to prove that the graph represents a linear ordering using the facts that the graph is complete and acyclic. That should be a good exercise for you (proving a relation is a ...

1

Here is what it boils down to. There are $n$ vertices $v_1,\ldots,v_{n-1},v$. For each $i$, there is an edge either from $v$ to $v_i$ or from $v_i$ to $v$. We need to show that one of the following holds: There is an edge from $v$ to $v_1$. There is $i < n-1$ such that there are edges from $v_i$ to $v$ and from $v$ to $v_{i+1}$. There is an edge from \$v_{...

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The class of graphs that contain a Hamiltonian cycle is known as Hamiltonian graphs. Being Hamiltonian doesn't really imply much: a wide range of well-known graph parameters (diameter, treewidth, chromatic number, ...) can still be unbounded. At least partly, this is the reason there aren't many non-trivial subclasses of Hamiltonian graphs known. As an ...

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