CS500 Homework #1, Spring 2008
Solution.
The graph on the left is colored so that each edge connected a light vertex
to a dark one. Formally, it is
bipartite
. Any Hamiltonian cycle would have to have an
equal number of light and dark vertices, but there are an odd number of vertices, and
one more dark one than light ones. More generally, no bipartite graph with an odd
number of vertices can be Hamiltonian.
For the Petersen graph, we use the fact that it is highly symmetric. In particular, all
the paths of four vertices are equivalent to each other. So, we can avoid most of the
search tree by choosing one of these paths and trying to complete it. Let’s take four
vertices going clockwise around the outside of the graph. The ﬁfth vertex in this cycle
has to be on the Hamiltonian cycle, so it has to connect to the ﬁrst or last vertex in
the path of four. Then there is no way to visit all ﬁve of the inner vertices.
Solution.
Euler’s argument, that every time we enter a vertex we have to leave it along
a diﬀerent edge, shows that every vertex must have an equal number of incoming and
outgoing edges.
Solution.
There are many ways to solve this problem. Here’s one. Suppose the
original graph
G
has
n
vertices. If the oracle says “yes” to the question of whether
G
is Hamiltonian, remove one of
G
’s edges, and ask her about the resulting graph
G
0
.
If she says “no”, remove a diﬀerent edge from
G
instead, until she says “yes” to
G
0
.
Then remove an edge from
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 Spring '08
 Moore,C
 Equivalence relation, equivalence class, equivalence classes, Hamiltonian path

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