Coloring Warm-Up
A graph is 2-colorable iff it has no
odd length cycles
1: If G has an odd-length cycle then G is not 2colorable
Proof: Let v0, …, v2n+1 be an odd length cycle (n≥1).
Suppose G can be colored with two colors red and
blue. Without loss of generality [WLOG] color v0
red.
Then v1 must be colored blue since v0 and v1 are
adjacent. Then v2 must be colored red, and in
general, all even-indexed vertices must be colored
red and all odd-indexed vertices must be colored
blue.
But v0=v2n+1 must get both colors, contradiction.
A graph is 2-colorable iff it has no
odd length cycles
2: If G has no odd-length cycles then G is 2colorable.
How to define the coloring?
Helpful Def. The distance of one vertex from
another is the length of the shortest path
between them.
Proof: Pick any connected component.
Pick an arbitrary vertex v. Color all vertices at
even distance from v with one color and all
vertices at odd distance from v with the other
color.
• It suffices to prove that for every n, there are
no nodes in G that (a) are distance n apart
but (b) also have a path between them of
length m, where one of m, n is even and the
other is odd.
• Suppose there were. By WOP there is a least
such n, and such a pair of nodes v, w.
• Then the two paths have no vertices in
common except the beginning and end.
• But the round trip out on one path and
returning on the other path is a cycle of
length m+n, which is odd, contradiction. QED