Chapter 11 Simple Graphs352
Inductive step:
We may assumeP.n/. To proveP.nC1/, letGnC 1 be a graph withnC 1 vertices
whose vertex degrees are allkor less. Also, supposeGnC 1 has a vertex,v, of
degree strictly less thank. Now we only need to prove thatGnC 1 isk-colorable.
To do this, first remove the vertexvto produce a graph,Gn, withnvertices. Letu
be a vertex that is adjacent tovinGnC 1. Removingvreduces the degree ofuby 1.
So inGn, vertexuhas degree strictly less thank. Since no edges were added, the
vertex degrees ofGnremaink. SoGnsatisfies the conditions of the induction
hypothesis,P.n/, and so we conclude thatGnisk-colorable.
Now ak-coloring ofGngives a coloring of all the vertices ofGnC 1 , except for
v. Sincevhas degree less thank, there will be fewer thankcolors assigned to
the nodes adjacent tov. So among thekpossible colors, there will be a color not
used to color these adjacent nodes, and this color can be assigned tovto form a
k-coloring ofGnC 1.
(c)With a slightly strengthened condition, the preceding proof of the False Claim
could be revised into a sound proof of the following Claim:
Claim.LetGbe a graph whose vertex degrees are allk. Ifhstatement inserted from belowi
has a vertex of degree strictly less thank, thenGisk-colorable.
Circle each of the statements below that could be inserted to make the proof correct.
Gis connected and
Ghas no vertex of degree zero and
Gdoes not contain a complete graph onkvertices and
every connected component ofG
some connected component ofGProblems for Section 11.9
Class Problems
Problem 11.25.
Then-dimensional hypercube,Hn, is a graph whose vertices are the binary strings
of lengthn. Two vertices are adjacent if and only if they differ in exactly 1 bit. For
example, inH 3 , vertices 111 and 011 are adjacent because they differ only in the
first bit, while vertices 101 and 011 are not adjacent because they differ at both
the first and second bits.
(a)Prove that it is impossible to find two spanning trees ofH 3 that do not share
some edge.