Math 454/553, Fall 2008, Instructor: Pelsmajer
Solutions to Assignment #11, version 2. We assume that graphs are simple.
5.1.29 It’s easy to find a proper 4-coloring of G, so χ(G)≤4. Let u, v be any pair of antipodal vertices, and
let {x, y}=V(G)−N(u)− {u, v}; then let H=G−u−xy. Copies of K4−ein Hwould force a proper
3-coloring to color vwith the same color as xand y; thus His not 3-colorable. Since χ(G)≥χ(H)≥4, we
have χ(G) = χ(H) = 4. Hmust have a 4-critical subgraph H′. Any proper subgraph of Hhas a vertex of
degree less than 3, which cannot be 4-critical; therefore H′=H.
5.1.31 (I use “×” for graph product.) Suppose that fis a proper m-coloring of G. Let S={(u, f (u)) :
u∈V(G)}; then |S|=n(G), and with V(Km) = [m] we get S⊆V(G×Km). If (u, f (u)) and (v, f (v)) are
adjacent then since u6=v, we have uv ∈E(G) and f(u) = f(v), contradicting that fis a proper coloring of
G. Therefore Sis an independent set, so α(G×Km)≥ |S|=n(G).
Now suppose that α(G×Km)≥n(G). Then there is an independent set S⊆V(G×Km) of size at
least n(G). Any two vertices of G×Kmwith the same first coordinate are adjacent, so for each v∈V(G),
Shas at most one vertex of the form (v , u). Since |S| ≥ |V(G)|, there is exactly one such vertex u(for each
v), let f(v) = u. Observe that fis a proper coloring of G.
Alternatively: Suppose that Gis m-colorable. Then χ(G)≤m, so by Proposition 5.1.11, χ(G×Km) =
χ(G). By Proposition 5.1.7, χ(G×Km)≥n(G×Km)/α(G×Km), and since n(G×Km) = n(G)m, we get
α(G×Km)≥n(G)m/χ(G)≥n(G).
Now suppose that α(G×Km)≥n(G), and let Sbe a maximum size independent of G×Km. Let the
vertices of Kmbe labeled y1, . . . , ym, and for each i∈[m], let Sibe the vertices of Swith second coordinate
equal to yi. Then Sihas the form {(x, yi) : x∈Xi}for some set Xi⊆V(G). Since Siis an independent
set, Ximust be an independent set.
For distinct i, j ∈[m], yiand yjare adjacent in Km, so Shas no pair of vertices of the form (x, yi),(x, yj);
hence Xi∩Xj=∅. Since X1, . . . , Xmare pairwise disjoint subsets of Gwhose sizes sum to α(G×Km)≥n(G),
they partition V(G). Since each is an independent set, they can be used as color classes of an m-coloring of
G.
5.1.48 By Brooks’s Theorem, each component of Gis 3-colorable; hence Gis 3-colorable. Let fbe a proper
3-coloring, and let S1, S2, S3be its color classes (independent sets). We may assume that S1is smallest.
Pick a vertex v∈S1; if it has no edges to S2or S3, move it to that set; if vhas exactly one edge eto S2or
S3, delete eand then move vto that set. One of those cases must apply, since d(v)≤3. After the move,
each Siis still an independent set, and S1is smaller, so this can be repeated until S1is empty. At the end,
there are at least n− |S1| ≥ n−m/3 edges left, and the graph is bipartite with partite sets S2, S3.
5.2.7 If not, then there is some color i, such that for each vertex vof color i, there is some color cvthat does
not appear on any neighbor of v. Let Sbe the set of vertices of color i. Recolor each v∈Sby the color cv.
Since Sis an independent set, for each v∈S, no vertex of N(v) is recolored; therefore the new coloring is
still a proper coloring. It uses less than kcolors, a contradication.
5.2.9 Let G′be the graph created from G, with Uand was specified in class and in the book, and let
k=χ(G). The proof of Mycelski shows that χ(G′) = k+ 1, so we need to show that for any e∈E(G′),
G′−eis k-colorable. Sketch:
If e∈E(G), we can properly color G−ewith colors [k−1], give color kto all of U, and use any color
from [k−1] for w.
If e=uiw, we can k-color Gsuch that viis the only vertex with color 1, copy the colors from V(G) to
the corresponding vertices in U, then color wwith color 1.
Otherwise e=viujwhere vivj∈E(G). Then we do a (k−1)-coloring on G−vivj, and copy the colors
to corresponding vertices in U. This is a proper coloring except for the edges vivjand vjui, so we recolor vj
with a new color, and use that color for w, too.
