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Solutions to assignment #11 in math 454/553 for the fall 2008 semester, focusing on graph coloring. The solutions include proofs that the chromatic number χ(g) of a graph g is at most 4 and that the product of two graphs g and km has an independent set of size at least n(g). The document also discusses the relationship between graph coloring and the graph product.
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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 K 4 − e in H would force a proper 3-coloring to color v with the same color as x and y; thus H is not 3-colorable. Since χ(G) ≥ χ(H) ≥ 4, we have χ(G) = χ(H) = 4. H must have a 4-critical subgraph H′. Any proper subgraph of H has 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 f is 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 u 6 = v, we have uv ∈ E(G) and f (u) = f (v), contradicting that f is a proper coloring of G. Therefore S is 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 × Km with the same first coordinate are adjacent, so for each v ∈ V (G), S has 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 f is a proper coloring of G. Alternatively: Suppose that G is 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 S be a maximum size independent of G × Km. Let the vertices of Km be labeled y 1 ,... , ym, and for each i ∈ [m], let Si be the vertices of S with second coordinate equal to yi. Then Si has the form {(x, yi) : x ∈ Xi} for some set Xi ⊆ V (G). Since Si is an independent set, Xi must be an independent set. For distinct i, j ∈ [m], yi and yj are adjacent in Km, so S has no pair of vertices of the form (x, yi), (x, yj ); hence Xi∩Xj = ∅. Since X 1 ,... , Xm are pairwise disjoint subsets of G whose 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 G is 3-colorable; hence G is 3-colorable. Let f be a proper 3-coloring, and let S 1 , S 2 , S 3 be its color classes (independent sets). We may assume that S 1 is smallest. Pick a vertex v ∈ S 1 ; if it has no edges to S 2 or S 3 , move it to that set; if v has exactly one edge e to S 2 or S 3 , delete e and then move v to that set. One of those cases must apply, since d(v) ≤ 3. After the move, each Si is still an independent set, and S 1 is smaller, so this can be repeated until S 1 is empty. At the end, there are at least n − |S 1 | ≥ n − m/3 edges left, and the graph is bipartite with partite sets S 2 , S 3.
5.2.7 If not, then there is some color i, such that for each vertex v of color i, there is some color cv that does not appear on any neighbor of v. Let S be the set of vertices of color i. Recolor each v ∈ S by the color cv. Since S is 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 k colors, a contradication.
5.2.9 Let G′^ be the graph created from G, with U and w as 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′^ − e is k-colorable. Sketch: If e ∈ E(G), we can properly color G − e with colors [k − 1], give color k to all of U , and use any color from [k − 1] for w. If e = uiw, we can k-color G such that vi is the only vertex with color 1, copy the colors from V (G) to the corresponding vertices in U , then color w with color 1. Otherwise e = viuj where 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 vivj and vj ui, so we recolor vj with a new color, and use that color for w, too.
5.2.15 Let G be a triangle-free graph. If ∆(G) < 2
n, then we can apply Brooks’s Theorem unless G is K 2 or an odd cycle, and in each case we have χ(G) ≤ 2
n. So we may assume that ∆(G) ≥ 2
n, and let v be a vertex with d(v) ≥ 2
n. N (v) is an independent set and G − N (v) is triangle free, so we can apply induction to color with at most 1 + 2
n − |N (v)| colors. 2
n − 2
n ≤ 2
n − 1 is true (square both sides), so this suffices. Alternatively: Repeatedly choose a vertex v of maximum degree, give N (x) a single color, then delete N (x); repeat this ⌊
n⌋ times, and let H be the subgraph that remains. (Each N (x) is an independent set since G is triangle-free.) For each of the first ⌊
n⌋ steps, we pick a vertex x of degree at least ∆(H). Therefore n(H) ≤ n − ∆(H)⌊
n⌋ > n − ∆(H)(
n − 1). If ∆(H) ≥
n then n(H) ≤
n, so H is ⌊
n⌋-colorable. H is also ⌊
n⌋-colorable if ∆(H) ≤ ⌊
n⌋ − 1 by greedy coloring, or if ∆(H) = ⌊
n⌋ and Brooks’s Theorem applies. A proper ⌊
n⌋-coloring of H suffices, so we may assume that ∆(H) = ⌊
n⌋ and H is an odd cycle or n(H) ≤ 2. So ∆(H) ≤ 2, so n < 9, and these cases can be handled somehow...
5.2.25 (a) Suppose that G is simple and
v∈V (G)
(d(v) 2
(m − 1)
(n 2
. The first quantity is the number of subgraphs isomorphic to P 3 (counted according to their center vertices). By counting P 3 -subgraphs according to their pairs of endpoints, this must equal the sum, over all pairs of distinct vertices {u, v}, of the number of common neighbors of u and v. If G does not contain Km, 2 as a subgraph, then every pair of vertices has at most m − 1 common neighbors. so the latter sum is at most (m − 1)
(n 2
(b) Note that 2
v∈V (G)
(d(v) 2
v∈V (G) d(v)^2 −^
v∈V (G) d(v). Balancing the degrees (by subtract- ing ∑ ǫ from d(u) and adding ǫ to d(v) when d(u) < d(v)) can only decrease the sum of their squares, so
v∈V (G) d(v)^2 ≥^
v∈V (G)(2e/n)^2 , which equals^ n(4e^2 /n^2 ) = 4e^2 /n.^ Also^
∑^ v∈V^ (G)^ d(v) = 2e.^ Therefore v∈V (G)
(d(v) 2
≥ 12 (4e^2 /n − 2 e) = e(2e/n − 1). (c) If e ≥ 12 (m − 1)^1 /^2 n^3 /^2 + n/4, then 2e/n ≥ (m − 1)^1 /^2 n^1 /^2 + 1/2 and 2e/n − 1 = (m − 1)^1 /^2 n^1 /^2 − 1 /2, so their product is (m−1)n− 1 /4. Therefore e(2e/n−1) ≥ (m−1)n^2 / 2 −n/8, which is more than (m−1)
(n 2
if and only if −n/ 8 > (m − 1)(−n/2), or m − 1 > 1 /4, which is true if m ≥ 2. Then parts (a) and (b) finish the argument. If m = 1, then 12 (m − 1)^1 /^2 n^3 /^2 + n/4 = n/4, and having more than n/4 edges isn’t even enough to guarantee a matching, let alone a copy of K 1 , 2 ; that is, the statement is false for m = 1. (d) Fix any n points in the plane, and let G be the graph where points are vertices and two vertices are adjacent when their distance in the plane is exactly 1. Note that two vertices can have at most two common neighbors. Then apply part (c) with m = 3.