graph theory lecture notes 6
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3/18/2015 Graph Theory Lecture Notes 6
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Graph Theory Lecture Notes 6
Chromatic Polynomials
For a given graph G, the number of ways of coloring the vertices with x or fewer colors is denoted byP(G, x) and is called the chromatic polynomial of G (in terms of x).
Examples:
G = chain of length n1 (so there are n vertices)
P(G, x) = x(x1)n1
G = K4
P(G, x) = x(x1)(x2)(x3) = x(4)
G = Star5
P(G, x) = x(x1)5
G = C4 ( or Z4)
P(G, x) = x(x1)2 + x(x1)(x2)2 = x4 4x3 + 6x2 3x
P(Kn, x) = x(n) = x(x1)(x2) ...(xn+1)
P(In, x) = xn, where In is the graph on n isolated vertices (no edges, called the empty graph).
Theorem: If G decomposes into k connected components, G1, G2, ..., Gk then P(G, x) = P( G1, x)P(G2, x) ... P(Gk, x).
Def: Reduced graph, Condensed graph
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Theorem: (Fundamental Reduction Theorem) Let ß be an edge of G, G'ß the reduced graph at ß, and G"ßthe condensed graph at ß, then
P(G, x) = P(G'ß, x) P(G"ß, x).
Example: Consider C4, and let ß be any edge. The reduced graph at ß is a chain of length 3 and thecondensed graph at ß is a K33, so by the theorem:P(C4, x) = x(x1)3 x(x1)(x2) = (x4 3x3 + 3x2 x) (x3 3x2 + 2x) = x4 4x3 + 6x2 3x.
Note: You can use the reduction theorem to either add or remove edges.
Properties of the Chromatic Polynomial
Theorem:
a) degree of P(G,x) = n = |V| (the number of vertices of G).b) coefficient of xn is 1. (a monic polynomial)c) the constant term of P(G,x) is 0.d) either P(G,x) = xn or the sum of the coefficients is 0.
Pf: Let P(G, x) = apxp + ap1xp1 + ... + a1x + a0.
c) Consider P(G,0) = a0 = the number of ways to color G with no colors, but this can't be done, so a0 = 0.
d) Consider P(G, 1) = ap + ap1 + ... + a1 = the number of ways to color G with 1 color. If G contains anedge, then it can not be colored with 1 color so this sum must be 0. If G contains no edge, then it is theempty graph In, which has chromatic polynomial xn.
Review Induction Proofs.
a) and b). Suppose that G has n vertices. We will proceed by induction on the number of edges.
First, consider the base case, where G has only one edge. Then the n2 isolated vertices can be colored inany way, and the two vertices on the edge can be colored in x(x1) ways, so, P(G,x) = xn2(x)(x1) = xn xn1. So, the degree of P(G, x) in this case is n and the polynomial is monic.
We now impose the induction hypothesis and assume that the statement is true for all graphs with fewerthan k edges (strong form of induction). Assume that G has k edges and let ß be one of them. By thefundamental reduction theorem,
P(G, x) = P(G'ß, x) P(G"ß, x).Now, since G'ß has fewer than k edges, by the induction hypothesis, P(G'ß, x) is monic of degree n. SinceG"ß has fewer than k edges and only n1 vertices, P(G"ß, x) is monic of degree n1. Upon subtraction ofthese polynomials, we see that there is no term in P(G"ß, x) that can remove the xn term in P(G'ß, x), soP(G, x) will be monic of degree n.
By induction, the statement is true for all graphs with n vertices and any number of edges.
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Theorem: (Whitney, 1932): The powers of the chromatic polynomial are consecutive and thecoefficients alternate in sign.
Proof: We will again proceed by induction on the number of edges of G. As in the proof of the abovetheorem, the chromatic polynomial of a graph with n vertices and one edge is xn xn1, so our statementis true for such a graph.
Now, assume true for all graphs on n vertices with fewer than k edges. Let G be a graph with k edges andconsider the fundamental reduction theorem. Since G'ß and G"ß both have fewer than k edges, by theinduction hypothesis we may assume that their chromatic polynomials have the form:
P(G'ß, x) = xn an1xn1 + an2xn2 + ... andP(G"ß, x) = xn1 bn2xn2 + bn3xn3 + ... .
Now, by the fundamental reduction theorem,P(G, x) = P(G'ß, x) P(G"ß, x)
= (xn an1xn1 + an2xn2 + ... ) (xn1 bn2x2n + bn3xn3 + ...)= xn (an1 + 1)xn1 + (an2 + bn2)xn2 + ...
and so, the signs alternate and none of the coefficients are zero.
By induction, the statement is true for all graphs with n vertices and any number of edges.
Theorem: |an1| = |E| (the number of edges of G).
Proof: We will again proceed by induction on the number of edges of G.
As in the proofs of the above theorems, the chromatic polynomial of a graph with n vertices and one edgeis xn xn1, so our statement is true for such a graph, |1| = 1.
Now, assume true for all graphs on n vertices with fewer than k edges. Let G be a graph with k edges andconsider the fundamental reduction theorem. Since G'ß and G"ß both have fewer than k edges, by theinduction hypothesis we may assume that their chromatic polynomials have the form:
P(G'ß, x) = xn an1xn1 + an2xn2 + ... andP(G"ß, x) = xn1 bn2xn2 + bn3xn3 + ... ,
where an1 is the number of edges in G'ß and bn2 is the number of edges in G"ß.
Now, by the fundamental reduction theorem,
P(G, x) = P(G'ß, x) P(G"ß, x)= (xn an1xn1 + an2xn2 + ... ) (xn1 bn2xn2 + bn3xn3 + ...)
= xn (an1 + 1)xn1 + (an2 + bn2)xn2 + ...and so, since G'ß is obtained from G by removal of just one edge, an1 = k 1, so an1 + 1 = k, and thestatement is true.
By induction, the statement is true for all graphs with n vertices and any number of edges.
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Theorem: The smallest exponent in the chromatic polynomial is the number of connected components ofG.
Proof: If G has k connected components, G1, G2, ..., Gk then P(G, x) = P( G1, x)P(G2, x) ... P(Gk, x).Thus, the smallest exponent in the chromatic polynomial of G is the sum of the smallest exponentsappearing in each of the P(Gi, x)'s. As these can be no smaller than 1, we have that the smallest exponentof P(G, x) >= k.
To show that this exponent must be k, we need to show that the chromatic polynomial of any connectedgraph has an x1 term. We proceed by induction on the number of edges of G.
As in the proofs of the above theorems, the chromatic polynomial of a graph with n vertices and one edgeis xn xn1. If the graph is connected, then n = 2 and our statement is true.
Now, assume true for all connected graphs on n vertices with fewer than k edges. Let G be a connectedgraph with k edges and consider the fundamental reduction theorem. We have two cases to consider:
Case I: The removal of ß leaves G'ß connected.Since G'ß and G"ß both have fewer than k edges and are connected, by the induction hypothesis we mayassume that their chromatic polynomials have the form:
P(G'ß, x) = xn an1xn1 + an2xn2 + ... +/ a1x andP(G"ß, x) = xn1 bn2xn2 + bn3xn3 + ... /+ b1x.
Now, by the fundamental reduction theorem,P(G, x) = P(G'ß, x) P(G"ß, x)
= (xn an1xn1 + an2xn2 + ... +/ a1x) (xn1 bn2xn2 + bn3xn3 + ... /+ b1x)= xn (an1 + 1)xn1 + (an2 + bn2)xn2 + ... +/(a1 + b1)x
and so, the statement is true.
By induction, the statement is true for all graphs with n vertices and any number of edges.
Case II: The removal of ß disconnects G'ß .G'ß breaks up into two connected components, but G"ß remains connected and both have fewer than kedges, so, by the induction hypothesis we may assume that their chromatic polynomials have the form:
P(G'ß, x) = xn an1xn1 + an2xn2 + ... /+ a2x2 andP(G"ß, x) = xn1 bn2xn2 + bn3xn3 + ... /+ b1x.
Now, by the fundamental reduction theorem,P(G, x) = P(G'ß, x) P(G"ß, x)
= (xn an1xn1 + an2xn2 + .../+ a2x2) (xn1 bn2xn2 + bn3xn3 + ... /+ b1x)= xn (an1 + 1)xn1 + (an2 + bn2)xn2 + ... +/ b1x
and so, the statement is true.
By induction, the statement is true for all graphs with n vertices and any number of edges.
These properties do not characterize the chromatic polynomials amongst all the polynomials with integer
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coefficients. For example,P(x) = x4 4x3 + 3x2 is not the chromatic polynomial of any graph.