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Chromatic Polynomial and Its Generating Function

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Chromatic Polynomial and Its Generating Function Chromatic Polynomial and its Generating Function An Ran Chen Supervised by Dr Anthony Licata Australian National University 1 1 Introduction The chromatic polynomial was first mentioned in Birkhoff[1] in 1912 in an attempt to prove the four-colour theorem. Although he did not prove the theorem, the chromatic polynomial became an object of interest and has connections with statistical mechanics and combinatorics. This report is structured into sections: section 2 will introduce the concept of the chromatic polynomial and section 3 will introduce generating functions. In section 4, I will investigate some basic properties of the generating function of the chromatic polynomial. Then in section 5, I will describe this generating function's link with combinatorics, namely, strict order polynomials. Section 6 is the concluding section which elaborates on the implications of this association and highlights some problems for the future. 2 The Chromatic Polynomial 2.1 Defining the Polynomial Let G be a simple graph, then the chromatic polynomial of the graph G, denoted fG(λ), is a function that given λ, counts the number of λ-colourings of G.A λ-colouring of G is a function g : V (G) ! f1; 2; 3; 4::; λg such that for any two vertices v1, v2 of G that share an edge, g(v1) 6= g(v2). Note that it is not neccessary to use all λ colours. Two colourings f and g are distinct if there exists a vertex v of G such that f(v) 6= g(v). Note that we have not yet justified why this chromatic counting function is in fact a polynomial. To prove this claim, we need to use the deletion-contraction relation: Theorem 2.1 ([2], Thm 1.3.1). Let x,y be two non-adjacent vertices of graph G, then: fG(λ) = fG+xy(λ) + fG=xy(λ) where G + xy is the graph obtained by adding the edge xy to G, and G=xy is the graph obtained by contracting the vertices x; y of G. Proof. Given a colouring c of G, either c(x) = c(y) or c(x) 6= c(y). The first case correpsonds to a colouring of G=xy where we let the colour of the contracted vertex xy be c(x) = c(y). The second case corresponds to a colouring of G + xy, since x and y will have different colours in G + xy as they are connected. We can also alternatively state this relation by setting F = G + xy: Corollary 2.2. fF (λ) = fF −xy(λ) − fF=xy(λ) Observe that the chromatic polynomial for the empty graph with n vertices and no edges n (denoted On) has the chromatic polynomial λ as we have λ colours to choose from for each vertex. Thus, given an arbitrary graph G we can keep applying the deletion-contraction relation until we reduce fG(λ) to a sum of fOi (λ). Thus fG(λ) is a polynomial. 2 2.2 Examples Here are some examples of the chromatic polynomials of common graphs. n 1. fOn (λ) = λ where On is the empty graph with n vertices as there are λ choices of colours for every vertex. 2. fKn (λ) = (λ)n = (λ)(λ − 1):::(λ − n + 1) where Kn is the complete graph with n vertices. n−1 3. Let G be a tree of order n, then fG(λ) = λ(λ − 1) . Proof by induction. Statement is trivially true when n = 1. Now consider the tree with a leaf removed, by the induction hypothesis there are λ(λ − 1)n−2 colourings. Now add the leaf and colour that new vertex. Since the only colour forbidden is that of its parent, there are λ − 1 ways of colouring it. n n 4. For n ≥ 3, fCn (λ) = (λ − 1) + (−1) (λ − 1). where Cn is the cycle of size n. Proof by induction using the deletion-contraction relation. Base case: fC3 (λ) = (λ)(λ−1)(λ− 3 3 2) = (λ − 1) + (−1) (λ − 1). By the deletion-contraction relation, fCn (λ) = fCn−e(λ) − n−1 n−1 n−1 fCn−1 (λ) By the induction hypothesis, = λ(λ − 1) − (λ − 1) − (−1) (λ − 1) = (λ − 1)n + (−1)n(λ − 1) 2.3 Expansion of the Chromatic Polynomial Using Spanning Subgraphs We can also compute the coefficients of the chromatic polynomial in terms of spanning subgraphs, that is, subgraphs that contain all vertices of the original graph. Theorem 2.3. Let G be a graph with n vertices and m edges. n m X X r p fG(λ) = ( (−1) N(p; r))λ p=1 r=0 where N(p; r) is the number of spanning subgraphs of G with p components and r edges. Proof. This proof is from [2] section 2.2. Proof by counting argument. Let V (G) = fV1;V2; :::; Vng and let a λ-map of G be a function which maps from V (G) to f1; 2; ::; λg. Then there are λn such maps. For any fi1; i2; :::; irg ⊂ f1; 2; ::; mg, define M(qi1 qi2 :::qir ) to count the number of λ-maps that satisfies the property that both vertices of the edge eik have the same colour for all k 2 [1; r]. Therefore fG(λ) counts the number of λ-maps that does not satisfy the criterion for M(qi1 qi2 :::qir ) for any choice of i1; :::; ir. By the principle of inclusion and exclusion: m n X X m−1 λ =fG(λ) + M(qi) − M(qiqj) + ::: + (−1) M(q1q2:::qm) i=1 1≤i≤j≤m m n X X m fG(λ) =λ − M(qi) + M(qiqj) − ::: + (−1) M(q1q2:::qm) i=1 1≤i≤j≤m 3 Now for a λ-map that satisfies M(qi1 qi2 :::qir ), consider the subgraph H of G consisting of edges fei1 ; ei2 ; :::; eir g, suppose it has p connected components, p 2 [1; n]. Note that each component is coloured according to the λ-map so each component has the same colour. Then H is a spanning subgraph with r edges and p components and there are λp ways of colouring the components. So: n X X p M(qi1 qi2 :::qir ) = N(p; r)λ 1≤i1≤i2≤:::≤ir ≤m p=1 And so, n n n n X p X p m X p fG(λ) = λ − N(p; 1)λ + N(p; 2)λ − ::: + (−1) N(p; m)λ p=1 p=1 p=1 Since N(n; 0) = 1 and N(p; 0) = 0 for 1 ≤ p ≤ n − 1 m n X X r p fG(λ) = ( (−1) N(p; r))λ r=1 p=0 Since we are dealing with finite sums, we can swap the summations and so we are done. 3 Generating Function 3.1 Definition Let f be a counting function, say f : N ! C, then we can `encode' f by the formal power series below: X f(n)xn n≥0 This is the (ordinary) generating function of the sequence ff(n)gn2N. By considering these objects as formal power series we are not concerned with the convergence or divergence of the power series at particular values, this allows us to add and multiply these objects. 3.2 Rational Generating Function To investigate what the generating function of the chromatic polynomial might look like, we first consider the generating function of a recursive sequence. Theorem 3.1 ([3]). Consider a sequence fangn2N given by the linear reccurrence relation of order k: an+k = c1an+k−1 + c2an+k−2 + ::: + ckan (1) where c1; c2; ::; ck are constant coefficients, and a0; a1; ::; ak−1 are given a priori. Then the generating function of the sequence is rational, that is: 4 1 X P (x) a xn = (2) n Q(x) n=0 where P is a polynomial of degree at most k − 1, and k X k Q(x) = 1 − ckx (3) n=1 Proof from [3]. We first show a lemma. k P1 n Lemma 3.2. For a polynomial c1x + ::: + ckx and a power series n=0 anx we have: 1 1 k X n X n+k (c1x + :: + ckx ) anx = R + (c1an+k−1 + ::: + ckan)x n=0 n=0 where R is some polynomial. Proof of lemma. Expanding the LHS of the equation and manipulating the power series we get (details on [3] pg.2): k−2 k−3 X n+1 X n+2 k−1 R = c1 anx + c2 anx + ::: + ck−1a0x n=0 n=0 Now, by the lemma and equation (1) we get: 1 1 X n X n+k (1 − Q) anx = R + (c1an+k−1 + ::: + ckan)x n=0 n=0 1 X n+k = R + an+kx n=0 1 X n = R + anx n=k k−1 k−1 1 X n X n X n = R − anx + anx + anx n=0 n=0 n=k k−1 1 X n X n = (R − anx ) + anx n=0 n=0 Pk−1 n Let P = −R + n=0 anx . Since R has degree at most k − 1, P has degree at most k − 1. Also: 1 1 X n X n (1 − Q) anx = −P + anx n=0 n=0 which yields equation (2). 5 Now consider the case when f is a polynomial. For any polynomial f(n) of degree d, we have the following relation: d+1 X d + 1 (−1)d+1−i f(n + i) = 0 i i=0 Alternatively, d X d + 1 f(n + d + 1) = − (−1)d+1−i f(x + i) (4) i i=0 Thus we know f(n) can be generated by a recursive relation of order d + 1 with the following coefficients.
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