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THE CHROMATIC POLYNOMIAL OF A GRAPH
A A ADAM
A thesis submitted to the Faculty of Science, University of the Witwatersrand. infulfilment of the requirements for the degree of Doctor of Philosophy,
Johannesburg 1990.
iii
ABSTRACT
In this work we study the chromatic polynomial P( G, x) of a graph G of order p inthe form Lf;~tt{Cp_i where Tp_i = x(x _l)p-i-l is the chromatic polynomial of a treeof order p,
Fistly we express the chromatic polynomials of some graphs in tree form. We thenstudy a special product that comes natural and is useful in the caculation of somechromatic polynomials. Next we use the tree form to study the chromatic polynomialof a graph obtained from a forest (tree) by "blowing up" or "replacing" the verticesof the forest (tree) by a graph. Then we give explicit expressions, in terms of inducedsubgraphs, for the first five coefficients of the chromatic polynomial of a connectedgraph. In the case of higher order graphs we develop some useful computationaltechniques to obtain some higher order coefficients. In the process we obtain someuseful combinatorial identities, some of which are new. We discuss in detail theapplication of these combinatorial identities to some families of graphs.We also discusspairs of graphs that are chromatically equivalent and graphs that are chromaticallyunique with special emphasis on wheels.
In conclusion, we mention some open questions and conjectures.
iv
DECL.A.RATION
I declare that the contents of this thesis are original except where due references havebeen made. It has not been submitted before for any degree to any other institution.
A A Adam
v
DEDICATION
To my Family and my Friends
vi
ACKNO'VLEDGEMENTS
Professor Izak Broere's insight and knowledge of graph theory has been a source ofinspiration for this thesis. He has been my mentor and guide throughout the courseof this research. I am immensely grateful to him for his critical appraisal of the thesisand for his constant encouragement.
I thank Dr Roland Walker for being a supportive supervisor.
I am greatly indebted to Fazal Mahomed for his invaluable technical skills and exper-tise, but more than that, for his friendship.
I am grateful to all my colleagues for their interest and encouragement.
To Zeenar, Zaheer and Zulekha, lowe more than I can express in words. Their lovemakes everything worthwhile.
vii
CONTENTS
Chapter
INTRODUCTION 1
1 PRELTh1INARIES
1.1 Introduction
1.2 Notation and Terminology
1.3 The chromatic polynomial of a graph
1.4 Motivation
1.5 Basic Properties of Chromatic Polynomials
3
3
7
811
2 OPERATIONS ON GRAPHS AND CHROMATIC POLYNOMIALS
2.1 Introduction
2.2 Computational Techniques
2.3 Operation on graphs and the chromatic polynomials
2.4 Combinatorial Formulae
19192532
3 CHRO~IATIC POLYNm.nALS OF GRAPHS OBTAINED BY REPLACINGA VERTEX WITH: A GRAPH
3.1 Introduction
3.2 Replacing the vertices of a tree by graphs
3.3 Replacing the vertices of a forest by graphs
37
37
42
4 THE r;OEFFICIENTS 01;i'THE CHRO}'fATIC POLYNOMIAL
4.1 Introduction 51
4.2 Interpretation of the coefficients
4.3 Relating Coefficients
5260
viii
4.4 Higher Order c.J', '( ;gt.s 65
·1.5The Coefficients a!1U Connr ivity 78
4.6 A New Combinatorial IdetltlLJ 81
4.7 The coefficients of the chromatic polynomial of a cascade of graphs 85
5 CHROMATICALLY EQUIVALENT GRAPHS
5.1 Introduction
5.2 Chromatic Equivalence
5.3 Chordal Graphs
5.4 Chromatically Unique Graphs
5.5 Wheels
9999
105
111116
COXCLUSION
REFEREXCES
APPENDIX
123127130
ABSTRACT
In this work we study the chromatic polynomial P(G, x) of a graph G of order pin
the form I:f;; ajTp-i where Tp-i = x(x _l)p-i-l is the chromatic polynomial of
a tree of order p.
Fistly we express the chromatic polynomials of some graphs in tree form. We then
study a special product that comes natural and is useful in the caculation of some
chromatic polynomials. Next we use the tree form to sn« 1" the chromatic polynomial
of a graph obtained from a forest (tree) by "blowing up" or "replacing" the vertices
of the forest (tree) by a graph. Then we give explicit expressions, in terms of
induced subgraphs, for the first five coefficients of the chromatic polynomial of
a connected graph. In the case of higher order graphs we develop some useful
computational techniques to obtain some higher order coefficients. In the process
we obtain some useful combinatorial identities, some of which are new. We discuss in
detail the application of these combinatorial identities to some families of graphs.We
also discuss pairs of graphs that, are chromatically equivalent and graphs that are
chromatically unique with special emphasis on wheels.
In conclusion, we mention some open questions and conjectures.
INTRODUCTION
In this thesis we write the chromatic polynoruial of a graph ill tf'rms vi the chromatic
polynomial of a tree. We now introduce the concept of colouring the vertices of a
graph which is the underlying theme of this thesis. Let a graph G, , natural number
a; and any set S with z elements (called colours) be g5ven. We define an z-colounng
of G as a mapping of the vertex-set V(G} of G into "he colour-set S, subject to
the restriction that each edge of G must join vertices of two different colours. \'Fe
denote the number of z-colourings of G by P(G,x). In section 1.5 we prove that
P(G,x) is i:l.deed a polynomial in x and henceforth we refer to it as the chromatic
polynomial cf a graph G.
In more detail the outline of this work is as follows:
Chapter 1 deals with preliminaries. In section 1.2 the notation and terminology in
gi "ph theory is given. In section 1.3 we give some intuitive results on the chro-
matic polynomials of complete graphs and trees. In section 1.4 we prove Whitney's
Reduction Theorem and give a motivation for studying the cl.romatic polynomial
of a graph in tenus of chromatic polynomials of trees; Wf' conclude this section by
illustrating where Stirling numbers of the first and second kind ..re used. In section
1.5 we present several easily proved results on chromatic polynomials in terms of
chromatic polynomials of trees.
C'rapter 2 deals with operations on chromatic polynomials. In section 2.2 we incro-
:1 ~~.i!the special operations 0, ® and ® and we use them to simplify calculations
ir. computing some chrometic polynomials. We also use th(· special operation @ in
section 2.3 to go from normal form to tree form for some special f~l.i of graphs. In
1
section 2.4 we illustrate combinatorial identities which make use of Stirling numbers
of the fIrst and second kind.
Chapter 3 is devoted ';0 replacing vertices of a tree or forest with a connected graph.
In section 3.2 \7;; describe the chromatic polynomial of a graph obtained by replacing
vertices of a tree with connected graphs using the special operations 0 and ®. We
generalize the above concept to forests in section 3.3.
Chapter 4 is concerned with the coefficients of the chromatic polynomial, In section
4.2 we begin by interpreting the first five coefficients of a chromatic polynomial in
tree form; we then relate these coefficients to that given in normal form in section
4.3. lNe use these five coefficients to obtain some higher order coefficients in section
4.4. In section 4.5 we discuss the concept of connectivity and how the coefficients of
a chromatic polynomial behave in tree form. 'rVethen present a new combinatorial
identity in section 4.6 and use it in section 4.7 to interpret the coefficients of the
chromatic polynomial of an rn-gon-tree,
In Chapter 5 we discuss the chromatic equivalence of pairs of graphs. In section
5.~!we construct pairs of graphs that are chromatically equivalent. The theory of
chordal graphs and their chromatic polynomials are discussed in section 5.3. In
section 5.4 we describe some family cf graphs that are chromatically unique and in
section 5.5 we discuss the chromaticity of wheels.
Finally, we conclude with a chapter describing some open problems and conjectures.
2
CHAPTER 1PRELIMINARIES
1.1 Intrcductlon
The basic notions from graph theory used are all defined in section 1.2 - this makes
this thesis self-contained. \Ve continue to define and ill' .b:ate chromatic polynomials
of graphs, the main issue at stake in this thesis, in section 1.3. In section 1.4 we give
a motivation for studying the chromatic polynomial of a graph in terms of chromatic
polynomials of trees, and in section 1.5 we discuss some elementary properties of
chromatic polynomials.
1.2 Notation and Terminology
Most of the definitions in this section are taken from [4].
A graph G is a finite nonempty set V(O) together with a (possible empty) set B(G)
(disjoint from V( G» of two-element subsets of (distinct) elements of V( G). The set
V(G) is called the »eriez set of G and its elements are called vertices, while B(G) is
called the ed,qe set of G and its elements are called edges, By an element of a gl'aph
we shall mean a vertex or an edge.
The edge e = [u, v} is said to join u and v, and wiDhenceforth be denoted by uv or
V'1<.. If e = uv is an edge of a graph G, then u and v are adjacent vertices, while the
edge e is incident with the two vertices 1/ and v. Furthermore if e and f are distinct
edges of G incident with a common vertex, then e and f arc adjacent edges.
A graph G1 is iSJmarphic to a graph Gz if there exists a one-to-one mapping rf> from
'V(Gl) onto V(G2) such that two vertices Ul and Vi are adjacent in G1 if and only
3
if ¢(Ul) and ¢(Vl) ate adjacent in G2; it is written as G1 ~ G'}..
A graph H is a s1tbgraplr of a graph G if VCH) ~ 'V(O) and B(H) ~ EeG). If
V(H) C \f(G) or E(H) C E(G) then H is a proper $·ubgraph. of G. A (p, q) - graph
G has p vertices and q edges; p is referred to as the order of G and q as the .;;ze of
G. Whenever a subgraph H of a graph G has the same order as that of G, then H
is called a spanning 8nbgraph of G.
If U is a nonempty subset of VeG) thon the subgraph (U) of G induced by U is the
graph having vertex set U and whose edge set consists of those edges of G incident
with two vertices in U. If V E: V( G) and V( G) ;:::2 then G - V denotes the subgraph
induced by the set V"(G) - {v}. If e E EeG), then G - e denotes the subgr ...ph with
vertex set V(G) and edge set E(G) ._ [e},
The degree of a vertex v in G, denoted by degv, l.sthe number of edges of G incident
with 7.'. The minimum degree of a vertex in G is denoted by S(G), and the maximum
degree by .6.(G). A vertex of degree 0 is called an isolated vertex and a vertex of
degree 1 is an end-vertex. The edge incident to such a vertex is called a pendent
edge.
The complement of a graph G, denoted by G, is the graph with vertex set V( G) and
such that two vertices are adjacent in G if and only if these vertices are not adjacent
in O. A (p, q) - graph is complete, denoted by [(p, if every two of its vertices are
adjacent. The complement E.-p of the complete graph J(p has p vertices and no edges
and is referred to as the empty graph of order p. The graph ](1 is called the trivial
graph.
By a u - v walk Jf a graph G is meant a finite, alternating sequence of vertices and
edges of G, beginning with u and ending with. t., such that every edge is immediately
4
preceded and succeeded by the two vertices with which it ls incident. A It ~ v irail
is a u - 1J walk in which no edge is repeated. A u - 1J path is a,U - v walk in which
no vertex is repeated. A vertex u is said to be connected to a vertex v in a graph G
if there exists au - v path in G. A graph G is connected if every two of its vertices
are connected. A graph which is not connected is disconnected.
A component of a graph G is a connected subgraph of G not properly contained in
any other connected subgraph of G. The number of components at a graph G is
denoted by k( G). Thus k( G) = 1if and only if G is connected.
A nontrivial closed trail of G is referred to as a circuit of G. A circuit VI, 112, ••• ,
'U", Vb n ;:::3, of a graph G whose n vertices '0;' 1 :; i ::;n, are distinct is called a
cycle of G. A cycle of length n is denoted' by On; it is unique up to isomorphism.
An acyclic graph has no cycles. A tree is any acyclic connected graph. A [crest is
an acyclic graph. Each component of a forest is a tree. A tree of order p is denoted
bj Sp; it is not unique up to isomorphism if p ;:: 4.
A vertex v of a graph G is called a cut-vertex of G if k( G - v) > k( G). An edge e
of a graph G is called a cut·edge or bridge of G if keG - e) > keG). A nontrivial
connected graph G with no cut-vertices is called a block. A block of a graph G is a
subgraph of G which is a block and which is not a proper subgraph of any subgraph
of G with this property. The blocks of a. graph G partition the edge set of G and
two blocks have at most one vertex in common; this must then be a cut-vertex of
G' An end-block of a graph G is a block of G which contains exactly one cut-vertex
ofG'
The uni(ll1 G1 U GZ of two graphs Gl and G2 is the graph with vertex set V( G1) U
V(G2) and edge set E(G1) U E(G?). It is usually formed if the graphs G1 and G2
5
are disjoint, that is, V(Gl) nV(G:!) = ifi. The join G1+G2 of disjoint graphs G1
and Gz is the graph obtained from G1 U G2 by joining each vertex or G1 1;0 each
vertex of Oz.
'I'ho vertex-connectivity or simply connectzvity I\:(G)of a graph G is the mini-cum
m.mber of vertices whose removal from G results in a disconnected graph or the
~rivial graph. The edge-connectivity II:l(G) of a graph G is the minimum number of
edges whose removal from G results in a disconnected graph or the trivial graph. A
graph G is saiel to be n-connected, n ~ 1, if II:(G) ~ n.
Let a graph G, a natural number n and any set S with rz elements (called colours) be
given, An n-colouring of G is a functior f of V( G) into S such that, for every edge
e = uv of G, feu) ::f f( v). If G has an a-colouring, it is called n-coltnirable - note
that every graph has an n-colouring for some n (take n = IV(G)I and j arty one-
to-one function for example). The minimum n for which a graph (J is n-colourable
is called the vertex chrattuiiic number or simply the chromatic number of G and is
denoted by X(G). Note that a graph G is n-colourable if and only if X(G) :::;nj it is
called n-ehromaiic if \(G) = n.
A graph Gis n-pariite, n;:: 1, if it is possible to partition V(G) into n subsets Vb
1'2, ... , Vn (called partite aets) such that every element of E( G) joins a vertex of ~~
to a vertex of i·j, i i= j. (Of course Gis n-partite if and only if G is n-colourable:
the notation of this definition is, however, convenient in what follows.) For n = 2,
such graphs are called bipartite graphs. A complete ti-pariiie graph G is an n-partite
graph with partite sets VI, Vi, ... , v~ having the added property that if u E Vi and
v E Vj, i i= j, then Ut' E E(G). A complete bipartite graph with partite sets li and
Vi, where 11'11= m and 11'21 = n, is then denoted by Kim, n).
6
The incIependence numbt;r PCG) of a graph G is the maximum number of mutually
non-adjacent vertices in G. The clique number w(G) of a graph G is the maximum
number ()f vertices in any complete subgraph of G.
The join](l + en-I; n ;;::3, is called the whe'11 on n vertices and is denoted by Wn•
An elemen tary subdivision of a nonempty graph G is a graph obtained from G by
the removal of some edge e = uv and the addition of a new vertex tv and edges Uta
and vw,. A subdivision of G is a graph obtained from G by a succession of elementary
subdivisions. A graph H is defined to be h.omeomorphic from G if either H ~ G or
H is isomorphic to a subdivision of G. A graph GI is homeomorphic with a graph
Gz if there exists a graph Ga such that each of G1 and G2 is homeomorphic from
G:;.,
1.3 The chromatic polynomial of a graph
Let a graph G and a natural number n be given. Then it is possible to define
P( G, n) as the trtmber of n-colourings of G. It is possible to prove (and we will do
so in Lemma 1.5.1) that. for every graph G, PCG. n) is a polynomial in n. Hence we
will immediately change notation to P( G, z}, thinking of .7: as a, complex variable
(although x will almost always be real), and refer to this function as the chromatic
polynomial of G. By way of illustration consider
G;
The miu.,«, vertex can be coloured in any of the x colours. When this has been
done, this colour is no longer available for colouring the end vertices. Hence the end
vertices can be coloured independently each in (x -1) ways. Thus
7
This result can be extended (see Lemma 1.5.3) to show that if G is any tree on p
vertices then peG,x) = x(x _1)P-l denoted by Tpo Next consider
There are x ways of colouring, say, the top vertex. There arc then (x - 1) ways of
colouring an adjacent vertex, and (x - 2) ways of colouring the remaining vertex.
Thus
P(I<3' x) = x(x -1)(x - 2).
This result can be extended (see Lemma 1.5.2) to
P(Kp,x) = x(x - 1)(x - 2) ... \x - p+ 1),
denoted by x {p} •
1.4 Motivation
In the remainder of this thesis we study the chromatic polynomial P( G, x) of a
graph G in its expression as a linear combination of the chromatic polynomials of
trees, where Tp = x(x _l)P-l is the chromatic polynomial of a tree of order p (see
section 1.5). In this section, we first answer two questions: Why this can be done
and how it can be done.
III [14] the chromatic polynomial P(G,x) of a graph G is often expressed as Liajxi
or as I:ib;.7;{i} where xli} = x(x - 1)(a; - 2) ... (x - i+ 1). This is always pos-
sible (see for example §4 (If [61) since the set {I, x, ... ,xP, ... } 15, well as the set
{I, xil}, ... , x{p}, .•. } are bases for Z[a;], the Z-module of all polynomials in x over
the ring of integers Z. Similarly, {I, TI,T2••.. , Tp, ... }, is a basis for this Z-module
8
and hence can be exploited in a similar fashion. This answers the "wb.y it can be
done" question.
vVenow turn to the "how it can be done" question. In computing chromatic poly-
nomials, one can make use of Whitney's Reduction Formula stated in the following
theorem.
Theorem lA.l. Let G be II, graph, and u, v be two non-adjacent vertices of G. Let
G' be the graph obtained frcrn G by joining u and v by au edge, and Gil be the
graph obtained by identifyin« u and v, then
P(G,x) = P(G',x) +P(G",x).
Proof. In any e-colouring of th.e vertices of G, either u and v have different colours
or they have the same colour. The number of z-colcurings in which u and v have
different colours is un(hE.."'"'jtr:If an edge joining 'U and v is added to G, and is
therefore equal to P( G', x). Similarly, the number of e-colourings in which u and v
have the same colou:: is unchanged if u and v are identified, and is therefore equal
to P(G", x). Hence
P(G,x) = P(G',x) + P(GII,x). o
Repeated application of the above formula to any graph Gwill result in an expression
of the form peG, x) = I:ibix{i} which we shall refer to as the factorial form of
P(G,x).
The above recursive formula can also be used in the reverse process namely
where G is obtained from G' by deleting any edge ttV and Gil is obtained from G by
identifying c and e.
9
Repeated application of this formula to any graph G will result in an expression of
the form PCG,:x) = 2:; a;x; which Weshall refer to as the normal form of PCG, x).
If one is removing edges in chromatic reduction of a connected graph G, one might
decide to stop the process once all the graphs involved are trees. This then results
in an expression of the form P(G,x) = 2:i ciT;, which we shall refer to as the tree
form of P( G, x). If on the other hand G is disconnected, we will later show how
P( G, x) can be obtained in tree form.
In the example which follows, we will (as is customary, see [14]) abuee notation by
using a drawing of the graph for its chromatic polynomial.
Example 1.4.1.
We know that P(J{3,x) == .r(x -l)(x - 2) == ::c{3} in factorial form.
= Ta - T2 in tree form
~/ 0 / (.
• 0 • 0 _ (.
•OJ
-)- •+ •
• •= :1;3 - 3:1;2 + 201:in normal form.
\Vc note. in conclusion of this section, that Stirling numbers are used to go from
10
factorial form to normal form and vice Versa. The particulars are a!' follows:
Stirling numbers of the jirst kind, denoted by s(p, k), are used to convert a chromatic
polynomial from factorial form to normal form:
zIp} =pl(;) =x(x-l)(x-2) ... (x-p+l)
= s(p,p)xP + s(p,p -l)xP-l + ... + s(p, O)xP-11
P
= Ls(p,k)x'"k=O
where s(O, 0) = 1 and s(p, k) = 8(p-1, k -1) - (p-l)s(p -1, k), k = 1,2, ... ,p-1.
Stirling numbers of the second kind, denoted by S(p, k), are used to convert a
chromatic polynomial from normal form to factorial form:
xn = S(p,p)x{p} + S(p,p -l)x{P-l} + ...+ S(p, O)x{Oj
P:::::ES(p, k)x{k}k=O
where S(O,O) = 1 and S(p, k) = kS(p -1, k) +S(p -1, k -1), k ::::1,2, ... ,p -1.
The reader is referred to [16]for more information about these numbers.
1.5 Basic Properties Of Chromatic Polynomials
The first thing we need to show is that P( G, x) is indeed a polynomial in x. The
following Lemma appears in [15]:
Lemma 1.5.1. If G is a graph on p vertices, then PCG, x) is a polynomial of degree
p.
n
Proo£ For each nonnegative integer k, let a{G, k) denote the number of partitions of
VeO) into exactly k nonempty subsets, such that no edge of G joins two vertices in
the same subset. From a set of e colours, there are x(x-1)(x-2) ... (x-k+l) = x{k}
wc.ys of allocating a different colour to each of the subsets, and each of these gives
a colouring of G. It follows that
P(G,x) = Ex(x -1)(x - 2) ... (x - k + 1)a(G, k)k
which is clearly a polynomial of degree p. o
Thus far we have calculated the chromatic polynomial of the empty graph on p
vertices, denoted by p(Rp, x), to be xp• We now establish a few more results on the
chromatic polynomials of familiesof graphs beginning with the family of complete
graphs which appears in [14].
Lemma 1.5.2. The complete graph ](p of order p has chromatic polynomial
P(J{p, x) :::;x(x -1)(x - 2) ... (x - p + 1) = x{p} .
Prooi. Let J(p be the complete graph on p vertices. Choose any vertex of Kp and
colour it; this can be done in x ways. Choosing another vertex we have x-I
colours with which it can be coloured, since it is adjacent to the first vertex. Choose
another vertex; it is adjacent to both vertices already coloured, and can therefore
be colour. 1 in x - 2 ways. Vfe continue in this way; the last vertex can be given
any of the remaining x - (p -1) colours. Hence
P(Kp,x) = x(x -l)(x - 2) .•. (x - p + 1) == x{p}. o
'Ve next consider the family of trees which also appears in (14J.
12
Lemma 1.5.3. The chromatic polynomial of any tree Sp of order p is
P(Sp,x) =: xex _l)p-l.
Proof. We employ induction on p, the result being obvious for p = 1 and p =
2. Assume the chromatic polynomial of all trees with p - 1 vertices is given by
x( x-I )p-2. Let v be an endvertex of G and suppose uv is the edge incident with
v. By hypothesis, th", tree G' = G - v has ;2:( x -1 )p~2 for its chromatic polynomial.
The vertex v can be assigned any colour different from that assigned to 1!, so that
v may be coloured in any of x _. 1. ways. Thus
P(G,x) == (x -l)P(G',x) == x(x _l)p-l o
In section 4.5, we prove the converse of this lemma too. \Ve will henceforth denote
We next consider the family of cycles which appears in [14]in normal form as
P(Cp,x) = (x -l)P + (-l)P(x -1).
Lemma l,.5.4. For p ;:::3, the chromatic polynomial of the cycle Cp of order p is
1)-2
P(Cp,x)::; '2)-l)iT!'_i';=0
Proof. \Ve employ induction on p. For p == 3 we have for any cycle on 3 vertices
3-2P(C'3,X) = TJ - T2 = 2..::( -1)iT3_i,
;=0
by Example 1.4.1. Assume that for some fixed integer p, p ;:::3, the result is true for
all cycles on n vertices, n < p. Let e be ai .' edge of Cp. By Whitney's Reduction
13
Formula, applied to the edge e, we have:
p-1-2=P(Sp,x)- I: (-1)iTp_1_i by the inductive hypothesis
j=O
p-3
= Tp -2:) -1)iTp_1_i by Lemma 1.5,3.j;O
p-2
= 2.:.:(-l\'Tp_i.;=0
Hence the lemma is true for all cycles. o
We note that for p = 2 this result gives the chromatic polynomial for the complete
graph on 2 vertices, [\2. Our l\ext result appears in [2]:
A cascade of triangles is defined recursively as any graph Gk that can be obtained
by starting with Go = ](2 and, if Gj is defined, then Gi+l can be obtained by adding
a new vertex Vi to Gj and joining Vi to any two adjacent vertices of Gi. Note that,
in the terminology of [13] a cascade of triangles is {J(2}- constructible while in [21]
it is called a 2-tree.
We now consider the family of cascades of triangles which appears in [21] as
P(G,:v) = x(x -1)(3; - 2)P-z.
Lemma 1.5Jj. If G is a cascade of triangles on p vertices then
p-2 (, p-?P(G,x) = 2.:.:(-1)' i -)Tp-i'
,,,,0
14
Proof. We employ induction on p. For p = 2, the only cascade is ](2 with
Assume that for some fixed integer p, p ;:::3, the formula holds for all cascades of
triangles on n vertices, n < p. Now let G be any cascade of triangles on p vertices
and let e = uv be an edge of G with degv = 2. By 'Whitney's Reduction Formula,
applied to the edge e we have
..-'(G,x) = peG - e,x) - peG - v, x)
where G - e consists of a cascade of triangles on p - 1 vertices and a pendent edge,
and G - v is a cascade of triangles on p - 1 vertices. The vertex v can be assigned
any colour different from that assigned to tt, so that v may be coloured in any of
x - 1 ways. Thus
peG, x} = (x -1)P(G - v,x) - P(G - v,x)
p-3 () p-3 ( ). p-3 . . p-3= (x -1) ?=(-1)' i Tp-1-i - ?=(-I)' - i Tp-1-i.=0 ,=0
by the inductive hypothesis
p-3 () p-3~ .' p-3 p-3= ?-'(-1)' i Tp-i - ?=(-1)' ( i )Tp-1-i.=0 .=0
(P-3) (P-3) (P-3) . 3=Tp- 1 Tp-1+ 2 Tp-2- 3 Tp_3+···+(-1)P-13
{ , (p - 3\ . (p - 3) . -3 (p - 3) }- lp-l- 1 )Tp-2+ 2 Tp-3- ... +(-1)P \,-3 T2
15
= Tp _ { ~ ~ 3) + 1}Tp-l + { (p; 3) + (p; 3) }Tp-2 _ { (p; 3)
+ (P;3) }Tp_3+«<+C-1)P-a{ ~:D+~=!)}T3+(-1)P-2C=DT2(P-2) (P-2) (P-2) 2(P-2)= Tp - 1 Tp-1 + 2 Tp-2 - 3 Tp-3 + < < < + (-l)P- \p_ 2 T2
p-2 ( 2)i P- ,= .?,:C-1) i Ip-i'
1=1)
Hence the formula holds for all cascades. o
Example 1.5.1.
Consider G; and H:
Then both G and H are cascades of triangles on 5 vertices, therefore
P(G,J.') = P(ll,x) = t(-1)i(~)T5_i,=0
\Ye next consider a special class within the family of complete bipartite graphs; in
factorial form these chromatic polynomials are described in [18] as
m n
P(E(m,n),x) =L.L S(m,r)S(n,s)x{r+.} <
,.=1.=1ie
Lemma 1.5.6. The chromatic polynomial, P(J(2,n),x), of the complete bipartite
graph ](2, n) on n + 2 vertices is
P(I«2,n),x) =Tn+1+ t(-l);(~)Tn+2-i';=0 2
Proof. Let u and v be the vertices in the class containing only two vertices. Then by
joining 7t and v by an edge e and applying Whitney's Reduction Formula, Theorem
1.4.1, we have:
P(J(2, n),x) = P(G',x) +peG", x)
Where G' = J«2, n) + e is a cascade of triangles on n+ 2 vertices and Gil = Sn+! is
a tree on n + 1 vertices. Hence by Lemma 1.5.5 and Lemma 1.5.3 we have:
o
We now consider the family of wheels.
Lemma 1.5.7. The chromatic polynomial of the wheel Wp of order p is
Proof. We employ induction on p, For p = 4 the formula is easily checked since
lV4 =: 1(4' Assume that for some fixed integer p, p ;?: 5, the result is true for all
wheels on n vertices with n < p. Now let G be the wheel on p vertices and let 7tV be
any edge on the rim of G. By 'Whitney's Reduction Formula, applied to the edge
U1J, we have:
P(HTp,X) = peG',x) - P(GIJ,x)
17
where G' is a. cascade of triangles on p vertices and Gil is a wheel on p -1 vertices.
Hence by Lemma 1.5.5 and the inductive hypothesis Wehave:
p-2 ( ) p-3 (.... . p-2 2 . p-2P(Wp,x) = ~(-1)' . i Tp-i - (-1)"- T2 - ~(-1)' i )Tp-1-i
1 ~ '(P-2) 8 '(P-2\::::(-l)P- T2+ L.,,( -I)' . Tp-i - L,,( -·1)' . J T.n-l-i;=0 t ;=0 Z I
Hence the result is true for all whee's. o
This result can be compared to the result given in [19] as:
The chromatic polynomial of more classes of graphs are computed elsewhere in the
thesis.
18
CHAPTER 2
OPERATIONS ON GRAPHS AND CHROMATIC POLYNOMIALS
2,1 Introduction
v'Vedescribe the operations x, *, 0, EEl and ® on chromatic polynomials and use
the join operation, +, on graphs to introduce some useful computational techniques
whereby it will become easier to calculate chromatic polynomials of certain complex
graphs, Finally we give useful formulae, for the chromatic polynomiais of certain
standard graphs, wherein the Stirling numbers of the first and second kind are the
coefficients,
2.2 Computational Techniques
We begin with a basic result on trees which uses the operation x.
Lemma 2,2.1. If Tp and Tq are the chromatic polynomials of trees on p and q vertices
respectively, then Tp x Tq ::::Tp+q + Tp+q-1'
Proof.
Tp x Tq = x(x _l)p-l Xx(x _1)q-l
= x2(:;; _1)p+q-2
= x(x -1)p+q-2{(x -1) + I}
= x(x _l)P+q-l +x(x _1)p+q·-2
= Tp+q +Tp+q_1' o
The next result appears as Theorem 3 in [14]:
19
Theorem 2.2.1. If a graph G consists of two subgraphs HI and H2 which overlap in
a complete graph on k vertices then
peG ) __P(Ht.x) X P(H2,x},x - x{.\:}
Proof. The number cf ways of colouring the common part is x {k}. If we fix the
colours ~h~se k vertices there will be p~~}x) ways of colouring the remaining
. £ H d P(lIz,:r,) f colouri h . . • f H Hvertices 0 1 an :r,{k} ways 0 co ounng t e remannng vertices o 2. ence
the total number of colourings is
o
To simplify some calculations, we use
. P(H1,x) X P(H2,X)Lemma 2.2.2. Any expression of the form Tr can be calculated
as {P(HI' x) 0 P(H2,.:c )}EBT,. where 0 and EB denote a type of multiplication and
division in which Tm 0 Tn = Tm+n and TmEBTn = Tm-n respectively.
m - p;:: rand n - q;;:: r. Using Lemma 2.2.1 we have:
!(Ht.x) x P(H2,x)T,.
_ O:':f=o(-l)iajTm-i) X n:::J=o(-l)ib_;Tn-i)- Tr
"."[(Tm+t• + Tm+lI-d - (al + bt){Tm+n-l + Tm+n-2) + (a2 + alb] + b2)x
20
+ ... + (-1)p+Q-l(apbq __1 + ap_lbq)(Tm+n-p-q+l + Tm+n-p-q)
But
Tp + Tp-1 = X(X _l)P-I + ;e(x - 1)P-2
= x(x -1)P-2[x -1 +1]
Hence
P(HbX) X P(H2,x)t;
= [x2(x _1)m+n-2 - (al + bl)X2(X _1)m+n-3 + (a2 + (llbl + bz)x2(x _1)rn+"-4
- (a3 +a2bl +c1b2 +b3)X2(X _1)rn+n-5 + .. + (-1)p+q-'(apbq_1
= [<t<-l)iaiTm-;) 0 et( -ljiOjTn_i)]®Tr1=0 }=o
21
Hence
o
Example 2.2.1.
With G:
and T; = Tz:
•I
I•P(G,x) = [P(K"x) ~_P(I{4'X)J
..L', by Theorem 2.2.1.
\Ve stress that the operations e· and EEl can only be used in the type of expression
specified in Lemma 2.2.2.
Corollary 2.2.1. If a graph G consists of two subgraphs Hl anti H2 which overlap
o
Another way of calculating these products is contained in Our next result.
22
Preof.
m-2 n-2L a;Tm_i 0 L.: bjTn_ji=O j=O
+ .
m-2 n-2
=L aiCL.: bjTtn+n-i-j).i=O j=O
o
Note that, since 0 is a commutative product, it follows that
, rn-2 'T1 ",,-2 b 'T1 "n-2 b ("m-2 m )L..ii=O aj.Lm-i 0 L..ij-::O j.Ln-j = L..ij=O j L..ii=O aj.Lm+,,-i-j·
Lemma 2.2.2 can be effectively applied to determine the chromatic polynomials of
certain special graphs. One such graph is the theta graph. A theta graph, 0, has
two vertices v and w of degree 3, between which are three paths whose other vertices
all have degree 2. A typical theta graph, 0, is shown below see [15].
23
Lemma 2.2.4. The chromatic polynomial of the theta graph, €I, for which the paths
have l, m and n edges is p(e,x) = ~P(Cl+l'X) <:) P(Cm+I,x) <:) P(Cn+1,x)](t)T4 +[P(C/,x) o P(Cm, x) <:)P(Cn,x)1®T2•
Proof. By applying Whitney'» Reduction Formula to the theta graph we have
€I:
p(e,x) = P(G',x) +P(G",x), where G' is a graph in which the three cycles CHI,
Gm+1 and Cn+1 overlap in a complete graph on two vertices and Gil is a graph in
which the three cycles Gl, Gm and Gn overlap in a cut vertex v. Hence by repeated
application of Theorem 2.2.1 and Corollary 2.2.1, we have by Lemma 2.2.2
p(e,x) = [P(G/+1,x) o P(Gm"!-l,x) o P(Gn+l,x)1®T4
o
Let 0 denote the operation on polynomials in which each term aixl'-i in P(x) =1:f.:-;ajxp-i is replaced by a;x(x -l)P-", that is, P®(x):::: 1:f;g aix(x _1)1'-;,
Lemma 2.2.h. Suppose the polynomial P(x) is a sum of terms of the form ax(x-
1)(x -2) ... (x - k)' ..... en, for each such term, p0(x) has ax(x -1), ..(x -k)(x-
k - 1); as a term.
Proof. Since the opertion ® satisfies [P(x) +Q(x)] ® = p0(x)+Q0(x), we need
only to show it for one term, that is, [ax(x -1) ... (x - k);]0 = ax(x -1) ... (x -
k)(x - ft· _l)i. This we do now by induction on k: For /.;= 0, our claim is nothing
24
but the definition of 0. Hence suppose the result is true for.le -1 and consider
[ ]0 0C~(x -1) ... (x - k + 1)(x - k)i = [ax(x -1) ... (x - k + 1)(:&- k + 1-1)J
~ [t,aG)( -')'-'x(. -1) (x- k+ 1)'Hf"
= i>G)( _1)i-Ix(x -1) (x - k + 1)(x _k)l+l1=0
= ax(x -1) ... (x - k)t(~}_1)i-/(x - ki1=0
= ax(x -1) ... (x - k)(x - k _1)i . o
The operation ® will be used in section 2.3.
2.3 Operation on graphs and the chromatic polynomials.
We begin with a result which appears as Theorem 2 in [14):
Theorem 2.3.1. If a graph has connected components GIl Gz, ... , Gk> then
P(G,x) = P(G1,x) X P(G2' x) X ••• X P(Ck,:C)'
Proof. Since the components Gi, 1 :s; i :::;k, are disjoint, the colouring of each
component is independent of the colouring of the others. Hence the number of
ways of colouring the whole graph G is simply the product, by the multiplication
principle, of the numbers of colouzings of the separate components. o
Example 2.3.1,
I we have
P(G,x) = (Ta - T2) X T2
= Ts +T4 - (T4 +Ta)
= 15 -Ta
The next result appears 8.1, Theorem 4 in [14]:
Theorem 2.3.2. The chromatic polynomial of the join, H; +H2, of two graphs HI
and H2 is P(HI,x) * P(H2,x) where * denotes a type of multiplication in which
x{m} *3;{n} =x{m+n}.
Proof. Let the two graphs HI and have manti n vertices respectively. By re-
P(' ted application ofVlhitney's Reduction Formula, P(G,x) = peG', x)+P(G", x),
to HI and H2, separately, we express the chromatic polynomials of HI and Hz in
terms of the chromatic polynomials of complete graphs, namely
P(H},x) = P(]{m,x) +a1P(J(m-l,X) + a2P(J(m-2,X) +... (2.1)
and
P(H2'x) = P(J(n,x) + bIP([(n-1>x) + b2P(J{n-2,X) + ... (2.2)
Abc by repeated application of Whitney's Reduction Formula to HI +H2, the Ht
and Hz portions of the join H; +H2 can be reduced in exactly the above way. In this
process every vertex of each graph used in the reduction of H1 will be adjacent to
every vertex of each graph in the reduction of Hz. Thus we shall end up by express-
26
ing P(Ht +H«, x) In terms of all possible joins of a. complete graph from (2.1) and
a. complete graph from (2.2). But the join ](p +Kg = ]{p+q, where Kp and Kg are
complete graphs on p and q vertices respectively. Thus corresponding to a. term x{p}
of P(l:ll,X) and a. term x{q} of P(H2,x), there will be a. term x{p+q} = x(p} * x{q}
o
Example 2.3.2.
With s. i•
P(1I1, x) = x{2} and P(1I2, x) = xIS} + x{2}, we have
P(l:lt + H2,X) = P(Ht,x) * 1'(H2,x)
= x(2} * (x{3} +X{2})
= x(x - l)(x - 2)(x - 3)(:;:- 4) + x(x -l)(x - 2)(x - 3)
= x(x -l)(x - 2)(x - 3)2
We now consider a special case of Th''''rem 2.3.:
Lemma 2.3.1. Let ao be any graph of order p with chromatic polynomial P(Co,x) =
Er;~(-l)iaixP-'. If Gt is the graph obtained from Go byndding a new vertex v
:'.'7
to Go and joining v to every vertex of Go, then
p-1=L( -l)ia;z(x _l)p-i.
i=O
Psooi. Let P(Go,x) = (,lOXP - a1xp-1 + ... + (-1)p-1ap_1x• Since v is adjacent
tc- ,_,veryvertex of Go in. G1, each empty graph term xk of P( Go, x), 1 :5 k :5 p, is
replaced by the polyno:nial of a tree on Ie + 1vertices, that is, by x( x-l)k. Hence
p-1P(G1,x) = L(-1)ia;x(x -l)P-;
;=0
o
Example 2.3.3.
If Go: o , where
= :r(x _1)3 - x(x -I? + :r(x - 1)
Hence
peG1,x) = P®(Go,x)
:: ;rCx -l)(x - 2)3 ._ x(x -1)(x - 2)2 + x(x -l)(x - 2}by Lemma 2.2.5.
28
As a special case of Lemma 2.3.1 let Go be the cycle on p vertices. Then
P(Go,x) == P(Cp,x)
=Tp-7;,-1 + ... + (-ly-2Tz
== xCx _ly-1 - xCx _ly-2 + '" + (-ly-2x(:c -1)
and
P(G1,:c) = P@(Go,x)
::::::x(x -l)(x - 2)1'-1 - xCx --l)(x - 2y-2
+ ... + (-1)P-2x(x -l)(x - 2)
which is the chromatic polynomial of the wheel on p + 1 Vertices. This proves
o
Since the wheel IVp of order p (which can aJso he considered as a pyramid) consists of
the cycle Cp-1 with an additional vertex joined to all the other vertices, we consider,
in a similar way two other families of graphs.
The biwheel Up of order p consist" of the wheel IVp_1 with an additional vertex
joined to all the other vertices, that is Up = 1(1 + Wp-1, P ~ 5.
The bipyramid Bp of order p is obtained from Up by deleting the edge joining the
two added vertices.
'We note that Us is J{s and that B6 is the graph of the octahedron. We give
equivalent formulae for Up and Bp, (see (15]), in the following lemma:
29
---------,-----------
Lemma 2.3.2.
(a) P(Jp,x):::: (-I)P-2x(x -1)(x - 2) + I:( ._I)i (p ~ 2)X(X - 1)(x - 2)P-2-i,=0
p-4::::2:)-1)ix(x -1)(x - 2)(x - 3Y-'3-i;=0
(b) peEp, z ) = x(x -1)(x - 2)1,-2 + (_1)P-2x(x -1)(:0 - 2)
Lemma 1.5.7, we have from Lemma 2.3.1 (by adding a new vertex and joining it to
all the vertices of Wp-1) that
P(Up, x) ::::(-I)p-2x(x - 1)(x - 2)+ I:f':-;( _1)i (Pi2)x(x - 1)(~;- 2)p··2-i .
Similarly from Corollary 2.3.1 since P(Wp-1, x) = I.:f;~(_1)ix(x -1)(x - 2)P-3-i
we have P(Up,x) :::: I.:f;~(-I)ix(x-1)(x - 2)(x - 3)1'-3-;. We prove (b) using
Whitney's Reduction Formula
p-3::::(-ly-2x(x -l)(x - 2) + ?=(_1)i (1' ~ 2).1:(X -.I)(x _ 2)P-2-i
1=0
1'-4+ '2)-lix(x -1)(x - W-3-i
i",O
o
Our next result generalizes Lemma 2.3.1.
30
Theorem 2.3.3. Let Go be any graph of order p with chromatic polynomial
P(Go,x) = L:f;;(-l)iuixp-i. lfGm is the join of Go and Km., for any m = 1.2, ... ,
then P(Gm,x} = I:f;~(-l)iaix(x -l)(x - 2) ... (x - m)p-i,
Proof. Repeated application of Lemma 2.3.1 and Lemma 2.2.5. o
Example 2.3.4.
With Go:
we have P(Go, x) = x3 - X2,
and
-P®®(G )- Qpr
= .1'\.1' -l)(.r - 2)3 - .1·(x -1)(.1' _ 2)2
Our next result is a corollary of Theorem 2.2.1 and is closely related to
Lemma 2.3.1,
Lemma 2.3.3. Let Go be any graph with chromatic polynomial P( Go.:1:). If G1 is
a graph obtained from Go by adding a new vertex t' to Go and joining t' to every
vertex of a complete subgraph of Go on p vertices, then PI.G1 • .1') == (x -n )Ft Go• .r I.
31
Proof. Since v is a vertex on a complete graph on n+ 1 vertices which overlaps with
Go in Kn we have by Theorem 2.2.1
_ P(Go, x) X x(x --1)(:1: - 2) (x - n + 1)(:1: - n)-- x(x-1)(x--2) (x-n+l)
:::::(x - n)P(Go,x). o
Example 2.3.5.
With Go:
we have
P(Gl> x) :::::(x - 3)P(Go,x)
:= (x - 3)x(x - l)(x - 2?
2.4 Combinatorial Formulae
Our next result is a corollary of Theorem 2.3.2.
Proof. The chromatic polynomials of the empty graphs on m and n vertices, re-
spectively, can be expressed in terms oi the chromatic polynomials of the complete
32
graphs, using Stirling numbers of the second kind, by
m
xm ,=LS(m,r)x{r}r=l
andn
:;;"= 2:S(n,s)x{a}8=1
Now by Theorem 2.3.2 we have
P(f{(m,n),x) = P(xm,x) *P(x",x)
m n= 2:S(m,r)x(r} * 2:S(n,.!')x{'J=1 3=1
== [scm, 1).x{l} + SCm,2)x{2} + ... + SCm,m)x{rnj]
n
* L:.S'(n,s)x{3}3=1
n n
= SCm, 1)2:Sen, s)x{l+s} + S(m, 2)LSen, S)X{2+3}.=1 .=1
n
+ ... + S(m,m) L .S'(n,s}.r{II1-!-.}.=1
m n
=L SCm, r)L S(n,s)x{r+s}.r=1 .==1
o
A cascade oj l(m '8 is defined recursively as any graph Gk that can be obtained by
starting with Go = J(m-l and, if G; is defined. then Gi+l can be obtained by adding
a new VCl't€'X t'i to G, and joining 1'1 to any m - 1 pairwise adjacent vertices of G,
33
As a special case of Theorem 2.3.3, let Go be the empty STaph, tc; on n vertices,
then Gm = Km + l{n which is a cascade of ](m+1's. In this case P(Km + I{n, x) =
x(:t'~1)(x-2) ... (x-m)", which we denote by xhm}, so that x~l} = x(x-l)" = Tn+l
which is the chromatic polynomial of a tree on n + 1 vertices.
Lemma 2.4.2. x~m} = 2::=1 S(n,s)x(m+s}.
Prooi. Since xim} = P(Km + tc; x) is the chromatic polynomial ~f the join of J(m
and s; we have by Theorem 2.3.2
n::::x{m} *2:S(n,8)X{s}
8=1
n
= 2: S(n,s)x{m+s}.s=l
o
As a special case of Lemma 2.1.2 we can express the chromatic polynomial of a tree
on n + 1 vertices in terms of the complete graph hasis, that is
n
= L:S(n,s)x{I+S} .•=1
We now apply Theorem 2.3.3 to bipartite graphs.
Let Go be an empty graph on n vertices with P(I{n,x) = xn, then P(G1,x) =P(K(l, n), x) :::::x(x _l)n = Tn+1, since G1 is It tree on n+ 1 vertices. To compute
P(1{(2,n),x) we apply 'Whitney's Reduction Formula: P(G,x) = P(l«2,n),x) :::::
3·1
PCG', x) + P( Gil, X) where G' is a cascade of triangles on n + 2 vertices and G" is
a tree on n + 1vertices. In fact G' ~ G2 and G" ~ G1. Therefore
P(!{(2, n), x) = x(x -l)(x - 2)n + x(x _l)n
n n
= 2::S(n,s)x{2+8} + 2:: S(n,s)x{I+s}3=J 0=1
2:::>. S(3, r)x~rJ,<....J,.=1
We now obtain a formula for the chromatic polynomial of the bipartite graph,
]{(m, n), in terms of cascades:
Lemma 2.4.3. P(J((m, n), x) = 2:;;'=1 SCm, r)x~r}
m nP(I«m,n),x) = 2::S(m,r)LS(n,s)x{r+s} byLemma2.4.1.
r=1 .=1
m
=LS(m,r)x~r}. bv Lemma 2.4.2.r=1
o
Example 2.4.1.
4
P(I«4,5),x) = LS(4,r)x~r},.",1
C'(4 ) {I} s(' ) {2} 's( ) {3} s( ) {4}:=,;} ,1 x5 + 4,2 X5 -r 4,3,X5 + 4,4 X5
We now express the chromatic polynomial of the complete graph on m vertices in
terms of the chromatic polynomial of cascades.
35
Lemma 2.4.4. x{m} = E~~nsCm - n,i)x/n}.
Proo£ By Theorem 2.3.3 and induction on m. o
As a special case of Lemma 2.4.4, with n = 1, we can express the chromatic poly-
nomial of the complete graph on m vertices in terms of the chromatic polynomial
of trees, that is
m-l
;rem}= 2: sCm -l,i)xf!J;=1
m-l
= L sCm -l,i)Ti+l'i=l
36
CHAPTER 3
CHROMATIC POLYNOMIALS OF GRAPHS OBTAINED BY
REPLACING A VERTEX WITH A GRAPH
3.1 In! -.Jductiol1
If T is a tree and v is a vertex of T and e is an edge containing v, then the branch of
T with respect to (v,e) is the maximal subtree of T which contains e and has v as,an end vertex. If v is incident to the edges el, e2, .•• , e"" then there are k branches
in T containing v as an end vertex. If B and H ate graphs and v is a vertex of
B then we say that B is attached to H at v if' a vertex of H is identified to v. If
H is a connected graph and T is a tree we use the notation G(H,T) for any graph
that results if we choose any vertex v of T of degree k and attach the k branches
of T containing l' to (any k vertices of) H. One may think of G(H, T) as a graph
obtained from T by replacing the vertex v with H or "blowing up" the vertex v to
H. vVestart by studying the chromatic polynomial of a graph obtained by replacing
the vertices of a tree by graphs in section 3.2 and extend this concept to forests in
section 3.3.
3.2 Replacing the vertices of a tree by graphs
Most of the results of this section appear in [2,3J (see also [20]):
Theorem 3.2.1. If H is a connected graph on Tn vertices with P(H, x) = .L::~2(-1);a.T,m-i and T is any tree on r vertices and G is any connected (p, p + n)-graph of
the form G = G(H,'J.'n), then p(G,x) = E:~2(-1)iaiTp_i'
Proof. The result is clearly true if r = 1 and hence we assume T to be nontrivial.
We employ induction on the number of edges IE(G)I of G. If G is a tree then
37
P(G,x) = Tp. But then H is a tree and P(H,x)::::: Tm. Assume that for some fixed
integer k, the result is true for all connected graphs G with IE(G)I :::; k -1. Now
consider G with p vertices and p+n :::::k edges and assume that G = G(H, T) with
F(H, x) = I:~~2( -1)'aiTm-i. Let w be any end-vertex of T and let u be adjacent
to w in G. Then, by Whitney's Reduction Formula applied to the edge 'UW, we
have P(G,x) = P(G',x) - P(G",x) where G' is a (p,p+ n -I)-graph and Gil is
a (p - l,p + n -I)-graph. Now by Theorem 2.3.1 and the inductive hypothesis we, m 2 •have P(G',x) = P(G",x) x Tl and P(G',x) = L,i=~(-I)'ajTp_1_j. Therefore
peG, x) = P(G',a:) - P(G",x)
m-2
= 2:(-l)'aiT/._i.i=O
Hence the result follows. o
Note that the result of this theorem can also be stated 0$:
P(G.x) = P(H,x) ::;)Tr-1•
38
Example 3.2.1.
Consider G=G(....H_,_T_):-1~_.......__L:'re H, [21
Then P(G,x) = TlO - 2T~+ T8 with P = 10.
In the above example T is a tree on 7 vertices. V.remay replace .:mY vertex of T
with the graph H and obtain the same chromatic polynomial of Gj that is we may
attach the branches of T to JI in any fashion.
"'i"'i-enow consider graphs that overlap in a cut vertex.
Theorem 3.2.2. If G is any connected graph with blocks HI. Hz, .... Hp, then
Proof. "'i'i'eemploy induction on p. For p = 2 the result follows from Theorem 2.3.1
and Lemma 2.2.2. Assume that the result is true for all connected graphs H with
k blocks. k ::; p - 1. Now consider any connected graph G with p blocks. Assume.
without loss of generality, that IIp is an endblock of G, v is the cut-vertex on Hp
and let H = G - ('V(Hp) - v). By Corollary 2.2.1 we have
P(G. r) == [P(H.x) ~ P(Hp,x)18 TI
= [{(P(H) . .r) -;_P\H1 ..z ) ~ ... ~ P(Hp-l,.x))8 Tp_2} -;:P(Hp,.rI]EB T:
3D
Hence the result follows. o
Note that the result of Theorem 3.2.1 can be obtained from Theorem 3.2.2 by
choosing HI = H and 1'2 = H3 = ...= Hp = [(2'
Example 3.2.2.
then HI:
If G:
•
Therefore
P{G, x) ::::[P(I(z, x) 0 P(I(4, x) 0 P(K3' x)] EB T2
= [T2 ::: (T~ - 3T3 + 2T2) 0 (T.1 - Tz)] EB Tz
Theorem 3.2.3. If HI is a 2-connected graph on m vertices with
P(Hl!x) ::::L:~2(-1)laiTm_l and Hz is a 2-coonected graph on n vertices with
P(}lz,x} = 'Lj';;;(-l)jbjTll-J and Gis any graph formed hi connecting HI and
H2 by a cut edge e. then P(G.x) = P(Hj, x) 2' P(Hz, x).
Proof. By Theorem 3.2.2 we have
rtc.»: = [PtHl'X', :: P(E2,.r) ! P(Hz,x)18 T2
= [P(H1 • .r) :::,T~ :::P(H~ . .r)leT2
= P(Hl ..rl ::PfH;;.x). o40
If G is a tree with vertex set {1, 2, ... ,p} and HIIH2, .•• , Hp are connected graphs
then a graph G(H1,H2, ... ,Hp) is formed by blowing up each vertex i of G to Hi
while one vertex of Hi is adjacent to one vertex of Hj iff ij E E(G).
Theorem 3.2.4. If G is any tree of order p, then the chromatic polynomial of
G(Hl, Ti2, .•. ,Hp)is
Proof. We employ induction on p, For p = 2 the result follows from Theorem 3.2.3.
Assume that the result is true for all trees of order k with k :5 p - 1. Now consider
a tree .'J of order p, Assum-. without loss of generality, that an end vertex of G is
blown up with Hp. Let H = G(Hl' H2, ••• , Hp) - V(Hp). Then by Theorem 3.2.3
we have
P(G(HllH2"" ,Hp),x) = P(H,x) 0 P(Hp,x)
= P(H1, x) 0 P(H2' x) 0 ... 0 P(Hp-llx) 0 P(Hp,x)
Hence the result follows. o
Example 3.2.3.
Ha: •
41
with G a tree on three vertices.
'Therefore
P(H,x) ;:: P(HI, e) 0 P(K3,x) 0P(J{I,X)
Theorem 3.2.4 can also be obtained as a special case of Theorem 3.2.2 by replacing
the chromatic polynomial of each cut edge by T2•
3.3 Replacing the vertices of a forest by graphs
The above concepts were confined to connected graphs, we now concentrate our
efforts on the disconnected case. We begin by generalising Theorem 3.2.3.
Lemma 3.3.1. If HI and H2 are disjoint graphs and G is any graph formed by
connecting H, and H2 by an edge e, then
Proof. Let GJ and G2 be graphs formed by adding the edge e to HI and Hz
respectively. Now by Theorem 2.2.1 and Lemma 2.2.2 we have:
PCG' ) _ P(llj,x) X Tz"x - x{l}
= [P(H;, x) 01'2J Ef;) Tl
;::P(Hi,x)0T1 for i= 1,2.
42
Also by Theorem 2.2.1 and Lemma 2.2.2 we have
= [P(G],x) 0 P(G2,x)]EB T2
= {[P(Hl>x) 0 TIl 0 [P(H2,x) o T1]}EBT2
= {[P(HbX) 0 P(H2,x)]0T2} EBTz
;.:::P(HI'X) 0P(H2,x}. o
The result of Theorem 2.3.1 can also be given in terms of 0 and EB. \Ve first
consider the special case when k = 2.
Lemma 3.3.2. If a graph G is the union of a 2-connected graph G1 and a graph G2,
then
Prooi. By Whitney's Reduction Formula we have
P(G,x) = P(G',x) +P(G",x)
where G' is any graph formed by connecting G1 and G2 by a cut edge uv and GIf
is obtained from G by identifying u and v. Hence PCG, x) = P( G1, x) 0 PCG2, x) +[P(G1,x) 0 P(G2,x)]EBTl by Theorem 3.2.3 and Corollary 2.2.1. o
Theorem 3.3.1. If a graph G has connected components Gll G2, ... , Gk, then
where Tu= 1.
43
Proof. We employ induction on k the number of components of G. For ~ = 2 we
hU'\I\1 by Lemma 3.3.2 that
Assume that the result is true for all graphs H with at most k -1 connected compo-
nents. Now consider G with connected components Gl, G2, ••• , Gs, By Whitney's
Reduction Formula se have
P(G,x):::: peG', x) + P(G",x)
where G' is a graph formed by connecting the graph }J with k - 1 components
G1, G2, ••• , Gk-J to Gk by an edge uv and G" is obtained from G by identifying u
and u, Therefore
P(G,x) = P(N,x) 0 P(Gk'X) + (P(H, :r.) 0 P(Gk. x)lEf:) TJ
by Lemma 3.3.1 and Corollary 2,2.1
where
by the inductive hypothesis. Hence
i.!:l P1Gk-t.X)}Ef:) Tj} 8P(GkIX)] Ef:)Tl
44
ov-o P(Gk-l,X)}EB Tk-2] 0P(Gk,X)
+ [ (k ~ 2){p(GbX) 0P(G2,x) 0"'0 F(Gk_1,X)} 0P(Gk'X)] EB Tl
•
(k-2)+ ... + k _ 2 {P(Gl, x) 0 P(G2, x) 0'" 0 P(Gk-l, x) <:V peG!;, x)} EB Tk-2
45
using the fact that (k;i) + (tD = (k-;+1). Hence the result follows. 0
Corollary 3.3.1. If each c, ~ [(I, 1::; i ::;k, then G ~ Rk and
oIn the normal form this polynomial is known to be peRk, x) = xk.
Corollary 3.3.2. If each G; is a tree on Pi vertices and P == I:~=1Pi, thenpeG ) ",k-l (k-l)'T'• ,x =LJj=O j -p-j' oIn the normal form this polynomial is
k-l
= I:(k ~ 1).1:(:1: _l)P-l-jj=O J
k-l (k -1)= 2: . Tp_j'j=Q J
ExampJe 3.3.1.
With G: LAl we have
= G) [Ts - T7) EF> To + G) [18 - T1J EB Tl + G) [Ts - T7J EF> Tz
= Ts - T1 + Z[T7 - Tel + [Ta - T5J
We now consider graphs obtained by replacing a vertex of a forest by a graph.
Tlieosexa 3.3.2. If a graph G is the disjoint union of two subgraphs G1, a connected
graph on n vertices with peGbX] = Ef:02( -l)iaiTn_i' and G2 a forest on p.- n
vertices, n < p, and k components, then
n+k-2 ["-2 . ( k )]P(G,x) == ?= 2:)~-l)'ai . _ i Tp_j.;=0 ,,,,0 J.
Proof. By Whitney's Reduction Formula we have PC G. x) == P(G',.r) + P( Gil, x)
4i
where G' is any graph formed by connecting G1 and G2 by an edge 'Uv and Gil is
obtained from G by identifying 'U and v. Therefore by Lemma 3.3.1 and Corollary
2.2.1 we have:
[
n-2, k-l (k _ 1) ]+ ~(-1)'aiTn-i0?= . Tp-n-j (BTl.=0 1",0 J
n-2, [k-l (k _ 1) k-l (1.: - 1) ]==: I:C-1)lajTn-i 0 L . Tp-n_j +I: . Tp-n-j-l
.=0 ],=0 J J=O J
n-2, [ ('k - 1) (k - 1)::::;t;(-1)'ajTn_i 0 Tp_n + 1, Tp-n-1 + ... + k -1 Tp-n-k+l
(k - 1) (k - 1) (k - 1) ]+ . 0 Tp_n_1 + ... + 1.:_ 2 Tp-n-k+l + k _ 1 Tp-n-k
n-2, [(1.:) (' k ):: ~(-l)'aiTn_i 0 Tp-n + 1 Tp-n-1 + ... + k -1 Tp-n-k+1
(1.:) 1. (1.: - i) (k - i) (1.: - i+ 1)+ k TI'-n-k J since i + i-I = i
48
n.+k-2 [n-2 'G k )]= 2.: ?=(-l)'ai '_ i Tp_j'
;=0 .=:oo
Corollary 3,3.3. If G is the disjoint union of two subgraphs Gil a cycle on n vertices,
and G2 a forest on p - n vertices, n < p, and k components, then
n+k-2 [1.-2 (k)]P(G,:c) = I: ?=C-1)i '~i Tp_j.
)=0 .=0 Jo
Example 3.3,2.
With G: [SJ /\ I we have
G1 G2
peG!, z) = T4 - 2Ta + T2; n::.::; 4; k = 2; lJ = 0; ao = 1; III = 2; and a2 = 1. Hence
4+2-2 [4-2 "G 2 )]P(G,x):::: L ?=<-l)'ai .'-i TO_j
]=0 ,=0
4 [2 . G 2 )]=L ?=(-l)'a; . _ i 7h-iJ=O ,=0
40
= Tg + [ao G) - al G)]Ts + [ao G) - al G) + a2G)]~+ [ao G) - al G) + a2G)]Ts + lao G) - al G) + az G)]Ts= Tg + [2 - 2]Ts + [1 - 4 + 1]T7 + [0 -:t + 2]TG + [0 - 0 + l]Ts
50
CHAPTER 4
THE COEFFICIENTS OF THE CHROMATIC POLYNOMIAL
4.1 Introduction
In this chapter we study the coefficients CIa, ClI, ••• , ap-l of the chromatic polynomial
pl'P(G, x) = L;':-o (-1)'a;Tp_i of a connected graph G of order p.
In the proof of Lemma 1.5.1 it was shown that P(G,x) = Ek a(G,k)x{q where the
coefficient aeG, k) denotes the number of partitions ofV(G) into exactly k nonempty
subsets, such that no edge of G joins two vertices in the same subset. This gives us
an interpretation of the coefficients of P( G, x) in factorial form. The interpretation
of the coefficients in the normal form of the chromatic polynomial P( G, x) requires
the inclusion-exclusion principle; it is due to Whitney [2~J (see also [14]): that
P(G,x) = L~;::o(-1)rN(p,r)xP where N(p,r) is the number of subgraphs of G
with p components and r edges, and k is the number of edges in G. This result was
used by Farrell [9] to interpret the first five coefficients of the chromatic polynomial
in normal form. In section 4.2 we start by interpreting the first five coefucients of
P(G,x) in tree form. The same results are obtained by using Farrell's results in
section 4.3. In section 4.4 we discuss techniques of computing some higher order
coefficients and in section 4.5 we consider the signs of the coefficients of P(G,x)
in tree form in terms of the connectivity of G. \Ve present a new combinatorial
identity in section 4.6 and apply it in section 4.7 to determine the coefficients of {he
chromatic polynomial of a cascade of graphs.
51
4.2 Interpretation of the coefficients
The results of this paragraph correspond to the results of Farrell in [9] in the sense
that he obtains formulas for the coefficients bo, bl, 112, b3 and b4 in a chromatic poly-
nomial in the normal form PC G, x) = 2:f,:g(-1)ibixP-i• More than that our results
can be deduced from his by equating 2:f~( -l)ia;Tp_; =Ef,:g( -l)ia;x(x-l)P-i-land peG, x) = 2:f,:g(-l)ibixP-i and expressing ao, aI, a2, aa and a4 in terms of the
bi'S; we do so in the next section. We give, however, independent proofs in this
section (see also [1]).
Throughout this section, let G be a connected (p,p + n) - graph with
p-l
P(G,x);:: L(-l)iaiTp_i.;==0
Theorem 4.2.1. ao = L; the coefficient of Tp in PtG, x) is 1 = (!If).
Proof. By the nature ofWhitney's Reduction Formula there is, at each stage, exactly
one graph having p vertices. This is therefore true at the final stage when a tree on
p vertices is reached. Hence the coefficient of Tp is 1. o
Proof. We employ induction on n. Fer n = -1we have for every tree G of order p
peG, x) ee Tp - O~'-l + OTp_2'"
Assume that for some fixed integer 11, n 2:: 0, the result is true for all connected
(p,p+nl) - graphs with n' ::;n-1. Now consider G with p vertices and p+n edges.
~~-~----~--~.-----.--------------
By Whitney's Reduction Formula
P(G,x) == P(G',x) - P(G",x)
where G' has p vertices and p + n - 1 edges and Gil has p - 1 vertices and at most
p +n - 1edges. Since the result is true for G' and Gil we can write
and
P(G",x) ==
Hence
Now by the inductive hypothesis
Therefore
I (n+1)al == al + 1= n + 1= 1 .
Since the result is true for all p and all connected (p,p + 71) - graphs, it is true for
all connected graphs. D
Let Cn be the number of induced cycles of •.-ngth TI, n ;:::3.
Proof: We employ induction on n, For n = -1we have for every tree G of order p
peG, x) = Tp - OTp_1 + OTp-2 - OTp-3 + ...53
i.e,
(-1+2)az= 2 -0=0-0=0.
Assume that for some fixed integer n, n ;::0, the result is true for all connected
(p,p + n') - graphs, with n' ~ n - 1. Now consider G, with p vertices and p +n
edges. Let uu be an edge !.L G and suppose that uv is on fl triangles, fl ~ Ca.
(Hence uv is a common edge of these triangles).
~=
~
By Whitney's Reduction Formula, applied to the edge uv, we have
P(G,x) = peG', x) - P(G",x)
where G' has p vertices, p + n - 1edges and c~= Ca - fl triangles; G" has p - 1
vertices, p + n - 1-1'1 edges and c~ = C3 - fl triangles. Since the result is true for
(i' and Gil we can write
and
Hence
P(G,x)::: Tp - (ai + l)Tp-l + (a~+ anTp-2 - (a~+ a~)Tp_J + ...
where
54
Now by the inductive hypothesis
and by Theorem 4.2.2
II _ r: +1)al - 1 .
'I'herefo- ,
(n+1)= 2 - C3 + 7'1 +n - 7'1 + 1
= (n; 1) + (n~ 1) _ C3
= (n;2) -~J'
Since the result is true for all p and all connected (p, p + n) - graphs. it is true for
all connected graphs. o
Let kn be the number of complete subgraphs of order n in G, n ~ 3.
Proof: 'We employ induction on n. For n = -1 we have for every tree G of order p
P( G, x) = Tp - OTp_1 + OTp_2 - OTp_3 + 01p_4 - •••
(-1+3) (-1+ 1)i.e, a3= 3 - 0 1 - (0 - 2(0)) = O.
Assume that for some fixed integer n, n ~ 0, the result is true fOI all connected
(p,p + n') - graphs, with nl :::; n -1. Now consider G, with p vertices and p + n
55
edges. Let uv be an edge in G and suppose that uv is on Tl triangles, i1 :5 C3i 1'2
induced cycles of length 4, i2 :5 C4; and tl complete subgraphs of order 4, tl :5 k4_.
e~·== C3-rl
c~ =- c" - r, + (r;) - t}
k~ = k. - I,
By Whitney's Reduction Formula, applied to the edge uv, we have
peG, x) = P(G',x) - P(G",x).
Now G' has p vertices, l! ;, n - 1. edges, c& = C3- 'I triangles, c~ = C,l -1':2 + (i) - tl
induced cycles of length 4 and k~ = k4 - il complete subgraphs of order ,1. Also Gil
has p - 1 vertices. p - 1+ n - 7'1 edges and c~ = C3 - il + I'l - tl triangles. Since
the result is true for G' and Gil we can write
and
P(GII.x) =
Hence
where
56
Now by the inductive hypothesis
and by Theorem 4.2.3
Therefore
where
n2 - 2nrl +3n - 3rt +rr + 2 rl(rl -1) 2rl 2rln=---- -----+-+-2 2 2 2
_ nZ - 2nrl + 3n - 3rI + rr + 2 - rr + rl + 2rI + 2rIn- 2
=n2 tan + 2
2
57
'lherefore
(n+2) (n+2\ (n+l)a3 = 3 + 2) - Ca 1 - (C4 - 2k4)
(n+3) (n+1\= 3 - C3 1) - (C4 - 2k4J .
Since the result is true for all p and all connected (p, p +n) - graphs, it is true for
all connected graphs. o
Example 4.2.1. (6,lO)-graphs
ws(l):::; 1
C3::' 5
Cs :::; 1
With G:
C3 = 5
C4 = 2
k4:::; 1and H:
we note that both G and II are (6,10)-graphs containing 5 triangles. Therefore
ao :::;1, al = n + 1 = 4+ 1 :::;5 and a2 = (nt2) - C3:::; (4t2) - 5 :::;15 - 5 :::= 10. Now
for graph G
= G) -sG) -[2-2(1)]
= 35-25- 0
= 10
and for graph H a3 = G) - 5(~)= 10. In fact, G and H have the same chromatic
polynomial: P(G,.r) =: P(H,::c) = T6 - 5Ts + 10T4 - lOT3 + 4T2•
(Such graphs are called chromatically equivalent and are studied in Chapter 5).
58
A broken wheel Hp(n) is a graph obtained from the wheel Wp on p vertices by
deleting n consecutive spoke edges, n :.::;p - 1.
Let k(2,3) be the number of induced subgraphs isomorphic to K(2,3); 'U'5(1) be
the number of induced subgraphs isomorphic to Ws(l) and W5 be the number of
induced subgraphs isomorphic to Ws,
Theorem 4.2.5.
- 2ws(l) - 3W5+ 3k4 + 6.~s}.
Proof. A tedious calculation which involves induction on p and n. o
Example 4.2.2. (6,1l).graphs
With G:
C3 =7
Ci:= 3
k4 :;: 2 and H:
ws(l) = 3
C3 = 7
C.I = 1
k., = 1
1.('.5 = 1
we note that both G and Hare (6,11)-graphs containing 7 triangles. Therefore
ao = 1, al = n + 1 = 5 + 1 = 6 and a2 = G) - 7 = 21 - 7 = 14. Also for G
a3 == (~) -7(~) -[3-2(2)] = 56-42+1 = 15 and for H a3 = m -7(~) -[1-2(1)J =5G -42 + 1 = 15. Now for graph G a1= (5;1) _7(5t2) - (-lWt1) - [0- G) - 0-
2(3) - 3(0) + 3(2) + 6(0)] = 126 -147 + 6 - [-21- 6 + 6]= 6 and for graph H a4 =m-7G)+(~)-[0-G)-O--2(0)-3(1)+3(1)+6(O)] = 126-147+6-[-21-3+3] = 6.
In fact P(G,x) = P(H, x) = T6 - 6Ts + 14T4 -15T3 + 6T2•
59
Abo. fOIthe graphs G ;''1dH from Example 4.2.1, we have for graph G a4 = (4t4)_
5(4t2)_(0)(4tl)_[0-(~)-0-2(1)-3(0)+3(1)+6(O)] == 70-75-0-F-IO-2+3] =70- 75+ 9= 4 and for graph H a4 = (414) - 5(4t2) - (O)et1) - [1- @ - 0--
2(0) - 3(0) + 3(0) + 6(0)J = 70 - 75 - [1 - 10J = 70 - 75 + 9 = 4.
The above technique can be used to obtain expressions for as,a6, .. ' of P(G,:c)
in terms of induced sub graphs of G. However, as sean above, the exercise would
become more tedious and the expressions more complicated.
4.3 Relating Coefficients
We now consider the relationship between the coefficients of a connected graph in
normal form with that of a connected graph in tree form.
Lemma 4.3.1. If G is a connected graph on p vertices with
P(G,x) = L:f':;(-l)ibp_ixP-i in normal form and
peG, x) = Ef':;(-1 'p_jx(x _1)p-1-i in tree form, then
k ( • 1)p-t -bp-k = ~ k _ i / ap-i and
k (. 1)-, k-i P - t-ap-k = ~(-l) k _ i bp-i.
,=0
=ap[;vp _ (P~l)xP-l + (P; 1)XP-2 -"'+(-l)P-lx]60
[
n_1 (P - 2) p-2 (p - 2) p-3 ()P-2 ]- ap-l X'- - 1 x + 2 x _ ... + -1 x
- + (-l)P-ltllX
[ (p - 1) (p - ?) ] 1 [(p - 1) (p - ")== apxP - 1 ap + 0 - Up-l xP- + 2 ap + J - ap-l
(p - 3) ] p-2 [(P -1\ (P - 2)+ () ap_2:r; - 3) ap + 2 ap-l
Now on equating coefficients we have
(P-l) (p- 2)bp-1 = 1 ap + 0 ap-l
(P-l) (P-2) (P-3)bn-2 = 2 ap + 1 ap-l + 0 ap-2
So that in general
(P-1) ('p-2\ (P-3)I ap + I. 1)ap-l + 1 ap-2 + ...
\ /'I.. '!" - tz - 2
61
k ( • 1)P-t-=?= k -i ap-i.=0
Similarly
(P-l)ap-l = bp-1 - 1 bp
(P-2) (P-l)ap_2 = bp_2 - 1 °p-l - 2 ap
(p- 2) (P-1):::::bp_2 - 1 bp_-1 + 2 bp
I al ",k ( l)k-i(P-i-l)bn gener, ap-k = ,,-,i=O - k-i . p-i· o
\Ve now prove Theorem 4.2.5 using Farrell's results. For completeness, we prove
Theorems 4.2.1 to 4.2.4 too in this way.
Let P(G,x) ee Li(-l)ibjxp-i• Then the following results appear in [OJ:
00 = 1
b1= q, the number of edges
(4.1)
- Cs - Oks + k(2, 3) + 2ws(1) + 3U's.
J2
ing coefficients, as before, in the two expressions for P(G,z), we have
bo = ao = 1
(P-l)bl = 1 ao+ al = q
Since G is a (p,p + n)-graph we bye
al ::::q - (p -1)
=p+n-p+l
(.I7+rt) ( ?)( +1) (p-l)(p-2)
:::: 2 - P - - n . - 2 - Ca
(p + n)(p -1- n - 1) - 2(pn + p - 21t - 2) - (p2 - 3p + 2):::: 2' - C3
(n+?):::: 2 - ca·
Also ba = (I';l)ao + (P;2)al + (1'~3)a2 + G3 :::::m - (q - 2)C3 - C,j + ,].7,.',1' Hence
63
(p + n)(p +n -l)(p +n - 2) (p -1)(p·_ 2)(p - 3)::;';':_-~---:::6--'-;:;__-_';_ 6
3(1- - 2)(p - 3)(n + 1) 3(p - 3)(n + 2)(n + 1)6 6
- C3(P +n - 2+p +3) - Col + 2k4
:.:: (p + n )(p2 + 2pn - 3p +n2 - 3n +2) ._ (p - 1)(p2 - 5p + 6)+6
-3(n + 1)(p2 - op + 6) - '(p - 3)(1.,2 + 3n + 2)6
- c3(n + 1) - (1.'4 - 2k,1)
(n+3)= 3. -- c3(n + 1) .~ (1.'4 - 2k4).
Also
(q) (q - '» (1.'3)b4::; 4 - 2 - C1 + 2 - (q - 3)C.1 + (2q - 0)k4
- 1.'5 - Gks + 1:(2,3) + 2ws(1) + 3ws
Hence
64
_. (p + n - 3)c{ + (2p+ 2n - 9)k4 - Cs - 6ks + k(2, 3) + 2ws(1) +3ws
== (p:n) _ (p~ 1)_ (p ;2) (n~'l) _ (p~J) (n~2)
_ (p ~ 4) (n; 3) _ Ca [_ (n {I) (p ~ 4) _ ~; 3) + (p +; - 2)]
- [-(p~ 4)(C4 - 2k4) + (p+ n - 3)C4 - (2p+ 2n - 9)k4]
- [CS - (c;) - k(2,::;) - 2w;,(1) - 3ll'5 +OkS]
The above expressions of the five coefficients can be used to compute chromatic
polynomials of all the graphs of order at most six (since each of them has at most
five non-zero coefficients). 'Ye now use other techniques to compute chromatic
polynomials of some gcaphs of higher order.
4.4 Higher Order Coefficients
Lemma 4.4.1. Let G be a connected graph. If G contains a complete subgraph Km
then
P(ll."m,.r)IP(G,x) or x{mJIP(G,x).
65
Proof. The number of ways of colouring Km. using x colours, ,., x{ml. If we fix
the colouring of these m vertices then there will be P;~,,:;)ways of colouxing the
remaining vertices of G. Hence xh'lIP(G.x). o
Remember that, by Lemma 2.4.4 with n = 1, the chromatic polynomial of the
complete graph can be expressed as P(J{m,x) = L~~lSm--iTi+l where Sm-i =sCm - 1, i) are Stirling numbers of the first kind. This will be used in our next
theorem. together with the fact given by Lemma 4.4.1: if the connected graph G
contains](m as a subgraph, then P(G,x) can be written as P(E'm' x) x f(x) where
f(x} is a polynomial with integer coefficients.
Theorem 4.4.2. Let G be a connected (p, p + n )-graph with
P(G,:r) = E;';;-;(-l)i ajTp_ .. If G contains a complete subgraph Ks, and
P(G,x) = P(Km'x) x f(x) with .f(.1:) = E~,:~mbjTp-m-i, then
bo = 1.
b _ (' al + 52 + 1)1 - - J '
and
_ ~r.! _ 2(h·.1 - (7) ))(a1 + ~2 + i) _ [c5 _ (C3 ~ (7))
66
-k(2, 3) - 2ws(1) - 3ws +3(k4 - (:)) + 6(ks - (~))]
where To= 1.
ProoE
P(G,x) = P(Km,x) x f(x)
p-2
== 2) -l)kakTp_kk""O
Hence
Now for
k = 0; we have slbo = ao = 1
Thus bfi == 1 since "1 = s( m - 1.m - 1) = 1
67
k = 1; we have slbo + Slbl + 52bO = -al
1+ b1 + 52 = -al
Thus b1 = -(at +S2 + 1) = _(01+:.+1)
k = 2: we have slb1 +S2bO +sIb2 +S2bl +S3bO = a2
bi + S2 + b2 + S2bl + S3 = az
Thus b2 = a2 - S3 + a1 - S2bl + 1since bI + S2 = -a1 -1.
Therefore
(Cl + 1) (-82 + 1) 2 [(m)]= 2 - 2 + a] + a1 S2 + 82 + 82 + 1 - C3 - 3
k = 3: we have 51 b2 + S2b] + S3hO + 8]b1 + 8202 + "3b] + S4bO = -a3
b2 + ~2bl + 83 + 1)3+ s2h2 + S3ht + "'4 = -«3
711U5 b] = -«3 - 84 - a2 - 8ah! - a1 - "2hz - Lsince b2 + S2bl .,.. 83 = a2 + al + 1.{is
But by Theorem 4.2.4
Therefore
-al(aj + l)(al +2) (-S2)(-S2 + 1)(-s2 +2)= 6 + 6
al(al + 1) (-82)(-S2 + 1)(al +S2 + 1)_..l. -a12' 2
= --ai - :Jai - 2al -- s~ + 3s~ - 282 - ..;ai - 3al + 3als~ + 3s~+6
-3<t1 $2 _- 3S2 - Gal - 3ai 52 - 6al S~ - 9a182 - 38~ - 95~ - 682 - 66
+ [ca- (~)][al+82+1] + [C4-2(k4- (:))]
= _ [ (at + ~2 -I- 3) _ (e3 _ c;}e1 + ~2 + 1) _ (C4 _ 2(l'.j _ (~)))]69
k = 4; we have
slb3 + 82b2 +S3bl + 84bO+81b4 +S2b3 +83b2 +S4bl +ssbo = a4
b3 + 82b2 + 83bl + 84+ b4 + S2b3 + 83b2 + S4bl + 85 = a4
Thus b4 = a4 - 85 + aa - S4bl + a2 - S3bZ + al - s2ba + 1 since
b3 + 82bZ + B3bl + 84,= -a3 - dZ - al - 1
But by Theorem 4.2.5 a4 = (41:3) - C3 (4Iil) - (C4 - 2k4)al - [cs - (~) - k(2, 3)-
2ws(1) - 3ws + 3k4 + Oks] and 85= (-81+3)- (';) (-'~+1) + 2(7)( -82) - [_(Cp) +
3(7) + 6(';)].
Therefore
[ (C3 \ ] (-82 + 3)- Cs - 2) -k(2,3)-2w5(1)-3ws+3k4+6ks - 4
+ (;) (-82/1) + 282 (':) _ (~)) +3(;)+6(';)+ (a1 t2) - c:}al - (C4 - 2k4) - r (-823+2) + (;)82
\.
+ 2(~) ] [al + 82 + 1J+ (al ; 1) - Ca - [ (-822+ 1) - (;) 1x
[(al + ;2 + 2) _ Ca ~ (;) ] + al + 82 [ (al + ~2 + 3)
70
= (a1+ 3\ _ (-82 + 3) (al + 2) _ (-82 + 2) ( 1)4) 4 + \ 3 3 at + 82+
(m) (m) [(-82 + 1) (-S2 +1)-2 4 (al+s2+1)+ 3 2 -s2(al+s2+1)- 2
- 2w5(1) - 3wtl+ 3k4. - 3(7) + 6k5 -. 6(;)]
+ ((~)) - k(2,3) - 2ws(1) - 3ws + 3k~ - 3(7) + 6ks- 6(;) ]= (al +:2 +4) _ (C3 _ (;))(a1 +;2 +2)
i1
+ 6 (k5 - C;))] since
_ (~) + ((~)) _ Ca (;) + (;) (;) = _(C3 -/;)). o
Again the same technique can be repeated to find expressions for bs, b6, ••• in the
situation of Theorem 4.4.2 but the details clearly become too involved.
Example 4.4.1.
With G:
C3 = 7C.1 =2
k4 = 1
W5 = 1
we have that G is a (8,14)-graph that contains a complete graph on 4 vertices so
that P(]{.1,X) == T4 - 3T3 + 2T2• Thus al == n + 1 == 6 + 1 = 7, 82 = -3 and
P(G,x) = (Xt - 3T3 + 2T2) X (boT4 + blTa + b2T2 + h3Tt + b4) where
bo = 1
hI == -Cal -I- 82 + 1) = -·(7 - 3+ 1) == -5
b _ (al +- 82 + 2) ( (m)' )2 - \ - ,C3-\ 2 3
(7-3+") (4)= 2 ~ - (7 - 3 ) = 15 - 3 = 12
b3 == - [ (a1 +;2 + 3) _ (C3_ (;) )(a1 + ~2+ 1) _ (C4 _ 2{k4 _ (~)})]
=_.[ C-~+3) -(7- e}C-i+1) -(2-2{1- (!)))]
::::-~35-15 - 2) == -18
i2
== C-!+4) - (7- (:)) C~~+2) -2(7-3+1)- [-C;4) -3(1)+3(1-1)]= (!) -3G) -2(5)-[-3-3]
= 70 - 45 - 10+ 6 = 21
Hence P(G,x) = (T.J.- 3T3 + 2T2) X (T4 - 5T3 + 12T2 -lSTl + 21).
Note that Sm-lbp-m-l + sm-1bp-m ::: (-1)p·-2ap_2, so that in the above example
for m ::: 4 and p = S we have
(2)( -18) + 2(21) ::: a6
a6 =42-36
=0
If the clique number of a graph is high then we can use Theorem 4.4.2 to compute
higher order coefficients.
73
Recall from Section 3.1 that G(Km' T) is a graph obtained from a tree T by replacing
the vertex v with Km. We will use this notation in our next theorem. Consider a
graph G of order n wherein if the chromatic number of G X(G) ;:::m ;:::3, m < n,
then P(G,x) canbe written asa linear combination of P(G(Km,T),x).
Theorem 4.4.3. Let G be a graph of order n with X(G) :::::m ;:::3, andm < n. Then
P(G,x)::: I;;;o\-l)iaiTn-i canbe written as
I;~;:( -l)ibi(l:;;,,,:;_l sCm -1, m_.k)Tn-i-k+l) iff
a}:= l:;:~2( -l)ibk_is(m -1, m - i-1) for 0 :::;k :s; n - 2
with bn-j = 0 for 2 :::;i :::;m-l.
Proof.
n-2I)-1)iaiTn_i;=0
= bo(.s(m - 1,71' - l)Tn + -1,m-2)Tn_1 t,,·+s(m-l.l)Tn __m+2)
- b1(s(m - 1,In -l)Tn-l + sCm -1, m - 2)Ta_2+ ... + s(m - 1, l)Tn-m+l)
+ ... + (-l)n-mbn_m(s(m -I,m -l)Tm + sCm -I,m - 2)Tm-1
+ ... +s(m -l,lm)
{:=} au=.s(m-·l,m-l)bo,
al =s(m-l,m-l)bt-s(m-1,m-2)bo,
az =: sCm-1, m - 1)b2 - sCm-1, In - 2)b1+ sCm -1, m - 3)bo•...
(-It-2an_2 = (-l)n-mbn_ms(m -1, 1)
74
m-2
{:::::? ak = 2:) -l)ibk_is(m -1, m - i -1)i=O
for 0 :::;k :::;n - 2 with bn-j = 0 for 2 :::;j :::;m - 1. o
Example 4.4.2. (7,ll)-graph
and P(]{4,X) = T4 - 3T3 + 2T2; so that
P( G(J(.l, T), x) = Tn - 3Tn-I + 21~_2 for 11= 7,6,5 and 4.
Now
P(G,x) = bo(T7 - 3T6 + 2Ts) - bI(Tf, - 3Ts +2T4)
where
1= ao = sCm --I,m -l)bo,
bo = 1
a1 = sCm -l,m -l)bI - sCm -I,m - 2)bo = bI - s(3,2)
(4+1)1 = bi + 3,
bi = 2
a2 = 8(3,3)b2 - 8(3, 2)bJ + 8(3, 1)bo•
(13 = 8(3,3)b3 - 8(3, 2)b2 + 8(3, 1)b1,
(4+3'
3 ) - e(5) - (1 - 2) = b3 - ( -3)(2) + 2(2),
Therefore P( G. x) = (T7 - 3Ts + 2T5) - 2(Ts - 3Ts + 2T4) + 2(T:; - 3T4 + 2T3) -
(T4 - 3T3 + 2T2)'
Theorem 4.4.4. Let G be a connected graph of order n with
P(G,x) = 2:~~1(-1);aix~~L m ~ 1. IfX(G) ~ m+2 then L:~ol(-l)jaj = o.
Proof
n-lP(G,x) = L(-l)iaix;~f
;==0
= aox(x -l)(x - 2) ... (x - m)" - alx(x -l)(x - 2) ... (x _ m)n-l
+a2x(x - l)(x - 2) ... {x - m)R-2
+ ... + (-1),,-lan_lX(x - l)(x - 2) ... {x - m)
= x(x -l)(x - 2) ... (x - m)[ao(x - m)n-l - alex _ m)n-2
+ ... + (_l)n-l(Ln_l]
Now since X(G) ~ m + 2 we have Pi G, m + 1) = O. Theretore
i6
o
~s a special case of Theorem 4.4.4 let m ;::::1 then
n-1'" . {l}P(G,x);:::: L.,,(-l)'aiXn_ii=O
n-1= :L:(-l)iuiX(x _1)"-;
;=0
n-1
= L(-I)ia;Tn+1_;i=O
and we have P(, . ,x) expressed in terms of the chromatic polynomials of trees where
Example 4. ;/..3.
~
~3 =4
With G: c.! = 1
cs;:::: 3
W5 = 1
we know that Gis:). (7,11)-graph with n ;::::4 so that
al;:::: (4~1) =5,a2 ;::::GY - 4 = 15 - 4 = 11,
77
a3 = G) - 4G) -1 = 35- 21= 14,
a4 := (!) - 4G) - 1G) - [3 - G) - 3(1)] = 70 - 60 - 5 - 3 + 6+ 3 = 11.
Now since X(G) ;:::3 we have I:~;g(-l)iai = 0 = Go - al + a2 - aa + G4 - as.
Therefore as = ao - al + a2 - a3 -l- a.j = 1- 5 + 11-14 + 11 = 4. Hence P(G,x) =
T7 - 5T6 + llT5 -14T4 + llT3 - 41.2.
This example illustrates how we can calculate the chromatic polynomial of any graph
G of order 7 with xCG) ;:::3 using the remark just before the example and the results
of section 4.2.
4.5 The Coefficients and Connectivity
If PCG, x) is the chromatic polynomial, in normal form, of a graph G on p vertices,
then it is known, see [14], that the multiplicity of the root 0 in P( G, x) is the number
of connected components of G.
Furthermore it is krown that the coefficients of PC G,.,;) alternate in sign. In tree
form, these matters are re! .ted by our next result (which appears in [3]).
Lemma 4.5.1. The coefficients of a chromatic polynomial P( G, x) = 2:f;; ajTp_i
alternate in sign. if and only if G is connected.
Pxooi: Let G be a connected (p,p+n)-graph. \Ve employ induction on p and n. For
n = -1,G must be a tree on p vertices and p -1 edges and hence P(G,x) = Tp.
Assume that for some fixed integer n, n ;:::0, the result is true for all connected
(p'.p' + n'l-graphs with p' :5 p and n' :5 n -1. Now consider any connected graph
78
G with p vertices and P+n edges" By Whitney's Reduction Formula applied to any
edge !lV, which is not a cut-edge (which exists because n ;::::0), we have:
P(G,x) = P(G',x) - P(G",x)
where G' is a connected graph onp vertices and p-l-rr=-L edges and Gil is a connected
graph OD p - 1 vertices and less than p + n - 1 edges. Since the result is therefore
true for G' and Gfl we can write
peG', x) = Tp - a~Tp_l -I- a2TI'-:l - a~11-3 +... and
P( Gil, X) = Tp_1 - a~Tp_2 + a~Tp_3 - ...
where all the a~, a~' are positive integers. Hence P(G,x) = Tp -_ (a~ + l)Tp_1 +(ll~ + arJTp_2 - (a~ + aq)Tp_3 + ... in which the coefficients alternate in sign. On
.he r eher hand, if G has k ;::::2 components and P(G,x) = Ef':;(-l)ibiTp_i =Ef;;(-l)ibjx(x - l)p-i-l with each bi ;::::0 and at least one bi positive, then
xkIP(G,x). But P(~,x) = E;;~(-l)ibi(X - l)p-l-i and with x = 0 this is
±Ef;; b, which is not zero. This contradiction proves the converse. o
We now aim at characterizing, in tree form, the chromatic polynomials of forests.
The result (Theorem 4.5.1) €xt.f'nds the characterization of the chromatic polyno-
mials of trees given in The 13 of [14].
Lemma 4.5.2. Let G be a graph with p vertices and p- k edges. Then G is a forest
if and only if G has k components.
Proof. If G is a forest with 1 components, Pi vertices and qi edges in the ith
component, then qi = Pi-1 for each i. Hence p-k = E:=o qi = E:=o(Pi-1) =:= p-l,
that is, I = I.'. If G has k components but is not a forest. then, in the above notation,
79
qi ~ Pi - 1 for each i with strict inequality for at least one i. Hence q, the number
of edges of G, satisfies q ;:: I:~=oqi > I:7=o(Pi -1) = p - k, a contradiction. 0
Theorem 4.5.1. ITG is a graph, then P(G,x) = I:;;:; (kjl)Tp_i if and only u G is
a forest with p vertices and k components.
Procd. ITG is a forest wi;h p vertices and k components, then
P(G,x) = I:;;:~(kjl) ...p_j by Corollary 3.3.2. Suppose P(G,x) = E;;:~(kjl)Tp_j.
Then P(G,x) = xk(x - 1)p-k (as shown after Corollary 3.3.2). Hence G has p
vertices (by Lemma 1.5.1), p- k edges (by section 4.3) and k components and hence
is a forest by Lemma 1.5.2. 0
Corollary 4.5.1. G is a tree of order p if and only if peG, x) = Tp. (k = 1) 0
We note that forests are not the only graphs which have no negative coefficients.
This can be seen from th~following example:
Example 4.5.1. If G is a graph on p = n + m + 1 vertices with G = en u I?m+l
where m + 2 ~ n + 1 ;:: 5 and n even, then by Corollary 3.3.3 we have
p-2 [n-2 ( )]" ,- i m+l ,P(;,x)=?:~(-1) '-i 1p_j)=0 .=0 J
=Tp+ [(m;l) _ (m;l) ]Tp_1+ [(m:1) _ (m;l)
+ (m;l) ]Tp-2 + ... + [(m; 1) - (::~) ]T3 + (::DT2
= Tp+ (7)Tp_~ + (~Jlj,_2 + ... + (7)Tm+l-i
80
To illustrate the above general formula consider n = 6 and m + 1 = 6, then p =
n +m +1= 12 and m := 5. Hence
peG,x) = T12 + G)Tu + G)TlO + G)T9 + G)Ts + G)T7
+T7 + G)T6 + G)T5 + G)T4 + C)T3 + G)T2= T12+ 5Tu + lOTIo + 10T9 + 5Ts + 27;'
4.6 A New Combinatorial Identity
We prove a combinatorial identity which, to us, is new and apply it in section 4.7 to
obtain the coefficients of a chromatic polynomial of some special family of graphs.
Theorem 4.6.1. For all nonnegative integers 11 and k with k 5 n + 1 we have
(n+1) ~lS.J( 1)i(n+i\(n+k'.2i) h L J' h .k == L...-i';O +». i}. k-2. ,were x IS t e greatest integer :5: $.
Proof. In order to prove this we note that the binomial coefficients (ntl),
k = 0, 1.... , n + 1 can be obtained as the unique solution of the recursion formula
fen + 1. k) = f(n, k) + fen, k - 1) with the boundary values fen + 1,0) :;;::;f(n +
81
only prove that f satisfies these equations, First, consider
lJJ ()( k-2') LfJ ( \( "_?')="<-1);~ n+ ." +).(_I';,n I n+h ,~z!-- l k - 2t ~__,) t - 1/' k - 2t.=0 .=0
, ('1'1. + 1) (n) (n)SInce i = i + i-I
Li}J ( ) ( + k 1 ?') LtJ ( ) ( k· 1 ?')="(-1);~ n '-, .:" +'\'(_l)i~ n+'- -,_I
!-- l k - 2l !-- z k - 1- 2z.=0 .",0
LLJ+ t<-l)i(, n ) (71 ' k -,2i)
;=0 Z- 1 Ie- 2z
L'}J ( ) ( k 2(' 1)) Li}J ( ) ( k 2')+ "( _1); ~ n + .- , l + + "(-1); ,n n + . -, t~ t k - 2(z + 1) ~. z - 1 le - 2z,=0 ,,,,0
=f(n,I~)+f()l,k-l)® +ot
= fen, k) + fen, k -1)
Notes:
® The second term of the previous step is f(n,k -1) if lk;lJ = lfJ and this
happens only if k is odd. For k even, however, the last term. of
"l f J I _l)i ('~)(n+k-~-~i) contains (n+k-2-~i) which is 0 since then k - 1 - ? lE.J =L.t,=o \ I {;-1-2. /,-1-2LtJ - 2
-1 < 0, Hence this summation effectively stops at l!J - 1 = l k;l J, '/ The terms
of the last two summations in the previous step cancel in pairs except two, which
82
are both zero. Finally
f(n+l,O)=(_l)o(n;l)(~) =1 and
f(n+ 1,n+ 1) = f(n,n+ 1) + f(n,n)
= G)f(n-l,n+l)+G)f(n-l,n)+G)J(, 1,n-l)
= G)f(n -2,n + 1)+ G)J(n - 2,n)+ (~)f(n - 2,n-1)
+ G)f(n -, 2,n-2)
= (~)f(l,n + 1)+ G)f(1, n)+ ... + (~)f(l,n + 1 - i)
+...+ f(l, 1)
=1
smce f(l, k) = I:ll~(_1)im (~:;Dhas only non-zero terms where i == 0 and i= 1
and they cancel: mm - G) (L~)= 0 llnl('~ k = 1, ill which rase i = 0 produces
o
Since fen+ 1 n+ 1) = ,,~~J(_1)i(ntl) (2n+1-2,i) = 1 we have, L.t.=o • n+1-21
1= L~J(_1)i(nt1)en+~-2i)
83
(-l)l~j (n+ 1)(2n+ 1- 2l~J)+ lnt1J n
= (~) en: 1) _ [(~) + (~)] en~-1)+ [G) + (7)] en; 3) -".+(-l)l~j[(l~J) +C~J-1)] en+l:2l~)
+(-l)l~j (l~J) [en+ 1~ 2l~J) _ en -1 ~ 2l~J)]
. l!!j-.!J(2n - 1 - 2l!!.;l:! J)+.,,+(-1) " 2, . n-l
lE-tlJ L~J= ~ (_1)i(;)(2~=:i)+ ta (_1)i(7)Cn~~~2i)
84
Hence
Corollary 4.6.1.
l~J ('J ?= >. (-1 \i (~) zn - _2 .) == O.7-' 'z n + 1 2z.=0
o
4.7 The coefficients of the chromatic polynomial of a cascade of graphs
We now use the combinatorial identity, presented in Theorem 4.6.1, which reminds
us of the inclusion-exclusion principle to interpret the coefficients of the chromatic
polynomial of a cascade of graphs. In Lemma 1.5.5 it was shown that if G is a
cascade of triangles on p vertices and (p + n )-edges then
Pi G, z ) = 2:;;;;( _1)i (Pi2, Tp_i = x(x -1)(.2: - 2)p-2.
Since G is a (p.p + n)-graph then the number of triangles in G is C3 = P - :2 and
.) Now if it P(G·) ,",p-2{ I)' '" ,",p-~( I)' (p-2)T.n = p -.." .,ow J we W1'1 e ,x = L..J,=o - Ui.Lp_, = L.t,=o - i p-i
then by equating the coefficients we have Uk = (Pk"2) = (c:) = (nil), Hence we
have:
Lemma 4.7.1. If G is a cascade of triangle!'! on p vertices and (p + n )-edges then
ak = r:1) = ~(-1)' (n 71) (n; ~;i2i)85
where lxJ is the, greatest integer :5 z ,
Proof. By Tl. eorem 4.6,7. o
We r.ote that if G is a (p, p + n)-gl.'l:Lphwhich is a cascade of triangles then
= (p - 2)' =~ _ i (p - 2) (p - 3 -I- k - 2i)ak k .{-I< 1) i k - 2i.=0
since n = P - 3 and
since C3 = n -I- 1.
Using this latter formula we have
ao = (~) = 1
al =c:1)a2 = (n;2) -C3
_ (n + 3) (n + 1)a3- 3 -C3 1
= (n +4) _ (n +2) (C3)a4 4 C32 +2
af = (n;5) _C3(n;3) + (~) (n{l)a6 =: (n;o) _C3(n;4) + (~) (n;2) _ (~) ...
80
If we now compare these coefficients with the general coefficients given in section
4.2 we note that some of these terms also appear in the general term, for example
("t4) - C3("t2) + (c2') also appeers in the general term a4 given in Theorem 4.2.5,
that is,
(n+4) (n+2) (n+l) (n+l)a4 = 4 - C3 2 - C4 1 + 2k4 1 - C5
+ (~) +k(2,3) +2w5(1) +3ws - 3k4 - 6k5•
In section 2.4 we defined the chromatic polynomial of a cascade of l{m+l 's as
xim} = x(a; - l)(x - 2) ... (x - rn)", Now a cascade of triangles, 1(3'5, on p vertices
and (p + n)-edges has the chromatic polynomial
x(X - l)(x - 2)p-2 = X~:!2
p-2
= 2)-1)iaiTp_ii=O
p-2"'" i {I}= 6(-1) aixp_l_i;=0
p-2
= L(-l)iaix(x _l)P-l-i;=0
where
( )L~J (). )ak= p-;2 = "'(_l)i k~ (P-3+f.,.-2i .
": 6 t k -. 2t,=0
If we now apply Lemma 2.3.3 by adding a new vertex and joining it to every vertex
of a cascade of triangles on p vertices we obtain the chromatic polynomial of a
87
----------------------
on (p -l 1[-vertices. Hence a cascade of ](4 's on p vertices has chromatic polynomial
"P-3(_1)ia'x{2} . where ak = (p-3) = ,,~tJ(_1)i(k1)(P-4+k:-2i).L.,,::-o , p-2-. \ k L.,,=O • I k-2.
Repeated application of Lemma 2.3.3 gives:
Theorem 4.7.1. A cascade of J(m's on p vertices has chromatic polynomial
"1;'-m+l(_1)ia'x{m-2} . where a'. = (p-m+l) = ,,~"'J(_1)i(k,?,)(p-m+k.-2i).L.J,=o • p-m+2-. " k L.J.=o , k-2.
We now generalize the combinatorial identity given in Lemma 4.7.1. Let m be
a positive integer where m 2:: 3. The graphs called m-qon-irees are defined by
recursion. The smallest m-gon-tree is the m-cycle, Cm, which is the only two-
connected graph containing m vertices and m edges. An rn-gon-tree with k + 1
m-gons (Cm 's) is obtained from an m-gon-tree with 10; m-gons by adding a new m-
gon which has Q.!!f edge in common with any m-gon of an m-gon-tree with k m-gons.
A 3-goo.-trce is the same as a cascade of triangles.
We now consider the chromatic polynomial of an m-gon-tree G with k rn-gons which
appears in [22] in normal form as P(G,x) = x(x -l){Q(Cm,x)}k where
Q(em, x) = P(Cm, x) -+ xCx -1) with P(Gm, x) = (x _l)m + (--l)m(x -I,.
If G is an m-gon-tree with k rn-gons then
factors.88
{m-2 }k
P(G,x) = ~(-l)iTm_i EB12I:-2,=0
{ T. r- {( 1)m-2m }km_l "" m- m-3 ••• - .1.2 'iD .L2k-2
where the sum is taken over all non-negative integers kl' k2, .•. , km-1 with kl +k2 +
... + km-1 = k, and
by the Multinomial Theorem. For m = 3 we have
P(C3,;z:) = {i:(-1)iT3_i}k @T2k-2.=0
where
which are the binomial coefficients. As an illustration consider the following
example:
89
Example 4.7.1. Let k :::::3 and Tn :::::4 then
P(G,x) = {t<-1)iT4-if@T4
:::::'\' ( 3 '){T4}kL{-T3}k2{T2}/;3@T4.'LJ klk2k3
Hence the coefficient of:
Ts::::: {T4}3{-T3}O{T2}O@:"4 is 313!0!0! = 1
Ti = {T4}2{-T3}1{T2)O@T4. is3!
2!1!0! :::::3
1'.,; = {T4}2{-T3}O{T2P@T43!
andis 2!0!1! :::::3
Ts = {T4P{-T3P{T2P EST.1 i.J 3tand1!111! = 6
= {T4}O{-T3}3{T2}O@T4 !'3!__ --1
013!0! - ~
T4 = {T4}O{-T3V{T2}1@T4 jo, 3!and012!1! :::::3
:::::{T.tP{-T3}O{T2}2@T4 i!3!
1!0!2! = 3
3!s 0!1!2!::::: 3
Thus P(G,.!:) :::::Ts - 3T7 +GT6 -7Ts +GT4 - IT3+ T2•
As can be seen from Example 4.7.1 it is tedio us working out the coefficients of an
rn-gcn-tree using the multinomial formula gin n above. 'Wenow generalize
90
Lemma 4.7.1 to give a useful computational technique to calculate these coefficients.
We begin with the following lemma:
Lemma 4.7.2. If a graph H consists of two subgraphs: G of order p with P(G, x) =Ef;;(-l)iajTp __ i and the cycle Cm of order m and they overlap in a 1(2 to form
H. Then P(H, x) = Ef,;;;,-4( -1)ibiTp+m_2_i with bk = ak +ak-l +...+ak-m<-2'
Proof By Lemma 2.2.2 we have
P(H,x) = [P(G,x) 0P(Cm,x)1@T2
p+m-4,= L (-1)ibjTp+m_2_i
j=O
. h b ,""m-2WIt k = £,.,i=O ak-i· o
Corollary 4.7.1. If H is as in Lemma 4.7.2 with G £:: Cm then {bk} is the sequence of
natural no-nbers bk =m-1-!k-m+21, that is, 1,2,3, ... ,m-2,m-1,m-2, ... ,2, 1
with a maximum coefficient of m-1.
'We notice that the coefficients described in this corollary first increase in absolute
value, and then decrease; two successive coefficients may be equal (as in Example
1..5.1 ~,-, P(G,x) = T~ - 3T4 + 3T3 - T2;. A sequence of numbers with this
91
property is called unimodal.
The Unimodal Conjecture. Let G be a connected graph with
P(G,x) = I:f':-;(-l)iajTp-i. Then the statement
aj > ai+l and ai+1 < ai+2
is false for every j = 0,1, ... ,p - 4.
This conjecture asserts that one never finds a coefficient flanked by larger coefficients.
The above conjecture can be replaced by a stronger assertion, see [15], namely:
Logarithmic Concavity. Let G be a connected graph with2 .
P(G,x) "" Ef':-o t-1)'ajTp_i' Then the statement
ajaj+2 :::;aJ+l holds for all j = 0,1, ... ,p - 4.
\Ve remark that the above conjecture does not hold in general for the chromatic
polynomials of disconnected graphs expressed in tree form: see Example 4.5.1.
Theorem 4.7.2. If G is a (p, p + n )-graph which is a rn-gon-tree where
P(G,x) = L:f':-;(-1)i'lj1p_j then
(4.1)
Proof. \Ve emp.v.) induction on n. For n = ° it is easy to check; lor n = 1 it is
harder to see and hence we do it. If n = 1, then G consists of two copies of Cm which
overlap in a 1(2. Hence, by Corollary 4,i.1 we have that ak = m -l-Ik - m + 21.
N id "L ':'lJ( 1);(2)· (1+k-(m-l)i) ii k 0 1 <) 4• ow cons] er LJ.=o - ; k-(m-l)i or' = , .. ,. ,_m-92
First: For k = 0,1, ... ,m - 2, this sum is
=k+l;
that is, the terms are 1,2, ,m - 1.
Next: For k =m-I,m, ,2m - 4 this sum is
~(-1);(~)(1+ k - (m - ~)i)L.t z k - (m - 1)'i==O "
= (_1)0 (2) (1+ k) + (_11 (2) (1+ k - (m -1))Ok) I; k - (m - 1)
::::k+ 1-2C + k-1(m-1))
::: k + 1- 2(1+ k - m + 1)
:::2m-k-3;
that is, the terms are m - 2, m - 3, ... , 1.
But these are exactly the values of ak for k = 0,1, ... , 2m - 4. Assume the result
is true for all m-gon-trees with p vertices and p + n + 1 edges, that is, with at
most n + 1 cycles on m vertices (em = n + 1). Now let II be an rn-gon-tree
on p vertices and p + n + 2 edges, that is, with n + 2 cycles where P(H,x) =
2:f;;-4( -I);b/Z;+m-2_i. By Lemma 4.7.2 we have that bk = ak + ak-l + ... +_ L~J i n+l n+k-i-(m-l)i· ?ak-m+2 where ak-i - 2:i=O (-1) ( ; )( k-j-{m-I); ) for) = 0.1, ... ,m --.
93
Hence we need only prove that
lm~'rJI:(_l)i(lI -: 2) [ri + 1+ k - (m-l)i),,,0 t i\ k-(m-l)i
lm:,J= L, (_l);(n-:l)(n+k-Cm-l)i);=0 ~ '. k-(m--l)i
L!;:."J+ I:C_1)i(n-:1)(n+k--1-(m-1)i)+ ...;=0 ~ .~-l-(m-l)i
l'-L~~2)J+ L (_1)i(n-:l)(·n~k-(m--2)-(m-~)i).
;=0 t k-(m-2)-(m-l)~ j
LbJNow L (_1)i(t),~2\.'n+1+k-(m-:-1)i)
;=0 \ ) \ k - err. - l)~
lm:"tJ= 2.:: (_1)1 (n -: 1) (n -1-1 + k - (m -l)i)i=O g. k - (m - l)i
lmkJ (+ L (_1)i ~+1)(n+1+k-(m-:-1)i)i--=O g - 1 k - ('n -. l)t
Lm:'tJ= L <_1)i(n-:1)(n+k-(m-1)i)
;=0 t k-(m-1)i
Lm:"1J+ :E (_l)i (n -:1) (n + k - (m -_ l)i)'
;"0 t k-1--(m-1)i
+ L~J(_ni(~ + 1\) (n + ~+ k - (m -:-l)i);=0 g - 1 k - (m _- 1}t
94
;::l~\ _1)1 (n -:1)(n + k - (m -1)i),=0 2 k - (m -l)i
lm':.,J+ I:(-1)' (n -:1) (n + k - 1- (m- l)i)
;=0 2 k - 1- (m - 1)i
Lm:,J+ I:(_1)i(n-:1)(n+k-1-(m-l)i)
1=0 ~ k - 2 - (m - l)i
+ LI:\_1)i(~+11(n+1+k-(m-:-1)i\\=0 ! - 1J k - (m - l)z )
l*rJ;::I:(-1)' (n :,1) (n -I- l~- (m - l)i\;=0 ~ k - (m - l)i )
+ l~J(_1);(n-:1)(n+k-1-(m-1)i);=0 2 k -1- (m - l)i
+2)-1)' (n -:1)(n + k - 3 - (m -l)i)i z k - 3 - (m - l)i
lm..l!:..r.l+ L (_1)i(~+1)(n+1+k-(m-l)i)
i=() 2-1 k-(m-1)i
L.n:,J== L (_1)i(n-:l)(n+~~-(m-.1)i)
1=0 z k - (m - l)i
l*-=\-J+ L C_1)i(n-:1)(n+k-1-. (m-1)i)
;=0 z k-l-(m-l)i
05
+ ... + 2)-1)i(n:- 1) (n+ k - (m - 2) - (m- ~)i)i ~ / k - (m - 2) - (m - .t)~
+ ""( _l)i(n:- 1) (n + k - (m - 2,)- (m -1)~)~' ,z t~-1-(m-2)-(m-1)t•
+ lI:\_l)i(~ + 1) (n + 1+~- (m -: l)i);=0 ~- 1 k - tm - l)z
The terms of the last two summations cancel. Hence
l~J _ i1n+2)(n+1+k-cm--1)i)t;;' ( 1)" i k - (m - l)i
Lm:,J ( 1) (' 1)')= 'C"' (_I)' n:- n+t: tm-. l~ l k-(m-1)z.=0
L k-(m-2) J+ ~ (_1)i(n+1)(n+k-(m-2)-(m-~)i).!:::- ~ k-(m-2)-(m-1)t.--0
As an illustration, let m := 4 and remember that C.l = n+ 1. Then
a . =~ ~tJ(_l)i (c:!)(n+k-~i) so thatk 6.=0 • k-3.
(n+o)ao = = 1,0,
96
o
(n+5) (n+2)as = 5 -C4 2
(On +6) (n + 3) (C4)an = 6 - C4 3 + 2 ...
Concerning Example 4.7.1 with k ;: 3:::: C4 and n + 1= 3, we have
ao = (~) = G) = 1
al::: eil) = 3
(2+4) (2+ 1)a4 = 4 - 3 1 = 15 - 9 ::: 6
(2+ 5) (2+2)as = 5 - 3 2 ::: 21 - 18= 3
(2+6) (2+3) (3)as = 6 - 3 3 + 2 = 28 -~30+ 3 ::: 1
Hence P(G,x) ::: Ta - 3T7 + 6T6 -7Ts + 6T4 - 3T;) + T2•
The above coefficients can be calculated without recourse to the formula (4.1). For
97
m = 3 we obtain the recursion formula bk = ak + ak-l, which is similar to the
recursion formula f(n + l,k) = f(n,k) + f(n,k -1) described in Theorem 4.5.1,
and use Pascal's triangle to obtain the coefficients of a 3-gon-tree or a cascade of
triangles. For m == 4 we obtain the recursion formula bl,; = ak + ak-l +ak-2 and use
it to obtain the following array of numbers, starting with the coefficient of T4 and
the absolute value of the coefficients of P(G4'X)::: T4 - T3 + Tz in the second row.
1
1 1 11 2 3 2 1
1 3 6 7 6 3 11 4 10 16 19 16 10 4 1
1 5 15 30 45 51 45 30 15 5 1
To illustrate how each entry is obtained in the above array we begin with the
second row with ao = 1, al = 1 and a2 = 1. Using the recursion formula bk ==
ak + ak-l + ak~·2 we obtain bo = ao = 1, b1 = a1 + ao = 2, bz := a2 + a1 + ao ""~3,
b3 = a3 +a2 +a1 = a2+a1 == 2, and b4 == a4+a3 +a2 == a2 == 1 which are the entries of
the third row. Now to illustrate how the coefficients can be calculated for Example
4:,.1 we begin with the third row with ao = 1, al = 2, a2 = 3, a3 = 2 and 04 = 1.
Again using the recursion formula bk = af,; + ak-1 + al<-2 we obtain bo = ao = 1,
b1 = a1 +ao = 3, b2:::: a2 +al +ao == 6, ba= a3 +a'2 +al = i, b4 = a4 -I-aa ta2 == 6,bs = as + a4 + a3 = a4 + a3 == 3 and b6 = a6 + as -I- a4 ::; a4 = 1, which are the
entries of the fourth row. The above can be repeated for any rn-gcn-tree.
98
CHAP;rER 5
CHROMATICALLY EQUIVALENT GRAPHS
5.1 Introduction
Two graphs G and H are said to be chromatically equivalent if they have the same
chromatic polynomial, that is, P(G,x) = P(H, x). A graph G is said to be chro-
matically uniqueif peG, e) = P(H,x) implies that G is isomorphic to H. In section
5.2 we dicuss families of graphs that are chromatically equivalent; in section 5.3 we
discuss chordal graphs; in section 5.4 we discuss chromatically unique graphs and
in section 5.5 we try to address the question: 'Which of the wheels Wp,p 2:: 4, are
chromatically unique?
5.2 Chromatic Equivalence
It is very clear that a chromatic polynomial does not generally belong exclusively
to one graph. There are many pairs of graphs which are chromatically equivalent
and nonisomorphic, For example two nonisomorphic trees of order p 2:: 4 have the
same chromatic polynomial- namely, Tp = x(x _l)p-lj n-gon-trees with n 2:: 3 are
chromatically equivalent but need not be isomorphic if they have more than two
n-gon cycles; cascades of J(m with me 3 ale chromatically equivalent but need not
be isomorphic if they have more than two J(m'S.
n seems natural to ask whether there is some way of proving that. two graphs are
chromatically equivalent without actually calculating the chromatic polynomial of
either of them. For some classes of graphs the answers are at hand.
Suppose G is a connected graph with blocks HI, H2, .•• , Hp and G has two nontrivial
blocks Hi and Hj which share a cut vertex v. Consider the graph F' obtained from
99
G by splitting v into two and p,,""inv; an edge of H; to an edge of Hi and growing
.a new edge onto this graph w +.i· ,'MS a new end-vertex, Then by Theorem 3.2.2
Pt G, x) = [P(HI, x)0P(H2, x)0" '0P(Hi,X)c:)P(Hj,x)0" '0P(Hp,x)] EB Tp-l.
Now in F we have that Hi and Hj overlap in ](2" Thus by Lemma 2.2.2 and Theorem
3.2.2 we have P(F,x) = [P(HI,x)0P(H2,x)0 ... (:::({P(ll"x)0P(Hj,x)} EB T2)0
T2 0 ... 0 P(Hp, x)] EB Tp-l = peG, e) while G ~ t' because F' has one more end-
vertex than G. Hence we have:
Lemma 5.2.1. If G is a connected graph with at least two nontrivial blocks then G
is not chromatically ..nique.
Example 5.2.1.
If G: then F:
By Example 3.2.2 peG, x) == T7 - 4T6 + 5T5 - 2T4
Let H:
then by Example 2.2.1, P(H,x) = Ts - 4T4 + 5T3 - 2T2• Hence by Theorem 3.2.1
P(F,x) = T7 - 416 + 5Ts - 2T4, so that P(G,x) = P(F,x) but G ~ F.
100
o
A cactus is a connected graph in which any two cycles are edge disjoint. We shall
show that all cacti with a given number of vertices, edges and cycles of each length
I = 3,4,5, ... are chromatically equivalent.
Lemma 5.2.2. All cacti with a fixed number of vertices, edges and cycles of sach
length 1= 3,4,5, ... are chromatically equivalent.
Proo£ Apply Theorem 3.2.2 to the cacti and their blocks. o
Example 5.2.2.
I£G, Nthen by Theorem 3.2.2 we have
P(G,x) = [P(Ca,x) 0 P(C4,X) 0 P(C3,x)lEB Ta_1
= [(T3 - T2) 0 (T4 - Ta + T2) 0 (T3 - T2)] EB T2
= Ts - 3T7 + 4T6 - 3T5 + T4
= P(H,x)
We now consider the case of disconnected graphs.
Theorem 5.2.1. If a graph G has nontrivial 2-connectcd components GI•G2 •••• ,
Gk+l and m trivial components. then G is chromatically equivalent to each graph
HU F where H is constructed from these nontrivial components of G by overlapping
a nontrivial G; with a nontrivial Gj, i =1= j, in 1(2 in any fashion until no more
nontrivial components remain and F is any forest with k + m components and
101
2k + m vertices.
Proof. P(lIUF,x) = P(1I.x)0P(F,x)+ [P(1I,x)0P(F, x)] EBTI by Lemrna 3.3.2.
But P(lI,x) = [P(Gl, x} 0P(G2,x)0" '0P(Gk+l,X)] EBT2k by repeated applica-
tion of Theorem 2.2.1 and Lemma 2.2.2, and P(F, x) = 2:7;:-1 (k+T1)T2k+II'_i
by Corollary 3.3.2. Therefore
[P(Gl'X) 0 P(G2,X) 0 •.• 0 P(Gk+l, x) 0Tm] ® TJ+l
102
· (k +m - 1,\ (k +m -. 1) _ (k +m)smce . J + . 1 - .~; ~- t
= P(G,x)
by Theorem 3.3.1. o
vVe remark that the condition that the Gi'S in Theorem 5.2.1 are 2-connected is
somewhat artificial. The same result can be proven without this condition but with
a longer proof.
Example 5.2.3.
If G: then
by Theorem 3.3.1, with k = 5,
we have:
P(G,x) =~ e j 1) [P(Ka,a:) 0 P(C4,x) 0 P(l(4,X) 0 T2] EBr,
=~ G) [(T3 - T2) 0 (T4 - T3 +T2) 0 (T4 - 3T3 +2Tz) 0TZ] EBTj
= ~ C) [TI3.- 5TI2 + lOTn -llT1o + 7Tg - 2TS] (BTj
103
= (T13 - 5T12 + lOTu -·llT10 + 7Tg - 2T8)
+ G) (T12 - 5Tu + lOTIO - nz, + 7Ts - 22"1)
+ G) (Tu - 5TIo + lOTg - llTs + 7T7 - 2T6)
+ G) (1'10 - 5Tg + lOTs -llT7 + 7Ta - 2Ts)
+ (!) (Tg - 5Ts + lOT7 -llT6 + 7Ts - 2T4)
= T13 - T12 - 4TH + 3TlO + 4Tu - 5Ts + 5T6 - T, - 2T",
Now consider
H: with F, I I " then•
I I e with k == 4•
104
where
and
4-1 (4-1)P(F,x) =L . T6-j
]=0 J
= t: (~)T6-jJ=O \j
Therefore
P(H UF,x) = P(H,x) 0 P(F,x) + [P(H, x) 0 P(F,x)] EBTJ
== (1;' - 5T6 + 10T5 -llT4 + 7Ta - 2T.) <:) (T6 + 3Ts + 3T4 + T3)
+ [(T7 - 5T6 + 10T5 -llT4 + 7T.~- 2Tz) 0 (Tn + 3Ts + 3T4 + T3)] EB Tl
5.3 Chordal G..aphs
ITG is a graph with X(G) = k + 1, then peG, x) =- 0 for x == 0,1, .. , I.:. Hence the
chromatic polynomial P(G,x) can be written as
= xmO(x _l)mt(x·_ 2)m2 ... (x - k)mk fex)
105
where mo,m!, •.. ,mk,n are positive integers and al,a2, ... ,an are integers with
mo +ml + ...+mk +n = p, the number of vertices of G and
ml + 2m2 +3m3 + '" + kmk +al =- q, the number of edges of G.
We now look at a class of graphs for which f(x) = 1, that is, P(G,:c) is completely
factorable over the nonnegative integers.
A graph G is called chordal (or triangulated) if every cycle of G with length greater
than 3 has a chord, that is, an edge joining two non-consecutive vertices of the cycle.
It can be shown, see [20J,that a graph is a chordal graph if and only if it can be built
up step by step, starting with a single vertex, each new vertex being joined to every
vertex of a complete subgraph of the existing graph; if the complete subgraph has n
vertices, then :c - n colours are available to colour the new vertex. Hence each step
of the construction contributes a factor x - n to the chromatic polynomial, which
must therefore be of the form:
peG, x) = xmO(x _1)ml(x - 2)"'2 •.. (x - k)m~ ,
for some integer k, where mo, ml! ... ,mk are positive integers. Clearly, k + 1 is the
size of the largest clique in G and the chromatic number of G. Conversely, given
....set of positive integers mo, mil'" ,mk we can usually construct several graphs
having P( G, x) as the chromatic polynomial.
Example 5.3.1.
G: H:
G and H are two of the graphs that have chromatic polynomial x2(:r. -1)3(x - 2).
106
It is an open problem to characterize those graphs G whose chromatic polynomials
P(G,x) have the form
vVenow make some remarks on this problem:
After expanding the :'ight hand side of P(G,x), the highest exponeus of x is
mo +ml + +m" which is, by Lemma 1.5.1, the number of vertices p of G; that
is mo +ml + + T7,,, =p, Also
= zrrlO ((xml" (7)xm1-1 + (~1)xml-2( _1)2 +...]
[xm• _ 2(~2)xm2-l _H(~2)xm'-2 + ]
[xma -3C~3)xm3-1 + 9(~3)xm3-2 + J
Therefore the coefficient of xmO+ntl+ ...+mk-l is -(ml + 2m2 + 3m3 + ... + km,,)
which by equation (4.1) in section 4.3 is the number of edges q of G; that is
ml + 2m2 + 3m3 + ...+ km" = q •
It is tempting to think, in fact it Was conjectured, see [5J, that
peG, x) = xmO(x - l)ml(x - 2)m .... (x - k)mk if and only if G is a chordal graph,
but this is not so. \Ve will now show how some non-chordal graphs whose chromatic
polynomials have the same form as P(G.x) described above can be constructed.
107
Lemma 5.3.1. Let H be an elementary subdivision of Kp. Then
P(H, x) = x(x -1)(x - 2) ... (x - p..l- 2)(x2 - 1)X + 2p -3).
Prooi. Let v be the "new" vertex (of degree two) of H. By 'Whitney's Reduction
Formula applied to any edge adjacent to v we have
P(lf,;,r) = T(Hl' x) - P(lI2' x)
h P(H ) P(Kp_1:x).P(J(p_llX).(X -1) b Thwere I,X = P(T.') y eorem 2.2.1.L\p_2,X
Therefore
)';)(H c) _ :rCx -1)(x - 2) ... (x - p + 2).xlx -1)(x - 2) ... (x - p + 2).(x -1)J. , ... - x(x-l)(x-2) •.. (.v-p+3)
- x(x -1)(x - 2) (x - p + 1)
= x(x - 1)2(x - 2) (x - p + 2)2 - x(x - 1)(a: - 2) ... (x - p + 1)
= x(x - 1)(a: - 2) (x - p + 2)[(x - l)(x - P + 2) - (x - p+ 1)]
= x(x -1)(x - 2) (x - p + 2)(x2 - px + 2p - 3). o
Example 5.3.2. For p = 6 in Lemma 5.3.1 we have
P(H,x) = xex -1)(x - 2)(x - 3)(x - 4)(x2 - 6x + 9)
= x(x -l)(x - 2)(x - 3)3(x - 4)
Hence
H: which is not chordal.
lOS
This example is given in [5J.We now generalize Lemma 5.3.1 to:
Theorem 5.3.1. Let G be a connected graph on p vertices with two complete sub-
graphs HI, on p - 1 vertices, and H2, on p - n vertices, which overlap in a complete
graph on p - n - m vertices, p > n+m. IfH is a graph obtained from G by adding
a new vertex v to G and joining v to two nonadjacent vertices of HI and H2• then
P(H, x) = x(x - 1)(x - 2) ... (x - p+ 2)(x - (p - n -1»)(x - (p - n - 2»)
... (x -- (p - n - m + 1») [x2 - (p - n - m + 2)x
+ 2(p - n - m + 2) - 3].
Proof By 'Whitney's Reduction Formula applied to any edge adjacent to v we have
P(H,x) = P(G1,x) -P(G2,x) where
P(G1,x) = (x -1)P(C z ) and P(G2,x) = peG + e,x)
where e is the edge between the two nonadjacent vertices of HI and H2• By Whit-
ney's Reduction Formula applied to the edge e we hwe
P(H,x) = (x -1)P(G.x) _ [P(G,X) _ P(J{P_~'X)P(J{p_n'x)]P(!\p-n-m+1,x)
(:..= I)P(J{p_t,x)P(J(p_n,x) _ [P(J(P-bX)P(J(p-n, x)p(J(p-n-m,X) P([{p-n-m,x)
_ P(J(P_bX)P(J(p_n,x)]P(J(p-n-m+l, x)
_ (x -1)P(J(p_t,x)P(J(p_n,x)- P(J...',,-n-m,x)
P(J(p-l,X)P(I{p-n,X)[x - (p- n -tIt + 1)]- . P(J{p-n.-m+ll x)
109
= ;rex -1)2(x ._ 2) ... (x - p + 2)(x - (p - n -1» (x - (p - n - 2»
· .. (x - (p - n - m») - xCx - 1)(x - 2)
· .. (x - p + 2) (x - (p - n - 1») (x - (p - n - 2» ... (x - (p - n _ m+ 1)2
= x(x -l)(x - 2) ... (x - p+2)(x - (p- n '1»)(x - (p -n - 2»
· .. (x - (p - n - m + 1»){(x - 1)(x - p + n + m) - (x - p + n + m - I)}
= x( x-l)( x - 2) ... (x - p + 2)( x - (p - n - 1)) (x - (p - n - 2» ...
(x - (p - n - m + 1»){,r2 - (p - n - m + 2).1: + 2(p - n - m + 2) - 3}. 0
Example 5.3.3.
For p = 7, n = 2 and m = 1 in Theorem 5.3.1 we have
P(H,x) = x(x -l)(x - 2) ... (.2: - (7 - 2 -1 + 1)){x2 - (7 -2 -1 +2).1:
+ 2(7 - 2 - 1+ 2j - 3}
= .r(.r -1)(x - 2)(x - 3)(.7: - 41(x - 5Hx'.! - 6x + 9}
= XIX -1)(x - 2)tx - 3)3(x - 4)(.r - 5)
Hence
H: which is not chordal,
110
5.4 Chromatically Unique Graphs
We now consider graphs which have a chromatic polynomial all to themselves, that is
if P(G,x) = P(H, x) then G ~ H, such a graph is said to be chromatically unique.
For example I;he empty graphs, the empty graphs with one edge, the complete
&"1:'?1·hsand the complete graphs without one edge are chromatically unique. In [7}
it is shown that the cycle en on n vertices if chromatically unique. Later in [12]
it is shown that all theta graphs are chromatically unique. Also in [17] it is shown
that the bipartite graphs Ktm, n) are chromatically unique; and in [24] it is shown
that the wheels, TVp on p vertices, are chromatically unique for odd p. The latest
result in [8] is that the broken wheel, Wp(1) on p vertices, is chromatically unique
if p is even.
,\Venow describe two families of graphs which are chromatically unique. For a proof
of the uniqueness, see [7]. The gcaphs in these families can be obtained from the
wheels by deleting some consecutive spoke edges, that is they are broken wheels (see
section 4.2),
The first family consists ot' the graph ~(p - 4), for p ~ 4, where the first graph
in this family W4(O) is the wheel rv.: (or K.t).
P 5
111c
,-----~----
By Whitney's Reduc+ton Formula applied to the edge 13 we howe
P(Hp(p - 4),x) = (:: - 2)P(Cp_bx) - (x - 2)P(CI'_2,X)
= (::- 2) [(x - 1)1'-1 - (;2: _1)P-2 + (-1)P-12(x -1)]
using P(Cp,x) = (x -1)1' -I- (-l)P(x -1) see section 1.5.
= (;;;- 2) [Tp_1 - 2Tp_z + ... + (-1)P-32T2] (5.1)
= (x - 2) [xex - 1)P-2 - 2x(x _1)P-3 + ... + (-1)P-32x(x -1)]
= x(x -l)(x - 2) [(x _1)P-3 - 2(x _1)P-4 + ... + (_1)P-32]
The second family consists of the graphs Wp(p - 5), for p ;:::5, where the first graph
in this family W5(O) is the wheel Ws.
Ws(O): lV.(I), l(5J
112
p 6
By Whitney's Reduction Formula applied to the edge 24 we have
p(Wp(p - 5), x) = (.1' - 2? P(Cp-2, e) - p(Wp-1(p - 5), x)
= (x - 2)2 [(x _1),,-2 + (_:...1),,-2(x -1)] - P(W,,-l(p - 5), x)
Also
,,-4P(Wp(p - 5), x) = (x -~ 2)2 2:( -ljiTp_2_i - P(Wp_l(p - 5), x)
;=0
::: (x - W [x(x _1),,-3 - x(x _1),,-4 + ... + (-l)"-'ix(x -1)]
- p(YV;"'l (p - 5), x)
= x(x - l)(x - 2) [{(x - 1) - 1}{ (x - 1),,-4 - (x - 1),,-5
+ ...+ (_1)"-4}] - P(Wp-1(p - 5),x)
= x(x -I)(x - 2) [ex - 1))>-3 - 2(x _1),,-4 + ...
+ (-1)P-·12(x -1) + (_1)1'-3]
- x(x -1)(;.: - 2) ffa: _1)1'-4
- 2(x - 1),,-5 + ... + (_1)1'-42] by (5.1)
113
Hence
P(Wp(p-5), x) = x(x-l)(x-2) [x_l)V-3 -3(x-l)V-4+4(x-l)P-5 - ... +( -1)P-331.
For p ~ 5 each Wp(p-4) contains exactly two triangles and for p ;:::6 each Wp(p-5)
contains exactly three triangles. It is natural to consider the chromatic uniqueness
of the corresponding family that contains exactly four triangles for p ~ 7. The
following example shows that Wp(p - 6), for p ~ 7, is not chromatically unique.
Example 5.4.1.
Consider
W7(l): andG: £JBy applying Whitney's Reduction Formula, we find that
p(n'7(1),x) = x(x - 1)(x - 2)[(x - 2)4+ (x - 2)2 - (x - 2) + 1] = P(G,x).
Clearly W7(1) ~ G.
We also describe two other classes of chromatically unique graphs. For a proof of
the uniqueness, see [11J.
Let rand s be integers such that r, s ~ 3. Let el be an edge of a complete graph,
K; on r vertices, and e2 be an edge of a cycle, C. on s vertices. Denote by G(r, s)
the graph obtained by identifying el and e2 in the disjoint union or K r and Cs• The
graph G(l',S) is chromatically unique for all r,s;::: 3.
114
G(r,s):
One may extend the structure of the graph G(r, s) to obtain certain new graphs in
the following way. Let 1(3 be a triangle in K«, r ;:::3, and let X be a Kt homeomorph
with a triangle J(~ as shown in (a) below. Denote by 971(1'), n ::::r + 1, the lass of
graphs of order n obtained by identifying ](3 and 1(£ in the disjoint
and X. All graphs, except two, in the class 971(1'), n ;::r + 1::::s, are chromatically
unique.
(a) (b)
The only graph 96(4) is the graph F.
F:
It can be verified, see Lemma 5.4.1, that
PCP, x) ::: x(x -l)(x - 2)(x - 3)(x2 - 4x + 5) ::: P(W6,;r) but We ~ P.
115
The graph H is one of the two graphs in g7( 4).
H: oIt can be verified that P(H,.r) == .r(x -l)(x - 2J2(x - 3)(x2 -_ 3,r + 4) = P( \" .. 1')
but X~H.
5.5 \Vheels
In [24] it is shown that the wheel Trp is chromatically unique for all odd integers p ;.::
5. The proof technique involved the fact that odd wheels are uniquely 3-colourable
graphs. ,,'c now consider the question whether the wheels Wp are chromatically
unique for even integers p ;.::4? W4 (which is isomorphic to 1(4) is chromatically
unique. lYe will use P(Wp+1,.r) ::: 2:f::;;( -1 )i.1'(x - 1)(,1: - 2)P-l-i from Corollary
2.3,1 to show that neither WG nor IVs is chromatically unique. The following result
also appears in [7]:
Lemma -5.5.1. Tr,; is not chromatically unique.
Pmo£.
1
PI n~j.x) = :L)-1)',r(J' - 1)(,!, - :W-ii:::::tl
= .r(.r -1)(.r - 2)1 - x(x ·-1)(.r - 2)3 + x(.r -1)(1' - 2f~
- ,r(.r -l)(.r - 21
== .r(.r - 1)( J' - 2) [(.z' - 21;j - lX - 2J~ + i.r - 2) - Ij
116
= :r(x - 1)(x - 2)[(x - 2?(x - 3) + (x - 3)]
= x(x -l)(x - 2)(x _. 3)[(x - 2)2 + 1]
= :r(x -l)(x - 2){x - 3)(x2 - 4x + 5)
~Th", if we start with lV,(I), "v7 which
by Lemma 5.2.2 with p = 4 has chromatic polynomial
P(W5(1~ e) == :I:(X -l)(x - 2)(x2 - 4x + 5), and add a new vertex and join it to
three pairwise adjacent vertices we obtain the graph
which is not isomorphic to iVa but satisfies P(.F, x) = (x - 3)P(Ws(1),x) by
Lemma 2,3.3. Also by Theorem 5.2.2 with p = 5, n = 2 and m = 1 we have
P(F, x) =: :c(x - 1)(a; - 2)(x - 3)(:v2 - 43:+ 5).
Renee
P(WG,x) = (x - 3)P(W~(1),;;;) = P(F,x)
where iVG ~ F. o
The following result also appears ill [24]:
Lemma 5.5.2. Wa is not chromatically unique,
117
Proof.
5
P(WS.X) = 2::)--1)ix(x -l)(x - t!t-i;=0
= :vCX- l)(x - 2)6 - X(X - l)(x - 2)5 + ... - :r(X -l)(x - 2)
= :r{X - l)(x - 2)[(x - 2)5 - (..: - 2)4 + ... -1]
= :rCx -l)(x - 2)[(x - 2)4(X - 3) + (x - 2)2(x - 3) + (x - 3)]
= x(x - l)(:r - 2)(x - 3)[(x - 2)4 + (x - 2}'l + 1]
= :r(x -l)(x - 2)(x - 3)[{:r - 2)2 - (x - 2) + l][(x - 2)2 + (x - 2) + J]
= x(x - 1)(x - 2)(x - 3)(x2 - 5x + 7)(x2 - 3x + 3)
But
2
P(H'5,X) = 2:)-1)iX(X --I)(x _ 2)3-;;=0
= xCX-1)(x - 2)3 - X(X - l)(x - 2)2 -l- .r{x - 1)(x - 2)
= x(x -1)(a: - 2}l(.c - 2)2 - (x - 2) + 1]
= xCX -I)(x - 2)(x2 - 3x + 7)
and
= x(x - 1)3 - .r(X _1)2 + x(x - 1)
= :rCX-l)[(x - I? - (x -1) + 1]
= ;r(x _1)(x2 - 3x + 3)
Thus P(TV!,;r) = (x - 3)P(Ws.x)(x2 - 3x + 3). Let HI be the graph which is
obtained from Kt and Irs by overlapping in a triangle
118
so that P(H1.x) = (x - 3)P(Ws.x) by Lemma 2.3.3, and Fl be one of the graphs
which are obtained from C4 and Hl by overlapping in an edge.
Thus by Theorem 2.2.1 we have:
P(F1.x) = P(Ht.X)P(C4.x)
;1:(x -1)
_ (x - 3)P(Ws.x)x(x -1)(x2 - 3x + 3)- x(x-l)
:::(x-3)P(W5.x)(X2- 3x+3)
= P(Ws.x)
Thus Fl is chromatically equivalent to We. Clearly Fi and We are not isomorphic
because they have different degree sequences. Also by Lemma 5.2.2 with p = 5 we
have
P(H2.x) = x(x - l)(x,- 2)(x - 3)(x2 - 5a;+ 7) where
}I"~~ 110
Now if the graph F2 is obtained from C4 and Hz by overlapping in an edge where
then by Theorem 2.2.1 we have
P(Fz,::c) = P(H~,x) X P(C4,x)x(x -1)
x(x -I)(x - 2)(x·- 3)(x2 - 5x + 7) X x(x -1)(x2 - 3x + 3)x(::c - 1)
== x(::c -1)(x - 2)(x - 3)(::c2 -- 5a:+ 7)(x2 - 3x + 3)
= P(H-'8,X)
Clearly F2 ~ TVa. Hence TVa is not chromatically unique. o
A computer investigation, see [15] revealed that TVa is not chromatically unique
since it has the same chromatic polynomial as 10 other graphs. It was conjectured
in [2-1]that the wheels lVj, are not chromatically unique for even p ;:::6.
We now consider lVlO and apply the same techniques used in Lemma 5.5.1 and
Lemma 5.5.2.
P(w]O, x) ::.:2:(-1)ix($ - 1)(x _ 2)8-;i=O
== xCx -1)(x - 2)8 _-x(x - 1)(x - 2)'7+ + (-I? x(x -- 1)(a: - 2)
:::::;xCx -l)(x - 2)[(x - 2)7 - (x - 2)6+ + (x -- 2) -1]
120
= x(a: -1)(x - 2) [(x - 2)6(x - 3) + (x - 2)4(x - 3)
+ ex - 2)2(X - 3) + ex - 3)]
= x(x -1)(a: - 2)(x - 3)[(x - 2)6+ (x - 2)4 + (x - 2? + 1]
= x(x - l)(x _. 2)(x - 3)[(x - 2)2 + 1][(x - 2)4 + 1]
= P(lV6,x)[(x - 2)4 + 1]
There is no graph with polynomial x(x-1) [(x-2)4+1] or $(.1:-1)(x-2)[(x-2)4+1].
Hence by this method we cannot obtain a graph that is chromatically equivalent to
~VlO'A computer investigation also in [15]revealed that lV'l) is indeed chromatically
unique. It remains an unsolved problem to determine whether or not the wheels Wp
a.e chromatically unique for even integers p ~ 12.
We used the above techniques to tackle thi problem for p ~ 12. Unfortunately
we found no graphs that are chromatically equivalent to IYp' We now give the
chromatic polynomials of some of these graphs in various forms.
P(W12,X) = xex -1)(x - 2)(x - 3)[(x - 2)8 + (x - 2)6 + (x - 2)4 + (x - 2)2 + 1]
= x(x -1)(x - 2)(x - 3) [(x - 2)4 - (x - 2)3 + ... + 1] [(x _ 2)4
+(x-2?+".+1)
= P(W7,X)(X - 3)[(x - 2)4 + (x - 2)3 + (x - 2)2 + (x - 2) + 1J
P(W14,x) = x(:e -l)(x - 2)(x - 3) [(x - 2?O + (x - 2)8 + ... + (x - 2)2 + 1]
= x(x -1)(x - 2)(x - 3) [(x - 2)4 + (x - 2)2 + 1] [(x - 2}6+ 1]
= peWs, x) [(x - 2)2 + 1] [(x - 2)4 - (x - 2)2 + 1]
121
P(W16,X) = X(X - l)(x - 2)(x - 3) [(x - 2)12 + (X - 2)10 + ... + (x - 2)2 + 1]
=: P(W9,x)(x - 3)[(x - 2)6 + (x - 2)5 + (x - 2)4 + (x - 2)3
+ (x - 2)2 + (x - 2) + 1)
P(fV18,X) =: x(x -l)(x - 2)(x - 3)[(x - 2)14 + (x - 2)12 + ...+ \x - 2)2 + 1]
=P(WlO,x)[(x-2)8 +1]
P(W20, x) = x(x -l)(x - 2)(x - 3) [(x -. 2)16 + (X - 2i4 + ... + (X - 2)2 + 1]
=: P(Wll'X)(X - 3)(x2 - 3x+ 3) [(x - 2)6 + (x - 2)3 + 1]
122
CONCLUSION
There are many unsolved problems relating to chromatic polynomials, see [14], [15]
and [22]. \Ve present a few in the following sections:
Characterization of Chromatic Polynomials
We begin with the question: What is a necessary and sufficient condition for a
polynomial to be the chromatic polynomial of some graph? In other words, is there
some way to determine whether a given polynomial belongs to a graph? We have
derived various necessary conditions for a polynomial to be the chromatic polynomial
of some graph - the polynomial must be monic. has terms that alternate in sign in
normal form (for connected graphs in tree form), has zero coastant term and so on-
but these conditions taken together Bl'C not sufficient. For example the polynomial
(see section 5.5 on wheels)
satisfies these conditions but is not the chromatic polynomial of any graph. From
the given polynomial we can determine the number of vertices, edges, triangles etc.
of a graph (if it exists) by looking at the degree of the polynomial, the first few
coefficients and applying the knowledge described in Chapter 4. We then generate
all such graphs and find their chromatic polynomials. This could work but it is a
tedious methcd and certainly not an effective way to clef-erminewhether a given
polynomial is chromatic.
Chromatic Equivalence and Chromatic Uniquer.ess
It is clear that two or more distinct graphs may have the same chromatic polynomial.
This prompts the question: What is a necessary and sufficient condition for two or
123
more graphs to have the same chromatic polynomial? It is easy to compute whether
two graphs are chromatically equivalent, but to determine whether ther- are other
graphs with the same chromatic polynomial is quite another problem. We have seen
in Chapter 5 how we can take some graphs apart at cut-vertices and reassemble them
by pasting edges eo get a different graph with the same chromatic polynomial. This
prompts speculation on the possibilty of an algorithm which will generate all the
graphs with a r,iven chromatic polynomial. This last problem is clearly an extension
of the first problem in this section, that of characterizing chromatic polynomials.
The same can be said about chromatic uniqueness. Is there a method of recognizing
whether a given graph is chromatically unique? The known results, see Chapter
5, were obtained by carefully wringing from a chromatic polynomial every drop d'
information about the gmph, until it can be shown that there is only one graph, At
present there seems to be no other way of resolving this problem,
The Unimodal Conjecture
A property that is very noticeable when one looks at lists of chromatic polynomials is
that the coefficients first increase in absolute value and then decrease; two successive
coefficients may be equal, but it seems that one never finds a coefficient flanked by
.arger coefficients, and it is natural to conjecture: If bj are the coefficients of the
chromatic polynomial in normal form, that the statemen;
is false for all j, This conjecture is replaced by the property of the strong logarithm.tc
con'!avity, namely that
The coefficients that arise when a chromatic polynomial is expressed in [acioriu!
124
form also seem t...·exhibit a unimodal distribution, and it bas been conjectured that
they too have the property of strong logarithmic concavity.
The coefficients that arise when a chromatic polynomial is expressed ill tree form
the situation is rather different. The unimodal property does not hold for many dis-
connected graphs, see Example 4.5.1, and examples are known. of connected graphs
for which three or more COlt. :ve coefficients ate equal, see appendix, thus contra-
dieting the property of'strong logarithmic concavity. However, no counter-examples
are known to the slightly weaker conjecture: ITaj are the coefficients of the chro-
matic polynomial of a connected graph in tree form, that these coefficients have the
property of (weak) /ogal'ithmic cf)ncavity, namely that:
Chromatic Polynomials of Families of Graphs
There appears to be no connection between the chromatic polynomials of a graph
and its complement. Hence the complements of any family of graphs can form a
fertile ground for further research. For further information see [15J.
Even within the area of finding general formulae for chromatic polynomials of fami-
lies of graphs, it is remarkably easy to formulate exceedingly difficult problems. For
example,
The m x n chessboard or lattice graph.
The graph Gmxn with vertices at the integer lattice points (u,v) with 1 .$. u :::;n,
1 .$. v .$. m, on the cartesian plane, and in which each vertex is joined to the four or
fewer vertices at distance 1 from it, is called the m X n chessboard 01' lattice graph.
Find a general formula for the chromatic polynomial P(Gmxll, x) of the graph Gmxn•
125
Arbitrary Trees of Polygons
Chao and Li, see [22J,proved the following characterization of m-gon-trees: A graph
G is an rn-gon-tree with k m-gons (where m ~ 3k ~ 1) if and only if P(G,:c) =x(x -1){QCCm, x)}k where Q(Cm,x) :::::P(Cm, x) -;-xCx -1), see also section 4.7.
It is an open problem to generalize the above characi .zation to arbitrary trees of
polqons, By arbitrary trees of polygons, we mean trees; of polygons where the length
of the polygons are not required to be the same.
Finally in section 5.3 we stated an open problem to characterize those graphs G
whose chromatic polynomials P(G,x) have the form
and in section 5.5 it remains an unsolved problem to determine whether or not the
wheels lVp are chromatically unique for even integer p ~ 12.
126
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129
130
APPENDIX
Tp = $(:' _1)1'-1 Tree FormCp = (x -l)P + (-l)P(x - 1)Cycle Form:z;{v} = x(x ~ l)(x - 2) ... (x - p + 1) Factorial Form
(pjq) Normal Form Tree Form Cycle Form Factorial Form(ljO) x T1 C1 x{l}
..==
(pjq) Normal Form Tree Form Cycle Form Factorial Form
(2;0) z2 T2+Tl C2+C1 x{2} + x{l}
•
•(2;1) z2 - x T2 C2 x{2}
l = x(2: - 1)
-
(pjq) Normal Form 'Iree Form Cycle Form Factorial Formx3 - X{3}+ 3:v{2}+ x{l}(3jO) T3+2T2 +Tl 03+302 +01
•
0$
(3;1) a;3_ X2 T3+T2 03+202 x{3} + 2x{2}
• =x2(x-1)
I--(3j2) x3_ 2x2+X T3 03+02 X{3}+ X{2}
1\= x(x - 1)2
(3;3) x3 - 3x2 + 2x T3-T2 03 x{3}
~
= X(3; - 1)(:c- 2)
(pjg) Normal Form Tree Form Cycle Form Factorial Form
(4;0) x4 74 +3T3 + 3T2 +Tl C4 + 4C3 + 6C2+ Cl x{"} + 6x{a} + 7x{2} + x{l}
• •
• •(4; I} x4 _ x3 T4+2T3+T2 04 +3C3+302 a:{4}+ 5x{S} + 4X{2}
• ;:::x3(x - 1)•
• ..(4j2) $4 _ 2x3+:r;2 T4+T3 C4+2C3+02 ${4} + 4a:{3}+2x{2}--. ;:::$2(a: _1)2
II
L(4;3) a:4_ 3:1:3+ 2X2 T4-T2 C4+03 -C2 a:{4}+ 3x{3}
V.;:::X2($ - l)(x - 2)
(PiQ) Normal Form Tree Form Cycle Form Factorial Form(4;3) x4 _ 3x3 + 3x2 - X T4 C4+C3 x{4} + 3x{3} + x{2}
n = x(x _1)3
~
I
(4;4) x4 - 4:v3+ 6:v2 - 3x T4-T3+T:z C4 X{4} + 2x{3} + x{2}
D= :v(x - 1)($2 - 3$ + 3)
(pjq) Normal Form Tree Form Cycle Form Factorial Form(4j4) x4 - 4x3 + 5x2 ...:.2x T4-T3 04 -02 x{4} + 2x{3}
1Z::::x(x - 1)2(x - 2)
(4;5) X4- 5x3 + 8x2 - 4x T4-2T3+TI) 04 -03-02 x{4} + :r.{3}
[SJ::: x(:r. -- l)(x - 2)2
(4;6) x4 - 6x3 + llx2 - 6x T4 -3Ts+2T2 04 -203- 02 x{4}
r:gJ::::x(x - 1)(x - 2)(x - 3)
~
(Pig) Normal Form Tree Form Cycle Form Factorial Form
(5; 0) • X5 Ts +4T4 + 6Ta +4T2 + TJ Co + 5C4+ 10C3 + 10C2 + C1 X{5} + lOx{4} + 25a;{3} + 15:1){2}+ :1){1}
• •
to •(5;1;-
xG _ 11:4 Ts +3T4 + 3T3 + T2 Cs + 4C4+ 6C3 + 4C2 :1){~~+ 9X{-II}+ 19x{3} +8X{2}
= a;4(x -1)I •
• "
'7 :v5 _ 2x4 + x3 Tu+2T4+T3 Cs + 3C4+ 3C3 + C2 x{5} +Bx{4}+ 14x{3} +4a:{2}.~
::::~3(a: _ 1)2
I• I1\• •
- -- -------- ------
(pjq).- ---
Normal Form free Form Cycle Form Factorial Fo- n
<6 xS - 3:v4 + 2:v3 Ts + T4 - Ta - T2 G5+2G4 - 2G2 x{5} + 7x{4} + 9a:{3}
== (l)3(x - 1)(:v - 2)
.. ~
<511\ (l;5 _ 3x4 + 3x3 _ x2 Ts+T4 G5+2G4 +Ga xiS} + 7x{4} + lOx{3} + 2x{2}
: x2(:1: _1)3
a
I)_J
"
u
NormalForm Tree Form CycleForm Factorial Form:1:5 _ 4x4+ 6x3 - 4x2 + X T5 Cs + C4 :z;{5}+ 6x{4} + 7x{3} + x{2}
=:o(x-1)4
!
VJ
I,
,_,.,,'---'(Pi q) Normal Form Tree Form Cycle Form Factorial Form1-----.-
;1:5 - 4x4 + 5:<:3- 2x2 xiS} + 6x{4} + 'iZ{3}(5;4) " T5-Ta Cs+ C4 - 03 - C2
L~== x2(:z; - 1)2(:r, - 2)
Q.__ .... I
/\•
1-,(5'1\ :c5 _ 4x4 + 6x3 _ 3:c2 Ts+T2 C5+C4+C2 xiS} + 6x{'l} + 7x{3} + 2X{2}
= x2(x - 1)(:v2 - 3x + 3)
V.(57 x5 - 5x4 + ]'ox3 - 10x2 + 4x T5-T4+T3-T~ Cr. xiS} + 5:v{'1}+ 5:;;{3}
= ;t;(:c - 1)(x - 2)(X2 - 2x + 2)
CJ I
,.....,--.(pjg) Normal Form Tree Form CycIeForm Factorial Form
(5;5) a;s ~ 5a;4 + lOa;3 - 9a;1l+ 3x T5~T4+T3 c; +C2 x{s} + 5x{4} + 5x{3} + :z:{2}
::::x(w ~ 1)2(x2 - 3x + 3)
--(5j5) x5 _ 5x4 + 9x3 - 7rr;2 + 2x Ts-T4 C5 -C3 x{G} + 5x{4} + 4x{3}
:= x(x - 1)3(X - 2)
/\
/\
(PiQ) - --Normal Form Tree Form Cycle Form Factorial Form(51\ :c5 _ 5:c4+ 8:u3 - 4x2._
1-:1; ;5}+ 5x{4} +3x{3}T5-T4-T3+T2 C5 - 2C3== x2(:I: -l)(x _ 2)2
IV I(/\ XU - 6,'1;4 + 14m3 - 15x2 + 6x T5 -2T4 +2T3 -12 C5 -C4 +C2 xiS} + 4x{4} + ax{3}
::::x(x - 1)(x - 2)(x2 - 3x + 3)
(v x5 - 6x4 + 15x3 -17x2 + 7x T5 -2:l~+ 3T3 - T2 C5 - C4+ C3 + 2C2 x{s} + 4x{4} + 4x{3} + x{2}
= x(x - 1)(x3 -- 5x2 + lOx - 7)
~~
(PiQ) Normal Form 'I!f;e Form Cycle Form Factorial Form
(5;6) \.to _ 6x4 + 13x3 - 123:2 + 4x 15 -2T,,+T3 Cs - C4 - Cs + C2 xiS} + 4x{4} + 2x{3}
= x(x _lyl(x - 2)2
V...:~
(iL_\ :to _ 6x4 + 11a:3 - 6x2 Ts- 2T4 - T3+ 2T2 Cs - C4 - 3C3 +C2 :c{S}+ 4X{4J
= x2(:c - 1)(x - 2)(x - 3)
f! //r-:
(PiQ) Normal Form Thee Form Cycle FJrID Factorial Form
(71\ rr;'5-7:c" + 19$3 - 23rr;2 + 10:c Ts - 3T4 + 4T3 - 2T2 Gs - 2G4 + G3 + 2G2 :c{5}+ 3rr;{4} + 2rr;{3}
= $(rr; - 1)(rr; - 2)(rr;2 - 4:c + 5)
V(5)\ rr;5- 7$4 + 18:c3 - 20.:r2 + 8x Ts -3T4+3T3-T2 G5 - 2G4 +2G2 rr;{5}+ 3x{4} + :c{3}
= :cCx -l)(x - 2)3
V/\[>< I
(PiQ) Normal Form Tree Form Cycle Form Factorial Form I(5; 7) Z5 _7x4 + 17x3 - 17x~+ 6x T5 -3T4 +2T3 C5 - 2C4 - C3+ 2C2 xiS}+ 3x{4}
= xCx -1)2(X - 2)(x - 3)
rg(5~
x5 - 8:1:4 + 24x3 - 31a:2 + 14x Ts - 4T4+ 6T3 - 3T:! Co - 3C4 + 2C3+ 3C2 xiS} + 2x{'I} + :x{3}
= x($ - 1)(:1:- 2)($2 - 5a:+ 7)
II \--,
(51A a:5 - 8x4 + 23x3 - 28x2 + 12x 75 - '!T4 + 5Ta - 2T2 Cs - 3C4 + Ca + 3Ca :ll{5} -I- 2x{4}
= :t:(:ll-l)(x - 2)2(x - 3)
~ --
ZO - 10z4 + 35;1;3- 50z2 + 24% Ts - 6T4+ llT3 - 6T2 Cs - 504 + 50a + 502 z{5}
:: z(x - 1)(2: - 2}(z - 3)(z - 4)
Normal Form
:£5 _ 9z4 + 29z3 - 39x2 + 18x
::::x(z - l)(x - 2)(x - 3)2
'free Form Cycle Form Factorial Form
(PiQ) Normal Form Tree Form Cycle Form Factorial Form(6;0) x6 x{6} + 15:v{s}+ 65w{4}
-~Ta+ 5Ts + 10T4 + IOTa + 5T2 +T1 Cf)+ 6Cs + 15C4+ 20C3 + 15C2 + C1
OIl +90x{3} + 31x{2} , x{l}II ....
'"Ii'
(6;1) $0 _$6 TlI+4T5 + 6T4 + 4T3 +T2 CG+ 5Cs + 10C4 + 10C3 + 5C2 X{6} + 14x{S}
". = x6(x -1) +55x{4} +65x{3} + 16x{2}"II ....(6;2) :c6 _ 2:e5 + x4 T· 3Ts + 3T.1+ T3 C6 +4C5 + 6C4 +4C3 +C2 X{6} + 13x{S}
~= x4(x _ 1)2 +46x{4} +46x{3} +8x{2}
" ..e
..
~I.. I
(pjq) Normal Form Tree Form Cycle Form Factorial Form
(6j3) ro6- 3ro5+ 3ro4 _ ro3 T6 + 2T5 +T4 GI)+ 3G5 + 30.1 + G3 ro{6} + 12:r.{S}
= ro3(ro-1)3 +38:t{4} + ,12ro{3}+ 4x{2}
:J•
11•
I -,___.
1': _j
(Pig) Normal Form Tree Form Cycle Form Factorial Form(6;3) reG_ 3re5 + 2:1:4 T6 +2Ts - 2T3 - T2 C6 + 3C5 + 2C4 -- 2C3 - 3C2 a;{6} + 12re{5}
~= re4(re -1)(:v - 2) +37x{4} + 27x{3}
• •• I
(6;4) a;6 _ 4:c5+ 6re4 _ 4:&3+ x2 T6+T5 C6+2C5 +C4 :I){6}+ l1X{5}
= re2(a: - 1)4 +31x{4} + 22re{3} + 2:v{2}
1':/1\
<0
(1•
(Pig) Normal Form Tree Form Cycle Form Factorial Form
(6;4) x6 _ 4:t5 + 6:t4 _ 4x3 + :1:2 16+T5 06+205+04 a:{6} + 1l:::{5}
= :t2(:t: _1)4 +31:t{4} + 22:t:{3} + 2:t:{2}
(:11
•
/ ........
~ I
(p;q) Normal Form Tree Form Cycle Form Factorial Form(6;4) m6 - 4x5 + 6x4 - 3m:j Ta+ Ts + Ta + T2 C6 + 2C5 + C4 + C3 + 2C2 m{6} + Ib{S}
• = m3(3:-1)(m2 - 3x + 3) +313:{4} + 23x{3} + 4m{2}
D0'
(6;4) :c6 - 4:1:° + 5x4 - 2x" Ts+T5 -74 - Ta C6 + 2C&- 2C3 - G2 X{6} + llx{5}
~
== $3(m - 1)2(x - 2) +30x{4} + 18x{3}
•
~
.......'" to
{Pig) Normal Form Tree Form Cycle Form Factorial Form
(6;5) :z:6 - 5:z:5 + 10:z:4 - 10x3 + 5:z:2 - :z: TG C6+CS :c{G} + lOx{5}
= xC:!: - 1)5) +25x{4.} + 15x{3} + x{2}
:;)rj):) I
I_.
(pj q) Normal Form Tree Form Cycle Form Factorial Form I(6j 5) m6_ 5x5 + 101:4 - 10x3 + 5:r.2- x T6 C6+CS ~{6} + lOx{S}
0 :::x(x _1)5) +25m{4} + 15;t;{3}+ x{2}
rS1)\
(6j5) 3)6 - 5x5 + 1Ox4 - 10x3 + 4x2 Ta -·T2 Co +C5 -C2 x{G} + lOx{S}
Q :::;1)2(x-1)(x - 2)(x2 - 2x + 2) +25:v{4} + 15x{3}
"
(pjq) Normal Form Tree Form Cycle Form Factorial Form
(6;5) x6 _ 5x6 + 1Ox4 _ 9x3 + 3x2 T6+T3 CG+c,+ C:, + C2 X{6} + lOx{5}
0 = x2(x -1)2(:t2 - 3x + 3) +253;{~}+ 16x{3} + 2X{2}
•
ID(6j5) x6 _ 5x. + 9x4 -7x3 + 2x2 Ts-T4 Cs + C5 - C4 - CS x{6} •.~ lOx{5}
0e- x2(x - 1)3(x - 2) +24:t{4} + 12x{3}
~ I LL'v
q) Normal Form Tree Form Cycle Form Factorial Form
5) x6 _ 5x5 + 9z4 _. 7x3 + 2:1)2 T6-T4 Co + C5 - C4 - C3 x{6} + lOx{5}
;::.:x2(x -1)3(x - 2) +24x{4} + 12x{3}
I
It
•
•
l(P;(6;
A
(pjq) Normal Form 'free Form Cycle Form Factorial Form
(6;5) :1:6 - 5x5 + 3;1;4 - 4x3 T6-2T4+T2 C6 + Cs - 2C4 - 2C3 + G2 X{6} + 10x{S}
It = x3(x -l)(x - 2)2 +23x{4} + 9x{3}
[ZJ•
(6j6) x6 - 6xs + 15x4 - 20x3 + l5x2 - 5x T6 - Til+ T4 ~. T3 + T2 G6 x{G} + 9x{5}
0 = x(x -1)(x4 - 5x3 + 10x2 - lOx + 5) +20x{4} + 10x{3}+ x{2}
T'-T'+"~-C'(6;6) x6 - 6x· + l5x4 - 20x3 + l4a;2 - 4x ziG} + 9X{5}
Q= x(x - 1)2(x - 2)(x2 - 2x + 2) t20x{4} + 10x{3}
I
-(pjq) Normal Form Tree Form Cycle Form Factorial Form(6; 6) x6 - 6x5 + 15x4 - 1()x3 + 12x2 -. 3x Ts-TIi+T4 06+03 x{6} + 9x{5}
1)= x(a; _1)3(a;2 - 3'c + 3) +2!lx{4} +l1x{3} + x{2}
t:J/
C1IClJ
(pjq) Normal Form Tree Form Cycle Form Factorial Form.(6j6) :v6 - 6x5 + 14:v4 - 16x3 + 9x2 - 2x T6-TS C6-C~ x{6} + 9x{5}
l}::: xCx·- 1)4(a: - 2) +I9x{4} + 8x{3}
I
VTlr)
I~
--(Pig) Normal Form Tree Form Cycle Form Factorial Form
(6j6) :uG - 6x5 + 14x4 - 16x3 + 9x2 - 2;1; T6-TS CO-04 x{G} + 9x{5}
== :::(x - 1)4(x - 2) +l9X{4} + 8X{3}
CI\j
~ I_.__L
-_ .._------------ .~---
(pjq) Normal Form Tree Form Cycle Form Factorial Form
(6;6) x6 - 6x5 + 15x4 -17x3 + 7x2 T6 - T5 +T4 +2Ta - T2 Cat 3C3 +C2 x{6} + 9x{5}
0 == x2(x - 1)(m3 - flx2 + lOx - 7) t20m{4} + 13m {a} + 2:v{2}
I•(6;6) m6 _ 6m5 + 14x4 - 15x3 + 6m~ T6 - Ts +23 - T2 C6- C4 +C3 :v{S}+ 9x{5}
0== 3:2(m - l)(m - 2)(m2 - 3x + 3) +l9x{4] + 9x{3}
•(6;6) ::;6 _ 6x5 + 13x4 _ 12m3 + 4m2 T6 - Ts ." T4 +Ta C6 -2C4 +C2 miG}+ 9m{5}
~= :l)2(m - 1)2(m _ 2)2 +18x{4' + 6m{3}
I =:>
tJ I--
, ,
jq) Normal Form Tree Form Cyc1eForm.._
Factorial Form
;6) x6 _ 6x5 + 13x4 - i',2:;:3 + 4x2 T6-Ts-T4+Ta 06 •• 204 +C2 x{G} + 9X{5}
= x2(x - 1)2(21- 2)2 +18x{4} + 6a:{3}
"I
..
iii
(p(6
(pjq) Normal Form _ Form ~Ycl' Form Factorial Form
Ts - Ts - 3T4 +T3+ 22~~ C6 - 4C4 - 2C3 + 3C2(6;6) :vS_ 8<1:6_ 11z4 - 6a;3 xiS} + 9:l;{S}+ 16a;{4}
" ;: x3(x -l)(x - 2)(a; - 3)
I:gJ II
0 I(6; 7) x3 - 7xs + 21$4 - 34x3 + 29$2 ':10;- T6 - 2Ts + 3T4 - 3T3 + T2 06 - Cs+C4 - 2C2 x{6} + 8x{li}
() = x(x - 1)(x - 2)(x3 - 4x2 + 7x - 5) +16x{4} + 7x{3}
(6; 7) ;1:6 - 7x5 + 20$4 - 30:v3+ 24$2 - 8x Ttl - 2Ts -:-2T4 - 2T3 + T2 C6- Os -C2 xiS} + 8x{5}
()= x(x - 1)(a; - 2)2(X2 - 2x + 2) +15x{4} + 5x{3}
(6;7) x6 - 7m5+ 21x4 - 33x3 + 27x2 - 9:;: Ts - 2Ts + 3T4 - 2T3 +T2 On - Cs + 04 + C3 - C2 x{G}+ 8x{S}
CD= :v(x -1)(a:2 - 3x + 3)2 +16X{4} + 8x{3} + x{'l}
(pjq) Normal Form Tree Form lJycle Forn:. Factorial Form(6;7) ! x6 - 7x5 + 21x4 - 32x3 + 24x2 - 7x Ta-2T5+3T4-Ts Cs - Cs + C4+ 2C3 - C2 x{6} + 8xiS}
~
= x(x - 1)2(:&3- 5x2 + lOx - 7) +16x{4} + 9x{3} + x{2}
©(6;7) reG _ 7x5 + 20x4 - 29x3 + 21x2 - I1x TG- 2Ts + 2T" - T3 C6 - C5 + Ca - C2 x{6} + 8X{5}
~
= x(x - 1)2(x - 2)(x2 - 3x + 3) +15x{4} + 6x{S}
/1]) I_j
~
(ViQ) Normal Form Tree Form Cycle Form Factorial Form1-----
xG -7x5 + 20x4 - 29x3 + 2l:l:2- 6x x{G}+ 8X{6}(6;7) T6-2Tst2T4-T3 C6 - Cs + C3 - C2
= :c(x - l)2(x ~ 2)(:1:2 - 3:1:+ 3) t15:c{4} + 6:c{3}
~
kj)
(pjg) Normal Form Tree Form Cycle Form F...ctorial Form(6;7) x6 - 7x5 + 19x4 - 25x3 + 16x2 - 4x Ts-2Ts+T4 Cs - Cs - C4+ C3 X{6} + 8:r.{5}
eJl = x(x - 1)3(x -~)2 +14X{4} + 4X{3}
vt~
~
(PiQ) Normal Form Tree Form Cycle Form Factorial Form(6; 7) x6 - 7$5 + 19:t4 ~ 25(1)3 + 16:c2 - 4:c T6 - 2Ts + T4 Ca - Cs ~ 0.1+ C3 x{6} + 8:c{&}
~
== :c(:c .-1)3(z - 2)2 +14:c{4} + 4:c{3}
.....
@q\J
(Pig) NormalI'orm Tree
(6i7) :r;6 - 7:&5+ 19z4 - 25z3 + 16:1:2- 4:& T6-= z(z -1)3(a: _ 2)2
r$11[21Ql
I
Fbrm Cycle Form21'c+ T4 C6- Cs - C4 + C3 ziG} + e:l:{5}
+14x{4}+ 4z{3}
(Pig) Normal Form Tree Form Cycle Form Factorial Form
(6i 7) x6 - 7x5 + 19z4 - 23x3 + 10x2 To - 2Ts + T4+ 2T3 - 2T2 Ca - Cs - C4 +2C3 Z{6} + 8x{5}
~
:= x2(x -1)(x - 2)(x2 - 4z + 5) +14x{4} + 6X{3}
"(6i7) 3)6 _ 7315+ 18a:4 - 20x3 + 8.::2 To - 2Ts + 2Ta - T2 Oe - Cs - 2G.l+ 2C3 +C" 3){O} + 8x{5}
~
= x2(~ - 1)(x-2)3 +13x{4} + 3x{3}
•
rS1e
(Pig) Normal Form 'free Form Cycle Form Factorial Form
(6i 7) x6 - 7x5 + 17:c" -17:c3 + 6:c2 T6 -2Ts - T" + 2T3 ' G6 - Gs - 3G4 + G3 + 2G2 x{6} + 8:c{5} + 12:c{4}
~
= :c2(:c - 1)2(:c - 2)(:c - 3)
I~
(6;8) :.:6 _ 8rcS+ 28:c" - 51:1;3+ 473;2 - 17x T6 - 3To + 6T4 - 5T3 + 2T2 Gil - 2Gs + 3G4 + G3 - 3G2 :c{6} + 7:c{3}
t9 = :c(:c -1)(:C" - 7:c3 + 21:c2 - 30x + 17) +13x{4} + 7:c{3} + :I;{2}
-(6;8) x6 - 8:c5 + 28x4 - 50:c3+ 44:c2 - 15:c T6 - 3Ts + 6T" -4T3+T2 Gs - 2Gs + 3G4 + 2G3 - 302 :c{6} + 1X{5}
tF= x(x - 1)(x4 - 7:c3 + 21x2 - 29x + 15) +l3x(4} + 8z{3} + :c{2}
I
(pjq) Normal Form 'free Form Cycle Form Factorial Form(6j8) xo - 8x5 +27z. - 47x3+ 4lx2 - 14z To - 3T. + 5T. - 4T3+ Tz 06 - 205 +204 + G''3 - 302 z{G} + 7X{5}
~
= z(z - 1)(z - 2)(x3 - 5:;;2+ 10:;;-7) +12X{4} +5x{3}
(6j8) X6 - 8x5 +26x4 - 44x3 + 39x2 - 14x T6 - 3TG+4T4 - 4T3+ 27'2 06 - 205 + 04 - 202 :;;{6} + 7x{5}
0 = xCx -1)(x - 2)(x3 - 5x2 +9x - 7) +l1x{4} + 2x{3}
(6j8) xO - 8x5 + 27x4 - 48x3 + 44x2 - 16x To - 3Ts + 5T. - 5T3+ 2T2 06 - 2C&+ 204 - 302 x{6} + 7x{5}
rj)= z(x - 1)(x - 2)2(x2 - 3x + 4) +l2x{4} + 4x{3}
.
(pjq) Normal Form Tree Form Cycle Form Factorial Form(6;8) :;:6 _ 8:r:5+ 26:r.4 - 43:r:3 + 37:;:2 - 12:;: Ta - 3T5 +4T4 - 3T3+ T2 C6 - 2Cs + C4 + Ca - 2C2 :;:{6}+ 7:;:{5}
¢;J= ret:;: -1)(:;: - 2)2(:1;2 - 3x + 3) +l1x{4} + 3x{3}
0~
kj)
(pjg) Normal Form Thee Form I Cycle Form Factorial Form- :&{6}+ 7:&{5}(6j8) :&6_ 8:;:0 + 26:&4 - 42:&3 + 33:&2 - 10:& To - 3Ts + 4T4 - 2T3 C6 - 2C5 + C4 + 2C3 - "~""2
~
= :;:(:&- 1)2(X - 2)(a:2 - 4a: + 5) +1l:l;{4} + 4z{3}
©Q
.......
(p;q) Normal Form Tree Form Cycle Form Factorial Form(6;8) x6 _ 8x5 + 25x4 - 38x3 + 28::2 - lix T6 - 3Ts + 3T4 - T3 C6 - 2C5 + 2C3 - C2 x{6} + 7Z{5}
t1J:::. :&(3: - 1)2($ _ 2)3 +10x{4} + 2r.;{3}
~
$~
(pjg) Normal Form Tree Form Cycle Form Factorial Form(6j8) x6 - 8:z:5+ 25:z:4 - 38x3 + 28x2 - 8;;: TG-3Ts + 3T4 - T3 Cs - 2C5 +2C3 - C2 x{s} + 7X{5}
= x($- !)2(X -2)3 +10X{4} + 2:z:{3}
ev~
~
(p;q) Normal Form Tree Form Cycle Form Factoria 1Form
(6; 8) ,1;6- 8x5 + 24x4 - 34x3 + 23x2 - 6x T6 - 3T6+ 2T4 C6 - 2C5 - C4 + 2C3 xiS} + +7:c{5}+ 9x{4}
= x(x -1)3(x - 2)(x - 3)
~
~
~(6;8) x6 _ 8x5 + 24x4 - 3lx3 + 14x2 T6 - 3T6+ 2T4 + 3T3 - 3T2 Cs - 2C5 - C4 + 5C3 xl6} + 7x{5}
~
;: x2(z - 1)(x - 2)(X2 - 5x + 7) +9:r{4}+ 3z{3}
I• .-
(p;g) Normal Form Tree Form Cycle Form Factorial Form(6;8) :c6 - 8x5 + 23x4 - 28x3 + 12x2 T6 - 3Ts +T,,+ 3T3 - 2T2 C6 - 2C5 - 2C4 + 4C3 +O2 X{6} + 7X{5} + 8X{4}
c§j = X2(<<. - l)(x - 2)2(x - 3)
•(6; 9) :e6 -. 9x5 + 36x4 - 75x3 + 78x2 - 3lx Ta - 4T5 + lOT4- llT3 + 5T2 C6 - aC5 + 6C4 - 03 - 6C2 x{6} + 6x{5}
@= :r(x _1)(:z:4 - 8x3 + 28x2 - 47x + 31) +llx{4} + 6x{3} + x{2}
(6;9) :16 - 9x5 + 34x4 - 67x3 + 67x2 - 26x T6 - 4T5 + 8T4 - 9Ta + 4T2 C6 - 3C5 + 404 - 03 - 502 X{6} + 6x{5}
~
:= x(X - 1)(:z:- 2)(x3 - 6x2 + 14x - 13) +9x{4} + 2x{3}
(6;9) x6 - 9xs + 34x4 - 66x:J+ 64x2 - 24x TG - 4T5 + 8T4 - 8T3 + 3T2 C6 - 305 + 404 - 502 x{G} + 6X{5}
r(fJ = :..(x - l)(x - 2)2(x2 - 4x + 6) +9x{4} + 3x{3}
_.
(PiQ) Normal Form 'free Form Cycle Form Factorial Form
(6i9) x6 - 9xs + 34:1:4- 65:1:3+ 61x2 - 22x T6 - 4To + 8T4 - 7T3 + 2T2 C6 - 3Cs + 3C4 + 2Ca - 4C2 x{6} + 6x{S}
~
::;:z(x - 1)(:,; - 2)(Z3 - 6x2 + 14z - 11) +9Z{4} + 4x{3}
-(6i9) x6 _ 9x5 + 33x4 - 62:t3 + 59x2 - 22x Ts - 4Ts + 7T4 - 'iT3 + 3T2 C6 - 3Co + 3C4 - 4C2 Z{6} + 6x{S}
~
= z(x - l)(x - 2)(x3 - 6x2 + 13x - 11) +8X{4} + xis}
(6i9) x6 - 9xo + 33x4 - 6lx3 -I- 56x2 - 20x T6 - 4T5 + 7T4 - 6T3 + 2T2 C6 - 3Cs + 3C4 + C3 - 4C2 x{6} + 6X{5}
:: x(x -1)(x - 2)2(X2 - 4x + 5) +8x{4} + 2z{3}
©
, ,
(PiQ) Normal Form Tree Form Cycle Form Factorial Form
(6;9) x6 - 9x5 + 33x4 - 61x3 + 56x2 - 20:r. To - 4Ts + 7T4 - 6Ta + 2T2 Cs - 3Cs +3C" +Cs -4C2 x{6} + 6x{5}
@= x(a: -1)(a: - 2)2(X2 - 4a:+ 5) +8x{4} + 2x{3}
$(6;9) a:6 _ 9x5 + 32x4 - 57a:3+ 51a:2- 18$ To - 4T5 + 6T4 - 5T3 + 2T2 C6 - 3C5 + 2C4 + Ca - 3C2 a:{O}+ 6a:{5}+ 7X{4}
r]1 = a:(a:-1)(a: - 2)(a: - 3)(x2 - 3x + 3)
(p;q) Normal Form Tree Form Cycle Form l!'actorial Form(6;9) :c6 - 9:c5+ 32$4 - 56$3+ 48:z:2 - 16x To - 4Ts + 614 - 4T3 + T2 C6 - 3C5 + 2C4 + 2Co - 3C2 :c{ll}+ 6:c{S}
== :c(:c - 1)(.: - 2)4 +7:cH} + :c{3}
~
~
~
-
-(pjq) Normal Form Tree Form Cycle Form Factorial Form
(6j9) :uS - 9x5 + 32x4 - 56x3 + 48x7. - 16x T6 -4Ts + 6T4 -4T3+T2 C6 - 3Cs + 2C4+ 2C3 - 3C2 X{6} + 6X{5}
Zlb= xCx -l)(x ~ 2)4 +7X{4}+X{3}
I
© ,
-(6;9) x6 _ 9:vo+ 32x4 - 55x3 +45x2 - 14x To - -iTo + 6T4 - 3T3 Cs - 3C5 + 2C4+ 3Ca - 3C2 x{6}+ 6x{S}
~
=x(x -1)2(x - ~)(X2 - 5x + 7) +7x{4} + 2:c{3}
~
(pjg) Normal Fonn Tree Form Cycle Form Factorial Form--
(6;9) a;6 - i):v5 + 31a:4 - 51a:3+ 40a;2 - 12:1: Ta - 4Ts + 5T4 - 2Ta 06 - 30s +04 + 303 - 2C2 a;{6} + 6a;{SI "" 6a;{4}
~
= x(a: -1)2(a: - 2)2(a: - 3)
~
~
@
(pjq) Normal Form 'free Form Cycle Form Factorial Form(6j9) x6 - 9x5 +29x4 - 39x3 + 18x2 T6 - 4Ts +3T4 + 4T3 - 4T2 C6 - 3C5 - C4 + 7C3 x{G} + 6x{S} + 4${4}
~
:;: :1:2(:1:-1)(:c - 2)($ - 3)2
•(6; 10) x6 - 10:1:5+ 42x4 - 90x3 + 95$2 - 38x T6 - 5TE + 12T4 - 1/lT3+ 6T2 C6 - 4C5 + 7C4 - 2C3 - 8C2 x{6} + 5x{S}
*= :v(x ~. 1)(x - 2)(x3 - 7a:2 + 19x - 19) +7${4} + 2X{3}
(6; 10) x6 - lfJx5 +41:1:4 - 85:1:3+ 87x2 - 34x To - 5Ts + llT4 - 12T3 + 5T2 C6 - 4CG + 6C4 - C3 - 7C2 x{6} + 5:t:{5}
0 == x(x -1)(:1: - 2)(a:3 - 7:1:2+ 18x -17) +6X{4} + X{3}
---(PiQ) Normal Form Tree Form Cycle Fbrm I Factorial Form(6i10) x6 - lOx" + 41a;4- 84x3 + 84x2 -- 32x Ttl - 5Ts + llT4 - 11T.1+ 4T2 06 - 4C5 + 604 - 702 x{6} +5:z:{5}
~
:::x(x - 1) (a; - 2)2(:c2 - 5x + 8) +6:c{4} + 2x{3}
«)(6i 10) x6 _ 10x6 + 40x4 - 79a;3+ 76x2 -- 28:z: T6 ._ 5T5 + 10T4 - 9T.1 + J3T2 06 - 405 + 5C4 + 03 - 6C2 X{6}+ 5:z:{5}
®:::x(x - l}(x - 2)2«(1;2- 5x + 7) +5x{4} + x{3}
I
Q'--
(pjg) Normal Form Tree Form Cycle Form Factorial Form(6j 10) xG - 10xo + 40x~ - 80x3 + 79x2 - 30x Ta - 5Ts + lOT4 - lOTs + 4T2 C6 - 405 + 5C4 - 6C2 ;t:M + 5X{5}+ 5x{4}
~
= x(x -1)(;t: - 2)(x - 3)(x2 - 4x + 5)
II
r(fJ(6i10) x6 -10x5 + 39x4 - 74x3 + 68x2 - 24x T6 - 5T5+ 9T4 - 7T3+ 2T2 Co - 4C5 + 4C4 + 2C3 - 5C2 x{o}+5x{S}+ 4x{4}
® = :v(x - 1)(a: - 2)3(X - 3)
,$
(Pig) Normal Form Tree Form Cycle Form Factorial Form(6i10) ;1:6 - 10$5 of 39;r.4 - 74x3 + 68x2 - 24:v To ._ 5Ts + 9T4 - 7T3 ;. 2T2 C6 - 4Cs + 40.1+ 2C3 - 5C2 x{G} + 5X{5} + 4X{4}
= x(x -1)(:v - 2)3(:v - 3)
~
~
(pjg) Normal Form Tree Form Cycle Form Factorial Form(6j HI) x6 _ 10x5 + 38x4 - 68x3 + 57x2 - 18x T6 - 5T5 + 8T4 - 4Ta C6 - 4C5 + 3C4 + 4C3 - 4C2 X{6}+ 5X{5}+ 3x{4}
~
= x(x - 1)2(x - 2)(:c - 3)2
~
(6j 10) :c6 _ 10x5 + 35x4 - 50x3 + 24x2 Ta - 5T5 + 5T4 + 5Ta - 6T2 C6 - 4C5 + 10G3 - C2 X{6}+ 5x{5}
® = x2(x -1)(x - 2)(x - 3)(x - 4)
•(6; 11) x6 _ 11:z:5 + 49:z:4 - 109x3 + 118x2 - 48:z: T6 - 6T5 + 15T4 - l8Ts + BT2 Co - 5C5 + 9C4 - 3G3 - 10C2 x{6} + 4x{5} + 4:c{4}
®== x(x - 1)(x - 2)(x - 3)(x2 - 5x + 8)
(PiQ) Normal Form Tree Form Cycle Form Factorial Form.(6; 11) x6 - l1xs + 49x4 - 108x3 +"115x2 - 46x T6 - 6T5+ 15T4 -17T3 + 7T2 06 - 505 + 904 - 20a - 1002 x{G}+ 4x{5}
~
= x(x - 1)(x - 2)(x3 - 8x2 + 23x - 23) +4x{4} + X{3}
*(6; 11) a:6 - llxG + 48x4 - 103:u3 + 107x2 - 42x T6 - 6T5 1- 14T4 - 15T3+ 6T2 06 - 505 + 804 - 03 - 902 :u{6}+ 4x{5} + 3a:{4}
~
= xCx - 1)(x - 2)(x - 3)(X2 - 5x + 7)
~
. IFactorial Form(pjq) Normal Form Tree Form Cycle Form(6;11) x6 - 11x5 + 47a;-t - 97x3 + 96x2 - 36x T6 - 6T5+ 13T4 - 12Ts+ 4T? (..'6 - 505 + 704.+03-8C2 x{G}+ 4~{5} + 2x{4}
~
= x(x -l)(x - 2)2(x - 3)2
~
~
(6j 11) x6 - 11x5+ 45:z:4 - 85:1)3 + 74:l)2 - 24:1: To - 6Ts + llT4 - 6T3 06 - 505 + 504 + 503 - 6C2 z{6} +4X{5}
~
::: :;:(:l; - 1)2(x - 2)(x - 3)(:c - 4)
(pjq) Normal Form Tree Form Cycle Form Factorial Form(6; 12) :z:6-12:z:5 + 58:1:4- 137x3 + 154:z:2- 64:1: To - 7Ts + 20T4 - 25T3 + llT2 06 - 60s -I- 1304 - 50a - 1402 x{6} + 3x{S}
0 = x(!!) - 1)(:z:- 2)(x3 - 9x2 + 29:1:- 32) +3x{4} + :z:{3}
(6; 12) x6 ~12a:5 + 51X4 -132:1:3 + 146a:2 - 60x T6 - 7Ts + 19T4 - 23T3 + 10Tz 06 - 60s + 1204 - 403 - 1302 W{6} + 3x{5} + 2X{4}
*= x(:c - 1)(a: - 2)(a: - 3)(x2 - 6x + 10)
(6; 12) ;1)6 -12a:5 + 56a:1-126:1:3 -I- 135x2 - 54:1: T6 - 7Ts + 18T4 - 20T3 + 8Tz 06 - 605 + 1004 - 03 - 1102 x{6} + 3x{S} -I- :v{4}
~
= x(:z:-1)(x - 2)(x - 3)3
*1 I
(p;q) Normal Form Tree Form Cycle Form.-
Factorial Form(6;12) x6 _. 12x5 + 55x4 - 120x3 + 124x2 - 48x T6 -7Ts + 17T4 ·-17T3 + 6T2 (16 _- 605 + 1004 - 1102 x{6} + 3x{5}
@= x(x -l)(x - 2)2(X - 3)(x - 4)
(6;13) f.C6- 13x5+ 66x4 - 161x3 + 185x2 - 78x T6 - 8T., + 24T4 - 3lT3 + 14T2 06 - 70s+ 1504 - 603 - 1602 X{8} +2x{5} + x{4}
~
:::::x(x - 1)(a: - 2)(x - 3)(x2 - 7x + IJ)
(6; 13) $6 _ 13x5 + 65x4 - 155x3 + 174x2 - 72x T6 - 8T5+ 2374 - 28T3+ 121'2 06 - 705 + 1504 - 503 - 1602 X{6} + 2x{5}
$= x(x -l)(x - 2)(x ,'3)2(x - 4)
(pjq) Normal Form Tree Form Cycle Form Factorial Form(6j14) $6 ~ 14x5 + 75x4 - 190x3 + 224x2 - 96x Ts - 9T5+ 29T4 - 39Ts + l8T2 Cs - 8C5 + 20T4 - 10C3 - 21C2 a;{6} + x{6}
~
:::: X(X - 1)(x - 2)(x - 3)(a: - 4)2
(6;15) a;6 -15x5 + 85x4 - 225x3 + 274x2 -120x T6 - lOTs + 3ST4 - SOT:)+ 24T2 C6 ,_ 9C5 + 25C4 - IliC3 - 26C2 x{6}
•:::: a;(a; -l)(x _ 2)(x - 3)(x - 4)(x - 5)
-
Author: Adam, A. A.Name of thesis: The Chromatic Polynomial Of A Graph.
PUBLISHER:University of the Witwatersrand, Johannesburg©2015
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