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COMBINATORIALMATHEMATICS

Y fAR

Febl'Ua:I']J 1969 - Juns 1910

. .

LEC'i.'UI'ES

on

CHftJdATIG POLYdO~·lU.LS

Itby I-1. '£. TU1'TE

University of WaterZoo~ Ontario

Department of StatisticsUniversity of North CapoUna at Chapel Hill

Institute of Statistics :t.'-;imeo Series tJo. 600.25

ApJr.J1, 1970

Guest lecturep during the Combinatorial Mathematics Year suppopted bythe u.s. Air Fopoo OJ:fic,? of Scientific ReseaPch under Contract No. AFOSR­68-1406.

LECtURES

on

CHf01ATIC Pot.YfOo1IALS

by W.T. Tutte+University of Waterloo, Ontario

1. INTRODuCtION

Consider a graph G with edge-set E and vertex-set V, both finite.

A >.-colouring of G, where A is a positive integer, is a mapping of V

into the set I A of integers from 1 to A, with the property that the

two ends of any edge are mapped onto distinct integers.

The integers 1 to A are commonly called "colours". The number of

A-co1ourings of G will be denoted by P(G,A). Recursion formulae are

known which permit the determination of the function P (G, A) of A for

any graph G. We proceed to list some simple observations, including the

statements of these recursion formulae.

1.1. If G is an edge1ess graph with k > 0 vertices, then

P(G,>.) = Ak •

For in this case, every mapping of V into I>. satisfies the defi­

nition of a colouring.

tGuest lecturer cJ.ur.ing the Combinatorial Mathematics Year supported by

the U.S. Air Force Office of Scientific Research under Contract No. AFOSR­88-1406.

2

It is convenient to postulate a null graph with no edges and no ver­

tices. For this graph, we make the obvious extension of formula 1.1 and

postualte that P{G,A)· 1.

1.2. Let k be a positive integer and let ~ be a complete k-graph.

This means that ~ has exactly k vertices, that each pair of vertices

are joined by exactly one edge, and that there are no loops. Then

P{G,A) • A{A-l){A-2) ••• (A-k+l) •

We can extend this rule to the case k· O. KO

is interpreted as

the null graph, so that

formula gives P{KO,A)

taken as 1.

P (KO

' A) is, by convention, unity. The above

as an empty product, and such a product is usually

1.3. If G has a loop, then

P{G,A) ... 0 for all >..

This is because the two ends of a loop coincide and so cannot have

different colours.

Suppose G has a complete k-graph ~ as a subgraph (k ~ 1).

Each A-colouring A of G induces a A-colouring ~ of ~, and may

be called an extension of Ak

• We note that

1.4. Each A-colouring of ~ has the same number of extensions.

To prove this, consider two A-colourings A and B of ~. Since

~ is complete, its vertices are all of different colours in A, and also

in B. It is, therefore, possible to find a permutation e of I A that

transforms the colour in A to the colour in B for each vertex of ~.

Evidently e transforms each extension of A into an extension of B,

and e-1 transforms each extension of B into an extension of A. Thus

3

e sets up a 1 - 1 correspondence between the extensions of A and the

extensions of B. The theorem follows.

We note the corollary

1.4.1. The number of extensions of any given A-colouring of ~ is

peG, A)P{"k,A) •

1.5. Let G be the union of two subgraphs H and K such that

HnK is a complete .f.-graph (.f. ~ 0). Then

P{G,X) ... P(H,X)P(K,X)P(HnK,X)

Pro>F: Let us deal first with the case in which .f. ... o. Then H

and K have no edge or vertex in counnon. Each X-colouring of P{G,A} is

then obtained by combining a A-colouring of H with a A-colouring of K,

and each combination of a A-colouring of H with a A-colouring of K

yields a A-colouring of G. The theorem follows for this case, since HnK

and P(HnK,A). 1.

For .f. > 0, the argument is similar. We start by constructing a A-

colouring Au of H. This induces a A-colouring Ao of HnK. To complete

a X-colouring of G, we have to combine '1I with an extension ~ of Aoin K. Clearly every A-colouring of G can be obtained in this way. But

we can choose ~ in P(H, A) 1"ays, and for each of these we can choose

~ in

P{K,X)P{HnK,X)

ways, by 1.4.1. The theorem follows.

!

4

Consider an edge A of G, not a loop. Let its ends be x and y.

Let GA be the graph obtained from G by deleting A but leaving x

and y. Let GA, be the graph obtained from G by contracting A to a

single vertex z. More precisely, we delete A and replace x and y

by a single new vertex z. An edge incident with x or y in G is

made incident with z in GA,. Incidences of an edge of GA, with vertices

other than z are the same as in G. In particular, an edge, other than

A, joining x and y in G becomes a loop on z in GA'

1.6.

PRooF: The ),-colourings of G can be identified with those ),-001-

ourings of GA in which x and y have different colours.

there is a 1 - 1 correspondence between the ),-colourings of

Moreover,

Gil andA

those A-colourings of GA. in which x and y have the same colour. This

1 - 1 correspondence is as follows: vertices common to the two graphs

receive the same colour in corresponding colourings, and z in Gil re­A

ceives the same colour as x and y in GA: The theorem follows.

1.7. Let two distinct vertices x and y of G be joined by two

distinct edges A and B. Then

P(G,).) = P(GA,A) •

This follows at once from the definition of a A-colouring. It can

also be regarded as a consequence of 1.3 and 1.6.

1.8. Let G be a non-null loopless graph. Then P{G,).) can be

expressed as a polynomial in A with the following properties.

(i) The coefficient of is an integer (all j) •

5

(ii) aj ~ 0 if and only if PO(G) S j ~ lvi, where PO(G) is the

number of components of G, and Ivl is the cardinality of V.

(1v) The non-zero coefficients alternate in sign.

PRooF: We proceed by induction over lEI. If lEI· 0, we have

P(G,A) • "Ivl

by 1.1. Moreover, PO(G). Ivl. Hence the four conditions are satisfied,

trivially in the case of (iv).

Assume as an inductive hypothesis that the theorem holds whenever

lEI is less than some positive integer q, and consider the case lEI =q.

Choose an edge A of G. It is not a loop, by hypothesis. By 1.6, we

have

P{G,A) • P(GA,A) - P{GA,A) •

By the inductive hypothesis, the theorem is true for P{GA,A). It is also

true for P(GA,A) unless GA has a loop, in which case P{GA,A). 0 by

1.3.

We deduce at once that P(G,A) can be expressed as a polynomial in

satisfying (1). We note also that (iii) holds for G, since G' hasA

one more vertex than GA.

The non-zero coefficients of GA extend from that of AIvl to that

of At, t. po(Gl), with alternation of sign, the coefficient of "Ivibeing positive. There are two possibilities for po(Gl); it is either

PO(G) or PO{G) + 1 according as the deletion of A fails or does not

fail to separate its ends. In the latter alternative, Gil has no loop.A

Gil has no loop, its non-zero coefficientsA

lto that of '" where l = PO(GA) = po{G),

sign, the coefficient of A1vl - l being positive.

extend from that of.

with alternation in

6

It now follows from the above equation that the non-zero coefficients

of P(G,~) extend from that of A/vi to that of APO(G), with alter-

nation in sign. The theorem thus holds when lEI a q. It follows, in

general, by induction.

When G has a loop, we can think of P(G,A) as a zero polynomial

in A. In all cases, therefore, we refer to P (G, A) as the Chromatic

pol.ynomia1. of G.

2. CHROMATIC POLYNOMIALS OF SOME SPECIAL GRAPHS

A k-arc ~ is a graph having k+l vertices aO,a1,a2 , ••• ,~, and

k edges A1'~' ••• ,~, the ends of Aj being a j _l and aj

(l S j S k). We identify LO

with the edgeless graph of one vertex.

2.1. P(~,A)k

• A(A-l).

PRooF: We have A choices for the colour of aO' If k > 0, we

then have A-l choices for the colour of ale If k > 1, we have A-1

choices next for the colour of and so on.

Let k be a positive integer. A k-circuit Ck

is a graph having k

vertices aO,al""'~_l' and k edges A1,A2""'~' the ends of Aj

being aj

_1 and aj

if j < k, and ~-1 and aO 1f j CI k. We note

that Ck

has a loop if k = 1 but not if k > 1.

k k(A-1) + (-1) (~-1).

7

PRa>F: If k· 1, then P(Ck' A) • 0, by 1. 3, and so the above

formula holds.

Assume as an inductive hypothesis that the formula holds when k is

less than some positive integer q ~ 2, and consider the case k" q.

Putting A· l\' we have

P(Ck,A) .. P(~_l,A) - P(Cg_l,A),

by 1.6,

..by 2.1 and the inductive hypothesis,

.. (A_l)k + (-l)k(A-l).

The theorem follows, by induction.

We note that P(C2 ,A). P(LI,A) by 1.7, and C3

• K3•

The wheel Wk

of k spokes, where k ~ 1, is formed from Ck

by

adjoining a new vertex v called the hub and joining v to each vertex

of ~ by an edge called a spoke.

We can obtain P(Wk,A) by using the following theorem.

2.3. Let a graph G be obtained from a graph H by adjoining a new

vertex v and then joining v to each vertex of H by a new edge. Then

P(G,A) • AP(H,A-l) •

PROOF: When v has been coloured, in anyone of the A colours,

only A-I colours are available for the vertices of H. On the other

h snd, any colouring of H in these A-I colours can be used to complete

a A-colouring of G. The theorem is now evident.

2.4. ...

8

To prove this, we apply 2.3 wi th G = Wk

and v the hub of Wk.

Then B" <1t, and we obtain P(H,A-l) from 2.2.

Having obtained the chromatic polynomial for the wheel Wk

, with hub

v we can carry the construction one step further as follows. We adjoin a

new vertex w and we join w to each vertex of Wk

, including v, by a

new edge. Let us denote the resulting double wheel by Uk. By 2.3 and

2.4, we have

2.5. ..k k

D ).().-1){().-3) +(-1) ().-3)}.

The double wheel Uk has an "axle" vw. When we delete this edge in

Uk' we obtain a bipyramid Bk • This is conveniently drawn on the sphere

with v and w at the North and South poles respectively. The circuit

St occupies the equator. v and ware joined to the vertices on the

equator by great-circular segments. When k ~ 2, this representation of

Bk

defines a map on the sphere whose faces are all triangles. Such a map

we call a triClYLfJUl,ation of the sphere.

2.6. ..

PRx:>F: We can identify Bk

with (Uk)A where A is the axle vw.

The graph (Uk)~ resembles a wheel, with a hub resulting from the identi­

fication of v and w, except that each spoke is doubled. However,

P(Bk,A) = P(Uk,A) + P(Wk,A), by 1.6

.. ).(A-l){(A-3)k+(-l)k(A-3)}

9

+ ~{{A-2)k+(_1)k(A_2)}

= ~{{A-l)(A-3)k+(A-2)k+(-1)k(~2_3A+l)}.

3. GENERAL A

As a polynomial, P{G,~) has a definite value when any given real or

complex number is substituted for ~. From now on, therefore, we shall

regard A as a complex variable, though usually we shall be concemed only

wi th its values on the real axis.

Polynomial identities that are valid for an integral variable remain

valid when this integral variable is replaced by a complex one. (Otherwise,

we could construct a polynomial with infinitely many zeros but not iden-

tically zero.) Thus the polynomial identities 1.3, 1.5, 1.6, 1.7 and 2.3

extend to complex A. In other words, we can apply our basic recursion

formulae without requiring that A be an integer.

3.1. Let G be a loop1ess graph. Let A be real and negative.

Then P(G,~) is non-zero. It is positive or negative according as the

number Ivl of vertices of G is even or odd.

I'ImF: This is clearly true when G is null and therefore P(G,~).

1. When G is non-null, the theorem is a consequence of 1.8.

A stronger result can be obtained by considering the quotient:

P(G.A)A

When G is non-null, this quotient is a polynomial in A and as such can

be assigned a value even when A = 0 (see 1. 8) •

10

3.2. Let G be a connected non-null loopless graph. Let A be real

and less than 1. Then

P(G,A)A

is non-zero. It is positive or negative according as the number Ivl of

vertices of G is odd or even.

PR:DF: We first dispose of two very simple cases.

Suppose that G is the edgeless graph of one vertex. Then peG, A)

... A and

The theorem is thus verified.

P(G,A)A

.. 1.

Next suppose G.~, i.e., G· Ll

• Then P{G,A)" A(A-l) by 1.2

or 2.1, and so

P(G,A)A

... A-I < 0 .

Again the theorem holds.

We proceed by induction over the number IEI of edges. If IEI • 0,

then G is the loopless graph of one vertex, for which we have already

verified the theorem. Assume the theorem to hold for IEI < q and consider

the case lEI ... q. (q is a positive integer.)

Suppose first that G is the union of two subgraphs H and K with

only a single vertex v in common, and that each of H and K has an

edge. Then H and K must be loopless, non-null and connected. Each

has fewer than q edges, and, therefore, each satisfies the theorem. By

1.5,

We deduce that

P(G,A)A

• P(B,A) • P(K,A)A A

11

P(G,>.)>.

is non-zero, and has the same sign as

h+k(-1) • (-1) lvi-I,

where h and k are the numbers of vertices of H and K respectively.

Thus G satisfies the theorem.

In general., choose an edge A of G.

It may happen that A is an isthmus, that is GA has two distinct

components each containing one of the ends of A in G. If each of these

components consists of a single vertex, then G I: K2

, and we have already

dealt with this graph. In every other case, G is a union of two sub-

graphs Hand K of the kind just considered.

It remains to consider the case in which A is not an isthmus. Then

GA and G~ are both connected non-null graphs, GA being 100p1ess. But

P(G,A) = P(GA,A) - P(G~,A)

by 1.6. If G" h 1 th P(G" ') • 0A as a oop, en A,A , by 1.3. Accordingly

P(G,A)A

But the theorem holds for GA, which has only q-l edges. Hence it holds

for G.

If G~ has no loop, the theorem holds for both GA and GA. Then

P(G,A)A

But GA has one fewer vertex than GA. Hence

and

have opposite signs. We deduce that the theorem is true for G.

The theorem now follows in general, by induction.

12

At the University of Waterloo, the zeros, both real and complex, of

the chromatic polynomials of some graphs triangulating the sphere have been

determined. This work was, of course, done with a computer. Polynomials

previously calculated by Ruth Basi and Dick Wick Hall were used.

The most striking regularity appearing in the results was that each of

the graphs studied has a zero close to

3 + 152

This is the number 2T D T+l, T being the usual symbol for the "golden

section" (1+/5)/2. Another way of describing the regularity is to say

that P(G,T2

) tends to be small when G is a triangulating graph of the

sphere.

An example of the effect is provided by the bipyramid Bk whose chro­

2matic polynomial is given by 2.6. When we substitute ;\. 0= T in 2.6,

2 2the quadratic form A -3A+l takes the value zero. Since T is approxi-

mately 2.618, the numbers A-2 and A-3 become less than 1 in abso-

lute value. We deduce that

3.3. co o.

4. THE VERTEX-ELIMINATION THEOREM

When a connected graph is drawn on the sphere, without crossings, we

obtain a spherical, or "planarn map. The connected regions into which the

graph separates the rest of the sphere are the nfaces" of this map. The

map is propera if the bO\mdary of each face 1s a simple closed curve. A

13

proper map in which each face is a triangle, i.e., is bounded by a 3-cir-

cuit of the graph, is a triangulation of the sphere.

The simplest example of a triangulation has two faces, three vertices

and three edges. Another example is the bipyramid Bk

, with k ~ 2.

There is an interesting theory of chromatic polynomials of planar

triangulation which so far applies only to the special value It

was constructed in an attempt to explain the computer results described in

§3. Its basic theorem is the subject of the present section.

A planar map M is triangutar at a vertex v if all the faces inci-

dent with v are triangles. The sides of these triangles opposite v

make up a connected graph L, though not necessarily a simple closedv

curve. The graph obtained from a graph G by deleting a vertex v and

its incident edges will be denoted by Gv

• In the present case, G isv

connected and defines a map M on the sphere. M is obtained from Mv v

by uniting the triangles incident with v so as to form a single new face.

It is,of course, not necessarily a triangulation.

When a map M is determined by a graph G, we refer to the chromatic

polynon:.ial of G also as the chromatic polynomial of M, and denote it

by P(M,).).

4.1. Let M be a map on the sphere, triangular at some vertex v.

Then

where m is the number of edges incident with v.

PRooF: If possible, choose M so that the theorem fails and the

number lEI of edges has the least value consistent with this condition.

14

Suppose first that the graph G of M is separable, that is it can

be represented as the union of two graphs H and K having only a vertex

x in CODlDlOll, and each having at least one edge. Then H and K are con-

nected graphs, by the connection of G, and x is distinct from v by

the connection of the graph L. Suppose v to be a vertex of K. Thenv

K defines a map N that is triangular at v, ,and G is the union ofv

H and K • By the choice of M, the map N satisfies the theorem. Hencev

2 -22m 1-m 2P(G,. ) ... • P(H,. )(-1) T P(K,. ),v

P(Gvt • 2) -2 2 2... • P(H,.) P(K ,. ),v

by 1.5. Combining these results, we find

p(G,.2) ... (_l)m T1- m P(Gv,t2).

From now on we may assume that M is non-separable. Hence it has no

isthmus.

Suppose next that M has an edge A that is not the side of a tri-

angle incident with v. If A is a loop, the theorem is trivially true,

by 1.3. From now on, therefore, we may assume M to be loop1ess. If the

ends x and y of A are joined by a second edge B, the theorem is

again trivially true for M, by 1. 7. We may, therefore, assume M to

have no 2-circuits whose edges do not both belong to the boundaries of

triangles incident with v.

Consider the graphs GA and GA. They are both connected, since A

is not an isthmus, and they define maps M' and M" on the sphere.A A We

form Ml from M by fusing the two faces of M incident with A into a

single new face, and we form MilA

from M by identifying all the points

of the edge A to form a single new point. Both M' and M" are tri-A A

angular at v, and both satisfy the theorem, by the choice of M. Using

1.6, we find

•

222P(M,T) 0 P(MA,T) - P(MA,T )

(_l)m Tl-m {P«MA>v'T2)_P( (MA)V,T2)}

(_l)m Tl-m {P«Mv>A,T2>-P«Mv)A,T2)}

15

•

by 1.6. Thus M satisfies the theorem, a contradiction.

From noW on we may assume that the only edges of M are the sides of

the triangles incident with v.

Suppose two distinct edges X and Y of M incident with v have

the same second end w. They make up a simple closed curve separating the

sphere into two Jordan domains, each containing some of the triangles in-

cident with v. Call these domains Dl and D2 •

\~Y

16

If we delete from the graph the edges and vertices inside Di and treat

Di as a single new face. we obtain a map Ni having a 2-goo incident with

v

x

w

17

Apart from the 2-gon, v is incident only with triangular faces in Nl

and Nr Let us suppose it is incident with~

triangular faces in NI'

and m2

in N2• Then m • m1+m2•

Let Hi be formed from Ni

by deleting X from its graph and fusing

the digon with the triangle incident with X so as to give a new tri-

angle.

w w

The maps Ml

and H2

are triangular at v. They satisfy the theorem by

the choice of H.

Let N be the map derived from M by deleting X from its graph and

by 1.5,•

fusing the two adjacent faces into a single new face. Then

P(H,T 2) • P{N,,2), by 1.7,

2 2P{Ml " )P(~,T )

2T • T

=

3 ml l-ml 2 m2 l-m2 2ell T- (-1) T P{(Ml)v''T )(-1) T P{(M2)v''T )

m -l-m 2 2(-1) T P«~)v,'T )P«M2)v''T ) •

But the graph of Mv is the union of the graphs of (Ml)v and (M2)v'

and these have only the vertex w in common. Hence

2 -2 2 2P{MV,T) = 'T P«Ml)v,T )P{(M2)v,T ).

18

Combining the above results, we find

We may now assume that all the m edges radiating from v in M

have distinct ends. Hence L is either a circuit or a complete 2-graph;v

the graph G of M is either a wheel of m spokes, or a complete 3-graph.

If M iii W, we have M III C. Hencem v m

2 2 -m m -1P(M,T) iii T (T +(-1) T ), by 2.4

•

... by 2.2 •

If H'" K3, we have Hv '" K2

•

P(M,T2) ... T2 • T

Hence, by 1.2,

• T-1 .,. T-1 P(M ,T2)v

In every case, we have shown that M satisfies the theorem, contrary

to its definition. We conclude that the theorem is true.

5. THE MAGNITUDE OF P(il, T2

)

In this section, we offer a theory to explain the empirical observation

that P(M, T2) is small when M 1s a triangulation of the sphere. We use

the vertex-elimination theorem to establish the following result.

5.1. If M is a triangulation of the sphere with vertex-set V,

then

5-lvlT •

19

Here Ivi is the cardinality of V as usual, but the vertical bars en­

2closing P(M, 'T) denote absolute value.

PRooF: If possible, choose a triangulation M so that the theorem

fails and Ivi has the least value consistent with this.

Suppose first that M has two edges X and Y with the same ends

v and w. H being a triangulation, it has no loops. The circuit made

up by X and Y separates the rest of the sphere into two residual domains

Dl and D2• As in the proof of the vertex elimination theorem, we define

2P(M, T )

and have

Let Vi be the vertex set of Mi " We note that

But Ml and HZ satisfy the theorem, by the choice of H. Hence

Ip(M,T2>! a T-3 \PCM1 ,T2>I • \PCM2,T2>I

-3 5-IV11 5-IV21T • T • t

=5-\V\

T •

From now on, we can assume that the graph G of M has no 2-circuit.

It is a well-known consequence of the Euler polyhedron formula that a

triangulation of the sphere has a vertex v of valency m ~ 5. The least

possible value of m is of course 2. Choose such a vertex v in M.

If m = 2, the two triangles incident with v have all three sides

in common, since there is no Z-circuit. Hence G is a complete 3-graph,

and

20

2 2 -1P(M,T) = T • T • T

5-lvlT •

If m Q 3, then M is also a triangulation, withv Ivl-l vertices.

I t satisfies the theorem by the choice of M. Hence

2 -2 2Ip(M,T ) I .. T Ip(M ,T >1, by 4.1,v

-2 5-( lVI-I).. T • T

.. 4-IVI 5-lvlT < T •

If m· 4, we consider the map Mv ' It is a proper map (since M

has no 2-circuit). It has one quadrilateral Q and its other faces are

triangular faces of M. It has Ivl-1 vertices.

Let the vertices of Q be, in their cyclic order a, b, c, d. Since

M is a map on the sphere, the pairs (a,c) and (b,d) cannot both be

joined by edges of Mv

• We may assume that a and c are not so joined.

We construct a triangulation T by subdiViding Q into two triangles by

a new diagonal edge A ClI (ac). We note that Mv '" T~ and that T~ is

also a triangulation of the sphere, except for the doubling of two edges.

We make it a true triangulation TO by retaining only one of the edges ab

and bc of Mv ' and only one of cd and da. We have P(TO,A). P(T~,A>,

by 1. 7. Hence

by 4.1,

-3 Ip(T,T2> + P(TO,T2)1, by 1.6,= T

S-3 {lp(T,T2>1+lp(TO,T

2>1} •T

But T and TO satisfy the theorem, by the choice of M. Hence

21

Ip(M 2)1 ' t-3 (t 5-( IVI-l)+r5-( IVI-2».t ~

• 5-lvlt •

Note that Ip(M .t2)1 S t

S- IVI by the above results.v

It remains to consider the case m· 5. In this case. M is a properv

map. It has one pentagon Q and its other faces are triangular faces of

M.

Let the vertices of Q be. in their cyclic order a, b. c. d, e.

Again. the pairs (a,c) and (b,d) cannot both be joined by an edge of

M and we assume that a and c are not so joined. We construct a mapv

N by subdividing Q into a triangle and a quadrilateral by a new diag-

onal edge A. (ac). We can identify N with the map denoted by M inv

the preceding argument, thereby deducing that

s-ivit •

If, in the map N~, we retain only one of the edges ab and be of

M, we obtain a triangulation T. It has only Ivl-2 vertices and sov

satis fies the theorem. Thus

by 4.1,

-4Ip(N,t

2) + P(T,t2>1, by 1.6,=- t"

-4Ip(N,/) I + t

-4Ip(T,t

2) IS t

S4-lvl 3-IVIt t

• 5-IVIt .We find that in all cases M satisfies the theorem, contrary to the

choice of M. We conclude that the theorem is true.

22

6. RECURSION FORKULAE FOR TRIANGULATIONS

The recursion formulae of Section 1 enable us to calculate the chro-

matic polynomial of any graph G. But if G is fairly complicated, they

require us to work out the chromatic polynomials of many smaller graphs

before we get to that of G.

The graphs whose chromatic polynomials have aroused most interest so

far are those of the triangulations of the sphere. If G is such a

graph and A is one of its edges, not a loop, then GA does not usually

represent a triangulation, though G" does so wIess it has a loop.A

It

1s desirable to find a recursion formula that works with triangulations

only.

One such formula is 1.5, in the case l· 3, and this is very helpful

in practice. Another, concerned with a separating 2-gon, has already been

used in the proofs of 4.1 and 5.1.

There is another formula for triangulations that can be derived from

1.6.

Suppose A is an edge of a triangulation M. Let Tl

and T2 be

its incident triangles. We denote the ends of A by x and z, the third

vertex of Tl

by Y and the third vertex of T2

by t.

Y

I~X\;jZt

We assume that x, y, z and t are distinct.

23

From M we can derive a number of other maps. We write N for the

map ob tained by deleting A from the graph so as to replace T1 and T2

by a quadrilateral Q. We write 6A

M for the graph obtained from N by

subdividing Q into two triangles by the diagonal B. (yt). It is often

said that eAM is obtained from M by "twisting" A.

In M we can contract A to a single vertex and delete the edge yz

of T1 and the edge zt of T2 • This operation yields a map ~AM, which

is a triangulation unless the contraction of A introduces a loop. We

denote the map ~B (6AM) also by JIJAM.

Using 1. 6 and 1. 7, we find

(i) P(M,A) + P('AM,A) • P(N,A).

Similarly

(ii) P(6AM,A) + P(J1JAM,A) • P(N,A) •

Hence we have the recursion formula

This formula involves triangulations only, except that loops may ap­

pear in J1JAM or $AM. In theory, we can use it to construct a list of chro­

matic polynomials of triangulations with vertex-valences ~ 3. Suppose we

have listed the polynomials up to a particular value of lvi, say Ivl· q.

Choose a map M with Ivl· q+l and let 8 ~ 3 be its'sma1lest vertex

valency. If 8· 3, we use 1.2 or 1.5 to calculate P(M,A}. If 8 = 4,

we apply 6.1 with A incident with a tetravalent vertex. Then 8· 3 in

eAM and so P (M, A) is expressed in terms of known polynomials. Similarly,

if 8· 5 in M, we can arrange that 8· 4 in eAM. Of course, 1f M

has a separating 2-gon or 3-gon. we can apply a simpler formula.

24

The recursion formulae used in practice are based on 6.1.

It is a curious fact that when A Ilt T2 the four polynomials of 6.1

satisfy a second linear relation

6.2. 2 2P(M,T ) + P(9AM,T )

2 2For A Q T, we can, of course, eliminate P(9AM,T) from 6.1 and

6.2, and so obtain P(M, T2

) directly in terms of triangulations with

fewer vertices.

PRooF OF 6.2: We can obta:tn from N a map N by subdividing the1

quadrilateral Q into four triangles by introducing a new central vertex

v and joining it by new edges to x, y, z, and t.

y

x

t

Z

The map (Nl

) liB' where B D (xv), has the same chromatic polynomial as

M, by 1.7. The map «Nl)B)C' where C a (vt) has likewise the same

25

chromatic polynomial as 6AM. The chromatic polynomial of «NI)B)C can

be computed from 1.5 as

(A-2)P(N,A) •

Hence

III

Now, by the vertex-elimination theorem,

So by substituting

2 -3 2P(NI,T) • T P(N,T).

2A • T in the above equation, we have

-1 -3 2 2 2(T -T )P(N,T) • P(M,T) + P(8AM,T ) ,

that is

2 2 -2 2P(M,T ) + p(eAM,T) • T P(N,T).

But by (i) and (ii)

2Eliminating P (N , T ), we find

-2But 2-T • T. The theorem follows.

(ii)

26

7. THE "GOLDEN IDENTITY"

For a triaugulation M, there is a curious relation between P(M,T2)

and p(M,IST). (We note that t2

"" t+l and Is T" T+2.)

PIu:J:: If possible, choose M so that the theorem fails, Ivl has

the smallest value consistent with this condition, and then the minimum

vertex-valency q of M is as small as possible. We have, of course,

(i) 2 ~ q ~ 5.

Suppose M has a 2-circu1.t with edges X and Y. We define Nl' N2 ,

~ and M2 as in the proof of 4.1. tTsing 1.5 and 1. 7, we find

P(Ml' A)P{~, ><)P{M,A) = >«><-l) •

Since HI and ~ are triangulations with fewer vertices than H,

they satisfy the theorem. Denoting the vertex set of HI by VI' we have

Hence

"" Ivi + 2.

But, by (ii)

Hence

27

From now on, we can assume that M has no 2-circuit. Let x be a

vertex of valency q.

If q "" 2, then, in the absence of a 2-circuit, the two triangles

incident with x have all their sides in common. Hence the graph G of

P(K3

,IST) Is T 2 Is 4• • T • T "" T

2 2 -1 2P(K3

,T ) ... T • T • T = T

p(M,IST) "" IS T31VI-9 • P2(M,T2) •

Suppose next that q > 2. Let A be an edge incident with x and

let its other end be z. Let Tl and T2

be the triangles incident with

A and let their third vertices be y and t respectively. Then y ~ t,

since there is no 2-circuit. We accordingly use the notation of §6.

since the theorem holds for 1/JAM and 4>AM, by the choice of M,

• 15 T31VI-12 {P(~AM,T2)+p(4>AM,T2)}.

2 2• {P(~AM,T )-P(4)AM,T )}

"" Is T31VI-9 {P(6A

M,T2)+P(M,T2)}.

2 2• {P(M,T )-P(6AM,T )},

by 6.1 and 6.2

But the theorem holds for 6AM since this has a vertex x of valency

q-l. That is

28

Combining this with the preceding result, we obtain

p(M,IS .)

In every case, H satisfies the theorem, contrary to its definition. The

theorem follows.

Combining Theorems 3.2 and 7.1, we can establish

7.2. p(M,ls.) is positive.

PJmr:: Write1-15.* • --2- Then the nlTt,lber 1+.* lies between 0

and 1. Hence P(M,l+T*) is non-zero, by 3.2. Since P(M,~) is a poly-

nomia! with integer coefficients, we must have also

P(M, l+t) ~ 0,

i.e.,

2P(M, .) ~ O.

The required result now follows, by 7.1.

The number 15. is approximately

3.61803398874989484820 •

It has been remarked that Theorem 7.2 is of interest because it shows

P(M,~) to be positive at a value of A near 4, and many people are

specially interested in the value ~ a 4. There is a long-standing con-

jecture that P(M,4) is positive, for every triangulation M. The obvious

suggestion that P(M,4) is always positive from 15. to 4 inclusive

turns out to be wrong. At Waterloo, a few cases have been found of tri­

angular polynomials kaving two rea! zeros between 15 T and 4.

29

An independent set of vertices in a graph G is a set W of vertices

such that each edge has one end in Wand one not.

One way of constructing a A-colouring of G is to begin by choosing

an independent set W of vertices and assigning the colour A to its

members. The other vertices are to be coloured in the remaining A-I

colours. Clearly in any >.-colouring, the vertices of anyone colour must

constitute an independent set.

Let the graph obtained from G by deleting W and its incident edges

by Cw. By the above considerations, we have

8.1. P(G,>.) • I P(Gw, >"-1).w

This is a polynomial id.entity: it remains valid when we substitute

A • '[2+1 • 15 '[. Hence

p(G,ls'[) • I p(~,'[2).\-1

We can use the vertex-elimination formula to express P(~, '[2) in

terms of P(G,t2), in the case in which G is the graph of a triangulation

M. For ~ is formed from G by eliminating one by one the vertices of

W, and at each stage the map is triangular at the vertex being eliminated.

So if meW) is the sum of the valences in G of the vertices of W we

have

p(M,ISt) • r P(Cw,t2)W

= ~ (_l)m(W) '[m(W)-IWI P(M 2)l. ,t ,W

by 4.1. But

30

Hence

8.2. P(M,T2) = 15 t 9- 3!VI L (_l)m(w) Tm(w)-Iwl.W

As an example, let us calculate2P(M,T ) for the regular icosahedroa.

In this triangulation, there is as usual one null independent set. There

are 12 independent sets of one vertex, and 6 independent sets consisting

of two diametrically opposite vertices. For other independent sets of two

vertices if we choose one member, such as X in the diagram, we have five

choices for the other. Hence the number of independent sets of two non­

opposite vertices is i<12 XS) = 30.

7 to Y

It is evident that an independent set of 3 vertices cannot contain

two diametrically opposite ones. It is thus symmetrically equivalent to a

set containing the vertices X and Z of the diagr8lll. The third member

must be 5 or its symmetrical equivalent T. A set W of three independent

31

vertices can thus be characterized by giving a triangle U and postulating

that the members of W are the remote vertices of the three triangles

sharing sides with U. The number of such sets of 3 vertices is thus 20.

Clearly a set of three independent vertices cannot be extended as a set of

four.

We tabulate this information about independent sets W as follows.

Iwi m(W) No. of sets W

0 0 I

1 5 12

2 10 36

3 15 20.

So by 8.2, we have

2P(M, T ) = 4 8 12{1 - 12T + 36T - 20T }.

1 9--36·rs T

We now need a table of powers of T. This is given below. We put

Tn = -21(A + Is B )n n

and tabulate A and B. Each row of the table is obtained as the sum ofn n

the two preceding ones, (except for the first two).

therefore

therefore

32

n A Bn n

0 2 0

1 1 1

2 3 1

3 4 2

4 7 3

5 11 5

6 18 8

7 29 13

8 47 21

9 76 34

10 123 55

11 199 89

12 322 144

13 521 233

14 843 377

36T8 • i (1692 + 756 15)

1 + 36T8

• i (1694 + 756 15)

12t4 a ~ (84 + 36 15)

20112a i (6440 + 2880 15)

1214 + 20T12a t (6524 + 2916 15)

1 - 1214 + 3618 - 20112 • ~1 (4830 + 2160 15)

Hence

• - 5(483 + 216 (5)

a - lS(161 + 72 IS)

12= - 1ST •

2 1 -27 12P(M, T) ... Ts T • (-1ST )

I -15=~- 3 ~S T •

33

(This has been checked by substitution in the known polynomial P(M,A).)