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COMBINATORIAL
MATHEMATICS
Y fAR
5
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5 6 0 1 7 8 9
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Febl'Ua:I']J 1969 - Juns 1910
LEC'i.'UI'ES
on
CHftJdATIG
.
.
POLYdO~·lU.LS
It
by I-1. '£. TU1'TE
University of WaterZoo~ Ontario
Department of Statistics
University of North CapoUna at Chapel Hill
Institute of Statistics :t.'-;imeo Series tJo. 600.25
ApJr.J1, 1970
the
Guest lecturep during the Combinatorial Mathematics Year suppopted by
Air Fopoo OJ:fic,? of Scientific ReseaPch under Contract No. AFOSR-
u.s.
68-1406.
LECtURES
on
CHf01ATIC Pot.YfOo1IALS
by W.T. Tutte+
University of Waterloo, Ontario
1.
Consider a graph
INTRODuCtION
G with edge-set
A >.-colouring of
G,
into the set
of integers from
I
A
where
E
and vertex-set
V,
both finite.
A is a positive integer, is a mapping of
1
to
A,
V
with the property that the
two ends of any edge are mapped onto distinct integers.
The integers
A-co1ourings of
1
to
A are commonly called "colours".
G will be denoted by
P(G,A).
Recursion formulae are
known which permit the determination of the function
any graph
G.
The number of
P (G, A)
of
A for
We proceed to list some simple observations, including the
statements of these recursion formulae.
1.1.
If
G is an edge1ess graph with
P(G,>.)
k > 0
=
For in this case, every mapping of
vertices, then
k
A •
V into
I>.
satisfies the defi-
nition of a colouring.
t
Guest 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 verFor this graph, we make the obvious extension of formula 1.1 and
tices.
P{G,A)· 1.
postualte that
1.2.
Let
k
This means that
be a positive integer and let
~
has exactly
k
~
be a complete k-graph.
vertices, that each pair of vertices
are joined by exactly one edge, and that there are no loops.
P{G,A)
•
A{A-l){A-2) ••• (A-k+l) •
k · O.
We can extend this rule to the case
the null graph, so that
formula gives
taken as
1.3.
P{KO,A)
Then
P (K ' A)
O
K is interpreted as
O
is, by convention, unity.
The above
as an empty product, and such a product is usually
1.
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
Each A-colouring
A of
as a subgraph
G induces a A-colouring
be called an extension of
1.4.
~
A •
k
Each A-colouring of
~
of
(k
~,
~
A and
~.
B of
is complete, its vertices are all of different colours in
in
B.
It is, therefore, possible to find a permutation
e
and may
has the same number of extensions.
~
Evidently
1).
We note that
To prove this, consider two A-colourings
transforms the colour in
~
A to the colour in
transforms each extension of
e
of
Since
A, and also
I A that
B for each vertex of
A into an extension of
~.
B,
3
and
e
e-1
transforms each extension of
1 - 1
sets up a
extensions of
B.
B into an extension of
A.
Thus
A and the
correspondence between the extensions of
The theorem follows.
We note the corollary
1.4.1.
~ is
The number of extensions of any given A-colouring of
peG, A)
P{"k,A) •
1.5.
HnK
Let
G be the union of two subgraphs
is a complete .f.-graph
(.f. ~ 0).
...
P{G,X)
Pro>F:
and
P(H,X)P(K,X)
P(HnK,X)
Let us deal first with the case in which
K have no edge or vertex in counnon.
and each combination of a A-colouring of
yields a A-colouring of
colouring
o.
Then
Each X-colouring of
H
H
H
P{G,A}
with a A-colouring of
with a A-colouring of
is
K,
K
The theorem follows for this case, since
Au
the argument is similar.
of
a X-colouring of
K.
H.
HnK
G,
~
in
we have to combine
P (H, A)
'1I
A
o
of
HnK.
with an extension
To complete
~
G can be obtained in this way.
of
Ao
But
1"ays, and for each of these we can choose
in
P{K,X)
P{HnK,X)
ways, by 1.4.1.
We start by constructing a A-
This induces a A-colouring
Clearly every A-colouring of
we can choose
~
.f. ...
P(HnK,A). 1.
For .f. > 0,
in
G.
K such that
Then
then obtained by combining a A-colouring of
and
H and
The theorem follows.
!
4
Consider an edge
A be
Let
G
and
y.
A of G,
the graph obtained from
GA,
Let
single vertex
z.
made incident with
z
G
z
GA,.
x
or
Incidences of an edge of
are the same as in
and
A but leaving
A
by contracting
An edge incident with
in
x
A and replace x
More precisely, we delete
z.
Let its ends be
G by deleting
be the graph obtained from
by a single new vertex
other than
not a loop.
y
GA,
x
to a
and
in
G
y.
y
is
with vertices
In particular, an edge, other than
G.
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-
A in which
y
have different colours.
ourings of
G
there is a
1 - 1
and
GA.
in which
correspondence is as follows:
and
x
ceives the same colour as
x
and
y
Let two distinct vertices
distinct edges
A and B.
and
Gil
A
have the same colour.
y
This
vertices common to the two graphs
receive the same colour in corresponding colourings, and
1.7.
Moreover,
correspondence between the ),-colourings of
those A-colourings of
1 - 1
x
in
G:
A
z in Gil
A
re-
The theorem follows.
x and y
of
G
be joined by two
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.
expressed as a polynomial in
(i)
The coefficient
Then
P{G,).)
A with the following properties.
of
is an integer (all
j) •
can be
5
(ii)
~
aj
0 if and only if PO(G)
number of components of
G,
and
Ivl
PRooF:
Moreover,
PO(G).
Ivl.
lvi, where PO(G)
is the
alternate in sign.
We proceed by induction over
P(G,A)
~
is the cardinality of V.
The non-zero coefficients
(1v)
by 1.1.
S j
•
lEI.
If
lEI·
we have
0,
"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,
Choose an edge
A of G.
and consider the case
It is not a loop, by hypothesis.
lEI = q.
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).
true for
P(G ,A)
A
unless
It is also
G has a loop, in which case P{GA,A). 0 by
A
1.3.
We deduce at once that
satisfying (1).
P(G,A)
can be expressed as a polynomial in
We note also that (iii) holds for G, since G'
has
A
one more vertex than GA.
of
The non-zero coefficients of G extend from that of AIvl to that
A
At, t . po(Gl), with alternation of sign, the coefficient of "Ivi
being positive.
PO(G)
There are two possibilities for
po(Gl);
it is either
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 coefficients extend from that of
A
l
'"
where l = PO(GA) = po{G),
sign, the coefficient of A1vl - l being positive.
to that of
.
with alternation in
6
It now follows from the above equation that the non-zero coefficients
P(G,~)
of
A/vi
extend from that of
nation in sign.
to that of
APO(G),
lEI
The theorem thus holds when
q.
a
with alter-
It follows, in
general, by induction.
When
in
has a loop, we can think of
G
In all cases, therefore, we refer to
A.
pol.ynomia1. of
k
CHROMATIC POLYNOMIALS OF SOME SPECIAL GRAPHS
~
is a graph having
A1'~' ••• ,~,
edges
(l S j S k).
2.1.
then have
A-l
Aj
k
being
that
a _
j 1
A
choices for the colour of
choices for the colour of
C
k
a
j
and
aj
if
has a loop if
< k,
j
k
=1
(A-1)
ale
a '
O
If
If
k > 1,
k > 0,
we have
we
A-1
and so on.
be a positive integer.
and
aj _l
and
A(A-l).
aO,al""'~_l' and
vertices
being
,~,
L
with the edgeless graph of one vertex.
O
choices next for the colour of
Let
a ,a ,a , •••
O 1 2
vertices
k
•
We have
k+l
the ends of
We identify
P(~,A)
PRooF:
as the Chromatic
P (G, A)
G.
2.
A k-arc
as a zero polynomial
P(G,A)
k
A k-circuit
edges
and
C
k
is a graph having
A1,A2""'~' the ends of
~-1
but not if
k + (-1) k (~-1).
and
k > 1.
a O 1f
j
CI
k.
Aj
We note
k
7
PRa>F:
If
k · 1,
then
by 1. 3, and so the above
P(Ck' A) • 0,
formula holds.
Assume as an inductive hypothesis that the formula holds when
q
less than some positive integer
Put ting
A·
l\'
~
2,
and consider the case
k
is
k" q.
we have
P(Ck,A)
by 1.6,
..
P(~_l,A) - P(Cg_l,A),
.
by 2.1 and the inductive hypothesis,
..
(A_l)k + (-l)k(A-l).
The theorem follows, by induction.
We note that
The wheel
P(C ,A). P(LI,A)
2
~
k
spokes, where
adjoining a new vertex
v
called the hub and joining
~
2.3.
v
P(Wk,A)
is formed from
v
C
k
by
to each vertex
Let a graph
by using the following theorem.
G be obtained from a graph
and then joining
v
P(G,A)
PROOF:
only
1,
by an edge called a spoke.
We can obtain
vertex
k
C • K •
3
3
of
of
W
k
by 1.7, and
A-I
When
v
•
H by a new edge.
colours are available for the vertices of
a A-colouring of
G.
H
in these
A-I
H.
A colours,
On the other
colours can be used to complete
The theorem is now evident.
...
Then
AP(H,A-l) •
has been coloured, in anyone of the
h snd, any colouring of
2.4.
to each vertex of
H by adjoining a new
8
To prove this, we apply 2.3 wi th
Then
B"
<1t,
and we obtain
G= W
k
P(H,A-l)
and
v
the hub of
from 2.2.
Having obtained the chromatic polynomial for the wheel
v
W,
k
we can carry the construction one step further as follows.
new vertex
new edge.
w and we join
w
Wk.
to each vertex of
W,
k
with hub
We adjoin a
including
Let us denote the resulting double wheel by
Uk.
v,
by a
By 2.3 and
2.4, we have
.
2.5.
).().-1){().-3) k+(-1) k ().-3)}.
D
The double wheel
Uk'
with
St
Uk
has an "axle"
we obtain a bipyramid
v
B •
k
This is conveniently drawn on the sphere
and w at the North and South poles respectively.
occupies the equator.
v
The circuit
and ware joined to the vertices on the
equator by great-circular segments.
B
k
When we delete this edge in
vw.
When
k
~
2,
this representation of
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:
The graph
fication of
We can identify
(Uk)~
v
B
k
with
(Uk) A where
A
is the axle
vw.
resembles a wheel, with a hub resulting from the identi-
and w,
P(Bk,A)
except that each spoke is doubled.
=
P(Uk,A) + P(Wk,A),
..
).(A-l){(A-3)k+(-l)k(A-3)}
by 1.6
However,
9
+ ~{{A-2)k+(_1)k(A_2)}
=
~{{A-l)(A-3)k+(A-2)k+(-1)k(~2_3A+l)}.
3.
As a polynomial,
P{G,~)
A
has a definite value when any given real or
complex number is substituted for
regard
GENERAL
~.
From now on, therefore, we shall
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 identically 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
3.1.
Then
Let
P(G,~)
number
Ivl
I'ImF:
1.
When
A be an integer.
G be a loop1ess graph.
is non-zero.
Let
A be real and negative.
It is positive or negative according as the
of vertices of
G is even or odd.
This is clearly true when
G is null and therefore
P(G,~).
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
be assigned a value even when
A= 0
(see 1. 8) •
A and as such can
10
3.2.
Let
G be a connected non-null loopless graph.
1.
and less than
A be real
Let
Then
P(G,A)
A
vertices of
PR:DF:
Ivl
It is positive or negative according as the number
is non-zero.
of
G is odd or even.
We first dispose of two very simple cases.
Suppose that
G is the edgeless graph of one vertex.
Then
peG, A)
... A and
P(G,A)
..
A
1.
The theorem is thus verified.
G.~,
Next suppose
i.e.,
G· L •
l
Then
by 1.2
P{G,A)" A(A-l)
or 2.1, and so
P(G,A)
A
...
A-I < 0 .
Again the theorem holds.
IE I
We proceed by induction over the number
then
G
lEI ...
q.
0,
Then
H
has fewer than
and
q
v
K
G is the union of two subgraphs
in common, and that each of
H
H
and
and
K
must be loopless, non-null and connected.
K
P(G,A)
A
•
P(B,A) • P(K,A)
A
A
with
has an
Each
edges, and, therefore, each satisfies the theorem.
1.5,
We deduce that
and consider
< q
is a positive integer.)
(q
only a single vertex
IE I
Assume the theorem to hold for
Suppose first that
edge.
IE I •
If
is the loopless graph of one vertex, for which we have already
verified the theorem.
the case
of edges.
By
11
P(G,>.)
>.
is non-zero, and has the same sign as
h+k
(-1)
where
h
and
k
(-1)
•
lvi-I,
are the numbers of vertices of
In general., choose an edge
It may happen that
A
of
G.
A in G.
components consists of a single vertex, then
G
dealt with this graph.
G
graphs
Hand
K
In every other case,
G~
If
and we have already
K ,
2
is a union of two sub-
A is not an isthmus.
are both connected non-null graphs,
=
P(G,A)
by 1.6.
I:
If each of these
of the kind just considered.
It remains to consider the case in which
and
GA has two distinct
is an isthmus, that is
A
components each containing one of the ends of
A
K respectively.
G satisfies the theorem.
Thus
G
H and
P(GA,A) -
A being
G
Then
100p1ess.
But
P(G~,A)
GA
" h as a 1oop, then P(G"A,A') • 0 ,
by 1.3.
Accordingly
P(G,A)
A
But the theorem holds for
for
A,
G
which has only
q-l
edges.
Hence it holds
G.
If
G~
has no loop, the theorem holds for both
GA and
GA.
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
The theorem now follows in general, by induction.
G.
Then
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
+ 15
2
This is the number
(1+/5)/2.
section"
2
P(G,T )
that
T
2
D
T+l,
T
being the usual symbol for the "golden
Another way of describing the regularity is to say
tends to be small when
G
is a triangulating graph of the
sphere.
An example of the effect is provided by the bipyramid
matic polynomial is given by 2.6.
the quadratic form
mately
2.618,
lute value.
2
A -3A+l
the numbers
When we substitute
takes the value zero.
A-2
and
A-3
;\.
Since
0=
B
k
T
T
become less than
whose chro-
2
in 2.6,
2
is approxi1
in abso-
We deduce that
3.3.
co
4.
o.
THE VERTEX-ELIMINATION THEOREM
When a connected graph is drawn on the sphere, without crossings, we
obtain a spherical, or "planar n 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-circuit 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
B ,
k
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
dent with
v
M is triangutar at a vertex
are triangles.
make up a connected graph
curve.
if all the faces inci-
The sides of these triangles opposite
L,
v
its incident edges will be denoted by
M
v
v
though not necessarily a simple closed
The graph obtained from a graph
connected and defines a map
v
G
by deleting a vertex
v
G•
In the present case,
G
v
M
v
on the sphere.
by uniting the triangles incident with
v
and
v
is obtained from
is
M
so as to form a single new face.
It is,of course, not necessarily a triangulation.
When a map
polynon:.ial of
by
M is determined by a graph
we refer to the chromatic
G,
G also as the chromatic polynomial of
M,
and denote it
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
PRooF:
number
lEI
If possible, choose
v.
M so that the theorem fails and the
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
x
in
and each having at least one edge.
CODlDlOll,
nected graphs, by the connection of
the connection of the graph
K defines a map
H
and
K •
v
G,
and
Suppose
L.
v
N that is triangular at
M,
By the choice of
...
...
P(G,. 2)
P(Gvt • 2 )
by 1.5.
and
H
the map
x
v
K having only a vertex
Then
H
and
is distinct from
to be a vertex of
v, ,and G
v
N
K are conv
K.
by
Then
is the union of
Hence
satisfies the theorem.
- 2 2 m 1-m
2
P(H,. )(-1) T
P(K,. ),
v
2 P(K ,. 2 ),
• -2P(H,.)
v
•
Combining these results, we find
p(G,.2)
...
(_l)m T1 - m P(G ,t 2 ).
v
From now on we may assume that
M
is non-separable.
Hence it has no
isthmus.
Suppose next that
angle incident with
v.
M has an edge
If
A
A
that is not the side of a tri-
is a loop, the theorem is trivially true,
by 1.3.
From now on, therefore, we may assume
ends
and
x
y
of
A
again trivially true for
M to be loop1ess.
are joined by a second edge
M, by 1. 7.
If the
B, the theorem is
We may, therefore, assume
M
to
have no 2-circuits whose edges do not both belong to the boundaries of
triangles incident with
Consider the graphs
v.
G
A
and
GA.
is not an isthmus, and they define maps
form
Ml
from
angular at
A
v,
1.6, we find
M' and M" on the sphere.
A
A
M by fusing the two faces of
single new face, and we form
of the edge
They are both connected, since
Mil
A
from
M incident with
A
A
We
into a
M by identifying all the points
to form a single new point.
Both
M'
A and M"A are tri-
and both satisfy the theorem, by the choice of
M.
Using
15
2 2 2
P(M,T)
0
P(MA,T) - P(MA,T )
(_l)m Tl-m {P«MA>v' T2)_P( (MA)V,T 2)}
•
(_l)m Tl-m {P«Mv>A,T 2 >-P«Mv)A,T 2)}
•
by 1.6.
Thus
M satisfies the theorem, a contradiction.
From noW on we may assume that the only edges of
the triangles incident with
v.
Suppose two distinct edges
the same second end w.
M are the sides of
X and
Y of
M incident with
v
have
They make up a simple closed curve separating the
sphere into two Jordan domains, each containing some of the triangles incident with
v.
Call these domains
\~
D
l
and
Y
D •
2
16
If we delete from the graph the edges and vertices inside
D
i
as a single new face. we obtain a map N
i
Di and treat
having a 2-goo incident with
v
x
w
17
Apart from the 2-gon, v
and
N
and
m
2
r
Let
is incident only with triangular faces in
Let us suppose it is incident with
in
Hi
N
l
triangular faces in
~
NI'
N2 • Then m • m1+m2 •
Ni
be formed from
by deleting
the digon with the triangle incident with
from its graph and fusing
X
so as to give a new tri-
X
angle.
w
The maps
M
l
and
the choice of
Let
H
2
w
are triangular at
v.
They satisfy the theorem by
H.
N be the map derived from
M by deleting
fusing the two adjacent faces into a single new face.
2
P(H,T )
•
•
2
T
3
ell
T-
=
(-1)
But the graph of
M
v
2
)P(~,T )
2
(-1)
by 1.5,
• T
ml
l-ml
T
m -l-m
T
2
P{(Ml)v''T )(-1)
2
m2
2
P«~)v,'T )P«M )v''T )
2
is the union of the graphs of
and these have only the vertex w
2
Then
by 1.7,
P{N,,2),
P{Ml "
from its graph and
X
P{MV,T)
=
'T
-2
in common.
2
l- m2
T
2
P{(M2 )v''T )
•
(Ml)v
Hence
2
P«Ml)v,T )P{(M2 )v,T ).
and
(M2 )v'
18
Combining the above results, we find
We may now assume that all the
have distinct ends.
the graph
If
G of
Hence
L
v
m edges radiating from
we have
m
2
P(M,T)
M
v
iii
III
in
M
is either a circuit or a complete 2-graph;
M is either a wheel of
M iii W,
v
C.
m
m spokes, or a complete 3-graph.
Hence
T2 (T -m+(-1) mT-1 ),
by 2.4
•
...
If
H'" K ,
3
we have
P(M,T 2 )
by 2.2 •
H '" K • Hence, by 1.2,
v
2
1
2
2
... T • T • T- 1 .,. T- P(M ,T )
v
In every case, we have shown that
to its definition.
M satisfies the theorem, contrary
We conclude that the theorem is true.
5.
THE MAGNITUDE OF
2
P(il, T )
In this section, we offer a theory to explain the empirical observation
that
P(M, T 2 )
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
then
T
5-lvl
•
V,
19
Ivi
Here
closing
is the cardinality of
2
P(M, 'T)
PRooF:
denote absolute value.
If possible, choose a triangulation
Ivi
fails and
and w.
up by
D
l
M has two edges
X and
Y with the same ends
H being a triangulation, it has no loops.
X and
and
M so that the theorem
has the least value consistent with this.
Suppose first that
v
V as usual, but the vertical bars en-
D •
2
The circuit made
Y separates the rest of the sphere into two residual domains
As in the proof of the vertex elimination theorem, we define
and have
P(M, T
Let
But
M
l
Vi
and
2
)
be the vertex set of
HZ
Mi "
We note that
satisfy the theorem, by the choice of
2
Ip(M,T >!
a
T-
T
=
T
3
-3
H.
Hence
2
\PCM ,T 2 >I • \PCM2 ,T >I
1
• T
5-IV11
• t
5-IV21
5-\V\ •
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
possible value of
If
m
= 2,
m is of course
2.
5.
Choose such a vertex
v
of valency
the two triangles incident with
in common, since there is no Z-circuit.
and
~
v
Hence
v
m
The least
in
M.
have all three sides
G is a complete 3-graph,
20
2
P(M,T)
2
=
T
-1
T
• T •
T
If
m
Q
3,
then
is also a triangulation, with
M
v
I t satisfies the theorem by the choice of
.
.
.
2
Ip(M,T ) I
If
m· 4,
T
-2
2
v
M.
T-2 • T5-( lVI-I)
T4-IVI
It has
Let the vertices of
<
T5-lvl •
M '
Ivl-1
v
It is a proper map (since
v
We construct a triangulation
a new diagonal edge
A
vertices.
(a,c)
We may assume that
M •
ClI
T
(ac).
M
Q and its other faces are
Q be, in their cyclic order
M is a map on the sphere, the pairs
joined by edges of
vertices.
by 4.1,
>1,
It has one quadrilateral
triangular faces of
Ivl-l
Hence
M.
Ip(M ,T
we consider the map
has no 2-circuit).
5-lvl •
and
a
We note that
(b,d)
and
by subdiViding
M
v
a, b, c, d.
Since
cannot both be
c
are not so joined.
Q into two triangles by
'" T ~
and that
T~
is
also a triangulation of the sphere, except for the doubling of two edges.
We make it a true triangulation
and
bc
by 1. 7.
of
Mv '
TO by retaining only one of the edges
and only one of
cd
and
da.
We have
P(TO,A).
Hence
by 4.1,
But
=
T-3 Ip(T,T 2> + P(T O,T 2)1,
S
T-3 {lp(T,T 2>1+lp(T O,T 2>1} •
by 1.6,
T and TO satisfy the theorem, by the choice of M.
Hence
ab
P(T~,A>,
21
Ip(M .t 2)1
Note that
Ip(Mv . t 2)1
S t
'~
t- 3 (t 5 -(
•
t
S IVI
-
5-lvl
IVI-l)+r 5 -( IVI-2»
•
by the above results.
It remains to consider the case m· 5.
map.
In this case.
M
v
is a proper
It has one pentagon Q and its other faces are triangular faces of
M.
Let the vertices of Q be. in their cyclic order
Again. the pairs
M
v
N
(a,c)
and we assume that
by subdividing
onal edge
and
a
(b,d)
and
c
a, b. c. d, e.
cannot both be joined by an edge of
are not so joined.
We construct a map
Q into a triangle and a quadrilateral by a new diag-
A. (ac).
We can identify
with the map denoted by M
N
v
in
the preceding argument, thereby deducing that
t
If, in the map
M,
v
N~,
we retain only one of the edges
we obtain a triangulation
satis fies the theorem.
s-ivi •
Ivl-2
It has only
T.
ab
and be of
vertices and so
Thus
by 4.1,
=-
t"
S
t
S
t
•
t
-4
-4
Ip(N,/)
4-lvl
5-IVI
We find that in all cases
choice of M.
2
Ip(N,t ) + P(T,t
t
I
+ t
-4
2
>1,
by 1.6,
2
Ip(T,t )
I
3-IVI
.
M satisfies the theorem, contrary to the
We conclude that the theorem is true.
22
6.
RECURSION FORKULAE FOR TRIANGULATIONS
The recursion formulae of Section 1 enable us to calculate the chromatic polynomial of any graph
But if
G.
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.
graph and
If
is one of its edges, not a loop, then
A
represent a triangulation, though
G"
A
is such a
G
A
G
does not usually
does so wIess it has a loop.
It
1s desirable to find a recursion formula that works with triangulations
only.
One such formula is 1.5, in the case
in practice.
l · 3,
and this is very helpful
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
its incident triangles.
vertex of
T
l
by
We denote the ends of
Y and the third vertex of
Y
I~
X\;jZ
t
We assume that
x, y, z
and
t
are distinct.
M.
Let
A by
T
2
by
T
l
x
t.
and
and
T
2
be
z, the third
23
From M we can derive a number of other maps.
map ob tained by deleting
Q.
by a quadrilateral
T1
We write
6 M
A
N for the
T1
and
T2
N
by
for the graph obtained from
B. (yt).
It is often
eAM is obtained from M by "twisting" A.
said that
of
from the graph so as to replace
Q into two triangles by the diagonal
subdividing
In
A
We write
M we can contract
and the edge
zt
of
to a single vertex and delete the edge
A
T •
2
is a triangulation unless the contraction of
~B (6AM)
denote the map
also by
~AM,
This operation yields a map
A
introduces a loop.
yz
which
We
JIJAM.
Using 1. 6 and 1. 7, we find
P(M,A) + P('AM,A)
(i)
•
P(N,A).
Similarly
P(6AM,A)
(ii)
+ P(J1JAM,A) • P(N,A) •
Hence we have the recursion formula
This formula involves triangulations only, except that loops may appear in
J1JAM or
$AM.
In theory, we can use it to construct a list of chro-
matic polynomials of triangulations with vertex-valences
have listed the polynomials up to a particular value of
Choose a map
M with
valency.
8 · 3,
If
we apply 6.1 with
eAM
if
and so
8· 5
P (M, A)
in
M,
Ivl·
q+l
and let
8 ~ 3
~
3.
lvi,
Suppose we
P(M,A}.
A incident with a tetravalent vertex.
If
Then
is expressed in terms of known polynomials.
8· 4
in
q.
be its'sma1lest vertex
we use 1.2 or 1.5 to calculate
we can arrange that
Ivl·
say
eAM.
8
= 4,
8· 3
Similarly,
Of course, 1f
has a separating 2-gon or 3-gon. we can apply a simpler formula.
in
M
24
The recursion formulae used in practice are based on 6.1.
It is a curious fact that when
A
Ilt
T
2
the four polynomials of 6.1
satisfy a second linear relation
6.2.
2
2
P(M,T ) + P(9AM,T )
For
A
2
we can, of course, eliminate
T,
Q
6.2, and so obtain
2
P(M, T )
2
P(9 M,T)
from 6.1 and
A
directly in terms of triangulations with
fewer vertices.
PRooF OF
quadrilateral
v
6.2:
We can obta:tn from
N a map
N
1
by subdividing the
Q into four triangles by introducing a new central vertex
and joining it by new edges to
x, y, z,
and
t.
y
x
Z
t
The map
M,
(N ) liB'
l
by 1.7.
where
The map
B
D
(xv),
«Nl)B)C'
has the same chromatic polynomial as
where
C
a
(vt)
has likewise the same
25
chromatic polynomial as
6 M.
A
The chromatic polynomial of
be computed from 1.5 as
(A-2)P(N,A) •
Hence
III
Now, by the vertex-elimination theorem,
2
P(NI,T)
So by substituting
(T
A•
-1
-T
T
-3
2
•
T
-3
2
P(N,T).
in the above equation, we have
2
)P(N,T)
2
2
P(M,T) + P(8AM,T ) ,
•
that is
2
2
P(M,T ) + p(eAM,T)
But by (i) and (ii)
Eliminating
But
2-T
-2
•
T.
P (N , T
2
),
we find
The theorem follows.
•
T
-2
2
P(N,T).
«NI)B)C
can
26
7.
For a triaugulation
p(M,IST).
and
PIu:J::
THE "GOLDEN IDENTITY"
M,
(We note that
P(M,T 2)
there is a curious relation between
t
2
If possible, choose
"" t+l
Is
and
T"
T+2.)
Ivl
M so that the theorem fails,
has
the smallest value consistent with this condition, and then the minimum
vertex-valency
2
(i)
Suppose
~
and
M2
of
q
~
q
~
M is as small as possible.
5.
M has a 2-circu1.t with edges
as in the proof of 4.1.
(ii)
P{M,A)
Since
HI
and
=
Hence
and
We define
Y.
Nl' N ,
2
tTsing 1.5 and 1. 7, we find
~
>«><-l)
•
are triangulations with fewer vertices than
Denoting the vertex set of
""
But, by (ii)
X
P(Ml' A)P{~, ><)
they satisfy the theorem.
Hence
We have, of course,
Ivi
+
2.
HI
by
VI'
H,
we have
27
From now on, we can assume that
vertex of valency
If
q "" 2,
incident with
Let
x
be a
q.
then, in the absence of a 2-circuit, the two triangles
x
have all their sides in common.
T • T2 • T ""
P(K ,IST)
3
•
Is
P(K ,T 2 )
3
...
T • T • T
p(M,IST)
""
IS
Suppose next that
let its other end be
A
has no 2-circuit.
M
Let
T
l
and let their third vertices be
4
Is
T
2
T
A
be an edge incident with
and
T
y
and
x
and
be the triangles incident with
2
t
respectively.
Then
y
We accordingly use the notation of §6.
since there is no 2-circuit.
since the theorem holds for
=
and
1/JAM
•
by the choice of
4>AM,
M,
15 T31VI-12 {P(~AM,T2)+p(4>AM,T2)}.
•
"" Is
{P(~AM,T
2
)-P(4)AM,T
2
)}
T31VI-9 {P(6 M,T 2 )+P(M,T 2 )}.
A
• {P(M,T 2)-P(6 M,T 2 )},
A
by 6.1 and 6.2
But the theorem holds for
q-l.
That is
6AM
of
G
T31VI-9 • P2 (M,T 2 ) •
Let
q > 2.
z.
-1
2
Hence the graph
since this has a vertex
x
of valency
~
t,
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.)
PJmr::
and
1.
is positive.
Write
Hence
1-15
.* • --2-
P(M,l+T*)
Then the nlTt,lber
is non-zero, by 3.2.
1+.*
Since
lies between
P(M,~)
0
is a poly-
nomia! with integer coefficients, we must have also
~
P(M, l+t)
0,
i.e.,
2
P(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
specially interested in the value
jecture that
P(M,4)
suggestion that
4.
4,
and many people are
There is a long-standing con-
is positive, for every triangulation
P(M,4)
turns out to be wrong.
~ a
A near
is always positive from
M.
The obvious
15. to 4 inclusive
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
such that each edge has one end in
W of vertices
Wand one not.
One way of constructing a A-colouring of
an independent set
G is a set
G
is to begin by choosing
W of vertices and assigning the colour
A
to its
A-I
members.
The other vertices are to be coloured in the remaining
colours.
Clearly in any >.-colouring, the vertices of anyone colour must
constitute an independent set.
Let the graph obtained from
by
Cw.
8.1.
G by deleting W and its incident edges
By the above considerations, we have
P(G,>.)
•
I
w
P(Gw, >"-1).
This is a polynomial id.entity:
15 '[.
A • '[2+1 •
it remains valid when we substitute
Hence
p(G,ls'[)
•
I
p(~,'[2).
\-1
We can use the vertex-elimination formula to express
terms of
2
P(G,t ),
~
in the case in which
G is the graph of a triangulation
M.
For
W,
and at each stage the map is triangular at the vertex being eliminated.
So if
meW)
is formed from
P(~, '[2) in
G
by eliminating one by one the vertices of
is the sum of the valences in
G of the vertices of W we
have
p(M,ISt)
•
r P(Cw,t2)
W
=
l.~ (_l)m(W) '[m(W)-IWI P(M ,t 2) ,
W
by 4.1.
But
30
Hence
8.2.
P(M,T 2 )
=
15
t
L (_l)m(w)
9- 3 !VI
Tm(w)-Iwl.
W
P(M,T 2 )
As an example, let us calculate
for the regular icosahedroa.
In this triangulation, there is as usual one null independent set.
are
12
independent sets of one vertex, and
of two diametrically opposite vertices.
opposite vertices is
independent sets consisting
For other independent sets of two
vertices if we choose one member, such as
choices for the other.
6
X in the diagram, we have five
Hence the number of independent sets of two non-
i<12 XS)
= 30.
7
It is evident that an independent set of
two diametrically opposite ones.
set containing the vertices
must be
There
3
to Y
vertices cannot contain
It is thus symmetrically equivalent to a
X and
Z
5 or its symmetrical equivalent
of the diagr8lll.
T.
A set
The third member
W of three independent
31
vertices can thus be characterized by giving a triangle
that the members of
sharing sides with
U and postulating
W are the remote vertices of the three triangles
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
Iwi
m(W)
0
0
I
1
5
2
10
12
36
3
15
20.
No. of sets
W as follows.
W
So by 8.2, we have
P(M, T
2
)
=
{1 - 12T 4 + 36T 8 - 20T 12 }.
1
·rs T
We now need a table of powers of
Tn
and tabulate
An
and
B.
n
=
T.
9--36
This is given below.
1 (A +
-2
n
We put
Is Bn )
Each row of the table is obtained as the sum of
the two preceding ones, (except for the first two).
32
n
B
A
n
n
2
0
1
2
3
4
5
11
6
18
7
29
8
9
10
47
76
123
199
322
521
843
0
1
1
1
3
4
2
7
11
12
13
14
3
5
8
13
21
34
55
89
144
233
377
therefore
36T
1 + 36T
12t
201
8
8
4
12
4
12
121 + 20T
•
i (1692 + 756 15)
i (1694 + 756 15)
a
~ (84 + 36 15)
•
a
a
i (6440 + 2880 15)
t
(6524 + 2916 15)
therefore
4
8
12
1 - 121 + 361 - 201 •
~1 (4830 + 2160 15)
33
•
- 5(483 + 216 (5)
a
-
lS(161 + 72 IS)
= - 1ST 12 •
Hence
2
P(M, T)
...
1
Ts
=~-
T
3
-27
I
~S T
• (-1ST
-15
12
)
•
(This has been checked by substitution in the known polynomial P(M,A).)