colorings of plane graphs

colorings of plane graphs

Colorings of plane graphs

This is the first draft of this write-up. Corrections of its contents and suggestions are welcomed.

A

planar graph

is a graph that can be embedded on a sphere or plane.
A

plane graph

is such a graph together with a choice how to embed it; its

faces

are the resulting regions of surface

separated

by edges.

A (face, edge, vertex etc.)

coloring

is just a

mapping

from the relevant
set of items to a small set of other things traditionally given the names of
colors.

In

graph theory

however there is usually an extra condition a coloring has to

satisfy

to be valid. Unless specified otherwise, the condition is that

adjacent

items must be given distinct colors — two edges meeting at a vertex
must have different colors, two faces with a common border must too.

Face colorings

Because of the history of the subject, being couched in terms of coloring
maps, faces have often been called “countries” and the plane graph a “map”.

A word on “border” in the

context

of face coloring. For countries to be
considered adjacent it is not enough to just meet at one point. The reason is
that that would make

coloring problems

uninteresting: divide a region into as
many countries as you want, arranged like a pie chart, and any number of colors
would be required if touching at a point did qualify.

So in a map which has a portion shaped like

×

, both the North and
South countries border both the East and West countries, but South does
not border North so they’re allowed to have the same color, and West is
likewise allowed to have the same color as East. Now consider a small border
adjustment from

×

to

>

⁣

-

⁣

<

where we’ve added one
extra constraint: now North and South must have different colors. If it was
possible to color the old map with a certain number of colors we can’t be
certain we can still color the new map.

Conversely

however, if we can color
the new map we can certainly color the old map (using the same color
assignment even).

If we are given any bridgeless planar graph

G

(not necessarily trivalent),
we can give any vertex with

valency

4 this treatment, splitting it into two
new vertices of valency 3 and a short new

stretch

of border, adding a
constraint on coloring. Vertices with valency

v

>

4

can be given the same
treatment in

v

-

3

steps, creating

v

-

2

vertices of valency 3 and

v

-

3

new
pieces of border which only makes the problem harder: if we can still color
this, we can also color the original.

2-valent vertices are simply funny features along an ongoing border
between two countries, and 0-valent vertices funny features within the
landscape. Removing them does not alter the coloring exercise. And finally
1-valent vertices create a bridge; we do not want them nor do we want any
other bridges, because at such an edge a country meets

itself

on the
other side of the border, making it impossible to have a color change there.
This is why face coloring

theorems

are stated for bridge-free graphs only.

By now we have only vertices left of valency 3 so our new graph

G

𝖸

is
trivalent (cubic). And if it can be

n

-colored (for any

n

), then so can

G

.
This

completes

the proof of

Lemma 0

Any bridge-free planar graph can be 4-colored

⟺

any bridge-free planar trivalent graph can be 4-colored.

as the

⟸

direction was argued above, and the

⟹

direction is
immediate (

any

includes all the trivalent ones).

∘

∘

∘

Note the lemma does not state either side of the “iff” is true, merely
that they are

equivalent

. The left-hand side is known as the

four-color
theorem

; having the right-hand side equivalent to it gives statements on
trivalent graphs a more general importance.

One last point: we will consider the graph drawn on a sphere. Or, what is the
same thing: if we draw it on a plane, we will also consider the

infinite

outside
area to be a face. If you were thinking of your graph as a map subdividing a
finite continent into countries just call the outside Oceania and color it
blue.

This convention too only makes the coloring harder: considering the outside to
be a colorless sea allows countries with a shoreline to be of all four colors,
while coloring Oceania (say) blue only allows those countries three different
colors (just like any other set of countries with a common neighbour). Again,
if we can color this, we can color plane maps of any other convention.

Edge colorings

A

Tait coloring

of a trivalent graph is a coloring of its edges with
only three colors, such that at each vertex the colors of the three edges
there are different. After Peter G. Tait (1831–1901), Scottish physicist (

http://www-history.mcs.st-andrews.ac.uk/Mathematicians/Tait.html

bio at St Andrews), who in 1880 proved the following

Theorem 1 (Tait)

A bridge-free trivalent plane graph can be
face-colored with 4 colors if and only if it can be edge-colored with 3
colors.

We will not give our four colors such poetic names as red or green, instead
they will be given two-bit labels (0,0) and (0,1) and (1,0) and (1,1), formally
vectors in

𝖨𝖥

2

2

, the vector space of

dimension

2 over the finite field
of order 2.

Edge colors used will be the

difference

between the face colors on
either side of the edge. Because

𝖨𝖥

2

2

has characteristic 2, differences
are the same thing as sums (so it doesn’t matter which we subtract from which).
They are carried out (mod 2) on the

individual

coördinates (in terms of bit

patterns

:

addition

without internal carry, that is, the

xor

operation

).

Note that the assignment of labels to

face

colors is quite
arbitrary (e.g.

(

0

,

0

)

has no special

significance

) but the
assignment to

edge

colors is less arbitrary: here

𝟎

=

(

0

,

0

)

really
means something. It means a border between faces of the same color, exactly
the thing that shouldn’t happen. You could say face colors are
“position vectors” (where

(

0

,

0

)

is an arbitrary point) while
edge colors are “free vectors”, differences between position vectors.

Proof of

⟹

: let

F

and

E

be the sets of faces and edges,
let

f

:

F

→

𝖨𝖥

2

2

be a valid face-4-coloring, and

g

:

E

→

𝖨𝖥

2

2

the

edge coloring

we are trying to construct from

f

.
Let

e

be any edge, and

α

and

β

the faces it separates. Now

f

being valid means

f

⁢

(

α

)

≠

f

⁢

(

β

)

so

g

⁢

(

e

)

=

f

⁢

(

α

)

-

f

⁢

(

β

)

≠

𝟎

. This means

g

really is an edge-3-coloring

g

:

E

→

𝖨𝖥

2

2

∖

{

𝟎

}

.

It is also easy to see

g

is a valid coloring: let

d

and

e

be any two
edges meeting at vertex

p

say, and let

α

,

β

and

γ

be the three faces found around

p

such that

e

separates

α

and

β

, and

d

separates

α

and

γ

. By

f

being valid we
have

f

⁢

(

β

)

≠

f

⁢

(

γ

)

from which

g

⁢

(

e

)

=

f

⁢

(

β

)

-

f

⁢

(

α

)

≠

f

⁢

(

γ

)

-

f

⁢

(

α

)

=

g

⁢

(

d

)

.

∘

∘

∘

Note in passing that the three edge colors around

p

all being different
(and nonzero) means they must be (0,1), (1,0), (1,1) in some order. That means
they sum to

𝟎

in a valid edge-3-coloring, a result we will use.

To construct

f

from

g

you might pick a color, give it to the country
you’re in, and travel the world. At each border, add its edge color to the
previous face color and assign that to the new country. I called edge colors
differences rather than sums for that reason: if a journey

Γ

visits faces

α

⁢

β

⁢

γ

⁢

…

⁢

ψ

⁢

ω

, and the face colors everywhere
have to be such that the edges crossed have colors

f

⁢

(

β

)

-

f

⁢

(

α

)

,

f

⁢

(

γ

)

-

f

⁢

(

β

)

, …

f

⁢

(

ω

)

-

f

⁢

(

ψ

)

then the color

f

⁢

(

ω

)

reached should be

f

⁢

(

α

)

+

∑

colors of edges crossed.

So we construct

f

by “integrating”

g

(and indeed, “plus a

constant

”,
the arbitrary color of the first country). If we

can

do that without

contradiction

between all possible journeys, this “potential”

f

will
automatically have

g

as its “gradient” as it should. Remains to prove
the “integral” is

independent

of path taken, in other words that

circular

paths “

integrate

” to zero.

Thus far we haven’t used the fact that we’re in a sphere or plane!
Of course, the use of “faces” at all means we have an

embedding

of a graph
in some surface; the fact we didn’t use yet is that the surface has genus 0.
Indeed, the forward half of Tait’s theorem (valid

f

implies valid

g

)
is true in surfaces of any genus. For the reverse half (valid

g

implies
valid

f

) we need to use the topological properties of the plane or sphere.

There are various ways of doing that. One wonderfully simple proof found in
Wilson

[

Wil02

]

,

adapted

here to use the

𝖨𝖥

2

2

notation, relies on
the graph being embedded in a plane. For a graph embedded in a sphere that
entails choosing a country and puncturing the sphere in the

interior

of
that country; it corresponds to the choice of giving that country the face
color (0,0).

Proof of

⟸

: the

subgraph

G

′

formed by edges of the two colors (1,any)
is

regular

2-valent. So is the subgraph

G

′′

formed by edges of the two
colors (any,1). Such graphs consist of

disjoint

cycles. In a plane, such
cycles have a

well-defined

interior and

exterior

. Every face

α

has a
number

g

′

counting the number of (nested) cycles of

G

′

it is inside, and
similarly a number

g

′′

with respect to

G

′′

.

Let

r

′

and

r

′′

be the residues of

g

′

and

g

′′

(mod 2). Give the face
the color

(

r

′

,

r

′′

)

. This is clearly a face-4-coloring. And it is valid:
to reach any neighbouring face

β

we cross an edge with color

(

e

′

,

e

′′

)

where

e

′

=

1

iff

β

is inside a number of

G

′

cycles that differs by
1 from

g

′

, likewise

e

′′

=

1

iff

β

is inside a number of

G

′′

cycles
that differs by 1 from

g

′′

. So

β

has color

(

r

′

+

e

′

,

r

′′

+

e

′′

)

and we
know

e

′

and

e

′′

are not both zero.

∘

∘

∘

Tait himself hoped to prove face-4-colorability, via edge-3-colorability, in
the following way.

Lemma 2 (Tait)

A trivalent plane graph that has a

Hamiltonian cycle

can be edge-colored with 3 colors.

A Hamiltonian cycle is a

closed path

that visits every vertex exactly once
(and yes,

that

Hamilton). We dropped the “bridgeless” qualifier
because having a cycle in which every vertex is involved means no bridges:
for any vertices

p

and

q

you can travel between them via the
Hamiltonian cycle, and back via all different edges using the other half of
the cycle. That would be impossible with a bridge: pick

p

and

q

either side of the bridge, now two routes between them would have to use the
same edge (that bridge).

Note also that in a trivalent graph the number of edges

m

equals

3

⁢

n

/

2

where

n

is the number of vertices. But

m

is an integer, so

n

is even.

Proof of lemma: the Hamiltonian cycle, having

n

vertices, also has

n

edges,
which is even. Color them alternatingly red and green. Each vertex now has one
red and one green edge; color the third edge blue.

∘

∘

∘

So Tait next tried to prove every planar graph has a Hamiltonian cycle.
Unfortunately, some don’t.

A vertex coloring

Curiously, the

property

that a graph is face-4-colorable, which turned out to
be equivalent to being edge-3-colorable by Tait’s theorem, is also equivalent
to admitting a certain kind of vertex-2-coloring. This seems to have been
discovered independently a few times, most notably by Percy J. Heawood (

http://www-history.mcs.st-andrews.ac.uk/Mathematicians/Heawood.html

bio at St Andrews).

We can call the vertex colors black and white, but think of them as representing values

+

1

and

-

1

interpreted as residues (mod 3).
This is not the kind of coloring where adjacent items have to be of different
color! It does have another more global constraint though: around any face, the vertex colors must sum to 0 (mod 3).

Theorem 3

A bridge-free trivalent plane graph can be edge-colored
with 3 colors if and only if it can be vertex-colored in the way just
explained.

Rename the three edge colors to 0, 1 and 2 interpreted as residues (mod 3).
These are again “position vector”-like; there’s no special significance
to being 0. Let

g

3

be this new relabeled verion of the old

g

.

We construct a

vertex coloring

h

3

:

V

→

{

+

1

,

-

1

}

from

g

3

as follows:
assign color

+

1

to a vertex if the edges there have colors 0, 1, 2 going
clockwise;

-

1

if counterclockwise. If the edge coloring

g

3

is valid,
it is possible to define an

h

3

this way.

Note this makes the vertex colors

differences

of edge colors in the
following sense. If we travel around a face counterclockwise (i.e. always
turning left) and

d

and

e

are two successive edges joined at vertex

p

, then

g

3

⁢

(

e

)

-

g

3

⁢

(

d

)

=

h

3

⁢

(

p

)

. If instead we turn right to edge

x

say, we get

g

3

⁢

(

x

)

-

g

3

⁢

(

d

)

=

-

h

3

⁢

(

p

)

.

Proof of

⟹

: Based on

g

3

, define

h

3

as above. To prove it is valid.
Pick any face

α

, and any edge

e

in it, let

g

3

⁢

(

e

)

=

c

. Go round
the face counterclockwise one whole cycle. Now

c

+

∑

V

∈

α

h

⁢

(

v

)

=

c

(mod 3). Which proves the sum of

h

3

values around the cycle is zero.

∘

∘

∘

The same

argument

works in reverse: if

h

3

is a valid vertex coloring then,
in any one cycle, we could assign an arbitrary color

c

to an edge

e

and,
going round counterclockwise finding vertices

p

,

q

… we could
assign to the edges after

e

the colors

c

+

h

3

⁢

(

p

)

,

c

+

h

3

⁢

(

p

)

+

h

3

⁢

(

q

)

, and so on. Because

h

3

sums to zero
around the cycle, it consistently assigns edge colors (e.g. edge

e

again
gets color

c

once around the cycle) and because

h

3

is never zero, we
never get the same edge color for successive edges.

But this isn’t yet enough. Given an

h

3

satisfying its rule, we can start
reconstructing

g

3

by this “integration” process. We just saw it works
locally on the level of single face

boundary

cycles. As with the previous
theorem however, the problem with “integrating” is to prove this can be done
unambiguously on a global scale, for which the topological properties of
genus 0 surfaces must again crop up somehow.

I will choose to use the following property of planes and spheres:

Lemma 4

With a plane graph of

u

faces on a genus 0 surface, it is possible to enlarge a

collection

of 1 face to

u

-

1

faces, one face at a time, in such a way that the area covered remains simply connected throughout.

Simply connected means that the area covered is

topologically equivalent

to a
disk, that it doesn’t enclose any

isolated

area. Its boundary is a

simple
closed curve

. Imagine the area covered by the collection of faces as shaded.
On a sphere, the area covered being simply connected means there is a single

connected

shaded area inside the boundary and a single connected unshaded area
outside, and that the boundary is

one

closed curve. Of course, while
the outside starts off big, it is

reduced

to just one face by the time the
inside has reached

u

-

1

faces.

Proof of lemma: start with one face shaded. At each stage, there will be
other faces immediately bordering the area shaded thus far. Choose one,
call it

α

. If accreting it to the side of the growing shaded area
would leave it simply connected, take it and move to the next step.

Now suppose

α

misbehaves, that adding it would enclose (one or more)
unshaded region(s)

ℜ

. Of course on a sphere we are already
enclosing an unshaded region, the whole outside. So what is really happening
if

α

misbehaves is that adding it would split the remaining outside
into two (or more) pieces. Choose any one of those pieces and let that be
our region

ℜ

.

Note that each of those pieces, including

ℜ

, contain some faces that
border the already shaded area. The way it works is this: The boundary of the
shaded region has

k

stretches where it borders

α

, and

k

⩾

1

because

α

was chosen as a face that borders the shaded area. If

k

=

1

it doesn’t
misbehave; if

k

>

1

then adding it would split the new outside into

k

regions: in between the

k

stretches where the shaded area borders

α

there are

k

stretches where each time it would border one of those regions.

Choose a face

β

from the faces of

ℜ

that border the shaded
region, and try adding it. If that would fail too it would be by splitting
the outside in more than one region. Choose one of those that does not contain

α

and call it

ℜ

′

. Because it does not include

α

, it is
wholly

contained

within one of the regions adding

α

would have created,
namely

ℜ

. In fact

ℜ

′

⊆

ℜ

∖

{

β

}

i.e. it has at least one fewer country than

ℜ

.

Choose a face

γ

from the faces of

ℜ

′

that border the shaded
region, if this doesn’t work either choose a

ℜ

′′

that does not contain

β

, and so on.

Now

ℜ

only has finitely many faces, and

|

ℜ

|

>

|

ℜ

′

|

>

|

ℜ

′′

|

⁢

…

.
So things can go wrong only a finite number of times. We end up with a country
that can be added without budding off a piece of outside.

∘

∘

∘

Now we can finish our proof of Theorem 3.

Proof of

⟸

: Pick one face. We saw we could pick an arbitrary color for one
of its edges and then continue assigning

g

3

colors based on

h

3

going
round, in a valid way. Shade the face to mark it as “done”.

Now pick additional faces along the lines of lemma 4. Keeping the boundary of
the shaded area simply connected means that each new face borders the previous
shaded area along a single stretch of boundary (say edges

a

⁢

…

⁢

m

going
counterclockwise), and this means the portion of the boundary of the new face
where the edges (

n

⁢

…

⁢

z

say) don’t have colors yet is also a single stretch.

This means we can edge-color the additional face validly: give

a

the color
it already has, and go round turning left each time assigning edge colors
based on

h

3

. This must agree with the colors for

b

⁢

…

⁢

m

we already
assigned consistently, and gives new colors for

n

⁢

…

⁢

z

(which are now also

consistent

with the rest) and gives the same color again to

a

because

h

3

is valid.

Once we have covered

u

-

1

faces, the last face already has all its edges
colored, validly.

∘

∘

∘

It’s a pity there doesn’t seem to be an easy way to find such a vertex
coloring from scratch (thereby making the four color theorem much more simple to prove). There cannot be a very

obvious

way to find one. For example, valid
face-4-colorings and edge-3-colorings of the

dodecahedron

correspond to a
vertex-2-coloring that uses 4 of one color and 16 of the other color. It’s hard to see how any automatic scheme could spontaneously break the

symmetry

and
treat 4 of those 20 vertices differently.

The four-color theorem says every bridgeless trivalent plane graph has a valid
face-4-coloring. It is equivalent to saying it has a valid edge-3-coloring.
And now it is equivalent to saying it has a valid vertex-2-coloring. The next
thing would be a valid something-1-coloring, and it shouldn’t be too hard to
prove that things can all be colored with 1 color! Unfortunately, we seem to
have run out of graph

elements

to take the place of that “something”.

A corollary:

3

⁢

n

-gon faces

Consider a valid vertex-2-coloring

h

3

, with some black (

+

1

) and some
white (

-

1

) vertices. Around each face, they sum to 0 (mod 3).

Now consider the following

modification

: pick any vertex, and replace it
by a tiny triangular country. If we give the three new vertices the opposite
color from the old vertex, the coloring is still valid (in each of the three
adjoining faces, one

±

1

is replaced by two

∓

1

which counts as the same).

Conversely, if the graph had any

triangle

face we could replace it by a single
vertex. The only way three

+

1

or

-

1

can add to zero (mod 3) is if they
were all of the same sign, so we replace it by one vertex of the opposite
sign.

Now suppose we expand all white vertices to black triangles. Or vice versa. We
saw each single modification leaves a valid vertex coloring valid. So once
we’re done and only have vertices of the same color, the coloring must
still be valid. That can only mean one thing: every face now has a number of
vertices divisible by 3. Not only the triangles we added, but also the
original faces, which must have grown to 6-, 9-, 12- etc. -gons.

Theorem 5

A valid vertex coloring as in Theorem 3 exists if and only if it is possible to modify the plane graph, turning some vertices into triangles, in such a way that we end up with only faces whose size is divisible by 3.

Proof of

⟹

:

We just did it.

∘

∘

∘

Proof of

⟸

:

We are given such a

sequence

of modifications where every face
ended up as a

3

⁢

n

-gon. It is very easy to give this a valid vertex-2-coloring:
just give every vertex the same color. Now undo the modification step by step,
each time coloring the new single vertex the opposite color from the triangle
it replaced. Modifications and their reversals

preserve

the

validity

of
vertex-2-colorings, so by the time we have undone every modification the graph
is still validly colored.

∘

∘

∘

This gives yet another corollary of the four color theorem: that we can always
find such a modication to

3

⁢

n

-gon faces.

Duals

The

dual

G

*

of a bridgeless plane graph

G

is formed as follows:
place one vertex of

G

*

inside each face of

G

. For each edge

e

of

G

,

separating

faces

α

and

β

, draw one edge of

G

*

from its vertex
in

α

to its vertex in

β

, in such a way that the only edge of

G

it crosses is

e

, and so that it crosses

e

at a point other than its

endpoints

.

This construction creates faces of

G

*

that each enclose one vertex of

G

and every one of those vertices is thus enclosed, making

G

likewise a dual
of

G

*

.

To prove that assertion: let

p

be a vertex of

G

, surrounded
by faces

α

,

β

⁢

…

and with edges

d

,

e

⁢

…

radiating from
it. All those edges will be crossed by edges of

G

*

forming a closed cycle
because they connect the vertices of

G

*

placed inside

α

,

β

⁢

…

going round. There are no other edges of

G

*

crossing our edges, and no
edges of

G

nearer to

p

to cross, so the cycle forms the boundary of a
face. This shows every vertex of

G

has a dual face of

G

*

.

To show every face of

G

*

is now accounted for, we can invoke Euler’s

formula

.

G

has

n

vertices,

m

edges,

f

faces, with

n

-

m

+

f

=

2

. For

G

*

we have by construction

n

*

=

f

vertices,

m

*

=

m

edges, and an unknown
number

f

*

of faces. By the previous argument there are at least the

n

faces with vertices of

G

in it. But by

n

*

-

m

*

+

f

*

=

2

(and

n

*

=

f

and

m

*

=

m

and

n

-

m

+

f

=

2

)

f

*

must equal that

n

.

∘

∘

∘

Modern formulations of the four color theorem usually take the dual view. In
stead of a face-4-coloring, a vertex-4-coloring. Note this is different from
the vertex-2-coloring in the previous

sections

; it has the standard constraint
that

adjacent vertices

must have different colors.

The dual

expression

of lemma 0 is now that it is

sufficient

to consider plane
graphs where every face is triangular (the dual of the

×

to

>

⁣

-

⁣

<

machinations is tiling up a

v

-gon into

v

-

2

triangles).

History

The subject has colorful beginnings. Francis Guthrie was born in England in
1832, obtained firsts in both mathematics and law, taught mathematics in South
Africa until his death in 1899 and also found the time to get involved with
railroad construction and to make his name in botany. One day he noticed
something when coloring a map of the counties of England — the original

four-color conjecture

. And on 23 October 1852, in London, his

brother

Frederick
asked his own math professor Augustus de Morgan about that

observation

(“With my brother’s permission”). De Morgan consulted Hamilton (“A
student of mine asked me to day to give him a reason for a fact which I did
not know was a fact — and do not yet.”). The latter didn’t think much of
it and quipped he didn’t have time for that “quaternion of colours”.
The problem didn’t go away though, Cayley dug it out again, Kempe gave his
flawed proof that stood for 11 years, and the rest is history

[

FW77

,

FF94

]

.

The four-color conjecture (now four-color theorem) has loomed large over graph theory. The story how Kenneth Appel, Wolfgang Haken and John Koch found the proof by

computer

in 1976 is well known. Perhaps it has been fortunate the theorem resisted proof so long, because it has been a motivating force for many exciting

developments

in graph theory. This includes important results in

graph coloring

such as Vizing’s theorem, but also work that transcends the bounds of graph theory, such as that by Whitney and Tutte on what is now called

matroids

.

And the future? Hadwiger’s

conjecture

can be regarded as a generalisation of the four-color theorem. And it is still open.

References

1

Ore67

Oystein Ore

,

The

Four-Color Problem

,

Acad. Pr. 1967 without

ISBN  0 12 528150 1

Long the standard work on its subject, but written before
the theorem was proven. Has a wealth of other graph theory
material.

FW77

S. Fiorini

and

R. J. Wilson

,

Edge-colourings of graphs

(Research notes in math.

16

), Pitman 1977,

ISBN  0 273 01129 4

The first ever book devoted to edge-colorings,
including material previously found only in Russian

language

journal articles.

SK77

Thomas L. Saaty

and

Paul C. Kainen

,

The Four-Color Problem: assaults and conquest

,

McGraw-Hill 1977; repr. Dover 1986,

ISBN  0 486 65092 8

Wonderfully broad, not only focussing on the now standard
route to the Appel-Haken proof but also giving a wealth of
other material related to colorings.

FF94

Rudolf Fritsch

and

Gerda Fritsch

,

Der Vierfarbensatz

, Brockhaus 1994;

The Four-Color Theorem

, tra.

Julie Peschke

,

Springer 1998,

ISBN  0-387-98497-6

History of the theorem with biographical cameos by
G. Fritsch, thorough topological treatment of plane
graph embedding and details of the Appel-Haken proof by
R. Fritsch.

Wil02

Robert A. Wilson

,

Graphs, Colourings and the Four-colour Theorem

,
Oxford Univ. Pr. 2002,

ISBN  0 19 851062 4

(pbk),

http://www.maths.qmul.ac.uk/ raw/graph.htmlhttp://www.maths.qmul.ac.uk/ raw/graph.html

(errata &c.)

A good general course in graph theory, with special focus on
the four-color theorem and details of the Appel-Haken proof.

Title

colorings of plane graphs

Canonical

name

ColoringsOfPlaneGraphs

Date of creation

2013-03-22 15:10:25

Last modified on

2013-03-22 15:10:25

Owner

marijke (8873)

Last modified by

marijke (8873)

Numerical id

8

Author

marijke (8873)

Entry type

Topic

Classification

msc 05C15

Related topic

FourColorConjecture

Related topic

TaitColoring

Related topic

KempeChain

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