# The Four-Color Theorem

```The Mathematics of Map
Coloring
–
The Four-Color Theorem
Roger House
Scientific Buzz Café
French Garden
Sebastopol, CA
2013 Jan 17
The Problem
You have a map of some sort, say a map of
the counties of England, and you wish to
color each county in such a way that no
two adjacent counties are the same color.
Of course, you could use a different color for
each county, but there's a condition:
Use as few colors as possible.
Counties of England
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The Counties, Colored
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Why Counties of England?
Some time before 1852 October 23, Francis
Guthrie (1831-99) was coloring a map of
the counties of England when several
questions occurred to him:
What is the minimum number of colors that
would work for the counties of England?
What is the minimum number of colors that
would work for any map?
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1852 October 23
How do we know Francis Guthrie came up
with the problem before 1852 October 23?
Because on that date his brother, Frederick
Guthrie (1833-86), sent a letter to
Augustus de Morgan (1806-71) stating the
problem and crediting Francis with it.
On the very same day, de Morgan sent a
letter to Sir William Rowan Hamilton
(1805-65) telling him of the problem.
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How to begin?
Mathematicians present their results in tidy,
logical, step-by-step, pristine papers in
learned journals.
They virtually never arrive at their results in
the manner suggested by their papers.
computers) and try this and try that, and
offer sacrifices to the Math Mistress.
We will do the same.
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1 country – how many colors?
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1 country – 1 color
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2 countries – how many colors?
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2 countries – 2 colors
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3 countries – how many colors?
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3 countries – 3 colors
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3 countries – how many colors?
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3 countries – 2 colors
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4 countries – how many colors?
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4 countries – 2 colors
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4 countries – how many colors?
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4 countries – 3 colors
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4 countries – how many colors?
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4 countries – 4 colors
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Important fact
The previous example proves something
The minimum number of colors required to
color a map so that no two adjacent
countries are the same color is four.
We now have a lower bound on the number
we seek.
Can we find an upper bound?
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Do the math
Useful facts leading to an upper bound:
1. The Euler Characteristic of a map is two.
2. Every map contains at least one country
with five or fewer neighbors.
3. At most six colors are needed to color
any map.
Note that this last result gives us an upper
bound on the number of colors needed.
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Leonhard Euler (1707-83)
everything:
Euler angles
Euler totient function
Euler's formula Euler constant
Eulerian path
Euler-Lagrange equation
Euler method
Euler identity
His collected works: 76 volumes, so far...
We focus on the Euler Characteristic.
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F+V–E=?
2+4-4=?
3+6-7=?
4+8-10=?
5+10-13=?
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Euler Characteristic
F+V-E is called the Euler Characteristic.
It certainly looks like it is always 2.
In fact, this is the case for maps drawn in the
plane or on a sphere.
We will sketch an informal proof of this.
For a map drawn on a doughnut, things are
different:
The Euler Characteristic on a torus is zero.
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F+V–E=2
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5 or fewer neighbors
Theorem: Every map has at least one
country with 5 or fewer neighbors.
Proof: Assume otherwise, i.e., a map exists
in which every country has 6 or more
neighbors. A typical face:
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5 or fewer neighbors
Each face has at least 6 edges, so 6F ≤ E.
But each edge is counted twice, because
there is a face on each side of the edge.
So 3F ≤ E.
Each vertex has at least 3 edges incident to
it, so 3V ≤ E.
But each edge is counted twice, once for
each end of the edge.
So 3V/2 ≤ E, or 3V ≤ 2E.
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5 or fewer neighbors
3F ≤ E
3V ≤ 2E
3F + 3V ≤ E + 2E = 3E
F+V≤E
F+V–E≤E–E=0
F+V–E≤0
What's wrong with this?
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5 or fewer neighbors
We just proved that
F+V–E=2
so it cannot be that
F+V–E≤0
Thus the assumption 'every face has at least
6 neighbors' must be false.
So, at least one face has 5 or fewer
neighbors.
█
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Six Colors Suffice
Theorem: Every map can be colored with
no more than 6 colors in such a way that
no two countries sharing a common
boundary have the same color.
Proof: Assume otherwise, i.e., maps exist
which require more than 6 colors.
From the set of such maps pick one with the
fewest countries.
By the theorem we just proved, one country
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has 5 or fewer neighbors.
Six Colors Suffice
Here's such a country (whose neighbors
may be more complicated than shown).
We remove one edge:
How many colors
are needed?
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Six Colors Suffice
The new map can be colored with 6 colors
because it has fewer countries than a
minimal map requiring more than 6 colors.
This part of the map requires
at most 5 colors:
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Six Colors Suffice
Add back the edge we removed.
Use the sixth color.
We have colored the map with 6 colors.
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Six Colors Suffice
Wait a minute!
We started with a map which required more
than 6 colors.
But we have just shown how to color it with
6 colors.
Which proves that all maps require no more
than 6 colors.
█
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Pick a number: 4, 5, or 6?
So now we know that the minimum number
of colors required to color any map so that
no two adjacent countries have the same
color is 4, 5, or 6.
Well, which is it? 4? 5? or 6?
It looks as if 4 is the answer, but is it?
For a mathematician, this is an
unsatisfactory situation.
“Wir müssen wissen!”
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Break time
While on break, you might try coloring the
following map to see if it requires 4, 5, or 6
colors:
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Except for a couple of letters and a book
review, there is no mention of the Four
Color Problem from 1852 to 1878.
On 1878 July 13, Arthur Cayley (1821-95)
inquired of the mathematical section of the
Royal Society if a solution had been
submitted.
None had, but this focused attention on the
problem.
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Solved, at last
In 1878 Sir Alfred Bray Kempe (1849-1922)
announced in Nature that he had found
“the solution of a problem which recently
achieved some renown.”
In 1879 his proof was published in the
American Journal of Mathematics Pure
and Applied.
Problem solved: Four colors suffice.
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How did he do it?
We will give a rough idea of Kempe's
approach.
We won't examine a lot of picky cases which
the actual proof dealt with.
We'll focus on example maps, not the
general case.
But, you'll get the idea ...
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Is this approach familiar?
Assume that some maps exist which require
more than four colors.
From these maps, pick one that has the
fewest number of countries.
There may be more than one such map.
Never mind – pick any map from the set of
those with the fewest countries.
Say this map:
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Minimum map for which 4 colors
are not enough
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Get rid of one country
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Now we can 4-color the map
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Restore the removed country
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Change gray to what?
We want to color the gray country red,
green, blue, or yellow.
In order to do this, we must change the
colors of some of the countries bordering
the gray country.
Consider a red-yellow chain:
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Red-yellow chain
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Swap B-G inside R-Y chain
`
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Red-green chain
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Swap B-Y inside R-G chain
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Color the gray country blue
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Here's a 4-coloring of the map
What's wrong with this picture?
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Four Colors Suffice!
The map requires more than 4 colors.
But we have just shown that it can be
colored with 4 colors .
So the theorem is proved: There are no
maps which need more than 4 colors.
Ah (a sigh of relief), all is right with the
world.
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█
Hold it a second
We haven't actually proved the theorem.
We've looked at one example map and
showed how Kempe argued that it could
be 4-colored, using Kempe chains.
His actual proof dealt with a number of
cases which we've ignored.
It dealt with general cases, not examples.
But we've seen the main idea of his method.
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Kempe's proof settled things – after 1879
not much was heard about map coloring.
Until …
In 1890, eleven years after Kempe's proof
came out, Percy John Heawood (18611955) published a paper entitled Mapcolour theorem.
The paper contained this map:
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Ouch!
Heawood's example demolished Kempe's
proof, i.e., Kempe's proof had a hole in it.
Kempe had not realized that applying his
chains twice simultaneously could lead to
an improper coloring.
So, suddenly, the Four Color Theorem once
again became the Four Color Conjecture.
However, all is not lost ...
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The Five Color Theorem
Heawood showed that a modification of
Kempe's method resulted in a proof of the
Five Color Theorem.
So now we know that the minimum number
of colors needed is 4 or 5.
What is needed to decide if 4 colors suffice?
A proof that works for all maps, i.e., a proof
that works for an infinite number of maps.
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Maybe 4 colors aren't enough?
What is needed to decide if 5 colors suffice?
Just one map!
And, of course, an accompanying argument
showing that the map requires 5 colors.
On 1975 April 1, Martin Gardner, editor for
many years of the Mathematical Games
column in Scientific American, published
this map:
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Gardner: 5 colors are needed
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80+ years had gone by ...
What happened between 1890 and 1975?
Lots.
Many mathematicians tried their hands at
proving the Four Color Conjecture.
But none succeeded in finding a proof.
Is that because 5 colors are needed?
Is Gardner's map an example?
Note again the date on his map: April 1.
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Garner's map: 4 colors work
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At last!
But a solution was near at hand.
In 1976, Kenneth Appel (1932-) and
Wolfgang Haken (1928-), two
mathematicians at the University of Illinois
at Urbana-Champaign, announced a
proof.
In 1977 in the Illinois Journal of Mathematics
they published their proof: "Every Planar
Map is Four Colorable”.
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Some reactions
“I am willing to accept the Appel-Haken
proof – beggars cannot be choosers.”
“To most mathematicians, however, the
proof of the four-color conjecture is deeply
unsatisfactory.”
“God wouldn't let the theorem be proved by
a method as terrible as that!”
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What's the problem?
A key part of the proof was a computer
program.
It analyzed 1476 cases, requiring 1200
hours of machine time.
arguments presented in papers.
How could they verify that the proof was
correct?
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Eventually: Acceptance
“To reject the use of computers as what one
may call 'computational amplifiers' would
be akin to an astronomer refusing to admit
“The things you can prove may be just tiny
islands, exceptions, compared to the vast
sea of results that cannot be proved by
human thought alone.”
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In conclusion ...
To commemorate the proof, the University of
Illinois adopted this postage meter stamp:
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