The five color theorem is a result from graph theory that given a plane separated into regions, such as a political map of the countries of the world, the regions may be colored using no more than five colors in such a way that no two adjacent regions receive the same color. Adjacent means that two regions share a common boundary of non-zero length (i.e., not merely a corner where three or more regions meet).
The five color theorem is implied by the stronger four color theorem, but is considerably easier to prove. It was first proved by Percy John Heawood in 1890, based on a purported proof of the four color theorem given by Alfred Kempe in 1879. That proof had been considered correct for 11 years, but Heawood found an error in it; he did manage to give a correct proof of the weaker five color theorem, based on Kempe's work. The first correct proof of the four color theorem was found only in 1976.
Contents
Outline of the proof by contradiction
First of all, one associates a simple planar graph
G
{\displaystyle G}
to the given map, namely one puts a vertex in each region of the map, then connects two vertices with an edge if and only if the corresponding regions share a common border. The problem is then translated into a graph coloring problem: one has to paint the vertices of the graph so that no edge has endpoints of the same color.
Because
G
{\displaystyle G}
is a simple planar graph, i.e. it may be embedded in the plane without intersecting edges, and it does not have two vertices sharing more than one edge, and it does not have loops, then it can be shown (using the Euler characteristic of the plane) that it must have a vertex shared by at most five edges. (Note: This is the only place where the five-color condition is used in the proof. If this technique is used to prove the four-color theorem, it will fail on this step, since there are some graphs that do not have any vertex shared by at most four edges. A notable counterexample is icosahedral graph: as a 5-regular and planar graph it does not have any vertex shared by at most four edges). Find such a vertex, and call it
v
{\displaystyle v}
.
Now remove
v
{\displaystyle v}
from
G
{\displaystyle G}
. The graph
G
′
{\displaystyle G'}
obtained this way has one fewer vertex than
G
{\displaystyle G}
, so we can assume by induction that it can be colored with only five colors. If the coloring did not use all five colors on the five neighboring vertices of
v
{\displaystyle v}
, it can be colored in
G
{\displaystyle G}
with a color not used by the neighbors. So now look at those five vertices
v
1
{\displaystyle v_{1}}
,
v
2
{\displaystyle v_{2}}
,
v
3
{\displaystyle v_{3}}
,
v
4
{\displaystyle v_{4}}
,
v
5
{\displaystyle v_{5}}
that were adjacent to
v
{\displaystyle v}
in cyclic order (which depends on how we write G). So we can assume that
v
1
{\displaystyle v_{1}}
,
v
2
{\displaystyle v_{2}}
,
v
3
{\displaystyle v_{3}}
,
v
4
{\displaystyle v_{4}}
,
v
5
{\displaystyle v_{5}}
are colored with colors 1, 2, 3, 4, 5 respectively.
Now consider the subgraph
G
1
,
3
{\displaystyle G_{1,3}}
of
G
′
{\displaystyle G'}
consisting of the vertices that are colored with colors 1 and 3 only and the edges connecting them. To be clear, each edge connects a color 1 vertex to a color 3 vertex (this is called a Kempe chain). If
v
1
{\displaystyle v_{1}}
and
v
3
{\displaystyle v_{3}}
lie in different connected components of
G
1
,
3
{\displaystyle G_{1,3}}
, we can swap the 1 and 3 colors on the component containing
v
1
{\displaystyle v_{1}}
without affecting the coloring of the rest of
G
′
{\displaystyle G'}
. This frees color 1 for
v
{\displaystyle v}
completing the task. If on the contrary
v
1
{\displaystyle v_{1}}
and
v
3
{\displaystyle v_{3}}
lie in the same connected component of
G
1
,
3
{\displaystyle G_{1,3}}
, we can find a path in
G
1
,
3
{\displaystyle G_{1,3}}
joining them that consists of only color 1 and 3 vertices.
Now turn to the subgraph
G
2
,
4
{\displaystyle G_{2,4}}
of
G
′
{\displaystyle G'}
consisting of the vertices that are colored with colors 2 and 4 only and the edges connecting them, and apply the same arguments as before. Then either we are able to reverse the 2-4 coloration on the subgraph of
G
2
,
4
{\displaystyle G_{2,4}}
containing
v
2
{\displaystyle v_{2}}
and paint
v
{\displaystyle v}
color 2, or we can connect
v
2
{\displaystyle v_{2}}
and
v
4
{\displaystyle v_{4}}
with a path that consists of only color 2 and 4 vertices. Such a path would intersect the 1-3 colored path we constructed before since
v
1
{\displaystyle v_{1}}
through
v
5
{\displaystyle v_{5}}
were in cyclic order. This clearly contradicts the planarity of the graph.
So
G
{\displaystyle G}
can in fact be five-colored, contrary to the initial presumption.
Linear time five-coloring algorithm
Multiple authors, beginning with Lipton and Miller in 1978, have studied efficient algorithms for five-coloring planar graphs. The algorithm of Lipton and Miller took time
O
(
n
log
n
)
{\displaystyle O(n\log n)}
, but subsequent researchers reduced the time bound to
O
(
n
)
{\displaystyle O(n)}
. The version below is from a 1996 paper by Robertson, Sanders, Seymour, and Thomas, which describes it briefly in connection with a slower
O
(
n
2
)
{\displaystyle O(n^{2})}
-time algorithm for four-coloring. The algorithm as described here operates on multigraphs and relies on the ability to have multiple copies of edges between a single pair of vertices. It is based on Wernicke's theorem, which states the following:
Wernicke's theorem: Assume G is planar, nonempty, has no faces bounded by two edges, and has minimum degree 5. Then G has a vertex of degree 5 which is adjacent to a vertex of degree at most 6.
Alternate proof
Kainen (1974) provides a simplified proof of the five color theorem, based on the non-planarity of K6 (the complete graph with 6 vertices) and graph minors. This proof generalizes to graphs that can be made planar by deleting 2 edges.