four colorin theorem

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The Four Color Theorem

This page gives a brief summary of a new proof of the Four Color Theorem and a four-coloring

algorithm found by Neil Robertson, Daniel P. Sanders, Paul Seymour and Robin Thomas.

History.

The Four Color Problem dates back to 1852 when Francis Guthrie, while trying to color the mapof counties of England noticed that four colors sufficed. He asked his brother Frederick if it was

true that any map can be colored using four colors in such a way that adjacent regions (i.e. thosesharing a common boundary segment, not just a point) receive different colors. Frederick Guthrie

then communicated the conjecture to DeMorgan. The first printed reference is due to Cayley in1878.

A year later the first `proof' by Kempe appeared; its incorrectness was pointed out by Heawood

11 years later. Another failed proof is due to Tait in 1880; a gap in the argument was pointed outby Petersen in 1891. Both failed proofs did have some value, though. Kempe discovered what

became known as Kempe chains, and Tait found an equivalent formulation of the Four ColorTheorem in terms of 3-edge-coloring.

The next major contribution came from Birkhoff whose work allowed Franklin in 1922 to prove

that the four color conjecture is true for maps with at most 25 regions. It was also used by othermathematicians to make various forms of progress on the four color problem. We should


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specifically mention Heesch who developed the two main ingredients needed for the ultimateproof - reducibility and discharging. While the concept of reducibility was studied by other

researchers as well, it appears that the idea of discharging, crucial for the unavoidability part ofthe proof, is due to Heesch, and that it was he who conjectured that a suitable development of

this method would solve the Four Color Problem.

This was confirmed by Appel and Haken in 1976, when they published their proof of the FourColor Theorem [1,2].

Why anew proof?

There are two reasons why the Appel-Haken proof is not completely satisfactory.

y Part of the Appel-Haken proof uses a computer, and cannot be verified by hand, andy even the part that is supposedly hand-checkable is extraordinarily complicated and

tedious, and as far as we know, no one has verified it in its entirety.

We have in fact tried to verify the Appel-Haken proof, but soon gave up. Checking the computerpart would not only require a lot of programming, but also inputing the descriptions of 1476

graphs, and that was not even the most controversial part of the proof.

We decided that it would be more profitable to work out our own proof. So we did and came up

with a proof and an algorithm that are described below.

Outline ofthe proof.

The basic idea of the proof is the same as Appel and Haken's. We exhibit a set of 633

"configurations", and prove each of them is "reducible". This is a technical concept that impliesthat no configuration with this property can appear in a minimal counterexample to the FourColor Theorem. It can also be used in an algorithm, for if a reducible configuration appears in a

planar graph G, then one can construct in constant time a smaller planar graph G' such that anyfour-coloring of G' can be converted to a four-coloring of G in linear time.

It has been known since 1913 that every minimal counterexample to the Four Color Theorem isan internally 6-connected triangulation. In the second part of the proof we prove that at least one

of our 633 configurations appears in every internally 6-connected planar triangulation (notnecessarily a minimal counterexample to the 4CT). This is called proving unavoidability, and

uses the "discharging method", first suggested by Heesch. Here our method differs from that of

Appel and Haken.

Main features of our proof.

We confirm a conjecture of Heesch that in proving unavoidability, a reducible configuration can

be found in the second neighborhood of an "overcharged" vertex; this is how we avoid"immersion" problems that were a major source of complication for Appel and Haken. Our

unavoidable set has size 633 as opposed to the 1476 member set of Appel and Haken, and our


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discharging method uses only 32 discharging rules, instead of the 300+ of Appel and Haken.Finally, we obtain a quadratic algorithm to four-color planar graphs (described later), an

improvement over the quartic algorithm of Appel and Haken.

Configurations.

A near-triangulation is a non-null connected loopless plane graph such that every finite regionis a triangle. A configuration K consists of a near-triangulation G and a map g from V(G) to the

integers with the following properties:

1. for every vertex v, G\v has at most two components, and if there are two, then the degreeof v is g(v)-2,

2. for every vertex v, if v is not incident with the infinite region, then g(v) equals the degreeof v, and otherwise g(v) is greater than the degree of v; and in either case g(v)> 4,

3. K has ring-size at least 2, where the ring-size of K is defined to be the sum of g(v)-deg(v)-1, summed over all vertices v incident with the infinite region such that G\v is

connected.

When drawing pictures of configurations we use a convention introduced by Heesch. The shapes

of vertices indicate the value of g(v) as follows: A solid black circle means g(v)=5, a dot (orwhat appears in the picture as no symbol at all) means g(v)=6, a hollow circle means g(v)=7, a

hollow square means g(v)=8, a triangle means g(v)=9, and a pentagon means g(v)=10. (We donot need vertices v with g(v)> 11, and only one vertex with g(v)=11, for which we do not use

any special symbol.) In the picture below 17 of our 633 reducible configurations are displayedusing the indicated convention. The whole set can be viewed by clicking here. (We refer to (3.2)

of our paper[7] for the meaning of "thick edges" and "half-edges" in those pictures.)

Any configuration isomorphic to one of the 633 configurations exhibited in [7] is called

a good configuration. Let T be a triangulation. A configuration K=(G,g)appears in T if G is aninduced subgraph of T, every finite region of G is a region of T, and g(v) equals the degree of v

in T for every vertex v of G. We prove the following two statements.


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THEOREM 1. If T is a minimal counterexample to the Four Color Theorem, then no goodconfiguration appears in T.

THEOREM 2. For every internally 6-connected triangulation T, some good configuration

appears in T.

From the above two theorems it follows that no minimal counterexample exists, and so the 4CTis true. The first proof needs a computer. The second can be checked by hand in a few months,

or, using a computer, it can be verified in about 20 minutes.

Discharging rules.

Let T be an internally 6-connected triangulation. Initially, every vertex v is assigned a charge of

10(6-deg(v)). It follows from Euler's formula that the sum of the charges of all vertices is 120; inparticular, it is positive. We now redistribute the charges according to the following rules, as

follows. Whenever T has a subgraph isomorphic to one of the graphs in the figure below

satisfying the degree specifications (for a vertex v of a rule with a minus sign next to v thismeans that the degree of the corresponding vertex of T is at most the value specified by the shapeof v, and analogously for vertices with a plus sign next to them; equality is required for vertices

with no sign next to them) a charge of one (two in case of the first rule) is to be sent along theedge marked with an arrow.


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A quadraticalgorithm.

The input to the algorithm will be a plane triangulation G with n vertices. (This is without loss ofgenerality, as any planar graph can be triangulated in linear time.) The output will be a coloring

of the vertices of G with four colors.

If G has at most four vertices color each vertex a different color. Otherwise if G has a vertex x ofdegree k < 5, then the circuit C surrounding it is a `k-ring'. Go to the k-ring analysis below.

Otherwise G has minimum degree five. For every vertex we compute its charge as explainedabove, and find a vertex v of positive charge. It follows from our proof of Theorem 2 that either

a good configuration appears in the second neighborhood of v (it which case it can be found inlinear time), or a k-ring violating the definition of internal 6-connection can be found in linear

time. In the latter case we go to the k-ring analysis below, in the former case we apply recursionto a certain smaller graph. A four-coloring of G can then be constructed from the four-coloring

of this smaller graph in linear time.

Given a k-ring R violating the definition of internal 6-connection a procedure developed byBirkhoff can be used. We apply recursion to two carefully selected subgraphs of G. A four-coloring of G can then be constructed from the four-colorings of the two smaller graphs in linear

time.

Discussion.

We should mention that both our programs use only integer arithmetic, and so we need not beconcerned with round-off errors and similar dangers of floating point arithmetic. However, an

argument can be made that our `proof' is not a proof in the traditional sense, because it containssteps that can never be verified by humans. In particular, we have not proved the correctness of

the compiler we compiled our programs on, nor have we proved the infallibility of the hardwarewe ran our programs on. These have to be taken on faith, and are conceivably a source of error.

However, from a practical point of view, the chance of a computer error that appears consistentlyin exactly the same way on all runs of our programs on all the compilers under all the operating

systems that our programs run on is infinitesimally small compared to the chance of a humanerror during the same amount of case-checking. Apart from this hypothetical possibility of a

computer consistently giving an incorrect answer, the rest of our proof can be verified in thesame way as traditional mathematical proofs. We concede, however, that verifying a computer

program is much more difficult than checking a mathematical proof of the same length.


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"Proof"ofthe4-ColorMapTheorem

Proof by Contradiction:

1) We assume that the 4-color theorem is false (i.e., that there exist finite maps that require at

least five colors), and show that that assumption leads to a contradiction. This means that the

assumption was incorrect and that four colors are therefore sufficient to color any finite planar

map.

2) There then must exist a counterexample of smallest possible size (i.e. a map with the smallest number

of regions that require 5 colors). Let M be such a "smallest" counterexample.

3) Removing any region from M therefore results in a 4-colorable map.

4) Every region must therefore have valence >= 4 (ie, have 4 or more neighbors) because if such

a region existed with a valence less than 4, you could:

a: remove that region from M,

b: 4-color the resulting map (via step 3), and

c: add the region back in, coloring it differently from its


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Figure 1: region v has valence = 4

6) Fix a 4-coloring of the map M-v (read "M minus v", the submap of M with v

removed). This must be possible because of the definition of M. The neighbors of v inM must all be different colors, say r, y, g, and b clockwise. (If they were not four

different colors, v could be colored using a remaining color to get a 4-coloring of M).

7) Consider the largest connected submap of M that contains region r and only red and green

regions. See figure 1.

8) If that submap does not contain g, then, within that submap, recolor all red regions green and

all green regions red. We now have a 4-coloring of M-v in which no neighbor of v is red.Therefore coloring v red results in a 4-coloring of M. Contradiction.

9) The contradiction in step 8 shows that the submap in step 7 must contain g. Therefore there

exists an alternating path of red and green regions that divides M-v into two parts such that y isin one and b is in the other. See figure 1.

10) Within the part containing y, recolor all yellow regions blue and all blue regions yellow. We

now have a new 4-coloring of M-v in which no neighbor of v is yellow. Therefore coloring vyellow results in a 4-coloring of M. Also a contradiction. CASE 1 leads to a contradiction eitherway, so it must be false.

CASE 2: There exists a region of valence 5. Call it v.


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11) As in step 6, fix a 4-coloring of the map M-v. By the same reasoning as in step 6, theneighbors of v must be four different colors, so now there are two possible unique types

of colorings of the neighbors of v: (clockwise) ryygb and rygyb.

12) The proof that the ryygb case cannot exist is identical to CASE 1 above.

Figure 2a: region v has valence = 5 and is surrounded by ryygb

3) The final possibility to examine is where the neighbors of v are colored rygyb.


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Figure 2b: region v has valence = 5 and is surrounded by rygyb

14) The alternating red-green path between r and g which isolates the first y region from b mustexist by the same reasoning as for CASE 1. This means that the yellow and blue regions in thearea containing the first y region can swap colors, turning that yellow region blue.

15) Likewise, the same argument can be used to show that there must exist a path of alternatingblue-green regions from b to g which isolate the second y region from r. This means that theyellow and red regions in that region can swap colors, turning that y region red.

16) Because steps 14 and 15 allow us to recolor both y regions to other colors, we can arrive at a4-coloring of M-v in which no neighbor of v is yellow. Therefore coloring v = yellow results in a

4-coloring of M. This proves that CASE 2 must also be false. Since both cases are false, M(which was any smallest counter-example) cannot exist proving that there can be no counter-

examples, and that all finite planar maps must be 4-colorable.

The Four Color Theorem

How many different colors are sufficient to color the countries on a map in such a way

that no two adjacent countries have the same color? The figure below shows a typical


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arrangement of colored regions.

Notice that we define adjacent regions as those that share a common boundary of non-zerolength. Regions which meet at a single point are not considered to be "adjacent".

After examining a wide variety of different planar graphs, one discovers the apparent factthat every graph, regardless of size or complexity, can be colored with just four distinct

colors. This "four-color conjecture" was first noted by August Ferdinand Mobius in 1840.In 1852 a young man named Francis Guthrie wrote about the problem in a letter to his

brother Frederick, then a student at University College in London. Neither of the brotherswas able to prove the conjecture, so Frederick asked one of his professors, Augustus

DeMorgan (1806-1871). DeMorgan too was unable to prove the conjecture, and afterrecognizing the difficulty of the problem, he wrote to Sir William Rowan Hamilton (1805-

1865) to ask for help. It might seem that this problem would be irresistible to Hamilton,since anything having to do with the number "four" immediately suggests a connectionwith his beloved quaternions. Also, Hamilton made contributions to graph theory (such as

the idea of a Hamiltonian circuit, i.e., a path along the edges of a graph that visits eachvertex exactly once), a subject that was developed largely through efforts to prove the four

color conjecture. Nevertheless, Hamilton immediately wrote back that he did not believehe would solve DeMorgan's "quaternion of colors" any time soon.

The coloring of geographical maps is essentially a topological problem, in the sense that itdepends only on the connectivities between the countries, not on their specific shapes,

sizes, or positions. We can just as well represent each country by a single point (vertex),

and the adjacency between two bordering countries can be represented by a line (edge)connecting those two points. It's understood that connecting lines cannot cross each other.A drawing of this kind is called a planar graph. A simple map (with just five "countries")

and the corresponding graph are shown below.


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A graph is said to be n-colorable if it's possible to assign one of n colors to each vertex insuch a way that no two connected vertices have the same color. Obviously the above

graph is not 3-colorable, but it is 4-colorable. The Four Color Theorem asserts that everyplanar graph - and therefore every "map" on the plane or sphere - no matter how large or

complex, is 4-colorable. Despite the seeming simplicity of this proposition, it was onlyproven in 1976, and then only with the aid of computers.

Notice that the above graph is "complete" in the sense that no more connections can be

added (without crossing lines). The edges of a complete graph partition the graph planeinto three-sided regions, i.e., every region (including the infinite exterior) is bounded by

three edges of the graph. Every graph can be constructed by first constructing a completegraph and then deleting some connections (edges). Clearly the deletion of connections

cannot cause an n-colorable graph to require any additional colors, so in order to prove theFour Color Theorem it would be sufficient to consider only complete graphs.

Although DeMorgan was unable to prove the four-color conjecture, he did observe that no

more than four regions on the plane can all be in mutual contact with each other. The

graph of a set of three mutually adjoining regions is simply a topological triangle, and ifwe add a fourth region, it is represented by a fourth vertex in the graph, which must belocated either inside or outside the triangle formed by the graph of the original three

vertices. In either case, we must then connect this fourth vertex with each of the threeoriginal vertices so that all four of the regions are mutually adjoining. Having done this,

we can slide the vertices around (if necessary) to bring the graph into the form shownbelow.

This is the unique graph of four mutually adjoining plane regions (V = 4, E = 6), and alsoof the tetrahedron. It clearly divides the graph plane into four isolated regions, one region

exterior to the big triangle, and the three interior regions. A fifth vertex added to thisgraph will be able to have edges that reach only three of the four existing vertices, because

each of the four regions is completely bounded by three edges that block access to one ofthe existing vertices. Hence there does not exist a plane graph of five mutually connected


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vertices, so there does not exist a set of five mutually adjoining regions on the plane.

Of course, the non-existence of a graph with five mutually adjoining vertices does not

automatically imply the non-existence of a graph requiring five distinct colors. Forexample, there exist graphs in which the largest subset of mutually adjoining vertices is

two, and yet for which three colors are required. This is the case with any simple loopwith an odd number of vertices, as illustrated in the left-hand figure below. Similarly there

are graphs containing no subset of four mutually adjoining vertices that neverthelessrequire four colors, such as the graph shown on the right.

Therefore, we must consider the possibility that five colors might be required for somegraph, even though it's not possible for a graph to contain any set of five mutually

connected vertices.

On the other hand, it could be argued that the two example shown above can each bereduced, clarifying the causal links between different parts of the graphs. For example, in

the left figure we have an alternating sequence of red, green, red, green, and if these werethe only two available colors, it's clear that the chain must continue to alternate, no matter

how long - or how short. Thus we could just as well replace any chain of the form rgrg...rgwith the simple pair rg, in which case the left-hand figure above consists of three mutually

connected vertices. Likewise if we replace the circumferential chain rgrg around the

central yellow vertex in the right hand figure with just the pair rg, we get a set of fourmutually connected vertices.

In view of this, it's tempting to think that DeMorgan's observation somehow, perhaps

indirectly, does ultimately account for the truth of the four-color theorem. Indeed if weconstruct a graph by adding vertices, one at a time, and make the maximum number of

connections at each stage, we will always find the graph plane divided into "triangular"regions, each of which has access to only three vertices. Thus it follows that four colors

suffice for any graph that can be constructed in this way. For example, consider the twographs shown below.


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These two graphs each have V = 6 vertices and E = 12 edges, and in fact the two graphsare topologically identical. The plane regions represented by the vertices "a" and "b" each

have five adjoining neighbors, whereas the vertices "e" and "f" each have four, and thevertices "c" and "d" each have three. These are complete graphs and, as noted above, a

complete graph divides the entire graph plane into "triangular" regions, i.e., regionsbounded by three edges connecting three vertices. Nevertheless, depending on how we

add the points, it is possible that when a vertex is added to a graph it has more than threeneighbors, so we cannot say automatically that four colors would suffice. For example, if

vertex "a" is the last to be added, it would have five pre-existing neighbors, and if fourcolors have already been used in those five vertices, the vertex "a" would require a fifth

color.

However, we need not add vertex "a" last. Another topologically equivalent way ofdrawing the above graph is shown below

This shows that we could first assign three distinct colors to the vertices e,b,f, and then

place the vertex "a" in this triangle, connect it to each of the three surrounding vertices,and give it a fourth color. Then we can place vertex d inside the triangle abe and give it

the same color as f. Then we can place vertex c inside the triangle abf and give it the colorof e. Hence the graph is 4-colorable. Moreover, any graph, or portion of a graph bounded

by a triangle such as ebf , and having this hierarchical pattern of nested triangles, is 4-

colorable. This is the case when the graph contains some vertices with only three edges,and when those vertices and edges are removed, the remaining graph has some verticeswith only three edges, which can be removed, and so on, until finally all that remains is a

single triangle.

Unfortunately (for the prospect of a simple proof), not every graph is of this hierarchical

form. For example, consider the complete graph shown below.


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This graph has V = 16 vertices and E = 42 edges. It is not hierarchical, because afterremoving the two vertices having only three edges each, the remaining graph has four or

more edges attached to each vertex. (We can also infer that this graph is not hierarchicalfrom the fact that no three mutually connected vertices are directly connected to a fourth

vertex.) Nevertheless, this graph is 4-colorable, as shown in the figure.

The above graph contains a "flat" patch consisting of a uniform hexagonal grid.

An infinitehexagonal grid is obviously 4-colorable by means of the alternating layeredpattern shown below

In a sense, two extreme types of graph structures are the purely hierarchical graphs and

the perfectly "flat" graphs, both of which are easily seen to be 4-colorable.


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Notice that we need not consider graphs containing any configurations of four mutuallyconnected vertices, because, as discussed previously, the edges of such a configuration

necessarily split the graph plane into four separate regions, each of which has access toonly three of the four vertices. Hence if we can 4-color the interior of such a triangle, we

can 4-color the tetrahedral "frame" as well. If we could reduce all the causal links in a

graph to their bare minimums, it might be possible to reduce every n-colorable graph to agraph with just n mutually connected vertices, and hence n could never exceed four.However, the reduction of graphs with more than two colors is not a trivial task.

If we denote the number of vertices, edges, and faces (i.e., the bounded regions) of aplanar graph by V, E, and F respectively, then Euler's formula for a plane (or a sphere) is

V E + F = 2. Furthermore, each face of a complete graph is bounded by three edges, andeach edge is on the boundary of two faces, so we have F = 2E/3, and Euler's formula for a

complete planar graph is simply E = 3V 6. Now, each edge is connected to two vertices,

so the total number of attachments (in a complete graph) is 2E = 6V 12, and hence the

average number of attachments per vertex is 6 12/V. For any incomplete graph, the total

number of attachments is less. Consequently the average number of attachments pervertex for any graph (with a finite number of vertices) is less than 6, which implies that atleast one vertex has only five or fewer attachments.

If we have six available colors, a vertex with only five neighbors obviously imposes no

constraint on the coloring of the other vertices, because, regardless of the colors of its five(or fewer) neighbors, we can assign it a color without exceeding the six available colors.

Therefore if we delete this vertex and all its connections from the graph, creating a graphwith one fewer vertices, it's clear that if the resulting graph is 6-colorable, then so was the

original graph. Moreover, Euler's formula assures us that this reduced graph also containsat least one vertex with five or fewer neighbors, so we can apply this procedure

repeatedly, reducing the graph eventually to one with just 6 vertices, which is obviously 6-colorable. Hence the original graph is 6-colorable.

So, we've seen that Euler's formula immediately implies that no graph can require morethan six colors. Furthermore, with just a little more work, we can also show that no graph

can require more than five colors. (Ultimately we will see that no graph can require morethan four colors, but it's worthwhile to begin with the proof of the 5-colorability of every

planar graph.) Obviously every graph with five or fewer vertices is 5-colorable, so if thereexists a finite graph that requires more than five colors, it must have more than five

vertices. Let the positive integer V6 denote the smallest number of vertices on which thereexists a graph that requires six (or more) colors. Conversely, every graph with fewer thanV6 vertices is 5-colorable.

Now, assuming the existence of a graph that requires more than five colors, we canconsider one that has exactly V6 vertices. By Euler's formula, this graph must contain at

least one vertex with five or fewer connections. However, it cannot contain any vertexwith just four (or fewer) connections, because if it did, we could delete such a vertex and

leave a graph with just V6 1 vertices, which is 5-colorable by definition. Re-inserting thedeleted vertex would clearly have no effect on the 5-colorability of the graph, because the


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vertex has only four (or fewer) neighbors, so the original graph must be 5-colorable,contradicting our assumption. Therefore, a graph with V6 vertices that requires more than

five colors cannot contain any vertex with just four or fewer neighbors.

Since Euler's formula implies that the graph contains at least one vertex with five or fewer

connections, the only remaining possibility is that the graph contains a vertex with exactlyfive connections. However, this too leads to a contradiction. To show this, it's helpful to

introduce the notion of a k-cluster, which is specified by a set of k distinct colors and oneparticular vertex that has one of those colors. The original vertex is included in the cluster,

and, in addition, every vertex with one of the k specified colors that neighbors a vertex inthe cluster is also in the cluster. By definition the only vertices outside a cluster that are

directly connected to vertices inside the cluster have colors that are not in the specified setof k colors. Therefore, we can apply any permutation of the k colors to the vertices in a

cluster without invalidating the coloration. In particular, a 2-cluster is a contiguous set ofvertices, each with one of two specified colors, and we can transpose these two colors

without upsetting the coloration of a graph.

Now consider a graph containing a vertex with exactly five immediate neighbors, of fivedistinct colors, as illustrated below.

Since the uncolored vertex in the center has neighbors of five distinct colors, it mightseem that a sixth color is required. However, notice that we can transpose the blue and

green colors in the blue/green 2-cluster attached to the upper left vertex of the centralpentagon, so that the upper left vertex is green instead of blue. Once we have done this,

the uncolored vertex in the center has neighbors of only four distinct colors. If we deletethe central vertex, the overall graph has V6 1 vertices, so it is 5-colorable, but re-inserting this vertex requires no sixth color (once we have transposed the blue and green

in the small 2-cluster as described), so the original graph is 5-colorable.

The only possible objection to the transposing of the colors in a cluster as a means of

reducing the number of distinct colors around the perimeter of the pentagon is that wecannot necessarily change the color of any vertex to the color of one of the opposite


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vertices of the pentagon, because two opposite vertices of a pentagon might be in the same2-cluster. This is the case, for instance, with the yellow and red vertices of the pentagon in

the figure above. If we transpose the red and yellow colors in this 2-cluster, there wouldstill be five distinct colors around the perimeter of the pentagon. However, if such a

cluster exists, connecting two opposite vertices of the pentagon, we are guaranteed that at

least one other pair of vertices are cut off from each other, i.e., they cannot be in the same2-cluster. For example, the blue/green 2-cluster outlined in red in the above figure cannotbe part of the blue/green 2-cluster attached to the upper right of the pentagon. In general,

for any vertex connected to exactly five other vertices, we can always (withoutinvalidating the coloration) change the color of at least one of the five neighbors so that

there are only four distinct colors. It follows, using the reduction argument described inthe preceding paragraph, that no graph can require more than five colors.

This still leaves open the question of whether five colors are ever actually required, orwhether four colors will always suffice. If there exist planar graphs requiring five distinct

colors, there must be a positive integer V5 that is the smallest number of vertices on which

such a graph can be constructed. Let's call a graph with V5 vertices that requires fivecolors a minimal 5-color graph. By the reduction argument described previously, such agraph obviously cannot contain any vertex with three or fewer connections. Moreover, by

considering 2-clusters again, we can prove that such a graph cannot contain any vertexwith just four connections. To see why, consider the portion of a graph illustrated below.

The uncolored vertex in the center has just four neighbors, which we've colored red, yellow,

green, and blue. This might seem to require a fifth color for the central vertex, but in fact bypermuting the yellow and blue colors in the 2-cluster (outlined in red) above the central

vertex we can transform this coloration so that the central uncolored vertex has neighbors of

only three distinct colors (red, green, and blue). Since the overall graph from which thisconfiguration was taken is assumed to have exactly V5 vertices, it follows that if we delete

the central vertex, the resulting graph has only V5 1 vertices, so it is 4-colorable, but we

can clearly add the central vertex back without requiring a fifth color, so the whole graphmust be 4-colorable, contradicting our original assumption. This shows that it's always

possible to modify a graph containing a vertex with only four connections in such a way thatthe vertex is connected to only three distinct colors. At most, only one of the two pairs of

opposite vertices can be linked by a common 2-cluster, because if one pair is so linked, the


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other cannot be linked.

To summarize the argument up to this point, we've shown that no graph can require more

than five colors, and we've also shown that if there exists a graph requiring five colors, theremust exist a minimal 5-color graph, and such a graph cannot contain any vertex with fewer

than five connections. Also, by adding connections, we can "complete" any graph on Vvertices so that all the faces are triangular and there are a total of exactly 6V12attachments. This graph must still be a minimal 5-color graph because we haven't increased

the number of vertices, and adding connections cannot reduce the number of colors required,whereas no graph requires more than five colors. Therefore, letting a5, a6, a7,... denote the

number of vertices with precisely 5, 6, 7,... attachments respectively, we must have aminimal 5-color graph such that

where V = a5 + a6 + a7 + a8 + a9 + ... Substituting this expression for V into the aboveformula and re-arranging, we have the condition

This places fairly severe constraints on any putative complete minimal 5-color graph. For

example, if we consider only such graphs with no more than six attachments at any singlevertex, then this formula implies a5 = 12. In other words, a complete minimal 5-graph with

no more than six attachments per vertex must have exactly 12 vertices with five attachmentseach. This suggests that these 12 vertices are arranged globally in the pattern of an

icosahedron, and the remaining vertices with six attachments per vertex are in a regularhexagonal pattern, filling in the "faces" of the icosahedron. (This pattern is the basis for

"geodesic domes"). The fundamental graph of this type, with a6 = 0, is shown below.

We can give explicit 4-colorings of every such pattern, so such graphs can be shownexplicitly to require no more than four colors. Therefore, if a complete minimal 5-color


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graph exists, it must contain at least one vertex with seven or more attachments.

It was not until 1976 that, with the help of modern computers, the 4-color conjecture was

finally proven to be true. The proof, developed by Appel and Haken based on ideas ofseveral earlier people (Kempe, Heawood, Birkhoff, etc) consisted of finding a set of

subgraphs, at least one of which must be contained in any normal planar graph (i.e., the setis unavoidable), and such that if a graph containing one of those sub-graphs is not 4-

colorable, then neither is a graph with one fewer vertices. This leads immediately to acontradiction, because it implies that, if we hypothesize the existence of a graph that is not

4-colorable, we can always construct another graph, with fewer vertices, that is also not 4-colorable, eventually arriving at a graph with only four vertices, but this graph

obviously is 4-colorable. Hence we have a contradiction, so we can conclude that theoriginal hypothesis was false, i.e., there does not exist a graph that is not 4-colorable. The

unavoidable set found by Appel and Haken consisted of nearly 1500 subgraphs, and many ofthese required considerable analysis to prove that they were "reducible". The development

of this unavoidable set, and the proofs of reducibility for its members, was carried out

largely on computers, so the proof is notable (and slightly controversial) as an early exampleof a mathematical proposition whose proof can only be carried out on a computer, and is, insome sense, beyond human mental capabilities to verify.

It's interesting to consider the problem of determining a 4-coloring for any given graph from

a purely algebraic standpoint (harking back to Hamilton's quaternions). We might stipulatethat each vertex is to be assigned one of four possible values, and the conditions imposed by

the connections would be represented by algebraic equations, one equation for eachconnection. Part of the difficulty of this approach is that the condition of adjacency is not an

equality, it is aninequality, i.e., we require that two vertices have unequal values (fromamong the four possible values). Also, there is necessarily a great deal of ambiguity in the

solution, because any permutation of the colors leaves a solution intact. In addition, there iseven more ambiguity, since in general there can be more than one distinct 4-coloration of a

graph.

Given the four allowable (distinct) values a,b,c,d, we could algebraically impose therequirement for every vertex to have one of these values by requiring that the value u

assigned to any vertex satisfies f(u) = 0, where the polynomial f is defined as

Then we could algebraically impose the required inequality on the values u,v assigned totwo connected vertices by stipulating that g(uv) = 0, where the polynomial g is defined as

The equation g(uv) = 0 is satisfied if and only if u and v have distinct values from the set{a,b,c,d}. For example, the set {-2,-1,1,2} gives


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To illustrate, consider again the octahedral graph (which is 3-colorable):

The algebraic conditions for the individual "color" values assigned to the six variablesA,B,C,D,E,F are

and the algebraic conditions imposed by the twelve connections are

This amounts to 18 equations in the 6 variables. We can, however, without loss of

generality stipulate the "colors" of three of the variables, say, A = 1, B = -1, C = 2, sincethese three must have mutually distinct values. This leaves us with the following twelve

equations in the three unknowns D,E,F

and

The three equations f(D) = 0, g(D) = 0, and g(-D) = 0 jointly have only two solutions,

namely, D = 2 and D = -2. Therefore those three equations can be replaced with the single

equation D2 4 = 0. Likewise the three equations involving only E can be reduced to E

2+

3E + 2 = 0, and the three equations involving only F can be reduced to F2

+ F 2 = 0.Thus we have six equations in the three unknowns.

A different approach would be to assign a four-dimensional vector to each vertex, and

then the connection between two given vertices would be represented by requiring the dotproduct of those two vectors to vanish, i.e., requiring that the two vectors be mutually

orthogonal. The three vectors assigned to the vertices of any triangle would then be an


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orthogonal triad with some orientation in 4-dimensional space. Likewise the four vectorsassigned to the vertices of a tetrahedral graph would be a complete set of four mutually

orthogonal vectors (a tetrad), but still with arbitrary orientation. For example, theconditions on the six vectors A,B,C,D,E,F assigned to the vertices of the octahedral graph

shown above would be simply

In this context the four color theorem tells us that a space of four dimensions is sufficient

to enable us to assign one of the four basis vectors to each vertex of a planar graph in sucha way that the vectors of every pair of adjacent vertices are orthogonal. We can assume

each vector is of unit length, so it has three independent components. Therefore, we have18 independent components in all, constrained by 12 equations. Naturally the system is

underspecified, because we can apply an arbitrary permutation to the vectors.

Since the edges of a complete graph partition the graph plane into three-sided regionswhose vertices are three mutually connected points, we can uniquely assign the fourth

color to each of these regions. The result is useful for visualizing the symmetries of thecoloring. For example, the 4-colored icosahedral graph discussed above looks like this:

four colorin theorem

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