The Four Color Problem

What is the minimum number of colors needed to color any map so that adjacent regions do not have the same color?

Lucia Maddalena Priulla Index: - History of the Four-Color Problem; - Kempe’s approach: the first proof of the Four-Color Theorem; - Heawood counterexample to Kempe’s proof; - The conjectures of Hajòs and Hadwiger; - The Four-Color Theorem as a corollary of Hadwiger’s Conjectures; - More result about coloring: Mycielski’s Construction; - Mader’s Theorem; - König’s Theorem; - Vizing’s Theorem; - Bibliography. Some mathematicians attribute the problem to August Ferdinand Möebius. It is said that during a lecture, the lecturer turn to his students the following problem:

“There was a king with five sons. In his will, he stated that after his death his kingdom should be divided into five regions in such a way that each region should have a common boundary with the other four. Can the terms of the will be satisfied?”

• Möebius said that this problem cannot be solved. • In 1878 Alfred Cayley, introduced this problem to London Mathematical Society. The Four Color Problem

•In 1852 Francis Guthrie (1831-1899), a former graduate of University College London, observed that the countries of England could be coloured with 4 colours so that neighboring countries are coloured differently. This, led him to ask if the previous statement holds for every map (real or imagined). •Francis mentioned this problem to his brother Frederick, who was taking a class from the famous mathematicians August De Morgan. •De Morgan was unable to solve this problem, and other mathematicians seemed to share this interest. •In 1879 there was an article written by the British lawyer Alfred Bray Kempe containing a proposed proof that every map can be coloured with 4 or fewer colours so that neighboring countries are coloured differently. •For the next ten years, the Four Color Problem was considered to be solved.

•An 1890 article by the British mathematician Percy John Heawood showed a counterexample to the technique used by Kempe; so it showed that Kempe’s method was unsuccessful. The idea of Kempe. Now, we show the steps of the Kempe’s proof, which it contains an error. To see this, we need to introduce some new definitions. Recall Definition: A graph is planar if it has a drawing without crossing. A plane graph is a drawing of a planar graph in the plan. Definition: a map is a planar graph G, whose faces have sides which are simple and closed curve, called regions, and it satisfies: 1) Every side divides different faces; 2) Every vertex v has degree greter or equal to 3. Every map is a planar graph, but it is not true the vice versa.

KEMPE CHAIN ARGUMENT Kempe chain argument in vertex colouring: Let G be a graph with a colouring using at least two different colours represented by i and j. For example i=1, and j=2. Let H(i, j) denote the subgraph of G induced by all the vertices of G coloured either i (=1) or j (=2) and let K be a connected component of the subgraph H(i, j).

If we interchange the colours 1 and 2 on the vertices of K and keep the colours of all other vertices of G unchanged, then we get a new colouring of G, which uses the same colours with which we started. This subgraph K is called a Kempe chain and the recolouring technique is called the Kempe chain argument.

Edge-colourings Definition: A k-edge-colouring of G is a labeling f : E(G) --> [k]; the labels are “colours”. A proper k-edge-colouring is a k-edge-colouring such that edges sharing a vertex receive different colours; equivalently, each colour class is a matching. A graph G is k-edge-colourable if it has a proper k-edge-coloring. The edge-chromatic number or chromatic index χ’(G) is the minimum k such that G is k-edge colourable. Kempe chain argument for edge colouring: it is the same topic of Kempe chain argument for vertex colouring. It’s enough to replace vertex with edge.

Definition: consider a map colored with four colors: a,b,c,d. A Kempe’s chain (or ac-chain) is a sequence of regions (faces) colored with colors a and c, without passing from regions colored with b or d.

ac-chain Kempe’s approach for solving the Four Color Problem. He considered a region R in a map M such that R is surrounded by five regions and showed that for every coloring of M with four colors, there is a coloring of the entire map M with four colors. If only three colors are used to color these five regions, then a color is available for R, so we have obtained a 4-coloring.

We consider a map with 4 colors: yellow, green, blue, red. Only the region R1 is colored yellow. Consider all the regions of the map M that are colored either yellow or red and that, beginning at R1, can be reached by an alternating sequence of neighboring yellow and red regions, that is, by a yellow-red Kempe chain. If the region R3 (which is the neighboring region of R colored red) cannot be reached by a yellow-red Kempe chain, then the colors yellow and red can be interchanged for all regions in M that can be reached by a yellow-red Kempe chain beginning at R1. This results in a coloring of all regions in M (except R) in which neighboring regions are colored differently and such that each neighboring region of R is colored red, blue, or green. We can then color R yellow to arrive at a 4-coloring of the entire map M. From this, we may assume that the region R3 can be reached by a yellow-red Kempe chain beginning at R1. See the figure : Let’s now look at the region R5, which is colored green. We consider all regions of M colored green or red and that, beginning at R5, can be reached by a green-red Kempe chain. If the region R3 cannot be reached by a green-red Kempe chain that begins at R5, then the colors green and red can be interchanged for all regions in M that can be reached by a green-red Kempe chain beginning at R5. Upon doing this, a 4-coloring of all regions in M (except R) is obtained, in which each neighboring region of R is colored red, blue, or yellow. We can then color R green to produce a 4-coloring of the entire map M. We may therefore assume that R3 can be reached by a green-red Kempe chain that begins at R5. (See Figure.) Because there is a ring of regions consisting of R and a green-red Kempe chain, there cannot be a blue-yellow Kempe chain in M beginning at R4 and ending at R1. In addition, because there is a ring of regions consisting of R and a yellow-red Kempe chain, there is no blue-green Kempe chain in M beginning at R2 and ending at R5. Hence we interchange the colors blue and yellow for all regions in M that can be reached by a blue-yellow Kempe chain beginning at R4 and interchange the colors blue and green for all regions in M that can be reached by a blue-green Kempe chain beginning at R2. Once these two color interchanges have been performed, each of the five neighboring regions of R is colored red, yellow, or green. Then R can be colored blue and a 4-coloring of the map M has been obtained, completing the proof. Heawood counterexample to Kempe’s proof.

Blue-yellow-chain and blue-green-chain Kempe’s proof is unsuccessful when it is applied to the Heawood map, because there are neighboring regions with the same color. So, Heawood had discovered a counterexample to Kempe’s technique, not to the Four Neither red-yellow-chain nor red- Color Conjecture green-chain are present. itself. •Nevertheless,Heawood was able to use Kempe’s technique to prove that every map could be colored with five or fewer colors. •The Four Color Problem can be stated strictly in terms of planar graphs, rather then in terms of maps. Definition: Let G be a graph. Then G is k-region-colorable if each region of G can be assigned one of k given colors so that neighboring (adjacent) regions are colored differently. The Four Color Conjecture: Every plane graph is 4-region-colorable.

There is another more popular statement of the Four Color Conjecture which involves the coloring of vertices. Definition: Let G be a graph. The planar dual G* of G can be constructed by first placing a vertex in each region of G. This set of vertices is V(G*). Two distinct vertices are then joined by an edge for each edge on the boundaries of the regions corresponding to these vertices of G*. Furthemore, a loop is added at a vertex of G* for each bridge of G on the boundary of the corresponding region. Each edge of G* is drawn so that it crosses its associated edge of G, but crosses no other edge of G or G*. Thus, G* is planar. Since G* may contain parallel edges and possibly loops, G* is a multigraph.

The dual G* has the properties that its order is the same as the number of regions of G, and the number of regions of G* is the order of G. Both G and G* have the same size. Definition: If each set of parallel edges in G* is replaced by a single edge and all loops are deleted, a graph G’ results, called the dual graph of G.

Observation: - If G is a connected plane graph, then (G*)*= G; - Every connected plane graph is the dual graph of some connected plane graph; - A plane graph G is k-region-colorable for some positive integer k if and only if its dual graph G’ is k-colorable. Hence, the Four Color Conjecture can now be refrased: Every planar graph is 4-colorable. Thus, what Heawood proved with the aid of Kempe’s proof technique is the following: THE FIVE COLOR THEOREM: every planar graph is 5-colorable. In 1976 Appel and Haken announced that they had been successful in providing a computer-aided proof of the . At the end, in 2000, Ashay Dharwadker (an indian mathematician), introduced a new and elegant proof of the Problem, that uses the Group’s Theory, without computers.

Steps of the proof (Ashay Dharwadker ): • First he proves the six colour theorem using Euler’s formula, showing that any map on the plane may be properly coloured by using at most six colours. • The whole proof works with the fixed number N (the minimal number of colours required to properly colour any map), the fixed map m(N) (which requires no fewer than N colours to be properly coloured), and the fixed proper colouring of the regions of the map m(N). • Then, he defines Steiner Systems and proves Tits’ inequality and its consequence that if a Steiner system S(N +1, 2N, 6N) exists, then N cannot exceed 4. In this way the Four Color Theorem is proved. Definition: A Steiner system S(t , k, v) is a set P of points together with a set B of blocks such that: i)There are v points; ii)Each block consists of k points; iii)Every set of t points is contained in a unique block. Lemma (J. Tits) “If there exists a nontrivial Steiner system S(t , k, v) then v ≥ (t +1)(k−t +1).” For more details you can see the book cited in the bibliography. The conjectures of Hajòs and Hadwiger. •We know that χ(G) ≥ ω(G) for every graph G. (where, χ(G) is the chromatic number, i.e. the minimum k so that G is k-colorable; and ω(G) is the number of vertices of a of maximum size, complete graph of G). Indeed: Theorem: for every integer k≥3, there is a graph G such that χ(G)=k and ω(G)=2. Observation:

KK is present in a k-chromatic graph for k=1,2. This is not true for k=3. Indeed, every odd cycle of order at least 5 is 3-chromatic but none of these graphs contains K3 as a subgraph. All of these contain a subdivision of K3, however. Definition: a graph H is a subdivision of a graph G if either H=G or H is obtained from G by inserting vertices of degree 2 into the edges of G. In 1952, Dirac showed that: Theorem: “if G is a graph with χ(G)≥4, then G contains a subdivision of K4.”

Subdivisions of the graph G. •Consequently, for 2 ≤ k ≤ 4, every k-chromatic graph contains a subdivision of Kk. In 1961 Hajòs conjectured that this is true for every integer k≥2. Hajòs Conjecture: if G is a k-chromatic graph, where k≥2, then contains a subdivision of Kk. Definition: a graph G is perfect if χ(H)=ω(H) for every induced subgraph H of G. Definition: for a graph G and a vertex v of G, the replication graph Rv(G) of G is that graph obtained from G by adding a new vertex v’ to G and joining v’ to every vertex in the closed neighborhood N[v] of v.

G: RV(G):

v v v’ Theorem: if G is perfect, then Rv(G) is perfect for every vϵV(G). Carsten Thomassen showed that there is a connection between perfect graphs and Hajòs’ conjecture: « a graph G is perfect if and only if every replication graph of G satisfies Hajòs’ Conjecture.» Definition: a graph H is a minor of a graph G if H can be obtained from G by a sequence of contractions, edge deletions, and vertex deletions (in any order).

A minor from G Theorem: if a graph G contains a subdivision of a graph H, then H is a minor of G. In particular, if a k-chromatic graph contains a subdivision of Kk, then Kk is a minor of G. Years before Hajòs’ conjecture, on December 15, 1942 Hugo Hadwiger made the following conjecture during a lecture he gave at the University of Zurich in Switzerland: Hadwiger’s Conjecture: Every k-chromatic graph contains Kk as a minor. ▪In 1937 Klaus Wagner had proved that : every planar graph is 4- colorable if and only if every 5-chromatic graph contains K5 as a minor. i.e. he showed the equivalence between the Four Color Conjecture and Hadwiger’s Conjecture for k=5 six years before Hadwiger stated his conjecture.

▪ In 1943 paper, Hadwiger mentioned that his conjecture for k=5 implies the Four Color Conjecture. So, Hadwiger’s Conjecture was considered as a generalization of the Four Color Conjecture. It was verified Hadwiger’s Conjecture for k≤6, but it is open for every integer k˃6. Definition: the had(G) of a graph G has been defined as the greatest positive integer k for which Kk is a minor of G. In this context, Hadwiger’s Conjecture can be stated as: «for every graph G, χ(G)≤had(G)». A PROOF OF THE GENERAL CASE OF HADWIGER’S CONJECTURE WOULD GIVE, AS A COROLLARY, A NEW PROOF OF THE FOUR COLOR THEOREM.

MORE RESULTS ABOUT COLORING •We have proved that χ(G) ≥ ω(G). There are graphs having arbitrarily large chromatic number, even though they do not contain K3. •Many constructions of such graphs are known, even though aren’t trivial. We give one here. MYCIELSKI’S CONSTRUCTION Mycielski (1955) found a construction that builds from any k-chromatic -free graph G a (k + 1)-chromatic triangle-free supergraph G’. Given G with vertex set V = {v1,…,vn} add vertices U = {u1,…,un} and one more vertex w. Beginning with G’[V] = G, add edges to make ui adjacent to all of NG(vi ), and then make N(w) = U. •An independent set is an empty induced subgraph in G. •The vertices having a given colour in a proper colouring must form an independent set, so χ(G) is the minimum number of independent sets needed to cover V(G). •Note that U is an independent set in G’. From the 2-chromatic graph K2, one iteration of Mycielski’s construction yields the 3-chromatic C5, and a second iteration yields the 4-chromatic Grӧtzch graph. These graphs are the triangle-free k-chromatic graphs with fewest vertices for k = 2,3,4. We see the following figures:

Theorem: “Mycielski’s construction produces a (k + 1)-chromatic triangle-free graph from a k-chromatic triangle-free graph.” Proof: Suppose G is triangle-free, χ(G) = k, V(G) = {v1,…,vn} and V(G’) = {ui}iϵ[n] U {vi}iϵ[n] U {w} as described above. By construction {ui}iϵ[n] is independent in G’. Hence the other vertices of any triangle containing ui belong to V(G) and are neighbors of vi. This would complete a triangle in G, which can’t exist. We conclude that G’ is triangle-free. A proper k-colouring f of G extends to a proper (k + 1)-colouring of G’ by setting f(ui) = f(vi) and f(w)=k+1; hence χ(G’) ≤ χ(G) +1. We prove equality by showing that χ(G) < χ(G’) . To prove this, we consider any proper colouring of G’ and obtain from it a proper coloring of G using fewer colors. Let g be a proper k-coloring of G’ by changing the names of colors, we may assume that g(w)=k. This restricts g to {1,…,k-1} on {ui}. On V(G) it may use all k colors. Let A= {vi : g(vi) = k}. We change the colors used on A to obtain a proper (k-1)-colouring of G. For each viϵA, we change the colour of vi to g(ui). Because all vertices of A have color k under g, no two edges of A are adjacent. Thus we need only check edges of the form (vi, v’) with v’ϵ V(G) - A. If (vi, v’) ϵ E(G), then we have constructed G’ so that (ui, v’) ϵ E(G’) , which implies g(ui) ≠ g(v’). Since we change the color on vi to g(ui), our change does not violate the edge viv’. We have shown that the modified coloring of V(G) is a proper (k-1)-coloring of G. We ignore possible conflicts between vi and uj , because we now delete {ui}iϵ[n] U {w} and have a proper (k - 1)-colouring of G.

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INDIPENDENT SET •Independent set: A set of vertices in a graph G is independent if no two of them are adjacent. The largest number of vertices in such a set is called the vertex independence number of G and is denoted by α0(G) or α0. Analogously, an independent set of edges of G has no two of its edges adjacent and the maximum cardinality of such a set is the edge independence number α1(G) or α1. For the complete graph Kn, α0= 1, α1 = n/2

Some observation about edge-coloring: (i) if v is a vertex in L(G) and u and w are the endpoints of the edge in G that corresponds to v, then dL(G)(v) = dG(u) + dG(w) - 2. Thus, the maximum degree Δ(L(G)) satisfies Δ(L(G)) ≤ 2Δ(G) – 2; (ii) An edge-colouring of a graph G is the same as a vertex-colouring of its line graph L(G). (iii) A graph G with maximum degree Δ(G) has χ’(G) ≥ Δ(G) since the edges incident to a vertex of degree Δ(G) must have different colours. (iv) Hence if G has maximum degree Δ(G) then L(G) has maximum degree at most 2Δ(G) - 2. So it follows from (i): χ’(G) = χ(L(G)) ≤ Δ(L(G)) + 1 ≤ 2Δ(G) - 2 + 1 = 2Δ(G) – 1. Theorem (Kӧnig, 1916): «if G is a bipartite multigraph, then χ’(G) = Δ(G).» Proof. An edge colouring is actually a partition of the edge set of a graph into matchings. We’ll use the theory of matchings. We have proved that every regular bipartite graph H has a perfect matching. This also holds for multigraphs, with the same proof. By induction on Δ(H), this yields a proper Δ(H)-edge-colouring. It therefore suffices to show that every bipartite multigraph G with maximum degree k has a k-regular bipartite supergraph H≥ G. We construct such a supergraph. If G have not the same number of vertices in each partite set, add vertices to the smaller set to equalize the sizes. If the resulting G’ is not regular, then each partite set has a vertex with degree less than Δ(G’) = Δ(G). Add an edge consisting of this pair. Continue adding such edges until the graph becomes regular. ▪

Definition: Let i be a colour used in the edge colouring of a graph G. If there is an edge coloured i incident at the vertex v of G, we say i is present at v, and if there is no edge coloured i at v, we say i is absent from v. The next theorem gives us the edge chromatic number of complete graphs. Theorem: if G=Kn is a complete graph, with n vertices, (n≥2), then: χ’(Kn )= n-1 if n is even; χ’(Kn )= n if n is odd. Proof. Let G = Kn be a complete graph with n vertices. Assume n is odd. Draw G so that its vertices form a regular polygon. Clearly, there are n edges of equal length on the boundary of the polygon. Colour the edges along the boundary using a different colour for each edge.

Now, each of the remaining internal edges of G is parallel to exactly one edge on the boundary. Each such edge is coloured with the same colour as the boundary edge.

So two edges have the same colour if they are parallel and therefore we have the edge colouring of G. Since it uses n colours, we have shown that χ’(G) ≤ n. Let G have an (n-1)-colouring. From the definition of an edge colouring, the edges of one particular colour form a matching in G, set of independent edges. Since n is odd, therefore the maximum possible number of these is (n-1)/ 2 .

This implies that there are at most [(n-1)] (n-1)/ 2 edges in G. This is a contradiction, as Kn has n(n-1)/ 2 edges. Thus G have not an (n-1)-colouring. Hence, χ’(G) = n. • Now, let n be even, and let v be any vertex of G. Clearly G-v is complete with n-1 vertices. Since n-1 is odd, G-v has an (n-1)-colouring. With this colouring, there is a colour absent from each vertex and different vertices having different absentees. Reform G from G-v by joining each vertex w of G-v to v by an edge and colour each such edge by the colour absent from w. This gives an (n-1)-colouring of G and therefore χ(‘G) = n-1. ▪ •Observation: Since in a complete graph Δ(G)= n-1, the previous theorem shows that χ’(Kn) is Δ(G) or Δ(G)+1.

Theorem (Vizing, 1964): Let G be a simple graph with maximum degree Δ(G). Then Δ(G) ≤ χ’(G) ≤ Δ(G) + 1. •Observation: this theorem divides the finite graphs into 2 classes, according to their chromatic index: graphs that satisfy χ’(G) = Δ(G) are called class 1, and graphs that satisfy χ’(G) = Δ(G)+1 are called class 2. Proof: Let G be a simple graph. We always have Δ(G) ≤ χ’(G). To prove that χ’(G)≤Δ+1 we use induction on |E|.

•Let Δ(G) = k. If G has only one edge, then k = 1 = χ’(G). •Therefore assume that G has more than one edge and that the result is true for all graphs having fewer edges than G. Let e =v1v2 be an edge of G. Then by induction hypothesis the subgraph G-e has (k + 1)-edge colouring and let the colours used be 1,2,…,k+1. Since d(v1)≤ k and d(v2)≤k, out of these k + 1 colours at least one colour is absent from v1 and at least one colour is absent from v2. If there is a common colour absent from both v1 and v2, then we use this to colour e and get a (k + 1)-colouring of G. Therefore in this case, χ’(G)≤k+1. We now assume that there is a colour, say 1, absent from v1 but present at v2 and there is a colour, say 2, absent from v2 but present at v1. We start from v1 and v2 and construct a sequence of distinct vertices v1, …, vj, where each vi for i ≥2 is adjacent to v1. Let v1v3 be coloured 2. This v3 exists, because 2 is present at v1. We observe that not all the k +1 colours are present at v3 and assume that the colour 3 is absent from v3. But the colour 3 is present at v1 and choose the vertex v4 so that v1v4 is coloured 3. Continuing in this way, we choose a new colour i absent from vi but present at v1, so that v1vi+1 is the edge coloured i. In this way, we get a sequence of vertices v1,…,vj-1,vj such that: (a) vi is adjacent to v1 for each i > 1, (b) the colour i is absent from each i = 1,2,…,j-1 and (c) the edge v1vi+1 is coloured i for each i = 1,2,…,j-1.

As d(v1)≤ k, (a) implies that such a sequence has at most k + 1 terms, that is, j≤k+1. Assume that v1,…,vj is a longest such sequence, that is, the sequence for which it is not possible to find a new colour j, absent from vj , together with a new neighbour vj+1 of v1 such that v1vj+1 is coloured j. We first assume that for some colour j absent from vj there is no edge of that colour present at v1. We colour the edge e = v1v2 by colour 2 and then recolour the edges v1vi by colour i, for i = 3,…,j-1. Since i was absent from vi , for each i = 2,…,j-1, this gives a (k + 1)-colouring of the subgraph G-v1vj . Now as the colour j is absent from both vj and v1, recolour v1vj by the colour j. This gives a (k + 1)-colouring of G. See the following figure: Now assume that whenever j is absent from vj , j is present at v1. If vj+1 is a new neighbour of v1 so that v1vj+1 is coloured by j, then we have extended our sequence to v1,…,vj+1, which is a contradiction to the assumption that v1,…,vj is the longest sequence. Thus one of the edges v1v3 ,…, v1vj has coloured by j, say v1vl , with 3 ≤ l ≤ j-1. Now colour e = v1v2 by 2, and for i = 3,…,l-1, recolour each of the edges v1vi by i while unaltering the colours of the edges v1vi , for i =l+1,…,j. Removing the colour j from v1vl , we have a (k + 1)-colouring of the edge deleted subgraph G-v1vl . Let H(1, j) represent the subgraph of G induced by the edges coloured 1 or j in this partial colouring of G. Since the degree of every vertex in H(1, j) is either 1 or 2, each component of H(1, j) is either a path or a cycle. As 1 is absent from v1 and j is absent from both vj and vl , it follows that all these three vertices do not belong to the same connected component of H(1, j). Therefore, if K and L represent the corresponding Kempe chains containing vj and vl respectively, then either v1ɇK or v1ɇL. ▪Let v1ɇL. Then interchanging the colours of L, the Kempe chain argument gives a (k + 1)-colouring of G-v1vl in which 1 is missing from both v1 and vl. Colouring v1vl by 1 gives a (k + 1)-colouring of G. ▪Now, let v1ɇK. Colour the edge v1vl by l, recolour the edges v1vi by i, for i = l,…,j-1 and remove the colour j-1 from v1vj . Then, from the definition of the sequence v1,…,vj, we get a (k+1)-colouring of G-v1vj without affecting two coloured subgraph H(1, j). Using Kempe chain argument to interchange the colours of K, we obtain a (k + 1)-colouring of G-v1vj in which 1 is absent from both v1 and vj. Therefore, again colouring v1vj by 1 gives (k + 1)-colouring of G. ▪ Bibliography: - , Reinhard Diestel, Springer. - CHROMATIC GRAPH THEORY, Gary Chartrand, Ping Zhang- crc press 2009 - GRAPHS, COLOURINGS AND THE FOUR-COLOUR THEOREM, Robert A. Wilson, Oxford University Press 2002. - GRAPH THEORY, Ashay Dharwadker, Shariefuddin Pirzada,2011. - INTRODUCTION TO GRAPH THEORY, Douglas B. West, 2002.