the odd cycles. Theorem If G is a k-critical graph, then every vertex in G has degree at least k 1.

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1 Colour critical graphs Colour critical graphs A graph G is (colour) critical if χ (H) <χ(g) for every proper subgraph H of G. G is said to be k-ciritical if G is critical and χ (G) =k. Small cases:1-critical graph. 2-critical graph is an edge. 3-critical graphs are the odd cycles. Theorem If G is a k-critical graph, then every vertex in G has degree at least k 1. Proof: Assume that G has a vertex v with degree at most k 2. G v can be coloured with k 1 colours. At least one of these colours is not used on the neighbors of v. Therefore there is at least one colour available for v. Then G can be coloured with k 1 colours. This is a contradiction. Theorem If G is a k-critical graph and X, Y is a partition of V (G), then there are at least k 1 edges between X and Y. Proof: Assume that there are at most k 2 edges between X and Y. We will show that this will lead to a contradiction. Our proof uses Lemma of the matching theory. Since G is k-critical, both X and Y can be colored with k 1 colours. Let the colour classes be X 1, X 2,...,X k 1 and Y 1, Y 2,...,Y k 1 respectively. We construct a bipartite graph H by letting V (H) ={X 1,..., X k 1,Y 1,..., Y k 1 } and X i Y j is an edge in H if and only if there is no edge joining a vertex in X i with a vertex in Y j in G. H hasatleast(k 1) 2 (k 2) = k 2 3k +3edges. Since every vertex has degree at most k 1 therefore covering at most k 2 edges, k 2 vertices can only cover at most (k 2) (k 1) = k 2 3k +2 edges. Therefore at least k 1 vertices are needed to cover all the edges. By Lemma 5.1.7, H has a matching of k 1 edges which is a complete matching of H. If X i and Y j are matched, there is no edge between vertices in X i and vertices in Y j. X i Y j is an independent set in G. Therefore V (G) can be partitioned into k 1 independent sets. That means that G can be coloured with k 1 colours. This contradiction shows that there must be at least k 1 edges between X and Y Edge density of 4-critical graphs and Constructions of ciritical graphs Join of two ciritical graphs. Theorem Let G be a k-critical graph and H a l-critical graph. The join of G and H, G H is a (k + l)-critical graph. Proof: We have seen that χ (G H) =k + l. We need to show that χ (G H e) k + l 1 for every edge e in G H. Case 1: e is an edge in G. We can colour H with colours 1, 2,..., l and G e with k 1 colours: l +1,l+2,..., l + k 1.

2 78 Graph coloring Case 2: e is an edge in H. We can colour G with colours 1, 2,..., k and H e with l 1 colours: k +1,k+2,..., k + l 1. Case 3: e is an edge joining u V (G) and v u (G). Since G u can be coloured with k 1 colours. G canbecolouredwithcolours1, 2,..., k such that u is the only vertex being coloured with colour 1. Similarly, H canbecolouredwithl colours 1, k +1,k+2,..., k + l 1 such that v is the only vertex being coloured with colour 1. This is a k + l 1 colouring of G H e. The 4-critical graphs that we can construct using joins of two critical graphs: K 2 K 2 = K 4. K 1 C 2l+1 = W 2l+1. In an odd wheel, there are 2l +2vertices and 4l +2edges. E =2 V 2. Conjecture (Gallai, 1964) If G is a planar 4-critical graph, then E (G) 2 V (G) 2. Gallai s conjecture implies an older conjecture of Dirac: Conjecture (Dirac, 1957) If G is a planar 4-critical graph, then G has a vertexwithdegreethree. However, both conjectures have been disproved by a graph discovered by Koester in 1985: The Koester s graph has 40 vertices and 80 edges. Every vertex has degree four. Grunbaum used Hajos s construction to show that there are planar 4-critical graphs with edge densities (that is the number of edges divided by the number of vertices) arbitrarily close to 79/39 = Grunbaum asked the question of determing the maximum edge density of planar 4-critical graphs. G. Koester: Note to a problem of T. Gallai and G.A. Dirac, Combinatorica, 5 (1985), Grunbaum: The edge density of 4-critical planar graphs, Combinatorica, 8 (1988),

3 Colour critical graphs 79 The Hajos s construction Let G and H be two graphs. Let uu be an edge in G and vv be an edge in H. The graph G H is obtained by identifying u and v, deleting the edges uu and vv and adding an edge u v. Theorem If both G and H are k-critical graphs (k 3), then G H is a k-critical graph. Proof: First we show that χ (G H) =k. (i) G H cannot be coloured with k 1 colours. G uu can be coloured with k 1 colours. In such a colouring, u and u must be coloured with the same colour (otherwise this would be a (k 1)-colouring of G as well). If G H were coloured with k 1 colours, then u has the same colour as u = v which has the same colour as v. Since there is an edge joining u and v, this impossible. (ii) G H can be coloured with k colours. We can colour G with k colours such that u is coloured with colour 1 and u is coloured with colour 2. Also we can colour H with k colours such that v is coloured with colour 1 and v is coloured with colour 3. This is a k-colouring of G H. Then we show that χ (G H e) can be coloured with k 1 colours for every edge e in G H. Case 1: e is in G. G e can be coloured with k 1 colours such that u has colour 1 and u has colour 2. H vv can be coloured k 1 colours such that v and v both are coloured with colour 1. Combining these, we have a (k 1)-colouring of G H e. Case 2: e is in H. Similar to Case 1. Case 3: e = u v. G uu can be coloured with k 1 colours such that both u and u are coloured with colour 1. H vv canbecolouredwithk 1 colours such that both vv are coloured with colour 1. Combining these, we have a (k 1)-colouring of G u v. Let G 1 be the Koester s graph and G i+1 = G i G 1. We have E (G i+1 ) = E (G i ) +79and V (G i+1 ) = V (G i ) +39. Therefore E (G n ) =79(n 1) + 80 and V (G n ) =39(n 1) The edge density of G n is 79 (n 1) (n 1) as n. Abbott and Zhou used a variation of the Hajos s construction to show that the edge density of a planar 4-critical graph can be arbitrarily close to 39/19 = H.L. Abbott and B. Zhou, The edge density of 4-critical planar graphs, Combinatorica, 11 (1991)

4 80 Graph coloring The general Hajos s construction Let G and H be two graphs. Suppose that u 1,u 2,...,u k generate a complete subgraph in G and v 1,v 2,...,v k generate a complete subgraph in H. u i u is an edge in G and v i v is an edge in H for some i. The graph G H is obtained by identifying u 1 with v 1, u 2 with v 2,..., u k with v k, deleting the edges u i u and v i v and adding the edge u v. Theorem If G and H both are k-critical graphs, then G H is a k-critical graph. In order for G H to be a planar graph, we can only identify two pairs of vertices at most. Let H 1 be the Koester s graph and H i+1 = H i H 1 (identifying one edge in each graph). We have E (H i+1 ) = E (H i ) +78 and V (H i+1 ) = V (H i ) +38. Therefore E (H n ) =78(n 1) + 80 and V (H n ) =38(n 1) TheedgedensityofH n is 78 (n 1) (n 1) = as n The upper bounds of edge density of planar 4-critical graphs Abbott and Zhou also showed that if G is a planar 4-critical graph then E (G) / V (G) 11/4 =2.75. Proof: Let F i be the number of regions that has i edges in its boundary. Let v, e, r be the number of vertices, edges, and regions in G respectively. Since every edge is in the boundaries of two regions, we have 2e = if i =3F 3 + if i (6.7) i 3 2e 3F 3 = if i 4 F i We have an upper bound for the number of regions that are not a triangle: F i 2e 3F 3 4 Thus r F 3 + 2e 3F 3, 4 r e 2 + F 3 4. (6.8) We need to estimate F 3, the number of triangles. We need the following lemma.

5 Colour critical graphs 81 Lemma It G is a 4-critical graph, then either G is an odd wheel or it does not contain any wheels. Proof : Suppose that G is not an odd wheel. Obviousely G cannotcontainanodd wheel. Suppose that G contains an even wheel with center w and v 1,v 2,..., v 2l on the rim. Since G is 4-critical, G v 1 v 2l can be 3-coloured. Without loss of generality, we assume that w is coloured with colour 3 and v 1 is coloured with colour 1. Then v 2 must be coloured with colour 2. v 3 must be coloured with colour 1. v 4 must be coloured with colour 2...v 2l must be coloured with colour 2. However, this is a 3-colouring of G as well. This contradicts to the assumption that χ (G) =4. If G is an odd wheel, the edge density of G is less than 2. We assume that G is not an odd wheel. Therefore, every vertex in G must be incident with a region that is bounded by at least 4 edges. Since every region bounded by i edges has i vertices in its boundary, we have if i v. Using (6.7), we have 2e = if i =3F 3 + if i 3F 3 + v i 3 That gives us F 3 2e v. 3 Substituting this into (6.8), we have r e 2 + 2e v 12 = 2 3 e 1 12 v. This and the Euler s formula gives us ( 2 v + 3 e 1 ) 12 v e v e 3 2. That gives e v 11 4 =2.75. Grunbaum raised four questions in his paper. We were able to answer three of them. The one question we could not answer was:

6 82 Graph coloring Problem Does every planar 4-critical graph G satisfy δ (G) 4 where δ (G) is the minimum degree of G. This is a modification of Dirac s conjecture. Recall that we already know that for a planar graph G, δ (G) 5. Later, Abbot, Katchalski and Zhou proved that for every planar 4-critical graph G, δ (G) 4. Koester proved that for such graphs, the edge density e < 2.5 therefore δ (G) < 5. Inhisproof,Koesterusedaresult v of Stiebitz (1987): If G is a 4-critical graph, then G contains at most v triangles. Indeed, the condition F 3 vtogether with (6.8) gives us Using the Euler s formula, we have r e 2 + v 4. v + e 2 + v 4 e 2 5v 4 e 2 +2 e < 5 2 v. Stiebitz s result is a special case of another conjecture of Gallai: Conjecture (Gallai) If G is a k-critical graph of n vertices then G contains at most n complete subgraphs of order k 1. Theorem (Stiebitz) If G is a 4-critical graph of n vertices then G contains at most n triangles. Proof : If G is an odd wheel, then the number of triangles is n 1. So we assume that G is not an odd wheel. Let the vertices of G be v 1,v 2,..., v n and the triangles of G be T 1,T 2,..., T t. We define a vector T i = [ T i 1 T i 2 T i n] for each T i where { 1 if T i j = vj T i 0 otherwise. We want to show that on the field of Z 2, the set of vectors {T i : i =1, 2,..., t} is linearly independent. Assume that it is linearly dependent. Then without loss of generality, we may assume that for some positive integer k, T 1 + T 2 + T k = 0. H.L. Abbott, M. Katchalski and B. Zhou: Proof of a conjeture of Dirac concerning 4-critical planar graphs, Discrete Mathematics, 132 (1994) G. Koester: On 4-critical planar graphs with high edge density, Discrete Mathematics, 98 (1991) M. Steibitz: Subgraphs of color critical graphs, Combinatorica, 7 (1987)

7 Colour critical graphs 83 That means that every vertex is contained in even number of triangles T 1,T 2,..., T k ( ). Since G does not contain a wheel, there is an edge that is contained in only one of the triangles. Say that v 1 v 2 is contained only in the triangle T 1 = {v 1,v 2,v 3 }. G v 1 v 2 can be coloured with 3 colours. Suppose that in this 3-colouring, both v 1 and v 2 are coloured with colour 1, v 3 is coloured with colour 3. Then every triangle except T 1 has one vertex of each colour and T 1 has two vertices of colour one and one vertex of colour 3. The sum of the entries in the columns corresponding to the vertices of colour 3 is k and the sum of the entries in the columns corresponding to theverticesofcolour1isk +1. If k is odd, then one of the vertices with colour 3 must be in odd number of the triangles; if k is even then k +1is odd and one of the vertices with colour 1 must be in odd number of the triangles. Either way, we have contradiction to ( ). This proves the set of vectors {T i : j =1, 2,..., t} is linearly independent. Therefore, t n. Adapting Stiebitz s technique, Abbott and Zhou proved Gallai s conjecture completely. We also showed that a 4-critical graph on n-vertices can have at most n 1 complete (k 1)-subgraphs. If G is not an odd wheel, then G can contain at most n 2 such subgraphs. H.L. Abbott and B. Zhou: On a conjecture of Gallai concerning complete subgraphs of k-critical graphs, Discrete Mathematics, 100 (1992)