By Bernhard Korte

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**Extra info for Combinatorics, Graphs, Matroids [Lecture notes]**

**Sample text**

K − 1} can be coloured with i colours, the graph Gk can be coloured with k-colours: for the colouring of all subgraphs G1 , . . , Gk−1 we need in total k − 1 colours and the vertices in Ak (which is a stable set) can get the same colour. On the other hand, Gk is not (k − 1)-colourable if none of the Gi (i = 1, . . , k − 1) is (i − 1)colourable. To prove this, assume that there was a k − 1-colouring of Gk . Then choose a vertex v1 in G1 with colour c1 . There must be a vertex v2 in G2 with colour c2 = c1 because G2 cannot be coloured with just one colour.

For n ∈ N \ {0}, we get: √ √ √ √ √ √ ( 2 + 3)2n = ( 2 + 3)2n−2 ( 2 + 3)2 √ √ = (an−1 + bn−1 6)(5 + 2 6)2 √ = (5an−1 + 12bn−1 ) + (2an−1 + 5bn−1 ) 6 This proves the claim. The proof also yields recursion formulas for an and bn for n ≥ 1: an = 5an−1 + 12bn−1 bn = 2an−1 + 5bn−1 Moreover, we have a0 = 1 and b0 = 0 We can solve this recursion by using the generating function A(z) = n n≥0 bn z . This gives us A(z) = a0 z 0 + an z n = a0 + n≥1 n≥0 an z n and B(z) = (5an−1 + 12bn−1 )z n = 1 + 5zA(z) + 12zB(z) n≥1 30 and B(z) = b0 z 0 + bn z n = n≥1 (2an−1 + 5bn−1 )z n = 2zA(z) + 5zB(z).

We call an edge e ∈ E(G) a bridge if G − e contains more connected components than G. Theorem 46 Let G be a 3-regular planar graph without bridges. Then χ (G) = 3. Proof: Since χ (G) ≥ 3 is trivial, we only have to show χ (G) ≤ 3. g. we can assume that G is connected (otherwise consider the connected components of G). 41 Consider a fixed planar embedding of G. The Four Colour Theorem implies that we can colour the faces of the embedding with four colours such that neighbouring faces get different colours.