By Gasper Fijavz

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**Example text**

6 (Four color theorem) Every planar graph G satisfies χ(G) ≤ 4. Four color theorem is the optimal result, K4 is a planar graph and χ(K4 ) = 4. It is however possible to give short proofs of weaker results. It is easy to see that χ(G) ≤ 6 for every planar graph G. Namely, every planar graph is 5degenerate. This implies that we can construct an ordering v1 , v2 , . . , vn of vertices of G so that vi has degree ≤ 5 in G[vi , vi+1 , . . 2. By refreshing the set of vertices of degree ≤ 5 in the remaining graph we can construct the sequence in linear time.

V5 , again c can be extended to v. Hence we may (without loss of generality) assume that c (vi ) = i. Observe the component C1,3 of G induced by vertices of colors 1 and 3 which contains v1 . We can recolor vertices of C1,3 by changing colors 1 and 3 within C1,3 , this procedure is called a Kempe change. If we are lucky and v3 ∈ C1,3 then the change results in exactly four colors 2, 3, 4, 5 being used on neighbors of v. Otherwise there is a v1 − v3 path consisting entirely of vertices colored with colors 1 and 3 in G − v, this is a Kempe chain.

Assume this is not the case. Let (G1 , G2 ) be a corresponding separation — a pair of graphs so that G1 ∪G2 = G and G1 ∩G2 = S. Let H1 = G1 − S + x + y, and simetrically H2 = G2 − S + x + y. Let P1 be a shortest x − y-path in H1 and P2 be a shortest x − y-path in P2 . As P1 is a shortest path it is induced, and a similar observation holds for P2 as well. Now P1 ∪ P2 is a cycle of length ≥ 4, and no internal vertex of P1 can be adjacent to an internal vertex of P2 , as S is a separator. The edge xy is then the only possible chord of P1 ∪ P2 .