Graph_Theory_Notes9.pdf

# If u 2 and u 4 are in the same component of g 24 then

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If u 2 and u 4 are in the same component of G 24 , then they are joined by a path Q in G 24 . Since G is a plane graph and u 2 int ( C ) and u 4 ext ( C ), Q has to pass through at least one vertex w on C other than v . So f ( w ) = 1 or 3 as P is in G 13 , but on the other hand we have f ( w ) = 2 or 4 as w G 24 . This contradiction shows that u 2 and u 4 are in different components of G 24 . Proof (continued). Similar to subcase 2.1, by swapping colours in the com- ponent of G 24 containing u 2 , we obtain a 5-colouring of G 0 under which both u 2 and u 4 are coloured 4. We can then assign 2 to v to obtain a 5-colouring of G . In each case we have proved that G admits a 5-colouring. By mathemat- ical induction, we have proved that every planar graph is 5-colourable. 6 On the proof of the 4-Colour Theorem (op- tional) Proof of the 4-Colour Theorem (optional) The proof of the 4-Colour Theorem by Robertson et al. has two parts: Unavoidability: Show that every planar graph has one of 633 ‘unavoidable configurations’ (20 minutes on a computer in 1996). Reducibility: Show that each of the unavoidable configurations is reducible; that is, if a planar graph contains a given unavoidable configuration then it can be 4-coloured by piecing together 4-colourings of certain smaller planar graphs (3 hours on a computer in 1996). A 4 -coloured planar triangulation How to prove 4CT? ... by contradiction If not every planar graph is 4-colourable, then choose a “smallest” counterexample 1. minimum # vertices 2. maximum # edges 3. etc. Such a planar graph is a plane triangulation (every face is a triangle)

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Unavoidable Configuration #1: A vertex v of degree 3 in a planar graph G . Proof of Reducibility. Then G - v is planar, and is 4-colourable by induction. There are at most 3 colours on the neighbours of v . Choose a colour for v that is not assigned to a neighbour of v . So G is 4-colourable. Unavoidable Configuration #2: A vertex v of degree 4 in a planar graph G . Proof of Reducibility. v has two nonadjacent neighbours x and y (otherwise K 5 is a subgraph of G ). Let G 0 be the graph obtained from G by deleting v and identifying x and y . So G 0 is planar, and by induction, G 0 is 4-colourable. Since xy is not an edge of G , a 4-colouring of G 0 determines a 4-colouring of G - v such that x and y receive the same colour. Thus at most three distinct colours are assigned to the neighbours of v . Choose a colour for v that is not assigned to a neighbour of v . So G is 4-colourable. Unavoidable Configuration #3: A cut-vertex v . Proof of Reducibility. There are planar graphs G 1 and G 2 with v as the unique common vertex such that G = G 1 G 2 . So G 1 and G 2 are 4-colourable by induction. The colours on G 2 can be permuted so that v has the same colour in G 1 and in G 2 . So G is 4-colourable. This reduction works for any graph. We can now assume that the given planar graph is 2-connected. Similar reductions show that the given planar graph is 5-connected. 7 Hadwiger’s conjecture Minor
Figure 1: A K 4 -minor. Each subgraph being contracted is called a branch set.

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