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Math 6700 Advanced Graph Theory Homework 3

# Math 6700 Advanced Graph Theory Homework 3 - Math 6700...

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Math 6700 Advanced Graph Theory Homework 3 HW 1.7 Let G be a 2-connected graph that is not a triangle, and let e be an edge of G . Show that either G e - or / G e is again 2-connected. Proof: If G e - is 2-connected, then there is nothing to show. Suppose G e - is not 2-connected, then we can prove / G e is 2-connected. Suppose e uv = , where , ( ) u v V G . Because G e - is not 2-connected, there is only one path p between u and v in G e - . So there are two paths, e and p , between u and v in G . Suppose 1 v and 2 v are any two vertices in G . Because G is 2-connected, there must be at least two disjoint paths, 1 p and 2 p , between 1 v and 2 v . Following are discussion in different cases: (a) e is edge-disjoint and vertex-disjoint with both 1 p and 2 p . In this case, 1 p and 2 p are still in / G e and still disjoint with each other . (b) e is edge-disjoint but not vertex-disjoint with 1 p and 2 p . Then there are two sub cases: (1) Only one of u and v is on 1 p and 2 p . Then 1 p and 2 p are still in / G e and still disjoint with each other . (2) One of u and v is on 1 p , and the other is on 2 p . This is impossible because then there will be three paths between u and v in G . (c) e is on one of 1 p and 2 p . Then 1 p and 2 p are still in / G e and still disjoint with each other . Because there are at least two disjoint paths between any two vertices 1 v and 2 v in / G e , / G e must be 2-connected. So if G is a 2-connected graph that is not a triangle, and e is an edge of G , then either G e - or / G e is again 2-connected.

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