07Ma6aHw7Sol - Math 6a - Solution Set 7 Problem # 1 Liu...

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Math 6a - Solution Set 7 Problem # 1 Liu 5.15 Let T 1 and T 2 be two spanning trees of a connected graph G. Let a be an edge that is in T 1 but not T 2 . Prove that there is an edge b in T 2 but not T 1 such that both ( T 1 - { a } ) ∪ { b } and ( T 2 - { a } ) ∪ { b } are spanning trees of G. Solution. Note that if the edge a = { x,y } , since spanning trees are minimal connected graphs, it follows that the removal of a from T 1 splits the vertices into two subsets A and B (without loss of generality, x A and y B ) such that A and B are connected, and a is the only edge between A and B. Also, note that T 2 being a spanning tree and a not in T 2 implies that if we take T 2 ∪{ a } we obtain a graph with a circuit. Clearly that circuit contains a and thus it contains the two points x and y. So, since it contains the edge a it must contain at least one more edge with one endpoint in A and one in B, since if it does not, without loss of generality all of its other edges have both endpoints in A which is a contradiction since there must be another edge with endpoint y which is in B. Thus, let b = { x 0 ,y 0 } be an edge in T 2 that is in that circuit and has endpoint x 0 A and the other endpoint y 0 B. Note that b cannot be in T 1 , since if it were, we would have a circuit in T 1 , simply by taking the path from x to x 0 in A, the edge b from x 0 to y 0 , then the path y 0 to y in B, followed by the edge a from y back to x. Note that ( T 1 -{ a } ) ∪{ b } and ( T 2 -{ b } ) ∪{ a } both have v - 1 - 1+1 = v - 1 edges (where v is the number of vertices in G ), so if we show both are connected, it will follow that both are spanning trees of G (since if not, we could remove edges from one to get a spanning tree with less than v - 1 edges, a contradiction). But note that in T 1 , A and B are connected, so in ( T 1 - { a } ) ∪ { b } they are both connected, and there exists a path from any z 1 in A to x 0 to y 0 (by b ) to any z 2 in B. Thus ( T 1 - { a } ) ∪ { b } is connected, and is a spanning tree of G. Also, since T 2 ∪ { a } contains a circuit containing the edge b, the removal of b from that circuit will not disconnect it and thus ( T 2 -{ b } ) ∪{ a } is connected and thus also a spanning
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This note was uploaded on 04/28/2010 for the course MATH 6A taught by Professor Wilson during the Fall '07 term at Caltech.

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07Ma6aHw7Sol - Math 6a - Solution Set 7 Problem # 1 Liu...

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