{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

07Ma6aHw7Sol

# 07Ma6aHw7Sol - Math 6a Solution Set 7 Problem 1 Liu 5.15...

This preview shows pages 1–2. Sign up to view the full content.

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 tree of G.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}