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A MBST tree is not always a MST. | A MBST tree is not always a MST. | ||
| − | Proof: Suppose we have a spanning tree <math>T</math> which is a MBST of <math>G</math> , and <math>T</math> is also a MST. The bottle neck is edge <math>e</math> with weight <math>w</math>. Then I have another vertex <math>X</math> that will be connected to the graph <math>G</math> to form a new graphs <math>G'</math>. <math>G' = G + X</math>. And the possible edges that connect graph <math>G'</math> and new vertex <math>X</math> has weights <math>w_1, w_2, ... W_x</math>, and each is no greater than the bottle neck weight <math>w</math>. Using any of the edges from <math>w_1, w_2, ... W_x</math>, I get a new tree <math>T'</math>. | + | Proof: Suppose we have a spanning tree <math>T</math> which is a MBST of <math>G</math> , and <math>T</math> is also a MST. The bottle neck is edge <math>e</math> with weight <math>w</math>. Then I have another vertex <math>X</math> that will be connected to the graph <math>G</math> to form a new graphs <math>G'</math>. <math>G' = G + X</math>. And the possible edges that connect graph <math>G'</math> and new vertex <math>X</math> has weights <math>w_1, w_2, ... W_x</math>, and each is no greater than the bottle neck weight <math>w</math>. Using any of the edges from <math>w_1, w_2, ... W_x</math>, I get a new tree <math>T'</math>. So <math>T'</math> is still a spanning tree, and its largest weight remains <math>w</math>, so <math>T'</math> is also an MBST. However, among these edges that I can choose to span the tree, only the one that has the smallest weight will be a MST, by definition. So an MBST for a graph <math>G'</math> is not a minimum spanning tree for <math>G'</math>. End of proof. |
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Revision as of 11:16, 24 July 2017
Computer Engineering(CE)
Question 1: Algorithms
August 2013
Solution 1
A MBST tree is not always a MST.
Proof: Suppose we have a spanning tree $ T $ which is a MBST of $ G $ , and $ T $ is also a MST. The bottle neck is edge $ e $ with weight $ w $. Then I have another vertex $ X $ that will be connected to the graph $ G $ to form a new graphs $ G' $. $ G' = G + X $. And the possible edges that connect graph $ G' $ and new vertex $ X $ has weights $ w_1, w_2, ... W_x $, and each is no greater than the bottle neck weight $ w $. Using any of the edges from $ w_1, w_2, ... W_x $, I get a new tree $ T' $. So $ T' $ is still a spanning tree, and its largest weight remains $ w $, so $ T' $ is also an MBST. However, among these edges that I can choose to span the tree, only the one that has the smallest weight will be a MST, by definition. So an MBST for a graph $ G' $ is not a minimum spanning tree for $ G' $. End of proof.
Solution 2
A MBST tree is not a minimum spanning tree for $ G $, ABST may not contain the lightest edge. A counter example is showing in figure below:
