Let $e$ be an edge of minimum weight in the connected weighted graph $G$. Every minimum spanning tree of $G$...
$begingroup$
Let $e$ be an edge of minimum weight in the connected weighted graph
$G$. Every minimum spanning tree of $G$ contains $e$.
I have been told that this is not true. But I also know this property :
Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?
Is it not true because that $e$ is not necessary the only minimum weight edge?
graph-theory
$endgroup$
add a comment |
$begingroup$
Let $e$ be an edge of minimum weight in the connected weighted graph
$G$. Every minimum spanning tree of $G$ contains $e$.
I have been told that this is not true. But I also know this property :
Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?
Is it not true because that $e$ is not necessary the only minimum weight edge?
graph-theory
$endgroup$
add a comment |
$begingroup$
Let $e$ be an edge of minimum weight in the connected weighted graph
$G$. Every minimum spanning tree of $G$ contains $e$.
I have been told that this is not true. But I also know this property :
Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?
Is it not true because that $e$ is not necessary the only minimum weight edge?
graph-theory
$endgroup$
Let $e$ be an edge of minimum weight in the connected weighted graph
$G$. Every minimum spanning tree of $G$ contains $e$.
I have been told that this is not true. But I also know this property :
Must a minimum weight spanning tree for a graph contain the least weight edge of every vertex of the graph?
Is it not true because that $e$ is not necessary the only minimum weight edge?
graph-theory
graph-theory
asked Jan 21 at 17:51
Marine GalantinMarine Galantin
860316
860316
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1 Answer
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$begingroup$
If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.
Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.
$endgroup$
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.
Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.
$endgroup$
add a comment |
$begingroup$
If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.
Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.
$endgroup$
add a comment |
$begingroup$
If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.
Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.
$endgroup$
If $G$ contains a cycle of edges each of the same minimum weight, then it is not possible for $G$ to have a spanning tree without omitting at least one of those edges.
Otherwise the answer is yes: Let $T$ be a spanning tree of $G$ NOT containing $e$. Then $T+e$ contains one cycle. By assumption, there is an edge in the cycle of non-minimum weight. Remove that edge, and we have a new spanning tree $T'$ of lower weight.
answered Jan 21 at 17:59
PMarPMar
461
461
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