proving that a graph has only one minimum spanning tree if and only if G has only one maximum spanning tree
$begingroup$
Is this claim true?
I thought about it and it seems true but for proving it i started with one direction by assuming that i have one minimum spanning tree and i want to show that from this i have also one maximum spanning tree.
Can i claim from having one minimum spanning tree the weight of every edge is different and then use this proof Show that there's a unique minimum spanning tree if all edges have different costs ?
graph-theory trees
asked Dec 29 '18 at 13:36
user3133165user3133165
1928
$endgroup$
add a comment |
$begingroup$
Is this claim true?
I thought about it and it seems true but for proving it i started with one direction by assuming that i have one minimum spanning tree and i want to show that from this i have also one maximum spanning tree.
Can i claim from having one minimum spanning tree the weight of every edge is different and then use this proof Show that there's a unique minimum spanning tree if all edges have different costs ?
graph-theory trees
asked Dec 29 '18 at 13:36
user3133165user3133165
1928
$endgroup$
add a comment |
$begingroup$
Is this claim true?
I thought about it and it seems true but for proving it i started with one direction by assuming that i have one minimum spanning tree and i want to show that from this i have also one maximum spanning tree.
Can i claim from having one minimum spanning tree the weight of every edge is different and then use this proof Show that there's a unique minimum spanning tree if all edges have different costs ?
graph-theory trees
asked Dec 29 '18 at 13:36
user3133165user3133165
1928
$endgroup$
Is this claim true?
I thought about it and it seems true but for proving it i started with one direction by assuming that i have one minimum spanning tree and i want to show that from this i have also one maximum spanning tree.
Can i claim from having one minimum spanning tree the weight of every edge is different and then use this proof Show that there's a unique minimum spanning tree if all edges have different costs ?
graph-theory trees
graph-theory trees
asked Dec 29 '18 at 13:36
user3133165user3133165
1928
asked Dec 29 '18 at 13:36
user3133165user3133165
1928
asked Dec 29 '18 at 13:36
user3133165user3133165
1928
asked Dec 29 '18 at 13:36
user3133165user3133165
1928
asked Dec 29 '18 at 13:36
user3133165user3133165
1928
1928
add a comment |
add a comment |
1 Answer
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$begingroup$
It's not true.
Let $G$ be $K_n$ and let $P$ be a Hamilonian path in $G=K_n$. [Say $i$ and $j$ are adjacent in $P$ iff $|i-j|=1$, then two vertices $i$ and $j$ are adjacent in $G setminus E(P)$ iff $|i-j| ge 2$. One can check that $G setminus E(P)$ is connected for $n ge 6$ and has many cycles.]
Give every edge in $P$ a weight of 100 and every edge in $G setminus E(P)$ a weight of 1. Then $G$ has one maximum-weight spanning tree--namely $P$, but many minimum spanning trees; indeed $G setminus E(P)$ is connected and has many cycles so any spanning tree of $G setminus E(P)$ is a minimum spanning tree, and $G setminus E(P)$ has more than one spanning tree.
answered Dec 29 '18 at 19:22
MikeMike
4,396412
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1 Answer
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$begingroup$
It's not true.
Let $G$ be $K_n$ and let $P$ be a Hamilonian path in $G=K_n$. [Say $i$ and $j$ are adjacent in $P$ iff $|i-j|=1$, then two vertices $i$ and $j$ are adjacent in $G setminus E(P)$ iff $|i-j| ge 2$. One can check that $G setminus E(P)$ is connected for $n ge 6$ and has many cycles.]
Give every edge in $P$ a weight of 100 and every edge in $G setminus E(P)$ a weight of 1. Then $G$ has one maximum-weight spanning tree--namely $P$, but many minimum spanning trees; indeed $G setminus E(P)$ is connected and has many cycles so any spanning tree of $G setminus E(P)$ is a minimum spanning tree, and $G setminus E(P)$ has more than one spanning tree.
answered Dec 29 '18 at 19:22
MikeMike
4,396412
$endgroup$
add a comment |
$begingroup$
It's not true.
Let $G$ be $K_n$ and let $P$ be a Hamilonian path in $G=K_n$. [Say $i$ and $j$ are adjacent in $P$ iff $|i-j|=1$, then two vertices $i$ and $j$ are adjacent in $G setminus E(P)$ iff $|i-j| ge 2$. One can check that $G setminus E(P)$ is connected for $n ge 6$ and has many cycles.]
Give every edge in $P$ a weight of 100 and every edge in $G setminus E(P)$ a weight of 1. Then $G$ has one maximum-weight spanning tree--namely $P$, but many minimum spanning trees; indeed $G setminus E(P)$ is connected and has many cycles so any spanning tree of $G setminus E(P)$ is a minimum spanning tree, and $G setminus E(P)$ has more than one spanning tree.
answered Dec 29 '18 at 19:22
MikeMike
4,396412
$endgroup$
add a comment |
$begingroup$
It's not true.
Let $G$ be $K_n$ and let $P$ be a Hamilonian path in $G=K_n$. [Say $i$ and $j$ are adjacent in $P$ iff $|i-j|=1$, then two vertices $i$ and $j$ are adjacent in $G setminus E(P)$ iff $|i-j| ge 2$. One can check that $G setminus E(P)$ is connected for $n ge 6$ and has many cycles.]
Give every edge in $P$ a weight of 100 and every edge in $G setminus E(P)$ a weight of 1. Then $G$ has one maximum-weight spanning tree--namely $P$, but many minimum spanning trees; indeed $G setminus E(P)$ is connected and has many cycles so any spanning tree of $G setminus E(P)$ is a minimum spanning tree, and $G setminus E(P)$ has more than one spanning tree.
answered Dec 29 '18 at 19:22
MikeMike
4,396412
$endgroup$
It's not true.
Let $G$ be $K_n$ and let $P$ be a Hamilonian path in $G=K_n$. [Say $i$ and $j$ are adjacent in $P$ iff $|i-j|=1$, then two vertices $i$ and $j$ are adjacent in $G setminus E(P)$ iff $|i-j| ge 2$. One can check that $G setminus E(P)$ is connected for $n ge 6$ and has many cycles.]
Give every edge in $P$ a weight of 100 and every edge in $G setminus E(P)$ a weight of 1. Then $G$ has one maximum-weight spanning tree--namely $P$, but many minimum spanning trees; indeed $G setminus E(P)$ is connected and has many cycles so any spanning tree of $G setminus E(P)$ is a minimum spanning tree, and $G setminus E(P)$ has more than one spanning tree.
answered Dec 29 '18 at 19:22
MikeMike
4,396412
answered Dec 29 '18 at 19:22
MikeMike
4,396412
answered Dec 29 '18 at 19:22
MikeMike
4,396412
answered Dec 29 '18 at 19:22
MikeMike
4,396412
4,396412
add a comment |
add a comment |
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