Mixing
Proof of a graph theorem
$begingroup$
Studying graph theory I found the following theorem for s,t vertex of the graph $G=(V,E)$ and a cut defined as "subset $C$ of $E$ an s−t cut if C = $δ ^{out}(U)$ for some subset U of V satisfying s $in$ U and t $notin$ U:
The minimum length of an s − t path is equal to the maximum number of disjoint s − t cuts.
Can someone prove it to me?
graph-theory network-flow network
asked Jan 10 at 22:25
phw.phw.
472
$endgroup$
add a comment |
$begingroup$
Studying graph theory I found the following theorem for s,t vertex of the graph $G=(V,E)$ and a cut defined as "subset $C$ of $E$ an s−t cut if C = $δ ^{out}(U)$ for some subset U of V satisfying s $in$ U and t $notin$ U:
The minimum length of an s − t path is equal to the maximum number of disjoint s − t cuts.
Can someone prove it to me?
graph-theory network-flow network
asked Jan 10 at 22:25
phw.phw.
472
$endgroup$
add a comment |
$begingroup$
Studying graph theory I found the following theorem for s,t vertex of the graph $G=(V,E)$ and a cut defined as "subset $C$ of $E$ an s−t cut if C = $δ ^{out}(U)$ for some subset U of V satisfying s $in$ U and t $notin$ U:
The minimum length of an s − t path is equal to the maximum number of disjoint s − t cuts.
Can someone prove it to me?
graph-theory network-flow network
asked Jan 10 at 22:25
phw.phw.
472
$endgroup$
Studying graph theory I found the following theorem for s,t vertex of the graph $G=(V,E)$ and a cut defined as "subset $C$ of $E$ an s−t cut if C = $δ ^{out}(U)$ for some subset U of V satisfying s $in$ U and t $notin$ U:
The minimum length of an s − t path is equal to the maximum number of disjoint s − t cuts.
Can someone prove it to me?
graph-theory network-flow network
graph-theory network-flow network
asked Jan 10 at 22:25
phw.phw.
472
asked Jan 10 at 22:25
phw.phw.
472
asked Jan 10 at 22:25
phw.phw.
472
asked Jan 10 at 22:25
phw.phw.
472
asked Jan 10 at 22:25
phw.phw.
472
472
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $P$ be a shortest $s-t$ path.
Then on the one hand, every cut $C$ has to contain at least one edge $e$ in $P$ [make sure you see why]. Thus in any family ${C_1,ldots, C_m}$ such that the $C_j$s are pairwise disjoint, it follows that $m le |E(P)|$ lest there is an edge in $P$ in at least two of the $C_j$s (which would contradict pairwise disjointness).
On the other hand, for each $j=1,2,ldots, |E(P)|$ let $C_j$ be the set of edges in $G$ cutting the vertices of precisely distance $j-1$ from $s$ from the vertices of distance precisely $j$ from $s$. Then the $C_j$s are disjoint and each is an $s-t$ cut [make sure you see why].
answered Jan 11 at 1:36
MikeMike
3,861412
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
Let $P$ be a shortest $s-t$ path.
Then on the one hand, every cut $C$ has to contain at least one edge $e$ in $P$ [make sure you see why]. Thus in any family ${C_1,ldots, C_m}$ such that the $C_j$s are pairwise disjoint, it follows that $m le |E(P)|$ lest there is an edge in $P$ in at least two of the $C_j$s (which would contradict pairwise disjointness).
On the other hand, for each $j=1,2,ldots, |E(P)|$ let $C_j$ be the set of edges in $G$ cutting the vertices of precisely distance $j-1$ from $s$ from the vertices of distance precisely $j$ from $s$. Then the $C_j$s are disjoint and each is an $s-t$ cut [make sure you see why].
answered Jan 11 at 1:36
MikeMike
3,861412
$endgroup$
add a comment |
$begingroup$
Let $P$ be a shortest $s-t$ path.
Then on the one hand, every cut $C$ has to contain at least one edge $e$ in $P$ [make sure you see why]. Thus in any family ${C_1,ldots, C_m}$ such that the $C_j$s are pairwise disjoint, it follows that $m le |E(P)|$ lest there is an edge in $P$ in at least two of the $C_j$s (which would contradict pairwise disjointness).
On the other hand, for each $j=1,2,ldots, |E(P)|$ let $C_j$ be the set of edges in $G$ cutting the vertices of precisely distance $j-1$ from $s$ from the vertices of distance precisely $j$ from $s$. Then the $C_j$s are disjoint and each is an $s-t$ cut [make sure you see why].
answered Jan 11 at 1:36
MikeMike
3,861412
$endgroup$
add a comment |
$begingroup$
Let $P$ be a shortest $s-t$ path.
Then on the one hand, every cut $C$ has to contain at least one edge $e$ in $P$ [make sure you see why]. Thus in any family ${C_1,ldots, C_m}$ such that the $C_j$s are pairwise disjoint, it follows that $m le |E(P)|$ lest there is an edge in $P$ in at least two of the $C_j$s (which would contradict pairwise disjointness).
On the other hand, for each $j=1,2,ldots, |E(P)|$ let $C_j$ be the set of edges in $G$ cutting the vertices of precisely distance $j-1$ from $s$ from the vertices of distance precisely $j$ from $s$. Then the $C_j$s are disjoint and each is an $s-t$ cut [make sure you see why].
answered Jan 11 at 1:36
MikeMike
3,861412
$endgroup$
Let $P$ be a shortest $s-t$ path.
Then on the one hand, every cut $C$ has to contain at least one edge $e$ in $P$ [make sure you see why]. Thus in any family ${C_1,ldots, C_m}$ such that the $C_j$s are pairwise disjoint, it follows that $m le |E(P)|$ lest there is an edge in $P$ in at least two of the $C_j$s (which would contradict pairwise disjointness).
On the other hand, for each $j=1,2,ldots, |E(P)|$ let $C_j$ be the set of edges in $G$ cutting the vertices of precisely distance $j-1$ from $s$ from the vertices of distance precisely $j$ from $s$. Then the $C_j$s are disjoint and each is an $s-t$ cut [make sure you see why].
answered Jan 11 at 1:36
MikeMike
3,861412
answered Jan 11 at 1:36
MikeMike
3,861412
answered Jan 11 at 1:36
MikeMike
3,861412
answered Jan 11 at 1:36
MikeMike
3,861412
3,861412
add a comment |
add a comment |
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