Find a partition of nodes into two sets $V_1$ and $V_2$ such that these two sets cover most edges
$begingroup$
Given an undirected connected graph $G$. Let $V$ be its node set.
The problem is to find a partition of $V$ ($V = V_1 +V_2$ and $V_1$, $V_2$ are disjoint) such that $|Cover(V_1)| + |Cover(V_2)|$ is maximized, where $Cover(V_i)$ is the set of edges incident to at least one node in $V_i$ (that is, the edges covered by $V_i$).
I wonder if the following hypothesis is correct:
$|Cover(V_1)| + |Cover(V_2)|$ is maximized if and only if one of $V_1$ and $V_2$ is a minimum vertex cover of the graph $G$.
combinatorics graph-theory computational-complexity
asked Nov 28 '18 at 17:04
ParadoxParadox
889
$endgroup$
add a comment |
$begingroup$
Given an undirected connected graph $G$. Let $V$ be its node set.
The problem is to find a partition of $V$ ($V = V_1 +V_2$ and $V_1$, $V_2$ are disjoint) such that $|Cover(V_1)| + |Cover(V_2)|$ is maximized, where $Cover(V_i)$ is the set of edges incident to at least one node in $V_i$ (that is, the edges covered by $V_i$).
I wonder if the following hypothesis is correct:
$|Cover(V_1)| + |Cover(V_2)|$ is maximized if and only if one of $V_1$ and $V_2$ is a minimum vertex cover of the graph $G$.
combinatorics graph-theory computational-complexity
asked Nov 28 '18 at 17:04
ParadoxParadox
889
$endgroup$
add a comment |
$begingroup$
Given an undirected connected graph $G$. Let $V$ be its node set.
The problem is to find a partition of $V$ ($V = V_1 +V_2$ and $V_1$, $V_2$ are disjoint) such that $|Cover(V_1)| + |Cover(V_2)|$ is maximized, where $Cover(V_i)$ is the set of edges incident to at least one node in $V_i$ (that is, the edges covered by $V_i$).
I wonder if the following hypothesis is correct:
$|Cover(V_1)| + |Cover(V_2)|$ is maximized if and only if one of $V_1$ and $V_2$ is a minimum vertex cover of the graph $G$.
combinatorics graph-theory computational-complexity
asked Nov 28 '18 at 17:04
ParadoxParadox
889
$endgroup$
Given an undirected connected graph $G$. Let $V$ be its node set.
The problem is to find a partition of $V$ ($V = V_1 +V_2$ and $V_1$, $V_2$ are disjoint) such that $|Cover(V_1)| + |Cover(V_2)|$ is maximized, where $Cover(V_i)$ is the set of edges incident to at least one node in $V_i$ (that is, the edges covered by $V_i$).
I wonder if the following hypothesis is correct:
$|Cover(V_1)| + |Cover(V_2)|$ is maximized if and only if one of $V_1$ and $V_2$ is a minimum vertex cover of the graph $G$.
combinatorics graph-theory computational-complexity
combinatorics graph-theory computational-complexity
asked Nov 28 '18 at 17:04
ParadoxParadox
889
asked Nov 28 '18 at 17:04
ParadoxParadox
889
asked Nov 28 '18 at 17:04
ParadoxParadox
889
asked Nov 28 '18 at 17:04
ParadoxParadox
889
asked Nov 28 '18 at 17:04
ParadoxParadox
889
889
add a comment |
add a comment |
1 Answer
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Your hypothesis is incorrect. Let $G = G_1 cup G_2 cup ldots G_m$ where each $G_i$ is a complete bipartite graph where the left side $L_{i}$ of $G_i$ has half as many vertices as the right side $R_{i}$ of $G_i$. The $G_i$'s are vertex-disjoint; so $G$ is $m$ connected components where $G$ on the $i$-th component is the graph $G_i$.
The unique minimum vertex cover of $G$ is $L_1 cup ldots L_m$.
However, construct $V_1$ and $V_2$ as follows: for each $i=1,ldots, m$ pick an arbitrary side of $G_i$ and put it in $V_1$ and put the other side into $V_2$. Then both $V_1$ and $V_2$ are disjoint and each cover all the edges, so $|Cov(V_1)| + |Cov(V_2)|$ is definitely maximized. But neither $V_1$ nor $V_2$ has to be a minimum vertex-cover--indeed $V_1$ can have the left side of some but not all of the $G_i$s while $V_2$ can have simultaneously the left side of some but not all of the $G_i$s.
If you insist on connectedness put an edge between $L_i$ and $L_{i+1}$ for each $i$. Then $L_1 cup ldots L_m$ is still the unique minimum vertex cover. But consider $V_1 = cup_{i odd} L_i + cup_{i even} R_i$ while $V_2$ is $V_1$'s complement.
answered Nov 28 '18 at 19:29
MikeMike
3,785411
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1 Answer
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$begingroup$
Your hypothesis is incorrect. Let $G = G_1 cup G_2 cup ldots G_m$ where each $G_i$ is a complete bipartite graph where the left side $L_{i}$ of $G_i$ has half as many vertices as the right side $R_{i}$ of $G_i$. The $G_i$'s are vertex-disjoint; so $G$ is $m$ connected components where $G$ on the $i$-th component is the graph $G_i$.
The unique minimum vertex cover of $G$ is $L_1 cup ldots L_m$.
However, construct $V_1$ and $V_2$ as follows: for each $i=1,ldots, m$ pick an arbitrary side of $G_i$ and put it in $V_1$ and put the other side into $V_2$. Then both $V_1$ and $V_2$ are disjoint and each cover all the edges, so $|Cov(V_1)| + |Cov(V_2)|$ is definitely maximized. But neither $V_1$ nor $V_2$ has to be a minimum vertex-cover--indeed $V_1$ can have the left side of some but not all of the $G_i$s while $V_2$ can have simultaneously the left side of some but not all of the $G_i$s.
If you insist on connectedness put an edge between $L_i$ and $L_{i+1}$ for each $i$. Then $L_1 cup ldots L_m$ is still the unique minimum vertex cover. But consider $V_1 = cup_{i odd} L_i + cup_{i even} R_i$ while $V_2$ is $V_1$'s complement.
answered Nov 28 '18 at 19:29
MikeMike
3,785411
$endgroup$
add a comment |
$begingroup$
Your hypothesis is incorrect. Let $G = G_1 cup G_2 cup ldots G_m$ where each $G_i$ is a complete bipartite graph where the left side $L_{i}$ of $G_i$ has half as many vertices as the right side $R_{i}$ of $G_i$. The $G_i$'s are vertex-disjoint; so $G$ is $m$ connected components where $G$ on the $i$-th component is the graph $G_i$.
The unique minimum vertex cover of $G$ is $L_1 cup ldots L_m$.
However, construct $V_1$ and $V_2$ as follows: for each $i=1,ldots, m$ pick an arbitrary side of $G_i$ and put it in $V_1$ and put the other side into $V_2$. Then both $V_1$ and $V_2$ are disjoint and each cover all the edges, so $|Cov(V_1)| + |Cov(V_2)|$ is definitely maximized. But neither $V_1$ nor $V_2$ has to be a minimum vertex-cover--indeed $V_1$ can have the left side of some but not all of the $G_i$s while $V_2$ can have simultaneously the left side of some but not all of the $G_i$s.
If you insist on connectedness put an edge between $L_i$ and $L_{i+1}$ for each $i$. Then $L_1 cup ldots L_m$ is still the unique minimum vertex cover. But consider $V_1 = cup_{i odd} L_i + cup_{i even} R_i$ while $V_2$ is $V_1$'s complement.
answered Nov 28 '18 at 19:29
MikeMike
3,785411
$endgroup$
add a comment |
$begingroup$
Your hypothesis is incorrect. Let $G = G_1 cup G_2 cup ldots G_m$ where each $G_i$ is a complete bipartite graph where the left side $L_{i}$ of $G_i$ has half as many vertices as the right side $R_{i}$ of $G_i$. The $G_i$'s are vertex-disjoint; so $G$ is $m$ connected components where $G$ on the $i$-th component is the graph $G_i$.
The unique minimum vertex cover of $G$ is $L_1 cup ldots L_m$.
However, construct $V_1$ and $V_2$ as follows: for each $i=1,ldots, m$ pick an arbitrary side of $G_i$ and put it in $V_1$ and put the other side into $V_2$. Then both $V_1$ and $V_2$ are disjoint and each cover all the edges, so $|Cov(V_1)| + |Cov(V_2)|$ is definitely maximized. But neither $V_1$ nor $V_2$ has to be a minimum vertex-cover--indeed $V_1$ can have the left side of some but not all of the $G_i$s while $V_2$ can have simultaneously the left side of some but not all of the $G_i$s.
If you insist on connectedness put an edge between $L_i$ and $L_{i+1}$ for each $i$. Then $L_1 cup ldots L_m$ is still the unique minimum vertex cover. But consider $V_1 = cup_{i odd} L_i + cup_{i even} R_i$ while $V_2$ is $V_1$'s complement.
answered Nov 28 '18 at 19:29
MikeMike
3,785411
$endgroup$
Your hypothesis is incorrect. Let $G = G_1 cup G_2 cup ldots G_m$ where each $G_i$ is a complete bipartite graph where the left side $L_{i}$ of $G_i$ has half as many vertices as the right side $R_{i}$ of $G_i$. The $G_i$'s are vertex-disjoint; so $G$ is $m$ connected components where $G$ on the $i$-th component is the graph $G_i$.
The unique minimum vertex cover of $G$ is $L_1 cup ldots L_m$.
However, construct $V_1$ and $V_2$ as follows: for each $i=1,ldots, m$ pick an arbitrary side of $G_i$ and put it in $V_1$ and put the other side into $V_2$. Then both $V_1$ and $V_2$ are disjoint and each cover all the edges, so $|Cov(V_1)| + |Cov(V_2)|$ is definitely maximized. But neither $V_1$ nor $V_2$ has to be a minimum vertex-cover--indeed $V_1$ can have the left side of some but not all of the $G_i$s while $V_2$ can have simultaneously the left side of some but not all of the $G_i$s.
If you insist on connectedness put an edge between $L_i$ and $L_{i+1}$ for each $i$. Then $L_1 cup ldots L_m$ is still the unique minimum vertex cover. But consider $V_1 = cup_{i odd} L_i + cup_{i even} R_i$ while $V_2$ is $V_1$'s complement.
answered Nov 28 '18 at 19:29
MikeMike
3,785411
edited Dec 2 '18 at 1:39
answered Nov 28 '18 at 19:29
MikeMike
3,785411
answered Nov 28 '18 at 19:29
MikeMike
3,785411
answered Nov 28 '18 at 19:29
MikeMike
3,785411
3,785411
add a comment |
add a comment |
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