Does a Vertex Cover exist?
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This should be a simple question, but I am a little bit confused.
A proof on page 556 of Algorithm Design says:
"Let $e=(u, v)$ be any edge of $G$. The graph $G$ has a vertex cover of size at most $k$ if and only if at least one of the graphs $G-{u}$ and $G-{v}$ has a vertex cover of size at most $k-1$."
But when I try the following example where the black ball shows a vertex cover, it seems to be faulty statement. Because $k$ remains as $k$ (not $k-1$) even after deletion of vertex $u$. Moreover, if I delete the vertex $v$, there would be no vertex cover left.
What is wrong about my deduction and example?
graphs graph-theory
asked Dec 4 at 4:20
Reza Hadi
324
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up vote
2
down vote
favorite
This should be a simple question, but I am a little bit confused.
A proof on page 556 of Algorithm Design says:
"Let $e=(u, v)$ be any edge of $G$. The graph $G$ has a vertex cover of size at most $k$ if and only if at least one of the graphs $G-{u}$ and $G-{v}$ has a vertex cover of size at most $k-1$."
But when I try the following example where the black ball shows a vertex cover, it seems to be faulty statement. Because $k$ remains as $k$ (not $k-1$) even after deletion of vertex $u$. Moreover, if I delete the vertex $v$, there would be no vertex cover left.
What is wrong about my deduction and example?
graphs graph-theory
asked Dec 4 at 4:20
Reza Hadi
324
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This should be a simple question, but I am a little bit confused.
A proof on page 556 of Algorithm Design says:
"Let $e=(u, v)$ be any edge of $G$. The graph $G$ has a vertex cover of size at most $k$ if and only if at least one of the graphs $G-{u}$ and $G-{v}$ has a vertex cover of size at most $k-1$."
But when I try the following example where the black ball shows a vertex cover, it seems to be faulty statement. Because $k$ remains as $k$ (not $k-1$) even after deletion of vertex $u$. Moreover, if I delete the vertex $v$, there would be no vertex cover left.
What is wrong about my deduction and example?
graphs graph-theory
asked Dec 4 at 4:20
Reza Hadi
324
This should be a simple question, but I am a little bit confused.
A proof on page 556 of Algorithm Design says:
"Let $e=(u, v)$ be any edge of $G$. The graph $G$ has a vertex cover of size at most $k$ if and only if at least one of the graphs $G-{u}$ and $G-{v}$ has a vertex cover of size at most $k-1$."
But when I try the following example where the black ball shows a vertex cover, it seems to be faulty statement. Because $k$ remains as $k$ (not $k-1$) even after deletion of vertex $u$. Moreover, if I delete the vertex $v$, there would be no vertex cover left.
What is wrong about my deduction and example?
graphs graph-theory
graphs graph-theory
asked Dec 4 at 4:20
Reza Hadi
324
asked Dec 4 at 4:20
Reza Hadi
324
edited Dec 4 at 4:35
asked Dec 4 at 4:20
Reza Hadi
324
asked Dec 4 at 4:20
Reza Hadi
324
asked Dec 4 at 4:20
Reza Hadi
324
324
add a comment |
add a comment |
1 Answer
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3
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If you delete the vertex $v$, then there are no edges left at all, and so there is a vertex cover of size $0 leq k-1$.
Why is the statement true in the first place? Suppose that $G$ has a vertex cover $C$ of size $k$. Since $(u,v)$ is an edge, $C$ must contain at least one of $u,v$, say $v$. I claim that $C setminus v$ is a vertex cover of $G setminus v$, from which the statement immediately follows. Indeed, let $(a,b)$ be any edge of $G setminus v$. Then $(a,b) in G$, and so $C$ contains one of $a,b$, say $a$. By construction, $a neq v$, and so $a in C setminus v$.
answered Dec 4 at 4:52
Yuval Filmus
188k12177339
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
If you delete the vertex $v$, then there are no edges left at all, and so there is a vertex cover of size $0 leq k-1$.
Why is the statement true in the first place? Suppose that $G$ has a vertex cover $C$ of size $k$. Since $(u,v)$ is an edge, $C$ must contain at least one of $u,v$, say $v$. I claim that $C setminus v$ is a vertex cover of $G setminus v$, from which the statement immediately follows. Indeed, let $(a,b)$ be any edge of $G setminus v$. Then $(a,b) in G$, and so $C$ contains one of $a,b$, say $a$. By construction, $a neq v$, and so $a in C setminus v$.
answered Dec 4 at 4:52
Yuval Filmus
188k12177339
add a comment |
up vote
3
down vote
accepted
If you delete the vertex $v$, then there are no edges left at all, and so there is a vertex cover of size $0 leq k-1$.
Why is the statement true in the first place? Suppose that $G$ has a vertex cover $C$ of size $k$. Since $(u,v)$ is an edge, $C$ must contain at least one of $u,v$, say $v$. I claim that $C setminus v$ is a vertex cover of $G setminus v$, from which the statement immediately follows. Indeed, let $(a,b)$ be any edge of $G setminus v$. Then $(a,b) in G$, and so $C$ contains one of $a,b$, say $a$. By construction, $a neq v$, and so $a in C setminus v$.
answered Dec 4 at 4:52
Yuval Filmus
188k12177339
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If you delete the vertex $v$, then there are no edges left at all, and so there is a vertex cover of size $0 leq k-1$.
Why is the statement true in the first place? Suppose that $G$ has a vertex cover $C$ of size $k$. Since $(u,v)$ is an edge, $C$ must contain at least one of $u,v$, say $v$. I claim that $C setminus v$ is a vertex cover of $G setminus v$, from which the statement immediately follows. Indeed, let $(a,b)$ be any edge of $G setminus v$. Then $(a,b) in G$, and so $C$ contains one of $a,b$, say $a$. By construction, $a neq v$, and so $a in C setminus v$.
answered Dec 4 at 4:52
Yuval Filmus
188k12177339
If you delete the vertex $v$, then there are no edges left at all, and so there is a vertex cover of size $0 leq k-1$.
Why is the statement true in the first place? Suppose that $G$ has a vertex cover $C$ of size $k$. Since $(u,v)$ is an edge, $C$ must contain at least one of $u,v$, say $v$. I claim that $C setminus v$ is a vertex cover of $G setminus v$, from which the statement immediately follows. Indeed, let $(a,b)$ be any edge of $G setminus v$. Then $(a,b) in G$, and so $C$ contains one of $a,b$, say $a$. By construction, $a neq v$, and so $a in C setminus v$.
answered Dec 4 at 4:52
Yuval Filmus
188k12177339
answered Dec 4 at 4:52
Yuval Filmus
188k12177339
answered Dec 4 at 4:52
Yuval Filmus
188k12177339
answered Dec 4 at 4:52
Yuval Filmus
188k12177339
188k12177339
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add a comment |
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