We have a connected graph with $2n$ nodes. Prove that exsist spanning subgraph each node with odd degree.
We have a connected graph with $2n$ nodes. Prove that exsist spanning subgraph each node with odd degree.
Idea: Let $M$ be an adjacency matrix and work all over field $mathbb{Z}_2$. Then if $M$ is nonsingular we know that equation $$Mvec{s} = (1,1,1,....1,1)$$ has notrivial solution. Suppose set $S$ has incidence vector $vec{s}$. Then if for each node $a$ we color each edge in $N(a)cap S$ we win.
There are quite a few questions here. First, where did I use conectivity, second, is it $M$ really nonsingular and do we even need that and last, if all this is true does this aproach actually work?
Note: I already asked initaly question and did get an answer, so please answer just this.
linear-algebra combinatorics graph-theory algebraic-graph-theory algebraic-combinatorics
asked 2 days ago
greedoid
37.7k114794
add a comment |
We have a connected graph with $2n$ nodes. Prove that exsist spanning subgraph each node with odd degree.
Idea: Let $M$ be an adjacency matrix and work all over field $mathbb{Z}_2$. Then if $M$ is nonsingular we know that equation $$Mvec{s} = (1,1,1,....1,1)$$ has notrivial solution. Suppose set $S$ has incidence vector $vec{s}$. Then if for each node $a$ we color each edge in $N(a)cap S$ we win.
There are quite a few questions here. First, where did I use conectivity, second, is it $M$ really nonsingular and do we even need that and last, if all this is true does this aproach actually work?
Note: I already asked initaly question and did get an answer, so please answer just this.
linear-algebra combinatorics graph-theory algebraic-graph-theory algebraic-combinatorics
asked 2 days ago
greedoid
37.7k114794
Sorry I am a bit confused, do you want an answer to the initial quoted question or are you only asking about your approach?
– Vincenzo
2 days ago
@Vincenzo second one.
– greedoid
2 days ago
I don't understand the question - clearly a simple path on $4$ vertices has an even number of vertices, but the only spanning subgraph (being the graph itself) has two vertices of even degree.
– Math1000
2 days ago
Take a look here. math.stackexchange.com/questions/3047513/… @Math1000
– greedoid
2 days ago
1
If you want a linear algebra solution, you should instead look at the incidence matrix of the graph and try to show that $Me={bf 1}$ has a solution. I do not know if this makes things any easier.
– Mike Earnest
21 hours ago
add a comment |
We have a connected graph with $2n$ nodes. Prove that exsist spanning subgraph each node with odd degree.
Idea: Let $M$ be an adjacency matrix and work all over field $mathbb{Z}_2$. Then if $M$ is nonsingular we know that equation $$Mvec{s} = (1,1,1,....1,1)$$ has notrivial solution. Suppose set $S$ has incidence vector $vec{s}$. Then if for each node $a$ we color each edge in $N(a)cap S$ we win.
There are quite a few questions here. First, where did I use conectivity, second, is it $M$ really nonsingular and do we even need that and last, if all this is true does this aproach actually work?
Note: I already asked initaly question and did get an answer, so please answer just this.
linear-algebra combinatorics graph-theory algebraic-graph-theory algebraic-combinatorics
asked 2 days ago
greedoid
37.7k114794
We have a connected graph with $2n$ nodes. Prove that exsist spanning subgraph each node with odd degree.
Idea: Let $M$ be an adjacency matrix and work all over field $mathbb{Z}_2$. Then if $M$ is nonsingular we know that equation $$Mvec{s} = (1,1,1,....1,1)$$ has notrivial solution. Suppose set $S$ has incidence vector $vec{s}$. Then if for each node $a$ we color each edge in $N(a)cap S$ we win.
There are quite a few questions here. First, where did I use conectivity, second, is it $M$ really nonsingular and do we even need that and last, if all this is true does this aproach actually work?
Note: I already asked initaly question and did get an answer, so please answer just this.
linear-algebra combinatorics graph-theory algebraic-graph-theory algebraic-combinatorics
linear-algebra combinatorics graph-theory algebraic-graph-theory algebraic-combinatorics
asked 2 days ago
greedoid
37.7k114794
asked 2 days ago
greedoid
37.7k114794
edited 2 days ago
asked 2 days ago
greedoid
37.7k114794
asked 2 days ago
greedoid
37.7k114794
asked 2 days ago
greedoid
37.7k114794
37.7k114794
Sorry I am a bit confused, do you want an answer to the initial quoted question or are you only asking about your approach?
– Vincenzo
2 days ago
@Vincenzo second one.
– greedoid
2 days ago
I don't understand the question - clearly a simple path on $4$ vertices has an even number of vertices, but the only spanning subgraph (being the graph itself) has two vertices of even degree.
– Math1000
2 days ago
Take a look here. math.stackexchange.com/questions/3047513/… @Math1000
– greedoid
2 days ago
1
If you want a linear algebra solution, you should instead look at the incidence matrix of the graph and try to show that $Me={bf 1}$ has a solution. I do not know if this makes things any easier.
– Mike Earnest
21 hours ago
add a comment |
Sorry I am a bit confused, do you want an answer to the initial quoted question or are you only asking about your approach?
– Vincenzo
2 days ago
@Vincenzo second one.
– greedoid
2 days ago
I don't understand the question - clearly a simple path on $4$ vertices has an even number of vertices, but the only spanning subgraph (being the graph itself) has two vertices of even degree.
– Math1000
2 days ago
Take a look here. math.stackexchange.com/questions/3047513/… @Math1000
– greedoid
2 days ago
1
If you want a linear algebra solution, you should instead look at the incidence matrix of the graph and try to show that $Me={bf 1}$ has a solution. I do not know if this makes things any easier.
– Mike Earnest
21 hours ago
Sorry I am a bit confused, do you want an answer to the initial quoted question or are you only asking about your approach?
– Vincenzo
2 days ago
Sorry I am a bit confused, do you want an answer to the initial quoted question or are you only asking about your approach?
– Vincenzo
2 days ago
@Vincenzo second one.
– greedoid
2 days ago
@Vincenzo second one.
– greedoid
2 days ago
I don't understand the question - clearly a simple path on $4$ vertices has an even number of vertices, but the only spanning subgraph (being the graph itself) has two vertices of even degree.
– Math1000
2 days ago
I don't understand the question - clearly a simple path on $4$ vertices has an even number of vertices, but the only spanning subgraph (being the graph itself) has two vertices of even degree.
– Math1000
2 days ago
Take a look here. math.stackexchange.com/questions/3047513/… @Math1000
– greedoid
2 days ago
Take a look here. math.stackexchange.com/questions/3047513/… @Math1000
– greedoid
2 days ago
1
1
If you want a linear algebra solution, you should instead look at the incidence matrix of the graph and try to show that $Me={bf 1}$ has a solution. I do not know if this makes things any easier.
– Mike Earnest
21 hours ago
If you want a linear algebra solution, you should instead look at the incidence matrix of the graph and try to show that $Me={bf 1}$ has a solution. I do not know if this makes things any easier.
– Mike Earnest
21 hours ago
add a comment |
1 Answer
1
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There are at least two gaps in the approach. We need an equation $Mvec{s} = (1,1,1,....1,1)$ to have a solution (and I don’t know what means “non-trivial” here). If the matrix $M$ is non-singular then this soluttion exists and is unique. But why in this case $vec{s}$ is an incidence vector of some set? Next, the adjacency matrix is singular for any graph which has two vertices with the same set of neighbors. For instance, for a cycle $C_4$ on four vertices.
edited 2 days ago
greedoid
37.7k114794
answered 2 days ago
Alex Ravsky
38.8k32079
Anyway, do you have a perhaps an idea how to atack this algebraicly
– greedoid
2 days ago
You think this is hopless try with linear algebra?
– greedoid
yesterday
add a comment |
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1 Answer
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1 Answer
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There are at least two gaps in the approach. We need an equation $Mvec{s} = (1,1,1,....1,1)$ to have a solution (and I don’t know what means “non-trivial” here). If the matrix $M$ is non-singular then this soluttion exists and is unique. But why in this case $vec{s}$ is an incidence vector of some set? Next, the adjacency matrix is singular for any graph which has two vertices with the same set of neighbors. For instance, for a cycle $C_4$ on four vertices.
edited 2 days ago
greedoid
37.7k114794
answered 2 days ago
Alex Ravsky
38.8k32079
Anyway, do you have a perhaps an idea how to atack this algebraicly
– greedoid
2 days ago
You think this is hopless try with linear algebra?
– greedoid
yesterday
add a comment |
There are at least two gaps in the approach. We need an equation $Mvec{s} = (1,1,1,....1,1)$ to have a solution (and I don’t know what means “non-trivial” here). If the matrix $M$ is non-singular then this soluttion exists and is unique. But why in this case $vec{s}$ is an incidence vector of some set? Next, the adjacency matrix is singular for any graph which has two vertices with the same set of neighbors. For instance, for a cycle $C_4$ on four vertices.
edited 2 days ago
greedoid
37.7k114794
answered 2 days ago
Alex Ravsky
38.8k32079
Anyway, do you have a perhaps an idea how to atack this algebraicly
– greedoid
2 days ago
You think this is hopless try with linear algebra?
– greedoid
yesterday
add a comment |
There are at least two gaps in the approach. We need an equation $Mvec{s} = (1,1,1,....1,1)$ to have a solution (and I don’t know what means “non-trivial” here). If the matrix $M$ is non-singular then this soluttion exists and is unique. But why in this case $vec{s}$ is an incidence vector of some set? Next, the adjacency matrix is singular for any graph which has two vertices with the same set of neighbors. For instance, for a cycle $C_4$ on four vertices.
edited 2 days ago
greedoid
37.7k114794
answered 2 days ago
Alex Ravsky
38.8k32079
There are at least two gaps in the approach. We need an equation $Mvec{s} = (1,1,1,....1,1)$ to have a solution (and I don’t know what means “non-trivial” here). If the matrix $M$ is non-singular then this soluttion exists and is unique. But why in this case $vec{s}$ is an incidence vector of some set? Next, the adjacency matrix is singular for any graph which has two vertices with the same set of neighbors. For instance, for a cycle $C_4$ on four vertices.
edited 2 days ago
greedoid
37.7k114794
answered 2 days ago
Alex Ravsky
38.8k32079
edited 2 days ago
greedoid
37.7k114794
edited 2 days ago
greedoid
37.7k114794
edited 2 days ago
greedoid
37.7k114794
37.7k114794
answered 2 days ago
Alex Ravsky
38.8k32079
answered 2 days ago
Alex Ravsky
38.8k32079
answered 2 days ago
Alex Ravsky
38.8k32079
38.8k32079
Anyway, do you have a perhaps an idea how to atack this algebraicly
– greedoid
2 days ago
You think this is hopless try with linear algebra?
– greedoid
yesterday
add a comment |
Anyway, do you have a perhaps an idea how to atack this algebraicly
– greedoid
2 days ago
You think this is hopless try with linear algebra?
– greedoid
yesterday
Anyway, do you have a perhaps an idea how to atack this algebraicly
– greedoid
2 days ago
Anyway, do you have a perhaps an idea how to atack this algebraicly
– greedoid
2 days ago
You think this is hopless try with linear algebra?
– greedoid
yesterday
You think this is hopless try with linear algebra?
– greedoid
yesterday
add a comment |
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Sorry I am a bit confused, do you want an answer to the initial quoted question or are you only asking about your approach?
– Vincenzo
2 days ago
@Vincenzo second one.
– greedoid
2 days ago
I don't understand the question - clearly a simple path on $4$ vertices has an even number of vertices, but the only spanning subgraph (being the graph itself) has two vertices of even degree.
– Math1000
2 days ago
Take a look here. math.stackexchange.com/questions/3047513/… @Math1000
– greedoid
2 days ago
1
If you want a linear algebra solution, you should instead look at the incidence matrix of the graph and try to show that $Me={bf 1}$ has a solution. I do not know if this makes things any easier.
– Mike Earnest
21 hours ago