Show that a graph with $n$ vertices and $n + 2$ edges must contain two edge-disjoint circuits.
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Show that a graph with $n$ vertices and $n + 2$ edges must contain two edge-disjoint circuits.
I'm a bit confused by what an edge-disjoint circuit means here.
graph-theory
asked Nov 26 at 18:35
cosmicbrownie
1016
add a comment |
up vote
0
down vote
favorite
Show that a graph with $n$ vertices and $n + 2$ edges must contain two edge-disjoint circuits.
I'm a bit confused by what an edge-disjoint circuit means here.
graph-theory
asked Nov 26 at 18:35
cosmicbrownie
1016
"Edge-disjoint" means that you have two circuits with no edges in common.
– Milo Brandt
Nov 26 at 18:36
Hint: Use what you know about trees and the relationship between the number of edges and vertices in a tree.
– JMoravitz
Nov 26 at 18:39
4
Take $G = K_4$, which has 4 vertices and 6 edges, but for any circuit on three vertices, removing it leave a tree (which has no circuits), so this isn't true for $n=4$.
– B. Mehta
Nov 26 at 19:23
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Show that a graph with $n$ vertices and $n + 2$ edges must contain two edge-disjoint circuits.
I'm a bit confused by what an edge-disjoint circuit means here.
graph-theory
asked Nov 26 at 18:35
cosmicbrownie
1016
Show that a graph with $n$ vertices and $n + 2$ edges must contain two edge-disjoint circuits.
I'm a bit confused by what an edge-disjoint circuit means here.
graph-theory
graph-theory
asked Nov 26 at 18:35
cosmicbrownie
1016
asked Nov 26 at 18:35
cosmicbrownie
1016
asked Nov 26 at 18:35
cosmicbrownie
1016
asked Nov 26 at 18:35
cosmicbrownie
1016
asked Nov 26 at 18:35
cosmicbrownie
1016
1016
"Edge-disjoint" means that you have two circuits with no edges in common.
– Milo Brandt
Nov 26 at 18:36
Hint: Use what you know about trees and the relationship between the number of edges and vertices in a tree.
– JMoravitz
Nov 26 at 18:39
4
Take $G = K_4$, which has 4 vertices and 6 edges, but for any circuit on three vertices, removing it leave a tree (which has no circuits), so this isn't true for $n=4$.
– B. Mehta
Nov 26 at 19:23
add a comment |
"Edge-disjoint" means that you have two circuits with no edges in common.
– Milo Brandt
Nov 26 at 18:36
Hint: Use what you know about trees and the relationship between the number of edges and vertices in a tree.
– JMoravitz
Nov 26 at 18:39
4
Take $G = K_4$, which has 4 vertices and 6 edges, but for any circuit on three vertices, removing it leave a tree (which has no circuits), so this isn't true for $n=4$.
– B. Mehta
Nov 26 at 19:23
"Edge-disjoint" means that you have two circuits with no edges in common.
– Milo Brandt
Nov 26 at 18:36
"Edge-disjoint" means that you have two circuits with no edges in common.
– Milo Brandt
Nov 26 at 18:36
Hint: Use what you know about trees and the relationship between the number of edges and vertices in a tree.
– JMoravitz
Nov 26 at 18:39
Hint: Use what you know about trees and the relationship between the number of edges and vertices in a tree.
– JMoravitz
Nov 26 at 18:39
4
4
Take $G = K_4$, which has 4 vertices and 6 edges, but for any circuit on three vertices, removing it leave a tree (which has no circuits), so this isn't true for $n=4$.
– B. Mehta
Nov 26 at 19:23
Take $G = K_4$, which has 4 vertices and 6 edges, but for any circuit on three vertices, removing it leave a tree (which has no circuits), so this isn't true for $n=4$.
– B. Mehta
Nov 26 at 19:23
add a comment |
1 Answer
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Well but this isn't true though. Take $K_4$ and replace each edge by a path with $k+1$ vertices, where $k$ an arbitrarily large integer. This graph has $4+6k$ vertices and $6+6k$ edges.
I got this idea from @B. Mehta 's comment.
answered Nov 26 at 20:23
Mike
2,764211
add a comment |
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1 Answer
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Well but this isn't true though. Take $K_4$ and replace each edge by a path with $k+1$ vertices, where $k$ an arbitrarily large integer. This graph has $4+6k$ vertices and $6+6k$ edges.
I got this idea from @B. Mehta 's comment.
answered Nov 26 at 20:23
Mike
2,764211
add a comment |
up vote
1
down vote
accepted
Well but this isn't true though. Take $K_4$ and replace each edge by a path with $k+1$ vertices, where $k$ an arbitrarily large integer. This graph has $4+6k$ vertices and $6+6k$ edges.
I got this idea from @B. Mehta 's comment.
answered Nov 26 at 20:23
Mike
2,764211
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Well but this isn't true though. Take $K_4$ and replace each edge by a path with $k+1$ vertices, where $k$ an arbitrarily large integer. This graph has $4+6k$ vertices and $6+6k$ edges.
I got this idea from @B. Mehta 's comment.
answered Nov 26 at 20:23
Mike
2,764211
Well but this isn't true though. Take $K_4$ and replace each edge by a path with $k+1$ vertices, where $k$ an arbitrarily large integer. This graph has $4+6k$ vertices and $6+6k$ edges.
I got this idea from @B. Mehta 's comment.
answered Nov 26 at 20:23
Mike
2,764211
edited Nov 26 at 20:38
answered Nov 26 at 20:23
Mike
2,764211
answered Nov 26 at 20:23
Mike
2,764211
answered Nov 26 at 20:23
Mike
2,764211
2,764211
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"Edge-disjoint" means that you have two circuits with no edges in common.
– Milo Brandt
Nov 26 at 18:36
Hint: Use what you know about trees and the relationship between the number of edges and vertices in a tree.
– JMoravitz
Nov 26 at 18:39
4
Take $G = K_4$, which has 4 vertices and 6 edges, but for any circuit on three vertices, removing it leave a tree (which has no circuits), so this isn't true for $n=4$.
– B. Mehta
Nov 26 at 19:23