Clarification of Edge-connectivity Menger's theorem
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Edge-connectivity Menger's Theorem: A graph is k-edge-connected if and only if every pair of vertices has k pairwise edge-disjoint (p.e.d.) paths in between.
I can see that the later "k" is actually a lower bound, i.e. there might be more than k p.e.d. paths in between some pairs of vertices. But then, if between all pairs of vertices there are more than k p.e.d. paths, wouldn't the graph be not k-edge-connected? Or is it the case that this construction is simply not possible?
graph-theory graph-connectivity
asked Nov 20 at 21:34
ensbana
265113
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Edge-connectivity Menger's Theorem: A graph is k-edge-connected if and only if every pair of vertices has k pairwise edge-disjoint (p.e.d.) paths in between.
I can see that the later "k" is actually a lower bound, i.e. there might be more than k p.e.d. paths in between some pairs of vertices. But then, if between all pairs of vertices there are more than k p.e.d. paths, wouldn't the graph be not k-edge-connected? Or is it the case that this construction is simply not possible?
graph-theory graph-connectivity
asked Nov 20 at 21:34
ensbana
265113
The construction is not possible. What you are saying is merely that if G is k edge connected then G has at most k edge disjoint paths. So you are “proving by contradiction” one direction of the theorem. In fact you are stating that it is obvious and I totally agree. However, the difficulty and beauty of this problem is to show that the converse is also true.
– Sorin Tirc
Nov 20 at 21:53
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Edge-connectivity Menger's Theorem: A graph is k-edge-connected if and only if every pair of vertices has k pairwise edge-disjoint (p.e.d.) paths in between.
I can see that the later "k" is actually a lower bound, i.e. there might be more than k p.e.d. paths in between some pairs of vertices. But then, if between all pairs of vertices there are more than k p.e.d. paths, wouldn't the graph be not k-edge-connected? Or is it the case that this construction is simply not possible?
graph-theory graph-connectivity
asked Nov 20 at 21:34
ensbana
265113
Edge-connectivity Menger's Theorem: A graph is k-edge-connected if and only if every pair of vertices has k pairwise edge-disjoint (p.e.d.) paths in between.
I can see that the later "k" is actually a lower bound, i.e. there might be more than k p.e.d. paths in between some pairs of vertices. But then, if between all pairs of vertices there are more than k p.e.d. paths, wouldn't the graph be not k-edge-connected? Or is it the case that this construction is simply not possible?
graph-theory graph-connectivity
graph-theory graph-connectivity
asked Nov 20 at 21:34
ensbana
265113
asked Nov 20 at 21:34
ensbana
265113
asked Nov 20 at 21:34
ensbana
265113
asked Nov 20 at 21:34
ensbana
265113
asked Nov 20 at 21:34
ensbana
265113
265113
The construction is not possible. What you are saying is merely that if G is k edge connected then G has at most k edge disjoint paths. So you are “proving by contradiction” one direction of the theorem. In fact you are stating that it is obvious and I totally agree. However, the difficulty and beauty of this problem is to show that the converse is also true.
– Sorin Tirc
Nov 20 at 21:53
add a comment |
The construction is not possible. What you are saying is merely that if G is k edge connected then G has at most k edge disjoint paths. So you are “proving by contradiction” one direction of the theorem. In fact you are stating that it is obvious and I totally agree. However, the difficulty and beauty of this problem is to show that the converse is also true.
– Sorin Tirc
Nov 20 at 21:53
The construction is not possible. What you are saying is merely that if G is k edge connected then G has at most k edge disjoint paths. So you are “proving by contradiction” one direction of the theorem. In fact you are stating that it is obvious and I totally agree. However, the difficulty and beauty of this problem is to show that the converse is also true.
– Sorin Tirc
Nov 20 at 21:53
The construction is not possible. What you are saying is merely that if G is k edge connected then G has at most k edge disjoint paths. So you are “proving by contradiction” one direction of the theorem. In fact you are stating that it is obvious and I totally agree. However, the difficulty and beauty of this problem is to show that the converse is also true.
– Sorin Tirc
Nov 20 at 21:53
add a comment |
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The construction is not possible. What you are saying is merely that if G is k edge connected then G has at most k edge disjoint paths. So you are “proving by contradiction” one direction of the theorem. In fact you are stating that it is obvious and I totally agree. However, the difficulty and beauty of this problem is to show that the converse is also true.
– Sorin Tirc
Nov 20 at 21:53