A Complete Digraph is an Undirected Graph?
$begingroup$
Can I consider the Undirected Graph as a special case of Digraphs where all edges points for both directions?
A complete (completely connected) digraph turns to an undirected graph?
directed-graphs
asked Dec 10 '18 at 23:26
Henrique TannúsHenrique Tannús
112
$endgroup$
add a comment |
$begingroup$
Can I consider the Undirected Graph as a special case of Digraphs where all edges points for both directions?
A complete (completely connected) digraph turns to an undirected graph?
directed-graphs
asked Dec 10 '18 at 23:26
Henrique TannúsHenrique Tannús
112
$endgroup$
$begingroup$
Yes, and yes. You should maybe also consider how you handle self-loops (technically these edges point in both directions), but this is not really a big deal.
$endgroup$
– platty
Dec 10 '18 at 23:28
$begingroup$
Thank you! In my case study there will be no self-loops
$endgroup$
– Henrique Tannús
Dec 10 '18 at 23:34
add a comment |
$begingroup$
Can I consider the Undirected Graph as a special case of Digraphs where all edges points for both directions?
A complete (completely connected) digraph turns to an undirected graph?
directed-graphs
asked Dec 10 '18 at 23:26
Henrique TannúsHenrique Tannús
112
$endgroup$
Can I consider the Undirected Graph as a special case of Digraphs where all edges points for both directions?
A complete (completely connected) digraph turns to an undirected graph?
directed-graphs
directed-graphs
asked Dec 10 '18 at 23:26
Henrique TannúsHenrique Tannús
112
asked Dec 10 '18 at 23:26
Henrique TannúsHenrique Tannús
112
asked Dec 10 '18 at 23:26
Henrique TannúsHenrique Tannús
112
asked Dec 10 '18 at 23:26
Henrique TannúsHenrique Tannús
112
asked Dec 10 '18 at 23:26
Henrique TannúsHenrique Tannús
112
112
$begingroup$
Yes, and yes. You should maybe also consider how you handle self-loops (technically these edges point in both directions), but this is not really a big deal.
$endgroup$
– platty
Dec 10 '18 at 23:28
$begingroup$
Thank you! In my case study there will be no self-loops
$endgroup$
– Henrique Tannús
Dec 10 '18 at 23:34
add a comment |
$begingroup$
Yes, and yes. You should maybe also consider how you handle self-loops (technically these edges point in both directions), but this is not really a big deal.
$endgroup$
– platty
Dec 10 '18 at 23:28
$begingroup$
Thank you! In my case study there will be no self-loops
$endgroup$
– Henrique Tannús
Dec 10 '18 at 23:34
$begingroup$
Yes, and yes. You should maybe also consider how you handle self-loops (technically these edges point in both directions), but this is not really a big deal.
$endgroup$
– platty
Dec 10 '18 at 23:28
$begingroup$
Yes, and yes. You should maybe also consider how you handle self-loops (technically these edges point in both directions), but this is not really a big deal.
$endgroup$
– platty
Dec 10 '18 at 23:28
$begingroup$
Thank you! In my case study there will be no self-loops
$endgroup$
– Henrique Tannús
Dec 10 '18 at 23:34
$begingroup$
Thank you! In my case study there will be no self-loops
$endgroup$
– Henrique Tannús
Dec 10 '18 at 23:34
add a comment |
1 Answer
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$begingroup$
This is often, but not always a good way to apply a statement about directed graphs to an undirected graph.
For an example where it does not work: plenty of connected but undirected graphs do not have an Eulerian tour. But if you turn a connected graph into a directed graph by replacing each edge with two directed edges, then the resulting directed graph will always have an Eulerian tour.
The problem in this example is that an Eulerian tour in the undirected graph can traverse each edge $vw$ only once, whereas an Eulerian tour in the directed graph we obtain would have to take the directed edge $(v,w)$ once and the directed edge $(w,v)$ once.
On the other hand, connectivity is preserved: having a path from one vertex to another in the undirected graph is equivalent to having such a path in the directed graph constructed from it.
Whenever you have a tool for dealing with directed graphs, you can always try to apply it to an undirected graph by using this double-edge construction, but you should check to see what the effect is before proceeding.
answered Dec 11 '18 at 23:16
Misha LavrovMisha Lavrov
47.9k657107
$endgroup$
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
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$begingroup$
This is often, but not always a good way to apply a statement about directed graphs to an undirected graph.
For an example where it does not work: plenty of connected but undirected graphs do not have an Eulerian tour. But if you turn a connected graph into a directed graph by replacing each edge with two directed edges, then the resulting directed graph will always have an Eulerian tour.
The problem in this example is that an Eulerian tour in the undirected graph can traverse each edge $vw$ only once, whereas an Eulerian tour in the directed graph we obtain would have to take the directed edge $(v,w)$ once and the directed edge $(w,v)$ once.
On the other hand, connectivity is preserved: having a path from one vertex to another in the undirected graph is equivalent to having such a path in the directed graph constructed from it.
Whenever you have a tool for dealing with directed graphs, you can always try to apply it to an undirected graph by using this double-edge construction, but you should check to see what the effect is before proceeding.
answered Dec 11 '18 at 23:16
Misha LavrovMisha Lavrov
47.9k657107
$endgroup$
add a comment |
$begingroup$
This is often, but not always a good way to apply a statement about directed graphs to an undirected graph.
For an example where it does not work: plenty of connected but undirected graphs do not have an Eulerian tour. But if you turn a connected graph into a directed graph by replacing each edge with two directed edges, then the resulting directed graph will always have an Eulerian tour.
The problem in this example is that an Eulerian tour in the undirected graph can traverse each edge $vw$ only once, whereas an Eulerian tour in the directed graph we obtain would have to take the directed edge $(v,w)$ once and the directed edge $(w,v)$ once.
On the other hand, connectivity is preserved: having a path from one vertex to another in the undirected graph is equivalent to having such a path in the directed graph constructed from it.
Whenever you have a tool for dealing with directed graphs, you can always try to apply it to an undirected graph by using this double-edge construction, but you should check to see what the effect is before proceeding.
answered Dec 11 '18 at 23:16
Misha LavrovMisha Lavrov
47.9k657107
$endgroup$
add a comment |
$begingroup$
This is often, but not always a good way to apply a statement about directed graphs to an undirected graph.
For an example where it does not work: plenty of connected but undirected graphs do not have an Eulerian tour. But if you turn a connected graph into a directed graph by replacing each edge with two directed edges, then the resulting directed graph will always have an Eulerian tour.
The problem in this example is that an Eulerian tour in the undirected graph can traverse each edge $vw$ only once, whereas an Eulerian tour in the directed graph we obtain would have to take the directed edge $(v,w)$ once and the directed edge $(w,v)$ once.
On the other hand, connectivity is preserved: having a path from one vertex to another in the undirected graph is equivalent to having such a path in the directed graph constructed from it.
Whenever you have a tool for dealing with directed graphs, you can always try to apply it to an undirected graph by using this double-edge construction, but you should check to see what the effect is before proceeding.
answered Dec 11 '18 at 23:16
Misha LavrovMisha Lavrov
47.9k657107
$endgroup$
This is often, but not always a good way to apply a statement about directed graphs to an undirected graph.
For an example where it does not work: plenty of connected but undirected graphs do not have an Eulerian tour. But if you turn a connected graph into a directed graph by replacing each edge with two directed edges, then the resulting directed graph will always have an Eulerian tour.
The problem in this example is that an Eulerian tour in the undirected graph can traverse each edge $vw$ only once, whereas an Eulerian tour in the directed graph we obtain would have to take the directed edge $(v,w)$ once and the directed edge $(w,v)$ once.
On the other hand, connectivity is preserved: having a path from one vertex to another in the undirected graph is equivalent to having such a path in the directed graph constructed from it.
Whenever you have a tool for dealing with directed graphs, you can always try to apply it to an undirected graph by using this double-edge construction, but you should check to see what the effect is before proceeding.
answered Dec 11 '18 at 23:16
Misha LavrovMisha Lavrov
47.9k657107
answered Dec 11 '18 at 23:16
Misha LavrovMisha Lavrov
47.9k657107
answered Dec 11 '18 at 23:16
Misha LavrovMisha Lavrov
47.9k657107
answered Dec 11 '18 at 23:16
Misha LavrovMisha Lavrov
47.9k657107
47.9k657107
add a comment |
add a comment |
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$begingroup$
Yes, and yes. You should maybe also consider how you handle self-loops (technically these edges point in both directions), but this is not really a big deal.
$endgroup$
– platty
Dec 10 '18 at 23:28
$begingroup$
Thank you! In my case study there will be no self-loops
$endgroup$
– Henrique Tannús
Dec 10 '18 at 23:34