Example of a nowhere-zero 4-flow
I'm trying to understand the concept of nowhere-zero-flows.
I have this example graph that's supposed to have a nowhere-zero-4-flow (since it has a Hamiltonian cycle).
So by one of the theorems by Tutte, it should also have a nowhere-zero $mathbb{Z}_2 times mathbb{Z}_2$-flow.
If I understood it correctly, the flows assigned to the edges need to have a value out of ${(0,1), (1,0), (1,1)}$ and have to sum up to an even number.
For some reason I don't manage to assign the flows correctly. Help would be appreciated!
graph-theory
asked Nov 26 at 8:42
Lydia
134
add a comment |
I'm trying to understand the concept of nowhere-zero-flows.
I have this example graph that's supposed to have a nowhere-zero-4-flow (since it has a Hamiltonian cycle).
So by one of the theorems by Tutte, it should also have a nowhere-zero $mathbb{Z}_2 times mathbb{Z}_2$-flow.
If I understood it correctly, the flows assigned to the edges need to have a value out of ${(0,1), (1,0), (1,1)}$ and have to sum up to an even number.
For some reason I don't manage to assign the flows correctly. Help would be appreciated!
graph-theory
asked Nov 26 at 8:42
Lydia
134
Is the linked graph supposed to be directed?
– Servaes
Nov 26 at 8:46
i'm trying to find an orientation for the graph such that it has a nowhere-zero-4-flow
– Lydia
Nov 26 at 8:49
add a comment |
I'm trying to understand the concept of nowhere-zero-flows.
I have this example graph that's supposed to have a nowhere-zero-4-flow (since it has a Hamiltonian cycle).
So by one of the theorems by Tutte, it should also have a nowhere-zero $mathbb{Z}_2 times mathbb{Z}_2$-flow.
If I understood it correctly, the flows assigned to the edges need to have a value out of ${(0,1), (1,0), (1,1)}$ and have to sum up to an even number.
For some reason I don't manage to assign the flows correctly. Help would be appreciated!
graph-theory
asked Nov 26 at 8:42
Lydia
134
I'm trying to understand the concept of nowhere-zero-flows.
I have this example graph that's supposed to have a nowhere-zero-4-flow (since it has a Hamiltonian cycle).
So by one of the theorems by Tutte, it should also have a nowhere-zero $mathbb{Z}_2 times mathbb{Z}_2$-flow.
If I understood it correctly, the flows assigned to the edges need to have a value out of ${(0,1), (1,0), (1,1)}$ and have to sum up to an even number.
For some reason I don't manage to assign the flows correctly. Help would be appreciated!
graph-theory
graph-theory
asked Nov 26 at 8:42
Lydia
134
asked Nov 26 at 8:42
Lydia
134
asked Nov 26 at 8:42
Lydia
134
asked Nov 26 at 8:42
Lydia
134
asked Nov 26 at 8:42
Lydia
134
134
Is the linked graph supposed to be directed?
– Servaes
Nov 26 at 8:46
i'm trying to find an orientation for the graph such that it has a nowhere-zero-4-flow
– Lydia
Nov 26 at 8:49
add a comment |
Is the linked graph supposed to be directed?
– Servaes
Nov 26 at 8:46
i'm trying to find an orientation for the graph such that it has a nowhere-zero-4-flow
– Lydia
Nov 26 at 8:49
Is the linked graph supposed to be directed?
– Servaes
Nov 26 at 8:46
Is the linked graph supposed to be directed?
– Servaes
Nov 26 at 8:46
i'm trying to find an orientation for the graph such that it has a nowhere-zero-4-flow
– Lydia
Nov 26 at 8:49
i'm trying to find an orientation for the graph such that it has a nowhere-zero-4-flow
– Lydia
Nov 26 at 8:49
add a comment |
1 Answer
1
active
oldest
votes
Take an entirely clockwise orientation; label all edges outside the triangle $(1,1)$, and label the edges of the triangle $(0,1)$ and $(1,0)$ accordingly.
There are of course many variations on this solution.
answered Nov 26 at 8:55
Servaes
22.3k33793
oh - i understand what i did wrong all the time now!!! thanks a lot!!
– Lydia
Nov 26 at 8:58
You're welcome, I've added a picture for clarity.
– Servaes
Nov 26 at 9:02
1
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
– Lydia
Nov 26 at 10:26
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
– Lydia
Nov 26 at 10:27
Indeed the direction can be removed, because the group has exponent $2$, i.e. $2g=0$ for all $gin G$, or equivalently $g=-g$ for all $gin G$.
– Servaes
Nov 26 at 11:24
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Take an entirely clockwise orientation; label all edges outside the triangle $(1,1)$, and label the edges of the triangle $(0,1)$ and $(1,0)$ accordingly.
There are of course many variations on this solution.
answered Nov 26 at 8:55
Servaes
22.3k33793
oh - i understand what i did wrong all the time now!!! thanks a lot!!
– Lydia
Nov 26 at 8:58
You're welcome, I've added a picture for clarity.
– Servaes
Nov 26 at 9:02
1
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
– Lydia
Nov 26 at 10:26
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
– Lydia
Nov 26 at 10:27
Indeed the direction can be removed, because the group has exponent $2$, i.e. $2g=0$ for all $gin G$, or equivalently $g=-g$ for all $gin G$.
– Servaes
Nov 26 at 11:24
|
show 2 more comments
Take an entirely clockwise orientation; label all edges outside the triangle $(1,1)$, and label the edges of the triangle $(0,1)$ and $(1,0)$ accordingly.
There are of course many variations on this solution.
answered Nov 26 at 8:55
Servaes
22.3k33793
oh - i understand what i did wrong all the time now!!! thanks a lot!!
– Lydia
Nov 26 at 8:58
You're welcome, I've added a picture for clarity.
– Servaes
Nov 26 at 9:02
1
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
– Lydia
Nov 26 at 10:26
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
– Lydia
Nov 26 at 10:27
Indeed the direction can be removed, because the group has exponent $2$, i.e. $2g=0$ for all $gin G$, or equivalently $g=-g$ for all $gin G$.
– Servaes
Nov 26 at 11:24
|
show 2 more comments
Take an entirely clockwise orientation; label all edges outside the triangle $(1,1)$, and label the edges of the triangle $(0,1)$ and $(1,0)$ accordingly.
There are of course many variations on this solution.
answered Nov 26 at 8:55
Servaes
22.3k33793
Take an entirely clockwise orientation; label all edges outside the triangle $(1,1)$, and label the edges of the triangle $(0,1)$ and $(1,0)$ accordingly.
There are of course many variations on this solution.
answered Nov 26 at 8:55
Servaes
22.3k33793
edited Nov 26 at 9:01
answered Nov 26 at 8:55
Servaes
22.3k33793
answered Nov 26 at 8:55
Servaes
22.3k33793
answered Nov 26 at 8:55
Servaes
22.3k33793
22.3k33793
oh - i understand what i did wrong all the time now!!! thanks a lot!!
– Lydia
Nov 26 at 8:58
You're welcome, I've added a picture for clarity.
– Servaes
Nov 26 at 9:02
1
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
– Lydia
Nov 26 at 10:26
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
– Lydia
Nov 26 at 10:27
Indeed the direction can be removed, because the group has exponent $2$, i.e. $2g=0$ for all $gin G$, or equivalently $g=-g$ for all $gin G$.
– Servaes
Nov 26 at 11:24
|
show 2 more comments
oh - i understand what i did wrong all the time now!!! thanks a lot!!
– Lydia
Nov 26 at 8:58
You're welcome, I've added a picture for clarity.
– Servaes
Nov 26 at 9:02
1
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
– Lydia
Nov 26 at 10:26
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
– Lydia
Nov 26 at 10:27
Indeed the direction can be removed, because the group has exponent $2$, i.e. $2g=0$ for all $gin G$, or equivalently $g=-g$ for all $gin G$.
– Servaes
Nov 26 at 11:24
oh - i understand what i did wrong all the time now!!! thanks a lot!!
– Lydia
Nov 26 at 8:58
oh - i understand what i did wrong all the time now!!! thanks a lot!!
– Lydia
Nov 26 at 8:58
You're welcome, I've added a picture for clarity.
– Servaes
Nov 26 at 9:02
You're welcome, I've added a picture for clarity.
– Servaes
Nov 26 at 9:02
1
1
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
– Lydia
Nov 26 at 10:26
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
– Lydia
Nov 26 at 10:26
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
– Lydia
Nov 26 at 10:27
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
– Lydia
Nov 26 at 10:27
Indeed the direction can be removed, because the group has exponent $2$, i.e. $2g=0$ for all $gin G$, or equivalently $g=-g$ for all $gin G$.
– Servaes
Nov 26 at 11:24
Indeed the direction can be removed, because the group has exponent $2$, i.e. $2g=0$ for all $gin G$, or equivalently $g=-g$ for all $gin G$.
– Servaes
Nov 26 at 11:24
|
show 2 more comments
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Is the linked graph supposed to be directed?
– Servaes
Nov 26 at 8:46
i'm trying to find an orientation for the graph such that it has a nowhere-zero-4-flow
– Lydia
Nov 26 at 8:49