Example of a nowhere-zero 4-flow
$begingroup$
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 '18 at 8:42
LydiaLydia
134
$endgroup$
add a comment |
$begingroup$
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 '18 at 8:42
LydiaLydia
134
$endgroup$
$begingroup$
Is the linked graph supposed to be directed?
$endgroup$
– Servaes
Nov 26 '18 at 8:46
$begingroup$
i'm trying to find an orientation for the graph such that it has a nowhere-zero-4-flow
$endgroup$
– Lydia
Nov 26 '18 at 8:49
add a comment |
$begingroup$
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 '18 at 8:42
LydiaLydia
134
$endgroup$
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 '18 at 8:42
LydiaLydia
134
asked Nov 26 '18 at 8:42
LydiaLydia
134
asked Nov 26 '18 at 8:42
LydiaLydia
134
asked Nov 26 '18 at 8:42
LydiaLydia
134
asked Nov 26 '18 at 8:42
LydiaLydia
134
134
$begingroup$
Is the linked graph supposed to be directed?
$endgroup$
– Servaes
Nov 26 '18 at 8:46
$begingroup$
i'm trying to find an orientation for the graph such that it has a nowhere-zero-4-flow
$endgroup$
– Lydia
Nov 26 '18 at 8:49
add a comment |
$begingroup$
Is the linked graph supposed to be directed?
$endgroup$
– Servaes
Nov 26 '18 at 8:46
$begingroup$
i'm trying to find an orientation for the graph such that it has a nowhere-zero-4-flow
$endgroup$
– Lydia
Nov 26 '18 at 8:49
$begingroup$
Is the linked graph supposed to be directed?
$endgroup$
– Servaes
Nov 26 '18 at 8:46
$begingroup$
Is the linked graph supposed to be directed?
$endgroup$
– Servaes
Nov 26 '18 at 8:46
$begingroup$
i'm trying to find an orientation for the graph such that it has a nowhere-zero-4-flow
$endgroup$
– Lydia
Nov 26 '18 at 8:49
$begingroup$
i'm trying to find an orientation for the graph such that it has a nowhere-zero-4-flow
$endgroup$
– Lydia
Nov 26 '18 at 8:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 '18 at 8:55
ServaesServaes
23.1k33893
$endgroup$
$begingroup$
oh - i understand what i did wrong all the time now!!! thanks a lot!!
$endgroup$
– Lydia
Nov 26 '18 at 8:58
$begingroup$
You're welcome, I've added a picture for clarity.
$endgroup$
– Servaes
Nov 26 '18 at 9:02
1
$begingroup$
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
$endgroup$
– Lydia
Nov 26 '18 at 10:26
$begingroup$
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
$endgroup$
– Lydia
Nov 26 '18 at 10:27
$begingroup$
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 $gin G$.
$endgroup$
– Servaes
Nov 26 '18 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
$begingroup$
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 '18 at 8:55
ServaesServaes
23.1k33893
$endgroup$
$begingroup$
oh - i understand what i did wrong all the time now!!! thanks a lot!!
$endgroup$
– Lydia
Nov 26 '18 at 8:58
$begingroup$
You're welcome, I've added a picture for clarity.
$endgroup$
– Servaes
Nov 26 '18 at 9:02
1
$begingroup$
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
$endgroup$
– Lydia
Nov 26 '18 at 10:26
$begingroup$
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
$endgroup$
– Lydia
Nov 26 '18 at 10:27
$begingroup$
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 $gin G$.
$endgroup$
– Servaes
Nov 26 '18 at 11:24
|
show 2 more comments
$begingroup$
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 '18 at 8:55
ServaesServaes
23.1k33893
$endgroup$
$begingroup$
oh - i understand what i did wrong all the time now!!! thanks a lot!!
$endgroup$
– Lydia
Nov 26 '18 at 8:58
$begingroup$
You're welcome, I've added a picture for clarity.
$endgroup$
– Servaes
Nov 26 '18 at 9:02
1
$begingroup$
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
$endgroup$
– Lydia
Nov 26 '18 at 10:26
$begingroup$
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
$endgroup$
– Lydia
Nov 26 '18 at 10:27
$begingroup$
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 $gin G$.
$endgroup$
– Servaes
Nov 26 '18 at 11:24
|
show 2 more comments
$begingroup$
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 '18 at 8:55
ServaesServaes
23.1k33893
$endgroup$
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 '18 at 8:55
ServaesServaes
23.1k33893
edited Nov 26 '18 at 9:01
answered Nov 26 '18 at 8:55
ServaesServaes
23.1k33893
answered Nov 26 '18 at 8:55
ServaesServaes
23.1k33893
answered Nov 26 '18 at 8:55
ServaesServaes
23.1k33893
23.1k33893
$begingroup$
oh - i understand what i did wrong all the time now!!! thanks a lot!!
$endgroup$
– Lydia
Nov 26 '18 at 8:58
$begingroup$
You're welcome, I've added a picture for clarity.
$endgroup$
– Servaes
Nov 26 '18 at 9:02
1
$begingroup$
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
$endgroup$
– Lydia
Nov 26 '18 at 10:26
$begingroup$
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
$endgroup$
– Lydia
Nov 26 '18 at 10:27
$begingroup$
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 $gin G$.
$endgroup$
– Servaes
Nov 26 '18 at 11:24
|
show 2 more comments
$begingroup$
oh - i understand what i did wrong all the time now!!! thanks a lot!!
$endgroup$
– Lydia
Nov 26 '18 at 8:58
$begingroup$
You're welcome, I've added a picture for clarity.
$endgroup$
– Servaes
Nov 26 '18 at 9:02
1
$begingroup$
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
$endgroup$
– Lydia
Nov 26 '18 at 10:26
$begingroup$
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
$endgroup$
– Lydia
Nov 26 '18 at 10:27
$begingroup$
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 $gin G$.
$endgroup$
– Servaes
Nov 26 '18 at 11:24
$begingroup$
oh - i understand what i did wrong all the time now!!! thanks a lot!!
$endgroup$
– Lydia
Nov 26 '18 at 8:58
$begingroup$
oh - i understand what i did wrong all the time now!!! thanks a lot!!
$endgroup$
– Lydia
Nov 26 '18 at 8:58
$begingroup$
You're welcome, I've added a picture for clarity.
$endgroup$
– Servaes
Nov 26 '18 at 9:02
$begingroup$
You're welcome, I've added a picture for clarity.
$endgroup$
– Servaes
Nov 26 '18 at 9:02
1
1
$begingroup$
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
$endgroup$
– Lydia
Nov 26 '18 at 10:26
$begingroup$
the direction can also be removed because it is in $mathbb{Z}_2$ can't it?
$endgroup$
– Lydia
Nov 26 '18 at 10:26
$begingroup$
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
$endgroup$
– Lydia
Nov 26 '18 at 10:27
$begingroup$
is there a way to use the Hamiltonian cycle to construct the nowherezero-$mathbb{Z}_2 times mathbb{Z}_2$-flow?
$endgroup$
– Lydia
Nov 26 '18 at 10:27
$begingroup$
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 $gin G$.
$endgroup$
– Servaes
Nov 26 '18 at 11:24
$begingroup$
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 $gin G$.
$endgroup$
– Servaes
Nov 26 '18 at 11:24
|
show 2 more comments
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$begingroup$
Is the linked graph supposed to be directed?
$endgroup$
– Servaes
Nov 26 '18 at 8:46
$begingroup$
i'm trying to find an orientation for the graph such that it has a nowhere-zero-4-flow
$endgroup$
– Lydia
Nov 26 '18 at 8:49