Example of a nowhere-zero 4-flow












1














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







share|cite|improve this question






















  • 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
















1














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







share|cite|improve this question






















  • 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














1












1








1







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










1 Answer
1






active

oldest

votes


















0














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.



enter image description here



There are of course many variations on this solution.






share|cite|improve this answer























  • 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











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














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.



enter image description here



There are of course many variations on this solution.






share|cite|improve this answer























  • 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
















0














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.



enter image description here



There are of course many variations on this solution.






share|cite|improve this answer























  • 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














0












0








0






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.



enter image description here



There are of course many variations on this solution.






share|cite|improve this answer














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.



enter image description here



There are of course many variations on this solution.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 26 at 9:01

























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


















  • 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


















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