Example of a nowhere-zero 4-flow












1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















0












$begingroup$

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.



enter image description here



There are of course many variations on this solution.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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.



enter image description here



There are of course many variations on this solution.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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.



enter image description here



There are of course many variations on this solution.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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.



enter image description here



There are of course many variations on this solution.






share|cite|improve this answer











$endgroup$



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.



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 '18 at 9:01

























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


















  • $begingroup$
    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


















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