Question on Vertex Labeling (Related to Lucky Labeling of Graphs)
$begingroup$
Suppose that for any bipartite planar graph $G=(V,E)$, we can find a vertex labeling $c:Vto {1,2,3}$ such that for any two adjacent vertices $u$ and $w$:
$$c(u)-sum_{vin N(u)}c(v)neq c(w)-sum_{vin N(w)}c(v)$$ where $N(v)$ denotes the neighborhood of the vertex $vin V$. My question is: is it true that there also must exist a labeling of $G$ with labels ${1,2,3}$ such that for any two adjacent vertices $u$ and $w$, we have:
$$sum_{vin N(u)}c(v)neq sum_{vin N(w)}c(v)?$$
It is possible for two adjacent vertices to satisfy the first equation, but not the second in some labeling $c$. My idea was to modify the initial labeling in a way that makes the second inequality hold. That is, for some adjacent vertices $u$ and $w$ satisfying the first equation, if $sum_{vin N(u)}c(v)=sum_{vin N(w)}c(v)$ holds then $c(u)neq c(w)$. Without loss of generality, we may assume that $c(u)<c(w)$. Then, change the label of $u$ to $c(w)$ thus obtaining the new labeling $c'$. However, this may affect the relationship of $u$ with its other neighbors and my attempts to analyze those weren't successful. I would really appreciate some help.
This question comes from reading the paper of Lason where he seems to claim that the second result is a consequence of the first (if I understand correctly what he means by the "special case").
combinatorics graph-theory algebraic-combinatorics
asked Dec 10 '18 at 5:55
Yulia AlexandrYulia Alexandr
646
$endgroup$
add a comment |
$begingroup$
Suppose that for any bipartite planar graph $G=(V,E)$, we can find a vertex labeling $c:Vto {1,2,3}$ such that for any two adjacent vertices $u$ and $w$:
$$c(u)-sum_{vin N(u)}c(v)neq c(w)-sum_{vin N(w)}c(v)$$ where $N(v)$ denotes the neighborhood of the vertex $vin V$. My question is: is it true that there also must exist a labeling of $G$ with labels ${1,2,3}$ such that for any two adjacent vertices $u$ and $w$, we have:
$$sum_{vin N(u)}c(v)neq sum_{vin N(w)}c(v)?$$
It is possible for two adjacent vertices to satisfy the first equation, but not the second in some labeling $c$. My idea was to modify the initial labeling in a way that makes the second inequality hold. That is, for some adjacent vertices $u$ and $w$ satisfying the first equation, if $sum_{vin N(u)}c(v)=sum_{vin N(w)}c(v)$ holds then $c(u)neq c(w)$. Without loss of generality, we may assume that $c(u)<c(w)$. Then, change the label of $u$ to $c(w)$ thus obtaining the new labeling $c'$. However, this may affect the relationship of $u$ with its other neighbors and my attempts to analyze those weren't successful. I would really appreciate some help.
This question comes from reading the paper of Lason where he seems to claim that the second result is a consequence of the first (if I understand correctly what he means by the "special case").
combinatorics graph-theory algebraic-combinatorics
asked Dec 10 '18 at 5:55
Yulia AlexandrYulia Alexandr
646
$endgroup$
add a comment |
$begingroup$
Suppose that for any bipartite planar graph $G=(V,E)$, we can find a vertex labeling $c:Vto {1,2,3}$ such that for any two adjacent vertices $u$ and $w$:
$$c(u)-sum_{vin N(u)}c(v)neq c(w)-sum_{vin N(w)}c(v)$$ where $N(v)$ denotes the neighborhood of the vertex $vin V$. My question is: is it true that there also must exist a labeling of $G$ with labels ${1,2,3}$ such that for any two adjacent vertices $u$ and $w$, we have:
$$sum_{vin N(u)}c(v)neq sum_{vin N(w)}c(v)?$$
It is possible for two adjacent vertices to satisfy the first equation, but not the second in some labeling $c$. My idea was to modify the initial labeling in a way that makes the second inequality hold. That is, for some adjacent vertices $u$ and $w$ satisfying the first equation, if $sum_{vin N(u)}c(v)=sum_{vin N(w)}c(v)$ holds then $c(u)neq c(w)$. Without loss of generality, we may assume that $c(u)<c(w)$. Then, change the label of $u$ to $c(w)$ thus obtaining the new labeling $c'$. However, this may affect the relationship of $u$ with its other neighbors and my attempts to analyze those weren't successful. I would really appreciate some help.
This question comes from reading the paper of Lason where he seems to claim that the second result is a consequence of the first (if I understand correctly what he means by the "special case").
combinatorics graph-theory algebraic-combinatorics
asked Dec 10 '18 at 5:55
Yulia AlexandrYulia Alexandr
646
$endgroup$
Suppose that for any bipartite planar graph $G=(V,E)$, we can find a vertex labeling $c:Vto {1,2,3}$ such that for any two adjacent vertices $u$ and $w$:
$$c(u)-sum_{vin N(u)}c(v)neq c(w)-sum_{vin N(w)}c(v)$$ where $N(v)$ denotes the neighborhood of the vertex $vin V$. My question is: is it true that there also must exist a labeling of $G$ with labels ${1,2,3}$ such that for any two adjacent vertices $u$ and $w$, we have:
$$sum_{vin N(u)}c(v)neq sum_{vin N(w)}c(v)?$$
It is possible for two adjacent vertices to satisfy the first equation, but not the second in some labeling $c$. My idea was to modify the initial labeling in a way that makes the second inequality hold. That is, for some adjacent vertices $u$ and $w$ satisfying the first equation, if $sum_{vin N(u)}c(v)=sum_{vin N(w)}c(v)$ holds then $c(u)neq c(w)$. Without loss of generality, we may assume that $c(u)<c(w)$. Then, change the label of $u$ to $c(w)$ thus obtaining the new labeling $c'$. However, this may affect the relationship of $u$ with its other neighbors and my attempts to analyze those weren't successful. I would really appreciate some help.
This question comes from reading the paper of Lason where he seems to claim that the second result is a consequence of the first (if I understand correctly what he means by the "special case").
combinatorics graph-theory algebraic-combinatorics
combinatorics graph-theory algebraic-combinatorics
asked Dec 10 '18 at 5:55
Yulia AlexandrYulia Alexandr
646
asked Dec 10 '18 at 5:55
Yulia AlexandrYulia Alexandr
646
edited Dec 10 '18 at 6:36
Yulia Alexandr
asked Dec 10 '18 at 5:55
Yulia AlexandrYulia Alexandr
646
asked Dec 10 '18 at 5:55
Yulia AlexandrYulia Alexandr
646
asked Dec 10 '18 at 5:55
Yulia AlexandrYulia Alexandr
646
646
add a comment |
add a comment |
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