Mixing
Magic labeling of the octahedron graph
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A magic labeling of a graph $G$ with $q$ edges is an edge labeling by the numbers $1, 2, 3, ldots, q $ so that the sum of the labels of all the edges incident with any vertex is the same.
We need to find a magic labeling of the octahedron graph. I found this problem in Pearls in Graph Theory by Hartfield and Ringel in the chapter Labeling Graphs.
I tried to find such a labeling by solving a system of linear equations with some conditions since we know that each vertex will have a total of $frac{2cdotsum_{i=1}^{12}i}{6}$.
Are there other ways of finding such a labeling?
combinatorics discrete-mathematics graph-theory
asked Jan 29 at 14:53
GeekGeek
113
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add a comment |
$begingroup$
A magic labeling of a graph $G$ with $q$ edges is an edge labeling by the numbers $1, 2, 3, ldots, q $ so that the sum of the labels of all the edges incident with any vertex is the same.
We need to find a magic labeling of the octahedron graph. I found this problem in Pearls in Graph Theory by Hartfield and Ringel in the chapter Labeling Graphs.
I tried to find such a labeling by solving a system of linear equations with some conditions since we know that each vertex will have a total of $frac{2cdotsum_{i=1}^{12}i}{6}$.
Are there other ways of finding such a labeling?
combinatorics discrete-mathematics graph-theory
asked Jan 29 at 14:53
GeekGeek
113
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1
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I couldn't find the other question by clicking on your user name. Is it under another account? There are instructions on the FAQ on how to merge accounts.
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– Ross Millikan
Jan 29 at 15:05
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I have added the link now.
$endgroup$
– Geek
Jan 29 at 15:19
add a comment |
$begingroup$
A magic labeling of a graph $G$ with $q$ edges is an edge labeling by the numbers $1, 2, 3, ldots, q $ so that the sum of the labels of all the edges incident with any vertex is the same.
We need to find a magic labeling of the octahedron graph. I found this problem in Pearls in Graph Theory by Hartfield and Ringel in the chapter Labeling Graphs.
I tried to find such a labeling by solving a system of linear equations with some conditions since we know that each vertex will have a total of $frac{2cdotsum_{i=1}^{12}i}{6}$.
Are there other ways of finding such a labeling?
combinatorics discrete-mathematics graph-theory
asked Jan 29 at 14:53
GeekGeek
113
$endgroup$
A magic labeling of a graph $G$ with $q$ edges is an edge labeling by the numbers $1, 2, 3, ldots, q $ so that the sum of the labels of all the edges incident with any vertex is the same.
We need to find a magic labeling of the octahedron graph. I found this problem in Pearls in Graph Theory by Hartfield and Ringel in the chapter Labeling Graphs.
I tried to find such a labeling by solving a system of linear equations with some conditions since we know that each vertex will have a total of $frac{2cdotsum_{i=1}^{12}i}{6}$.
Are there other ways of finding such a labeling?
combinatorics discrete-mathematics graph-theory
combinatorics discrete-mathematics graph-theory
asked Jan 29 at 14:53
GeekGeek
113
asked Jan 29 at 14:53
GeekGeek
113
edited Jan 29 at 16:11
Geek
asked Jan 29 at 14:53
GeekGeek
113
asked Jan 29 at 14:53
GeekGeek
113
asked Jan 29 at 14:53
GeekGeek
113
113
1
$begingroup$
I couldn't find the other question by clicking on your user name. Is it under another account? There are instructions on the FAQ on how to merge accounts.
$endgroup$
– Ross Millikan
Jan 29 at 15:05
$begingroup$
I have added the link now.
$endgroup$
– Geek
Jan 29 at 15:19
add a comment |
1
$begingroup$
I couldn't find the other question by clicking on your user name. Is it under another account? There are instructions on the FAQ on how to merge accounts.
$endgroup$
– Ross Millikan
Jan 29 at 15:05
$begingroup$
I have added the link now.
$endgroup$
– Geek
Jan 29 at 15:19
1
1
$begingroup$
I couldn't find the other question by clicking on your user name. Is it under another account? There are instructions on the FAQ on how to merge accounts.
$endgroup$
– Ross Millikan
Jan 29 at 15:05
$begingroup$
I couldn't find the other question by clicking on your user name. Is it under another account? There are instructions on the FAQ on how to merge accounts.
$endgroup$
– Ross Millikan
Jan 29 at 15:05
$begingroup$
I have added the link now.
$endgroup$
– Geek
Jan 29 at 15:19
$begingroup$
I have added the link now.
$endgroup$
– Geek
Jan 29 at 15:19
add a comment |
1 Answer
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oldest
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A possible solution would be:
$x_1 = 1$
$x_2 = 2$
$x_3 = 9$
$x_4 = 6$
$x_5 = 11$
$x_6 = 12$
$x_7 = 7$
$x_8 = 8$
$x_9 = 10$
$x_{10} = 4$
$x_{11} = 3$
$x_{12} = 5$
One way to make things easier is to compute which has to be the sum of the edges incident to each vertex. In this case, the sum of all edges $1+2+3+4+5+6+7+8+9+10+11+12 = 78$.
If we add the sum of the edges incident to each vertex, we will be counting each edge twice, meaning that we will get $2cdot 78 = 156$. Since there are 6 vertices, the sum of the edges incident to each vertex is $frac{156}{6}=26$. This may be helpful when trying to solve this kind of problems.
answered Jan 29 at 15:45
maxbpmaxbp
1467
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
A possible solution would be:
$x_1 = 1$
$x_2 = 2$
$x_3 = 9$
$x_4 = 6$
$x_5 = 11$
$x_6 = 12$
$x_7 = 7$
$x_8 = 8$
$x_9 = 10$
$x_{10} = 4$
$x_{11} = 3$
$x_{12} = 5$
One way to make things easier is to compute which has to be the sum of the edges incident to each vertex. In this case, the sum of all edges $1+2+3+4+5+6+7+8+9+10+11+12 = 78$.
If we add the sum of the edges incident to each vertex, we will be counting each edge twice, meaning that we will get $2cdot 78 = 156$. Since there are 6 vertices, the sum of the edges incident to each vertex is $frac{156}{6}=26$. This may be helpful when trying to solve this kind of problems.
answered Jan 29 at 15:45
maxbpmaxbp
1467
$endgroup$
add a comment |
$begingroup$
A possible solution would be:
$x_1 = 1$
$x_2 = 2$
$x_3 = 9$
$x_4 = 6$
$x_5 = 11$
$x_6 = 12$
$x_7 = 7$
$x_8 = 8$
$x_9 = 10$
$x_{10} = 4$
$x_{11} = 3$
$x_{12} = 5$
One way to make things easier is to compute which has to be the sum of the edges incident to each vertex. In this case, the sum of all edges $1+2+3+4+5+6+7+8+9+10+11+12 = 78$.
If we add the sum of the edges incident to each vertex, we will be counting each edge twice, meaning that we will get $2cdot 78 = 156$. Since there are 6 vertices, the sum of the edges incident to each vertex is $frac{156}{6}=26$. This may be helpful when trying to solve this kind of problems.
answered Jan 29 at 15:45
maxbpmaxbp
1467
$endgroup$
add a comment |
$begingroup$
A possible solution would be:
$x_1 = 1$
$x_2 = 2$
$x_3 = 9$
$x_4 = 6$
$x_5 = 11$
$x_6 = 12$
$x_7 = 7$
$x_8 = 8$
$x_9 = 10$
$x_{10} = 4$
$x_{11} = 3$
$x_{12} = 5$
One way to make things easier is to compute which has to be the sum of the edges incident to each vertex. In this case, the sum of all edges $1+2+3+4+5+6+7+8+9+10+11+12 = 78$.
If we add the sum of the edges incident to each vertex, we will be counting each edge twice, meaning that we will get $2cdot 78 = 156$. Since there are 6 vertices, the sum of the edges incident to each vertex is $frac{156}{6}=26$. This may be helpful when trying to solve this kind of problems.
answered Jan 29 at 15:45
maxbpmaxbp
1467
$endgroup$
A possible solution would be:
$x_1 = 1$
$x_2 = 2$
$x_3 = 9$
$x_4 = 6$
$x_5 = 11$
$x_6 = 12$
$x_7 = 7$
$x_8 = 8$
$x_9 = 10$
$x_{10} = 4$
$x_{11} = 3$
$x_{12} = 5$
One way to make things easier is to compute which has to be the sum of the edges incident to each vertex. In this case, the sum of all edges $1+2+3+4+5+6+7+8+9+10+11+12 = 78$.
If we add the sum of the edges incident to each vertex, we will be counting each edge twice, meaning that we will get $2cdot 78 = 156$. Since there are 6 vertices, the sum of the edges incident to each vertex is $frac{156}{6}=26$. This may be helpful when trying to solve this kind of problems.
answered Jan 29 at 15:45
maxbpmaxbp
1467
answered Jan 29 at 15:45
maxbpmaxbp
1467
answered Jan 29 at 15:45
maxbpmaxbp
1467
answered Jan 29 at 15:45
maxbpmaxbp
1467
1467
add a comment |
add a comment |
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$begingroup$
I couldn't find the other question by clicking on your user name. Is it under another account? There are instructions on the FAQ on how to merge accounts.
$endgroup$
– Ross Millikan
Jan 29 at 15:05
$begingroup$
I have added the link now.
$endgroup$
– Geek
Jan 29 at 15:19