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 $\frac{2\cdot\sum_{i=1}^{12}i}{6}$.

Are there other ways of finding such a labeling?

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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 '19 at 15:05
  • $\begingroup$ I have added the link now. $\endgroup$ – Geek Jan 29 '19 at 15:19

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 $2\cdot 78 = 156$. 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.


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