Magic labeling of the octahedron graph

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?

• 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. – Ross Millikan Jan 29 at 15:05
• I have added the link now. – Geek Jan 29 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.