1
$\begingroup$

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

Are there other ways of finding such a labeling?

enter image description here

$\endgroup$
  • 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
2
$\begingroup$

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 $2\cdot 78 = 156$. 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.