I did not try to do the problem. But I was just thinking. Perhaps, it is truly impossible. For example, I can create a similar problem which is impossible, likewise here.
I was attempting to prove the impossibility of a solution by considering this to be a di-graph and then to demonstrate you cannot have an generated abelian group of two elements having order 8??? rgep this is what you know best.
Consider a node of this figure. If it is not a starting or ending node, everytimeOriginally Posted by nertil1
you draw an arc that enters the node, you must also draw an arc leaving a
Therefore every node other than the start and stop nodes must have an even
degree (number of arcs meeting at the node), and at most two nodes (the
start and stop nodes) can have an odd degree.
Now the figure has four nodes of odd degree, and so cannot be drawn without
lifting the pen.
Ooops, The damn coffee is not working yet!Originally Posted by rgep
Now I can't count- though I suppose I was not paying much attention
to the counting as there were obviously too many nodes of odd degree.
While walking the dog after replying I was thinking about this problem
(It was clearly related to Euler's Königsberg bridges problem) I recalled
that this is in Rouse Ball, so I thought I would look it up when we got back
and give it as a reference: Rouse Ball W. W, Mathematical
Recreations, Chapter 9, "Unicursal Problems".
Its nice to be able to refer to such a classic.