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- April 30th 2006, 06:14 PM #1

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## how do you do this?

- April 30th 2006, 07:35 PM #2

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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.

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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.

- April 30th 2006, 10:35 PM #3

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## Error corrected version

Originally Posted by**nertil1**

you draw an arc that enters the node, you must also draw an arc leaving a

node.

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.

RonL

- April 30th 2006, 11:01 PM #4
Four nodes of odd degree actually (the number of nodes of odd degree must be even). But that is still sufficient to prove the graph is not Eulerian: see this Wikipedia article.

- April 30th 2006, 11:15 PM #5

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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.

RonL

- May 1st 2006, 05:34 AM #6

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