# how do you do this?

• Apr 30th 2006, 05:14 PM
nertil1
how do you do this?
I have been trying to solve this puzzle for a while, and I'm starting to think it's impossible. The question is to draw the picture below without lifting your pen off the paper and without crossing the same lines, anybody have any ideas?

http://img166.imageshack.us/img166/8416/untitled9ty.png
• Apr 30th 2006, 06:35 PM
ThePerfectHacker
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.
• Apr 30th 2006, 09:35 PM
CaptainBlack
Error corrected version
Quote:

Originally Posted by nertil1
I have been trying to solve this puzzle for a while, and I'm starting to think it's impossible. The question is to draw the picture below without lifting your pen off the paper and without crossing the same lines, anybody have any ideas?

http://img166.imageshack.us/img166/8416/untitled9ty.png

Consider a node of this figure. If it is not a starting or ending node, everytime
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
• Apr 30th 2006, 10:01 PM
rgep
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.
• Apr 30th 2006, 10:15 PM
CaptainBlack
Quote:

Originally Posted by rgep
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.

Ooops, The damn coffee is not working yet! :mad:
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.