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This question maybe ridiculous but I always found it interesting... Here it is : (I cannot put image so I put you the link of the pictures) When I was in school I used to draw houses when I was bored :

http://www.imagup.com/data/1174491066.html

You can draw it without lifting the pen. But then I tried to draw two houses side-to-side :

http://www.imagup.com/data/1174491088.html

Then I realized you cannot actually do it without lifting the pen. I tried any ways I could. You end up all the time with one line missing.

That's why I was wondering whether there existed a proof that I cannot actually do it.

I heard that we have to count the total number of lines and the number of intersections and that this number tells us something but I am not sure at all.

I am a first year maths student so I haven't studied graph theory yet !

Thanks !

Martin Sleziak
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ALM
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2 Answers2

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This sounds like a great excuse to study (just a little) graph theory to me! Check out the conclusions of Euler in this famous problem. It should tell you all you need to know (and more). In particular, the second figure has $4$ nodes of odd degree (that is, an odd number of segments having that common endpoint)--the outer bottom corners and the two nodes where the "houses" meet--so it cannot be traced without lifting the pen or retracing at least one segment of the path.

rschwieb
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Cameron Buie
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    Perfect, thanks ! I'll go and study a bit of graph theory, it sounds interesting ! – ALM Feb 02 '13 at 17:43
  • The first one also has two nodes of odd degrees(bottom corners), but still it is possible to draw that image without lifting the pen. "An undirected graph has an Eulerian trail if and only if at most two vertices have odd degree" – Sanmoy Mar 14 '15 at 19:42
  • @Sanmoy: Correct. – Cameron Buie Mar 15 '15 at 04:14
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Take a vertex with three edges. Assume you do NOT start there. Then there is a first time you will reach this vertex, through one of the possible paths leading to it. It won’t be the end of your drawing because two other paths are not yet traversed. Thus, you will leave this vertex, through a second adjacent path. As a result, when you traverse the third path some time, you won’t be able to leave this vertex because no untraveled paths will be available. In other words, your drawing will have to end here.

So, we showed that if you don’t start at a vertex (with odd number of edges), you will have to end there.

This is similarly true about vertices with 5 edges. Thus, if any graph has more than two vertices with odd degrees, you won't be able to draw it.

Behnam Esmayli
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