Solve the Infamous Puzzle: One Continuous Line Challenge | Step-by-Step Guide

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The discussion revolves around a puzzle that requires drawing a continuous line through every segment without crossing or retracing any part. Participants highlight that it is impossible to solve this puzzle on a standard plane due to graph theory principles. A previous thread provided a proof supporting this claim, emphasizing that while the puzzle cannot be completed on a plane, it can be solved on a Mobius strip. The Mobius strip's unique properties, such as its Euler characteristic of 0, differentiate it from a standard plane, which has an Euler characteristic of 1. The conversation also touches on the relationship between the Mobius strip and other surfaces, like cylinders and Klein bottles, questioning how dimensionality affects these properties. Overall, the consensus is that the puzzle is unsolvable in a traditional sense but can be approached differently on non-planar surfaces.
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Hi everyone. I've been stumped trying to figure this one out for the past week, I think some may have seen it before. It looks like this:

*EDIT* The picture I attempted to draw didn't post right, I'll have one up soon.

*EDIT 2* Here we go:
http://members.lycos.co.uk/evilx22/hpbimg/Untitled-1%20copy.jpg

What you have to do is draw one continuous line through every line segment on the puzzle, without going through one twice and the line can never cross itself. I'm hoping some have seen this before, so any help would be appreciated.
 
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Its impossible... you can prove it with graph theory.

This was posted previously and there was a link to a site that had a general proof that it can't be done under certain conditions and this puzzle fit the conditions...
 
You could draw this on a Mobius strip.
That would fix the problem.
 
There's an infuriatingly long thread dedicated to this problem here

The short answer is that it is impossible to do this, if the picture is drawn on a plane, unless you cheat by going through a corner or using a giant marker, or some such thing. The proof is found in post #5 (by NateTG) in the above linked thread.
 
Gokul43201 said:
if the picture is drawn on a plane.
Isn't the Mobius surface considered to be a plane?
A toroid or sphere, two solutions in the thread you referred to, would not be.
 
NoTime said:
Isn't the Mobius surface considered to be a plane?

No, a mobius strip is not homeomorhic to a (projective) plane.

(Additional Info : the mobius strip has an Euler Characteristic = 0, while this is 1 for a plane)
 
Gokul43201 said:
No, a mobius strip is not homeomorhic to a (projective) plane.

(Additional Info : the mobius strip has an Euler Characteristic = 0, while this is 1 for a plane)
Interesting.
So it would be related to a cylinder somehow?
Does this remain true if the dimensionality changes?
I vaguely recall something special about this as well as klein bottle for N other than 3.
 

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