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May 28, 2019 at 4:00 comment added ppgdev @Moo-Juice, you cannot solve this puzzle on the surface of a sphere. If you were able to do it, you could then select a point on the sphere not covered by your drawing, use the point to make a hole in the sphere and then continuously transform the sphere into a plane. Now you sphere solution would turn into a plane solution. But we already know that plane solution does not exist. Contradiction. Q.E.D.
May 27, 2019 at 11:40 comment added Moo-Juice @Gnudiff this puzzle is entirely solveable when drawn on, say, an orange :)
May 27, 2019 at 6:13 vote accept Navid2132
May 26, 2019 at 14:56 comment added Roman Odaisky All of the above is well summarized in this 3Blue1Brown video: youtube.com/watch?v=VvCytJvd4H0
May 25, 2019 at 23:01 comment added Euro Micelli @Gnudiff, not a cylinder. You need a torus like this one. (no financial interest)
May 25, 2019 at 6:20 comment added Gnudiff Can it be solved if you join the sides of the paper in a cylinder?
May 25, 2019 at 3:33 comment added user2357112 Proving that $K_{3,3}$ is nonplanar isn't that hard. Even without the Jordan curve theorem, the Wikipedia page on the three utilities problem includes a short proof based on Euler's formula for planar graphs.
May 25, 2019 at 3:17 comment added user2357112 Kuratowski's theorem is way stronger than the statement that this puzzle is impossible, and way harder to prove than proving that this puzzle is impossible. Kuratowski's theorem is a characterization of all finite graphs, while this puzzle only requires proving one specific graph nonplanar.
May 24, 2019 at 12:46 comment added hexomino I would say it is not too difficult to prove once you can assume the Jordan curve theorem. I've essentially outlined the argument in my answer which I think is what they are referring to here
May 24, 2019 at 12:23 history answered Glorfindel CC BY-SA 4.0