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You are attempting to hang a picture on the wall. There are two nails in the wall. You want to use a string to hang the picture, so the string will be attached in two places to the picture, and wrapped around the nails in some way. Is there a way to hang the picture in such a way that the removal of either one of the nails causes the picture to fall down? What about with three nails? n nails?

If the picture hangs x above the floor then wrap 2*x+epsilon of string around each nail. If we neglect friction the picture will touch the floor after removing one nail. I guess it's not the solution you were looking for...

a is the path clockwise around the nail Aa' is the path counterclockwise around the nail Ab is the path clockwise around the nail Bb' is the path counterclockwise around the nail Be is elementary path hanging out in the airnote that aa' = a'a = ebb' = b'b = eyou can represent a way of hanging by a string formed of a and b.if you remove the nail A then you will have a=a'=e (similar for B)so if you hang with a combination like aba'b' and remove the nail A you get bb' = efor the case of 3 nailsput s2 = aba'b'c and c' is as above for the nail Cyou have to hang using the combination s3 = (s2)c(s2)'c'for n nail you can getsn+1 = (sn)n(sn)'n'for further details see homotopy groups

nice solution arkol. i haven't seen anybody solve this at the level of three nails without rephrasing it in terms of homotopy groups. i think its a great demonstration of algebraic topology.