Solution to Problem 21: Hexasperation

The rule about no sharp corners has a whole mess of corollaries, which are at the heart of every one of the assumptions that may look at first like a leap of logic.

	If either of the A's were in the path, then B would have to be as well. This would violate the 1, so neither of the A's are on the path. Therefore, only the B is on the path, and both of the C's must be its neighbors. If D were not in the path, then neither of the E's could be either, but this would violate the 4. Therefore, both the D and the E's are in the path. The two F's are either both on or off the path. To satisfy the 5, they must both be on. If either of the G's were on the path, one of the H's would be as well, which would violate the 1. Therefore, neither of the G's are on the path. Whichever of the H's is, the path must go through J. Obviously, all of the K's are on the path, but we're not going to put them in quite yet. If L were on the path, then M would have to be too. To satisfy the 2 on the right, L's other neighbor would have to be N. This satifies all of the 2's around the L, so none of the P's are on the path. Therefore, none of the Q's would be on the path since there would be no exit for that path. There is no way to legally satisfy the 3 by the Q's, so our assumption that L was on the path was erroneous.
	The A's are either both on or both off the path. If they were off, then the B's would be on. The C's would have to be on as well, and the path would continue through the D's. This satisfies the 3, so the E's are not on the path. The F cannot be on the path either, but this doesn't satisfy the 5. Therefore, the A's are on the path and the B's are not. The path continues to the G and to the path already laid out at the bottom, so the H's are not on the path. The 4 would not be satisfied if J were on the path, so K is on the path instead. L is not on the path.
	Assume that A is not on the path. Then B and C are on the path to satisfy the 5, and the D's are to continue the path to the left of the A. The path through B must continue to E, which satisfies the 3 and forces the two F's to be off the path. The path from E must continue through G, H, and J. The only way to legally satisfy the 4 is to have L on the path, which connects to B and M. But C now has no legal way to be on the path. Therefore, A is on the path. The D's cannot be on the path, so the path from the bottom must continue through the F's. This satisfies the 3, so E is not on the path and B and G are. This satisfies the 2 below the F and G, so H is not in the path and N is. The P's are either both in the path or neither are. If they are, then the path must continue through Q. If neither are, then Q is in the path to satisfy the 4. In either case, Q is in the path.
	If A were on the path, then B would not be since the 4 would be satisfied. The paths would continue through the cells marked C, which would satisfy the 3 and make D not be on the path. But now there is no way for the path around the 5 to not be a cycle without overloading the 1 above it. So A is not on the path, and B must be. The path continues through D, so C cannot be on the path. Therefore, E is on the path to complete the 5. The path through E continues to F and the two previously marked cells. G cannot be in the path, so H is. The only way to satisfy the 2 legally is for the path to continue through J and for K to not be on the path. If L were on the path, then M would be too and N wouldn't be. The paths would have to continue through the P's, but this overloads the 3. Therefore, L is not on the path, and M and N are. The three Q's are either all on the path or all off. They cannot be on, because they would satisfy the 3 but would not be able to connect to L or R. So the three Q's are off the path and the other three neighbors of the 3 are on.
	The A's are not on the path. The only way to satisfy the 3 is to have B on the path and C off. B's neighbors on the path are the D's. To ensure that the path in the lower left does not become a premature cycle, we need to have all of the E's be in the path. This satisfies the 2 and 4, so all of its other neighbors are off the path. The only way to legally satisfy the 1 is to have the F's on the path. This satisfies the 3, so the top path must continue through the G's. This satisfies the 4, so the path must also continue through the H's. The F's must connect on the path to the H and the G's on one end and the J's and E's on the other. The path on the far right must continue through the K's and into the L's. The only way to satisfy the 2 is with the M's.
	The answer we're looking for is 1 0 0 0 1 0 0 1 0 1 1 0 1.

Notes: I didn't touch this problem during the test. The night after the test, I finished it in about 45 minutes. 2 points every three minutes is the speed I was going during the test, but I was concerned that difficult puzzles like this might take all of your time and you still might not get a solution.

I also didn't research this problem the night before the way I wanted. I know that there was a World of Puzzles last summer that had a half-dozen of these problems, where the easiest of the problems was the problem used in last year's WPC qualifier. But I couldn't find my copy of the magazine. I'm sure it would have been a great help in increasing my confidence with this problem.