3 ms·
For some reason that I can't explain that reminded me of a diabolical trick someone once did on the 15 puzzle [1]. They made a 15 puzzle where instead of numbe
by tzs 2y ago
For some reason that I can't explain that reminded me of a diabolical trick someone once did on the 15 puzzle [1].
They made a 15 puzzle where instead of numbers the tiles had letters on them, and the goal was to make a particular sentence. That sentence had 13 letters that appeared once and one letter that appeared twice.
That letter that appeared twice was the first letter of the sentence, and its other appearance was at or near the end. Let's call the two tiles with the same letter T1 and T2, where T1 is the tile that is the tile that is first.
They would show someone the solved puzzle, then scramble it, and challenge the person to solve it. When they were scrambling it they made sure to move T2 to the first position and T1 to somewhere near the end.
Nearly everyone who tried to solve it would see the right letter at the first position and never move it, because most people try to solve left to right and/or top to bottom.
On the 15 puzzle exactly half of the possible permutations of the 15 tiles are reachable from a given starting position. Exchanging two tiles in a reachable permutation produces an unreachable permutation.
That means that the right sentence but with T1 and T2 swapped is not reachable and so people who left T2 in front were doomed to failure.
[1] https://en.wikipedia.org/wiki/15_puzzle https://en.wikipedia.org/wiki/15_puzzle
- mjcohen 2y agoWow - that is fiendish!
- hinkley 2y agoThat's eeeeevil.
- jcl 2y agoHeh… that exact situation came up here a few years ago, when someone posted their software implementation of the puzzle: https://news.ycombinator.com/item?id=10540014 https://news.ycombinator.com/item?id=10540014
- vikingerik 2y agoThe same thing can happen on some Rubik's cuboids - it seems unsolvable with a parity problem, then the answer is to swap two seemingly identical elements. I know it happens on the 3x3x4 cuboid - each of the 3x4 faces has two centers that appear identical, but if you have exactly two such unsolved pieces, it's unsolvable because it's impossible to exchange exactly two of them - what you need to do is cycle 3 of them where 2 appear identical. I remember this happening on some other shapes and sizes as well. (Pic for reference: https://www.grubiks.com/images/puzzles/30/small.png https://www.grubiks.com/images/puzzles/30/small.png . This puzzle exists as do many other NxMxH cuboids. Rectangular but non-square faces can only make 180º turns.)
- hinkley 2y agoIf memory serves, it also happens with the hollow 3^3 cube, because there’s a translation of the center which also pulls two of the edges with it. And since you can’t see the center you don’t realize you fucked up until you’re solving the top face and something is wrong.