Solve the crossword then slide the blocks to create another configuration with eight new words.
Across
1. Lure
5. Before long
6. Frosted
7. Permit
Down
1. Scold
2. Formerly
3. Frost, for example
4. Limit
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$\begingroup$ what are the clues for the second configuration? thanks! $\endgroup$ – Omega Krypton Jun 16 '19 at 1:32
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1$\begingroup$ So basically just find an even permutation of the letters that creates eight new words? I'd wager that there are plenty of ways to do that. $\endgroup$ – greenturtle3141 Jun 16 '19 at 2:09
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3$\begingroup$ @greenturtle3141 But the no-computers restriction makes it difficult (and kinda tedious) to do. (I've been trying for about an hour, and have come very close, but never got it.) $\endgroup$ – Deusovi♦ Jun 16 '19 at 2:15
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$\begingroup$ @Deusovi Same... $\endgroup$ – Rubio♦ Jun 16 '19 at 3:04
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3$\begingroup$ Does the post-sliding configuration have to have the gap at the bottom right corner? $\endgroup$ – Gareth McCaughan♦ Jun 16 '19 at 10:35
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The solution to the crossword is:
And this can be arranged as:
My strategy:
After several failed attempts, I started with
P?LEandP?Nin the upper left, andTENDalong the bottom: the first two because they could be filled with many vowels (giving me freedom to move the vowels around if necessary), and the latter because all of those letters were very common as ending letters. Next cameALOE, because I'd need vowels in the middle-right area, and thenIRONwas the best word ending inON. After that, I placedCAT/CERO/NOED, and was disappointed thatPELEwas a proper noun, not a word (andCEROwas obscure too)... then didn't realize until the next day that I could just swap theEandOto fix it.
But is this solvable as a 15-puzzle?
Yes: The 15-puzzle is solvable in exactly half of its positions, based on parity of swaps. If you can't get to a particular position, you can always get to that position with any two tiles of your choice swapped. So, if your correspondence of numbers to letters fails, simply swap two of the tiles with the same letter and it will be solvable.
An explicit solution:
Number the tiles 1-15 so that the puzzle is "solved" at the very start, in the given configuration. (You can use this applet to try it out, as I did.) Then move the following tiles. (Parentheses represent a sequence that should be repeated once.)
12 8 (7 6 10 9 13 14 15 11 8): TEND is now in the bottom row.
13 14 5 (9 10 3 2 1) 3 14 5 10 (14 3 9) 14 3 10 5: PCAT is now in the left column.
13 7 6 4 1 14 9 2: IOLE is now in the second column.
4 6 7 4 6 1 (14 6 1) 4 7 14 4 1 9 3: The puzzle has been rearranged into the new grid.
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1$\begingroup$ Nice job! In the version with 15 distinct squares I believe some positions are actually unreachable by sliding. I assume it's not an issue here when we have multiple instances of the same letter in the grid? $\endgroup$ – jafe Jun 16 '19 at 18:13
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$\begingroup$ @jafe Yep, it's parity-based -- if you can't solve the 15 puzzle exactly, you can get it with any two tiles of your choice swapped. If the first attempt at assigning correspondences to letters fails, you can just swap positions of any two identical letters. (I should probably add a solution for completeness' sake, though.) $\endgroup$ – Deusovi♦ Jun 16 '19 at 19:36
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Initial grid as follows:
rope
anon
iced
let
answered Jun 16 '19 at 1:12
