Custom rule: Prime Takuzu — No contiguous span of three digits in any row or column may be all prime or all composite. (1 is neither prime nor composite.)
My first puzzle featured on Cracking the Cryptic! Played by Mark on Feb 3, 2025.
I like this. It requires several moderately clever steps, rather than one big galaxy-brain step. It also contains a sort of hint for itself, something I appreciate in a weird puzzle.
I was stumped on making a "3" puzzle because I couldn't come up with a good theme that seemed based around 3s, other than just using a bunch of modular/entropic lines, which isn't terribly interesting on its own. I was finally inspired by a CtC takuzu sudoku, Two to Tango by Scott Strosahl (which refers to the rule as "Tango", apparently the name used on LinkedIn??) — I realized that using primality as the condition would effectively create three categories of digits, since 1 stands alone.
I struggled with this for a couple days, trying to make the rule interesting — initially with modular lines, then with German whispers. Juggling two different partitioning schemes at once was a mess and didn't lend itself to very interesting interactions, so I switched to the old standby: killer cages. That gave me a bit more control over the starting digits, and the initial part of the break-in basically created itself.
About a third of the effort here was writing the custom JS to teach Sudoku Maker about the constraint. There's not really any documentation, and the solver runs in a worker which seems to fuck up browser dev tools a bit, so it was... awkward. But I did it! I hope!
Spoiler commentary
The break-in requires a series of escalating observations.
The 6-cage at the bottom is 123, with the 1 on the right side, so r9c4 and r9c5 are 23 and both prime. The 1 has already been placed in the row, so r9c3 must be composite.
The 21-cage in box 7 can be any of 489, 579, or 678. But all of 489 are composite, so no matter how those digits were arranged, there would be a run of three composite digits along the bottom of box 7. Thus, the 21-cage is 579 or 678.
Next, the 10-cage in column 5 cannot be 235, as all those digits are prime, so it must contain a 1. That 1 cannot go in box 5, or the 9-cage becomes 234, which, in combination with the 6-cage, would take too many small digits away from the 10-cage. (Specifically, it would create a 234 triple in column 5, forcing the 10-cage to be 156 = 12 at minimum.)
Lemma: If all four digits of type A are clumped at the far ends of a row or column, the 1 and the four digits of type B form a contiguous span of five digits in the middle. The only way to avoid violating the takuzu rule would be to have the 1 in the center of that span, forming BB1BB.
The 23-cage on row 4 can only be 689. Those are all composite digits. If the 21-cage on row 4 also contained all composite digits, then by the lemma, the 1 in row 4 would have to go in r4c5. But we just proved a 1 cannot go there. Thus the upper 21-cage cannot be 489.
So both 21-cages are 579 or 678, and thus both contain a 7. Furthermore, they can't both be the same combination — that would mean four cells in column 2 have only three digits between them. So one is 579 and one is 678.
This forces the 7 in box 1 into the 11-cage, which can now only be 137. From there it's a matter of chipping away until the puzzle is solved.
