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I'm wondering whether there is a general approach to solving cross-numbers, such an algorithm-like sequence of steps. I understand that for the majority of puzzles like this, trial and error serves best. I was also wondering how to tackle missing clues. An example is shown below:

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Assume that all entries are unique and none begin with a zero. The bold, thicker lines show the separation between entries. Is there a way to potentially brute-force this partially, e.g. using an algorithm to try for example all the two digit primes (for 8 across), and then building upon that to satisfy the other clues? Or is there simply not enough information to do this, and trial and error is the only route?

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    $\begingroup$ I think that "brute force" and "trial and error" are basically the same thing here. Your example of trying all the two digit primes is just a good, logical way of limiting what you have to try in your brute force. To go further, you know that the 2-digit prime in 8 is 50 or less, because it is a factor of another 2-digit number (12 across). So it seems you've answered your own question: The sequence of steps is to start with the clues that are the most restrictive, and intersect with other restrictive cluse, and try (brute force) until you find ones that work. $\endgroup$
    – Stevish
    Dec 1, 2023 at 19:32
  • $\begingroup$ It's a bit wicked to have 8) "Factor of 12 across, prime" and 12) "multiple of 8 across." I would write down the possibilties for each, and whittle it down. But of course there is an algorithmic approach if you care to formulate the conditions for each clue, and if there is not enough information, then there will be multiple solutions. $\endgroup$ Dec 1, 2023 at 20:19
  • $\begingroup$ I understand what you guys are trying to say, but do you have any other tips as in this example, there are so many different possibilities for each of the different clues, and it would surely be impossible to try each and every one? $\endgroup$
    – Developer
    Dec 1, 2023 at 20:31
  • $\begingroup$ @Developer you can whittle down the possible range of solutions, and each of their digits. And look, 9 and 10 intersect and their clues are related too. So use similar techniques as with Kakuro. $\endgroup$ Dec 1, 2023 at 20:44
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    $\begingroup$ Is it intentional that some clues are missing? 3d, 11a, 13a $\endgroup$ Dec 1, 2023 at 21:44

1 Answer 1

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There are a number of things you can look at here to try to narrow down the number of possible solutions for a given clue. It will probably still involve a bit of guesswork, but hopefully if you can narrow it down enough, you will only have a small set of numbers to guess from.

  • As @Stevish noted in the comments, you know that 8a is a 2-digit prime less than 50, because it is also a factor of the 2-digit 12a. That gives you 12 possibilities: {11, 13, 17, 19, 21, 23, 29, 31, 37, 41, 43, 47} This also means the last digit of 8a must be one of {1, 3, 7, 9}
  • 2d is a 3-digit triangular number, of which there are 31, but we know that it's middle digit must be in {1, 3, 7, 9}, because of 8a. That narrows the list down to only 11 possibilities: {136, 171, 210, 231, 276, 378, 435, 496, 595, 630, 990}
  • 10d is a factor of 9a, and the first digit of 10d is the last digit of 9a, so again that means that the first digit of 10d must be $\le $ 4.
  • Wikipedia's page on triangular numbers lists some interesting properties, including these, which might be useful:
    • The final digit of a triangular number is 0, 1, 3, 5, 6, or 8, and thus such numbers never end in 2, 4, 7, or 9.
    • A final 3 must be preceded by a 0 or 5; a final 8 must be preceded by a 2 or 7.
  • (Also, there is no 2-digit triangular number ending in 3, so for the 2-digit ones, you only have 5 possibilities for the last digit.)

When you've narrowed down a particular clue (or digit) to a smallish set of possible values, keep a list of them, and keep cross-referencing. If one clue must not end in 2, 4, 6, or 9, then maybe you can eliminate some of the possibilities from the clue(s) it intersects with.

Keep narrowing things down, and hopefully you will be able to find a solution. If you are stuck and can't seem to narrow it down further, you can always try one of the possible values and see if you can get everything else to fit. If not, cross it off and try another.

You might find it helpful to write all the possible values for each digit in each square, and then erase/cross them out when you have eliminated a possibility (a la sudoku).

Good luck!

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