You are supposed to create a new $5$x$5$ Futoshiki puzzle with the least amount of clues possible, but this puzzle needs a unique answer. In other words,

What is the minimum number of clues for $5$x$5$ Futoshiki puzzle having a unique solution?

Note that

Each number and each sign is a clue.

  • 2
    $\begingroup$ If you're interested, there are the very similar Mainarizumu puzzles $\endgroup$ Commented Oct 12, 2017 at 14:28
  • 1
    $\begingroup$ NB: There are only 30960 possible solutions (up to some measure of equivalence) and 22600736 different ways of putting down up to 5 clues, so brute-force is entirely feasible. $\endgroup$
    – Veedrac
    Commented Oct 13, 2017 at 11:46

1 Answer 1


To provide a starting point: The following puzzle with 6 clues should be uniquely solvable.

This was mostly found by trying out different strategies. The basic idea was starting with a chain of 4 signs to get 5 numbers. The chain was constructed to put the numbers in as many different columns and rows as possible to provide the most restrictions to the puzzle. This leaves us with a column and a row with 3 numbers already determined. For each of those a sign was added to be able to deduce the order of those two missing numbers while also providing a clue for one other number. Thus each of those signs immediately gives us 3 new numbers and are together enough to solve the puzzle.

  • $\begingroup$ Perhaps an explanation of why those symbols at the specified locations (meaning how these are arrived at) would provide much needed clarity. $\endgroup$ Commented Oct 12, 2017 at 13:50
  • $\begingroup$ @MeaCulpaNay I've added a small explanation how I came up with this solution. I have no proof that this is the best possibility. Let me know if the explanation is understandable. $\endgroup$
    – w l
    Commented Oct 12, 2017 at 14:38
  • $\begingroup$ You say "This leaves us with a column and a row with 3 numbers already determined." Surely your chain of four signs gets you 5 numbers with at most 2 in any column or row... Am I missing a logical step here? $\endgroup$
    – Chris
    Commented Oct 12, 2017 at 14:40
  • $\begingroup$ @Chris With those five numbers we can deduce another 3 due to unique digits in each row/column. $\endgroup$
    – w l
    Commented Oct 12, 2017 at 14:42
  • 1
    $\begingroup$ @Veedrac I know that from my code, I tried all 5 pairs, couldn't find any unique solution. $\endgroup$
    – Oray
    Commented Oct 13, 2017 at 20:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.