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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.

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    $\begingroup$ If you're interested, there are the very similar Mainarizumu puzzles $\endgroup$
    – Engineer Toast
    Commented Oct 12, 2017 at 14:28
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    $\begingroup$ 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

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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.

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  • $\begingroup$ Perhaps an explanation of why those symbols at the specified locations (meaning how these are arrived at) would provide much needed clarity. $\endgroup$
    – Mea Culpa Nay
    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
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    $\begingroup$ @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

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