13
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Goal of Hidato:

Fill in a grid with a series of consecutive numbers that connect each other horizontally, vertically, or diagonally.

So I have been working on figuring out the minimum number of clues needed for an $n\times m$ Hidato puzzle such that it had a unique solution. (up to the $8\times8$ case)

Here is the table I have made which lists the lower bounds of clues needed for a Hidato puzzle of size $n\times m$ to have a unique solution:

enter image description here

I am currently trying to figure out the lower bounds for $4\times m$ Hidato puzzles where $m\in\mathbb N$ and $4\le m\le8$, and would like to know the minimum amount of clues needed for a $4\times4$ Hidato puzzle to have a unique solution.

I have managed to create one with 8 clues

enter image description here

however I would like to know:

Is 8 is truly the minimum needed, or is it possible to have a uniquely solvable 4x4 Hidato puzzle with 7 clues?

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10
  • 1
    $\begingroup$ Is the solution unique? It looks there are two solutions where '6' and '16' could be swapped to get from one to the other. $\endgroup$
    – hexomino
    Mar 5 at 17:44
  • 1
    $\begingroup$ Wait a sec, does that mean I have accidentally answered my own question? because then we could remove the 5 to get a 4x4 Hidato puzzle with a unique solution with only 7 clues? $\endgroup$
    – CrSb0001
    Mar 5 at 17:48
  • 2
    $\begingroup$ For $3\times3$, the minimum is $2$ (not $3$): \begin{matrix} . & . & . \\ . & . & 9 \\ 7 & . &. \\ \end{matrix} $\endgroup$
    – RobPratt
    Mar 6 at 17:41
  • 2
    $\begingroup$ For $3\times4$, the minimum is $2$ (not $4$): \begin{matrix}.&.&.&.\\.&.&.&.\\.&.&11&6\\ \end{matrix} $\endgroup$
    – RobPratt
    Mar 7 at 0:01
  • 2
    $\begingroup$ I'm pretty sure this works for 3 x 7: \begin{array} {c} . & . & 16 & . & 5 & . & . \\ . & . & . & . & . & . & . \\ . & 18 & . & . & . & . & 3 \\ \end{array} $\endgroup$
    – Nitrodon
    Mar 7 at 15:05

3 Answers 3

22
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Update: I think 3 is possible:

3 soln

Solving this is as follows:

First, add the diagonals and the 8 is then forced.
s1
Next, there's only one way to connect the 12 and the 16 without creating an island:
s2
Next, there's a deadend, that must have a 1
s3
And you're done
done


I think 4 is possible. I believe this has a unique solution. (Oh, just saw that someone posted something similar)

4 soln

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4
  • 3
    $\begingroup$ Both the 4-clue puzzle and 3-clue puzzle have a unique solution, good job! $\endgroup$
    – CrSb0001
    Mar 5 at 18:55
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    $\begingroup$ We're trying for the minimum, right? Not specifically 7-clue. $\endgroup$
    – Dr Xorile
    Mar 5 at 19:01
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    $\begingroup$ Yes, we are trying for the minimum. 7 clues was just my proposed lower bound. $\endgroup$
    – CrSb0001
    Mar 5 at 19:02
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    $\begingroup$ That solution with three is very nice. Good explanation +1 $\endgroup$
    – hexomino
    Mar 5 at 19:39
7
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Here is a 4x4 Hidato with 6 clues which I believe has a unique solution

enter image description here

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0
5
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What about this?

..A4
....
....
1..7

(A is 10)

For 3x4, what about this?

3..
c..
..1
...

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1
  • $\begingroup$ I have confirmed that both have a unique solution. $\endgroup$
    – CrSb0001
    Mar 5 at 18:55

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