What is the minimum amount of numbers needed to create a 4x4 Hidato with a unique solution?

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:

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

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?

• Is the solution unique? It looks there are two solutions where '6' and '16' could be swapped to get from one to the other. Commented Mar 5 at 17:44
• 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? Commented Mar 5 at 17:48
• For $3\times3$, the minimum is $2$ (not $3$): \begin{matrix} . & . & . \\ . & . & 9 \\ 7 & . &. \\ \end{matrix} Commented Mar 6 at 17:41
• For $3\times4$, the minimum is $2$ (not $4$): \begin{matrix}.&.&.&.\\.&.&.&.\\.&.&11&6\\ \end{matrix} Commented Mar 7 at 0:01
• I'm pretty sure this works for 3 x 7: \begin{array} {c} . & . & 16 & . & 5 & . & . \\ . & . & . & . & . & . & . \\ . & 18 & . & . & . & . & 3 \\ \end{array} Commented Mar 7 at 15:05

Update: I think 3 is possible:

Solving this is as follows:

First, add the diagonals and the 8 is then forced.

Next, there's only one way to connect the 12 and the 16 without creating an island:

Next, there's a deadend, that must have a 1

And you're done

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

• Both the 4-clue puzzle and 3-clue puzzle have a unique solution, good job! Commented Mar 5 at 18:55
• We're trying for the minimum, right? Not specifically 7-clue. Commented Mar 5 at 19:01
• Yes, we are trying for the minimum. 7 clues was just my proposed lower bound. Commented Mar 5 at 19:02
• That solution with three is very nice. Good explanation +1 Commented Mar 5 at 19:39

Here is a 4x4 Hidato with 6 clues which I believe has a unique solution

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

(A is 10)