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A Suguru puzzle consists of a bunch of polynominoes fit together in a rectangular shape. The goal of the puzzle is to fill all the cells in all the polynominoes with the numbers $1$ to the amount of cells the polynomino has. A pentomino requires you to fill in $1$, $2$, $3$, $4$, and $5$, for example. A number is not allowed to touch the same number, not diagonally either!

Recently I was wondering if it is possible to construct a Suguru puzzle without any given clues (no given numbers) with exactly one solution. I was able to find some examples so I started wondering for which sizes this would work. So my question is:

For which $a$, $b$ is it possible to construct an $a\times b$ Suguru puzzle with no given clues and exactly one solution?

I was able to prove some cases, but not all cases.

  • I was able to prove that if $a$ and $b$ are both odd, than there exists a possible arrangement with no given clues and exactly one solution unless both $a$ and $b$ are equal to $3$, in which case it is impossible to construct one.

  • If one side is even and the other one is odd, I was able to proof that there exists an arrangement if the even side is at least $6$ and the odd side at least $3$. If the odd side is equal to $1$ than no arrangement with the other side being even exists.

  • If the even side is equal to either $2$ or $4$, I am not sure whether it will ever be possible to construct a puzzle with no given clues and exactly one solution. So far I have not been able to find any working examples, so I believe there aren't any at all.

  • With the help of the 8x8 configuration which was provided by Richard Tobin, I was also able to prove that an arrangement exists if $a$ and $b$ are both even and at least 8, however, I still haven't figured out what happens if $a$ and $b$ are both even, but not both at least 8.

This brings all the unsolved cases down to three categories. $2\times n$ for all $n$, $4\times n$ for all $n$ and $6\times n$ for all even $n$

Does anyone know how I could prove all cases, or just some of the cases I have not yet already solved?

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    $\begingroup$ Please define a Suguru puzzle. $\endgroup$ Nov 29 '20 at 17:39
  • $\begingroup$ A Suguru puzzle consists of a bunch of polynominoes fit together in a rectangular shape. The goal of the puzzle is to fill all the cells in all the polynominoes with the numbers 1 to the amount of cells the polynomino has. A pentomino requires you to fill in 1, 2, 3, 4 and 5 for example. A number is not allowed to touch the same number, not diagonally either! I hope this helps! $\endgroup$
    – Rubenscube
    Nov 29 '20 at 17:57
  • $\begingroup$ You might be more likely to get an answer on puzzling.stackexchange.com $\endgroup$ Nov 30 '20 at 3:36
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Here is an 8x8 suguru with no clues and one solution (so your conjecture that there are none with both sides even is wrong):

+---+---+---+---+---+---+---+---+
|           |           |       |
|---+---+---+---+---+   +---+---|
|               |   |       |   |
|---+---+   +---+   +---+---+   |
|   |   |   |       |       |   |
|   +   +---+   +   +   +---+   |
|   |       |       |   |       |
|---+   +   +---+---+   +   +---|
|   |       |   |       |   |   |
|   +---+---+   +---+---+---+   |
|   |           |   |           |
|   +---+   +---+   +---+---+   |
|   |   |   |       |       |   |
|---+---+---+---+   +---+   +---|
|               |       |       |
+---+---+---+---+---+---+---+---+
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  • $\begingroup$ Can I ask how you were able to find this one? $\endgroup$
    – Rubenscube
    May 15 at 13:03
  • $\begingroup$ Using your 8x8 configuration I was able to prove that if a and b are both even and at least 8, it is also always possible to make a suguru with no given clues. So thank you for this! $\endgroup$
    – Rubenscube
    May 15 at 15:59
  • $\begingroup$ I wrote a program to generate puzzles and see if they were uniquely solvable. I have not been able to find a 6x6 - if there is one, it must contain a polyomino of size 6. $\endgroup$ Sep 13 at 23:53

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