Suguru puzzles with no given clues

Question

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 $8 \times 8$ 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.

• With the help of the $6 \times 6$ configuration provided by Kris van Bael I was able to prove all $even \times even$ are possible if both evens are at least 6, slightly improving the argument from before.

• With the help of Kris van Bael who also provided a working 5x4 I was able to prove all $4 \times odd$ work if the odd is at least 5. It is not hard to prove that $3 \times 4$ is impossible.

This brings all the unsolved cases down to two categories. $2\times n$ for all $n$ and $4\times n$ 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?

Answer 1

Here is an 8x8 suguru with no clues and one solution (so your conjecture that there are none with both sides even is wrong):

+---+---+---+---+---+---+---+---+
|           |           |       |
|---+---+---+---+---+   +---+---|
|               |   |       |   |
|---+---+   +---+   +---+---+   |
|   |   |   |       |       |   |
|   +   +---+   +   +   +---+   |
|   |       |       |   |       |
|---+   +   +---+---+   +   +---|
|   |       |   |       |   |   |
|   +---+---+   +---+---+---+   |
|   |           |   |           |
|   +---+   +---+   +---+---+   |
|   |   |   |       |       |   |
|---+---+---+---+   +---+   +---|
|               |       |       |
+---+---+---+---+---+---+---+---+