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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 $8 \times 8$ 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?

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  • $\begingroup$ You might be more likely to get an answer on puzzling.stackexchange.com $\endgroup$ Nov 30, 2020 at 3:36
  • $\begingroup$ A clueless 2-row Suguru must contain a region of size at least 5, as shown in my answer to the related question on Puzzling.SE, and cannot have any region spanning both rows. Even with those difficulties, it's difficult to prove anything concrete. $\endgroup$ Mar 3, 2023 at 21:27

3 Answers 3

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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, 2021 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, 2021 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, 2021 at 23:53
  • $\begingroup$ I wrote similar program. Biggest empty puzzles so far found are 9x9. Will take challenge to search 6x6. $\endgroup$ Mar 3, 2023 at 17:29
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I have a bit an obsession for empty Suguru puzzles. So happy to have found this old question and learn that I'm not alone.

Similar to Richard Tobin, I also wrote a program to search for such puzzles (and have a lot of fun to solve manually). But he did it almost 2 years earlier.

So, here are some more interesting sizes:

5x3 The smallest size for non-trivial puzzles:

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

5x4 Disproving Rubenscube's 3rd bullet.

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

6x6 As to your 4th bullet.

Rare, but they exist. Here's a 1 in 10 million find:

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

9x9 The biggest size I found so far.

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

See also: https://puzzling.stackexchange.com/questions/119003/empty-suguru-tectonic-puzzle/119969#119969

and

https://puzzling.stackexchange.com/questions/119245/happy-2023-another-empty-suguru-tectonic?rq=1

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  • $\begingroup$ Congratulations on finding the 6x6. Is it possible to find a 6x6 with one each of blocks size 1-8? That would be rather elegant. $\endgroup$ Mar 4, 2023 at 13:12
  • $\begingroup$ Yo, this is awesome. I can check off some more cases now! We are getting very close to finding them all! $\endgroup$
    – Rubenscube
    Mar 4, 2023 at 19:53
  • $\begingroup$ Also if you are looking for big sizes, I can make them in any very large size you want. For a lot of these you can inductively add 2x2 blocks on the sides (and for odd sides an additional 1x2) to extend them, that is also how I do the extension proofs. If all cases are found I might write a short paper on the findings $\endgroup$
    – Rubenscube
    Mar 4, 2023 at 20:04
  • $\begingroup$ @RichardTobin sizes running up to 8: I highly doubt it exist. I gave up after testing 10M arrangements. Most had multiple solutions. $\endgroup$ Mar 4, 2023 at 23:17
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We can rule out a number of cases by considering how many blocks are allowed and required.

An mxn puzzle cannot have more than ⌈m/2⌉⌈n/2⌉ blocks, otherwise there would be two adjacent ones. (⌈x⌉ is the ceiling or "round up" operator.) This is an example of the rather trivial "N kings" problem.

A clueless puzzle with k blocks will need one of each block size 1..k, using k(k+1)/2 squares, so this must be less than or equal to mn for those blocks to fit. And it will need at least a further ⌈(m*n - k(k-1)/2) / k⌉ blocks for the remaining squares.

Combining these, k + ⌈(m*n - k(k-1)/2) / k⌉ must be less than or equal to ⌈m/2⌉⌈n/2⌉.

For example, a 6x6 puzzle with a maximum block size of 5 requires at least 5+5=10 blocks, but cannot have more than 9, so this is impossible.

Applying this to 2xn, we find there is no possible maximum block size for n < 13.

For 4xn, 4x5 is not excluded (and can be done, as Kris's example shows), but 4x6 is impossible. 4x8 might be possible with a maximum block size of 6 or 7.

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