The New Sudoku Players' Forum

[SpAce]

I guess this is not the right place to ask about sudoku variants, but then again, I'm looking for generic rules of using uniqueness methods. I've never actually used them on other than plain vanilla sudokus, as I've only solved basic Killers, Sudoku Xs, and (only a few) Jigsaws so far, but started thinking about it the other day. Even when solving basic puzzles, URs etc can be really handy because they're often available and almost always easier to spot than triples and quads, especially if solving without pencil marks. Quite often they can be used to bypass such harder parts of basic puzzles, and I freely do that because I avoid using pencil marks with basic(+) puzzles as much as possible. A couple of times I've even spotted extended (2x3) URs without pencil marks, which solved those (non-basic) puzzles quite easily. BUGs, of course, are usually even easier.

Now, my question is about URs (and their generalizations) on some sudoku variants. It's quite obvious that the plain vanilla rules (four corners, two boxes) aren't enough because of the additional constraints. A quick-and-dirty analysis would suggest that a UR on a Killer puzzle would only work if the corners are not only in two boxes but two cages as well. Similarly I would imagine that a UR would only work on a Sudoku X if none of the corners is on a diagonal. Am I on the right track so far? I can't see similarly obvious restrictions on Jigsaws (no additional constraint sets), but I'm not very familiar with them anyway so I might easily miss something.

How would you generalize those observations (assuming they were correct) for any sudoku variant? What about BUGs and other larger DP structures? I would imagine that using them safely might be much harder than simple URs if additional constraint sets are in play.

[Leren]

Have a look here at the six cell DP in 2 digits about half way down the page in the post by Myth Jellies.

That would work in Sudoku X wouldn't it ? As I suspect, would many other fancy DP's with "diagonal" turning points that are on a puzzle diagonal.

Leren

[SpAce]

Have a look here at the six cell DP in 2 digits about half way down the page in the post by Myth Jellies.

That would work in Sudoku X wouldn't it ? As I suspect, would many other fancy DP's with "diagonal" turning points that are on a puzzle diagonal.

Thanks, Leren. I agree that it should work if both members of the pivot point are on the same diagonal. That's only possible with such pivoting DPs, though. A normal or extended UR which appears in a single chute is by definition rectangular and can't have a corner on either X-diagonal (because it can't have two on the same one), right? (Btw, I guess the same pattern would work in a Killer puzzle if all box-peers, including the pivot, were also cage-peers?)

In fact, you're answering the second part of my question which I hadn't even asked yet My primary concern is to avoid logical mistakes, so I want to first chart what definitely doesn't work (but would work in a normal sudoku). This was a good example of what does work in both, though. Can you think of anything uniqueness-related that would only work in some mentioned variant but not in plain vanilla? Or is it a strict subset, as it seems?

[eleven]

This was a good example of what does work in both, though. Can you think of anything uniqueness-related that would only work in some mentioned variant but not in plain vanilla? Or is it a strict subset, as it seems?

As a quick answer: For all variations, which have additional rules to vanilla, like X-Sudoku or Dsjoint Groups, it must be a subset of course.
E.g. for Jigsaw puzzles, the unique rectangles would have different locations, depending on the jigsaw pattern. The rectangle must be in 2 rows and 2 columns and here in 2 jigsaw fields.

[SpAce]

As a quick answer: For all variations, which have additional rules to vanilla, like X-Sudoku or Dsjoint Groups, it must be a subset of course.

Thanks for confirming that, eleven. Seemed logical, but I don't like to assume much.

E.g. for Jigsaw puzzles, the unique rectangles would have different locations, depending on the jigsaw pattern. The rectangle must be in 2 rows and 2 columns and here in 2 jigsaw fields.

That's what I thought also. Essentially it's not really different from normal if you just apply the two-box-rule to fields, right?

[SpAce]

Leren, this gets a bit off track, but thanks for bringing up the pivoting DPs. I've known their existence, of course, but never actually used them. Now that you mentioned them, I was able to spot some on my own in the very first puzzle I solved after your post. They weren't really needed in this case because there were simpler methods available (a basic puzzle anyway), but they do seem like a useful addition especially to no-pm solving (which I was doing).

The first example:

Code: Select all

+---------+---------+---------+
| 1       |         | 6  5  2 |
| 6       | 2  5  3 | 8  7  1 |
| 2  5  7 |    8    | 4  9  3 |
+---------+---------+---------+
| 5     1 |         |    2    |
| 3  2  6 |         |    4  8 |
| 4       |    2    |    6  5 |
+---------+---------+---------+
|    6  4 |       2 | 5  8    |
|    3  2 |    6    |    1  4 |
| 8  1  5 | 7  9  4 | 2  3  6 |
+---------+---------+---------+

1.....6526..253871257.8.4935.1....2.326....484...2..65.64..258..32.6..14815794236

Even without pencil marks it's easy to see the 79 pairings in b79, r78, and c9. From that it's clear that (379)r4c7 can't hold 7 or 9 or it would complete a 6-long DP (r4c9-r7c9-r7c1-r8c1-r8c7-r4c7); thus it must be 3. (In reality I continued with the obvious singles (4,9 r2c23) and got a naked pair (79)r4c29, which did the same trick -- but I was happy to have predicted the result beforehand.)

The same puzzle a bit later:

Code: Select all

+---------+---------+---------+
| 1  8  3 |         | 6  5  2 |
| 6  4  9 | 2  5  3 | 8  7  1 |
| 2  5  7 |    8    | 4  9  3 |
+---------+---------+---------+
| 5     1 |         | 3  2    |
| 3  2  6 |    1    |    4  8 |
| 4     8 | 3  2    | 1  6  5 |
+---------+---------+---------+
|    6  4 | 1  3  2 | 5  8    |
|    3  2 |    6    |    1  4 |
| 8  1  5 | 7  9  4 | 2  3  6 |
+---------+---------+---------+

183...652649253871257.8.4935.1...32.326.1..484.832.165.6413258..32.6..14815794236

Now there are even more obvious 79 pairings and another potential (6- or even 10-long) DP (79)r6c6-r6c2-r4c2-r4c9-{{r7c9-r7c1-r8c1-r8c7}}-r5c7-(79+5)r5c6. So we know r5c6 must be 5. Of course that observation had no practical value as it was already stte, but it was fun to notice.

My point: These kinds of DPs seem very easy to spot even without pencil marks if the pairings are as obvious as here.

Original puzzle: 1.....6..6....3.7.....8..935.1....2.3......484...2.....6....5....2....1....794...

[Leren]

If you look at the second page of the same thread here it looks like the eight cell Bug Lite Patterns T1 - T5 should also work. Whether or not you can find real world examples of these exemplars, and which are also not contrived, is another matter.

It's also worth a careful look through this thread here for some patterns that might work.

Leren

[SpAce]

Thanks for the tips, Leren! I have a question about this example under your first link:

Here is one for the most general Multivalue Universal Grave theory...

Code: Select all

*48+1 *48   5    | 7    14   2    |  3    9    6     
 14    6    2    | 9    3    14   |  5    7    8     
 9     7    3    | 8    6    5    |  4    1    2   
-----------------+----------------+----------------
 3     1    9    | 2    8    6    |  7    4    5   
 5   **28   6    | 14   7    14   |**28   3    9   
*48  **248  7    | 5    9    3    |  6  **28   1     
-----------------+----------------+----------------
 26    9    8    | 346  24   7    |  1    5    34   
 27    5    4    | 13   12   8    |  9    6    37   
 76    3    1    | 46   5    9    |**28 **28   47   


The double-starred cells form a simple uniqueness pattern which forces r6c2 = 4. The set of starred and double-starred cells all form a MUG+1 pattern forcing r1c1 = 1. Combined, these two solve the puzzle.

The first partial pattern forcing the 4r6c2 is very similar to what I just found in the previous example, so that's easy. Forcing the 1r1c1 using the full pattern is more interesting. I can accept that it works but I'd have a hard time spotting or applying the pattern with confidence. I'd probably just use the simpler pattern to place 4r6c2 first and then the resulting AR(48) to place 1r1c1 -- two moves but less risk. But my question...

Is the rectangle (48)r16c12 some kind of named UR type? It seems a bit like Hodoku UR Type 6 (a variation of Type 4) or a Hidden Rectangle but not quite. In any case it would seem to allow eliminating 8r6c2 (uselessly but anyhow). Neither Hodoku nor SudokuWiki seems to pick it up as any UR/Hidden type.

[Leren]

Don't think there is anything too mysterious about this pattern. It's an overlap of the 6 cell DP (28) of the type I mentioned in my first post in this thread, and a 4 cell UR (48) with a common cell r6c2. The 4 cell UR has 2 internal outs r1c1 = 1 or r6c2 = 2, but the 6 cell DP forces r6c2 = 4 <> 2, so that then forces r1c1 = 1. I think most fancy DP's with an odd number of cells can be broken down into 2 smaller DPs with an overlapping cell.

Leren

[SpAce]

Thanks, Leren. So it's in fact two overlapping patterns and essentially two moves (whether you actually place the 4r6c2 or not), as I preferred to see it? That surely makes it pretty simple to understand.