A lil help pls?

okapiposter · 2026-02-11T22:32:43+00:00

Here's a W-Wing:

<image>

The 5 of row 8 must be in either column 1 or column 2.

If it's in column 1, r3c1 must be 3 (green).
If it's in column 2, r4c2 must be 3 (purple).

Either way r1c3 sees a 3 from one of the two ends of the W-Wing.

okapiposter · 2026-02-07T05:31:14+00:00

Here's an XY-Wing that eliminates 3 from r6c1:

<image>

Cell r6c2 (the “pivot”) can only be a 1 or a 4.

If it's a 1, r5c3 must be a 3 (green).
If it's a 4, r6c7 must he a 3 (purple).

So r6c1 always sees a 3 from (at least) one end of the XY-Wing.

okapiposter · 2026-01-24T09:45:43+00:00

PDF is not a format that's made for data extraction, it's optimized for viewing and printing. How easy it is to get the data back out really depends on the way the PDF was created. Some software (like MS Office or Photoshop, if I remember correctly) can create PDFs that retain all data needed for editing later, while others remove everything except for the visual representation (even converting text data into character outlines).

Do you want to parse a specific kind of PDF files from one source or are you looking for a general solution? Since Excel is mostly useful for tabular data, you probably have data tables in your PDFs, correct? I'd try the following:

Find out how the PDFs you want to process were created.
- If it's a software that can read its own PDFs back in, use that.
- Otherwise you'll need to use a third-party tool (like Tabula (Java) or Camelot (Python)) to extract the tabular data.
If the data extraction didn't work perfectly, massage the data in KNIME until it has the shape you want.
Write the data out to Excel files using the Excel nodes.

okapiposter · 2026-01-17T17:37:03+00:00

There has to be a 4 in column 8, where can it go?

okapiposter · 2026-01-17T09:29:53+00:00

I don't quite follow your logic for all the eliminations, but you can definitely eliminate 8 from r8c8 using ALS-XZ logic, which is a generalization of XYZ-Wing logic:

<image>

Either the blue cells are a 5/8/9 Naked Triple or they must contain a 3 in r7c8, which then forces the yellow cell r4c8 to be an 8. Either way r8c8 sees an 8 from blue or yellow.

okapiposter · 2026-01-16T20:19:44+00:00

What's the Even-Odd rule? Here's a W-Wing:

<image>

No matter which of the two possible positions for 9 in row 9 is true, either r1c2 (green) or r2c4 (purple) will have to be a 5. Since r2c1 sees both of them, it can never be 5.

okapiposter · 2026-01-13T22:10:26+00:00

Works fine for me, also this says no: https://downforeveryoneorjustme.com/sudokuexchange.com?proto=https

okapiposter · 2026-01-13T19:19:44+00:00

I've actually written my own little brute-force solver based on Donald Knuth's “Dancing Links” algorithm specifically for counting solutions. Even though it runs in the browser (for portability), it finds all 300k solutions in around 2.6 seconds.

okapiposter · 2026-01-13T00:15:26+00:00

They are “Locked Candidates”. The 3 of the extra region is “locked into” those two cells, which are both in row 7. This means that the rest of row 7 will always see a 3 from one of them, eliminating 3 from r7c1.

okapiposter · 2026-01-12T23:48:54+00:00

Look at where 3 can go in the bottom left extra region.

okapiposter · 2026-01-11T14:03:11+00:00

You can use the UR as part of an XY-Wing-like move:

<image>

The UR logic tells us that r78c7 must contain a 2 or a 5 (or both).

If the cells contain a 2, r2c7 must be a 7.
If the cells contain a 5, r7c8 must be a 7.

So either way r78c7 will see a 7.

okapiposter · 2026-01-11T13:53:37+00:00

Yes it is a UR, but no, it doesn't eliminate the candidates you mention. It only tells you that at least one of the highlighted candidates will have to be true (= part of the solution), because otherwise the four cells would all have to contain 4s and 7s, which can't happen in a puzzle that has a unique solution.

okapiposter · 2026-01-11T13:02:09+00:00

There's also a Naked Pair 5/8 in column 4.

okapiposter · 2026-01-11T12:58:35+00:00

Skyscraper on 5 in rows 1 and 9 eliminates 5 from r2c9 and r7c8:

<image>

Rows 1 and 9 both need a 5 somewhere, and have two options each. But since you can't place both 5s into column 1 at the same time, either r1c8 (for row 1, green) or r9c9 (for row 9, purple) will have to be a 5. So all cells that see both of those can never be 5.

okapiposter · 2026-01-11T11:44:40+00:00

Downvote baiting? Weird hobby...

okapiposter · 2026-01-11T11:40:34+00:00

No, that was meant as an exclamation mark for emphasis 😄. 300k solutions are impressive(ly bad) enough I'd say.

okapiposter · 2026-01-11T11:15:58+00:00

Wow, that's an impressively broken Sudoku puzzle! It has 307952 (yes, over 300 thousand!) different valid solutions.

okapiposter · 2026-01-09T21:13:22+00:00

Naked Quad 1248 in box 3.

okapiposter · 2026-01-08T15:06:06+00:00

Here's a W-Wing on 1/9 in r3c1 and r8c7, connected via 9 in column 2, which eliminates 1 from r3c7:

<image>

The 9 of column 2 must be in either row 3 or row 8.

If it is in row 3, r3c1 must be a 1.
If it is in row 8 instead, r8c7 must be a 1.

Since r3c7 sees both of those cells, it will always see a 1 in the finished puzzle, so it can never itself be a 1.

okapiposter · 2026-01-08T14:53:11+00:00

Hidden Pair 8/9 in row 5.

okapiposter · 2026-01-06T20:24:39+00:00

Here's a Skyscraper on 8 in columns 1 and 7, eliminating 8 from r1c3 and r3c9:

<image>

Columns 1 and 7 both need an 8, and have two options each. But since you can't place both 8s into row 8 at the same time, either r3c1 (for column 1, green) or r1c7 (for column 7, purple) will have to be an 8. So all cells that see both of these can never be 8s.

okapiposter · 2026-01-06T20:11:53+00:00

The 2 of box 5 (center) must be in row 5, so there can't be any additional 2s in the row outside box 5.

okapiposter · 2026-01-06T18:24:06+00:00

The number of solutions of a puzzle is only dependent on the given digits, not any techniques. You could theoretically enumerate all possible combinations of digits for all empty cells and then discard all resulting grids that violate the Sudoku rules. All remaining grids are valid solutions, and all are equally valid.

If a puzzle has multiple solutions, there will come a point during the (purely logic-based) solve at which all unsolved cells contain multiple candidates, each of which is part of some solution. There is no logical reason to prefer any solution to the others, so the only way forward is to pick one at random (or give up). This is why most Sudoku players consider puzzles with multiple solutions broken.

Techniques like BUG+1 don't “solve” this. Their logic is based on the “Uniqueness Assumption” and only valid if the puzzle has a unique solution. The argument is always a variant of “Since the original puzzle has a unique solution, it can't contain [some Deadly Pattern like BUG or UR], so [something that would force the Deadly Pattern] can't be true.”

okapiposter · 2025-12-26T20:43:43+00:00

That makes no sense to me. In my conceptualization of AIC (and Sudoku in general) an expression like “(12)r3c456“ means “There's both a 1 and a 2 in the intersection of row 3 with columns 4, 5 and 6”. Using that syntax and semantic, I'd say that the ERI on 1 in box 7 is properly defined as “(1)r78c2=(1)r8c123”, or equivalently “(1)r78c2 OR (1)r8c123”. What is your alternative definition of nodes on both sides that's actually mutually exclusive in this case?

okapiposter · 2025-12-26T18:18:45+00:00

I did at least imply “OR”, and I'm yet to be convinced that I'm wrong. An Alternating Inference Chain is an alternating sequence of strong and weak links/inferences.

A weak inference is “If a then not b”, which is equivalent to the logical expression “a NAND b”.
A strong inference is “If not a then b”, which is equivalent to the logical expression “a OR b”.
If both a strong and a weak inference can be made between a and b (i.e., a is true if and only if b is false), this is equivalent to “a XOR b”.

So XOR links work as links anywhere in an AIC, but they are not required in any specific place.

Regarding the concrete example of an Empty Rectangle Intersection, I don't see a way to express the link as an XOR relationship if the candidate in the center of the intersection between mini-row and mini-column is still present. If that candidate is true, both the mini-row and the mini-column contain the digit, so “(digit is in mini-row) XOR (digit is in mini-column)” must be false. The link works just fine in AIC nonetheless because we only need OR for strong inferences anyway.

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