The New Sudoku Players' Forum

[joedulong]

Can someone please explain the logic in the below puzzle using coloring.

IMG_2253.PNG (130.6 KiB) Viewed 1089 times

[Leren]

Here is a simple solution to your puzzle , which I've included in line format as follows : 8243.5..9613.2.45.5974.1.3274.5...2.2...4....9....2.84152..7346389614275476253.9.

Code: Select all

*--------------------------------------------------------------*
| 8     2     4      | 3     67    5      | 67-1 a16    9      |
| 6     1     3      | 789   2     89     | 4     5     78     |
| 5     9     7      | 4     68    1      |b68    3     2      |
|--------------------+--------------------+--------------------|
| 7     4     18     | 5     389   689    | 169   2     13     |
| 2     36    158    | 1789  4     689    | 15679 16    137    |
| 9     36    15     | 17    37    2      | 1567  8     4      |
|--------------------+--------------------+--------------------|
| 1     5     2      | 89    89    7      | 3     4     6      |
| 3     8     9      | 6     1     4      | 2     7     5      |
| 4     7     6      | 2     5     3      |c18    9     18     |
*--------------------------------------------------------------*

There is an XY Wing in cells a, b and c. What this shows is that if the 1 in r1c8 is False, the 1 in r9c7 is True and conversely if the 1 in r9c7 is False, the 1 in r1c8 is True.

So what this means is that at least one of the 1's in r1c8 and r9c7 must be True. They might both be True but they can't both be False.

Since r1c7 can see both of these cells, the 1 in it must be False, and this solves the puzzle via cascade of follow-on singles (the first of which is that r1c8 must be 1).

However, from the look of your diagram it looks like your program had a somewhat complex form of coloring in mind. The green and yellow cells are supposed to have opposite parity, but it's unclear what digit(s) is/are being considered.

It looks like the parity on digit 8 is being considered. I can explain some, but not all, of the coloring in the diagram. I can't see off-hand how the coloring in cells r4c3, r5c3 and r5c6 relates to the coloring in the other yellow and green cells.

There could be a link via some other digit or pattern but I can't see it at the moment. The idea is that an 8 in any cell that can see a yellow cell and a green cell can be eliminated.

This might solve the puzzle, but it's much more complex than the XY Wing. If I were you I'd stick with that. If you are unfamiliar with this move let me know and I'll explain it in more detail.

Leren

[JasonLion]

Looking at your coloring, you appear to have marked three disjoint groups of cells in yellow/green, even though they are not connected. Normally you would use a distinct pairs of colors for each disconnected set of cells.

Let's start at R9C79, that is a clear conjugate pair and a good place to start coloring 8. That group extends to R2C9 and R3C7 easily, and on from there to R3C5.

You can not link from there to either R2C4 or R7C45, so they need to be colored in a different set of colors. R7C45 is a good place to start our second group, and they link back up to R2C4 (in the new set of colors). The third disjoint group starts with R45C3, which links to R5C6.

Note: the order in which cells are connected together is arbitrary and could be done in a different sequence. But which cells can end up connected is fixed by the pattern of pencil marks in the various cells.

Marking the disjoint groups in yellow/green is misleading, as simple coloring rules can not be applied across disjoint groups. Note that I am using simple coloring in my explanation. There are more complex coloring systems that do allow the connection of disjoint groups under certain conditions. You don't appear to be using any of those more complex systems here, or if you were, you made a couple of mistakes along the way, to the point where I can't disentangle what might have happened.

With the coloring changed to show the three disjoint groups in different pairs of colors it would be correct, but no eliminations could be made. If the coloring was correct as you show it, the pencil mark for 8 in R2C6 could be eliminated, as it sees both yellow and green cells. However, that cell actually is 8 in the solution, confirming that the coloring is indeed wrong as shown.

By the by, I am not sure why you don't have a pencil mark for 8 in R5C4. That cell isn't 8, but I don't see any way you could have known that at this stage of the puzzle. This is not a big deal, as the elimination is correct, and perhaps you saw something I don't.

[Leren]

Hi Jason, actually I think that there are only two disjoint parity groups, although r2c4 shouldn't be colored yellow. r2c9 and r3c5 have the same parity via the Group Strong links on 8 in Row 3 and Box 2

Actually I thought that the coloring was done via a software hint and was correct, although I couldn't see why.

FWIW I think that coloring is a horrible technique and is only really suitable for computer solutions.

Here is my solver's Simple Coloring solution for this puzzle.

Code: Select all

*-----------------------------------------------------------------------*
| 8      2      4       | 3      67     5       | 167    16     9       |
| 6      1      3       | 789    2      89      | 4      5      78      |
| 5      9      7       | 4      68     1       | 68     3      2       |
|-----------------------+-----------------------+-----------------------|
| 7      4      18      | 5      389    689     | 69-1   2     *13      |
| 2      36     158     | 1789   4      689     | 5679-1 16     137     |
| 9      36     15      | 17     37     2       | 567-1  8      4       |
|-----------------------+-----------------------+-----------------------|
| 1      5      2       | 89     89     7       | 3      4      6       |
| 3      8      9       | 6      1      4       | 2      7      5       |
| 4      7      6       | 2      5      3       |#18     9     *8-1     |
*-----------------------------------------------------------------------*

Simple Coloring on 1. Parity Group details: | r4c9 r9c9 | r9c7 | Eliminate 1 from cells seeing both Parity Values.

I can't see why the Parity Groups for r4c9 and r9c9 have been joined because my solver doesn't backtrack on all the joins. If it did it would probably produce some messy gobledegoook equivalent to some complicated chain.

UGH !

Leren

[JasonLion]

Learn, I suppose it depends on what joedulong meant by "coloring". I was sticking with Simple Coloring, based on conjugate pairs, as I mentioned. That is a good place to start before exploring Coloring more generally. By the by, I love your solution using coloring on 1.

[keith]

FWIW I think that coloring is a horrible technique and is only really suitable for computer solutions.

Leren

Let's not forget that a basic M-wing is based on Medusa coloring. I still have a box of erasable colored pencils in my desk drawer, though I haven't done Medusa for years.

Keith