X-Colouring Question

Advanced methods and approaches for solving Sudoku puzzles

X-Colouring Question

Postby BryanL » Thu May 26, 2011 9:09 am

Hi all,

if there is a house with a cell that is a peer to an A coloured cell, and then subsequent steps show all the other cells in that house are peers to B coloured cells, can the A coloured peer cell be coloured B? Or must it be left alone?

thx,

Bryan
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Re: X-Colouring Question

Postby ronk » Thu May 26, 2011 11:01 am

BryanL wrote:if there is a house with a cell that is a peer to an A coloured cell, and then subsequent steps show all the other cells in that house are peers to B coloured cells, can the A coloured peer cell be coloured B? Or must it be left alone?

Re-coloring doesn't sound like a "safe" thing to do, in general. Do you have an example?
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Re: X-Colouring Question

Postby BryanL » Thu May 26, 2011 12:32 pm

Hi Ron,

thanks for the interest.

It's from a windmill diagonal windoku so a text example wasn't going to show the extra houses. Here's a graphic.

Image

Initial conjugate pair is r45c8. Small coloured dots are peers.

r3c5 is coloured as an exception cell (last cell in b2) with an extra peer at r6c5. This leaves r5c6 as a potential exception cell (last in b5), if it is legal to re-colour the already marked blue peer of r5c8 green, thus eliminating r5c23.

My guess is that it is OK since after posting I followed another path via r2c2, r3c3 to get the same result.

There are so many house intersections in these things that I have seen the same eliminations from different starting conjugate clusters, even when different cells of a house are coloured during the progression of exceptions.

Cheers,

Bryan
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Re: X-Colouring Question

Postby ronk » Thu May 26, 2011 1:12 pm

BryanL wrote:thanks for the interest.

It's from a windmill diagonal windoku so a text example wasn't going to show the extra houses. Here's a graphic.

Sorry BryanL, but I'm not familiar enough with Windoku to provide an experience-based answer. Hopefully someone else will chime in.

P.S. I may move this thread to the Sudoku variants forum.
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Re: X-Colouring Question

Postby RW » Thu May 26, 2011 1:20 pm

I'm not familiar enough with colouring, but I can tell that the elimination you suggest is valid, no matter if the colouring rules allow you to do so or not.

Btw. using your terminology, isn't r5c6 an exception cell (last cell in c6) even without coloring r3c5 first?

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Re: X-Colouring Question

Postby BryanL » Thu May 26, 2011 1:33 pm

RW wrote:I'm not familiar enough with colouring, but I can tell that the elimination you suggest is valid, no matter if the colouring rules allow you to do so or not.

Btw. using your terminology, isn't r5c6 an exception cell (last cell in c6) even without coloring r3c5 first?

RW


Hi RW,

yes I noticed that after I had posted - but my question still stands because it is already peer of the opposite colour...

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Re: X-Colouring Question

Postby BryanL » Thu May 26, 2011 1:42 pm

ronk wrote:Sorry BryanL, but I'm not familiar enough with Windoku to provide an experience-based answer. Hopefully someone else will chime in.

The same rules apply, except there are extra houses imposed...

P.S. I may move this thread to the Sudoku variants forum.

I hope you don't because my OP was regarding a solving technique far more than it applies to a particular variant - even if it is unlikely to be seen in regular sudoku.

And I have another question re the same technique which is more likely to be answered in this forum.


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Re: X-Colouring Question

Postby JasonLion » Thu May 26, 2011 2:14 pm

It doesn't matter that it is a peer of A, that has no effect on the outcome. What matters is that all other candidates in a shared house are peers of B, so it is an exception cell and can be colored when you get to the appropriate step in the X coloring sequence.
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Re: X-Colouring Question

Postby BryanL » Thu May 26, 2011 2:28 pm

JasonLion wrote:It doesn't matter that it is a peer of A, that has no effect on the outcome. What matters is that all other candidates in a shared house are peers of B, so it is an exception cell and can be colored when you get to the appropriate step in the X coloring sequence.

Thanks Jason,

it was what I suspected but wanted clarification.

RW wrote:I'm not familiar enough with colouring, but I can tell that the elimination you suggest is valid, no matter if the colouring rules allow you to do so or not.

From an X-Chain?

Can an X-Chain include two weak links as a set?

[r5c8] = 3 = [r4c8] - 3 - [r3c6] = 3 = [r3c5] - 3 - [r4c5 and r6c5] = 3 = [r5c6] ???

I can see that the logic is valid either way.

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Re: X-Colouring Question

Postby BryanL » Thu May 26, 2011 3:41 pm

BryanL wrote:
Can an X-Chain include two weak links as a set?

Which leads onto my other question.

In X-Colouring, can a group of cells remaining in a house, left after the other candidates of that house are noted as peers of a colour, be taken as one exception cell? So that they can be used together with a peer in an intersecting house to find more exception cells?

Code: Select all
Contrived and otherwise invalid example
+-------+-------+-------+
| . a a | . . . | A . . |
| x . . | . . . | . . . |
| x . . | . . . | . . . |
+-------+-------+-------+
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
+-------+-------+-------+
| . . . | . . . | . . . |
| . . . | . . . | B . . |
| y . z | . . . | . b . |
+-------+-------+-------+

When the cells marked x are taken as a group of A, y becomes a peer of A, allowing z to be an exception cell thus eliminating b. Is that valid?


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Re: X-Colouring Question

Postby RW » Thu May 26, 2011 4:40 pm

BryanL wrote:
RW wrote:I'm not familiar enough with colouring, but I can tell that the elimination you suggest is valid, no matter if the colouring rules allow you to do so or not.

From an X-Chain?

Can an X-Chain include two weak links as a set?

Sorry, I usually don't bother with formal definitions of techniques, so I can't answer anything about what is "allowed" in a chain. In the puzzle, all I see is that either r5c8=3 or r4c8 and r5c6=3. In any case the rest of r5<>3.

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Postby ronk » Thu May 26, 2011 8:28 pm

BryanL wrote:Can an X-Chain include two weak links as a set?

[r5c8] = 3 = [r4c8] - 3 - [r3c6] = 3 = [r3c5] - 3 - [r4c5 and r6c5] = 3 = [r5c6] ???

Since r3c5 and r5c6 each see both r4c5 and r6c5, grouping the latter two is valid. However, what is the basis for the weak link highlighted blue above?
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Re:

Postby BryanL » Fri May 27, 2011 2:00 am

ronk wrote:
BryanL wrote:Can an X-Chain include two weak links as a set?

[r5c8] = 3 = [r4c8] - 3 - [r3c6] = 3 = [r3c5] - 3 - [r4c5 and r6c5] = 3 = [r5c6] ???

Since r3c5 and r5c6 each see both r4c5 and r6c5, grouping the latter two is valid. However, what is the basis for the weak link highlighted blue above?

Ahh, the grey shaded 3x3 areas are extra houses (windows in the windoku world - hence the name I guess), superimposed on the traditional 9x9 grid.

So r3c6 sees r4c8 because they are both in the top right "window" house.

FYI, the windows then lead to 5 implicit houses. In this puzzle, in which I have only shown the central 9x9, there are 4 other 9x9s that overlap. Also the diagonals are another 2 houses in each 9x9.

The puzzle is by dyitto in this thread

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