The New Sudoku Players' Forum

[Sped]

Code: Select all

 *-----------------------------------------------------------*
 | 1     458   2     | 6     57    9     | 3     48    478   |
 | 58    589   6     | 45    457   3     | 2     1     789   |
 | 7     49    3     | 1     8     2     | 49    6     5     |
 |-------------------+-------------------+-------------------|
 | 6     35    1     | 9     345   45    | 8     7     2     |
 | 58    358   7     | 2     1     6     | 459   3459  49    |
 | 4     2     9     | 8     35    7     | 1     35    6     |
 |-------------------+-------------------+-------------------|
 | 9     7     8     | 45    6     1     | 45    2     3     |
 | 3     1     5     | 7     2     48    | 6     489   489   |
 | 2     6     4     | 3     9     58    | 7     58    1     |
 *-----------------------------------------------------------*


Firstly, I want to apologize if the notation I use is non standard. I haven't quite figured out the proper way to express XY chains.

4-(r3c2)-9-(r3c7)-4-(r1c8)-8-(r9c8)-5-(r7c7)-4

This is a discontinuous XY chain of 5 bivalue cells, starting at r3c2 and meandering to r7c7. It should allow the exclusion of 4s in cells that see both r3c2 and r7c7.

But.. this would mean that the 4 in r3c7 is excluded. r3c7 is part of the chain. Is the exclusion still valid?

If the exclusion is valid, and r3c7 is forced to 9, does this solve the other cells in the chain? r3c2=4, based on completing row 3. But does it also lead to r1c8=4, r9c8=8 and r7c7=5?

Those are all the correct values for the cells, but does the chain prove it? Does it even prove the exclusion of the 4 in r3c7?

[ravel]

Your xy-chain says, that either r3c2 or r7c7 must be 4, therefore you undoubtely can eliminate 4 from r3c7 safely (though it is part of the chain). But you would have needed only a part of the chain to get there:
4-(r1c8)-8-(r9c8)-5-(r7c7)-4
After the elimination you cannot follow, that all cells in the chain have to get the alternative value. E.g. the 4 in box 3 also could go to r1c9, or the 5 in column 7 to r5c7.

[Sped]

Your xy-chain says, that either r3c2 or r7c7 must be 4, therefore you undoubtely can eliminate 4 from r3c7 safely (though it is part of the chain). But you would have needed only a part of the chain to get there:
4-(r1c8)-8-(r9c8)-5-(r7c7)-4
After the elimination you cannot follow, that all cells in the chain have to get the alternative value. E.g. the 4 in box 3 also could go to r1c9, or the 5 in column 7 to r5c7.

Thanks for clearing that up for me.

And thanks for pointing out the XY Wing:

4-(r1c8)-8-(r9c8)-5-(r7c7)-4

That's more useful than my chain as it eliminates the 4 in r8c8 also.

[ravel]

You might be interested, that there are cyclic bivalue or bilocation chains, where you know that for the whole chain you only have 2 possibilities. You then can make eliminations based on that. See here an example posted by tso and a very sophisticated one by absolute beginner. I think, there is a thread by Jeff about that, but i cannot find it now.

[emm]

Here is Carcul’s link – Nice loop notation - and Jeff’s - Terminology & definition.

We don’t hear much about XY chains in this forum, Sped. I agree with you that it seems a simpler description when it leads to multiple eliminations than using two forcing chains or one closed loop for every single elimination.

No-one has replied to your query about your notation for an ‘open loop’ which I would take to mean that no-one has a better way to describe it. Your way makes sense to me.

[Sped]

Here is Carcul’s link – Nice loop notation - and Jeff’s - Terminology & definition.

We don’t hear much about XY chains in this forum, Sped. I agree with you that it seems a simpler description when it leads to multiple eliminations than using two forcing chains or one closed loop for every single elimination.

No-one has replied to your query about your notation for an ‘open loop’ which I would take to mean that no-one has a better way to describe it. Your way makes sense to me.

I think XY chains have the reputation of being difficult for a human solver to see, so most people don't often look for them. That's unfortunate because with a little practice it's not too hard to pick them out, and they turn up a lot.