The New Sudoku Players' Forum

[TopRank]

QBasicMac:

There are two xy-wings in this puzzle: r2c6(23) plays pivot for r3c5(12) and r4c6(13), so one of those is a 1. r4c5(13) cannot be 1 and r1c6(15) also cannot be 1 - the first was found by NateDogg but he did not use the term xy-wing.

Also r1c6(15) plays pivot for r3c5(12) and r5c6(52) so that one of those is a 2 and r5c5(27) cannot be a 2 and r2c6(23) cannot be 2 since both of those cells are influenced by both wings and one of the wings has to be a 2.

If you have a pivot xy and two wings that see the pivot, even if they don't see each other, xz and yz then one of those two wings has to have the value z. So any other cells that sees both wings cannot have the value z.

I expect you to see both xy-wings if you look over the diagram at the top.

[sweetbix]

QBasic Mac, sorry I missed the question.

The XY and XZ cells are in the same row, column, or box i.e. buddies
The XZ and YZ are also buddies but in a different group from above.
The candidates you eliminate are the Zs that intersect with the two Zs above.

Whichever way you label them, you eliminate from cells that intersect with two of the same label.

In this example TopRank had another XYwing - you could label this XZ r1c6(15), XY r3c5(12) and YZ r5c6(25).
Here you eliminate the Ys = 2s from the intersecting cells r5e5(27) and r2e6(23).

Someone else may be able to explain it better.



Mac[/quote]

[TopRank]

See if you can see how r3c5(12) can also be the pivot of an xy-wing

[QBasicMac]

sweetbix, toprank

The following is the best I can figure out from you two guys attempt to help me. It has to be wrong because when I apply it carefully to the puzzle for xyz=213 I incorrectly eliminate 2 from r2c5.

Mac

Code: Select all

Pattern to look for: In two boxes:
========
-- -- --
-- -- xz
-- xy --
========
-- -- yz
-- -- --
-- -- --
========
where xz and yz are in the same column (any of the three)
xy is not in the same row or column as xz
otherwise, which rows doesn't matter
other x's and y's can be anywhere in the two boxes or elsewhere.
The two z's must be the only two z's in the column.

stuff outside the 2 boxes doesn't count

(row/column reversed if pattern is rotated)

The possible end cases are
========          ========
-- -- --          -- -- --
-- -- x           -- -- z
-- y  --          -- x  --
========          ========
-- -- z           -- -- y
-- -- --          -- -- --
-- -- --          -- -- --
========          ========


Since in either case x falls in the same box, all other x's can be erased from that box.

In addition, the 2 intersections of the row and column containing y cannot have a y.

[ronk]

Code: Select all

Pattern to look for: In two boxes:
========
-- -- --
-- -- xz
-- xy --
========
-- -- yz
-- -- --
-- -- --
========

That a pattern for eliminating y. If you're trying to eliminate z, you have the yz misplaced. It must be in the column of the xy cell.

Code: Select all

========
-- *  --
-- *  xz
-- xy --
========
-- yz *
-- -- *
-- -- *
========

Then you can eliminate z in the cells marked '*'.

[TopRank]

You are looking for this pattern or an analogue

Code: Select all

========
-- -- **
-- -- xy
-- xz **
========
-- ** yz
-- ** --
-- ** --
========

xy is a pivot. It forces either xz or yz to be z. You do not have to know which one is z but you know that ** cannot be z because it can see two cells one of which is z.

You should see all three xy-wings in the first page, using a 12 as a pivot, using a 15 as a pivot and using a 23 as a pivot.