dailysudoku.com

[keith]

[Earl]

I solved this with two key moves;

After standard elimination, there is a UR (parallel pairs) at R2C1&7, R3C1&7 which yields a 3 in R3C1.

After subsequent elimination, there is a skyscraper of 9's R3C2&4, R8C3&4, which eliminates the 9 in R7C2. This opens up the puzzle for solution by simple eliminations.

Any one find a more direct solution?

Earl

[Steve R]

A w-wing with pincers (39) does the trick on this one.

Steve

[keith]

Basic methods get you here:

[Marty R.]

[keith]

[Steve R]

Curiously, the x- and w- wings are almost the same here.

This is Keith’s grid again with the finned x-wing for 9 in rows 2 and 7 marked with x:

[Mogulmeister]

Marty - take a look at this:

Using Colouring put in the conjugates for 9 in two different colours:



Notice that we use green and orange for the different polarities - notice also that we have two orange cells in the same house (box 7) QED orange MUST be false. What is also great is that orange is false throughout and therefore ALL orange cells here are eliminated..!



Hooray for colouring - strong links etc.

[Marty R.]

[Mogulmeister]

[Marty R.]

[Steve R]

Marty

In logic it does not matter what you call it: you are using two conjugates which are linked (= a fork argument) however you look at it. In terms of pattern recognition it is useful to make a distinction.

I think Keith may have meant that all the structures you mentioned solved the puzzle. All four structures below are based on 9:

(a) There is a finned x-wing in rows 2 and 7. It eliminates 9 from r8c3.

(b) There is a skyscraper in rows 2 and 8. It eliminates 9 from r2c3 and r7c2.

(c) There is (what I think is now called) a two-stringed kite in column 2 and row 8. It eliminates 9 from r2c3 just as you say.

(d) Simple colouring eliminates 9 from r2c3, r3c4, r7c2, r7c6 and r8c3 in one fell swoop.

(a), (b) and (c) are all turbot fish (grouped in case (a)).

I have an idea you were flying a kite.

Steve

[Asellus]

I would not have recognized the W-Wing because I thought of W-Wings based on the graphic posted elsewhere showing it in a row (or column) of Boxes with "Not X" in the "third row" of the "third box" eliminating Y from the buddies of the two bivalue cells.

This example's alternate geometry reveals a more concrete notion of the logic to me. Let me give it a try:

For a matching pair of bivalue cells, XY, if there exists a strong link in a third "House" for one of these digits, X, the two ends of which "see" the two bivalue cells, then Y can be eliminated from all "buddies" of the two bivalue cells. This is a W-Wing.

In the previous graphic (alluded to above), the "Not X" is not what's essential. It is that it implies a strong link in X in the same two rows as the bivalue cells but in the third Box (a separate house). But the third house can be anywhere. In the present example, it is Row 2.

Maybe these remarks will help clarify this concept for some others. Or... maybe not.

[Asellus]

PS:

I too saw the finned X-Wing on 9, but only in the rows. Interesting that it is also a finned X-Wing in columns. A two-finned mutant!!!

[Mogulmeister]

[keith]

[Mogulmeister]

I tend to agree and I always recommend Havard's excellent explanation
of strong links which helped me a great deal in the early days.

As a development aid and a shorthand, the patterns are helpful while the fundamental understanding is still sinking in and being assimilated - especially if, like me you are more visually orientated.