W Wings in June 16,2009 tough


The following illustrated proof for the relatively easy Tough Sudoku of June 16, 2009 serves as a reasonably good example of the power of one type of relatively simple Sudoku solving technique. This particular Sudoku solving strategy, tip or trick is sometimes referred to in Sudoku forums as the W Wing. This blog discusses this strategy numerous times, albeit under a different name. I previously have referred to this technique as a very common Y wing style. It is disussed in some detail on the Y Wing Styles page.

If this is your first visit to this blog, welcome! Unfortunately, if you are a first time visitor, this page may seem like it is written in a different language. Well, it is!! Previous blog pages may be helpful. Links to these pages are found to the right, under Previous Entries. The earliest posts are at the bottom, and if you have never perused the intricacies of our special coded language here, you may wish to start close to beginning.

In some of the illustrations, more than one step is shown at once. This reflects an attempt to shorten the page. However, it may contribute to some confusion. Hopefully, the confusion is managable.


The Puzzle

For those whom do not wish to struggle through the illustrations, I have written the steps taken in this proof at the bottom of the page in sudoku.com.au notation.


Puzzle at Start

Two Unique Possibilities (UPs) are available here.

  1. (6)b8 % box
    • translation: Hidden single (6) at b8, as it is the only (6) possible in the box
  2. (6)d1 % box & column
This brings the puzzle to UP 24 (Unique Possibilities to a total of 24 given plus solved cells).


Step 2b: A very common Y Wing Style, or W Wing

Below, two steps are shown. Step 2a in column b and box b5 is a very easy step:

  • Locked Candidate 5 (LC5) b46 => c56<>5
Other similar very easy steps are also possible. A step that, IMO, is almost as easy as the group of standard Sudoku solving techniques, strategies, tips or tricks is shown in the center three boxes. The common marker to look for this step is two cells limited to exactly the same two candidates. When these two cells do not share a common container, but do share a bi-local strong inference set that sees the two cells, one has a W Wing. Oftentimes, as illustrated below, the bi-local strong inference set uses one or more groupings to create the two condition state.


Steps 2a and 2b: A locked candidate plus a W Wing

Above, the fact that d123 is devoid of (5) as a possibility creates a grouped bilocal condition with candidate (5) both between boxes e5&e8 in column d and between columns e&f within box e2. I illustrated the latter above, although either one will do for the deduction:

  • (9=5)f4 - (5)f12 = (5)e12 - (5=9)e9 => e5,f9≠9


Steps 2c & 2d: Some Locked Candidates

Below, again two steps are shown.

  • (LC9)ac5 => ab4≠9
  • (LC9)b23 => ac12≠9


Steps 2c and 2d: Locked Candidates with candidate 9

Since candidate 9 is now limited to ef1 in row 1, we have another W Wing, this time using bilocal (9).


Step 2e: Another W Wing

Below, the same two cells considered in step 2b are part of another W Wing


Step 2e: W Wing again

The W Wing could be written:

  • (5=9)f4 - (9)f1 = (9)e1 - (9=5)e9 => e5,f9≠5
This leads to a short cascade of singles, both hidden and naked, that brings the puzzle to UP 35

Of some minor interest, (59)e9 and (59)f4 are remote pairs.


Step 3: Another W Wing

Below, yet another W Wing, this time using some new information


Step 3: W Wing using bilocal 2

(23)e5, (23)d3 are key to search for a bi-local link with either candidate 2 or 3. There is a grouped link in column f, and a simple bilocal link in row 8. The latter is shown as:

  • (3=2)d3 - (2)d8 = (2)e8 - (2=3)e5 => e2,d46≠3
This leads to 16 singles that bring the puzzle to UP 51


Step 4: A short bi-value AIC

Below, there are many ways to finish this puzzle. One could use a Unique Loop with candidates (59)at ef1, df4, de9. One could eventually use a BUG +2. ETC. The short chain illustrated below also suffices.


Step 4: A short bivalue chain

Above, find:

  • (1=2)e2 - (2=3)c2 - (3=4)c7 - (4=1)f7 => f23,e8≠1
A cascade of naked singles finishes the puzzle.


Solution


Solution

Steps taken in this proof:

  • 1) Start with 22 givens. UP 24.
  • 2a) (LC5)b46 => c56≠5
  • 2b) WWing: (9=5)f4 - (5)f12 = (5)e12 - (5=9)e9 => e5,f9≠9
  • 2c) (LC9)ac5 => ab4≠9
  • 2d) (LC9)b23 => ac12≠9
  • 2e) WWing: (5=9)f4 - (9)f1 = (9)e1 - (9=5)e9 => e5,f9≠5 UP 35
  • 3) WWing: (3=2)d3 - (2)d8 = (2)e8 - (2=3)e5 => e2,d46≠3 UP 51
  • 4) (1=2)e2 - (2=3)c2 - (3=4)c7 - (4=1)f7 => e8,f23≠1 UP 81
If I still used my old and often irrelevant rating system, this soltuion's rating spectrum would look like:
  • Sets: 1+3+1+1+3+3+4 = 16
  • Maximum depth: 4 at step 4
  • Rating: .01 +.07 +.01 +.01 +.07 +.07 +.15 = .39




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