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 Sudoku Techniques. 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.
Two Unique Possibilities (UPs) are available here.
- (6)b8 % box
- translation: Hidden single (6) at b8, as it is the only (6) possible in the box
- (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.
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
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
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
(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.
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
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