# Y Wing Style Again for March 20, 2007

The following is an illustrated proof for the Tough Sudoku of March 20, 2007. The primary key used in this proof is, once again, Y wing styles.

You may need to refer to previous blog pages to understand this proof. Links to these pages are found to the right, under Previous Entries.

At many times during this illustration, there are other steps available. It is not the goal of this page to show every possible step, but rather to illustrate steps that, taken together, unlock this puzzle.

The illustrations of forbidding chains used in this proof will share the same key:

• black line = strong inference performed upon a set (strong link)
• red line = weak inference performed upon a set (weak link)
• black containers define a mulit-native strong set Boolean
• candidates crossed out in red = candidates proven false
Strong and weak need not be mutually exclusive properties.

### Puzzle at start

A few Unique Possibilities are available here:

• a9 = 1% box (hidden single)
• i1 = 1% box
• c8 = 4% box and row
• g2 = 6% box and row
One may note that this puzzle is almost absurdly symmetrical regarding candidates 1,4,6. If it were just a bit more symmetrical regarding some other candidates, one could make Uniqueness of solution arguments before solving a single cell!

### Possibility Matrix at 26 cells solved (UP 26)

At this point in the puzzle, there are a few locked candidate steps one can take. Rather than list every possible step, the rest of this page will concentrate on only those steps that are required to illustrate the theme: Y Wing Styles.

The next few steps require only the lower third of this puzzle

### Locked Candidate 2

Since ef3 are the only locations left for 2 in row 3, 2 is excluded from the rest of box e2. Naturally, one could alternatively argue that since ac1 are the only locations left for 2 in box b2, that 2 is excluded from the rest of row 1. Either way, f1≠2.

### Locked Candidate 5

Since hi3 are the only locations left for 5 in box h2, 5 is excluded from the rest of row 3.

## Three ways to see the same Y Wing Style

Since the goal of this blog is to teach and explain, I have decided to present the same idea in three similar, but different, ways.

The following illustrations have many different names across the Sudoku Solving world.

### Almost Locked Set: Pair 47 and cell 57

Above, I have illustrated a Y Wing Style elimination using the following Almost Locked Sets:

1. b13 = 457
2. f1 = 57
As a pure ALS, the logic would go like this:
• Set 1 = {b1 = 457, b3=47}
• Set 2 = {f1 = 57}
• Since candidate 5 is a restricted common (weak across the sets) (can exist in only one of the two sets, Set 1 & Set 2,
• The 7s are at least locked into one of Set 1, Set 2
• Thus any cells that see all the possible 7s in both sets cannot contain 7
I much prefer the forbidding chain or Alternating Inference Chain (AIC) representation of the same idea:
• {pair 47 at b113} == b11=5 -- f1=5 == f1=7 => ac1≠7
No matter how one expresses the idea, the following proof by contradiction certainly exists:
• Suppose ac1 contain 7
• => (b3=4 & f1=5) => b1=7 => too many 7s in both box b2 and row 1

### Almost Locked Set: Pair 57 and cell 47

Above, the exact same eliminations can be viewed as relying upon the 4s being weak across two groups:

• {pair 57 at b11f1} == b11=4 -- b3=4 == b3=7 => ac1≠7
Notice that in either viewpoint, one is still considering the same native puzzle information:
1. f1 = 57
2. b1 = 457
3. b3 = 47
One can say that both eliminations rely upon 3 native sets upon which one can apply a strong inference (at least one of the items contained in the set must be true). I differ somewhat with other sites by calling all such sets strong. Many sites reserve strong for sets that have only two possibilities. This is somewhat absurd, as one can always partition the set into two pieces: for example, b1=4 == b1=57, as used above. Freely allowing such groupings is a main difference in my point of view from those that one may find elsewhere.

### Almost Y Wing

Above, I have illustrated yet another way to see the same elimination using the same information. Here, one would have a typical Y wing at b1, b3, f1 but for the 7 at b1. Clearly, since both the Y wing and b1=7 would forbid ac1=7, ac1≠7. As a forbidding chain:

• {Y Wing f1=75, b11=54, b3=47} == b11=7 => ac1≠7

### Booleans

Hopefully, these three illustrations have exposed the power of using Booleans (merely something that is True or False) as an argument within a forbidding chain. Doing so allows one to freely chunk, or use what one already knows about sudoku solving, in almost limitless combinations.

### Y Wing Style vertex presentation

If all three of these ideas are truly a Y Wing Style, then one should be able to define a vertex and two endpoints. In this case, the cell b1=457 is the vertex:

• b1=7 => ac1≠7
• b1=4 => b3≠4 => b3=7 => ac1≠7
• b1=5 => f1≠5 => f1=7 => ac1≠7

The original page on Y Wing Styles uses a matrix argument to clearly bring all Y Wing Styles under one umbrella.

### Locked candidate 7

The newly strong 7s in box b2 are limited to b13, thus eliminating 7 from the rest of column b.

Then, b7 = 6% cell begins a cascade of mostly naked singles (%cells), plus a few hidden singles, until the puzzle is complete.

### Proof

1. Start at 22 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
1. Locked 2s at ef3 forbids f1=2
2. Locked 5's at hi3 forbids bdef3=5
3. Y wing style: f1=57, b1=457, b3=47 forbids ac1=7
4. Locked 7s at b13 forbids b567=7 UP 81
• Sets: 3(1) + 3 = 6
• Max depth 3 at step 2.3
• Rating: 3(.01) + .07 = .10 (not tough)

### Notes

There are many other Y Wing Styles possible in this puzzle. Many can be used to help unlock the puzzle. There are also numerous steps that consider more than 3 native strong inference sets that directly unlock this puzzle. As such, this puzzle can be used as an excellent exercise in finding a great many paths to solution.

 Indicate which comments you would like to be able to see GeneralJokesOtherSudoku Technique/QuestionRecipes
 I can't find the explanation of this notation used in this solution. pair at b1(subscript 1)3 b1(subscript 1)=5 20/Mar/07 6:03 AM |  |
 Thanks Justin!I should add that to my definitions page, and soon I will! The subscripts are used to show when a strong inference involves more than two parts. All the parts of such a strong inference are marked with a subscript. 20/Mar/07 1:05 PM |  |
 Steve, sorry to post this comment here, but wasn't sure whether you revisit old blogs. Thank you for your explanation on the hidden pair in Nice Loops. It makes perfect sense now. Have been doing Toughs from the Archives and reading gb's notes with great interest. Also working on ones on Paulspages More... 20/Mar/07 3:09 PM |  |
 Hi Giblet!Thanks for the comment!You may post a comment on any blog page, as Gath has set the blog up so that I receive an automatic email telling me about any comment on any blog page. 20/Mar/07 7:50 PM |  |
 bpgryqcov jwlm jwmb wvehg kfugjbql rwnaljgs yzdaikhso 28/Apr/07 7:10 AM |  |
 pztmwbi xseanhfl fvcdalr clbwet vekdyb shqpcl rvdmy http://www.skngrdy.afuk.com 28/Apr/07 7:10 AM |  |

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