Y Wing Style and Coloring Proof April 10, 2007

The following illustrated proof for the Tough Sudoku of April 10, 2007 employs exactly three Sudoku techniques: Hidden Pairs, Very Common Y Wing Style, and Coloring . Any technique that uses exactly one type of candidate to justify an elimination I call Coloring.

Previous blog pages may be helpful. Links to these pages are found to the right, under Previous Entries. This list is getting long, so specifically, one may want to refer to the following previous blog pages:

Many steps not illustrated are possible. To show every possible step is, well, just overkill. However, this page illustrates sufficient deduced elimination steps to both solve the puzzle, and also to prove a unique solution.

The illustrations of forbidding chains, also called Alternating Inference Chains (AIC), shown below will share this 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 partioning of a strong set(s) (not used)
  • candidates crossed out in red = candidates proven false
Please be aware that, for me, strong and weak need not be mutually exclusive properties.

The Puzzle


Two Unique Possibilities are available here:

  • e2 = 7% row (hidden single in row)
  • i8 = 2% cell (naked single)
Typically, at this point, and more so if I am solving the puzzle on paper, I prefer to search for Hidden Pairs before filling in the Possibility Matrix. I am probably in the minority in this regard!

Hidden Pair 12

Hidden Pair 12

The relatively symmetrical placement of candidates 1 & 2 in column b, column i, and box e5 make the Hidden Pair 12 at ch6 relatively easy to spot. Since ch6 are the only two places left for candidates 1,2 in row 6, no other candidates can exist in those cells. Illustrated above is the forbidding chain representation of this technique:

  • c6=1 == h6=1 -- h6=2 == c6=2 => c6=12.
One can at this point make many other eliminations. However, cell f4=89 and cell a2=89 is a signal that a very easy to use and spot elimination may be available.

Very Common Y Wing Style using candidates 89

Very Common Y Wing Style

The 8s in row 5 ensure that at least one of a2,f4 must be 9. Thus, all cells that see both a2 and f4 cannot contain 9. The elimination is illustrated as a forbidding chain above:

  • a2=9 == a2=8 -- a5=8 == e5=8 -- f4=8 == f4=9 => a4,f2≠9
If you have trouble understanding this elimination, suppose:
  • a4=9 or f2=9
  • => a2 = f4 = 8
  • => row 5 has no possible locations for 8
  • There is always a proof by contradiction available for every possible elimination. However, the goal of most Sudoku techniques is to find eliminations by recognizing some sort of pattern. As such, this pattern is very easy to identify. Nevertheless, I have not seen this pattern well utilyzed in other sites describing tips, tricks and techniques to solve Sudoku puzzles. I call this pattern a Very Common Y Wing Style because the logic is analagous to a standard xy Wing or Y Wing, and the pattern is fairly prevalent. In my opinion, it is more prevalent than a standard Y wing, and it is easier to find!

    Many puzzles are designed to be solved using a Y Wing, but few are designed to be solved using the Very Common Y Wing Style. Nevertheless, despite the fact that the pattern is not designed to exist, it is still occurs with enough frequency that it should be standard issue in any Sudoku Toolbox.

    In my experience, I have found many puzzles designed to be very difficult that are trivialized with this technique. Hopefully, I have driven this point home!

    Perhaps, if this technique had a more catchy name, it could be included in the standard issue Sudoku solving list. Feel free to suggest such a name! It is quite likely, however, that this technique already has a name given to it by the Sudoku community. If anyone can identify that name, please let me know. Then, I can call it by its proper name.

    Hidden Pair 79

    Hidden Pair 79

    After knocking off the 9 at a4, both candidate 7 and candidate 9 are limited to the same to cells within box b5. Since these two cells are also aligned in column b, all the eliminations above are justified by this situation. As a continuous nice loop forbidding chain:

    • b4=9 == b6=9 -- b6=7 == b4=7
    • => b178≠9, b79≠7, b4≠8, b6≠3
    This elimination group causes a cascade of Unique Possibilities, beginning with b1 = 8% column. If you find all the hidden singles and naked singles that result from this cascade, you should arrive at 49 cells solved. (UP 49).

    Coloring with candidate 8

    Coloring with candidate 9

    The possible 8s in both row 7 and row 2 allow the eliminations shown above. As a forbidding chain with candidate 8, one could write:

    • i7 == f7 -- f2 == h2 => h9,i3 ≠8
    Again, one can arrive at a proof by contradiction by considering What if h9 or i3 are 8? However, most prefer to find this elimination by recognizing the pattern caused by the relative configuration of candidate 8 in rows 2 & 7.

    After making the eliminations noted above, the puzzle is reduced to another progression of Unique Possibilities, this time until the puzzle is complete. Start with i7 = 8% column and box, then h9 = 4% column and box. Naked singles are then available to the end.


    1. Starting with the given puzzle, Unique Possibilities advance it to 24 filled. (UP 24)
      1. Hidden Pair 12 at ch6 => c6≠379
      2. Very Common Y Wing Style: f4=98, ae5=8, a2=89 => a4,f2≠9
      3. Hidden Pair 79 at b46 => b178≠9, b79≠7, b4≠8, b6≠3 UP 49
    2. fc on 8s: i7 == f7 -- f2 == h2 => h9,i3≠7 UP 81
    • Native Strong Sets: 2 + 3 + 2 + 2 = 9
    • Maximum Depth, at step 2.2: 3
    • Rating based upon this proof: 3(.03) + .07 = .16 A very easy tough.


    If anyone decides to try to better name the main technique used in this puzzle, perhaps an alternative place to post such could be at either of the following Sudoku Forums:

    Indicate which comments you would like to be able to see

    How to make an easy tough tougher:
    1) Start at 22 filled. UP 24.
    2) b4=7 == g4=7 -- i6=7 == i6=6 -- d6=6 == d6=9 -- f4=9 == f4=8, forbids b4=8. UP 27.
    3) locked 3's at c13, forbid c5679=3. UP 49.
    4) d1=1 == d1=5 -- i1=5 == i1=4 -- h2=4 == h2=8 -- f2=8 == f2=5, forbids d1=5. UP 60.
    5) fc on 8's: i7 == i3 -- h2 == f2, forbids f7=8. UP 81. ☺
    10/Apr/07 11:57 PM
     |  |
    Not too much tougher Dave!
    By some criterion, you get to UP 49 more efficiently than I do, as you only consider 5 strong sets, and I use 8. On the other hand, I am at max depth 3, and you at 4. The rating of your proof to get to UP 49 is .16, while mine is at .13 - a very insignificant More...
    12/Apr/07 1:34 PM
     |  |
    prtna wmpf idsopamqw bqoyprk dblqfwr yivsw ufvq
    27/Apr/07 8:50 AM
     |  |
    mhubanej yraxf wntoz hnzm ugsxo tcewkgqz tpvwixu http://www.xzrtahnwv.xrpuyijq.com
    27/Apr/07 8:51 AM
     |  |
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