Y Wing Style Proof for March 27, 2007


The following is an illustrated proof for the Tough Sudoku of March 27, 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.

To understand this page, some previous blog pages that could be particularly helpful are:

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)
  • candidates crossed out in red = candidates proven false
Strong and weak need not be mutually exclusive properties.


Puzzle at start


Puzzle start

A few Unique Possibilities are available here:

  • g2 = 2% column & box (hidden single)
  • e7 = 3% row
  • b6 = 4% row



Hidden Pair 27 at 26 cells solved (UP 26)


Hidden Pair 27

Since abcf6 are already solved, and both 2 & 7 are located in box h5 outside of row 6, without considering the possibility matrix, one can limit cells de6 to only candidates 2,7.

This solves the rest of box e5:

  • e4 = 8% box
  • d5 = 5% box
  • f4 = 1% row
  • f5 = 9% box


Hidden Pair 19 at 30 cells solved (UP 30)


Hidden Pair 19

Once again, sans possibility matrix, the hidden pair 19 exists at e12. This is due to the solved cells d12, gh3, f45. Limiting e12 to only candidates 19 causes a cascade of Unique possibilities leading to 38 cells solved. Start the cascade with e3=5% box and column.

Finally considering the possibility matrix, there are many possible paths to take. One can use a group of standard easy techniques to advance the puzzle a bit further. This page illustrates only two more steps that are sufficient to completely unlock the puzzle.


Locked candidate 8 at 38 cells solved (UP 38)


Locked 8s

Above, since candidate 8 is limited to only gi7 in row 7, and since both gi7 are in box h8, 8 cannot exist in box h8 outside of row 7. One can write this easy step as a forbidding chain or Alternating Inference Chain (AIC>.

  • g7=8 == i7=8 => gh9≠8
One can alternatively consider the possible 8s at ac9 to eliminate the same 8s at gh9.


Y Wing Style using canidates 8 & 9


Y Wing Style

Eliminating the 8 at h9 has opened up a few Y Wing Style eliminations. One that immediately causes the puzzle to cascade into Unique Possibilities is shown above. Viewing hi6=9 as the vertex:

  • h6=9 => h6≠8 => h2=8 => i2≠8
  • i6=9 => i7≠9 => i7=8 => i2≠8
As a forbidding chain, one could write:
  • h2=8 == h6=8 -- h6=9 == i6=9 -- i7=9 == i7=8 => i2≠8
As a Y Wing style Strong Inference set list, one could write:
  • (h26=8) (hi6=9) (i7=98) => i2≠8

The puzzle is now nothing more than hidden and naked singles to the end.


Proof

  1. Start at 23 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
  2. Hidden pair 27 at de6 forbids ef4=2,df5=7, d6=59, e6=589 UP 30
  3. Hidden pair 19 at e12 forbids e89=19, e1=567, e2=57 UP 38
    1. Locked 8s at ac9 forbids gh9=8
    2. Y wing style: h26=8,hi6=9,i7=98 forbids i2=8 UP 81
  • Sets: 2+2+1+3 = 8
  • Max depth 3 at step 4.2
  • Rating: 2(.03) + .01 + .07 = .14
Not very tough!


Notes


Very Common Y wing style

Illustrated to the left, instead of the Y wing style illustrated, one could have used a Very Common Y Wing Style. These are easier to find than even the standard Y Wing - except that most computer aided sovlers will not find them for you!

Both column g and i have a cell limited to only candidates 58. All one needs is a strong link in either 5s or 8s to connect the dots. Here, column h has such a link in 8s, thus eliminating the 5s indicated. After these 5s are eliminated, the puzzle is again solved using only Unique Possibilities.

I see such Very Common Y Wing Styles often. They do not make it into the final proofs that I submit as often, since there often exists a more efficient manner to solve the puzzle. Nevertheless, if one is solving for speed, these are so easy to spot, they should be added to one's toolbox.

Today's puzzle thus has an alternate proof of equal rating, substituting step 4.2 above with:

  • Y Wing Style: g6=58. h62=8, i2=85 => g1,i6≠5 UP 81




5 Comments
Indicate which comments you would like to be able to see

Nice!

So step 5 in my solution could be vastly simplified by concentrating on the 8s in columns ghi. I had to arrive at the crucial contradiction using an assumption about the location of 5 in the bottom left region, and simply hadn't noticed that
one can use the fact that cell i2 is More...
28/Mar/07 10:35 PM
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Hi Jyrki!
Unfortunately, I suffer from the same confusion when I try to read about sudoku at other sites. The naming conventions used at most other sites are different than the ones that I learned here.
For that reason, I try to illustrate the grid that I am using constantly, so that at least More...
28/Mar/07 10:42 PM
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rihfmae zbtheiqao inxysdhgt mahzwc ldngrm pljiqr ujknrsml
27/Apr/07 3:30 PM
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ryiwe btzidfa yawdim zpdk tapigd ngyf egdzhq http://www.rnvmoc.gnxbs.com
27/Apr/07 3:30 PM
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A different solution path can be found by using the bilocal 9s in box h8, which result by using the hidden pair 16 at i58. Only simple steps are necessary after the AIC d7=4 == d7=9 -- i7=9 == h9=9 -- h6=9 == h6=8 -- g6=8 == g7=8 => g7<>4 is used.
29/May/11 9:58 AM
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