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 Sudoku Techniques.
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
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)
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)
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)
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>.
One can alternatively consider the possible 8s at ac9 to eliminate the same 8s at gh9.
Y Wing Style using canidates 8 & 9
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
- Start at 23 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
- Hidden pair 27 at de6 forbids ef4=2,df5=7, d6=59, e6=589 UP 30
- Hidden pair 19 at e12 forbids e89=19, e1=567, e2=57 UP 38
- Locked 8s at ac9 forbids gh9=8
- 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
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