Sudoku Proof for Tough of December 23,2007


The following illustrated proof for the Tough Sudoku of December 23, 2007 employs a small group of Sudoku techniques, tips and tricks: Hidden Pairs, Y Wing Style, Locked Candidates, Alternating Inference Chains, and an Almost Hidden Pair used in AIC.

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)
  • 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


Puzzle at Start

Two Unique Possibilities are available here:

  • r5c3 = 2% Column (hidden single in column)
  • r2c7 = 9% Cell (naked single)
  • Thus by Unique Possibilities, the puzzle is advanced to 24 cells filled, UP 24

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. There are 3 hidden pairs immediately available based upon the givens

  1. (89) r6c4,r4c6 indicated by (89) given in column 5 and 1257 given in box 5
  2. (23)r9c45 indicated by (23)given in column 6, (23) given in row 7, (69) given in box 8
  3. (23)r8c89 indicated by (23)given in column 6, (23) given in box 7, (569)given in row 8
Of these the proof only requires hidden pair 89.


Step 2 - Hidden Pair 89


Hidden Pair 89

As noted, (89)in column 5 and 1257 in box 5 reveal, before any Possibility Matrix, the Hidden Pair 89. As a forbidding chain, also called an Alternating inference chains or AIC, all hidden pairs form a continuous loop, (wrap around chain). One could write:

  • (8)r6c4=(8-9)r4c6=(9)r6c4 => r6c4≠34 & r4c6≠6
After making these eliminations, one can get to UP 26 noting:
  • (3)r9c4 % column
  • (2)r9c5 % box


Step 3 - Almost Hidden Pair 15 in a short chain


Hidden Pair 15 chained

Above, but for the 5 at r2c4, one would have Hidden Pair 15 at r2c89. Alternatively, one could use the Almost Naked Triple 247 at r2c124. The chain illustrated above could be written:

  • (5)r2c4=(Hidden Pair 15 - 7)r2c89=(7)r1c9-(7)r1c5=(7)r3c5 => r3c5≠5
Note that the 7's are strong in column 5, thus I did not need to use the Locked 7's in column 5 first. One could, and in most cases would, do exactly that (Locked 7's r13c5 => r13c6 ≠7). After making the elimination of 5 from r3c5, one can get to UP 32 noting:
  • (5)r7c5 % column
  • (1)r1c5 % column
  • (7)r3c5 % column (thus we did not need the Locked 7's)
  • (8)r3c3 % cell
  • (3)r3c1 % cell
  • (1)r7c6 % row


Step 4 - Y Wing Style 89


Y Wing Style 89

Y Wing Styles are nothing but short chains using exactly 3 native strong inference sets. The one shown above can be written as:

  • (9)r9c3=(9)r6c3-(9=8)r6c4-(8)r7c4=(8)r7c2 => r7c2≠9
After making this elimination, one can get one more solved cell (UP 33) noting:
  • (9)r7c9 % row


Step 5 - A very common Y Wing Style Plus one link

Typically, two disparite cells containing exactly the same two candidates cause me to look for a Very Common Y Wing Style, also called a semi-remote pair. Below, both cells r7c7 and r1c6 are limited to candidates 46. There exists a strong link using candidate 4 to connect the two cells. Although this directly leads to one elimination of 6 from r1c7, adding one more strong link using candidate 6 reveals an additional chain that serves to signifiantly unlock this puzzle.


Y Wing Style 46 Plus one link

The chain shown above could be written:

  • (6=4)r7c7 -(4)r7c4=(4)r12c4-(4=6)r1c6-(6)r3c6=(6)r3c8 => r1c7,r9c8≠6
After making these eliminitions, 11 more cells are solved. The puzzle is thus advanced to UP 44. Here is the list:
  • (4)r9c8 % cell
  • (6)r7c7 % cell
  • (7)r9c6 % cell
  • (9)r9c3 % cell
  • (1)r6c3 % cell
  • (7)r8c3 % cell
  • (4)r5c7 % column & box
  • (4)r6c5 % column, box & row
  • (1)r8c1 % column, box & row


Step 6 - Locked candidate 8


Locked 8s

Above, candidate 8 in box 7 is limited to r78c2, therefor 8 cannot exist at r456c2. After making these eliminations, the puzzle is solved, starting with:

  • (8)r5c8 % row
  • Naked singles (% cells) to end (UP 81)


Solution


Solution


Proof

Below, find the path illustrated above in shortened form:

  1. Start at 22 filled - the given puzzle. Unique Possibilities to 24 filled. (UP 24).
  2. HP 89 r4c6,r6c4 => r4c6≠6, r6c4≠34 UP 26
  3. (5)r2c4=(hp15-7)r2c89=(7)r1c9-(7)r1c5=(7)r3c5 => r3c5≠5 UP 32
  4. Y Wing Style: (8)r7c2=(8)r7c4-(8=9)r6c4-(9)r6c3=(9)r9c3 => r7c2≠9 UP 33
  5. (6=4)r7c7-(4)r7c4=(4)r12c4-(4=6)r1c6-(6)r3c6=(6)r3c8 => r9c8≠6 UP 44
  6. Locked 8s r78c2 => r456c2≠8 UP 81
  • Sets 2 + 4 + 3 + 4 + 1 = 14
  • Max depth 4, twice
  • Rating: (.03)+2(.15)+.07+.01 = .41

Notes

Probably this puzzle is just about right for a tough. It cannot be solved without resorting to something beyond the standard pairs, triples, locked candidates. However, it does not require much heavy lifting. The awareness of Y wing Styles, and techniques much like them, easily undoes the puzzle. Note that in step 3, if one uses the super cell idea to compress r2c89 into one cell, that step reduces also to a Y Wing Style. Familiarity with the structure and look of such short chains can be an enormous help in unlocking puzzles such as this one.




1 Comment
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Dear all members,
I hove just subscribed , I want to see some proof but I can't, receive "Error finding blog post file '122307.txt'"
Please help
Thanks
Thuydung
01/Feb/08 2:14 PM
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07/Jan/07 Ywing Styles
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