Final Page for Tough of November 28, 2007


The following is page four, the final page, of an illustrated solution for the Tough Sudoku of November 28, 2007. Their are no truly difficult steps on this page.

Page One of this solution was primarily about the use of an Almost Unique Rectangle in a chain. Page Two of this solution included one complex step linking several disparite types of Almost AICs. Page Three of this solution also included one complex step linking two Y Wing Styles. There are many ways to finish this puzzle. One path is illustrated below.

Previous blog pages may be helpful. Links to these pages are found to the right, under Previous Entries.


Step 5a - Locked candidate 9


Locked candidate 9


Above, candidate 9 is limited to column 5 within box 8, thus it cannot exist outside of column 5 within box 8.


Step 5b - a short chain using an Almost Naked Pair


Almost Naked Pair 13










On the left,

  • (3=4)r8c4-(4)r3c4=(4)r2c5-(4=pair13)r24c1
  • => r8c1≠3


Step 5c - a longer chain that could be viewed as two chains


Depth 6 AIC




On the left,

  • (8)r7c1=(8)r7c2-(8=3)r5c2-(3=1)r4c1-(1)r5c3=(1-4)r3c3
  • =(4)r23c1 => r7c1≠4
  • continue chain: -(4=5)r8c1 => r7c1≠5
Once one has done the work of finding the depth 5 chain forbidding (4) from r7c1, adding the strong inference set (sis) (4=5)r8c1 is very little extra work to eliminate (5) from r7c1. One might alternatively find:
  • (HP38)r7c12=(3)r89c3-(3=6)r1c3-(6=pair29)r12c2-(2)r1c1=(2-8)r6c1=(8)r7c1
  • => r7c1 limited to (38) => r7c1≠45
Again, there are always multiple ways to justify an elimination.


Step 5d - Locked candidate 5


Locked candidate 5


Above, candidate 5 is limited to row 7 within box 9, thus it cannot exist outside of row 7 within box 9.


Step 5e - Hidden Pair 59 - Hidden Pair 54 used in a chain


Dueling HP


The newly stronger 5's in box 9 row 7 form a super cell such that only one other candidate can fit into r7c79. Thus,

  • (9)r8c9=(HP59-HP54)r7c79=(4)r7c45-(4=3)r8c4 => r8c9≠3
  • => (9)r8c9 %cell
  • => (9)r7c5 %box (& %row, & %column)
  • => UP 31


Step 6 - Skyscraper with candidate 1


Skyscraper 1


The last step made candidate 1 stronger in column 5, revealing a simple coloring chain with candidate 1 that finishes the puzzle:

  • (1): r2c5=r5c5-r5c3=r3c3 => r2c1,r3c6≠1
  • => (9)r3c6 %cell
  • => (5)r6c6 %cell
  • => (5)r5c3 %box
  • => %cell (naked singles) to end => UP 81


Solution


Solution


Notes


  • Step 4b on page two is by far the most difficult step of this solution. It is depth 9, which is also the deepest step used in this solution.
  • Step 2d on page one used an Almost Unique Rectangle. I think that step is the primary puzzle breaker.
  • Step 4g on page three employed linked Y Wing Styles. I found that step to be the most fun.
All of the steps can be reduced to counting. Counting certainly works as an idea that catches everything. However, it seems to provide scant insight into how one might break a puzzle down into managable pieces. I may change my mind about that evaluation.....

The proof that counting can catch everything is trivial. Counting allows one to consider any arbitrary sis at any time. It also insures that each weak inference emanating from any arbitrary sis can be considered. Thus, there exists no sudoku axiom that counting cannot consider - except perhaps Uniqueness. However, uniqueness of solution is never required to solve a puzzle. It merely provides a convenient short-cut, at times.

Of course, this is nothing earth-shattering. It is not even sudoku world shattering! It really is only a brute force method of allowing complete axiomatic coverage. At this time, (for truly difficult puzzles), I primarily locate potential puzzle weaknesses using snippets of known techniques. Sans such devices for deciding where to look, I am not sure that counting saves time. In fact, it may just be too general to be of much use in deciding where to look.




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

Hi STEVE,

Thank you very very ... much for your works on this Site!
Can I have one (only one) question : How about your rating for this puzzle...?
05/Jan/08 2:52 AM
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Hi Steve,

This blog is far and away the most useful site for advanced methods (beyond the logic of sudoku). I have been able to solve about half a dozen 'unsolvables,' those Stuart's solver cannot crack using your advanced AIC's. One problem: I cannot locate your discussions of the 11/28 More...
09/Jan/08 12:39 PM
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Hi Tom!

There have been some server problems. The blog info was one of the casualities. In a few days, all should be restored.

Thanks so much for your kind words!

09/Jan/08 6:32 PM
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Hi ttt!

Blogged Solution rating for puzzle of 11/28/07:

10.39

Interestingly enough, the best rating that I can come up with for the puzzle of 11/26/07 is

10.38

When the blog problems get fixed, I will probalby also blog that one.
10/Jan/08 7:50 AM
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For whatever it is worth, I have made a list of sudokus from sudoku.com.au that Andrew's solver cannot crack. I have only been at it since late August and mayhave missed a few. I have been able to complete all but 10-3-07.

AU 8-26-07 More...
11/Jan/08 10:39 AM
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Gath, any idea when the sudoku blog will be back? Really miss it (and need it!)!! Thanks.
29/Jan/08 7:34 AM
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Thomas, What about the tough from Jan 14th, 2008? If I turn off the two guessing methods Andrew's solver does not solve it.
06/Feb/08 9:08 AM
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Jeff and Steve,

1. 1/14/08 is unsolvable as is 1/15/08. I think I have a solution to 1/14. 1/30/08 is the hardest sudoku I have ever seen. Sudocue.net rates it at over 20,000, making it harder than the Easter Monster or Escargot. I cannot even begin using any of the methods described More...
08/Feb/08 7:07 AM
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Hi Thomas!
I have not had much time lately, but I have looked at 1/14. Hopefully I will blog it soon. Guess that I have a lot of work to do....
08/Feb/08 12:19 PM
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Steve,

I saw your solution to 1/14. It is better than mine. I think I have 1/30 cracked, but the solution is a mess. It uses Dual Forcing AIC's. I would not have been able to get it had I not read your study in chain forging. I read somewhere that I could post solutions on the Tough More...
10/Feb/08 7:33 AM
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Thanks for the notice of these tough puzzles. I am not certain that I will find the time to crack all the really tough ones. I am not sure that I have the ability to crack the super tough ones, either!

On the tough pages, in the comment area - people (including once upon a time, me) have More...
10/Feb/08 1:23 PM
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Steve,

I finally got through the new pages. A couple of questions: At step 3d, you did not eliminate the 3 at r4c6. It seems to me that the 3 should have been eliminated by your chain. I found it easier to see by rewriting the chain as a continuous wrap around. The 3 then falls out of More...
14/Feb/08 2:44 AM
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I have only glanced at step 3d again. It seems that two things are of note:
As written, it is not really a wrap-around chain. (xcell)b-(pairxy)cellsab is not a true statement.
One may be able to eliminate the 3 at r4c6, but I think that it would require consideration of at least one sis not More...
14/Feb/08 7:54 PM
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typo above: but (8)r6c12-(8)r7c2 is not evident.
14/Feb/08 7:56 PM
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