Diabolical Proof for March 13 2007


The following is an illustrated proof for the Tough Sudoku of March 13, 2007. This is a very difficult puzzle to solve, unless either one resorts to guessing, or truly understands using strong sets.

There are many ways to tackle this one. The primary focus of this proof is to illustrate using Forbidding Chains, also called Alternating Inference Chains or AIC. Some of the Chains that are presented are complex forbidding chains - or advanced forbidding chains. Such chains merely use arguments that are Boolean variables (something that is either TRUE, or False). Some of these Booleans represent native puzzle conditions that are neither bivalue in a cell, nor candidates that are bilocation.

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. Some steps that are not required will be illustrated. The purpose of illustrating such steps is primarily to clue the potential solver on how to decide upon a plan of attack for truly diabolical puzzles.

The information on the following blog pages is required to understand this page:

The illustrations of forbidding chains used in this proof will share the same key:

  • black lines = strong links
  • red lines = weak links
  • black circles connected with black lines = multi-part strong sets
  • candidates crossed out in red = candidates proven false

Puzzle at start


PUzzle start

A few Unique Possibilities are available here.

They are illustrated in the next picture.

Puzzle at UP 26

26 cells solved

Completely illustrating each of the relatively easy steps will consume much space. Therefor, easy steps will be referenced, without complete illustrations.

To advance this puzzle a very small bit further, here is what I found:

  1. Locked 7's at hi8 forbids hi9=7
  2. Locked 8's at gh8 forbids gh7h9=8

Possibility matrix at 27 after some locked candidates

triple

Illustrated above is the naked triple 145, forbidding the indicated items.

i1 = 5% column & box

Puzzle at UP 27

triple

Above, the hidden pair 14 justifies the indicated eliminations

Two sixes solve

Puzzle at UP 29

hmmm

Above, one can make some easy eliminations. One step is rather obvious to me that unlocks this puzzle

The Step

The step

Above, the Almost Locked Set 259 at e346 would lock the 2's at e46 but for the 3 at e6. Since df5=2 == h5=2 and h5=8 == g6=8, there can be no 3 at g6.

Suppose that g6=3:

  • => g6≠8 => h5=8 => h5≠2 => df5 contain 2
  • => e46≠2 => pair 59 at e34 => e6≠25 => e6=3 => contradiction

As a forbidding chain, one could write:

e6=31 =={triple e61=25, e4=259, e3=59} -- df5=2 == h5=2 -- h5=8 == g6=8 => g6≠3

Most solving systems published elsewhere will miss this elimination and instead find the Death Blossom that forbids e79=9. The Death Blossom is a very nice technique, but it is no more complex than the elimination noted above. In fact, it considers many of the same strong sets.

Eliminations like the one above are precisely the reason that I present this blog in such a general fashion.

The puzzle mark-up only weakly indicates this elimination, as the ALS used is not part of the mark-up. That is why I have included the blog page on ALS.

After making this elimination, the puzzle is reduced to Hidden and Naked singles to the end

Done

Solution

Proof

  1. Start at 23 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
    1. Locked 7's at hi8 forbids hi9=7
    2. Locked 8's at gh8 forbids gh7h9=8
    3. Naked triple 145 at gh7h9 forbids i9=145 UP 27
  2. Hidden pair 14 at i46 forbids g46h4=14, i4=7, i6 = 3 UP 29
  3. e6=31 =={triple e61=25, e4=259, e3=59} -- df5=2 == h5=2 -- h5=8 == g6=8 forbids g6=3 UP 81
  • Sets: 1+1+3+2+5 = 12
  • Max Depth 5 at step 4
  • Rating: .02 + .07 + .03 + .31 = .43
This puzzle is thus not diabolical at all!




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

Please feel free to provide alternate proofs!

Of some minor interest, Andrew Stuarts solver resorts to Bowman's Bingo to unlock this relatively easy diabolically tough puzzle.

Ok, so perhaps it is both diabolical and relatively easy, depending on one's point of view!

13/Mar/07 3:36 AM
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BTW:
One can reduce the proof illustrated above to something a bit shorter:
1)Start at 23 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
2) e6=31 =={triple e61=25, e4=259, e3=59} -- df5=2 == h5=2 -- h5=8 == g6=8 forbids g6=3 UP 81
rating .31 - sets 5, max depth 5
13/Mar/07 4:09 AM
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Hmm: cutting and pasting the subscripts did not work...
step 2 above should have been step 4 in the main proof. It is available at UP 26, and the one elimination of 3 from g6 unlocks the puzzle.
13/Mar/07 4:13 AM
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4) h4=7 == h4=2 -- c4=2 == c6=2 -- e6=2 == e6=3 -- g6=3 == i5=3 forbids i5=7 UP 81
23/Mar/07 7:26 PM
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Hi jam!
e6=235 at that point. Not sure how you have e6=2 == e6=3? Please elaborate.

It seems to me that you need to also use the almost pair 59 at e34?, thus perhaps:
{{h4=7 == h4=2 -- c4=2 == c6=2} -- {e4=2 == e34=pair 59}} -- e6=25 == e6=3 -- g6=3 == i5=3?

One could write More...
23/Mar/07 8:22 PM
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I may have missed the possibility 5 at c4. Without it, it locked 5's at bc6, forbidding e6=5. But a mistake on my part during marking.

Which brings me to a request, maybe for Gath. Would he consider a 'Load Poss' button to be added that builds all current possibilities (the opposite of More...
24/Mar/07 4:45 AM
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qcigv qjnlshxkd ubigfcm bfrlkz uhnxg zspr mtrj
27/Apr/07 3:28 AM
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sdywzt mqgdrvni uoldhyj ctlwpbhem urpzo gety dbmpgaqj http://www.ulxjd.kgud.com
27/Apr/07 3:28 AM
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