Almost Locked Sets Proof for March 29,2007


The following is an illustrated proof for the Tough Sudoku of March 29, 2007. This proof illustrates both the power of using Almost Locked Sets within a Forbidding Chain, also called an Alternating Inference Chain or AIC, and using standard forbidding chains.

You may wish to refer to previous blog pages to properly 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.

The information on the following blog pages may be helpful:

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)
  • black containers define either:
    • a partioning of a strong set
    • the use of an almost locked set
  • 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:

  • b2 = 4% box (hidden singleton in box)
  • c6 = 7% box & column
  • b8 = 9% box & column


Hidden Pair 28


Hidden Pair 28

Quite often, Hidden Pairs are easier to find before entering the possibilities.

Illustrated to the left, d6=2 and f4=8 conspire to force only 28 at e28.

After noting this, f1 = 1% box

Also available at this time, but not illustrated, is the Hidden Pair 89 at h56.


Possibility matrix at 27 cells solved (UP 27)


Possibility matrix at UP 27

At this point, there are many standard eliminations available. The only one of these that this proof requires is locked 8s at h56 => h89≠8

The following standard eliminations are not used in this proof, but available at this time:

  • Hidden pair 89 at h56 forbids h89=8, h3=9, h5=1367, h6=136
  • Locked 1s at c79 forbids b7=1
  • Locked 2s at abc1 forbids c23=2
  • Locked 4s at i46 forbids g4=4
  • Locked 5s at hi3 forbids c3=5
  • Locked 5s at e45 forbids d45=5
  • Locked 6s at c12 forbids a1=6
  • Locked 6s at gh4 forbids ad4=6
  • Locked 8s at c23 forbids c789=8
  • Locked 9s at e56 forbids d5f56=9
Whew! That is many locked candidate eliminations!

After making these eliminations, if one tries a puzzle mark-up, (explained in some detail here), one will find that cells d9 and f3 are strongly indicated as potential forbidding chain pieces - three circles each. In fact, a number of forbidding chains use these cells. Most of them involve four strong inferences or more.

There are a great number of other forbidding chains also available. All of them that I found are at least partially indicated by the puzzle marks. I will illustrate only those that I found to be the most efficient path to solution. Regardless of the path chosen, this is an easy puzzle from which to extract forbidding chains, or AIC.

Of particular interest, however, are some items not commonly noted on my puzzle marks:

  • There are many Almost Locked Set configurations. So many, in fact, that to list them all could take quite some time!
  • The solved cells in box b2 force the 3's in box b2 into an easy grouped argument. Some may call such a configuration a hinge. This particular hinge is easy to spot even before the possibilities are entered!


Knocking off a seemingly harmless 3


depth 4 chain using 239


Illustrated above is a standard forbidding chain that uses the hinge with 3s:

  • c9=2 == f9=2 -- f9=9 == d9=9 -- d1=9 == d1=3 -- abc1=3 == c23=3 => c9≠3
Oftentimes, the candidate with the most possible locations on the puzzle grid is the easiest one to attack. Nevertheless, it is often the least valuable one to consider. With this puzzle, though, the 3's exist in many bivalue cells, such as c3=38, etc. For this reason, they might be important.


Another Locked candidate elimination


Locked 3s


The 3s are now locked in box b8 at cells ac8, forcing gh8 to contain no 3s. Alternatively, the 3s are locked in row 9 at ghi9, within box h8, eliminating the same 3s.


Almost Hidden Pair 67 used in a chain


depth 4 chain using 567


The chain illustrated above uses the Almost Hidden Pair 67 at df7. One could alternatively use the Almost Locked Naked Triple 2589 at df9 & e8. For me, the Almost Hidden Pair is much easier to see, but for others it is probably obscure.

The forbidding chain illustrated above can be written as:

  • f8=7 == {Hidden pair 67 at df7} -- d7=5 == d9=5 -- hi9=5 == h8=5 => h8≠7
The logic can be described otherwise as follows:
  • f8=7 => h8≠7
  • f8≠7 => both 67 are locked at df7, thus hidden pair 67 at df7
  • d7 limited to only 67 => d7≠5 => d9=5 =>
  • hi9 cannot contain 5 => h8=5 => h8≠7
If one were to instead use the almost naked triple, the forbidding chain could be:
  • h8=5 == hi9=5 -- d9=5 == {triple 289 at df9,e8} -- f8=2 == f8=7 => h8≠7
The logic could also be described as follows:
  • h8=5 => h8≠7
  • hi9=5 => d9≠5 => {d9=89, f9=29, e8=28, thus naked triple 289} =>
  • f8≠2 => f8=7 => h8≠7
If you are paying attention, one could write either chain differently and use the cell h8=57, plus the other information, to forbid hi9=5. This is quite common - that one can find many very similar eliminations. Interestingly enough, though: The elimination of the 5s is available only after the elimination of the 3, while the 7 elimination is available whether or not one has eliminated h8=3. It is because I saw this depth 4 elimination that I knew the previous depth 4 elimination of 3 from c9 would be particularly valuable.

After making this elimination, h8=5 and the puzzle becomes a cascade of Unique Possibities to the end. If one has made all the locked candidate eliminations listed above, one can find naked singles to the end. If, however, one only makes the steps listed below in my proof, one needs to find some hidden singles.


Proof

  1. Start at 22 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
  2. Hidden pair 28 at e28 forbids e2=1, e8=5 UP 27
    1. Locked 8s at h56 forbids h89=8
    2. c9=2 == f9=2 -- f9=9 == d9=9 -- d1=9 == d1=3 -- abc1=3 == c23=3 forbids c9=3
    3. Locked 3's at ac8 forbids gh8=3
    4. f8=71 == {hidden pair 671 at df7} -- d7=5 == d9=5 -- hi9=5 == h8=5 forbids h8=7 UP 81
  • Sets: 2 + 1 + 4 + 1 + 4 = 12
  • Max depth 4, twice
  • Rating: 2(.01)+.03 +2(.15) = .35
Not too tough, but at least interesting!


Notes

The value of understanding the concept of Almost Locked Sets is well recognized across the Sudoku solving community. One can certainly solve this puzzle without using ALS, but the easiest path, in my opinion, includes using at least one such step.




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

Hi Steve,

In your hidden chain f8=7 == {Hidden pair 67 at df7} -- d7=5 == d9=5 -- hi9=5 == h8=5 => h8≠7, why do we have to include the 6?
29/Mar/07 6:17 AM
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Hi Steve, what you say about precedence in fcs is interesting. While there is no trouble in finding fcs (sometimes almost too many!) I still find it difficult to judge which ones are needed first, but I suppose this comes with experience?Sometimes it seems like luck whether you get a cascade of UPs or keep chipping away.
29/Mar/07 8:50 AM
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Hi Six Stringer!
Suppose for a moment that we do no include the 6:
Then, f8=7 ==df7=7 -- ???
Without including the 6, we cannot forbid d7=5, as 7 could exist at either d7 or f7, but....
if we include the 6, then although 7 can still exist at either d7 or f7, so does 6, thus we have the conditional hidden pair 67 at df7, allowing the next part of the chain, -- d7=5.

29/Mar/07 5:15 PM
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Hi Giblet!
Your observation, that finding fc's is really not the problem, but rather evaluating which ones to use!
Proofs list only required steps - since if one is to evaluate a puzzle by listing all possible steps, then the evaluation will be made based solely upon how long one keeps More...
29/Mar/07 5:37 PM
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I see...Thanks, Steve!
30/Mar/07 4:03 AM
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The 'almost hidden pair' apparently moves to col f when following the fc in the CCW direction. Is there a construct that avoids such a repositioning?
25/Apr/07 1:55 AM
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Hi ronk!
Indeed, one of the quirks of using hidden sets in a chain is that they can appear to be unidirectional. However, if one allows that the necessary and sufficient condition for a hidden pair is: limiting two candidates to no more than two locations each in a large house, there really is More...
25/Apr/07 5:06 PM
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OK, got it.

d7=5 -- {hidden pair 67 at df7} == f8=7

R-to-L, the hidden pair is true (exists)
L-to-R, it's false (doesn't exist)

Thanks, Ron
25/Apr/07 10:50 PM
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