Taming Unsolvable 32

The following illustrated proof for the almost diabolical Unsolvable #32 is an attempt to illustrate the power using strong inference sets that are bivalue/bilocation in conjunction with strong inference sets that have more than two possible truth states. Additionally, I hope to illustrate how one can find the partitions of these multi-value inference sets that will be useful. If this is your first visit to this blog, WELCOME!!

I must apologize, as I often employ terminology that is different from that used in most other Sudoku sites. For this reason, plus many others, previous blog pages may be helpful. Links to these pages are found to the right, under Previous Entries. Specifically, one may want to refer to the following pages:

The illustrations of forbidding chains, also called Alternating Inference Chains (AIC), shown in this proof 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.

Although this puzzle is meant to be very difficult, it is quite easy for the label: unsolvable.

Unsolvable #32


No unique possibilities (hidden or naked singles) are available here. Typically, I would locate the Hidden pair 13 at gh5 before considering the possibility matrix.

Five steps combined in one illustration

5 steps

To save some space, I have chosen to illustrate a few steps simultaneously:

  1. Locked 5s at bc2 (also, b2=5 == c2=5, and in Eureka: (5)b2=(5)c2) =>bc1≠5
  2. Locked 9s at df7 (also, (9)d7=(9)f7) => i7≠9
  3. Hidden Pair 13 at gh5 (also, (1)g5=(1-3)h5=(3)g5) =>
    • h5≠6
    • bf5≠1
    • def5≠3
  4. Pair 56 at i46 (also, (5=6)i4-(6=5)i6) =>
    • h5,i2≠6
    • i8≠5
  5. Triple 268 at dfg9 <(also, (8)1d9={(2=6)1d9-(6=2)f9}-(2=8)g9) =>
    • bh9≠8
    • b9,e8≠6
    • c9≠2
Above, the proof probably only requires the last two of these five steps. However, I have chosen to note all of the steps because after making these eliminations, the puzzle becomes a bit problematic. To help unravel the puzzle, at this point I print out the puzzle with the possibilities and mark the page up. The mark-up I currently use is indicated below.

First puzzle mark-up

First Puzzle Mark-up

Since naked strengths (strength within a cell, such as a4=16, thus (1=6)a4) is made transparent by the possibility matrix, my mark-up primarily concentrates upon hidden strengths (strength within a large house - box, column, row). I merely circle the endpoints of each bilocation strength. Addtitionally, if within a row or column a candidate is limited to one location in one box, and a group of locations within exactly one other box, I underline the single candidate location in the first box. If the group of possibilities in the other box would cause some cell to evaluate uniquely, I point a V towards that box, and indicate what might occur with a short note. This latter process helps me to find the most difficult, but often very useful, of hidden bilocation partitions. Finally, if the bivalue cells are predominated by the occurence of one candidate, I indicate that. In the case above, I have circled in red all the bivalue cells containing candidate 2.

Any cell above with two or more marks (circles, underlines) is an intersection point for a chain piece. My searches will typically start from one of these cells. Also, some weight is given to the bivalue cells, with special emphasis on those circled red.

Note that the puzzle mark-up looks a bit messy. It takes some practice to utilyze it well. Also, the usefulness of the puzzle marks is greatest with fewer solved cells. In other words, as the puzzle progresses, the puzzle mark-up becomes so full that it is less useful. However, generally as the puzzle progresses there are so many chains available that inspection without the mark-up reveals chains more easily.

Before I jump into looking for chains, it is helpful to think of each circled group of bilocation strength as a Super Cell. Using this information, I quickly scan the puzzle for chain pieces. At this point, I do not care if they lead anywhere. I am merely trying to get into my head as many short pieces of derived infomation as I can. Thus, for example, I think: (9)i3=(9-8)supercellab3=(8)h3, therefor (9)i3=(8)h3. Also, (1=6)a4-(6)b5=(6)def5-(6=3)e4, therefor (1)a4=(3)e4. Also, (8):e6=d6-d9=g9 => (8)e6=(8)g9. Also, (3)e8=(hidden pair 36)ab8. Typically, after doing some of these, I sit back and look for something easy to eliminate.

Digression: If you like uniqueness techniques, the strong links circled often reveal some Almost Unique Rectangles.

Usually, such a quick analysis catches most depth 3 chains. I have called such chains Y Wing Styles. Some depth 4 chains will also be caught. Chains of greater depth take more time.

Having satisfied myself that depth 3 chains do not exist in this puzzle, my informal list of chain snippets yields one quick depth 5 chain:

  • (8=2)g9-(2=9)i8-(9)i3=(hidden pair 19-8)ab3=(8)h3 =>g1,h8≠8
I do not use this chain in my proof, but.... pieces of the shorter chains that I find often serve as building blocks for the larger chains. The new bilocation link with 8s in row 1 suddenly becomes more interesting.

Puzzles often seem to have a theme. This is sometimes by design. Often, however, it is just a factor of overall puzzle consistency. I have some theories on this, but they are mostly just informal. However, there is the following tendency:

If a group of chain snippets restrict the puzzle enough to prove an elimination, some of the snippet links will likely contribute to another elimination.

Because I have a preference for wrap-around chains, I often search fairly hard for those. Thus, any cell that is strong bivalue with candidate 2 that shares a large house with another such cell would get a special look. There is a depth 9 chain that I can find that uses such a relationship. However, it needs an appendage that makes it no longer a wrap=around chain. Nevertheless, it exists.

That particular chain is not what I found next. Instead, I found the equivalent of that chain in mostly hidden space. Oftentimes when I publish such chains at other locations, I translate them into their naked space counterparts, as it seems that others find ALS type arguments much easier than Hidden ALS arguments. This is not the case for me, however. In fact, it is often true that at this point in the puzzle, HALS arguments will generally be shorter then ALS arguments. Thus, the chain snippets that I must store in my head are significantly reduced looking at hidden strengths.

At this point, I now have in my head many Almost Hidden Pairs. They form a very powerful structure for connecting chain snippets. Also, and this is a very important point for puzzles that are labelled very difficult:

  • The box with the fewest givens is often the key.
What this means is that I will pay special attention to the center box. It is almost certain that some candidates can be eliminated from it with relative ease. Concentrating for a moment at that box, I store some chain snippets, such as the almost vanishing intersection with candidate 6: (6):{b5=df5-e46=e2}=e5. This relationship reveals a quasi Almost Hidden Pair 26 and Almost Hidden Pair 67 at e25. The super cell in 8s at e68 makes this Almost Hidden Pair relationship a bit stronger.

There is something to be found here, but I did not find it at this time. Rather, I found something else that returned me to the center box for analysis.

Depth 7 chain that is fundamentally revealed by the 8s in column a - Kraken Column 8s

Depth 7 Kraken Column 8s chain

The definition of the term, Kraken... can be found at various sudoku forums. For me, it basically means that the eights in column a are limited to 3 locations, and each of these 3 locations prove that 2 cannot exist at e8. However, I do not build my chain from that column explicity. Instead, I use chain snippets that I have already identified:

  • (2=9)i8-(9)i3=(hidden pair 19-2378)ab3
  • (3)e8=(hidden pair 36-258)ab8
All I need to link these together is the following new chain snippet:
  • (8-2)a7=(2)ac8
Now, it should be obvious what one can do. The 8s in column a must exist somewhere, and each possible location is linked to some chain snippet. Thus, putting it altogether:
  • (2=9)i8-(9)i3=(9-1)b3=(1-81)a3={(3)e8=(3&6)ab8-(81)a8=(81-2)a7=(2)ac8} => e8≠2
Another way to write this depth 7 chain is in pieces:
  1. Build the Almost depth 4 chain:
    • (2)ac8=(2-8)a7=(8-36)a8=(6-3)b8=(3)e8, which I think of as:
    • (2)ac8=(2-8)a7=(8-Hiddenpair36)ab8=(3)e8
    • Let the above chain be the Boolean Z. Clearly, Z=(8)a3.
    • In my head, since Z-(2)e8, I am thinking:
      • (38=2)e8-Z=(8)a3, thus
      • (38)e8=Z(8)a3
  2. Link the definite chain snippet above to something.
    • (38=2)e8-(2=9)i8-(9)i3=(9-1)b3=(1-8)a3=Z(38)e8
DPBird calls this a proving loop. Clearly, we have (38)e8=(38)e8. My preference is to merely write the elimination chain, which now could be written:
  • (2=9)i8-(9)i3=(9-1)b3=(1-8)a3=Z => e8≠2

For me, this elimination essentially unlocks the puzzle. Below I attempt to illustrate why and how.

As one further note, however. Looking at what I have proven above, I have as the conlusion: (3)e8={(2)ac8=(2)i8}. An additional chain snippet, therefor, would be: (3=6)e4-(6=1)a4-(1)a3=(1-9)b3=(9)i3-(9=2)i8. Clearly, if I were to link this to the original chain, g8≠2. However, this puzzle will solve without that longer chain. Nevertheless, one could see this relationship easily if one has already identified the original chain, since so many pieces overlap.

Second puzzle mark-up

Second Puzzle Mark-up

Above, I have updated the original puzzle mark-up with the new information derived from eliminating 2 from e8. The new super cell, e25, formed by the stronger 2s in column e, makes the next step easy. Note that it is fairly easy to update the puzzle mark-up. Usually, unless many cells solve, there is no reason to print out a new puzzle. Even if one prints out a new puzzle, the old mark-up can be used as a quick template for the new mark-up.

Recall some of the earlier chain snippets that were identified. Although the next step can be written using only Almost Naked Locked Sets, I found it considering the Almost Hidden Locked Sets 26, 27 in column e which are easily reduced, logically, using the concept of super cells. A very strong indicater that this will exist can be seen in cell e2, which contains a circled 2, an underlined 7 (relative to the center box), and underlined 6 (also relative to the center box), and the circled 6 (but this is relative to the possible 6 at h2, so the circled 6 is less interesting than the underlined 6!). The underlined canidates should not be undervalued. I think this is a common error in sudoku analysis. Oftentimes, as in this case, the underlines are very important!

A wrap-around chain using candidates 1,2, 6,7 - but candidates 1,2 are carriers or appendages

Pretty wrap around chain

The supercell formed by the 2's at e25 produces the following chain snippet:

  • (7)e6=(7-6)e25=(6)e46
  • Also written as: (7)e6=(2&7-6)e25=(6)e46
Note, however, that the concept of supercells allows one to think like the first representation. I find that this makes linking these types of snippets easier.

The ALS 167 at a46 is the naked equivalent of a super cell. (OK, that is turning most viewpoints on sudoku upside down, but - there is a tremendous symmetry between the ideas that is powerful to recognize!) Thus we also have this chain snippet:

  • (7)a6=(pair16)a46-(6)b5=(6)def5
Considering these two snippets, since (7)a6-(7)e6 we certainly can link the chains together. Alternatively, since (6)def5-(6)e46, we can certainly link the chains together. Finally, since we can link the snippets together on both ends, we have found a wrap-around chain:
  • (7)a6=(pair16)a46-(6)b5=(6)def5-(6)e46=(hidden pair 26 - hidden pair 27)e25=(7)e6
Note that I have changed the terminology slightly. This is done intentionally, as it is, in my opinion, better to think of the chain above conceptually in this fashion. Furthermore, it makes the eliminations that are justified more transparent. In any event, we have now proven the following six strengths:
  1. (pair 16)a46 = (6)b5 => b4≠6
  2. (6)def5=(6)e46 => df46≠6
  3. (Hidden pair 26=Hidden pair 27)e25 =>e2=267, thus e2≠3
  4. (Hidden pair 26=Hidden pair 27)e25 => e5=267, but we already had that
  5. (7)a6=(7)e6 => d6≠7
  6. (pair 16=pair17)a46 => b4,a3≠1
One may have to think about the last listed item for a moment, but a pigeonhole matrix representation of the chain makes this last strong inference set completely transparent. It is worthwhile to struggle through understanding the proven strong set. If you cannot follow the logic, let me know. I think, however, one will own this important concept best if one finds it independently.

After making the eliminations above, quite a few cells can be solved, beginning with b3 = 1% column & box.

Y wing style 78

Y Wing Style 78

If you like uniqueness arguments, one can easily unlock this puzzle considering the Almost Unique Rectangle 47 at hi27 => h2≠47 =>h2 = 6%cell. However, since cell a7 has been solved, and since I previously had proven (7)a6=(7)e6, it can be immediately obvious that:

  • (7)a6=(7-8)e6=(8)e8-(8)a8=(8)a3 => a3≠7 => a3 = 8%cell => naked singles to end
There are probably many easy ways to finish the puzzle besides those two.



I have more or less temporarily abandoned my rating system in an attempt to make a better one. However, if I were to use the old rating system, this puzzle would rate slightly above 2.0. This places it as monstrous, but not quite up to level of difficulty one might infer from the label unsolvable.

Hopefully, this step by step puzzle analysis will be helpful in finding complex chains.

Indicate which comments you would like to be able to see

Well you certainly made short work of U32 !
Another delightful solution.
If I may, a few points for your consideration and any comment

1) on chain appending :
I'd be interested in your comments on this :
Appending can be summarised as :
Suppose A=>X, and B=>X More...
12/Jul/07 10:01 AM
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Thanks Michael!

1) Chain appending: Everything you say is completely valid. All cases can be reduced to:


Here, it is understood that A,B can be as complex a Boolean expression as one wishes.

Nevertheless, the snippets, or pieces, one may experience More...
14/Jul/07 7:58 PM
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Many thanks for the reply.
I had a look again at the almost remote pair (e9/i6 : 69) used in the wrap around chain in your u13 solution.
Two points for your consideration :
1) on the many eliminations arising from the wrap-around :
- to me all bar one can be seen More...
14/Jul/07 10:48 PM
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we need more explanation
15/Jul/07 8:22 AM
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Hi michael!

Two thinkgs spring to mind:
1) I should write better
2) I often combine steps in ways that obfuscate rather that illuminate:

The following eliminates 6 from f8:
This is a discontinuous chain that just happens to be part of the More...
17/Jul/07 7:26 AM
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Steve, thanks again.
Yes my logic was flawed, and that had nothing at all to do with your drafting skills which are excellent !
The way I should have reasoned within the wrap around chain was this :
- chain begins : (7)e9=(pair69)(e9i5)-(69)i9=(4)i9 etc
- relevant weak link is More...
17/Jul/07 8:33 PM
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Probably, that is a better way to view (~6)f8.

As I noted, I tend to grab what I deem in my narrow view as efficient. This tendency tends to make some of my deductions opaque. I suppose that I need to rethink using elephant guns to kill mice!
17/Jul/07 11:34 PM
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I sit here and do the "Easy" and "Medium" standards. I feel rather pleased when I complete a complex "Medium".

Y'all are smart!
23/Sep/07 12:17 AM
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24/Sep/07 6:55 AM
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I'm with Dan-the-Man. I complete the hard and tough puzzles. I'm better than most people I know at the exercise, but those two (and even the explanation) have me completely lost.
30/Oct/07 1:17 AM
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Hmm, so when will the bugs be fixed? cause i keep losing my game before i finish it! lol i click on save and it's gone and i accidentally click delete and it's gone!
25/Nov/07 2:50 AM
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Your recent blog post on an "unsolvable" puzzle led me to review what I did on this one a few years back. I've cleaned it up slightly. It uses an AALS in a loop.

1. Basic techniques to UP = 25.
2. loop: (2=9)i8 - i3 = (9-1)b3 = (1)b4 - (1=67)a46 - (6|7)bc5 = More...
27/Jun/10 10:21 AM
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Correction: The UP count after step 2 is 29.
27/Jun/10 11:00 AM
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Oops again! In the note following step 2, the eliminations of 6 and/or 7 are valid at bdf4|df6.
27/Jun/10 12:48 PM
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