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 Sudoku Techniques. 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
To save some space, I have chosen to illustrate a few steps simultaneously:
- Locked 5s at bc2 (also, b2=5 == c2=5, and in Eureka: (5)b2=(5)c2) =>bc1≠5
- Locked 9s at df7 (also, (9)d7=(9)f7) => i7≠9
- Hidden Pair 13 at gh5 (also, (1)g5=(1-3)h5=(3)g5) =>
- Pair 56 at i46 (also, (5=6)i4-(6=5)i6) =>
- Triple 268 at dfg9 <(also, (8)_{1}d9={(2=6)_{1}d9-(6=2)f9}-(2=8)g9) =>
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
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
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:
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=(1-8_{1})a3={(3)e8=(3&6)ab8-(8_{1})a8=(8_{1}-2)a7=(2)ac8} => e8≠2
Another way to write this depth 7 chain is in pieces:
- 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
- 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
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
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:
- (pair 16)a46 = (6)b5 => b4≠6
- (6)def5=(6)e46 => df46≠6
- (Hidden pair 26=Hidden pair 27)e25 =>e2=267, thus e2≠3
- (Hidden pair 26=Hidden pair 27)e25 => e5=267, but we already had that
- (7)a6=(7)e6 => d6≠7
- (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
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.
Solution
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.