The following illustrated proof for the diabolical & extreme
Tough Sudoku of April 12, 2007 employs
a wide array of techniques. Of special import for this puzzle are Advanced Forbidding Chains
. By this I mean that a technique, including possibly a Forbidding Chain (also called
Alternating Inference Chain (AIC) is used as an argument in the chain. If this is
your first visit to this blog, you are strongly encouraged to research how I use Forbidding Chains,
as this puzzle is a slight bit hard.
Previous blog pages may be helpful. Links to these pages are found to the right, under
Previous Entries. The list is growing, so specifically, one may want
to refer to the following previous blog 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
Five Unique Possibilities are available here:
- f3 = 8% box & column(hidden single in both box and column)
- c7 = 8% box & row
- e7 = 7% box & column
- a3 = 7% column
- i1 = 7% box & row
Typically, at this point, I prefer to
search for Hidden Pairs before filling in the Possibility Matrix
Hidden Pair 78
In row 6, only two places remain for both candidate 7 and candidate 8. Thus, if any other
candidate were to exist there, too few places would remain for at least one of candidate 7,8.
I have illustrated the elimination as a wrap around, or continuous nice loop forbidding chain:
- b6=7 == h6=7 -- h6=8 == b6=8 => bh6=78.
Locked candidate 9
Since the 9s in box e2 exist only within column e, if a 9 were to exist in column e outside of
box e2, there could be no 9s in box e2. Thus, one can eliminate the possible 9s shown above. I have
also illustrated this elimination as a very short forbidding chain:
Naked Pair 19
Since cells i23 are both limited to only candidates 1 and 9, all the cells which see both i23
cannot contain 1 or 9. The logic is almost exactly the same as for the hidden pair explained earlier.
The wrap around forbidding chain is also very similar, but there are subtle differences. A very
good understanding of how forbidding chains work can be achieved by studying such simple
techniques. As a forbidding chain:
- i2=1 == i2=9 -- i3=9 == i3=1 => gh2,h3,i678 ≠ 19
After making these eliminations, one can solve another cell: c6 = 9% row.
Coloring with candidate 3
I call any technique which considers exactly one type of candidate coloring. Coloring is also
a great place to start learning about forbidding chains. Above, the 3's in columns a and e are
sufficient information to justify the eliminations shown. As a forbidding chain on only candidate
- a2 == a4 -- e4 == e1 => c1,d2≠3
A simple forbidding chain like this one reduces to: a2=3 == e1=3, which means that at least one
of a2,e1 must be 3. Thus, all the possible 3s which see both of the cells a2,e1 are eliminated.
Two more cells are now solved: c1 = 2% cell and then b8 = 2% row, column & box.
Typical forbidding chain using candidates 345
Most easy techniques have been exhausted. To advance the puzzle a bit further, I used the
chain illustrated above:
- c8=4 == a789=4 -- a4=4 == a4=3 -- a2=3 == c2=3 -- c2=5 == c3=5 => c3≠4
The first part of the chain, c8=4 == a789=4 is a type of grouped argument that is frequently
helpful. One may note that a789=4 == a4=4 would also be a true statement. The fact that the
relationship between a789=4 and a4=4 is both a strong inference and a weak inference is of
little concern. This is also the case with a4=3 & a2=3. Please review the blog pages on
forbidding chain theory if this causes confusion.
After making the elimination of 4 from c3, the following Unique Possibilities are available:
- b3 = 4% box & row
- b4 = 8% cell
- b6 = 7% cell
- b5 = 6% cell
- h6 = 8% cell
- g2 = 8% row & box
Illustrated above is the newly uncovered naked pair 34 at cd5. After making the indicated
eliminations, f5=9% cell.
Very Common Y Wing Style
The cells d5 = i7 = 34 are common markers for one to look for this elimination. The 4s
in row 6 serve as the vertex, or bridge, between these two cells, allowing the elimination
indicated above. As a forbidding chain, this very common and very easy to find technique
- d5=3 == d5=4 -- d6=4 == i6=4 -- i7=4 == i7=3 => d7≠3
After making this elimination, the 3s in box e8 exist only in column f, thus one can
- f7=3 == f8=3 => f4≠3
- or, alternatively: Locked 3s at f78 => f4≠3
Naked Pair 19
The naked Pair 19 at bi2 justifies the eliminations shown above.
This concludes the first page of this puzzle proof. Some possible headaches, and certainly
some very interesting stuff, can be found on the
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