The following is an illustrated proof for the
Tough Sudoku of March 30, 2007.
This proof uses only Forbidding Chains, also called
Alternating Inference Chains or AIC.
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. Some easy steps are
not taken, nor illustrated. This page only illustrates 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 a partioning of a strong set
- candidates crossed out in red = candidates proven false
Strong and weak need not be mutually exclusive properties.
Puzzle at start
A few Unique Possibilities are available here:
- h7 = 7% box & column (hidden singleton both in box and column)
- h8 = 4% box & column
- c7 = 4% box & row
Hidden Pair 49, Hidden Triple 279, Locked Candidates 3, 5, & 7
Quite often, Hidden sets are easier to find before entering the possibilities.
Illustrated to the left, g5=4 and c5=9 conspire to force only 49 at d4,f6.
Clearly, then, candidates 357 must be locked at def5, thus one has the eliminations
noted in row 5.
It is rare that one will find a hidden triple without considering the possibilities.
While looking for Hidden Pairs, I noticed the almost Hidden Pair 27, illustrated in orange, at
e23,f3. Since I can also see that the 9s are locked in box e2 because e23 are the only places
left for 9 in column e, not only can I eliminate 9 from the rest of box e2, but I also can
place the Hidden Triple at e23f3 as shown.
A few Unique Possibilities are available after a5≠357:
- a5 = 2% cell (naked single)
- c8 = 2% column & box
- i4 = 2% row & box
An alternate approach to finding the Hidden Triple above:
- Find the Locked 9s at e23 => d123f13≠9
- Fill in the possibilities
- Find the naked quad 1468 at d123f1 => e23,f2 are reduced as shown
There is some value to recognizing that one has the Hidden Triple indicated above based
only upon the information cited: c5,a7,i8=9(thus locked 9s at e23); bi1=27; d68=27. Very rarely,
one may use such a group of conditions within a chain to justify an Almost Hidden Triple,
eliminating a key value from one of the Hidden Triple cells. Admittedly, that concept is
advanced, and somewhat obscure. Fortunately, this puzzle does not require any sort of logic
Locked candidate 9
Above, the 9s at gh1 are the only 9s possible in row 1. Since gh1 are both in box h2, one
can safely remove 9 from cells g2,h2,g3. I have chosen to illustrate this elimination as a
Naked Pair 13
Above, the naked pair 13 at hi2 is illustrated as a continuous nice loop forbidding chain:
- h2=1 == h2=3 -- i2=3 == i2=1 =>
- h2=1 == i2=1 => bdg2, gh1, g3 ≠1
- h2=3 == i2=3 => bcg2,g3 ≠3
One now can solve g2 = 2% cell.
Coloring with candidate 4
The forbidding chain with candidate 4:
- b2 == d2 -- d4 == f6 => b6≠4
is a typical coloring elimination. The logic can also be viewed as follows:
- b2=4 => b6≠4
- b2≠4 => d2=4 => d4≠4 => f6=4 => b6≠4
This step is a set up for the fist truly difficult step.
Forbidding chain using candidates 4,7,9
The elimination above can be written as the following forbidding chain:
- b4=4 == b2=4 -- b2=9 == e2=9 -- e2=7 == bc2=7 -- a3=7 == a6=7 => a6≠4
The only difficult item about this chain is the grouped argument with the 7s in row 2. They are
grouped according to their location at cell e2 versus box b2 (at cells bc2). Using Eureka notation,
but maintaining the grid coordinate system that I use, the elimination could be written:
(corrected later, many thanks to David!)
- (7)a6 = (7)a3 - (7)bc2 = (7-9)e2 = (9-4)b2 = (4)b4 => a6≠4
Let me know if this notation is preferable here.
Regardless of the notation system, this elimination can also be understood as follows:
- a6=7 => a6≠4
- a6≠7 => a3=7 => bc2 do not contain 7 => e2=7 =>
- e2≠9 => b2=9 => b2≠4 => b4=4 => a6≠4
After making this elimination, a few cells solve:
- f6 = 4% row
- d4 = 9% cell & box
- f9 = 9% column & box
Forbidding chain using candidates 1, 2, 9
Illustrated above, once again the chain uses one grouped argument, this time involving
candidate 1. As a forbidding chain:
- e8=1 == e7=1 -- e7=2 == e3=2 -- e3=9 == b3=9 -- b3=1 == a13=1 => a8≠1
In my puzzle mark-up for this position, the 1 at b3 was underlined, with a V
next to it,
pointing at cell a8. This was a great clue that not only this chain existed, but also that it would
at least solve one cell: c8=3% cell.
One may note above that I did not bother to eliminate the 1s from d79 caused by the locked 1s
at e78. This elimination, and some other easy eliminations available here, are not significant,
one way or the other, to this puzzle proof.
Naked Pair 68
Illustrated to the left as a continuous nice loop forbidding chain,
the naked pair 68 at f18 forbids 68 from f7.
Coloring with candidate 8
Illustrated above as a forbidding chain using candidate 8:
- g3 == d3 -- f1 == f8 => g8≠8
This elimination is a required set up for the final non-native elimination step.
Forbidding chain using candidates 6, 7, 8
The forbidding chain:
- c2=6 == c1=6 -- f1=6 == f8=6 -- f8=8 == b2=8 -- c9=8 == c9=7 => c2≠7
causes a cascade of Unique Possibilities (mostly naked singles, a few hidden ones) until the
puzzle is finished.
This proof is not the exact path that I illustrated, but functionally the same.
- Start at 22 filled - the given puzzle. Unique Possibilities to 25 filled. (UP 25).
- Naked triple 357 at def5 forbids abhi5,d5f6=3 & ahi5,d4=5 & ab5f6=7 UP 28
- Hidden triple 279 at e23f3 forbids e23=1,f3=8,d132f13=9
- Locked 9s at gh1 forbids h2g23=9
- Pair 13 at hi2 forbids bdg2,g13h1=1 & bc2,g23=3 UP 29
- fc on 4s:b2 == d2 -- d4 == f6 forbids b6=4
- b4=4 == b2=4 -- b2=9 == e2=9 -- e2=7 == bc2=7 --a3=7 == a6=7 forbids a6=4 UP 32
- e8=1 == e7=1 -- e7=2 == e3=2 -- e3=9 == b3=9 -- b3=1 == b89=1 forbids a8=1, d79=1 UP 33
- Pair 68 at f18 forbids f7=68
- fc on 8s: g3 == d3 -- f1 == f8 forbids g8=8
- c2=6 == c1=6 -- f1=6 == f8=6 -- f8=8 == b2=8 -- a9=8 == a9=7 forbids c2=7 UP 81
- Sets: 3 + 3 + 1 + 2 + 2 + 4 + 4 + 2 + 2 + 4 = 27
- Max depth 4, three times
- Rating: .01 + 4(.03) + 2(.07) + 3(.15) = .72
This puzzle is quite tough, but not monstrous or diabolical.
One could fool around with the steps
and achieve a rating of .70 by using the Hidden pair 49 and 3 locked candidate eliminations versus
the naked triple 357. Since this took more space, I choose the latter. I suppose that now I have
consumed the space saved!
This puzzle is standard forbidding chain fare. Nothing really exceptional about it. A
workman-like approach to tackling this one will solve it.
If one prefers fewer steps of greater depth, there are quite a few other possible paths to
take. I found a large number of depth 5 and depth 6, and even some depth 7 or 8, forbidding chains.
In fact, an advanced forbidding chain of about depth 8 or 9 is possible at UP 29 that forbids
c2=7 and reduces the puzzle to Unique Possibilities. My preference is more chains of lesser
depth, but others may well prefer fewer chains of greater depth.
Please let me know of interesting alternative proofs to this one, as I strongly suspect many
are possible. I feel that I missed something easy here, and that this particular proof
does not represent my best work!
Please to post a comment.
Not a member? Joining is quick and free.
As a member you get heaps of benefits.