The following is an illustrated proof for the
Tough Sudoku of March 13, 2007.
This is a very difficult puzzle to solve, unless either one resorts to guessing, or truly understands
using strong sets.
There are many ways to tackle this one. The primary focus of this proof is to illustrate
using Forbidding Chains, also called Alternating Inference Chains or AIC.
Some of the Chains that are presented are complex forbidding chains - or advanced forbidding chains.
Such chains merely use arguments that are Boolean variables (something that is either TRUE, or False).
Some of these Booleans represent
native puzzle conditions that are neither bivalue in a cell, nor candidates that are bilocation.
You may need to refer to previous blog pages to understand
this proof. Links to these pages are found to the right, under Sudoku Techniques.
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. Some steps that are not required will be illustrated. The purpose of
illustrating such steps is primarily to clue the potential solver on how to decide upon a plan
of attack for truly diabolical puzzles.
The information on the following blog pages is required to understand this page:
The illustrations of forbidding chains used in this proof will share the same key:
- black lines = strong links
- red lines = weak links
- black circles connected with black lines = multi-part strong sets
- candidates crossed out in red = candidates proven false
Puzzle at start
A few Unique Possibilities are available here.
They are illustrated in the next picture.
Puzzle at UP 26
Completely illustrating each of the relatively easy steps will consume much space. Therefor,
easy steps will be referenced, without complete illustrations.
To advance this puzzle a very small bit further, here is what I found:
- Locked 7's at hi8 forbids hi9=7
- Locked 8's at gh8 forbids gh7h9=8
Possibility matrix at 27 after some locked candidates
Illustrated above is the naked triple 145, forbidding the indicated items.
i1 = 5% column & box
Puzzle at UP 27
Above, the hidden pair 14 justifies the indicated eliminations
Two sixes solve
Puzzle at UP 29
Above, one can make some easy eliminations. One step is rather obvious to me that unlocks
this puzzle
The Step
Above, the Almost Locked Set 259 at e346 would lock the 2's at e46 but for
the 3 at e6. Since df5=2 == h5=2 and h5=8 == g6=8, there can be no 3 at g6.
Suppose that g6=3:
- => g6≠8 => h5=8 => h5≠2 => df5 contain 2
- => e46≠2 => pair 59 at e34 => e6≠25 => e6=3 => contradiction
As a forbidding chain, one could write:
e6=3_{1} =={triple e6_{1}=25, e4=259, e3=59} -- df5=2 == h5=2 -- h5=8 == g6=8 => g6≠3
Most solving systems published elsewhere will miss this elimination and instead find the
Death Blossom that forbids e79=9. The Death Blossom is a very nice technique, but
it is no more complex than the elimination noted above. In fact, it considers many of the same strong
sets.
Eliminations like the one above are precisely the reason that I present this blog in such a
general fashion.
The puzzle mark-up only weakly indicates this elimination, as the ALS used is not part of
the mark-up. That is why I have included the blog page on ALS.
After making this elimination, the puzzle is reduced to Hidden and Naked singles to the end
Done
Proof
- Start at 23 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
- Locked 7's at hi8 forbids hi9=7
- Locked 8's at gh8 forbids gh7h9=8
- Naked triple 145 at gh7h9 forbids i9=145 UP 27
- Hidden pair 14 at i46 forbids g46h4=14, i4=7, i6 = 3 UP 29
- e6=3_{1} =={triple e6_{1}=25, e4=259, e3=59} -- df5=2 == h5=2 -- h5=8 == g6=8 forbids g6=3 UP 81
- Sets: 1+1+3+2+5 = 12
- Max Depth 5 at step 4
- Rating: .02 + .07 + .03 + .31 = .43
This puzzle is thus not diabolical at all!