When trying to prove this pattern by Mauricio to be invalid, I happened to notice the following fact:
If a group of all instances of N different digits in a chute is spread over max N+1 minirows/-columns, then the group will contain at least one unavoidable set.
(The proof for all but N=5 or N=6 are mentioned in the thread above. The proofs for N=5 and N=6 are very long and ugly T&E style proof by contradiction. If someone absolutely must see them, I can try to write them down in a readable form, but it'll be a while until I have time for that. In the meantime, anyone who wants may try (in vain) to contradict the rule.)
Then I thought, why not use this for solving some puzzles! So, how can I do this? Let's say we have a band where N digits don't appear at all. From the rule above we know that they must be spread over at least N+2 columns or there will be an unavoidable set not covered by any givens (=multiple solutions). For example in this puzzle:
- Code: Select all
*-----------*
|.3.|..1|..2|
|.4.|..2|..3|
|.2.|...|..1|
|---+---+---|
|2.5|..6|..7|
|..3|...|.5.|
|.6.|.8.|...|
|---+---+---|
|...|4..|71.|
|1..|..7|..8|
|...|.6.|9..|
*-----------*
SE 7.3
the top band contains only digits 1,2,3&4, digits 5-9 do not appear at all. The 5 missing digts must be spread over at least 7 different columns. From this we can conclude that r3c6<>34 and the puzzle solves with locked candidates.
And another example that also needs a very nice layered BUG-lite:
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*-----------*
|.7.|.4.|..5|
|...|.7.|..4|
|.4.|.5.|...|
|---+---+---|
|.8.|.6.|...|
|5..|4..|1.2|
|...|1.2|.8.|
|---+---+---|
|6..|5..|3..|
|...|...|..9|
|19.|..6|...|
*-----------*
ER 7.1
In the top band six different digits are missing. These six digits must be spread over all eight columns with available cells, so we can immediately eliminate 5 from r2c2 and 7 from r3c9. Unfortunately, this doesn't really help us at all and after basic technique we get stuck here:
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*-----------*
|.7.|.4.|..5|
|.15|.7.|..4|
|.4.|.5.|...|
|---+---+---|
|281|.65|...|
|5..|4..|1.2|
|...|1.2|58.|
|---+---+---|
|62.|5..|3..|
|.5.|...|..9|
|19.|..6|.5.|
*-----------*
Now we have a '1' in r2c2 and only five missing digits left. These five digits must be spread over the remaining 7 columns with empty cells, so we may eliminate candidate 1 from r3c9. This takes us here:
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*-----------------------------------------------------------------------------*
| 389 7 *26 |*23689 4 1389 |*2689 *12369 5 |
| 389 1 5 |*23689 7 389 |*2689 *2369 4 |
| 389 4 *26 |*23689 5 1389 |*26789 *123679 #368 |
|-------------------------+-------------------------+-------------------------|
| 2 8 1 | 79 6 5 | 479 3479 37 |
| 5 36 79 | 4 389 789 | 1 #679 2 |
| 47 36 479 | 1 39 2 | 5 8 -67 |
|-------------------------+-------------------------+-------------------------|
| 6 2 478 | 5 89 4789 | 3 47 1 |
| 47 5 3478 | 378 1 3478 |*26 *26 9 |
| 1 9 347 | 37 2 6 | 478 5 78 |
*-----------------------------------------------------------------------------*
Here we have a very beautiful 13 cell layered BUG-lite on candidates {26} with only two guardian cells, r3c9=6 or r5c8=6. R6c9 can see both these cells => r6c9<>6. Puzzle solved. (This last elimination may also be seen as a reverse BUG elimination).
RW
