Yet another crazy uniqueness technique

Advanced methods and approaches for solving Sudoku puzzles

Yet another crazy uniqueness technique

Postby RW » Sun Aug 12, 2007 3:21 pm

Ok, time for one more crazy uniqueness technique that you probably never will be able to use in a real puzzle!:D

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:
Code: Select all
 *-----------*
 |.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:

Code: Select all
 *-----------*
 |.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:
Code: Select all
 *-----------------------------------------------------------------------------*
 | 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
RW
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Joined: 16 March 2006

Postby Mauricio » Sun Aug 12, 2007 8:23 pm

Nice! I am glad my pattern helped to develop a new technique, even though it is very rare, as you say. Don't worry, I think the automorphism technique is rarer and harder to use.
Mauricio
 
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