Help needed!

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Help needed!

Postby pgoyette » Sun Mar 23, 2008 1:35 am

I've reduced the problem puzzle to the following:

819324765
763195000
020876193
058060401
201500006
607000050
082650000
000008600
076902508

I _know_ what the answer is by using brute force, and I know that if I force R7C7 = 3 then everything else just falls into place. (If I set R7C7 to its only other possible candidate, the puzzle becomes unsolvable.)

What I can't seem to find is a solution rule (other than brute force / guessing) that leads me to the correct solution.

Any clues
pgoyette
 
Posts: 2
Joined: 22 March 2008

Postby daj95376 » Sun Mar 23, 2008 3:30 am

Code: Select all
# Your information
 *-----------*
 |819|324|765|
 |763|195|...|
 |.2.|876|193|
 |---+---+---|
 |.58|.6.|4.1|
 |2.1|5..|..6|
 |6.7|...|.5.|
 |---+---+---|
 |.82|65.|...|
 |...|..8|6..|
 |.76|9.2|5.8|
 *-----------*

Code: Select all
# Assuming you found 2x Locked Candidate (1)
# A Unique Rectangle Type 1 elimination exists for 45 in [r8c1]
 *--------------------------------------------------------------------*
 | 8      1      9      | 3      2      4      | 7      6      5      |
 | 7      6      3      | 1      9      5      | 28     248    24     |
 |*45     2     *45     | 8      7      6      | 1      9      3      |
 |----------------------+----------------------+----------------------|
 | 39     5      8      | 27     6      379    | 4      237    1      |
 | 2      349    1      | 5      348    379    | 389    378    6      |
 | 6      349    7      | 24     1348   139    | 2389   5      29     |
 |----------------------+----------------------+----------------------|
 | 1349   8      2      | 6      5      137    | 39     134    479    |
 |*139-45 39    *45     | 47     134    8      | 6      1234   2479   |
 | 134    7      6      | 9      134    2      | 5      134    8      |
 *--------------------------------------------------------------------*

Code: Select all
# After 3x Singles, you are left with ...
# There's probably an elegant elimination here, but all I have is a loop
 *-----------------------------------------------------------*
 | 8     1     9     | 3     2     4     | 7     6     5     |
 | 7     6     3     | 1     9     5     | 28    248   24    |
 | 5     2     4     | 8     7     6     | 1     9     3     |
 |-------------------+-------------------+-------------------|
 | 39    5     8     |e27    6     379   | 4    d237   1     |
 | 2     349   1     | 5     348   379   | 389   378   6     |
 | 6     349   7     |f24    1348  139   | 2389  5     29    |
 |-------------------+-------------------+-------------------|
 | 1349  8     2     | 6     5     137   | 39    134   479   |
 | 139   39    5     |a7-4   134   8     | 6    c1234 b2479  |
 | 134   7     6     | 9     134   2     | 5     134   8     |
 *-----------------------------------------------------------*


Code: Select all
4=[r8c4]=7=[r8c9]=2=[r8c8]-2-[r4c7]=2=[r4c4]-2-[r6c4]-4-[r8c4]; => [r8c4]<>4

SSTS completes the puzzle.
daj95376
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Posts: 2624
Joined: 15 May 2006

Postby pgoyette » Sun Mar 23, 2008 2:24 pm

Thanks for the reply.

Yep, I had found the UR.

I had also found an elimination based on r7c7 = 9. In this case none of the remaining candidates for r4c8 result in a solution. (I have no idea of the notation that would be used to represent this!)

I guess my real question is, how does finding an elimination chain differ from guessing?:) It seems that both require you to pick some arbitrary cell and try all the possible remaining candidates to see if (all but) one result in a contradiction?:)

Is there some algorithmic way (rather than arbitrarily or randomly) selecting the cell to try?
pgoyette
 
Posts: 2
Joined: 22 March 2008

Postby daj95376 » Sun Mar 23, 2008 3:38 pm

pgoyette wrote:I guess my real question is, how does finding an elimination chain differ from guessing?:) It seems that both require you to pick some arbitrary cell and try all the possible remaining candidates to see if (all but) one result in a contradiction?:)

The question of guessing vs. technique has been discussed numerous times. I'm not going to reopen that Pandora's Box.

I'm a terrible manual Sudoku player. However, there are many talented manual players who could easily find my chain -- and others. All but one link in my chain conforms to a b/b plot, and that's considered a technique -- not guessing.

http://forum.enjoysudoku.com/viewtopic.php?t=2143
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