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Nasenbaer
Master
Joined: 20 Jul 2006
Posts: 167
Location: Fellbach (Deutschland)
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 Posted: Fri Nov 10, 2006 12:10 am Post subject: Stuck in a Nonconsecutive puzzle |
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Some months ago I found a nonconsecutive puzzle (sorry, don't know where) and printed it out. It rested on the todo-pile, until today. Here is the original puzzle:
| Code: | . . .|. . .|. . .
. . 7|. . .|6 . .
. 2 9|. 8 .|1 4 .
-----+-----+-----
. . .|. . .|. . .
. . 1|. 9 .|8 . .
. . .|. . .|. . .
-----+-----+-----
. 8 6|. 2 .|7 1 .
. . 2|. . .|3 . .
. . .|. . .|. . . |
Please remember: this a nonconsecutive puzzle, that means additionally to the usual sudoku rules neighboring cells are not allowed to have neighboring numbers.
Now I'm lost. Here is how far I got:
| Code: | .---------------------------.---------------------------.---------------------------.
| 1468 16 3 | 1679 17 14679 | 29 25 589 |
| 18 45 7 | 129 45 129 | 6 89 3 |
| 56 2 9 | 356 8 356 | 1 4 7 |
:---------------------------+---------------------------+---------------------------:
| 27 9 45 | 123678 136 123678 | 45 67 12 |
| 24 67 1 | 4567 9 2456 | 8 3 56 |
| 567 3 8 | 12456 4567 124567 | 245 5679 129 |
:---------------------------+---------------------------+---------------------------:
| 3 8 6 | 49 2 459 | 7 1 459 |
| 179 45 2 | 456789 4567 156789 | 3 5689 4689 |
| 179 17 45 | 136789 13567 13456789| 59 25689 245689 |
'---------------------------'---------------------------'---------------------------' |
Maybe someone could give me a hint how to progress from here.
Thanks in advance.
Peter |
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Myth Jellies
Hooked
Joined: 04 Apr 2006
Posts: 42
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| Posted: Fri Nov 10, 2006 2:17 am Post subject: |
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Note that in r4c78, you cannot have those two cells equal 5 and 6 together. Therefore you know that r4c7 = 4, r4c8 = 7, or both are true. This is a strong inference you can use in an AIC, and you write it as follows...
(4=5)r4c7 - (6=7)r4c8
This little AIC doesn't tell us anything yet, but you can continue this AIC pattern as follows (noting that r4c1 must be 7 or 2, and r5c1 must be 2 or 4)...
(4=5)r4c7 - (6=7)r4c8 - (7=2)r4c1 - (2=4)r5c1
At least one of the endpoints of an AIC must be true. Since r4c3 sees both endpoints (the 4 in r4c7 and the 4 in r5c1) we know that r4c3 cannot equal 4. Notice that you could eliminate that 4 without ever making any assumption about what candidate went into any cell.
Here is a post on AIC's
http://www.sudoku.com/forums/viewtopic.php?t=3865 |
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Nasenbaer
Master
Joined: 20 Jul 2006
Posts: 167
Location: Fellbach (Deutschland)
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| Posted: Fri Nov 10, 2006 10:58 pm Post subject: |
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Thanks for your help. I actually understood it! 
Also thanks for the link. I will have to find a quiet hour to fully understand it. 
Peter |
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