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Stuck in a Nonconsecutive puzzle

 
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Nasenbaer
Master
Master


Joined: 20 Jul 2006
Posts: 167
Location: Fellbach (Deutschland)

PostPosted: Fri Nov 10, 2006 12:10 am    Post subject: Stuck in a Nonconsecutive puzzle Reply with quote

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
Hooked


Joined: 04 Apr 2006
Posts: 42

PostPosted: Fri Nov 10, 2006 2:17 am    Post subject: Reply with quote

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
Master


Joined: 20 Jul 2006
Posts: 167
Location: Fellbach (Deutschland)

PostPosted: Fri Nov 10, 2006 10:58 pm    Post subject: Reply with quote

Thanks for your help. I actually understood it! Wink

Also thanks for the link. I will have to find a quiet hour to fully understand it. Very Happy

Peter
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