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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Sun Apr 24, 2011 1:19 am Post subject: Not just a shortcut? Shared peers |
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All,
I use the following technique all the time. Suppose we have three boxes: | Code: | +-----------+-----------+-----------+
| / / *e | / / / | *a *b *c |
| . . . | . . . | / / / |
| . . . | . . . | / *d / |
+-----------+-----------+-----------+
/ Cell solved
* Cell not solved
. Any value (not relevant)
a, b, c, ... Cell identifiers (labels) |
a,b,c have e as their only unsolved peer in R1. They also have d as their only unsolved peer in B3.
In the solution, d and e must be the same. Not only that, they must have the same candidates in the unsolved puzzle.
I think of this as a basic move, and often use it to eliminate a candidate in d or e. Simpler to see than a box / line interaction. No big deal.
Yesterday we found this in the Freep puzzle:
http://www.dailysudoku.com/sudoku/forums/viewtopic.php?p=30164#30164
| Code: | +-------------------+-------------------+-------------------+
| 8 7 69e | 1 23-6g 4 | 2356a 2369b 235c |
| 59 356 2 | 7 36 8 | 1 -369d 4 |
| 146 136 146 | 236 5 9 | 7 8 23f |
+-------------------+-------------------+-------------------+ |
There are two sets of peers: abcdf and abceg. Clearly, in the solution, d and f contain the same two values as e and g. (Edit: Note that the solved cells in R1 and B3 have the same values.)
There is, however, more, which I do not think is a basic move. 6 in e forces 6 in d, and 9 in e forces 9 in d. So, d <>3. If d and e are the same, then f and g must be the same, so g <>6. (By "same", I mean same in the solution AND in their candidates.)
(Or, 2 in f forces 2 in g, and 3 in f forces 3 in g, so g <>6.)
Amazing, I think.
Keith
Last edited by keith on Sun Apr 24, 2011 2:12 pm; edited 1 time in total |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Apr 24, 2011 1:46 am Post subject: |
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It is basically APE. It can also be viewed as a pair of ALS eliminations:
(6=9)r1c3 - ALSr1b3|r3c9[(9)r1c8=(6)r1c78]; r1c5<>6
and
(3=2)r3c9 - ALSr1c3789[(2=3)r1c789]; r2c8<>3
Perhaps best of all, it is a closed AIC loop:
(6=9)r1c3 - (9)r1c8=(9-6)r2c8=(6)r1c78 - Loop; r2c8<>3, r1c5<>6
Edit to add:
It's also a Sue de Coq |
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