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Not just a shortcut? Shared peers

 
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keith



Joined: 19 Sep 2005
Posts: 3182
Location: near Detroit, Michigan, USA

PostPosted: Sun Apr 24, 2011 1:19 am    Post subject: Not just a shortcut? Shared peers Reply with quote

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

PostPosted: Sun Apr 24, 2011 1:46 am    Post subject: Reply with quote

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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