The Type E3 Almost Unique Rectangle

Advanced methods and approaches for solving Sudoku puzzles

Postby ronk » Mon Feb 27, 2006 2:30 pm

Carcul wrote:Ok, so we have 54 starting grids, good job.

An additional word about this library: I've only looked at the few puzzles previously mentioned on this thread, so I can't vouch for their quality. My program wasn't designed to find eliminations ... so some puzzles may not have any. Some puzzles may also look so much like another, that you'll be like Yogi Berra saying ... "deja vu all over again".:D

But enjoy, Ron
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Postby ravel » Mon Feb 27, 2006 4:35 pm

ronk wrote:Some puzzles may also look so much like another, that you'll be like Yogi Berra saying ... "deja vu all over again".:D

Interesting, those similar puzzles. I looked at those in lines 26 and 27. Though the candidate lists are (necessarily) different, both can be solved with the same grouped colorings:
Let a be 3 or 8
[r8c7]-a-[r8c23]=a=[r7c1]-a-[r4c1]=a=[r5c23]-a-[r5c9]=a=[r4c7]-a-[r8c7] => r8c7=1
Then both puzzles have the same remote locked pair 38 in r5c9 and r2c6.
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Postby Wolfgang » Tue Feb 28, 2006 10:11 am

Hi Carcul, think i found a nice sample for this type:
Code: Select all
------+-------+------
. . . | 3 . . | 8 . .
. . . | . . . | . 6 .
. . . | . 6 9 | . 7 3
------+-------+------
. . . | . . 8 | . 4 .
7 . . | . 3 4 | 6 9 .
. . 4 | 6 .  7| 3 . 5
------+-------+------
. 4 2 | . 8 5 | . . .
. 8 5 | 7 . . | 9 . .
1 . 7 | . . 3 | 5 . 8
------+-------+------

46  279   69   3   47  12  8   5 129       
458 23579 39   458 457 12  124 6 129       
458 25    1    458 6   9   24  7 3         
56  1359  369  259 159 8   12  4 7         
7   15    8    25  3   4   6   9 12         
2   19    4    6   19  7   3   8 5         
9   4     2    1   8   5   7   3 6         
3   8     5    7   2   6   9   1 4         
1   6     7    49  49  3   5   2 8       

If r4c1=5:
1. r23c1<>5
2. r4c45<>5, r5c4=5,r23c4<>5
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Postby Carcul » Tue Feb 28, 2006 2:50 pm

Hi Wolfgang.

Very nice one. In chain notation we have:

[r4c1]-5-[r2c1|r3c1]=5=[r2c4|r3c4]-5-[r4c4|r5c4]=5=[r4c5]-5-[r4c1], => r4c1<>5 which solve the puzzle.

Good work.

Regards, Carcul
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