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
ronk
2012 Supporter
 
Posts: 4764
Joined: Wed Nov 02, 2005 5:00 pm
Location: Southeastern USA

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.
ravel
 
Posts: 998
Joined: Tue Feb 21, 2006 5:00 pm

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
Wolfgang
 
Posts: 208
Joined: Wed Jun 22, 2005 4:00 pm

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
Carcul
 
Posts: 724
Joined: Fri Nov 04, 2005 5:00 pm

Previous

Return to Advanced solving techniques

Who is online

Users browsing this forum: No registered users and 0 guests

cron