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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Tue Sep 21, 2010 4:56 am Post subject: Puzzle 10/09/21: C |
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| Code: | +-----------------------+
| 3 . . | 2 . 6 | . 8 . |
| . 7 . | . . . | . 3 6 |
| . . 6 | 3 . . | . 4 2 |
|-------+-------+-------|
| 9 . 7 | 6 . 4 | . 2 . |
| . . . | . 9 2 | . . . |
| 2 . . | 5 3 . | 6 9 4 |
|-------+-------+-------|
| . . . | . . 9 | 2 6 . |
| 6 4 5 | 7 . 8 | 3 1 . |
| . 2 9 | . . 3 | . . . |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Tue Sep 21, 2010 8:04 am Post subject: |
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| Quote: | 4-SIS AIC : (41)R1C3 1B3 1R5 3R5 : (4)r1c3=(3)r5c3 : => r5c3<>4
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Tue Sep 21, 2010 1:58 pm Post subject: |
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Took me two .. and a bit of time!
| Quote: | ANP(1=78)r79c1 - (8=4)r2c1 - (4=1)r1c3 ; r3c1<>1, r7c3<>1
xy-chain (4=1)r1c3 - (1=5)r1c9 - (5=3)r4c9 - (3=5)r4c2 - (5=4)r5c1 ; r2c1<>4, r5c3<>4 |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Tue Sep 21, 2010 4:32 pm Post subject: |
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| Quote: | 4-SIS AIC : (41)R1C3 1B3 1R5 3R5 : (4)r1c3=(3)r5c3 : => r5c3<>4 to be replaced by
XY Wing Style : (41)R1C3 1C9 3R5 : (4)r1c3=(3)r5c3 : => r5c3<>4 !!! |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Tue Sep 21, 2010 5:03 pm Post subject: |
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| JC Van Hay wrote: | | Quote: | | 4-SIS AIC : (41)R1C3 1B3 1R5 3R5 : (4)r1c3=(3)r5c3 : => r5c3<4> r5c3<>4 !!! |
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JC, nice spot.
[Edit. Withdrawn question about SIS - looking at wrong elimination. Doh!] |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Tue Sep 21, 2010 6:30 pm Post subject: |
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Peter : Notations can be confusing! I use capital letters with SIS and small letters with WIS. So the SIS (41)R1C3 stands for the entire (strong) cell R1C3, the WIS (13)r5c9 stands for a subset of the cell (1357)R5C9 - (1)r5c9 and (3)r5c9 cannot both be true, i.e. (1-3)r5c9 -, the same for rows, columns and blocks.
So the XY Wing Style move may be fully written in Eureka notation as [(4=1)r1c3] - [(1)r1c9=r5c9] - [(3)r5c9=r5c3]. The shorthand notation is therefore (41)R1C3 1C9 3R5. I eventually add the endpoints of the AIC to ease the reading of the AIC with the pencilmarked grid.
To "complete" the explanation, here is the Xsudo presentation of the move, where the SIS and the WIS are represented by solid and light lines, respectively:Regards, JC
PS : see storm_norm's proof of 10/09/20C for another Xsudo illustration and here for XY Wing Styles.
PPS : Never mind ...  ... You are welcome ! |
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