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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Sat Sep 04, 2010 5:19 am Post subject: Puzzle 10/09/04: B |
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| Code: | +-----------------------+
| 3 . . | . . . | 4 1 . |
| . 2 . | . 5 . | . . . |
| . . 7 | . . . | . 3 6 |
|-------+-------+-------|
| 2 . . | 6 . . | . . . |
| . . . | . 9 . | . . . |
| . . . | . . . | 6 8 . |
|-------+-------+-------|
| . . 6 | . 7 2 | 9 . 8 |
| . 4 . | . . 5 | 3 . . |
| 1 . 9 | . . . | 7 . 5 |
+-----------------------+
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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: Sat Sep 04, 2010 10:15 am Post subject: |
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| Quote: | X Wing (1)R25/c36 : r36c36<>1
Grouped Skyscraper (8)R14 : r3c6<>8
XY Wing (19-4), pivot at R3C2 : => r2c46,r3c1<>4 |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sat Sep 04, 2010 11:52 am Post subject: |
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A wing...
| Quote: | | m-wing(49) with pseudocell (9=1)r3c2-(1=4)r2c3 - r3c1=(4-9)r3c6=r1c6 ; r1c2<>9 |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Sat Sep 04, 2010 7:14 pm Post subject: |
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Peter, would you mind checking the cell R3C1. I have R3C1={489}. TIA.
JC |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sat Sep 04, 2010 7:43 pm Post subject: |
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Peter's pseudo-cell is in the first two cells.
| Code: | (Y =X)a - (X)b = (X-Y)r = (Y)s
(9=1)r3c2 - (1=4)r2c3 - r3c1 = (4-9)r3c6 = (9)r1c6 ; r1c2<>9
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| Code: | +-----------------------------------------------------------------------+
| 3 59 58 | 78 6 789 | 4 1 2 |
| 6 2 14 | 34 5 134 | 8 79 79 |
| 489 19 7 | 28 12 1489 | 5 3 6 |
|-----------------------+-----------------------+-----------------------|
| 2 579 3458 | 6 34 3478 | 1 579 3479 |
| 48 6 13458 | 578 9 13478 | 2 57 347 |
| 49 1579 1345 | 257 12 1347 | 6 8 3479 |
|-----------------------+-----------------------+-----------------------|
| 5 3 6 | 1 7 2 | 9 4 8 |
| 7 4 2 | 9 8 5 | 3 6 1 |
| 1 8 9 | 34 34 6 | 7 2 5 |
+-----------------------------------------------------------------------+
# 65 eliminations remain
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An ALS perspective:
| Code: | (Y =X)a - (X)b = (X-Y)r = (Y)s
(9=1=4)r3c2,r2c3 - r3c1 = (4-9)r3c6 = (9)r1c6 => r1c2<>9
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Sat Sep 04, 2010 8:06 pm Post subject: |
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Danny, Thanks a lot. In fact, it was R3C4 that I should have asked the checking. I had {248} because I overlooked the NP(34)C4. My bad  . Now, Peter's wing is crystal-clear 
JC |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Sat Sep 04, 2010 8:12 pm Post subject: |
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| daj95376 wrote: | Peter's pseudo-cell is in the first two cells.
| Code: | (Y =X)a - (X)b = (X-Y)r = (Y)s
(9=1)r3c2 - (1=4)r2c3 - r3c1 = (4-9)r3c6 = (9)r1c6 ; r1c2<>9
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Or drop the last SL for an h-wing instead: (9=1)r3c2 - (1=4)r2c3 - (4)r3c1 = (4)r3c6 ==> r3c6<>9 [edit: typos]
h-wing <=> hybrid-wing
Last edited by ronk on Sat Sep 04, 2010 11:42 pm; edited 1 time in total |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Sat Sep 04, 2010 9:41 pm Post subject: |
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Like many others, all happening around blocks 1&2 an ALS xz removing 8s (blue)
Set A {5,8,9} Green
Set B {1,2,8,9} Orange
x = 9
z = 8
(r1c46), r3c1 <> 8 Blue
Also viewed as a simple XY chain, pincers at r1c3 and r3c4 |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Sep 05, 2010 8:43 pm Post subject: |
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X-Wing (1), not needed
Finned X-Wing (8)
AIC, r3c1, r5c6<>4 |
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