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tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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Posted: Fri Dec 02, 2011 1:46 am Post subject: Vanhagen Extreme December 1,2011 |
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I was gone a couple of days but here is the current puzzle.
Code: |
+-------+-------+-------+
| 9 . 1 | . 3 . | . . . |
| . . 6 | 9 . 7 | . . . |
| . . . | . . 2 | . 9 3 |
+-------+-------+-------+
| . . 3 | . 2 . | 4 . 5 |
| . 5 . | . . . | . 8 . |
| 4 . 8 | . 7 . | 6 . . |
+-------+-------+-------+
| 5 6 . | 2 . . | . . . |
| . . . | 3 . 9 | 5 . . |
| . . . | . 5 . | 8 . 4 |
+-------+-------+-------+
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Play online
After basics:
Code: | *-----------------------------------------------------------*
| 9 27 1 | 8 3 5 | 27 4 6 |
| 23 234 6 | 9 14 7 | 12 5 8 |
| 78 478 5 | 146 146 2 | 17 9 3 |
|-------------------+-------------------+-------------------|
| 167 9 3 | 16 2 8 | 4 17 5 |
| 1267 5 27 | 146 9 146 | 3 8 127 |
| 4 12 8 | 5 7 3 | 6 12 9 |
|-------------------+-------------------+-------------------|
| 5 6 47 | 2 8 14 | 9 3 17 |
| 1278 1278 247 | 3 146 9 | 5 1267 127 |
| 123 123 9 | 7 5 16 | 8 126 4 |
*-----------------------------------------------------------* |
AUR(78)r38c12 Mixed SIS: r4c2=4, r78c3=7
AUR(78)r38c12[(7)r78c3=(4)r3c2]-r2c2=r2c5-r8c5=r7c6-(4=7)r7c3; r5c3<>7
BUG+3; r4c4=1
(2-4)r2c2=r3c2-(4=6)r3c4-(6=1)r4c4
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(4)r3c5-(4=6) r3c4-(6=1)r4c4
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(6)r5c4-(6=1)r4c4
Ted |
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peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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Posted: Fri Dec 02, 2011 4:37 pm Post subject: |
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Long time... Hi!
Ted! Erstwhile king of almost-wings...
Code: | xy-wing(12-7)r5c9,r5c3,r7c9
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(7)r5c9* - r4c8=(7-6)r4c1=r4c4 - r3c4=r3c5 - r8c5=r8c8 - *(7)r8c89=r8c123 ; r7c3<>7 |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Fri Dec 02, 2011 6:16 pm Post subject: |
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Peter, I have a problem with selective memory that ignores a contradiction. This section of your chain, r8c8 - *(7)r8c89, ignores the fact that (7=)r5c9 also results in (-7)r7c9. If this elimination is also remembered, then there's a contradiction where all of the 7s are eliminated in [b9].
I would have preferred two steps:
Code: | (7)r4c8 = (7-6)r4c1 = r4c4 - r3c4 = r3c5 - r8c5 = (6)r8c8 ; r8c8<>7
<12+7> xy-wing r5c9/r5c3+r7c9 ; r7c3<>7
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Where the first step is an embedded chain in your chain.
Regards, Danny |
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peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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Posted: Fri Dec 02, 2011 9:34 pm Post subject: |
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Selective memory doesn't worry me... best to ignore the collateral damage of chains and use the outcomes that are helpful!
Anyway it was only a bit of fun - I think I initially had an l-wing and then the xy-wing... |
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arkietech
Joined: 31 Jul 2008 Posts: 1834 Location: Northwest Arkansas USA
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Posted: Fri Dec 02, 2011 10:11 pm Post subject: |
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I have selective memory. I just wish I could do the selecting! Code: | *-----------------------------------------------------------*
| 9 a27 1 | 8 3 5 |b27 4 6 |
| 23 234 6 | 9 d14 7 |c12 5 8 |
| 78 478 5 | 146 146 2 | 17 9 3 |
|-------------------+-------------------+-------------------|
| 167 9 3 | 16 2 8 | 4 17 5 |
| 1267 5 g27 | 146 9 146 | 3 8 127 |
| 4 1-2 8 | 5 7 3 | 6 12 9 |
|-------------------+-------------------+-------------------|
| 5 6 47 | 2 8 14 | 9 3 17 |
| 1278 1278 f247 | 3 e146 9 | 5 1267 127 |
| 123 123 9 | 7 5 16 | 8 126 4 |
*-----------------------------------------------------------*
(2)r1c2=(2)r1c7-(2=1)r2c7-(1=4)r2c5-(4)r8c5=(4-2)r8c3=(2)r5c3 => r6c2<>2
*--------------------------------------------------*
| 9 27 1 | 8 3 5 | 27 4 6 |
| 23 234 6 | 9 14 7 | 12 5 8 |
| 8 47 5 | 46 146 2 | 17 9 3 |
|----------------+----------------+----------------|
| 67 9 3 |*16 2 8 | 4 #17 5 |
| 67 5 2 | 146 9 4-6 | 3 8 17 |
| 4 1 8 | 5 7 3 | 6 2 9 |
|----------------+----------------+----------------|
| 5 6 47 | 2 8 14 | 9 3 17 |
| 1 8 47 | 3 46 9 | 5 67 2 |
| 23 23 9 | 7 5 *16 | 8 #16 4 |
*--------------------------------------------------*
w-wing
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JC Van Hay
Joined: 13 Jun 2010 Posts: 494 Location: Charleroi, Belgium
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Posted: Fri Dec 02, 2011 10:38 pm Post subject: |
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Splitting of an equivalent of Peter's single step (Kraken 1R5) into 2 Wings ....
#1. ANP(12=6)r69c8-6r9c6=6r5c6-(6=1)r4c4 => -1r4c8
#2. (7=1)r7c9-ANP(1=27)r5c39 => -7r7c3 |
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