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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Thu May 12, 2011 1:46 am Post subject: Energizer |
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Keeps on going ...
| Code: | Puzzle: BB042911sh
+-------+-------+-------+
| . 3 . | . 4 . | . 2 . |
| 5 . 8 | . 6 . | 7 . 4 |
| . 4 1 | . . . | 3 8 . |
+-------+-------+-------+
| . . . | 8 . 3 | . . . |
| 1 7 . | . . . | . 9 2 |
| . . . | 2 . 7 | . . . |
+-------+-------+-------+
| . 1 4 | . . . | 2 6 . |
| 3 . 7 | . 2 . | 1 . 8 |
| . 5 . | . 8 . | . 3 . |
+-------+-------+-------+ |
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu May 12, 2011 4:59 am Post subject: |
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Yes, this was a little stubborn.
W-Wing (59), flightless with transport; r3c9<>9
XY-Wing (692); r4c3<>2
Hidden UR (46); r4c1<>6, r6c7<>4
AIC; r9c49<>9
BUG+1 |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Thu May 12, 2011 9:30 pm Post subject: |
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After basics: | Code: | +----------------+----------------+----------------+
| 7 3 69 | 1 4 8 | 59 2 569 |
| 5 29 8 | 3 6 29d | 7 1 4 |
| 269 4 1 | 59e 7 259d | 3 8 6-9 |
+----------------+----------------+----------------+
| 246 26 256 | 8 9 3 | 456 7 1 |
| 1 7 3 | 46 5 46 | 8 9 2 |
| 469 8 569 | 2 1 7 | 456 45 3 |
+----------------+----------------+----------------+
| 8 1 4 | 579 3 59c | 2 6 579b |
| 3 69 7 | 4569 2 46 | 1 45 8 |
| 269 5 269 | 4679 8 1 | 49 3 79a |
+----------------+----------------+----------------+ |
Note the triple abc. Any cell that sees all three cannot be 9. If c is 9, so is e, making the elimination shown.
Interesting, though it does not solve the puzzle.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu May 12, 2011 11:15 pm Post subject: |
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You sat toh-MAY-toh, I say toh-MAH-toh. I see it as an XYZ-Transport. But that elimination is the one I made with the flightless W-Wing on 59, using the same transport.
I had another solution, this one an eight-stepper, but was all pattern-based and didn't need an AIC. |
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susume
Joined: 13 May 2011
Posts: 36
Location: Southeastern US
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| Posted: Fri May 13, 2011 4:18 am Post subject: |
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Fun puzzle! It seems to me a UR and a chain of 9s also cracks it.
Starting from "after basics" as given by keith,
UR 46 requires either 5 or 9 in r8c4, and r7c6 takes the other; r79c4<>59
I hope I've remembered the right notation for an AIC:
r2c2=9=r2c6-9-r3c4=9=r8c4-9-r8c2=9=r2c2 => (9)r2c2
Singles to end.
I've been away from the forums for a couple of years. Please let me know if I've got the loop or its notation wrong. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Fri May 13, 2011 11:54 am Post subject: |
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FWIW, you can do it in one step if you think of the internal SIS in the UR rather than as a Type 1...
| Code: | | ur(46)r58c46[(9=5)r8c4] - (5=9)r7c6 - r2c6=r2c2 ; r8c2<>9 |
.. though clearly it amounts to the same solution. |
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