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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Fri Mar 18, 2011 4:35 pm Post subject: Free Press March 18, 2011 |
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Not yet done. Does require advanced moves.
| Code: | Puzzle: FP031811
+-------+-------+-------+
| . . . | . 3 9 | . 4 . |
| . . 7 | 8 . . | 3 6 . |
| . . . | . . . | . 7 . |
+-------+-------+-------+
| 5 . . | 4 . 1 | . . . |
| 4 . . | . 5 . | . . 2 |
| . . . | 9 . 8 | . . 7 |
+-------+-------+-------+
| . 1 . | . . . | . . . |
| . 3 . | . . 7 | 9 . . |
| . 2 . | 3 1 . | . . 4 |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Mar 18, 2011 6:46 pm Post subject: |
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| Quote: | | XY-Chain. The 9 in r5c8 proves a 1 in r8c9; r8c8<>1 |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Mar 18, 2011 8:17 pm Post subject: |
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After basics: | Code: | +-------------+-------------+-------------+
|-12 58 6 | 7 3 9 |125f 4 158 |
| 12d 59 7 | 8 4 25 | 3 6 -159c|
| 89 4 3 | 1 6 25 | 25 7 89 |
+-------------+-------------+-------------+
| 5 89b 2 | 4 7 1 | 68 39 36 |
| 4 7 89 | 6 5 3 | 18e 19a 2 |
| 3 6 1 | 9 2 8 | 4 5 7 |
+-------------+-------------+-------------+
| 78 1 58 | 25 9 4 | 567 23 36 |
| 6 3 4 | 25 8 7 | 9 12 15 |
| 79 2 59 | 3 1 6 | 57 8 4 |
+-------------+-------------+-------------+ |
M-wing: 9 in a forces 9 in c (coloring via b). a and d are pincers on 1. Transport a through e to f, making the eliminations shown. Bringing us here: | Code: | +-------------+-------------+-------------+
| 2 58 6 | 7 3 9 | 1-5 4 158@|
| 1 59 7 | 8 4 2 | 3 6 59 |
| 89 4 3 | 1 6 5 | 2 7 89 |
+-------------+-------------+-------------+
| 5 89 2 | 4 7 1 | 68 39 36 |
| 4 7 89 | 6 5 3 | 18 19 2 |
| 3 6 1 | 9 2 8 | 4 5 7 |
+-------------+-------------+-------------+
| 78 1 58 | 25 9 4 |567@ 23 36 |
| 6 3 4 | 25 8 7 | 9 12 1-5 |
| 79 2 59 | 3 1 6 | 57 8 4 |
+-------------+-------------+-------------+ |
Which is a BUG+2 - one or both of the cells marked @ is 5.
Keith |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Mar 18, 2011 9:40 pm Post subject: |
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In the second grid above, there are other ways to make the same eliminations:
| Code: | +-------------+-------------+-------------+
| 2 58 6 | 7 3 9 | 1-5 4 158 |
| 1 59d 7 | 8 4 2 | 3 6 59a |
| 89 4 3 | 1 6 5 | 2 7 89 |
+-------------+-------------+-------------+
| 5 89 2 | 4 7 1 | 68 39 36 |
| 4 7 89 | 6 5 3 | 18 19 2 |
| 3 6 1 | 9 2 8 | 4 5 7 |
+-------------+-------------+-------------+
| 78 1 58 | 25 9 4 | 567 23 36 |
| 6 3 4 | 25 8 7 | 9 12 1-5 |
| 79 2 59c | 3 1 6 | 57b 8 4 |
+-------------+-------------+-------------+ |
a and b are M-wing pincers on 5, since 9 in a forces 9 in c.
Another way to look at it is c and d are not just a W-wing - they are a remote pair since they are connected by strong links on 9. Then, make the one-link extension of both ends, d to a and c to b, to establish the pincers on 5 at a and b.
Keith |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sat Mar 19, 2011 1:31 am Post subject: |
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I initially found three different anp() moves that reduced the puzzle to the BUG+2 posted by Keith, but then I spotted a flightless w-wing(58)r1c2|r7c3 with SL (8)b4
(The w-wing (59)r2c2/r9c3 with SL b4 provides the same pincers on 5)
I could not find a transport that provided a deletion on (5), but with two pincer transports; r7c8,r8c4<>2
(5)r1c2-(8)r1c2=r4c2-(8=9)r5c3-(9=1)r5c8-(1=2)r8c8
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(5)r7c3-(5=2)r7c4
Fun puzzle.......
Ted |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Mar 19, 2011 2:23 am Post subject: |
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| tlanglet wrote: | Fun puzzle.......
Ted |
I thought so too. Had to break out the Sonoma juice.
It's too bad that these Friday Freep puzzles are so variable, when some of them are quite good.
Keith |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sat Mar 19, 2011 2:39 am Post subject: |
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| keith wrote: | | tlanglet wrote: | Fun puzzle.......
Ted |
I thought so too. Had to break out the Sonoma juice
Keith |
I agree about breaking out the juice, but I prefer the Zinfandels from the Northern Foothills of California where I live.
Ted |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sat Mar 19, 2011 3:52 am Post subject: |
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| Code: | after basics
+-----------------------------------------------------+
| 12 58 6 | 7 3 9 | #125 4 *158 |
| 12 59 7 | 8 4 25 | 3 6 #159 |
| 89 4 3 | 1 6 25 | 2-5 7 89 |
|-----------------+-----------------+-----------------|
| 5 89 2 | 4 7 1 | 68 39 36 |
| 4 7 89 | 6 5 3 | 18 19 2 |
| 3 6 1 | 9 2 8 | 4 5 7 |
|-----------------+-----------------+-----------------|
| 78 1 58 | 25 9 4 | #567 23 36 |
| 6 3 4 | 25 8 7 | 9 12 15 |
| 79 2 59 | 3 1 6 | 57 8 4 |
+-----------------------------------------------------+
# 35 eliminations remain
BUG+4:
(=5)r1c7|r2c9|r7c7 => r3c7<>5 -or-
(=1)r1c9 - (1=2)r1c1 - r2c2 = r2c6 - (2=5)r3c6 => r3c7<>5
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Now, where did I put the Skyy vodka and Bloody Mary mix? |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Mar 19, 2011 12:11 pm Post subject: |
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Danny,
That is amazing, I would never have thought to look for it.
Your second chain is a little simpler if you note that in R1, 1 in R1C8 forces 5 in R1C7, and then the first chain applies.
And, this takes us to the BUG+2, right?
Keith |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sat Mar 19, 2011 4:17 pm Post subject: |
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| keith wrote: | Your second chain is a little simpler if you note that in R1, 1 in R1C9 forces 5 in R1C7, and then the first chain applies.
And, this takes us to the BUG+2, right?
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It was late and I was tired. However, I didn't use any of the 5s from the BUG+4 in the second chain because they were presumed false from the "-or-" condition. What I should have done is show: (5=12)r1c71 - (1)r1c9 to exclude that possibility. Essentially, if the 5s are false, then r1c9=1 is also false. Maybe this is really a BUG+3 scenario?
There were so many XY-Chains that crack the puzzle, I was sure that any elimination in a bivalue cell would crack the puzzle. My oversight!
Thanks for cleaning up my sloppiness!!!
Regards, Danny |
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