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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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 Posted: Sun Nov 21, 2010 11:06 am Post subject: MM Diabolical 211110 |
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| Code: | *-----------*
|..1|5..|...|
|.35|97.|4..|
|...|.2.|.8.|
|---+---+---|
|...|..6|57.|
|7.6|...|2.1|
|.1.|2..|...|
|---+---+---|
|.6.|.8.|...|
|..8|.53|62.|
|...|..2|3..|
*-----------* |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun Nov 21, 2010 12:40 pm Post subject: |
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Certainly more of a challenge for me than the Friday Free Press MM (basics!) - two steps with a nice ADP/ALS sort of move 
| Quote: | skyscraper(9) r5, r9 ; r4c1<>9, r8c2<>9
ADP(79) forms a pincer (7)r1c9 and (7)r7c7...
(7=1)r7c7 - ADP(79)r13c27:[(1)r3c7=(248)r1c126] - (2=7)r1c9 ; r13c7<>7, r78c9<>7 |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Nov 21, 2010 2:42 pm Post subject: |
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Great move Peter !
I initially skipped looking for single digit patterns and found several steps with some value but definitely not great. I started again and spotted your first move and then your second move, which I handled in a different fashion but same result.
| Quote: | skyscraper (9)r59c8; r4c1,r8c2<>9
AUR(79)r13c27 external SIS r3c3=9,r6c7=7; r3c7<>79=1
AUR(79)[(9)r3c3=(7)r6c7]-(7=9)r1c7; r3c7<>9
AUR(79)[(7)r6c7=(9)r3c3]-(9=7)r3c2; r3c7<>7 |
Ted |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Nov 21, 2010 2:56 pm Post subject: |
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After te SS: | Code: | +----------------+----------------+----------------+
| 248 2789 1 | 5 6 48 | -79 3 27 |
| 28 3 5 | 9 7 18 | 4 16 26 |
| 6 79 49 | 3 2 14 | -179 8 5 |
+----------------+----------------+----------------+
| 2348 289 2349 | 148 149 6 | 5 7 349 |
| 7 89 6 | 48 3 5 | 2 49 1 |
| 5 1 349 | 2 49 7 | 8 469 3469 |
+----------------+----------------+----------------+
| 23 6 23 | 147 8 9 | 17 5 4-7 |
| 19 4 8 | 17 5 3 | 6 2 -79 |
| 19 5 7 | 6 14 2 | 3 149 8 |
+----------------+----------------+----------------+ |
The 79 DP means either
R1C2 is 28 > R1C9=7
or
R3C7 is 1 > 7C7=7.
Why not just eliminate the 7's as shown?
Keith |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun Nov 21, 2010 3:31 pm Post subject: |
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| keith wrote: | | Why not just eliminate the 7's as shown? |
 That's exactly what I thought I did - though wrote it as an AIC  |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Nov 21, 2010 4:29 pm Post subject: |
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OK. Thank you,
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Nov 21, 2010 4:30 pm Post subject: |
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Where is this puzzle from? I first thought it was the Sunday diabolical from here,
http://www.sudoku.org.uk/daily.asp
But it's not. |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Nov 21, 2010 5:50 pm Post subject: |
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Well, it's not the #1960 that shows, so I'll let someone else figure out what's what.
W-Wing (49); r5c2<>4
XY-Wing (894); r8c4<>4; transport; r9c1<>4
Chain; r8c2<>9
XY-Chain; r9c8<>4 |
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