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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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 Posted: Tue May 11, 2010 2:36 pm Post subject: Vanhegan Fiendish May 11 |
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| Code: |
+-------+-------+-------+
| 9 . 4 | 2 . 1 | . . . |
| . . . | . 8 . | . . . |
| . . 6 | 9 . 3 | . 5 . |
+-------+-------+-------+
| . . 9 | 7 . . | . 1 . |
| 5 . 7 | 1 . 9 | 3 . 4 |
| . 2 . | . . 8 | 7 . . |
+-------+-------+-------+
| . 4 . | 3 . 5 | 1 . . |
| . . . | . 4 . | . . . |
| . . . | 8 . 7 | 2 . 5 |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
I was specifically looking for "odd/unusual" moves and found two.
A MESS: HP(36) in r4c25 or r4c2=8 or r4c5=2
HP(36) - (6)r4c6 = r2c6 - (6=9)r2c7 - (9=8)r8c7; r78c8<>8,
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(8)r4c2 -r4c9 = r5c8; r78c8<>8,
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(2)r4c5 - (2=6)r5c5 - (6)r4c6 = r2c6 - (6=9)r2c7 - (9=8)r8c7; r78c8<>8,
Thus, r78c8<>8, which opens up an the next step.
"Almost" XYZ-wing(AXYZ-wing) 236 in r4c5 with (26) in r5c5, (36)in r4c2 plus (8 ) in r4c2
(2)r4c5 - (2=6)r5c5; r4c6<>6,
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(3)r4c5 - (3=6)r4c2; r4c6<>6,
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(6)r4c5; r4c6<>6,
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(8)r4c2 - (8=1)r3c2 - (1=28)r34c9 - (8)r78c9 = (8-9)r8c7 = (9-6)r2c7 = (6)r2c6; r4c6<>6
I did not see a lot of VH/VH+ moves in this puzzle but I think it should be fun to work on.
Ted
[Edited to correct typo in thread title.]
Last edited by tlanglet on Tue May 11, 2010 4:00 pm; edited 1 time in total |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Tue May 11, 2010 3:40 pm Post subject: |
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With all those bivalues there had to be an xy-chain somewhere!
(5=4)r2c4-(4=6)r2c6-(6=9)r2c7-(9=8)r8c7-(8=5)r8c3; r2c3<>5
On reflection you can also see it as an ALS (I admit I didn't!)..
ALS(4569) r2c467, ALS(589) r8c37, x=9, z=5 |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue May 11, 2010 4:36 pm Post subject: |
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| The 37 UR in boxes 39 will do it. R1c89=8 or r8c89=89 (forms pseudo cell). Either forces r8c3=5. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Tue May 11, 2010 4:52 pm Post subject: |
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| Marty R. wrote: | | The 37 UR in boxes 39 will do it. R1c89=8 or r8c89=89 (forms pseudo cell). Either forces r8c3=5. |
Marty, That's one impressive UR elimination!!! |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue May 11, 2010 5:28 pm Post subject: |
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| Marty R. wrote: | | The 37 UR in boxes 39 will do it. R1c89=8 or r8c89=89 (forms pseudo cell). Either forces r8c3=5. |
Extraordinary Find 
Ted |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Wed May 12, 2010 9:33 am Post subject: |
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Sublime UR Marty!
(Ed - You might very well be sublime Marty, but I was praising the move  ) |
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