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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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 Posted: Sun Jul 04, 2010 10:18 pm Post subject: Vanhegan Extreme July 4 |
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Here is another challenging Vanhegan Extreme.
| Code: |
+-------+-------+-------+
| . 2 7 | . 6 . | 8 . . |
| . 5 . | 7 . . | . 6 2 |
| 3 . . | 2 . . | . . 7 |
+-------+-------+-------+
| . . . | . . . | 1 3 . |
| 2 . . | . . . | . . 6 |
| . 8 4 | . . . | . . . |
+-------+-------+-------+
| 5 . . | . . 4 | . . 8 |
| 8 4 . | . . 3 | . 7 . |
| . . 1 | . 5 . | 6 4 . |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
Here is my rambling solution.
| Quote: | 1. xyz-wing (159)r8c9; r9c9<>9
2. "Weird" xyz-wing (169)r7c4
(1)r7c4; r8c4<>1
(6)r7c4-(6=3)r7c2-(3=1)r5c2-r5c5=(1)r8c5; r8c4<>1
(9)r7c4-(9=1)r8c5; r8c4<>1
3. ER1 with pivot in box3 and extended strong link(1)r6c1-r6c4=r7c4-(1)r7c8; r1c1<>1
4. xy-wing -489 with vertex (89)r2c3; r2c1<>4
5. Almost flightless xy-wing -159 with vertex 59 in r6c9 pincers (19)r6c1 & (15)r8c9 and fin (9)r8c9
If xy-wing is true: (1)r6c1-r5c2=(1)r5c5; r8c5<>1
If fin is true: (9)r8c9-r6c9=r5c7-(9=1)r5c5; r8c5<>1 |
Ted |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Mon Jul 05, 2010 5:42 am Post subject: |
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I'll start after ted's step number 1
and use the almost w-wing approach.
which solves it.
| Code: | +-------------------+--------------------+-----------------+
| 49-1 2 7 | 3 6 159 | 8 159 9(14) |
| 149 5 89 | 7 48 19 | 3 6 2 |
| 3 16 689 | 2 48 159 | 49 159 7 |
+-------------------+--------------------+-----------------+
| 6 7 (59) | 9(45) 2 8 | 1 3 59(4) |
| 2 3(1) 359 | 9-1(45) 9(1) 7 | 459 8 6 |
| -9(1) 8 4 | 19(5) 3 6 | 7 2 (59) |
+-------------------+--------------------+-----------------+
| 5 36 236 | 169 7 4 | 29 19 8 |
| 8 4 26 | 169 9(1) 3 | 259 7 59(1) |
| 7 9 1 | 8 5 2 | 6 4 3 |
+-------------------+--------------------+-----------------+ |
there would be a w-wing {5,9} if the 5 in r5c4 is false.
the pincers would be in r4c3 and r6c9
the w-wing would be
(5=9)r4c3 - (5)r4c4 = (5)r6c4 - (9=5)r6c9;
obviously this eliminates the 9 in r6c1.
if the 5 is true in r5c4 then
(5-4)r5c4 = (4)r4c4 - (4)r4c9 = (4-1)r1c9 = (1)r8c9 - (1)r8c5 = (1)r5c5 - (1)r5c2 = (1)r6c1;
the 9 is still false.
or as a diagram of a kraken column on 5's
(5)r4c4 - (5=9)r4c3
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(5-4)r5c4 = (4)r4c4 - (4)r4c9 = (4-1)r1c9 = (1)r8c9 - (1)r8c5 = (1)r5c5 - (1)r5c2 = (1)r6c1
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(5)r6c4 - (5=9)r6c9 |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Mon Jul 05, 2010 9:20 am Post subject: |
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Poetry.
I'm still struggling with xsudo Norm - it's a bit like my dishwasher - too many controls!  |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Mon Jul 05, 2010 11:04 am Post subject: |
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Hi Norm, good to see you posting.
Great move on that (5)r5c4. I also saw the almost w-wing but could not find a solution for the (5)r5c4. Oh well, having missed that step I had the opportunity for a couple of other interesting moves.
Ted |
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ronk
Joined: 07 May 2006
Posts: 398
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| Posted: Mon Jul 05, 2010 12:35 pm Post subject: |
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Seeing the damage storm_norm's r6c1<>9 could do ... inspired me to find this 2-stepper with a total of five strong inferences:
| Quote: | ER: (9)r8c5 = (9)r5c5 - (9)r5c7 = (9)r46c9 --> r9c9<>9
s-wing: (5)r6c4 = (5)r6c9 - (5=1)r8c9 - (1)r7c8 = (1)r7c4 --> r6c4<>1 |
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