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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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 Posted: Tue Dec 29, 2009 3:01 am Post subject: Dec 29 VH |
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Must be a mistake. This one is solved with basics, not a VH.
Early Earl |
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arkietech
Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA
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| Posted: Tue Dec 29, 2009 9:09 am Post subject: |
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What a bummer!  Hardest step I found was a couple of pairs. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Tue Dec 29, 2009 2:54 pm Post subject: |
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| I think you both got lucky - and that this IS a VH - and that a one-step solution is an x-wing on on 5. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Dec 29, 2009 3:12 pm Post subject: |
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Sudoku Susser says: | Quote: | Heuristics used:
1 x Intersection Removal
4 x Simple Naked Sets
8 x Pinned Squares |
So, by the definition of this site, this is not a VH.
Here are the critical steps:
| Quote: |
(edited - removing singles)
Deduction pass 7; 39 squares solved; 42 remaining.
* Squares R6C7 and R6C8 in row 6 form a simple naked pair. These 2 squares both contain the 2 possibilities <49>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the row.
R6C2 - removing <49> from <3469> leaving <36>.
R6C3 - removing <49> from <34679> leaving <367>.
Deduction pass 8; 39 squares solved; 42 remaining.
* Squares R8C7 and R9C7 in column 7 form a simple naked pair. These 2 squares both contain the 2 possibilities <57>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the column.
R1C7 - removing <5> from <2459> leaving <249>.
R2C7 - removing <5> from <245> leaving <24>.
Deduction pass 9; 39 squares solved; 42 remaining.
* Squares R8C7 and R9C7 in block 9 form a simple naked pair. These 2 squares both contain the 2 possibilities <57>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the block.
R9C8 - removing <5> from <358> leaving <38>.
Deduction pass 10; 39 squares solved; 42 remaining.
* Intersection of row 2 with block 2. The value <5> only appears in one or more of squares R2C4, R2C5 and R2C6 of row 2. These squares are the ones that intersect with block 2. Thus, the other (non-intersecting) squares of block 2 cannot contain this value.
R1C6 - removing <5> from <569> leaving <69>.
R3C4 - removing <5> from <1456> leaving <146>.
R3C5 - removing <5> from <2459> leaving <249>.
R3C6 - removing <5> from <569> leaving <69>.
Deduction pass 11; 39 squares solved; 42 remaining.
* Squares R1C6 and R3C6 in column 6 form a simple naked pair. These 2 squares both contain the 2 possibilities <69>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the column.
R4C6 - removing <6> from <567> leaving <57>.
R7C6 - removing <9> from <379> leaving <37>.
Deduction pass 12; 39 squares solved; 42 remaining.
(edited - removing singles)
Deduction pass 15; 81 squares solved; 0 remaining.
Solution found!
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Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Dec 29, 2009 4:42 pm Post subject: |
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| I obviously missed something as I also needed the X-Wing on 5. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Dec 29, 2009 5:06 pm Post subject: |
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| Marty R. wrote: | | I obviously missed something as I also needed the X-Wing on 5. |
I think this is it:
| Code: | +-------------------+-------------------+-------------------+
| 8 456 246 | 7 3 569 | 249 459 1 |
| 9 7 124 | 145 245 8 | 24 6 3 |
| 234 3456 12346 | 1456 2459 569 | 8 459 7 |
+-------------------+-------------------+-------------------+
| 47 469 4679 | 3 57 567 | 1 2 8 |
| 5 2 8 | 9 1 4 | 3 7 6 |
| 1 36 367 | 68 78 2 | 49 49 5 |
+-------------------+-------------------+-------------------+
| 347 1 5 | 48 4789 379 | 6 38 2 |
| 6 8 347 | 2 457 357 | 57 1 9 |
| 237 39 2379 | 58 6 1 | 57 38 4 |
+-------------------+-------------------+-------------------+ |
The X-wing is in C28. But, the intersection of R2B2 makes the same eliminations of 5 in B2.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Dec 29, 2009 8:56 pm Post subject: |
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| That's probably it Keith. |
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Wendy W
Joined: 04 Feb 2008
Posts: 144
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| Posted: Wed Dec 30, 2009 12:38 am Post subject: |
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| Marty and CGordon, I needed the x-wing too, and for me it was a tough one to spot. Excellent end-of-year puzzle! |
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