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| AMcK
| | Joined: 07 Apr 2005 | | Posts: 89 | | : | | Location: Cambridge |
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 Posted: Sun Oct 09, 2005 9:54 pm Post subject: Sunday Times "The superior version 11" October 9, |
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Anyone got a simple solution to today's Sunday Times 9x9 Sudoku
| Code: | 1 . . ¦ 2 . 7 ¦ . . 9
. 9 . ¦ . 4 . ¦ . 7 .
. . . ¦ 8 . 9 ¦ . . .
------+-------+-------
. 2 . ¦ . 9 . ¦ . 5 .
. . 9 ¦ . . . ¦ 2 . .
. 3 . ¦ . 7 . ¦ . 4 .
------+-------+-------
. . . ¦ 1 . 4 ¦ . . .
. 4 . ¦ . 3 . ¦ . 8 .
8 . . ¦ 9 . 6 ¦ . . 7
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I can unlock it with just two contradictions r2c6<>1 (13 steps) and r5c6<>1 (5 steps) and then it's easy.
Unless I missed an easier solution this seems a bit hard for a newspaper puzzle.
Regards
Andrew |
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| Nick70
| | Joined: 08 Jun 2005 | | Posts: 160 | | : | |
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| Posted: Sun Oct 09, 2005 10:08 pm Post subject: |
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| There's an X-Wing in the 1s. |
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| AMcK
| | Joined: 07 Apr 2005 | | Posts: 89 | | : | | Location: Cambridge |
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| Posted: Sun Oct 09, 2005 10:20 pm Post subject: |
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| Nick70 wrote: | | There's an X-Wing in the 1s. |
Thanks Nick. So there is. Silly me for not seeing it. There hasn't been an X-Wing in the Times since puzzle 4.
Regards, Andrew |
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| Ruud
Site Admin
 | | Joined: 17 Sep 2005 | | Posts: 708 | | : | | Location: Netherlands |
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| Posted: Sun Oct 09, 2005 10:33 pm Post subject: |
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There's a little more to be discovered in this Sudoku
These are the key log entries from my solver:
Row 6 found a naked pair with values {5,6}
Col 2 and Col 8 found an X-Wing for value 1
Box 5 found a naked pair with values {5,6}
Col 9 found a naked pair with values {1,8}
Col 5 and Col 8 found an X-Wing for value 6
Cell (9,8) causes an XY-Wing eliminating value 6
You still have to do the basic logic to make these discoveries |
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| kranser
| | Joined: 18 Aug 2005 | | Posts: 35 | | : | |
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| Posted: Mon Oct 10, 2005 8:32 am Post subject: Last Sunday's Superior version |
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Last Sunday's [20051002] Superior version seems to require some advanced techniques to solve as well:
....4...6..6...1..8...91.5..5..23.79.........69.75..3..3.16...5..7...3..2...7....
Does anyone know if this requires X-Wing also?
Kranser. |
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| AMcK
| | Joined: 07 Apr 2005 | | Posts: 89 | | : | | Location: Cambridge |
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| Posted: Mon Oct 10, 2005 9:46 am Post subject: Re: Last Sunday's Superior version |
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| kranser wrote: | Last Sunday's [20051002] Superior version seems to require some advanced techniques to solve as well:
....4...6..6...1..8...91.5..5..23.79.........69.75..3..3.16...5..7...3..2...7....
Does anyone know if this requires X-Wing also?
Kranser. |
| Code: | . . . ¦ . 4 . ¦ . . 6
. . 6 ¦ . . . ¦ 1 . .
8 . . ¦ . 9 1 ¦ . 5 .
------+-------+-------
. 5 . ¦ . 2 3 ¦ . 7 9
. . . ¦ . . . ¦ . . .
6 9 . ¦ 7 5 . ¦ . 3 .
------+-------+-------
. 3 . ¦ 1 6 . ¦ . . 5
. . 7 ¦ . . . ¦ 3 . .
2 . . ¦ . 7 . ¦ . . .
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Yes, there are x- and y-wings in 2 and 8
| Code: | x-Wing found p=2 n=2 columns 3 7 rows 3 6
Reduction in r1c3: before=1259 after=159
Reduction in r1c7: before=289 after=89
y-Wing found p=8 n=2 rows 5 9 cols 2 9
Reduction in r9c3: before=14589 after=1459
Reduction in r9c7: before=489 after=49
Reduction in r5c8: before=248 after=24
Reduction in r9c8: before=14689 after=1469
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... but these don't seem to be sufficient to solve using normal means.
It can be solved after proving the contradiction r1c2<>1 using implications.
I leave it to others to let us know what simpler techniques are applicable.
Regards, Andrew |
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| Bo
| | Joined: 02 Sep 2005 | | Posts: 27 | | : | |
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| Posted: Mon Oct 10, 2005 9:39 pm Post subject: Re: Last Sunday's Superior version |
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Hi,
there are 2 hidden pairs:
r8c1,r9c3 {1,5} and r8c8,r9c8{1,6}
and X-wings (as already mentioned) in r3c3,r3c7,r6c3,r6c7 {2}
Bo |
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