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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Mon Apr 05, 2010 5:26 pm Post subject: Puzzle 10/04/05 (A) |
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| Code: | +-----------------------+
| . . 7 | . . . | . 3 . |
| . . . | . 8 5 | . . . |
| 9 . . | . . . | . . . |
|-------+-------+-------|
| . . . | . 5 . | . 8 9 |
| . 3 . | 8 . 9 | 7 5 . |
| . 9 . | . 7 6 | 3 . 4 |
|-------+-------+-------|
| . . . | . 6 7 | . . . |
| 8 . . | 5 2 . | . . 7 |
| . . . | 9 . 8 | . 4 6 |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Apr 05, 2010 6:19 pm Post subject: |
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| Quote: | | XY 13-4; UR3 67; UR3 24. |
Keith |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue Apr 06, 2010 12:50 am Post subject: |
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Two steps.
| Quote: | Type 3 UR67 in r23c48 provides a pseudocell 134 which combines with r1c6 and r3c56 to form a quad 1234 thereby forcing r1c4=6,
UR 24 in r13c67. Looking at the implication outside the UR, we find
Digit 2: an x-wing overlay so no 2s to consider,
Digit 4: the only 4s in c67 are r48c6, one of which must be a 4 to prevent the deadly pattern.
(4)r4c6 - r5c5 = r3c5 - r3c7 = r1c7,
(4)r8c6 - r7c4 = r7c1.
Thus, r1c1<>4 to complete the puzzle.
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Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Apr 06, 2010 4:36 am Post subject: |
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Type 4 UR (67)
Coloring (1)
M-Wing (14)
UR (24) with pseudo cell forming 13 pair. |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Tue Apr 06, 2010 7:02 am Post subject: |
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V Nice Ted,
However, could we not also say that a 6 placed in in either of r23c4 will produce the deadly pattern hence r1c4 must be 6 ?
The nice logic following on from the second UR produced the two chains that proved 4 could not exist in r1c1.
I was looking at a colouring solution after the first UR and this seemed to link your 2 chains:
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue Apr 06, 2010 11:46 am Post subject: |
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| Mogulmeister wrote: |
However, could we not also say that a 6 placed in in either of r23c4 will produce the deadly pattern hence r1c4 must be 6 ? |
Absolutely! I just thought that posting the quad was more fun.
| Mogulmeister wrote: |
The nice logic following on from the second UR produced the two chains that proved 4 could not exist in r1c1.
I was looking at a colouring solution after the first UR and this seemed to link your 2 chains:
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l never notice/realized the connection, but given the strong inference of the two forcing cells, r48c6, the two chains are connected as you note.
Ted |
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