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| Iavor
| Joined: 14 Jun 2006 | Posts: 1 | : | Location: Berlin | Items |
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Posted: Wed Jun 14, 2006 7:57 am Post subject: Sudoku from Stern |
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The german magazin "Stern" publish sudokus. In the issue from 8.6.2006 (Nr 24, page 189) there was a sudoku "KAMIKAZE", where Simple Sudoku couldn't help:
Code: | *-----------*
|6..|...|..4|
|.3.|...|.9.|
|..1|479|3..|
|---+---+---|
|..3|6.5|8..|
|946|...|513|
|..2|1.3|9..|
|---+---+---|
|..7|824|6..|
|.2.|...|.8.|
|8..|...|..1|
*-----------* |
After several quite simple steps I come to (confirmed with the solving mode of Simple Sudoku):
Code: | *-----------*
|6..|...|..4|
|.3.|...|.9.|
|251|479|368|
|---+---+---|
|..3|695|842|
|946|782|513|
|582|143|976|
|---+---+---|
|..7|824|6..|
|.2.|...|.8.|
|86.|..7|..1|
*-----------* |
or
Code: | *-A----B----C------D----E----F------G----H----I----*
| 6 79 89 | 235 13 18 | 127 25 4 |
| 47 3 48 | 25 156 168 | 127 9 57 |
| 2 5 1 | 4 7 9 | 3 6 8 |
|----------------+----------------+----------------|
| 17 17 3 | 6 9 5 | 8 4 2 |
| 9 4 6 | 7 8 2 | 5 1 3 |
| 5 8 2 | 1 4 3 | 9 7 6 |
|----------------+----------------+----------------|
| 13 19 7 | 8 2 4 | 6 35 59 |
| 34 2 459 | 359 16 16 | 47 8 79 |
| 8 6 459 | 359 35 7 | 24 23 1 |
*--------------------------------------------------* |
Further no hint is available. Questions:
1. Is the following way legitimate or just guess (invalid)?
if A4 is 1, than
A7=3, A8=4, H7=5, I7=9, I8=7, G8=4
but A8 was already 4, therefore A4 cannot be a 1.
The final solution:
Code: |
*-----------*
|679|318|254|
|438|256|197|
|251|479|368|
|---+---+---|
|713|695|842|
|946|782|513|
|582|143|976|
|---+---+---|
|197|824|635|
|324|561|789|
|865|937|421|
*-----------* |
2. Is there another logical solution here?
Cheers
iavor |
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| rkral
| Joined: 21 Oct 2005 | Posts: 233 | : | | Items |
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Posted: Wed Jun 14, 2006 12:50 pm Post subject: Re: Sudoku from Stern |
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Iavor wrote: | 1. Is the following way legitimate or just guess (invalid)?
if A4 is 1, than
A7=3, A8=4, H7=5, I7=9, I8=7, G8=4
but A8 was already 4, therefore A4 cannot be a 1. |
Perfectly logical deduction. You don't need to include A4, however.
A7=3, H7=5, I7=9, I8=7, G8=4, A8=3, A7<>3, implies A7<>3
Iavor wrote: | 2. Is there another logical solution here? |
It can be shown that D8 cannot be a 3 by using cells A8 and the almost-locked-set in C9, D9, and E9. |
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| Carcul
| Joined: 29 Dec 2005 | Posts: 50 | : | Location: Coimbra, Portugal | Items |
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Posted: Thu Jun 15, 2006 4:24 pm Post subject: Another Solution |
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C9=9 or G1=2. But G1=2 => G9=4. So C9 cannot be 4 and the puzzle is solved.
Carcul |
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| maarten
| Joined: 16 Jun 2006 | Posts: 7 | : | Location: The Netherlands | Items |
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Posted: Mon Jul 10, 2006 1:25 pm Post subject: |
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Hehe, my solver finds a XY Chain of 15 steps, and of course a different number and cell. Cool
Found a Naked Single - added a 8 in cell (5,5)
Found a Naked Single - added a 4 in cell (6,5)
Found a Naked Single - added a 9 in cell (4,5)
Found a Hidden Single in row 4 - added a 4 in cell (4,8)
Found a Hidden Single in row 4 - added a 2 in cell (4,9)
Found a Hidden Single in column 2 - added a 6 in cell (9,2)
Found a Naked Single - added a 7 in cell (9,6)
Found a Naked Single - added a 2 in cell (5,6)
Found a Naked Single - added a 7 in cell (5,4)
Found a Hidden Single in row 6 - added a 8 in cell (6,2)
Found a Naked Single - added a 5 in cell (3,2)
Found a Naked Single - added a 2 in cell (3,1)
Found a Naked Single - added a 6 in cell (3,8)
Found a Naked Single - added a 8 in cell (3,9)
Found a Naked Single - added a 7 in cell (6,8)
Found a Naked Single - added a 5 in cell (6,1)
Found a Naked Single - added a 6 in cell (6,9)
Found a Locked Candidate 1 in row 7 - removing candidate from cells (8,1)
Found a Locked Candidate 5 in row 7 - removing candidate from cells (8,9),(9,8)
Found a Hidden Double 1/6 in row 8 - removing other candidate from cells (8,5),(8,6)
Found an XY-Chain-15 from cell (1,2) to cell (8,7), resulting in removal of candidate 7 in cells (1,7) |
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