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sweedy
Joined: 21 Oct 2005 Posts: 3 Location: Zurich
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Posted: Fri Oct 21, 2005 11:55 am Post subject: march 17- i'm stuck and the hint doesn't help |
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i'm stuck on the puzzle of march 17.. (yes i'm working through the archives..)
this is how far i got (in brackets are the possible canditates for each field). the hint this website gives to me is a 1 in r5c3 but that doesn't help because i can't retrace it.
(2,4,8.) (2,4,6) (4,8.) 9 1 (5,6,7)(3,5,7,8.) (3,5,7) (3,4,8.)
3 (4,6,9) (4,8,9) (6,7) 2 (5,6,7) (5,7,8.) 1 (4,8.)
1 7 5 (3,4) (3,4) 8 6 2 9
(7,8.) 5 2 (1,3,7) (3,7,8.)(1,3,7) 4 9 6
6 (1,4) (1,4,8.) 5 (4,8.) 9 2 3 7
(4,7,9) (3,4,9) (3,4,7,9) (2,3,4,7) 6 (2,3,7) 1 8 5
5 (1,2,3,9) 6 8 (3,7)(1,2,3,7) 9 4 (1,3)
(4,7,9) 8 (1,3,4,7,9)(1,3,6,7) 5 (1,3,6,7) (3,7) (6,7) 2
(2,7) (1,3) (1,3,7) (1,2,3,6,7) 9 4 (3,5,7,8.)(5,6,7) (1,3,8.) |
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Steve R Guest
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Posted: Fri Oct 21, 2005 3:13 pm Post subject: March 17 |
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4 and 9 are already in rows 7 and 9
In row 8, they must occupy r8 c1 or r8 c3
This reduces your possibilities for r8 c3 from 1, 3, 4, 7 or 9 to 4 or 9
Inspection of column 3 then shows 4, 8 and 9 fill r1 c3, r2, c3 and r8, c3
None of these entries can go in r5 c3
... which leaves just 1. |
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David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado
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Posted: Fri Oct 21, 2005 3:15 pm Post subject: Isolating that "1" |
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First you need to clean up your grid. There's a hidden pair {4, 9} in row 8. There's also a hidden pair {3, 7} in column 3. After you make the eliminations the matrix looks like this
Code: |
248 246 48 9 1 567 3578 357 348
3 469 489 67 2 567 578 1 48
1 7 5 34 34 8 6 2 9
78 5 2 137 378 137 4 9 6
6 14 148 5 48 9 2 3 7
479 349 37 2347 6 237 1 8 5
5 1239 6 8 37 1237 9 4 13
49 8 49 1367 5 1367 37 67 2
27 13 37 12367 9 4 3578 567 138 |
Now the "1" in r5c3 is clearly a necessity, and you can also place a "4" in r5c2. dcb |
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Guest
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Posted: Fri Oct 21, 2005 3:24 pm Post subject: |
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wow, thank you very much!! i forgot all about checking for hidden pairs! |
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