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Blas
Joined: 19 Jul 2005
Posts: 8
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 Posted: Tue Jul 19, 2005 6:52 am Post subject: Friday 18th July |
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| Can someone please help me? Even with the hint from this excellent site (row 1, col 1 = 6), I can't get anywhere and will go mad shortly. I've never been defeated before and am totally mortified. |
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Guest
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| Posted: Tue Jul 19, 2005 6:57 am Post subject: VERY HARD puzzle was Sunday 17-Jul not Friday 15-Jul!! |
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| Sorry - got the wrong day. |
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mike
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| Posted: Tue Jul 19, 2005 8:46 pm Post subject: Sunday 17 July |
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| I could only do this puzzle by trial and error (either 4 or 9 in the bottom right corner of the middle square - needless to say I tried the wrong option first). Surely there must be a way to avoid using trial and error though? |
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samgj
Site Admin
Joined: 17 Jul 2005
Posts: 106
Location: Cambridge
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| Posted: Wed Jul 20, 2005 4:03 pm Post subject: Re: Friday 18th July |
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| Blas wrote: | | Can someone please help me? Even with the hint from this excellent site (row 1, col 1 = 6), I can't get anywhere and will go mad shortly. I've never been defeated before and am totally mortified. |
Is this right from the start?
| Code: |
. * 2 . . . 1 9 8
* 4 . . 9 . . 3 6
3 . . . . . . . 4
. . . . 7 3 . 4 .
. . . 1 . 6 . . .
. 8 . 2 5 . . . .
8 . . . . . . . 3
9 3 . . 2 . . 5 .
1 6 4 . . . 9 . .
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The options for the top left corner are 567. But the two *-ed squares both have only 5 and 7 as options, so the top left cell must be the 6.
Hope this helps
Sam |
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gurdy
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| Posted: Wed Jul 20, 2005 8:33 pm Post subject: also having trouble |
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| with what I have, you could still have a 6 in any of r3c3, r1c5, or r4c1. meaning that you can't prove even the 6 it gives you as a clue. I am having the same problem, did anyone prove the puzzle w/out trial and error? |
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linoliverpool
Joined: 21 Jul 2005
Posts: 4
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| Posted: Thu Jul 21, 2005 9:33 am Post subject: This may help |
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If you have 2 boxes like you do in the top left quadrant like r1c2 and r2c1
That only have 2 options (5 & 7) and these are the same for both boxes you know that if one of them is a 5 the other must be a 7.
THe box r1c1 could only have had 5, 6 or 7 and you now know the 5 or 7 are taken by the other two boxes, these can no longer be an option in this box which only leaves the number 6. |
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