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PostPosted: Fri Oct 14, 2005 3:34 am    Post subject: help Reply with quote

after finishing one most hard sudoku from local news paper, I am confident of challenging today's most hard sudoku.

but Sad I am stuck for an hour and still not slove it

x9x xxx 328
3x6 2x4 975
xx2 x9x xx6

x63 951 7x2
279 x4x 561
1x5 726 x93

xxx x3x 6xx
63x 4xx 2xx
9x7 xxx x34
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Guest Too
Guest





PostPosted: Fri Oct 14, 2005 8:15 am    Post subject: Reply with quote

Consider Row 7 .. the only cells that the pair 7,9 can go are col 6 and col 9.
Therefore any other possibilities in those two cells can be eliminated.
That will leave one nr 2 as the only candidate in row 9, Box 8

Now over to you ...Good Luck!
Hope this helps get you going. Very Happy
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geoff h



Joined: 07 Aug 2005
Posts: 58
Location: Sydney

PostPosted: Fri Oct 14, 2005 8:19 am    Post subject: Reply with quote

Hi there,

You should consider Column 3 and its interaction with Box No. 7. Nr 8 MUST appear in either r7c3 or r8c3 as there is nowhere else it can go in Column 3. Therefore, you can exclude all other 8s in Box No. 7. This then leads you to a quadruple of 1,4,5,8 in Row 7 and a resulting pair of 7,9 in r7c6 and r7c9. You can then easily place the Nr 2 in r7c2 and solve the puzzle from there.

Hope this helps.

Cheers.
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Guest






PostPosted: Fri Oct 14, 2005 9:11 am    Post subject: Reply with quote

thx for yr instructions
I should have discovered the pair 7, 9 in row 7.
I am too bad Crying or Very sad
I was so stupidly spent an hour staring at the puzzle but not fine the pair in row 7. anyway thx guys
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kosjka
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PostPosted: Fri Oct 14, 2005 1:54 pm    Post subject: stuck Reply with quote

I read the opst and was stuck at the same point. Now i'm stuck again:
x9x xxx 328
3x6 2x4 975
xx2 x9x 416

863 951 742
279 x4x 561
145 726 893

x2x x3x 6xx
63x 4xx 2xx
9x7 xx2 x34
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Fri Oct 14, 2005 2:47 pm    Post subject: Reply with quote

You must be tired, Kosjka. Smile

First, you already have 8 different numbers entered in column 7, so clearly r9c7 = 1. And you have 7 different numbers in columns 8 and 9, so you can see two pairs in the lower right 3x3 box -- {5, 8} in r7c8 & r8c8, and {7, 9} in r7c9 & r8c9.

Now look at row 9. With the "1" filled in the only values missing are {5, 6, 8}. Because of the "6" at r4c2 it appears that the only candidates at r9c2 are {5, 8}. But in column 3, you still need an "8", and it can't go at r1c3 because of the "8" at r1c9. So the "8" in column 3 must appear either at r7c3 or at r8c3 -- since there can only be one "8" in the bottom left 3x3 box, you can now place a "5" at r9c2, and the rest should be easy. dcb
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Kosjka
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PostPosted: Fri Oct 14, 2005 3:18 pm    Post subject: `sleeping Reply with quote

Yes indeed i was sleeping. i did have the 1 in r9c7 but forgot the whipe out the 1 in r9c2. so now im on the move againe.
txs
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