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2005-10-14 (Fri - Very Hard)

 
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Mooie
Guest





PostPosted: Fri Oct 14, 2005 7:00 pm    Post subject: 2005-10-14 (Fri - Very Hard) Reply with quote

Okay, got stuck at this stage;

Code:
_ 9 _  _ _ _  3 2 8
3 _ 6  2 _ 4  9 7 5
_ _ 2  _ 9 _  _ _ 6

_ 6 3  9 5 1  7 _ 2
2 7 9  _ 4 _  5 6 1
1 _ 5  7 2 6  _ 9 3

_ _ _  _ 3 _  6 _ _
6 3 _  4 _ _  2 _ _
9 _ 7  _ _ _  _ 3 4


If I expand a little, I get this;

Code:

 _   9  _     _   _   _      3   2   8
 3  1/8 6     2  1/8  4      9   7   5
 _   _  2     _   9   _     1/4 1/4  6

4/8  6  3     9   5   1      7  4/8  2
 2   7  9    3/8  4  3/8     5   6   1
 1  4/8 5     7   2   6     4/8  9   3

 _   _  _     _   3   _      6   _  7/9
 6   3  _     4   _   _      2   _  7/9
 9   _  7     _   _   _      _   3   4


Okay, I can expand further, but that will make this cluttered view even more cluttered.. So.. Where do I go from here? I can of course guess, but that should not be necessary. I think I personally only need one little hint and hopefully an explanation to that one and I'll be able to solve the rest.


//Mooie
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Guest Too
Guest





PostPosted: Fri Oct 14, 2005 7:57 pm    Post subject: Reply with quote

Suggest you read the topic 'help' posted by 'Guest' ! That'll get you on the way. Very Happy
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chobans



Joined: 21 Aug 2005
Posts: 39

PostPosted: Fri Oct 14, 2005 8:01 pm    Post subject: Reply with quote

In column 3, 8 can only go in box 7(r7c1-r9c3) so you can eliminate 8 in other cells in box 7. So r7c1 - {4,5} and r7c2 - {1,2,4,5}

r7c1 - {4,5}, r7c3 - {1,4,8}, r7c4 - {1,5,8}, r7c8 - {1,5,8}

Naked quadruple. I don't know why there are so many of these lately. Basically 4 cells have possibilities made up with combination only 4 numbers, {1,4,5,8}. So you know these 4 numbers will go into these 4 cells. We don't know exactly what number goes into which cell but we know what numbers we can eliminate on OTHER cells in row 7.

So after eliminating 1, 4 and 5 from r7c2, we get 2 in that cell.

You can also say that this is hidden pair of 7 and 9 in row 7. Usually naked quadruple can be solved as hidden pair. But to me, it was always easier to spot the naked <something> than hidden <something>.
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Mooie



Joined: 14 Oct 2005
Posts: 1
Location: Blekinge, Sweden

PostPosted: Fri Oct 14, 2005 8:11 pm    Post subject: Yay! Reply with quote

Thanks both of you, but chobans really helped me explaining the naked quadruple (and triplet etc) mentioned on the page linked from http://www.dailysudoku.com/sudoku/forums/viewtopic.php?t=80.

Thanks lots!

Oh.. I have eliminated the eights from block seven. I guess I should've expanded the board fully. Anyway, not important since I now got another nifty "tool".. and I do understand it!


//Mooie
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someone_somewhere



Joined: 07 Aug 2005
Posts: 275
Location: Munich

PostPosted: Sat Oct 15, 2005 7:56 am    Post subject: Reply with quote

Hi,

Nacked quadruple - they are very nice.
You could solve it also without them:

4 not in r3c1, it is in r1c1 or r1c3 (Row on 3x3 Block interaction)
4 not in r3c2, it is in r1c1 or r1c3 (Row on 3x3 Block interaction)
8 not in r7c1, it is in r7c3 or r8c3 (Column on 3x3 Block interaction)
8 not in r7c2, it is in r7c3 or r8c3 (Column on 3x3 Block interaction)
8 not in r9c2, it is in r7c3 or r8c3 (Column on 3x3 Block interaction)
1 not in r3c2, it is in r3c7 or r3c8 of the 3x3 (3x3 Block on Row/Column interaction)
1 not in r3c4, it is in r3c7 or r3c8 of the 3x3 (3x3 Block on Row/Column interaction)
2 not in r7c6, Hidden Pair 7 9 in r7c6 and r7c9 (in Row)
5 not in r7c6, Hidden Pair 7 9 in r7c6 and r7c9 (in Row)
8 not in r7c6, Hidden Pair 7 9 in r7c6 and r7c9 (in Row)
2 in r7c2 - Unique Horizontal

and so on ...

see u,
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