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debbie
Joined: 27 Mar 2006 Posts: 7

Posted: Mon Mar 27, 2006 8:53 pm Post subject: need help with Sunday's very difficult 


I got stuck at:
2x6 xx3 x95
3x9 4xx xx6
x85 x69 xx3
xxx x7x 63x
x37 xx6 58x
x6x x3x 9x7
8x1 34x 76x
7x3 6x8 1xx
624 xxx 358
The hint was 9 in row 4/col 1 and I don't get it.. couldn't the 9 also be row 4/col4 and row 5 col 5? 

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jabejochke
Joined: 16 Mar 2006 Posts: 21 Location: Reading

Posted: Mon Mar 27, 2006 9:31 pm Post subject: 


I also found the puzzle difficult. Check the impacts of the 5's, r1c9 and r9c8, on column 4. That was the breaking point for me.
Regards,
Jack 

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George Wooods Guest

Posted: Tue Mar 28, 2006 12:01 am Post subject: 26 March difficulties 


I couldn't understand the chase the 5s argument. BUT adopting the convention That the Squares are ABCDEFGHI and the cells int hhe square are abcdefghi. So Aa is 2 De is 3and Ii is 8.
Ae is 17 as is Ce so Bf cannot be 7. Hence analysing col 6 Hi is 7. so Hg and Hh are 19 so Hc is 25 So Ec is 14 But so Db 14 so Fc must be 2 and 4 and 1 follow in F and hence can work on I ........ 

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

Posted: Tue Mar 28, 2006 12:06 am Post subject: The "hint" isn't always the most obvious ... 


Hi, Debbie! Welcome to the forum.
The "hint" from the draw facility isn't always the most obvious next move in a puzzle. That's the case here ... you need to develop quite a few "candidate profiles" to understand this move.
Start by looking at row 2  the only possibilities at r2c2 are {1, 7}. This is fairly obvious. Now concentrate on r2c8. The only possibilities here are also {1, 7}, because of the triplet {2, 4, 8} that must lie in column 7, rows 1  3.
Knowing that the pair {1, 7} lies in r2c2 and r2c8 you can see that the only possibilities at r2c6 are {2, 5}. This in turn matches up with the cell r7c6, which also has possibilities {2, 5}. So you can eliminate those two values  "2" and "5"  from every other cell in column 6.
This reveals what I regard as the most obvious next move  you can place a "7" at r9c6. The reason is fairly simple  it's the only place a "7" can fit in column 6. There can't be a "7" at r4c6, r6c6, or r7c6, obviously. And there can't be a "7" at r2c6 because of the {1, 7} pair we identified in r2c2 & r2c8. So the only place left for a "7" in column 6 is at r9c6.
But you asked me to explain the hint ... here's how that works. Continuing the analysis of column 6 you can see that the pair {1, 4} must lie in r4c6 & r6c6. And by eliminating those two values from r5c4 & r5c5 you can identify the "hidden pair" {2, 9} lying in those two cells.
Now we can understand the hint  there must be a "9" at r4c1 because that's the only place a "9" can fit in column 1. You can't put a "9" at r3c1 because of the "9" at r2c3. You can't put a "9" at r6c1 because of the "9" at r6c7. And you can't put a "9" at r5c1 because of the "hidden pair" {2, 9} lying in r4c5 & r5c5. The only place left in column 1 is r4c1.
I hope that helps. dcb 

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keith
Joined: 19 Sep 2005 Posts: 3321 Location: near Detroit, Michigan, USA

Posted: Tue Mar 28, 2006 12:08 am Post subject: Find the pairs and triples 


From the position you give, there are a lot of eliminations that involve finding naked and hidden sets.
R=Row
C=Column
B=Block
There is a pair in C2B7.
There is a triple in C7B3.
There is a triple in R9B8.
etc.
Best wishes,
Keith 

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Steve R
Joined: 24 Oct 2005 Posts: 289 Location: Birmingham, England

Posted: Tue Mar 28, 2006 2:48 am Post subject: Sunday's very difficult 


Jack had it right, I think.
The pair (58) in r46c4 leaves only one place for 9 in row 4. A neat observation ... and the cigar.
Steve 

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Guest

Posted: Tue Mar 28, 2006 11:45 am Post subject: 


The comment about the hint not always being the most obvious can be seen in this puzzle as simply a choice between 4 possible "next moves"once some preparatory work has been done. If we fill in all the "doublets" that count, then Ae and Be are 17 so Bf is 25, and with Hc being 25. Ec and Ei must be 14. Now in E  d and e must be 29 and a and g must be 58. (mind you it needs a bit of self discipline not to put the 7 in Hi after filling in the 14's)
So we have not filled in another cell with a single number yet, BUT Da9 or Dg5 or Fc 2 or Hi 7 are all valid and fairly obvious "next moves". So the hint that the computer chooses may well be the one of four nearest to "top left"
Once again can I commend my nomenclature
ABC Where the order is either used for "blocks" of 9 (Capitals)
DEF and lower case for cells in a block
GHI Further I record progress as I do the puzzle via a convention. for which I give an example
For this original Sudoku: a6,i3,g3,a3,b3,f69.g6h6.e6,a5,f7 which brings us to the crunch point of the original question (I believe) OK I have used lc for blocks when I have previously said Upper case. But a6 menas put the 6 (in the obvious cell) in A and f69 means put the 6 and 9 (in that order) into F. The beauty of this is that sometimes for instance I find I have entered 4and 6 into the grid instead of 6 and 4. Result  two 4's in the same row. But I can use my recorded data to erase latest entries to the point where the mistake was made! 

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Guest

Posted: Tue Mar 28, 2006 11:52 am Post subject: 


Anonymous wrote:  The comment about the hint not always being the most obvious can be seen in this puzzle as simply a choice between 4 possible "next moves"once some preparatory work has been done. If we fill in all the "doublets" that count, then Ae and Be are 17 so Bf is 25, and with Hc being 25. Ec and Ei must be 14. Now in E  d and e must be 29 and a and g must be 58. (mind you it needs a bit of self discipline not to put the 7 in Hi after filling in the 14's)
So we have not filled in another cell with a single number yet, BUT Da9 or Dg5 or Fc 2 or Hi 7 are all valid and fairly obvious "next moves". So the hint that the computer chooses may well be the one of four nearest to "top left"
Once again can I commend my nomenclature George woods
ABC Where the order is either used for "blocks" of 9 (Capitals)
DEF and lower case for cells in a block
GHI Further I record progress as I do the puzzle via a convention. for which I give an example
For this original Sudoku: a6,i3,g3,a3,b3,f69.g6h6.e6,a5,f7 which brings us to the crunch point of the original question (I believe) OK I have used lc for blocks when I have previously said Upper case. But a6 menas put the 6 (in the obvious cell) in A and f69 means put the 6 and 9 (in that order) into F. The beauty of this is that sometimes for instance I find I have entered 4and 6 into the grid instead of 6 and 4. Result  two 4's in the same row. But I can use my recorded data to erase latest entries to the point where the mistake was made! 


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debbie
Joined: 27 Mar 2006 Posts: 7

Posted: Wed Mar 29, 2006 6:55 pm Post subject: 


Just wanted to thank everyone! 

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Charles Guest

Posted: Thu Mar 30, 2006 11:06 pm Post subject: Mar 24 very hard 


Debbie, this is my first posting on these puzzles. From where you were, the only possibilites for 5,9 are R4C1 and R4C4. This forces an 8 into R4C3 and the puzzle can be finished without further help from an elimination grid. Good luck, Charles 

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