The New Sudoku Players' Forum

[Roger Scriven]

Have a "very hard puzzle"from the computer program and am stuck.
start as follows:
*** **2 9**
2*4 *5* *6*
**7 4** **3

*4* *6* *9*
*** 1*5 ***
*3* *4* *5*

1** **6 8**
*8* *1* 7*6
**6 8** ***
and from there to
*1* 6*2 9**
294 *5* 16*
**7 491 **3

*4* *6* *9*
*** 1*5 4**
*3* *4* 65*

1** **6 8**
*8* *1* 7*6
**6 8** *1*
Any suggestions

[Animator]

Column 9 looks intresting...

Candidates:

r1c9: 4, 5, 7, 8
r2c9: 7, 8
r4c9: 1, 2, 7, 8
r5c9: 2, 7, 8
r6c9: 1, 2, 7, 8
r7c9: 2, 4, 5, 9
r9c9: 2, 5, 9

Now look where the numbers 4, 5 and 9 can occur... There are only three cells that hold one of these numbers, which means that no other number is possible in those cells... (3 numbers, 3 cells)

And when you figured that out you should look at box 6...

[Roger Scriven]

Thanks

[Guest]

Call me stupid, but I can still not progress with this one.

[Guest]

Have a "very hard puzzle"from the computer program and am stuck.
start as follows:
*** **2 9**
2*4 *5* *6*
**7 4** **3

*4* *6* *9*
*** 1*5 ***
*3* *4* *5*

1** **6 8**
*8* *1* 7*6
**6 8** ***

Recheck what you typed above... somewhere there is an error with either one of the 4s or one of the digits in the 1,1 position of a 3x3 box.

In the one above, there is no way for the 4 to occupy the '1,1' position in any of the 9 3x3 boxes. Since each digit occupies the '1,1' in every box in a valid puzzle, logic dictates if it doesn't then the puzzle is invalid.

[simes]

...Since each digit occupies the '1,1' in every box in a valid puzzle...

This isn't a requirement of Sudoku. (I seem to remember someone suggesting a variation along these lines, but it's not standard Sudoku.)

[Animator]

At what point are you stuck? do you see the pairs in column 9? If you do, then you really should take a close look at the number 2

[Guest]

"At what point are you stuck? do you see the pairs in column 9? If you do, then you really should take a close look at the number 2" ....

My possibilities for column 9 are (r,c):
(1,9) = 4,5
(2,9) = 7,8
(3,9) = 3
(4,9) = 1,2,7,8
(5,9) = 2,7,8
(6,9) = 1,2,7,8
(7,9) = 4,5,9
(8,9) = 6
(9,9) = 5,9

[Pi]

As a huge sudoku fan i made a spreadsheet as a sudoku aid, the logoic is (I think) almost flawless although i know it doesn't include x wing detection and when i entered your sudoku results i found that it was not telling me any obvious answers, there are therefore no squares in your sudoku that can be filled with absolute certainty, unless my spreadsheet is wrong which i don't think it is as it enables me to do feindish sudoku's in under 2m 30s, i used trial and error by substituting possible numbers into A3 and if i substituted in 5 or 6 i soon found an error but if i substituted 8 i got to another stalemate, from this stalemate i again used trial and error (unusual for me) and substituted a 3 into A1, this worked and i completed the Sudoku, i was baffled, never before have i used trial and error twice in one sudoku although i have used in very occasionaly in a very hard feindish sudoku in the event of a stalemate.

[Animator]

My possibilities for column 9 are (r,c):
(1,9) = 4,5
(2,9) = 7,8
...

Which are correct...

Where can the 2 go in column 9? as in, in what boxes? Now look at box 6... and remember what you just said.

[Animator]

As a huge sudoku fan i made a spreadsheet as a sudoku aid, the logoic is (I think) flawless...

Your logic isn't complet.

It is missing atleast these three things:
a) X-wing detection, which could be required for Very Hard puzzles (not for Findish puzzles)
b) a combination of numbers in several cells (a pair, a triplet, ...)
c) a number that can only occur in one box for a particular column can be removed from all other cells in that box.

[Roger Scriven]

I have noticed some additional coments on this problem. I followed the advice I was given and did solve the puzzle but I had to think carefully about the candidates in box 6 in relation to the candidates in column 9 (Mainly about what could not be in column 9 of box six because of the requirements of column 9 of box one)
further explanation can be give if required.

[Guest]

Both R2-C5 and R2-C9 must be either 7 or 8 ... and they must also be the same. If both 7, R5-C8 must be 7, and both R2-C6 and R5-C5 must be 8. However, that means placing 3's into the middle tier is impossible. Ergo, only 8 works in positions mentioned first above. And the rest of the puzzle falls out from there. Cheers, Paul