Sudoku Players' Forum (clone)

The 6 in box 4 can be placed like this:

In box 9, only options for 6 is row 7. Therfore, for box 7, row 7 can be eliminated. Then there is only one option left for 6 in column 2, which is in box 4.

rrabbit wrote:Earlier in this thread, a number could be located in box 9.
Once that has been done, the 6 in box 9 must be in column 7.

Anette wrote: The 6 in box 4 can be placed like this: In box 9, only options for 6 is row 7. Therfore, for box 7, row 7 can be eliminated. Then there is only one option left for 6 in column 2, which is in box 4.

Okay, I didn't realize that rrabbit was using my clue that a number could be filled in box 9. I thought he/she was referring to the puzzle as it stood in the original post.

I agree that once a cell is filled in box 9 (based on eliminations in column 9) that a 6 can be placed in box 4.

su_doku wrote:I agree but in this particular case, the way box 2 is populated means r3c4 is 3 or 9.

It's true that r3c4 could be a 3 or 9, but if it's 9 - the 3 could still be in either r3c1 or r3c8. therefore, I disagree that the 3 is a valid candidate to make a triple with 2 and 9 in row 3.

scrose wrote:

su_doku wrote:Nick70, how does column 9 help?

Look for a triple in column 9. This will let you make some eliminations that permit a cell to be filled in block 9.

su_doku wrote:I think the clue is in row 3 - 3 numbers go exclusively into 3 cells which allows a number in box 2 to be filled.

As this point in the puzzle, I don't think there are any triples in row 3. Here are my candidates for row 3.
{135} {29} {1259} {39} [8] [4] [7] {135} [6]

Thanks scrose.
A triplet, naked, as naked as it can be. DOH!

There seem to be two possible grids for this (based on the original query)

386175942
471329856
592684731
815792463
243856179
769431285
927543618
138967524
654218397

and

386175942
475923816
192684537
518792463
243856179
769431285
924317658
831569724
657248391

Anyone agree?

Stuart
www.brightonandhove.org/sudoku/solver5c.xls

r3c9 should be 6.

A gin too many at lunch methinks. Thanks Simes.

Stuart

R2,C6, 3 or 9
R3,C4, 3 or 9
3 in R3, C1 can be eliminated leaving only R1, C1 for the 3
finish R1 - C3/6, C7/9, C9/2
pretty easy from there on

jayal wrote:R2,C6, 3 or 9
R3,C4, 3 or 9
3 in R3, C1 can be eliminated leaving only R1, C1 for the 3
finish R1 - C3/6, C7/9, C9/2
pretty easy from there on

You cannot remove the 3 in (r3c1).
The fact that (r3c4) can be 3 or 9 does not enable you to remove the 3 in (r3c1)

stuartn:

This one also works:
386175942
475629831
192384756
518792463
243856179
769431285
927543618
831967524
654218397

S o what do we reckon is the maximum number of KNOWN cells to limit the result to one grid only?....... I'll have a gin and consider it.

Stuart

www.brightonandhove.org/sudoku/solver5c.xls

triplet in col 9-- r1c9:2 or 3, r4c9:3 or 5, r6c9:2 or 5
==> in box9-- no 3 in col 9 or row 7==> 3 in r9c7
also in box 9:6 in r7c7 or r7c8==> no 6 in r7c2==> 2 in r7c2
rest is straightforward

solution:
3 8 6 | 1 7 5 | 9 4 2
4 7 5 | 6 2 9 | 8 3 1
1 9 2 | 3 8 4 | 7 5 6
----------------------
5 1 8 | 7 9 2 | 4 6 3
2 4 3 | 8 5 6 | 1 7 9
7 6 9 | 4 3 1 | 2 8 5
----------------------
9 2 7 | 5 4 3 | 6 1 8
8 3 1 | 9 6 7 | 5 2 4
6 5 4 | 2 1 8 | 3 9 7