The New Sudoku Players' Forum

[MCC]

For the last few nights I've been trying to solve this program generated v.hard

Code: Select all

  .    8    .  |  1    .   5  |  .    4    .
  4    7    .  |      2       |  .    .   .
  .    .    .  |  .    .    . |  .    .    6
  -------------+---------------+------------
  .    1    .  |  .   9    2  |  .    .    .
  .    .    3  |  8   .    6  |  1    .    .
  .    .    .  |  4   3    .  |  .    8    .
  -------------+---------------+------------
  9    .   .   |  .    .    . |  .    .    .
  .    .   .   |  .    6    . |  .    2    4
  .    5   .   |  2    .   8  |  .    9    .


I've got as far as this.

Code: Select all

  .    8    .  |  1    7    5  |  .   4    .
  4    7    .  |  6    2    .  |  .    .    .
  .    .    .  |   .   8   4   |  7   .    6
  -------------+---------------+------------
  .    1    .  |  7    9    2  |  4    .    .
  2    4    3  |  8    5    6  |  1    7   9
  .    .    .  |  4    3    1  |  .    8    .
  -------------+---------------+------------
  9    .    .  |  .    .   .   |  .    .    .
  .    3    .  |  .    6   .   |  .    2    4
  .    5    .  |  2    .   8   |  .    9    .


I cannot see any obvious x-wings that are useful.

It can be solved by T&E with the (3,9) pairs in box 2, but pappocom generated puzzles are not supposed to need T&E, so what am I overlooking?

[Nick70]

You are not missing a x-wing.

Look closely at column 9.

[su_doku]

Nick70, how does column 9 help?

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.

MCC - nice puzzle!

[scrose]

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.

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]

[su_doku]

scrose look at row 3, column 1, column 8 and box 2 together

you will note that 2 can only go into r3c2 and r3c3
also 9 can go only into r3c2, r3c3 and r3c4
but for box 2, r3c4 can additionally have a 3 only

hence r3c2, r3c3 and r3c4 give you a triple such that r3c4 is only 3.

[scrose]

hence r3c2, r3c3 and r3c4 give you a triple such that r3c4 is only 3.

I'm still unable to understand how you arrive at that conclusion. The candidate 3's in r3c1 and r3c8 prevent any eliminations from being made. As row 3 currently stands, it is still possible to place a 3 into r3c1 or r3c8.

I could place a 3 in r3c1 and get the following result.
[3] [2] {15} [9] [8] [4] [7] {15} [6]

Or I could place a 3 in r3c8 and get the following result.
{15} [2] {15} [9] [8] [4] [7] [3] [6]

[su_doku]

Put in a different way, for row 3 given that 2 and 9 as a pair are restricted to r3c2, r3c3 and r3c4 only, the cell r3c4 then provides the possibility of a 3 in addition to 9 that makes those 3 cells a triple. Hence the 1 and 5 of r3c2 become extraneous and can be eliminated.

[scrose]

Put in a different way, for row 3 given that 2 and 9 as a pair are restricted to r3c2, r3c3 and r3c4 only, the cell r3c4 then provides the possibility of a 3 in addition to 9 that makes those 3 cells a triple. Hence the 1 and 5 of r3c2 become extraneous and can be eliminated.

To form a triple you must have three candidates that are restricted to three cells in a group (be it a row, column, or box). In row 3, the candidates 2 and 9 are restricted to 3 cells (I agree with you there), and the candidate 3 is restricted to three cells, but the candidates 2, 3, and 9 together are not contained in the same three cells -- they are spread over five cells. Therefore you cannot form a triple, and you cannot eliminate candidates 1 or 5 from r3c2.

[rrabbit]

I think you can place the 6 in box 4.


Thomas

[su_doku]

To form a triple you must have three candidates that are restricted to three cells in a group (be it a row, column, or box).

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

2 of row 3 can only go in 2 cells (r3c2 and r3c3)
9 of row 3 can only go in 3 cells (r3c2, r3c3 and r3c4)

But because r3c4 can only be 3 or 9, the additional 3 provides the third number to add to the 2 and 9. This gives 3 numbers for 3 cells and hence we can eliminate the extraneous numbers.

Anyone else see it the way I do?

[rrabbit]

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

2 of row 3 can only go in 2 cells (r3c2 and r3c3)
9 of row 3 can only go in 3 cells (r3c2, r3c3 and r3c4)

But because r3c4 can only be 3 or 9, the additional 3 provides the third number to add to the 2 and 9. This gives 3 numbers for 3 cells and hence we can eliminate the extraneous numbers.

Anyone else see it the way I do?

If you put a 9 into r3c4, row 3 can still be completed.
Thus, your analysis is flawed.


Thomas

[scrose]

...the way box 2 is populated means r3c4 is 3 or 9.

I agree.

2 of row 3 can only go in 2 cells (r3c2 and r3c3)

I agree.

9 of row 3 can only go in 3 cells (r3c2, r3c3 and r3c4)

I agree.

But because r3c4 can only be 3 or 9, the additional 3 provides the third number to add to the 2 and 9.

This is where I disagree. If you add the 3 to the 2 and 9, you must take into consideration all of the cells in that row which contain the candidate 3, not just cell r3c4.

Anyone else see it the way I do?

I agree with this, too!

[scrose]

If you put a 9 into r3c4, row 3 can still be completed.
Thus, your analysis is flawed.

I agree! In the example I provided earlier, the 9 ends up in r3c4 whenever I place the 3 in r3c1 or r3c8.

Edit: Fixed spelling.

[scrose]

I think you can place the 6 in box 4.

I'm unable to see how you reached that conclusion. Could you please provide an explanation.

[rrabbit]

I think you can place the 6 in box 4.

I'm unable to see how you reached that conclusion. Could you please provide an explanation.

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.


Thomas