The New Sudoku Players' Forum

[crolan]

Every now and then I get stuck in something similiar to this one. Please help me if you can. There's some rule I am missing.
Thanks,
Claudia

3 6 49| 89 7 1| 2 48 5
8 7 49| 5 2 6|1 3 49
1 2 5| 3 4 89|78 6 79

4 9 7|6 3 2|5 1 8
5 1 6|947 8 947|47 2 3
2 8 3| 1 5 47| 6 9 47

7 4 8| 2 1 3| 9 5 6
9 3 2| 48 6 5|48 7 1
6 5 1| 748 9 748|3 48 2

[Kent]

Try using colouring technique!! I'm sure u'll be able to solve it

[crolan]

I am not familiar with that technique. Is there a place I can read about it?

[Kent]

http://angusj.com/sudoku/hints.php#colors
http://www.setbb.com/sudoku/viewtopic.php?t=311
THese are some useful threads

[tso]

Once a puzzle reaches a postion with this many bivalue cells, there will be a pletora of choices. For example:

A simple 5 cell forcing chain (there are several others):

Code: Select all

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

Code: Select all

+-------------+-------------+-------------+
| 3   6   49  |[89] 7   1   | 2   48  5   |
| 8   7   49  | 5   2   6   | 1   3   49  |
| 1   2   5   | 3   4  [89] | 7x8 6   79  |
+-------------+-------------+-------------+
| 4   9   7   | 6   3   2   | 5   1   8   |
| 5   1   6   | 479 8   479 | 47  2   3   |
| 2   8   3   | 1   5   47  | 6   9   47  |
+-------------+-------------+-------------+
| 7   4   8   | 2   1   3   | 9   5   6   |
| 9   3   2   |[48] 6   5   |[48] 7   1   |
| 6   5   1   | 478 9   478 | 3   48  2   |
+-------------+-------------+-------------+

Any value in any of the four cells in [brackets] will exclude the '8' from {r3c7}.

r3c6=8 -> r3c7=7
r3c6=9 -> r1c4=8 -> r8c4=4 -> r8c7=8 -> r3c7=7
Therefore, r3c7=7.

Code: Select all

+-------------+-------------+-------------+
| 3   6   49  |x89  7   1   | 2  [48] 5   |
| 8   7   49  | 5   2   6   | 1   3   49  |
| 1   2   5   | 3   4   89  | 78  6   79  |
+-------------+-------------+-------------+
| 4   9   7   | 6   3   2   | 5   1   8   |
| 5   1   6   | 479 8   479 | 47  2   3   |
| 2   8   3   | 1   5   47  | 6   9   47  |
+-------------+-------------+-------------+
| 7   4   8   | 2   1   3   | 9   5   6   |
| 9   3   2   |[48] 6   5   |[48] 7   1   |
| 6   5   1   | 478 9   478 | 3  [48] 2   |
+-------------+-------------+-------------+

Another lesser-used tactic is the "remote pair".

R8c4 and r1c8 form a remote pair -- eliminating '8' from r1c4.

An advantage of either of these tactics over coloring is that it doesn't require filtering and can be easier to spot. Coloring however can be much easier to use, allowing one to almost mindlessly find an exculsion.

[emm]

Looking for strong links can be a good way of finding these patterns.

Where you have two cells that share a group ( row, column or box) and also each have a conjugate (only one other of that number in another group) - then either of those conjugates must be true and so any cell that 'sees' both of those conjugates must be false.

Here the 8s in r3 share a group and also each have a conjugate (r9c6 and r8c7).
The cell that sees r9c6 and r8c7 must be false = r9c8

Code: Select all

*--------------------------------------------------*
 | 3    6    49   | 89   7    1    | 2    48   5    |
 | 8    7    49   | 5    2    6    | 1    3    49   |
 | 1    2    5    | 3    4    89*  | 78*  6    79   |
 |----------------+----------------+----------------|
 | 4    9    7    | 6    3    2    | 5    1    8    |
 | 5    1    6    | 479  8    479  | 47   2    3    |
 | 2    8    3    | 1    5    47   | 6    9    47   |
 |----------------+----------------+----------------|
 | 7    4    8    | 2    1    3    | 9    5    6    |
 | 9    3    2    | 48   6    5    | 48*  7    1    |
 | 6    5    1    | 478  9    478* | 3   !48!  2    |
 *--------------------------------------------------*

[Kent]

Any value in any of the four cells in [brackets] will exclude the '8' from {r3c7}.

r3c6=8 -> r3c7=7
r3c6=9 -> r1c4=8 -> r8c4=4 -> r8c7=8 -> r3c7=7
Therefore, r3c7=7.

What if we start from another cell like r3c7 instead of r3c6???Will we get the same result??
Another ques if we start from a cell and we tried both candidate but each candidate give a different result(for example, if r3c7=7 if r3c6=8 and r3c7= another number if r3c6=9) what does this imply??

[emm]

Forcing chains work two ways.

1. Chains from all numbers in a cell give the same result. The deduction is that that is the correct number for that cell as tso explained here

Code: Select all

r3c6=8 -> r3c7=7
r3c6=9 -> r1c4=8 -> r8c4=4 -> r8c7=8 -> r3c7=7
Therefore, r3c7=7.

2. A single chain ends in a contradiction. The deduction is that this number is not in that cell. This is what happens if you start from r3c7. As r3c7 cannot equal both 7 and 8, the deduction is that the original premise ie r3c7=8 is not correct.

Code: Select all

r3c7=8 -> r8c7=4 -> r1c4=9 -> r8c4=8 -> r1c4=9 -> r3c6=8 -> r3c7=7

Not all numbers will lead to a useful chain, not all chains will lead to exactly the same end point but it doesn't matter where you start a chain - if it's logical then it will give you a correct answer.

[Kent]

Thanks a lot em.So what ur trying to say is that it doesn't matter where u start.If the number contradicts then that means its the wrong number right?? So if you try with r3c7=7 1st, does that mean that u wont get the contrdicted numbers??

[emm]

It means you'll either get somewhere or you'll get nowhere. You'll either get a contradiction or you'll get a confirmation or you'll get nothing.

Try it and see.

[Kent]

so r3c7=7->r3c9=9->r2c9=4->r1c8=8->r1c4=9->r3c6=8 and therefore the cell r3c7=7.Is that how it works?? So since u start from r3c7=7 and at the end u also get r3c7=7, so that mean it's correct right?

[tso]

so r3c7=7->r3c9=9->r2c9=4->r1c8=8->r1c4=9->r3c6=8 and therefore the cell r3c7=7.Is that how it works?? So since u start from r3c7=7 and at the end u also get r3c7=7, so that mean it's correct right?

No. If you get a contradition, you know your assuption was wrong. If you don't, your assuption may or may not be wrong -- unless of course you take it all the way to the end and solve the puzzle.

The statement ALL CROWS ARE BLACK is disproved by a single WHITE crows, but 1000 black crows proves nothing. The white crow is proof by contradiction.

If all possibilities lead to the same conclusion, that conclusion must be true. That's a forcing chain.

If an assumption leads to a contradiction, that assumtion is false. That's proof by contradiction.

[Kent]

SO what if i add r3c7=8 -> r8c7=4 -> r1c4=9 -> r8c4=8 -> r1c4=9 -> r3c6=8 -> r3c7=7 and r3c7=7->r3c9=9->r2c9=4->r1c8=8->r1c4=9->r3c6=8 and therefore the cell r3c7=7 together then can i SAFELY conclude that r3c7=7??

Since the 2 possibilitise lead to the same conclusion like what u said

If all possibilities lead to the same conclusion, that conclusion must be true. That's a forcing chain.