The New Sudoku Players' Forum

[tso]

There's another more convoluted way to solve it from the same given position:

Code: Select all

  .    .    .  |  .    .    .  |  .    .    .
  .    .    .  | 2-9   .   129 |  .   1-2   .
  .    .    .  |  .    .    .  |  .    .    .
  -------------+---------------+-------------
  .    .    .  |  .    .    .  |  .    .    .
  .    .    .  |  .    .   1-4 |  .    .    .
  .    .    .  | 4-9  1-9   .  |  .    .    .
  -------------+---------------+-------------
  .    .    .  |  .    .    .  |  .    .    .
  .    .    .  |  .    .    .  |  .    .    .
  .    .    .  |  .    .    .  |  .    .    .


1) r5c6 = 1 or 4
2) if r5c6 = 4, then r6c4=9, then r2c4=2, then r2c8=1
3) if r5c6 = 1, then r2c6=(2-9) forming a twin with r2c4=(2-9), then r2c8=1

This proves r2c8=1. After some easy steps, the grid looks like this:

Code: Select all

 1-3   8    6  |  5    4   1-3 |  9    2    7
  5    4    7  | 2-9   6   2-9 |  8    1    3
  .   3-9  123 |  7   1-3   8  |  4    5    6
  -------------+---------------+-------------
  8    2    9  |  3    5    7  |  6    4    1
  7    6   1-4 |  8    2   1-4 |  5    3    9
 134   5   134 | 4-9  1-9   6  |  7    8    2
  -------------+---------------+-------------
 3-9   7    8  |  1   3-9   5  |  2    6    4
  6   3-9  2-4 | 249   8    .  |  1    7    5
 2-4   1    5  |  6    7   2-4 |  3    9    8


And now it is possible to see this:

Code: Select all

 1-3   .    .  |  .    .   1-3 |  .    .    .
  .    .    .  |  .    .    .  |  .    .    .
  .    .    .  |  .   1-3   .  |  .    .    .
  -------------+---------------+-------------
  .    .    .  |  .    .    .  |  .    .    .
  .    .    .  |  .    .    .  |  .    .    .
  .    .    .  |  .   1-9   .  |  .    .    .
  -------------+---------------+-------------
 3-9   .    .  |  .   3-9   .  |  .    .    .
  .    .    .  |  .    .    .  |  .    .    .
  .    .    .  |  .    .    .  |  .    .    .


1) r1c1 = 1 or 3
2) if r1c1 = 1 => r1c6=3 => r3c5=1 => r6c5=9
3) if r1c1 = 3 => r7c1=9 => r7c5=3 => r3c5=1 => r6c5=9

This proves r6c5 is 9. The rest follows easily.

[Yellowbelly]

I am always looking at other ways to improve my speed, when I came across you post. I wrote out the grid and then filled in the Pencil marks.

r3c3 pencil marks should be 289 (not 28), which would change the result in r7c3[/quote]

[tso]

I am always looking at other ways to improve my speed, when I came across you post. I wrote out the grid and then filled in the Pencil marks.

r3c3 pencil marks should be 289 (not 28), which would change the result in r7c3

[/quote]

R3c2 and r3c7 were both 5/9 -- so I removed the other 9s from that row.

[abrecher]

I'm not sure on this yet, but I'm begining to believe that for each 'Trial and Error'-case there is a 'Foreched chain'-case.

This isn't so. Up to now, I've compiled about 18,000 problems. About 12,000 can be solved using common techniques. About 2,000 can be solved using advanced coloring (which includes swordfish and foreched chains). About 4,000 my program still cannot solve without guessing (9 times in one case).

[/code]

I've always suspsected that all puzzles with a unique solution can be solved without "guessing"/T&E techniques. For puzzles that no one has solved without this, we just haven't found the right alternative technique, just like TSO found a technique that Pappocorn didn't know or someone else has a technique that nick70's program doesn't know.

I'm sure this has never been proven, though, and frankly, I don't think it ever will be. (I suspect this is an offshoot of Godel's Incompleteness Theorem somehow, though I've never looked at it formally.) But it's something to think about.[/i]