The New Sudoku Players' Forum

[susimuthu]

I am struggling to get it resolved. could anyone pls help me with the logic you use for resolving the below.

-9-/4--/7--
4--/26-/---
8-6/1--/---

-73/---/---
--2/---/8--
---/---/53-

---/--5/9-2
---/-42/--1
--9/--6/-8-

I could able to find out only 3 numbers. 7 on R2C3,2 on R6C5,9 on R8C4.

[wintder]

Code: Select all

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

This does involve advanced stuff.

[udosuk]

This is a rather inelegant solving process. After a few trivial moves:

Code: Select all

 *--------------------------------------------------------------------*
 | 1235   9     A15     | 4     F358   F38     | 7     E1256   3568   |
 | 4      135    7      | 2      6     F389    |D13    E159    3589   |
 | 8      235    6      | 1      3579   379    |D234    2459   3459   |
 |----------------------+----------------------+----------------------|
 | 1569   7      3      | 568    89     1489   | 1246   12469  469    |
 | 1569   1456   2      | 3567   379    13479  | 8      14679  4679   |
 | 169    1468   148    | 67     2      1479   | 5      3      4679   |
 |----------------------+----------------------+----------------------|
 | 1367   1346  A14     | 378    1378   5      | 9     B467    2      |
 | 3567   3568  -58     | 9      4      2      |C36    B567    1      |
 | 12357  12345  9      | 37     137    6      |C34     8      3457   |
 *--------------------------------------------------------------------*

If r8c3=5 => A=[14] => B={67} => C=[34] => D=[12] => E=[59]
Then F will become {38} for 3 cells, invalid.

Therefore r8c5 can't be 5, and the puzzle can be solved easily from there.

I'm sure others can find better ways.

[Sudtyro]

If r8c3=5 => A=[14] => B={67} => C=[34] => D=[12] => E=[59]
Then F will become {38} for 3 cells, invalid.

Therefore r8c5 can't be 5, and the puzzle can be solved easily from there.

That's a really productive elimination
How did you find the forcing chain? Was it done manually or by computer?

[udosuk]

If r8c3=5 => A=[14] => B={67} => C=[34] => D=[12] => E=[59]
Then F will become {38} for 3 cells, invalid.

Therefore r8c5 can't be 5, and the puzzle can be solved easily from there.

That's a really productive elimination
How did you find the forcing chain? Was it done manually or by computer?

Manually with some help from the computer.

Yes I just dropped by. Don't expect me to stay for long.

[Sudtyro]

Hi susimuthu,
You may not have gotten the answer(s) you were looking for, since this puzzle does indeed require some of the more advanced solving methods, such as what udosuk provided above. But, maybe one could start you off a little more slowly...

Code: Select all

  --------------------------------------------------------------------
 | 1235   9      15     | 4      358    38     | 7      1256   3568   |
 | 4      135    7      | 2      6      389    | 13     159    3589   |
 | 8      235    6      | 1      3579   379    | 234    2459   3459   |
 |----------------------+----------------------+----------------------|
 | 1569   7      3      | 568    89     1489   | 1246   12469  469    |
 | 1569   1456   2      | 3567   379    13479  | 8      14679  4679   |
 | 169    1468   148    | 67     2      1479   | 5      3      4679   |
 |----------------------+----------------------+----------------------|
 | 1367   1346   14     | 378    1378   5      | 9      467    2      |
 | 3567   3568   58     | 9      4      2      | 36     567    1      |
 | 12357  12-345 9      | 37     137    6      | 34     8      3457   |
  --------------------------------------------------------------------

One approach to a puzzle like this is to begin with a look at the unresolved single-digit grids, for which a number of powerful solving techniques are available. For example, the 5’s grid appears as follows:

Code: Select all

----------------------
5 . 5 | . 5 . | . 5 -5
. 5 . | . . . | . 5  5
. 5 . | . 5 . | . 5 -5   
------+-------+-------
5 . . | 5 . . | . .  .
5 5 . | 5 . . | . .  .
. . . | . . . | . .  .
------+-------+-------
. . . | . . . | . .  .
5 5 5 | . . . | . 5  .
5 5 . | . . . | . .  5
----------------------

Using the Eureka notation for an Alternating Inference Chain (AIC), one can form the short AIC,
(5): r9c9 = r8c8 – r8c3 = r1c3 => r1c9 <> 5.
As a pattern, this chain is called (among other names) a Turbot Fish.

By adding two more cells to the chain one then gets
(5): r9c9 = r8c8 – r8c3 = r1c3 - r1c5 = r3c5 => r3c9 <> 5.
This is known as a Turbot Chain (or X-Chain).

By arranging for certain cells to be grouped, one can actually get both of the above eliminations with one chain:
(5): r2c89 = r2c2 – r1c3 = r8c3 - r8c8 = r123c8 => r13c9 <> 5.
This is known as a Grouped Turbot Chain (with overlapping cells).

Another entirely different kind of pattern provides the same two eliminations. It’s called a finned Swordfish, in this case comprising columns 3,5 and 8, with the fin being cell r2c8. This “fish” pattern (with fin) implies r13c9 <> 5.

It’s certainly comforting (to me) to see that there often exist multiple methods to develop the same eliminations. And, thankfully, detailed descriptions and examples of all of these solving methods have been made available in Sudopedia (and in this forum).

OK, let’s see what we’ve got so far...looks like a grand total of two eliminations (excluding udosuk’s). That leaves only 130 more (pencil marks) to go to solve the puzzle!

At some point then, one must move on to the various multi-digit solving techniques. For example, using the puzzle’s full grid position above, one can form the grouped AIC,
(2)r9c2 = (2)r9c1 – (2)r1c1 = (2)r1c8 – (2)r4c8 = (2-6)r4c7 = (6-3)r8c7 = (3)r9c79 => r9c2 <> 3.
We get one more elimination, but in my opinion this chain is a little too long, is not very productive and is just plain scary! Fortunately, there always seem to be other (often simpler) solving techniques available, as udosuk has already demonstrated. Admittedly, though, I could not find an equivalent AIC for his r8c3 <> 5 elimination.

So much to learn...