The New Sudoku Players' Forum

[Amerzel]

Having trouble figuring out where to go from here. Filling in all the candidates doesn't yield any glaring solutions. For an easy way to see all the candidates, I've found importing it into http://www.paulspages.co.uk/sudoku/ works well. If anyone could point me in the right direction that would be great.

Code: Select all

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

Thanks,
James

[daj95376]

Since you didn't include pencilmarks, I don't know how far you progressed with eliminations.

Code: Select all

Locked Candidate (1)
X-Wing
X-Wing
X-Wing
XY-Wing
Naked  Quad
XY-Wing


[RW]

And here's the short solution:

Code: Select all

 *-----------------------------------------------------------*
 | 8     7     2     | 1     6     3     | 4     9     5     |
 | 9     14    146   | 5     2     48    | 68    3     7     |
 | 36    5     346   | 48    9     7     | 1     68    2     |
 |-------------------+-------------------+-------------------|
 | 367   2     1367  | 9     4     5     | 367   167   8     |
 | 5     8     69    | 3     7     1     | 69    2     4     |
 | 4    *13    1379  | 2     8     6     | 379   5    -139   |
 |-------------------+-------------------+-------------------|
 | 2    #34    5     | 6     1     48    | 3789  78    39    |
 |#37    6    #347   | 48    5     9     | 2     178  *13    |
 | 1     9     8     | 7     3     2     | 5     4     6     |
 *-----------------------------------------------------------*

All 3s in box 7 can see one of the '13'-pairs in r6c2 and r8c9. Any cell that can see both of those pairs must not be 1. Does this technique have a name?

RW

[Amerzel]

That gives me a good place to start.

Thanks!

[Ruud]

Does this technique have a name?

You have an AIC with a grouped node:

(1=3)r6c2 - (3)r7c2 = (3)r8c13 - (3=1)r8c9 => r6c9<>1

You can also use the same cells to form an AIC with an ALS in r8c13.

(1=3)r6c2 - (3=4)r7c2 - (4=7&3)r8c13 - (3=1)r8c9 => r6c9<>1

The shorter version wins.

[re'born]

All 3s in box 7 can see one of the '13'-pairs in r6c2 and r8c9. Any cell that can see both of those pairs must not be 1. Does this technique have a name?

This is an example of what Steve R called "transport".

[RW]

You have an AIC with a grouped node:

I see it more as a pattern than as a chain. Sort of a Empty Rectangle/XY-wing hybrid... It's not any harder to spot than a ER or grouped turbot. I wonder if there's more of these.

RW

[re'born]

You have an AIC with a grouped node:

I see it more as a pattern than as a chain. Sort of a Empty Rectangle/XY-wing hybrid... It's not any harder to spot than a ER or grouped turbot. I wonder if there's more of these.

Yep. Take your favorite pattern. Deduce which cells it gives you a choice of being true. Then tack on a weak then strong link going away from some or all of the cells which you know contain a true cell. You've now effectively transported the truth value to new cells.

I wrote up a thread about it over on the Eureka forum. It is a cool and ubiquitous technique.

Here is a related example in this grid, this time transporting without changing digits:

Code: Select all

 
 *-----------------------------------------------------------*
 | 8     7     2     | 1     6     3     | 4     9     5     |
 | 9     14    146   | 5     2     48    | 68    3     7     |
 | 36    5     346   | 48    9     7     | 1     68    2     |
 |-------------------+-------------------+-------------------|
 | 367   2     1367  | 9     4     5     | 367   167   8     |
 | 5     8     69    | 3     7     1     | 69    2     4     |
 | 4     13*   1379  | 2     8     6     | 379   5     139*  |
 |-------------------+-------------------+-------------------|
 | 2     34a   5     | 6     1     48    | 3789  78    39*   |
 | 37A   6     347A  | 48    5     9     | 2     178   13-   |
 | 1     9     8     | 7     3     2     | 5     4     6     |
 *-----------------------------------------------------------*


The xyz-wing in r6c29,r7c9 gives no eliminations but does say that one of those three cells is a 3. By transporting r6c2 to r8c13 via r7c2, we see that either r67c9 = 3 or r8c13 = 3. Hence r8c9 <> 3, solving the puzzle.

[udosuk]

Does this technique have a name?

From another perspective it's a Y-wing with a grouped strong link:

Code: Select all

 *-----------------------------------------------------------*
 | 8     7     2     | 1     6     3     | 4     9     5     |
 | 9     14    146   | 5     2     48    | 68    3     7     |
 | 36    5     346   | 48    9     7     | 1     68    2     |
 |-------------------+-------------------+-------------------|
 | 367   2     1367  | 9     4     5     | 367   167   8     |
 | 5     8     69    | 3     7     1     | 69    2     4     |
 | 4    @13    1379  | 2     8     6     | 379   5    -139   |
 |-------------------+-------------------+-------------------|
 | 2    #34    5     | 6     1     48    | 3789  78    39    |
 |#37    6    #347   | 48    5     9     | 2     178  @13    |
 | 1     9     8     | 7     3     2     | 5     4     6     |
 *-----------------------------------------------------------*

Grouped strong link of 3 in n7: r7c2=r8c13
One of these 2 "groups" must be 3.
If r7c2=3, r6c2=1.
If r8c13 have 3, r8c9=1.
So at least one of r6c2 or r8c9 must be 1.
Therefore r6c9 can't be 1.

For me, a easier alternative grouped strong link for this move would be:

[r6c2] - 3 - [r7c2] = 3 = [r7c79] - 3 - [r8c9]

[keith]

You can solve this by stringing together some simpler concepts:

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+----------------+----------------+----------------+
| 8    7    2    | 1    6    3    | 4    9    5    |
| 9    14#  146  | 5    2    48@  | 68   3    7    |
| 36   5    346  | 48@  9    7    | 1    68   2    |
+----------------+----------------+----------------+
| 367  2    1367%| 9    4    5    | 367  167% 8    |
| 5    8    69   | 3    7    1    | 69   2    4    |
| 4   -13   1379 | 2    8    6    | 379  5    139  |
+----------------+----------------+----------------+
| 2    34   5    | 6    1    48   | 3789 78   39   |
| 37   6    347  | 48@  5    9    | 2    18#% 13   |
| 1    9    8    | 7    3    2    | 5    4    6    |
+----------------+----------------+----------------+


In the <48> chain @, the ends R2C6 and R8C4 have the same value. You can make an extended XY-wing with the cells marked #. One of R2C2 and R8C8 is <1>.

Coloring on <1>, %, says that if R8C8 is <1>, so is R4C3. So, R6C2 cannot be <1>, it must be <3>, and the puzzle is solved.

Keith

[wintder]

udosuk and re'born, it would be easier to use the strong link in r7, wouldn't it?

Code: Select all

 *-----------------------------------------------------------*
 | 8     7     2     | 1     6     3     | 4     9     5     |
 | 9     14    146   | 5     2     48    | 68    3     7     |
 | 36    5     346   | 48    9     7     | 1     68    2     |
 |-------------------+-------------------+-------------------|
 | 367   2     1367  | 9     4     5     | 367   167   8     |
 | 5     8     69    | 3     7     1     | 69    2     4     |
 | 4    @13    1379  | 2     8     6     | 379   5     39-1  |
 |-------------------+-------------------+-------------------|
 | 2    #34    5     | 6     1     48    |#3789  78   #39    |
 | 37    6     347   | 48    5     9     | 2     178  @13    |
 | 1     9     8     | 7     3     2     | 5     4     6     |
 *-----------------------------------------------------------*