## Strong Corner Marker for X-Wing AIC Deduction

Advanced methods and approaches for solving Sudoku puzzles
I agree with daj95376, that the eliminations presented here, are much too complicated for this grid, where his relatively simple move (together with coloring) solves it. So i also did not look at Allan Barker's samples closer.

David P Bird's move is nice, but it does not help much either in this extra hard puzzle (i could place the 7 in r8c6 after that, because the right side lead to a contradiction, but that was all i got from it). Btw i dont like such hard puzzles. Most of them definitely dont have an elegant solution, and at the end i have the feeling, that i wasted my time.

So we are still far from having a puzzle, which can be solved nicely with strong or weak corners.
eleven

Posts: 1999
Joined: 10 February 2008

Maybe this is a better sample, a puzzle from here http://forum.enjoysudoku.com/viewtopic.php?p=62624#p62624:
Code: Select all
. . . | . . 1 | 2 3 .
4 . . | . 5 . | . . .
6 . . | 2 . . | . . .
-------+-------+-------
7 . . | . . . | 4 . .
. 8 . | . . . | . 1 .
. . 6 | . . . | . . 8
-------+-------+-------
. . . | . . 8 | . . 2
. . . | . 4 . | . . 6
. 1 2 | 3 . . | . . .
*-----------------------------------------------------------*
| 589   579   78    | 6     79    1     | 2     3     4     |
| 4     2     3     | 8     5     79    | 79    6     1     |
| 6     79    1     | 2     3     4     |*5789  5789  579   |
|-------------------+-------------------+-------------------|
| 7     3     59    | 1     8     6     | 4     2    *59    |
| 2     8    #59    | 4     79   #579   | 6     1     3     |
| 1     4     6     | 59    2     3     | 579   579   8     |
|-------------------+-------------------+-------------------|
| 59    6     4     | 7     1     8     | 3    *59    2     |
| 3     579   78    | 59    4     2     | 1    *589   6     |
| 589   1     2     | 3     6    #59    | 5789  4     579   |
*-----------------------------------------------------------*

Either r5c6=5 -> r9c6=9
Or r5c3=5 & r9c6=5 -> r4c9=5 & r78c8=5 -> r3c7=5 -> r3c8=8 -> r78c8=59
I.e. r9c79<>9 and the puzzle solves with coloring and w-wings.
Edit: corrected a typo.

Are there better solutions for it ?
Last edited by eleven on Sun Oct 19, 2008 7:54 am, edited 1 time in total.
eleven

Posts: 1999
Joined: 10 February 2008

eleven wrote:Are there better solutions for it ?

Just different.

Code: Select all
[r9c6]=9                                        => [r9c79]<>9
[r9c6]=5 [r6c4]=5 [r26c7]=79 *[r9c7]=8 [r9c1]=9 => [r9c79]<>9

Code: Select all
SIN:  9r9c7  5r9c6  5r6c4  [c7]~7  => [r9c7]<>9
SIN:  9r9c9  7r9c7  9r2c7  [b6]~5  => [r9c9]<>9
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

eleven wrote:Either r5c6=5 -> r9c6=9
Or r5c3=5 & r9c6=5 -> r3c9=5 & r78c8=5 -> r3c7=5 -> r3c8=8 -> r78c8=59
I.e. r9c79<>9

Do you mean: Or r5c3=5 & r9c6=5 -> r4c9=5 ? Otherwise everything seems fine.
Allan Barker

Posts: 266
Joined: 20 February 2008

Here is a simpler strong corner with 2 short AICs using a naked pair (79)r2c67 instead of an X-Wing, which makes it vertical or "hidden".

Either r2c7=9
Or r2c7=7 & r2c6=9 -> r6c8=7 & r8c4=9 -> r6c7=9
Thus r39c7<>9.

Code: Select all
Strong corner in 79r2c67
+-----------------------------------------------------------+
| 589   579   78    | 6     79    1     | 2     3     4     |
| 4     2     3     | 8     5     7#9   | #7#9  6     1     |
| 6     79    1     | 2     3     4     | 578-9  5789  579  |
+-----------------------------------------------------------+
| 7     3     59    | 1     8     6     | 4     2     59    |
| 2     8     59    | 4     79    579   | 6     1     3     |
| 1     4     6     | 5(9)  2     3     | 5(79) 5(79) 8     |
+-----------------------------------------------------------+
| 59    6     4     | 7     1     8     | 3     59    2     |
| 3     579   78    | 5(9)  4     2     | 1     589   6     |
| 589   1     2     | 3     6     5(9)  | 578-9  4    579   |
+-----------------------------------------------------------+

9R2: 9r2c7A===============9r2c6
|  A                 |
2N7: 9r2c7A=7r2c7           |
|      |             |
7R6:   |    7r6c7==7r6c8    |
|             |      |
9B8:   |             |    9r9c6==9r8c4
|             |             |
9R6: 9r6c7=========9r6c8=========9r6c4
|
X=9r3c7
X=9r9c7

= Strong link, | weak link, 'A' label for strong corner

Edit:Point out this is a vertical, or hidden strong corner. Thanks to eleven.
Last edited by Allan Barker on Sun Oct 19, 2008 9:03 am, edited 1 time in total.
Allan Barker

Posts: 266
Joined: 20 February 2008

daj95376 wrote:
eleven wrote:Are there better solutions for it ?

Just different.
Thanks for confirming that. Your solution is probably not more complicated, but i think harder to find for a manual solver.

Allan Barker wrote:Do you mean: Or r5c3=5 & r9c6=5 -> r4c9=5 ?
Yes, of course, thanks, i will correct it.

Allan Barker wrote:Here is an even simpler strong corner with 2 short AICs.
Nice move, but where is the strong corner ?
eleven

Posts: 1999
Joined: 10 February 2008

deleted
Last edited by StrmCkr on Sun Oct 19, 2008 2:28 pm, edited 3 times in total.
Some do, some teach, the rest look it up.

StrmCkr

Posts: 942
Joined: 05 September 2006

I am tired, but i cant follow that.Is it again, that you are seeing things, no one else can see ?
eleven

Posts: 1999
Joined: 10 February 2008

eleven wrote:
Allan Barker wrote:Here is an even simpler strong corner with 2 short AICs.
Nice move, but where is the strong corner ?

Good point! The corner is made from a naked pair instead of an X-wing, which would make it a hidden corner.

I have added a PM and explanation to the original post. I often work in 3D and the difference did not occur to me. Thanks.
.
Allan Barker

Posts: 266
Joined: 20 February 2008

deleted
Last edited by StrmCkr on Sun Oct 19, 2008 2:28 pm, edited 1 time in total.
Some do, some teach, the rest look it up.

StrmCkr

Posts: 942
Joined: 05 September 2006

StrmCkr wrote:if the 7 is in R8c3

it leaves a pair of 59's in the box which are linked to a chain of 59 pairing. ie. remote pair eliminations.

But that has an even number of links in the chain; it needs to be an odd number.

the other way to see it would be to assert R8C8 is an 8 and notice the contradiction in R9C78 (two 7's as the only choice)

That proves that r8c8<>8 ... and says nothing about r9c79.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

deleted garbage post...
Last edited by StrmCkr on Sun Oct 19, 2008 1:57 pm, edited 1 time in total.
Some do, some teach, the rest look it up.

StrmCkr

Posts: 942
Joined: 05 September 2006

StrmCkr wrote:why do i need an odd number of links?

Set [r8c3]=7 and then use Blue/Green to color all of the (59) cells in band 3. You'll discover that [r7c8] and [r9c6] are the same color -- and thus don't qualify as a Remote Pair.

Your use of [r8c3]=8 to force the Naked Pair [r9c16]=59 does work for the eliminations [r9c79]<>59.
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

StrmCkr, your deduction is invalid; r9c9=5 does not produce a contradiction anywhere in band 3 (rows 7, 8 and 9).
Code: Select all
5   6   4   | 7   1   8   | 3   9   2
3   9   7   | 5   4   2   | 1   8   6
8   1   2   | 3   6   9   | 7   4   5

Your secondary argument that r8c8=8 produces a contradiction is also false.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

ok thanks for the info i'll check my stuff again if its true i'll delete my posts.

thought i found something while at work...figured id get other imput on it if i missed anything and i did.
thanks!

.. just to clarify this is not a remote pair
Code: Select all

59*   .     .   |.     .     .     | .     59*   .    |
| .    59*   .  | 59*   .    .    | .     . .   |
| .   .     .     | .     .    59*   | x   x  x

i see it now.. the x's don't see both 5 & 9 from either side of the chain. they see "5"s x2 or 9 x 2
Some do, some teach, the rest look it up.

StrmCkr

Posts: 942
Joined: 05 September 2006

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