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Novajlija Rookie
Joined: 17 Nov 2007 Posts: 5

Posted: Sat Apr 12, 2008 9:08 am Post subject: Is this valid 2? 


Let's use this code from another topic.
Code:
....
 7 6 1  8 3 2  4 9 5 
 4 9 8  1 6 5  2 37 37 
 3 5 2  4 9 7  6 1 8 
:++:
 9 38 67  36 4 1 *378 5 2 
*58 4 35  2 7 9  1 *368 36 
 2 1 67  36 5 8  37 4 9 
:++:
 1 37 9  57 8 6  35 2 4 
 568 2 35  57 1 4  9 3678 367 
56 78 4  9 2 3 *58 67 1 
''''
For digit 6 we have linked candidates in R8C1 and R9C1.
Let's suppose: R8C1 is 6 or 5 and R9C1 is 6 or 5 then R9C2 is 8 and then R9C7 is 5 so we can eliminate 5 from R9C1.
The point is to take 2 cells with linked candidates (number 6 in box 7) and make naked pair with another candidate (number 5) and then make some chains in grid and eliminate second candidate (number 5).
24.4.2008. not linked candidates, bat locked candidates
Last edited by Novajlija on Thu Apr 24, 2008 8:20 am; edited 1 time in total 

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Glyn Major Major Major
Joined: 16 Jan 2007 Posts: 92 Location: London

Posted: Sat Apr 12, 2008 4:20 pm Post subject: 


The simplest way forward is the XYwing that Sudocue gives.
As a chain that reads (5=3)r8c3(3=7)r7c2(7=5)r7c4 => r8c4<>5. Singles from here.
I think you are looking for an ALS type elimination. Setting Sudocue to ignore simple chains and Medusa, it finds this one.
(3567=8)ALS:r8c3489(8=56)ALS:r9c17 so we either have a 6 in r8c89 or in r9c1 so r8c1<>6.
Or we could go mad with the elimination you were after, which is an AALS (Almost ALS).
(5&8=6)ALS:r58c1(6=378)AALS:r8c89(8=5)r9c7 so we either have a 5 in r58c1 or in r9c7 => r9c1<>5.
The ALS is 3 candidates {568} shared between 2 cells r58c1, the AALS is 4 candidates {3678} shared between 2 cells r8c89. _________________ I have 81 brain cells left, I think. 

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Steve Regular
Joined: 24 Jan 2006 Posts: 12 Location: England

Posted: Sun Apr 13, 2008 7:39 pm Post subject: 


Novajlija
I don’t think your argument is valid as it stands. What you have shown is that, if r89c1 = (56), r9c1 ≠ 5. That’s fine as far as it goes but you also need to allow for the case r89c1 ≠ (56). In this event r8c1 = 8 so you only need a further line:
If r8c1 = 8, r5c1 = 5 so r9c1 ≠ 5.
This is much the same argument as one of Glyn’s.
Steve 

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Novajlija Rookie
Joined: 17 Nov 2007 Posts: 5

Posted: Thu Apr 24, 2008 8:17 am Post subject: 


Hi Steve.
My point is: If r8c1 = 8 then r9c1=6 so r9c1 ≠ 5.
Generally:
There is two cells (in the box/ column/ row) with locked candidate A:
cell1 = ABCD
cell2 = ABCFG
If cell1 = B(or C or D) then cell2 = A ....
Bat my point is:
cell2 is not equal to B(or C or F or G)
And now about solving, lets take this 2 cells (in the box/ column/ row) with locked candidate A and make naked pair with another candidate (in this case B or C), so we can get few other candidates (in the box/ column/ row) and then make some chains in grid and eliminate second candidate.
This is very useful to eliminate some candidates.
My question: is this kind of guessing (bifurcation) valid strategy in the sudoku community. 

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Steve Regular
Joined: 24 Jan 2006 Posts: 12 Location: England

Posted: Thu Apr 24, 2008 1:08 pm Post subject: 


Well, any method which is logical and makes an elimination is good enough for me.
Steve 

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JeanChristophe Addict
Joined: 23 Apr 2007 Posts: 92 Location: Belgium

Posted: Fri Apr 25, 2008 6:37 am Post subject: 


Another similar pattern is the YWing (also called WWing in Sudopedia):
The 8 at r4 locked at r4c27. Whichever the location, either r5c1 = 5 or r9c7 = 5. Therefore r9c1 <> 5 = 6
Written as an AIC:
(5=8)r5c1  (8)r4c2 = (8)r4c7  (8=5)r9c7 > r9c1 <> 5
I personally find it easier to spot than an XYWing:
Highlight digit X, search for strong links (conjugate pairs), then two bivalue cells with XY which are buddies (peers) of the strong link. You can eliminate Y from buddies of these two XY cells. 

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