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Is this valid 2?

 
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Novajlija
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PostPosted: Sat Apr 12, 2008 9:08 am    Post subject: Is this valid 2? Reply with quote

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
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PostPosted: Sat Apr 12, 2008 4:20 pm    Post subject: Reply with quote

The simplest way forward is the XY-wing 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.
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Steve
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PostPosted: Sun Apr 13, 2008 7:39 pm    Post subject: Reply with quote

Novajlija

I donít think your argument is valid as it stands. What you have shown is that, if r89c1 = (56), r9c1 ≠ 5. Thatís 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 Glynís.

Steve
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Novajlija
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PostPosted: Thu Apr 24, 2008 8:17 am    Post subject: Reply with quote

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
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PostPosted: Thu Apr 24, 2008 1:08 pm    Post subject: Reply with quote

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

Steve
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Jean-Christophe
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PostPosted: Fri Apr 25, 2008 6:37 am    Post subject: Reply with quote

Another similar pattern is the Y-Wing (also called W-Wing 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 XY-Wing:
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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