Problem with AIC ...Need help

Post the puzzle or solving technique that's causing you trouble and someone will help

Problem with AIC ...Need help

Postby Gee » Mon Nov 12, 2007 8:12 pm

I have proceded to this point and I wonder why I cannot do what I thought was possible. I would appreciate any comments. Thank You.

The AIC starts with a (8) in r5c8 then a (2) r4c8 and then an (8) in r4c2. Now at this point, I would think that I could exclude all the (8's) in r5 c123. ....That doesn't happen and I don't know why since the (8) in r5c8 and the (8) in r4c2 "sees" the (8's) in r5c123. Help Please.

*-----------*
|..6|9..|4..|
|.3.|.5.|.6.|
|54.|...|.92|
|---+---+---|
|4..|7.5|..9|
|...|...|...|
|9..|8.2|..4|
|---+---+---|
|87.|...|.13|
|.9.|.1.|.5.|
|..5|..8|9..|
*-----------*


*-----------*
|..6|9..|4.5|
|.39|.5.|861|
|54.|6..|.92|
|---+---+---|
|4..|7.5|1.9|
|...|149|...|
|9..|8.2|..4|
|---+---+---|
|874|596|213|
|.9.|.1.|.58|
|..5|..8|94.|
*-----------*


*-----------------------------------------------------------*
| 127 128 6 | 9 278 137 | 4 37 5 |
| 27 3 9 | 24 5 47 | 8 6 1 |
| 5 4 178 | 6 78 137 | 37 9 2 |
|-------------------+-------------------+-------------------|
| 4 268 238 | 7 36 5 | 1 28 9 |
| 2367 258 2378 | 1 4 9 | 3567 28 67 |
| 9 15 137 | 8 36 2 | 3567 37 4 |
|-------------------+-------------------+-------------------|
| 8 7 4 | 5 9 6 | 2 1 3 |
| 236 9 23 | 234 1 347 | 67 5 8 |
| 1236 126 5 | 23 27 8 | 9 4 67 |
*-----------------------------------------------------------*
Gee
 
Posts: 50
Joined: 18 March 2007

Postby RW » Mon Nov 12, 2007 9:19 pm

Gee, first of all, there's no strong link on candidate 8 between r4c2 and r4c8. If r4c8<>8, then either r4c2 or r4c3=8. However, this is not what prevents your elimination. Your problem is that you have an eight in both ends of the chain. You have if r5c8=8 => r4c23=8. This cannot eliminate anything, because it does not consider the option of r5c8<>8. To succesfully use a chain like this, you must find one where either one end has a certain digit or the other end has that same digit. In that case you know that since one of the ends must be true, then any cell that can see both must be false. If you would have found a chain saying if r5c8<>8 => r4c23=8, then your elimiation would be valid.

RW
RW
2010 Supporter
 
Posts: 1010
Joined: 16 March 2006

Re: Problem with AIC ...Need help

Postby Sudtyro » Mon Nov 12, 2007 9:24 pm

Gee wrote:The AIC starts with a (8) in r5c8 then a (2) r4c8 and then an (8) in r4c2. Now at this point, I would think that I could exclude all the (8's) in r5 c123. ....That doesn't happen and I don't know why since the (8) in r5c8 and the (8) in r4c2 "sees" the (8's) in r5c123.
Code: Select all
 
 *-----------------------------------------------------------*
 | 127   128   6     | 9     278   137   | 4     37    5     |
 | 27    3     9     | 24    5     47    | 8     6     1     |
 | 5     4     178   | 6     78    137   | 37    9     2     |
 |-------------------+-------------------+-------------------|
 | 4     268   238   | 7     36    5     | 1     28    9     |
 | 2367  258   2378  | 1     4     9     | 3567  28    67    |
 | 9     15    137   | 8     36    2     | 3567  37    4     |
 |-------------------+-------------------+-------------------|
 | 8     7     4     | 5     9     6     | 2     1     3     |
 | 236   9     23    | 234   1     347   | 67    5     8     |
 | 1236  126   5     | 23    27    8     | 9     4     67    |
 *-----------------------------------------------------------*


Short answer: There's no AIC.
Can you show your AIC's components and their Alternating Inferences?

You would need an AIC of the form
(8)r5c8 = (?)r?c? - (?)r?c? = (8)r4c2 => r5c123 <> 8.

BTW, if you add the "code" tags to your grids they will display better, as above.

[Edit for following addition:]
Your grid has two nice same-digit AIC examples as shown below. They are actually Turbot fish.
(6): r5c9 = r9c9 – r8c7 = r8c1 => r5c1 <> 6
(2): r9c5 = r1c5 – r2c4 = r2c1 => r9c1 <> 2
Last edited by Sudtyro on Mon Nov 12, 2007 6:23 pm, edited 1 time in total.
Sudtyro
 
Posts: 68
Joined: 21 December 2006

Aic Help

Postby Gee » Mon Nov 12, 2007 9:38 pm

Thanks RW;

I was under the impression that since r4c8 was a strong link to r5c8, a strong link was not needed between r4c2 and r4c8. I thought AIC was a strong link followed by a weak link and ect. I do know that if I assumed r5c8 a (2) and followed through,all the (2's) in r5c123 could be eliminated.

I thank you for you help. It is appreciated. Also, do you know of a web site that explains AIC in detail? I still get confused with the srong link and weak link definitions. I thoght a a strong link was where there were only 2 candidates in two different cells in a row, column or box. And a weak link was where there were more than 2 candidates in cells in a row, column or box. I still have a way to go, obiviously.

Thanks again.
Gee
 
Posts: 50
Joined: 18 March 2007

Re: Aic Help

Postby Sudtyro » Wed Nov 14, 2007 11:39 pm

Gee wrote:...do you know of a web site that explains AIC in detail? I still get confused with the srong link and weak link definitions.

Gee,
For what it's worth, I think that Myth Jellies' best captures the whole AIC concept, along with definitions for strong/weak links (and inferences), in this definitive posting:
http://forum.enjoysudoku.com/viewtopic.php?t=3865

Another good source for definitions and examples is Sudopedia:
http://www.sudopedia.org/wiki/Main_Page

Hope this helps!
Sudtyro
 
Posts: 68
Joined: 21 December 2006

Aic Help

Postby Gee » Wed Nov 14, 2007 11:52 pm

Thank you, Sudtyro. I have printed them out. Thoughful of you to send those links.
Gee
 
Posts: 50
Joined: 18 March 2007


Return to Help with puzzles and solving techniques