AIC Problem

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AIC Problem

Postby Jasper32 » Mon Sep 01, 2008 12:18 pm

Code: Select all
 
 *--------------------------------------------------------------------*
 | 68     58     4      | 2      56789  3      | 67     1      689    |
 | 1368   13568  7      | 4      15689  1589   | 2      689    3689   |
 | 9      1368   2      | 67     1678   178    | 5      4678   3468   |
 |----------------------+----------------------+----------------------|
 | 2      79     1      | 5679   4567   579    | 8      3      469    |
 | 5      389    38     | 1      468    2      | 46     469    7      |
 | 378    4      6      | 379    3789   789    | 1      2      5      |
 |----------------------+----------------------+----------------------|
 | 468    68     5      | 79     179    179    | 3      4678   2      |
 | 137    13678  38     | 357    2      4      | 9      5678   68     |
 | 347    2      9      | 8      35     6      | 47     457    1      |
 *--------------------------------------------------------------------*


Above is a puzzle I ran through the Scanraid Solver and came up with this AIC.

Scanraid says:
"AIC Rule 2, on 8 (AIC , length 6:
7[H2] =7[D2] -7[F1] =8[F1] -8[E3] =8[H3]-
-Chain ends H2 cannot be 8 and H3 cannot be 7"

I can follow what is going on in that if either D2 or H2 (strong links) =7 that D2 must equal 9 and F1 must equal 7. That is as far as I can go and I do not understand the logic that make E3 = 3.

I don't understand the notation that Scanraid used to describe the AIC.

Perhaps there is a different notation that would explain this to me in a way that I could understand. Or, barring that, somebody could explain just how to go about understanding the notation used by Scanraid. Any help or comments will be appreciated.

Either way, many thanks to the members that have always come to my rescue.
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Re: AIC Problem

Postby hobiwan » Mon Sep 01, 2008 1:38 pm

Are you sure that you put the right PM into your message? I don't really understand the AIC either but with candidate 3 in r6c1 it is definitely invalid.
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AIC

Postby Jasper32 » Mon Sep 01, 2008 2:21 pm

My error. Sorry, I rechecked and below is how it should have appeared. I have triple checked it and this is what the Scanraid Solver showed.
Code: Select all

 *-----------*
 |..4|2.3|.1.|
 |..7|4..|2..|
 |9.2|...|5..|
 |---+---+---|
 |2.1|...|83.|
 |5..|1.2|..7|
 |.46|...|125|
 |---+---+---|
 |..5|...|3.2|
 |...|.24|9..|
 |.29|8.6|..1|
 *-----------*

 
 *--------------------------------------------------------------------*
 | 68     58     4      | 2      56789  3      | 67     1      689    |
 | 1368   13568  7      | 4      15689  1589   | 2      689    3689   |
 | 9      1368   2      | 67     1678   178    | 5      4678   3468   |
 |----------------------+----------------------+----------------------|
 | 2      79     1      | 567    4567   57     | 8      3      469    |
 | 5      389    38     | 1      468    2      | 46     469    7      |
 | 78     4      6      | 39     3789   789    | 1      2      5      |
 |----------------------+----------------------+----------------------|
 | 468    68     5      | 79     179    179    | 3      468    2      |
 | 137    13678  38     | 35     2      4      | 9      5678   68     |
 | 347    2      9      | 8      35     6      | 47     457    1      |
 *--------------------------------------------------------------------*



Again, I am sorry for the error on my part.
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Postby daj95376 » Mon Sep 01, 2008 5:12 pm

Using your last PM, Scanraid's AIC chain using NL notation:

Code: Select all
[r8c2]=7=[r4c2]-7-[r6c1]-8-[r5c3]=8=[r8c3]
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Postby Steve R » Mon Sep 01, 2008 5:18 pm

The chain, whether in Eureka or NL mutation, is a sophisticated way of writing

r8c2 ≠ 7 => r4c2 = 7 => r6c1 = 8 => r5c3 ≠ 8 => r8c3 = 8

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Re: AIC Problem

Postby Luke » Tue Sep 02, 2008 8:44 am

Jasper32 wrote:Scanraid says:
"AIC Rule 2, on 8 (AIC , length 6:
7[H2] =7[D2] -7[F1] =8[F1] -8[E3] =8[H3]-
-Chain ends H2 cannot be 8 and H3 cannot be 7"

I can follow what is going on in that if either D2 or H2 (strong links) =7 that D2 must equal 9 and F1 must equal 7. That is as far as I can go and I do not understand the logic that make E3 = 3.

I don't understand the notation that Scanraid used to describe the AIC.

How's about no notation at all? First, keep in mind:

* An AIC chain alternates between strong and weak inferences.
* The inference between values in a bivalue cell can be considered strong OR weak (as with all strong inferences.)

Your chain starts on a strong link. It ends with on a strong link in a different cell, and on a different value. If the start value can see the end value, then the logic inferred by the chain demands the start value cannot be in the end cell and the end value cannot be in the start cell.

Here's a graphic that might help, with the strong links in red and the weak links in green.
Image
Starting value = 7 in r8c2, ending with 8, r8c3. 8 cannot be in the start cell.

Compare this to the different notations offered and hopefully they'll start to make sense.

I do not understand the logic that make E3 = 3.

Either do I. Where'd that come from?
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Postby Jasper32 » Tue Sep 02, 2008 12:38 pm

Luke451 Wrote:

I do not understand the logic that make E3 = 3.

Either do I. Where'd that come from?


To answer that E3 is a (3). My question is why is it (3) since there is also a (3) in E2. Is it because e3 is a bivalve? All the cells except (1) are bivalves.

Part of the proble I was having was I didn't think about the (7) and the (8) in f1 as being strongly linked. Thanks for helping me on that. If you could tell me about my mix-up on E2 and E3, it would be appreciated.

Many thanks.
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Re: AIC Problem

Postby daj95376 » Tue Sep 02, 2008 6:12 pm

Jasper32 wrote:I do not understand the logic that make E3 = 3.

It appears that you are under the misconception that somehow the AIC has caused [r5c3]=3. This is not the case! What's occurred is that the chain says [r5c3]<>8 if the initial assumption [r8c2]<>7 is true. There is no guarantee that the initial assumption is true!!!
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Re: AIC Problem

Postby Luke » Tue Sep 02, 2008 8:22 pm

Jasper32 wrote:Scanraid says:
"AIC Rule 2, on 8 (AIC , length 6:
7[H2] =7[D2] -7[F1] =8[F1] -8[E3] =8[H3]-
-Chain ends H2 cannot be 8 and H3 cannot be 7"

I don't understand the notation that Scanraid used to describe the AIC.

You prob know by now that "=" means strong inference and "-" means weak.
W/this notation the AIC will always go back and forth between strong and weak (or vice versa) with the cell designations on either side.

See this page here for Scanraid's explanation of "AIC Rule 2." They say it's for conjugate pairs but your example proves it can do more than that.

Now for my questions: what does the NL stand for in "NL notation"?
Is a "PM" simply a grid showing the pencil marks, or is it something else?
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Re: AIC Problem

Postby daj95376 » Tue Sep 02, 2008 11:02 pm

Luke451 wrote:Now for my questions: what does the NL stand for in "NL notation"?
Is a "PM" simply a grid showing the pencil marks, or is it something else?

I've been using the following:

NL => Nice Loop
PM => Pencil Marks (grid)
pm => private message
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Re: AIC Problem

Postby Luke » Fri Sep 19, 2008 11:18 am

Jasper32 wrote:
Code: Select all
 
 *--------------------------------------------------------------------*
 | 68     58     4      | 2      56789  3      | 67     1      689    |
 | 1368   13568  7      | 4      15689  1589   | 2      689    3689   |
 | 9      1368   2      | 67     1678   178    | 5      4678   3468   |
 |----------------------+----------------------+----------------------|
 | 2      79     1      | 5679   4567   579    | 8      3      469    |
 | 5      389    38     | 1      468    2      | 46     469    7      |
 | 378    4      6      | 379    3789   789    | 1      2      5      |
 |----------------------+----------------------+----------------------|
 | 468    68     5      | 79     179    179    | 3      4678   2      |
 | 137    13678  38     | 357    2      4      | 9      5678   68     |
 | 347    2      9      | 8      35     6      | 47     457    1      |
 *--------------------------------------------------------------------*


Above is a puzzle I ran through the Scanraid Solver and came up with this AIC.

Scanraid says:
"AIC Rule 2, on 8 (AIC , length 6:
7[H2] =7[D2] -7[F1] =8[F1] -8[E3] =8[H3]-
-Chain ends H2 cannot be 8 and H3 cannot be 7"

I can follow what is going on in that if either D2 or H2 (strong links) =7 that D2 must equal 9 and F1 must equal 7. That is as far as I can go and I do not understand the logic that make E3 = 3.

I don't understand the notation that Scanraid used to describe the AIC.

Perhaps there is a different notation that would explain this to me in a way that I could understand. Or, barring that, somebody could explain just how to go about understanding the notation used by Scanraid. Any help or comments will be appreciated.

Either way, many thanks to the members that have always come to my rescue.

I recently got turned on to JSudoku (thanks tarek!) This program will produce a graphical representation of these chains that are easy to follow. JSudoku refers to these as "xy-x chains."
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