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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Sun Nov 09, 2008 3:58 am Post subject: |
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| nataraj wrote: | I used: UR (37) type 4 => r9c2=3
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Craig,
| Code: | after basics, but before XY-Wing, X-Wing, Kite/Skyscraper
(37) UR Type 4 (w/ X-Wing in 7) => [r9c13]<>3
followed by Hidden Single [r9c2]=3
*-----------------------------------------------------------*
| 1 26 8 | 9 7 246 | 46 5 3 |
| 24 2456 25 | 26 3 8 | 9 7 1 |
|*37 9 *37 | 1 46 5 | 46 8 2 |
|-------------------+-------------------+-------------------|
| 23 235 4 | 2357 9 27 | 8 1 6 |
| 6 8 125 | 25 24 124 | 7 3 9 |
| 9 7 13 | 36 8 16 | 5 2 4 |
|-------------------+-------------------+-------------------|
| 8 1 6 | 4 5 3 | 2 9 7 |
| 5 24 9 | 267 1 267 | 3 46 8 |
|*2347 234 *237 | 8 26 9 | 1 46 5 |
*-----------------------------------------------------------*
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Nov 09, 2008 4:04 am Post subject: |
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Yes. "A=B" means "not A implies B" when reading left to right and "not B implies A" when reading right to left. This condition is met when it is impossible for A and B both to be false. Either or both of them can be true and the inferences are still valid.
In the same way, we can use the strong inference links induced by a UR or other DP in an AIC. For instance, consider the UR:
AXY ... XY
BXY ... XY
I don't know whether A, B or both A and B is true. But, I know that A and B cannot both be false because that produces the DP. I can use this "A=B" in any AIC.
This and the ER we have been considering are examples of "strong only" links. Conjugate links are both strong (can't both be false) and weak (can't both be true), which is why they can be used in both the strong and weak inference positions in an AIC. An ER that includes a candidate in the ERI cell can only be used as a strong inference in an AIC since it is possible that both the box-row segment and the box-column segment are true (if the ERI cell is true). An ER without a candidate in the ERI is conjugate and can be used as strong or weak in an AIC. (When using it as a weak link, we aren't using it as an ER in the traditional sense.)
Does that help? |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Nov 09, 2008 4:59 am Post subject: |
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| storm_norm wrote: | (6=2)r1c2-(2)r8c2=(2)r9c123-(2=6)r9c5-(6)r3c5=(6)r3c7; r1c7 <> 6
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I'm going to jump into this discussion with both feet ... and probably end up with them in my mouth.
First off, I was (originally) under the impression that you were thinking of an ER pattern in [b7]. This seems to have been Asellus' position as well. Once I realized that you'd taken a different approach, I decided to review it. First off, I can barely read Eureka notation ... and I can't follow inferences/links at all in them. So the first thing I did was rewrite your chain using NL notation. Then I decomposed the NL chain into its AIC components.
| Code: | SI: Strong Inference
WI: Weak Inference
SL: Strong Link
WL: Weak Link
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| Code: | 6-[r1c2]-2-[r8c2]=2=[r9c123]-2-[r9c5]-6-[r3c5]=6=[r3c7] => [r1c7]<>6
6-[r1c2]-2 (bivalue SL; SI)
[r1c2]-2-[r8c2] (peers WL; WI)
[r8c2]=2=[r9c123] (grouped SL; SI)
[r9c123]-2-[r9c5] (grouped SL; WI)
2-[r9c5]-6 (bivalue SL; SI)
[r9c5]-6-[r3c5] (bilocation SL; WI)
[r3c5]=6=[r3c7] (bilocation SL; SI)
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As far as I can tell, your logic is an impressive way to get around dealing with the ER pattern in [b7]. It relies on the fact that chains are logical steps that don't have actual eliminations associated with them -- like forcing networks -- and so you can ignore that a real assignment of [r1c2]=2 would have also forced [r9c2]<>2.
You may have found the perfect way to deal with ER patterns -- even though that wasn't your original intent 
[Addendum:] Oops!!! I forgot about using your approach in a continuous loop. In that case, a possible emimination in the ER pattern would be overlooked. Make that an almost perfect way to deal with ER patterns.
Last edited by daj95376 on Sun Nov 09, 2008 5:19 am; edited 1 time in total |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Nov 09, 2008 5:17 am Post subject: |
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There's no reason you couldn't write
6-[r1c2]-2-[r89c2]=2=[r9c123]-2-[r9c5]-6-[r3c5]=6=[r3c7] => [r1c7]<>6
provided that "=" is understood as "strong inference" rather than as "conjugate link".
I feel rather the opposite about the notations: Eureka makes each link explicit. With NL, it is not so obvious that the links/inferences alternate without doing more work. I guess it's what one is used to. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Nov 09, 2008 5:26 am Post subject: |
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| Asellus wrote: | There's no reason you couldn't write
6-[r1c2]-2-[r89c2]=2=[r9c123]-2-[r9c5]-6-[r3c5]=6=[r3c7] => [r1c7]<>6
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This is, in fact, what I would have done ... had it been my chain. My intent was to validate Norm's chain. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sun Nov 09, 2008 5:37 am Post subject: |
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| Quote: | | "A=B" means "not A implies B" |
agreed.
| Quote: | AXY ... XY
BXY ... XY
I don't know whether A, B or both A and B is true. But, I know that A and B cannot both be false |
agreed.
----------------------------------
| Quote: | | An ER that includes a candidate in the ERI cell can only be used as a strong inference in an AIC since it is possible that both the box-row segment and the box-column segment are true (if the ERI cell is true) |
and this by Danny...
| Quote: | | like forcing networks -- and so you can ignore that a real assignment of [r1c2]=2 would have also forced [r9c2]<>2. |
this might be the crux of conflicting thoughts in my head.
here is why...
if I am writing the chain starting in r1c2, then if the 2 in r1c2 is true, then obviously the 2 in r89c2 is false... this makes r9c13 2's true. right?
from the other direction...
the 2 in r9c5 is true... obviously makes 2's r9c123 false... right?
but...
is this also true ( from what danny mentioned)...
that if I assume the 2 in r1c2 is true that the 2 in r8c2 is false (ignoring the fact that the 2 in r9c2 would also be false)... making r9c123 true?
and...
from the other way.... if I assume the 2 in r9c5 is true this would make the 2 in r9c13 false (and ignoring the fact that it would also make the 2 in r9c2 false) thus making the 2's in r89c2 true.... right?
or wrong? |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sun Nov 09, 2008 5:54 am Post subject: |
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| Quote: | | I forgot about using your approach in a continuous loop. In that case, a possible emimination in the ER pattern would be overlooked |
in NL notation maybe?  |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Nov 09, 2008 6:03 am Post subject: |
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Norm,
I believe that Danny and I have agreed that you are right. And, in most situations, notating the ER portion of the AIC with the ERI cell contained in the group of only one side of the strong link (instead of in the groups on both sides) will work fine.
There is a potential problem, however, since in the case that the ERI cell itself is actually true, an AIC with the ERI cell included on only one side of the link will only propogate in one direction. I'm not sure if this would lead to an error, though Danny seems to believe that it can in certain circumstances.
I'll try to illustrate what I mean. Below are the chains with the three possible ways of notating the ER link. Below each I show the true-false propogations. Notice that only the first chain, with the ERI on both sides of the ER link, propogates in both directions from the ERI TRUE condition.
| Code: | If r9c2=2:
(6 = 2)r1c2-(2)r89c2=(2)r9c123-(2 = 6)r9c5-(6)r3c5=(6)r3c7
T<- F <- T T -> F ->T -> F -> T
(6 = 2)r1c2-(2)r8c2=(2)r9c123-(2 = 6)r9c5-(6)r3c5=(6)r3c7
?<- ? <- F <- T -> F ->T -> F -> T
(6 = 2)r1c2-(2)r89c2=(2)r9c13-(2 = 6)r9c5-(6)r3c5=(6)r3c7
T<- F <- T -> F -> ? ->? -> ? -> ? |
This is the reason that I prefer the first approach. But, without seeing an example of how the others might lead to an error, I can only appeal to aesthetics!
Meanwhile, it is difficult to learn and absorb AIC techniques without thinking of "if this is true, then that is false and..." etc. However, the goal is to learn to identify the various types of links and the situations in which they occur and then exploit them as alternating implications without thinking about individual truth values (i.e., no forcing). We let the alternating strong/weak inferences take care of that for us and reveal the underlying contradictions so that we can make eliminations and/or placements. That's when the training wheels come off!  |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Sun Nov 09, 2008 6:37 am Post subject: |
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| Quote: | | I believe that Danny and I have agreed that you are right |
keith, marty, I want this topic signed by both Danny and Asellus, sealed and delivered.
| Quote: | | That's when the training wheels come off! |
watch it !!!
| Quote: | | However, the goal is to learn to identify the various types of links and the situations in which they occur and then exploit them as alternating implications without thinking about individual truth values (i.e., no forcing). |
I hope everyone got this !
how else on this green/blue planet can I find AIC without going through the motions? I have to go through the motions. I must. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Nov 09, 2008 6:45 am Post subject: |
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I hasten to point out that I never said that I don't rely on those training wheels at times! |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Nov 09, 2008 12:59 pm Post subject: |
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Oh boy, do I hate the ERI cell and trying to explain it in an AIC chain.
A forcing chain eliminates all confusion because it allows two different streams, and doesn't confuse logical eliminations with real eliminations. (Note: forcing chains are only read left-to-right.)
| Code: | Stream 1: [r1c2] -6-[r1c7]
Stream 2: 6-[r1c2]-2-[r89c2]=2=[r9c13]-2-[r9c5]-6-[r3c5]=6=[r3c7] -6-[r1c7]
Q.E.D: [r1c7]<>6
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The problem occurs when we try to write an AIC where the notation must be understandable in both directions.
For left-to-right, this reads the same as real eliminations:
| Code: | 6-[r1c2]-2-[r89c2]=2=[r9c13]-2-[r9c5]-6-[r3c5]=6=[r3c7]
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For right-to-left, this reads the same as real eliminations:
| Code: | 6-[r1c2]-2-[r8c2]=2=[r9c123]-2-[r9c5]-6-[r3c5]=6=[r3c7]
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They can't be made the same unless there's an agreement on how to represent it in the notation. Asellus and I (and others) include the ERI cell twice.
| Code: | 6-[r1c2]-2-[r89c2]=2=[r9c123]-2-[r9c5]-6-[r3c5]=6=[r3c7] => [r1c7]<>6
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I've seen others take the smart way out and use a macro (like Asellus does for an ALS):
| Code: | 6-[r1c2]-2-ER(b7)-2-[r9c5]-6-[r3c5]=6=[r3c7] => [r1c7]<>6
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Sun Nov 09, 2008 3:10 pm Post subject: |
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| ....Looking at the previous posts, the important thing here is that there are a pair of naked 4's in Row 2 and a pair of naked 7's in Col 5. |
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