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PT47, S4, G2 A lighting control panel has

by rob.schimmel Fri May 22, 2009 6:00 pm

I got smoked by this game. I thought I was close, but think I misunderstood the condition saying that the switch equal to the total number of switches on must be on as well. It would be very helpful to see this game worked. Thanks.
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Re: E47, S4, P 6-11

by noah Fri May 22, 2009 8:28 pm

Yeah, it's a tricky one. Two takeaways from this game:

1. Make it your standard to set up your chain elements as they come. It's too easy to get stuck laying them in the original order and then not noticing that the whole diagram could be a lot neater.

2. When there are only a few constraints and one of them is odd, expect that it'll play a role in every question, so focus on it when you're hunting for inferences.

Tell me how it goes when you replay it . . .
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Re: E47, S4, P 6-11

by rob.schimmel Tue May 26, 2009 11:02 am

The the cleaner diagram and pre-game thought process approach were very helpful...100% accuracy second time around vs. 80% inaccuracy first time through. One open ended question, would you say that your diagram on this one is a sort of happy marriage between the "original" and "modified" approach in the logic chain chapter of the LG Strategy Guide? I tried the purely modified approach but found myself too slow with it. With this approach, listing as they come but in the same order on both sides of the chain, I thought it was a good balance with respect to speed / neatness / accuracy. One not so open ended question - would you agree that if we see a logic chain question type like E50, S3, P 6-11, with some type of subsets defined (i.e. parents, students, teachers), then to go with the standard list approach and not the list as they come? Thanks.
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Re: E47, S4, P 6-11

by noah Tue May 26, 2009 7:17 pm

rob.schimmel Wrote:The the cleaner diagram and pre-game thought process approach were very helpful...100% accuracy second time around vs. 80% inaccuracy first time through. One open ended question, would you say that your diagram on this one is a sort of happy marriage between the "original" and "modified" approach in the logic chain chapter of the LG Strategy Guide? I tried the purely modified approach but found myself too slow with it. With this approach, listing as they come but in the same order on both sides of the chain, I thought it was a good balance with respect to speed / neatness / accuracy. One not so open ended question - would you agree that if we see a logic chain question type like E50, S3, P 6-11, with some type of subsets defined (i.e. parents, students, teachers), then to go with the standard list approach and not the list as they come? Thanks.


Yeah, I agree with your thinking. However, I have always put the two columns the same -- even if only the "In" version of an element is given, I put both down -- though I'll have to try out what you're suggesting. I fear not having them parallel will make it hard to read. If there are categories of elements, I agree, definitely list them first, by category, as given. Glad to hear that 80% = getting smoked!
 
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Re: PT47, S4, G2 A lighting control panel has

by nneamakaeze Mon Nov 30, 2009 12:39 am

Im having some trouble understanding what the last constraint means...
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Re: PT47, S4, G2 A lighting control panel has

by noah Mon Nov 30, 2009 11:36 am

Yeah, that's a tricky one! It means that if 4 switches are in the on position, than switch number 4 is one of those. If 5 switches are on, then switch number 5 is definitely on.

If there was only one switch on, it would have to be switch #1.

Does that clear it up?
 
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Re: PT47, S4, G2 A lighting control panel has

by nneamakaeze Tue Dec 01, 2009 4:36 am

Definitely does...thanks!
 
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Re: Diagram

by cdjmarmon Wed May 09, 2012 1:00 pm

Can we infer that the maximum number of switchs that can be on is 5 based on the fact we can't have 6 or 7 switchs on?

I did and it helped significantly by realizing, for instance, if switchs 1 and 3 are off then I know the only CL possibiliities is going to be 2, 4, and 5 because I can't have a CL of 6,7,1,or 3.

From there I just workout possiblities with the CL of 2,4, and 5.
 
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Re: Diagram

by timmydoeslsat Thu May 10, 2012 3:54 pm

We can infer that from the first rule that at least 1 switch must be off:

1 ---> ~3 and ~5

4 ---> ~2 and ~5

So for a maximum, I would start by considering if all can be in. Nope. Can 6 of them be in?

Well, that would not work because I would have to put 1 out. (If I did not put 1 out, 1 would be in, and 1 would force two variables out!)

But now I am left with my 4 ---> ~2 and ~5 rule to consider, which also forces at least one out. So six cannot work.

What about five. The easiest way to do this is to place 1 and 4 out. Now we have no rules to deal with other than the CL rule. (We have rules regarding 1 and 4 being in. We have no rules when 1 and 4 are out. We free to do anything when 1 and 4 are out.)

23567 ...-... 14

This works,
 
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Re: Diagram

by tzyc Wed Jul 25, 2012 12:02 pm

I'm still having trouble to see the third constrain :oops:
The correct answer for question #6 is B, but why not C??
Because it includes switch 7 and that would mean all number would be on? (but then I think that switch 6 is on would mean a similar thing...)
I feel the constraints contradict each other...?? (If switch 7 is on means all 7 switches are on, then isnt' it contradicts the other 2 constraints...? when can switch 7 be on? Or did I make "Mistaken negation"?? :( )
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Re: Diagram

by noah Wed Jul 25, 2012 12:53 pm

tz_strawberry Wrote:I'm still having trouble to see the third constrain :oops:
The correct answer for question #6 is B, but why not C??
Because it includes switch 7 and that would mean all number would be on? (but then I think that switch 6 is on would mean a similar thing...)
I feel the constraints contradict each other...?? (If switch 7 is on means all 7 switches are on, then isnt' it contradicts the other 2 constraints...? when can switch 7 be on? Or did I make "Mistaken negation"?? :( )

That last rule is a doozy!

It means, for example, that if 4 switches in total are on, switch #4 has to be on.

So, if there's only 1 switch in total on, which switch would it be?

Think before reading on.
.
.
.
.
.
.
It'd have to be switch #1!

There's a discussion above about the fact that there can't be seven switches on -- but that doesn't mean that switch #7 can't be on. You did reverse the logic there. It's if 7 total --> #7 on (which is impossible), and I think you incorrectly thought: #7 on --> 7 total, and since 7 total is impossible--because of the first two rules--then #7 can't be on. But, it's fine for #7 to be on, as we see in (B) to question 6.

For #6, the discussion is here if you have follow-up questions: q6-t6318.html

I hope that helps!