Diagram

[cyruswhittaker]

Can you please explain the setup and any inferences for this game?

The inference I came up with was that if H is selected, then I cannot be selected, and hence V must be selected, not allowing V to be selected.

Unfortuanetly, I'm still having trouble getting the answers to 20, 22, 24 with certainty. I can elminate a couple of the choices and resort to creating a possibility, but this becomes both time-consuming and not certain.

[ManhattanPrepLSAT1]

Here you go!

[mhb07]

I am really struggling to understand the diagram posted. In the game there are two years, and four people selected each year, one of whom is a chairman. But the frames in the diagram are split into two columns with only three people in each column, and no indication of chairmen. To me it doesn't seem to add up. Will some one please help me understand that diagram?

[timmydoeslsat]

I am really struggling to understand the diagram posted. In the game there are two years, and four people selected each year, one of whom is a chairman. But the frames in the diagram are split into two columns with only three people in each column, and no indication of chairmen. To me it doesn't seem to add up. Will some one please help me understand that diagram?

Good questions.

The diagram posted shows the two possibilities in this game. Either I is in or V is in. One and only one must be in of those two.

This diagram simply shows what happens that first year in both situations. Mainly because, I imagine, of the uncertainty surrounding the second year.

The diagram told us to keep in mind the factor of the chairperson. There is no practical or prudent fashion to frame chairman in this game.

The prompt of this game, in my opinion, revolves on this idea of 2 of 3 scientists. Once we determine a scientist who cannot be in, we know the rest are in.

I will go through the two frames listed.

#1: I is in.

This means that V is out since only one of those guys can (must) be in.

Since V is a scientist and he is out, that means that Y and Z are in! We know this because 2 L's and 2 S's are in each year for the panel.

So with Y now being in, H cannot be in. This leaves open F or G being in as the last L.


#2: V is in.

We know that I is out. We know that G is out. Those are 2 L's. That means the other two must be in!

F and H are both in. With H being in, that kicks Y out. We now have to place Z in since the out column is full!

[ManhattanPrepLSAT1]

Thanks Timmy for a great walk through of how I came up with the two scenarios for the possible combinations of lawmakers and scientists. It was perfect!

Taking the two sets of four a bit further though, we can see that if the panel includes F, H, V, and Z in the first year, the panel must include I, F/G, Y, and Z in the second year.

If the panel includes I, F/G, Y, and Z in the first year, it may either include I, G/F, Y, and Z or F, H, V, and Z in the second year.

All the while keeping in mind the rule that the chairman in the first year must not serve in the second year, and the chairman in the second year must have served in the first year.

[ptraye]

how did you come up with these wonderful diagrams? what was your thought process? why did you disregard the Chairperson in your setups, even though the prompt mentions it.

[ManhattanPrepLSAT1]

The first step is to realize that you're playing a Closed Binary Grouping game with subpools. These often have good framing opportunities which are typically based around a limited number of numerical distributions. This game has only one numerical distribution 2 lawmakers and 2 scientists.

If you play it with either I or V selected but not both, and then follow the rules, you'll find there's only a couple ways of selecting the panel. Then following the rules about the chairperson from year one leaving the panel in year two, you're able to constrain how the two possible selections work year to year.

[Nina]

Does every element has to appear at least once during the two years?

[hannahhodge]

Taking the two sets of four a bit further though, we can see that if the panel includes F, H, V, and Z in the first year, the panel must include I, F/G, Y, and Z in the second year.

If the panel includes I, F/G, Y, and Z in the first year, it may either include I, G/F, Y, and Z or F, H, V, and Z in the second year.]

I understand how you framing the first year, but am curious how your inferred the line-ups for the second year. Are you assuming that all of the elements must be used at least one time each? The game doesn't seem to require that this be the case.

Thanks for your help!

H

[maryadkins]

Great question!

You're right, it doesn't tell us that all of the elements must be used.

What Matt has done is this:

If V is in the first year, you have to have a set line-up for Year 1: F H V Z . (That's what you figure out just by applying the rules.) This means that if you were to have V in Year 2, you'd have that EXACT SAME line-up. But that would violate a rule...because someone in that line up in Year 1 was a chairperson! And the chairperson can't serve on Year 2.

So instead of V in Year 2, you have to have I. And once you have I, you know you have I, F/G, Y and Z.

The other option is to have I in Year 1, not V. If you have I in Year 1, again, you have I, F/G, Y and Z (by applying the rules you figure this out). So does this mean we can do what we did before—say that I can't be in Year 2 because we'd have the exact same line up as we did in Year 1 (and the chairperson has to sit out, remember!)? No, because of that "F or G" option. This means you could have F one year, G the other year, and all the other people could be the same, and it wouldn't violate the chairperson rule. So if you have I in Year One, you have two options for Year Two: 1) switch out F for G (or G for F...doesn't matter), or 2) replace I with V: F H V Z.

I hope this helps!

[ZoeL466]

I have a question about rule 3, which states "Each year, either Ivans or Vega, but not both, serves on the panel."

I interpreted the rule as meaning that per every year I/V could serve and in any given year IV together could not.

My interpretation allows for 3 frames. Frame 1: I I Frame 2: I V Frame 3: V I. These frames can be used without violating any other rules.

I've also seen the rule interpreted in a way that only two frames exist. Frame 1: I V and Frame 2: V I.

Could you explain if/why my interpretation is incorrect?

Thanks!

[ohthatpatrick]

You said:
My interpretation allows for 3 frames. Frame 1: I I Frame 2: I V Frame 3: V I. These frames can be used without violating any other rules.


It's hard for me interpret what you're typing there, but are you saying .... ?
Frame 1:
I is in, V is out

Frame 2:
V is in, I is out

Frame 3:
V and I are both in

If that's what you were saying, look at the rule again and see if you can see why Frame 3 would not be allowed:
"Each year, either Ivans or Vega, but not both, serves on the panel."

Incidentally, if someone were to consider this frame,
Frame 4:
V and I are both out
it would also violate the rule
"Each year, either Ivans or Vega, but not both, serves on the panel."


Hope this helps.

[ZoeL466]

Thanks, Patrick! I actually meant frames including I V (etc.) in year 1 and year 2 respectively. I think that the rules permit Frame 3, but just want to make sure.

Here are frames (Sorry about the formatting, read them like columns):

Frame 1

[Year 1] /// [Year 2]

[ I ] /// [ V ]

[Z/F/H] /// [ Z]

[ ] /// [ F]

[ ] /// [H]



Frame 2

[Year 1] /// [Year 2]

[ V ] /// [ I ]

[Z ] /// [Z/F/H (chair) ]

[ F ] /// [ ]

[H ] /// [ ]



Frame 3

[Year 1] /// [Year 2]

[ I ] /// [ I ]

[G/F (chair)] /// [ G/F]

[Y] /// [ Y]

[Z ] /// [Z]

[ohthatpatrick]

Yeah, those made sense to me.

I wasn't clear why sometimes you were inferring everything that goes with I being in, and other times not.

When we have I in, V is out, so Y and Z are in, so H is out, and so F/G is the remaining member.

So every time you put I into a group, wouldn't you have that group as
I, F/G, Y, Z
?

In frames 1 and 2, it didn't look like you had all that filled in.

Otherwise, looks great!