Diagram

[mkeat_theraptor]

Does anyone know how to diagram this game? I tried setting it up as 4 rows, with 2 vertical columns (one for shuttle stops, one for passengers), but I couldn't come close to finishing in time with this setup. The main issue I'm having is diagramming the rules about who is still on the van when it stops. Help?

[mkeat_theraptor]

Someone on TLS suggested one way to set it up:

http://www.top-law-schools.com/forums/v ... =6&t=84696

Any other ideas?

[noah]

Looking over old notes, when I played this game under real time pressure, I only used careful application of the rules (versus building up a set of overall inferences or frames). I just went back and re-played it and while I didn't come away with any huge inferences (though more than the first time), I still see this game's difficulty as being that it's easy to misinterpret the constraints and not understand that being on the bus when it reaches a stop either means you travel beyond it or disembark at that specific stop. Beyond missing that, I also imagine that some people reverse the idea of being on the bus when it reaches a stop.

For diagramming the constraints, I separated my diagram into two levels (it is a 3-D game after all) and kept my rules noted on the correct level. When it came to that last rule, as well as the second one, I wrote them out as relative constraints -- tree-style -- but I kept it clear whether elements were people or stops by placing each element on the correct level.

I think this game raises a good question about when to use frames. The L is in slot 1 or 2 might make one think to build frames around that, but there are not a bunch of other rules that will "connect" to that one. The more likely candidate is the last constraint, which creates two "if and only if" scenarios. In one, J travels to F and perhaps beyond and G travels to S and perhaps beyond. In the other, both J and G must get off before they reach F and S respectively. In this second option, we can eliminate J and G from the fourth slot for passengers. You should have eliminated V from that slot based on the third rule, leaving only R as a possibility for the last slot. That sounds useful. However, in the other frame, in which J makes it at least to F and G makes it at least to S, we cannot easily make any significant inferences without going into further "what ifs", i.e. sub-frames. Framing past the first division is only useful when you're really going to be able to quickly and thoroughly sketch out some seriously limited scenarios. The fact that the these frames didn't lead to many more inferences does not mean they prove useless; indeed, they helped me quickly attack some of the questions, as I noted in the attached diagram.

I'd love to hear if someone else came up with something else about this game or hear how the 2 options helped (or did not).

[farhadshekib]

I struggled with this game during the section, but luckily missed only 1 due to a couple of guesses. Upon review, however, the questions were pretty easy - they just required simple application of the rules.

Here is what I mean:

20)

Diagram:
M (1), L (2), (3) F/S, (4) F/S

Quick inferences: J can't get off first - (eliminate E).

Now, looking at the choices, I realized that V is the best option to test out. Why? because if V can get off first, then it eliminates (A) and (B).

M: V, L: J, F: G, S: R.

So, yes, this works - eliminate (A) and (B).

Now, deciding between (C) and (D), R is the best option to test out:

M: R, L: V, F: J, S: G.

Yup, eliminate and choose and (D).

21) F (1), L (2), M/S (3), M/S (4)

We know that J cannot get off first because V must get off before J; therefore, J is still on board when the van reaches Fundy.

This means that G must be on board when the van reaches Simcoe.

So V must be first (since R must be on board when the van reaches M).

This eliminates (A) and (B). (E) can also be eliminated because it has R getting off at L (2).

(C) can also be eliminated because it has G getting off at L (2), but we know that G has to be on board until the van reaches S (3 or 4).

22) has G getting off second, but he must be on board when the van reaches L and S.

We can create two templates:

S (1), L(2), (3) M/F, (4) M/F.

L (1), S (2), (3) M/F, 4 (M/F).

What does this tell us? R has to get off 3rd or 4th. In other words: R has to be on board when the van reaches both Simcoe and Los Atlos. Hence, (D) is correct.

23) If G is not board when the van reaches Simcoe, J is not on board when the van reaches Fundy. This also means that V is not on board when the van reaches Fundy because V gets off before J.

Hence, (D) is correct.

[cdjmarmon]

Im very confused on one aspect of this game.

Can a person get off of the bus at a particular stop and still be considered as still being on when they get to that stop?

For instance, can j get off at F but still be considered "on" when they get to F?

If so that would then trigger g to still be on board when the van reaches S. So could be get off at S and still be consider on? or does he have to get off sometime after S?

Other than that this game was easy, just too bad that little tid bit makes this game difficult.

[timmydoeslsat]

To get off at a particular stop means that you were on when you got to that bus stop. The fact that you got off of the bus at that stop does not change the fact that you were on the bus when that stop was made.

[genieb]

Could someone help me out on this one?

I don't understand the last rule, which says, "If Jasmine is still on board when the van reaches Fundy then Greg is still on board when the van reaches Simcoe;

*otherwise*, Greg is not still on board when the van reaches Simcoe."

What does the "otherwise" indicate? That "If J is *not* still on board when the van reaches F, then Greg is *not* still on board when the van reaches Simcoe?"

Would this be a good diagram?
(CAPS denote the stations and small letters denote the passengers)

If F-->j, then S-->g
(contrapositive:
If g-->S, then j-->F)

but (second half of rule 4)
if j --> F, then g-->S (negated)"?

Noah (first commentor) was mentioning that that constituted a double-arrow/ "if and only if" chain which I can see here as well (provided my diagramming above is correct). Would that be enough for us to make sense of that rule then?

When you look at Q19 answer choices D and E, since Fj and Sg are each paired up (and not before or after each other), does that mean rule 4 is not triggered, hence we can ignore that rule?

ANY HELP IS APPRECIATED!

[timmydoeslsat]

Yes, I would advise you to represent that as a double arrow situation.

This is a if and only if situation. We know that A is sufficient for B and it is also necessary.

A ---> B
~A ---> ~B

The thing about if and only if situations is that, in the context of #19 choices D and E, this rule will always be relevant. We will always either have it be true that J is still on when Fundy is reached or J is not still on when Fundy is reached.

For #19, I would go through the answer choices using the easiest rules first to apply for choices A through E. That would be the Los Altos rule being 1 or 2. Get rid of B. Next, R is still on when Mineola is reached. Get rid of D. J is on longer than V. Get rid of C.

We are now left with A and E. I would make the application of the last rule as simple as possible. Hunt for a situation where J is still on when Fundy is reached. This happens in A. Check to see if G is still on board when Simcoe is reached. He is not. Get rid of A.

E is correct.

[genieb]

Thanks a lot. It's much clearer now that I can see there are two double-arrow statements.

[ganbayou]

So the "otherwise" part mean "~A"? Otherwise, -B happen...
so -A=>-B?
The rules make the game really hard...

Thank you

[ohthatpatrick]

That's correct.

Picture me saying this
If it rains, the party will be in the house.
Otherwise, it will be in the yard.


That's giving you two different IF/THEN ideas
Rains --> party in da hizzy
~Rains --> party in the yard

[sengdykes]

Looking over old notes, when I played this game under real time pressure, I only used careful application of the rules (versus building up a set of overall inferences or frames). I just went back and re-played it and while I didn't come away with any huge inferences (though more than the first time), I still see this game's difficulty as being that it's easy to misinterpret the constraints and not understand that being on the bus when it reaches a stop either means you travel beyond it or disembark at that specific stop. Beyond missing that, I also imagine that some people reverse the idea of being on the bus when it reaches a stop.

For diagramming the constraints, I separated my diagram into two levels (it is a 3-D game after all) and kept my rules noted on the correct level. When it came to that last rule, as well as the second one, I wrote them out as relative constraints -- tree-style -- but I kept it clear whether elements were people or stops by placing each element on the correct level.

Referring to the solution posted for this problem, why is it that R must go last when G is off the bus before S and J is off before F (Option 2)?

[ohthatpatrick]

If g got off before S, we know that j got off before F (last rule).

so we have
v - j - F, demanding at least three of the four spots.

The latest you can put j is 3rd.
_ _ j _
_ _ _ F

v is 1 or 2. And g is before S, so g is also 1 or 2.

That makes r have to go in 4.

If we instead put j at its earliest spot, it would be 2nd.
_ j _ _
_ _ _ _

v would have to go first

v j _ _
_ _ _ _

g has to go before S, so g can't be last. g would go 3 and again r would go 4.

Does that make sense?

[MadisonE16]

Just in case people are still having trouble, this video really helped me understand the rules of the game visually.
https://7sage.com/lsat_explanations/lsa ... -4-game-4/

[mshinners]

Noah's Diagram

Note: There's a typo in the PDF. For question 21, the explanation states that G must go with M. That is incorrect: R must go with M. The explanation is correct; just the diagram is off.