Setup and Rule Diagrams

[Jon Denning]

Finally, some Grouping.

The fourth game on the June 2017 LSAT is a standard Grouping Game where four product managers—F, G, H, and I—visit three cities—M, S, and T. We're then told that each manager visits at least one city (we need to use each of F G H I) and each city is visited by two people.

So we'll end up with the three cities, M S T, as the base, with two stacks above each letter to represent that city's two visitors. The numbers here don't quite match up, and that deserves our attention straight away: in those six total spaces we need to use all four managers at least once—so that's four spaces filled—and then figure out how to fill the remaining two "empty" spots.

We have two distribution possibilities at this point then:

3-1-1-1 where one manager visits all three cities, and the other three managers visit one city apiece

2-2-1-1- where two managers visit two cities, and the other two managers each visit one city

Let's see if the rules can help us determine which distribution is at work (note: it's possible at this point that both distributions could be used for this game; narrowing it down to just one though, if we can, would be nice).

Here are the rules:

I visits exactly 2 cities. Well then. Problem solved. It's the 2-2-1-1 distribution of people to cities, with I as a 2.
Knowing how important numbers tend to be for Grouping I'd keep a close eye on the other three variables to
see if I can figure out who else is a double and who the two singles are (whether it's constant, like I, or question
to question). And always remember the overall distribution as you go!

F and H don't visit the same city. We could show this as a vertical Not Block (probably my choice), or with
a Double Not arrow: F H

GM HT , and the contrapositive HT GM (I showed the cities as subscripts, but it's
entirely up to you—do whatever you want as long as you have the order correct)

Lastly, a Not Law for G under the S column.

So what inferences can we draw from this?

In games with variable sets this small (four people, three cities, two spots on each city...) it's extremely common to find inferences anywhere you can begin to eliminate options. In short, when you have few options to begin with, and begin to reduce them even further, you often find yourself in situations where things MUST occur!

So the first thing I would do is look to my most restricted city, S. We know from the last rule that G cannot go there, so that leaves just F, H, and I available. But we also know that F and H cannot visit the same city as each other...that means at most they could take just one of the two S spots. What that means then is that I must visit S! G can't, F and H both can't...I must.

So one of I's two visits is known: S. The other S spot? Either F or H. That's absolutely worth showing:

__ F/H .....__
__ I_ __
M S T

Is there more that we could discover if we keep testing hypotheticals or toying with possibilities? Perhaps. But this is the stage where I'd forego all of that and eagerly move to the seven questions. It's also the final game of the test, so if this is the last game you're attempting (i.e. you've done the other three already), how much time you choose to spend setting it up can be in part determined by how much time you have remaining for the section. Assuming time is tight though, I'd move on.

[Etsevdos]

Time permitting, thoughts on setting up 3 scenarios with G in M, G in T, and G in M and T. It would establish three templates and establish I in S.

[Claire Horan]

Hi Etsevdos,

It would be fine to do templates, but I would advise just two templates: one with G in Manila and one with G in Tokyo. The reason is that this is an either/or. You know that if G is not in Manila then G must be in Tokyo. The purpose of templates is to show EVERY possible solution, so I would not make a third diagram with G in both places. The reason is that we don't know if G is used twice, and you are not going to draw a template where F is used twice and another where H is used twice. It would be better to just have the two templates I described, and leave it open which of G, H, or F is used twice.

I hope that is clear!

[ples2020]

I actually tried setting up two templates: one with both of II in M and S, and one with II in S and T. From here, you can deduce that in scenario 1, T must have G and one of F and H, and in scenario 2, M must have G and one of F and H. Because of the conditional rule, in scenario 2 you can also deduce that T must have I and H, since G is in M. You can't make any further deductions for scenario 1, but you just have to work with the one conditional rule remaining. I didn't have time to do this during the actual exam, but looking at it now I don't think it would have taken lots of time if you knew what to look for.

[Rachael Wilkenfeld]

I think those would be reasonable templates, ples. I'm not sure the payoff is enough here myself to find them useful. But it's a pretty low time investment, so you can certainly do it. The related inference you can pull is that one of M/T has G with F/H and the other has I. That's a bit helpful for the game overall, so it's certainly not a waste of time to run the templates here. You could also see that without templates though. We know that two cities have I. The last city can't have I (because there are only 2 Is) so it must have G and either F or H.

Excellent work!

[Vasilia]

HI, i am a little hesitant to do the contrapositive of -Ht ---> -Gm, because i remember in a similar scenario when each person can go more than once, that contrapositive was misleading. But i cannot be sure, and wondering what is the difference here that made this contrapositive work fine. Thanks!

[Adam Tyson]

Never hesitate to do a contrapositive, Vasilia, because it's always true! You may be thinking of certain situations involving two-value systems. In a true two-value system, if you are not one thing, you must be the other thing. For example, in a so-called "In/Out" game, every variable is either in or else out. So if you have a conditional rule that says "if X is in, Y is out," the contrapositive could be expressed as "if Y is not out, X is not in," but because it's a two-value system you can also say "if Y is in, X is out."

Or imagine a game with two teams, a Red team and a Blue team, where every variable is on exactly one team. Now consider "if X is on the Blue team, Y is on the Red team." The contrapositive could be written as "if Y is not on the Red team, X is not on the Blue team," but you could also say "if Y is on the Blue team, X is on the Red team."

But what if someone could be on both teams? Then we no longer have a true two-value system, because being on one team doesn't mean you are not also on the other team. In that case, the correct contrapositive would be the first one. The second one would be incorrect!

In short, the contrapositive is always correct, and you shouldn't be afraid to go there (and in fact, you absolutely SHOULD go there to really see the full relationship and make all the right inferences). You just can't always equate one thing occurring to another thing not occurring.