Setup and Rule Diagrams

[Dave Killoran]

Setup and Rule Diagram Explanation

This is a Grouping, Balanced, Defined-Moving, Numerical Distribution game.

This a Defined-Moving, Balanced Grouping game. It has been chosen for this section since it features two Numerical Distributions that control the placement of film buffs to films. Although virtually all games contain a Numerical Distribution, the distribution becomes a significant element if there are multiple distributions, or if the single distribution is unusual. This game is one of the former.

The information in the game scenario establishes that there are seven film buffs attending a showing
of three movies. Each film buff sees exactly one film. The first rule establishes two fixed distributions:

D98_Game_#3_setup_diagram 1.png

The two fixed distributions create two distinctly different scenarios. And since each scenario requires a different analysis, the best strategy is to create two templates, one for the 1-2-4 distribution, and another for the 2-4-1 distribution:

D98_Game_#3_setup_diagram 2.png

The rules can now be considered:

D98_Game_#3_setup_diagram 3.png

Of course, the rules impact each template:

D98_Game_#3_setup_diagram 4.png

Let us analyze the two diagrams. The fifth rule states that L sees the Hitchcock film. In the 1-2-4 this leaves one open space at Fellini, one open space at Hitchcock, and four open spaces at Kurosawa.
But the third rule states that V and Y see the same film as each other, so in the 1-2-4 it can be inferred that V and Y see the Kurosawa film. In the 2-4-1 we can infer that V and Y see either the Fellini film or the Hitchcock film, but at the moment it is uncertain which one. G cannot see the Hitchcock film, and so Not Laws are drawn on both templates, and G split-options are placed on Fellini and Kurosawa.

The two negative grouping rules help fill in both templates. In the 1-2-4, there are four remaining spaces for G, R, I, and M: one space at Fellini, one space at Hitchcock, and two spaces at Kurosawa. Since G and R, as well as I and M, cannot see the same film, it can be inferred that one of G and R must see the Kurosawa film, and one of I and M must see the Kurosawa film.

In the 2-4-1 there are still six open spaces: two at Fellini, three at Hitchcock, and one at Kurosawa. Of the six remaining variables, two—V and Y—must see the same film. If V and Y see the Fellini film, then there would be three open spaces at Hitchcock and one open space at Kurosawa. Since G and R cannot see the same film, one must see Hitchcock and the other must see Kurosawa. That leaves two spaces at Hitchcock for I and M. But wait, I and M cannot see the same film, and so this scenario causes a violation of the rules. Essentially, when V and Y see the Fellini film, there are not enough remaining spaces to properly separate G, R, I, and M. It can therefore be inferred that in the 2-4-1 distribution V and Y see the Hitchcock film.

In both templates there are still several possible solutions, and in the next section we will discuss Identifying the Templates further. For now, it is important to be aware of G, R, I, and M since they are the only variables still in play.

[reop6780]

I had problem with possibilities of numerical distribution.

From "each of the film buffs sees exactly one of the three film," I know 7 buffs choose one from 3 films.

*However, I do not see any indication that 3 films are chosen at least once*

Thus, I came up with several possibilities of numerical distribution such as,

016, 025, 034, 115, 124, 133, 223

including the possibility that 7 buffs choose only 2 of 3. (007, which implies all of 7 buffs choose exactly same movie is excluded due to additional rules)

016, 133, 223 are further excluded due to other rules, but the options given are still too many to work on.

I went to answer and explanation page, and the explanation started with 124 set only.

I want to know how to come up with only one numerical distribution that simply.
(I must have missed some valuable rules, which I am not aware of)

First of all I want to make sure whether "each of the film buffs sees exactly one of the three film," does or does not exclude possibility that 7 buffs chooses to watch the same one movie or two movies out of three.

Thank you

[Dave Killoran]

This is the game on page 479 in the 2013 editions of the LGB.

Hi Hyun,

Thanks for the question. I think the phrase "three films are shown" is intended by the test makers to indicate that all three films are ultimately reviewed. You see this type of language elsewhere in LG as well, and consider the difference in how this is stated vs something like "there are three films available to be reviewed." However, let's simply assume that that phrase was never intended to mean that each film was reviewed. Does that change anything? No, so let's look at why.

As I state in the book, the first rule establishes two fixed distributions. That first rule indicates that exactly twice as many film buffs see H as see F. What are the possibilities under that?

• If 1 buff sees F, then 2 see H. Where are the others? All 4 have to see K--they have no other choice. That yields a 1-2-4 distribution.
  
  If 2 buffs see F, then 4 see H. Where is the other? The remaining 1 has to see K. That yields a 2-4-1 distribution.
  
  If 3 buffs see F, then 6 would have to see H, but that is impossible since there are only 7 buffs.
  
  What about the idea of 0-1-6, or some of the other variants you mention? Well, mathematically, 1 is not exactly twice the number of 0, which eliminates that possibility. That reasoning also eliminates a lot of other possibilities.

So, regardless of how you look at it, you end up with just those two distributions.

Please let me know if that helps. Thanks!

[reop6780]

Apparently simple mathematical truth helped me a lot!

Thanks !

[DlarehAtsok]

"Three films are shown ... The following restrictions apply:
Exactly twice as many of the film buffs see the the H as see the F".

From the later rules, it is clear that the distribution 0-0-7 (with all the film buffs in the K group) is impossible. However, say that the other rules did not specify that. Is a 0 group exactly twice as big as another 0 group in LSAT Logic Games?

[Adam Tyson]

Interesting question! But no. While it is mathematically correct that 2x0=0, the end result is not one group that is twice as big as another, but two that are equal. When you get a rule like that, the smaller group must have at least 1 in it.

The fact that you are asking this question, as interesting as it is, makes me a little concerned about your approach to the test. Some students tend to overanalyze and overcomplicate things, looking for ways to argue that a wrong answer could be right or that a right answer looked at in a different way might be wrong. This is a road you should not go down, not even to explore it philosophically, for fear of getting lost. Having a creative approach to problem solving will serve you well as a lawyer, and somewhat as a law student, but it will do you nothing but harm in preparing for this test. The goal is to pick the best answer of the five you are given - not the right answer, not a perfect answer, not even a good answer, but the best of the ones presented. That means you should do nothing to help any of the answers - no "well, that kind of works if you look at it this way". No arguing for one answer or against another. No intellectual exercises. The authors of this test are, in a way, your enemy, and that sort of thinking gives aid and comfort to that enemy.

So, while I commend you for your creativity, and I actually kind of love that you thought of this, do your best to take a more boring and straight-forward approach. The reward will be a higher score, which, while it is not as stimulating as arguing the finer points, is, after all, the goal.

[srcline@noctrl.edu]

Hello

So This is a defined moving balanced game with a numerical distribution. My biggest issue with this game is the numerical distribution ascpet. So I know this game is in the LG Bible and I have been staring at this game for 15 minutes trying to figure it out. I know the numerical distribution has to add up to 7. The first rule is giving me the most trouble

"Exactly twice as many of the film buffs see the Hitchcock film as see the Fellini film.

So the distribution would be F1 H2 and K4
So the distribution would be 1-2-4 totaling 7.
I understand this b/c of the first rule.

The second distribution is where I am running into problems. How do we get the 2-4-1 distribution as its described in the LG bible.

I diagrammed the rules as follows:

GR vertical not block
IM vertical not block
VY vertical not block

pg 480 of LR bible

Thankyou
Sarah

[Dave Killoran]

Hi Sarah,

It might help to think about this distribution in a totally different sense: what if you had 7 pieces of candy that you were giving to 3 friends (F, H, and K)? The main rules are that everyone gets at least one piece of candy and that H must get exactly twice as many pieces of candy as F. Using that scenario, what distributions can you make?

If you found that you were able to make the distributions above more easily, that tells you that it's not the numerical idea that's bothering you, but the context of the numbers. Sharing candy with friends is easy to understand; directors and film buffs less so.

Regardless of the context, the way to create these distributions is to be systematic. We know that each director has a film shown, which creates a fairly simple initial setup:

• F H K
  
  1 1 1
  
  Remainder = 4

Now, the rule about H being twice F comes into play:

• F H K
  
  1 2 1
  
  Remainder = 3

Well, with F and H fixed at 1:2, that means the 3 remaining films go to K:

• F H K
  
  1 2 4
  
  Remainder = 0

Is that our only scenario? Maybe, maybe not. Let's double the films that F sees, and see what happens:

• F H K
  
  2 1 1
  
  Remainder = 3

Now, the rule about H being twice F comes into play:

• F H K
  
  2 4 1
  
  Remainder = 0

Well, with F and H fixed at 2:4, that means that only 1 film is assigned to K. Since there are no remaining films, that is it for this distribution. But, 2-4-1 works just fine and violates no rules.



Are those the only two scenarios? Maybe, maybe not. Let's again increase the films that F sees, and see what happens:

• F H K
  
  3 1 1
  
  Remainder = 2

Now, the rule about H being twice F comes into play:

• F H K
  
  2 6 1
  
  Remainder = -2

Uh oh, this doesn't work—we don't have enough variables. So, this one fails and we are done. But we did establish the 1-2-4 and 2-4-1 distributions were just fine. Now we have those two to work with, and since they seem to "split" the game in two separate directions, we can attack it by making diagrams for each distribution scenario.

Please let me know if that helps. Thanks!