Setup and Rule Diagrams

[Dave Killoran]

Setup and Rule Diagram Explanation

This is a Grouping: Defined Moving, Balanced, Identify the Templates. game.

The game scenario indicates that six doctors are at exactly one of two clinics:

PT34-June2001_LGE-G1_srd1.png

Because each doctor must be at either Souderton or Randsborough, this game features a two-value system. Thus, if a doctor is not at Souderton, he or she must be at Randsborough; if a doctor is not at Randsborough, he or she must be at Souderton. The two-value system leads to a number of interesting contrapositives. Ultimately, the game is sufficiently restricted to lead to four templates. First, let’s examine each rule, with the rule diagram presented first, followed by the contrapositive, which will be translated in order to account for the effects of the two-value system:

PT34-June2001_LGE-G1_srd2.png

Of course, several of these rules and contrapositives can be combined, such as in the following:

PT34-June2001_LGE-G1_srd3.png

In that process of linking the rules together, you might notice a curious feature of the relationships: the third rule can never be enacted. The third rule indicates that if L is at S, then N and P are at R. But from the fourth rule when N is at R, then O is at R, and from the fifth rule when P is at R, then ultimately O is at S. So, attempting to put N and P at R is impossible since it requires O to be at both clinics. Thus, we can draw one of the challenging inferences of the game: L can never be at Souderton, and thus L must always be at Randsborough. We can also ignore the third rule going forward since it is effectively dead.

With L assigned to R, there are only five variables remaining to track, and four active rules (all the rules except #3). Notably, O appears in three of the four "live" rules, making O very powerful. We can take the relationships a bit further, and make four templates based on O's placement, which ultimately reveal there are only six total possibilities:

PT34-June2001_LGE-G1_srd4.png

[pcrandall]

Hi! I was wondering if anyone could post the game set up for this...I was having some trouble with this one. Thanks

[Adam Tyson]

I'll do my best!

First, the base is the two groups - R and S. The variables are JKLNOP.

The rules are:

J (s) -> K (r)
J (r) -> O (s)
L (s) - > N (r) and P (r)
N(r) -> O(r)
P(r) -> K (s) and J (s)

Now we just do our inferences, which will be conditional chains. The big one here is this:

L(s) -> N(r) -> O(r) -> J(s) (contrapositive of the second rule) ->K(r) ->P(s) (contra of the last rule).

So, if L is in Group S, we have a problem - P is supposed to be in R, but the chain forces him into S. That means L must not be in S, and must therefore be in R.

We can also do the smaller chain that starts with P in R and yields J in R (by way of K being in S) as well as O in R.

Check the various chains and contrapositives on the remaining rules and you will have full command of this game. Hope that helps!

Adam M. Tyson
PowerScore LSAT Instructor

[pcrandall]

Thanks a lot!

[PowerSteve]

Game 4:
I didn't have the time to even attempt this game while taking the June '01 LSAT under testing conditions, but when I did attempt it afterward, I found it extremely annoying and I was not able to finish it in a reasonable amount of time. It seemed to me that all I could do was list a bunch of conditional reasoning statements (I had thirteen which included contrapositives that I wrote out since there was dual-options going on) and then attack the problems. The local questions weren't so bad because they basically guided me to look at the appropriate conditional statements, but the global questions (namely, numbers 21 and 23) were extremely difficult for me. Please let me know how I should have actually went about this game in general, and especially how I should have attacked questions 21 and 23.

Thanks,
Steve

[Nikki Siclunov]

Regarding Game 4 in June 2001, indeed this is a rather annoying Undefined Grouping game. However, the important thing is to connect the rules in a way that forms an easily readable conditional chain:

N(r) --> O(r) --> J(s) --> K(r) --> P(s)

Also, make the connection (above) that O(r) --> P(s) irrespective of any other rules/inferences.

Now consider the last remaining rule:

L(s) --> N(r) + P(r)

Based on the chain above, you can never have both N(r) and P(r). Consequently, L cannot be at (s).

You are correct in that due to the dual value system in this game, each contrapositive produces a positive statement. If you like, you can run the contrapositive chain backwards, but personally I would prefer not to clutter my space with rules that mean the same thing. Know the contrapositives and use them whenever necessary, but whether you decide to write them down is a matter of personal preference.

Onto the Global questions. In Question 21, they ask us to identify the minimum number of doctors that could be at (s). If we want to put as many doctors at (r) as possible, the conditional chain reveals that we still need to put at least 2 doctors at (s) - J and P.

The chain also shows that the correct answer to Question 23 is (E): if N(r) --> P(s). Answer choice (A) is incorrect, because although J and K cannot both be at (s), they can both be at (r) without breaking the chain.

As long as you understand the relationships between the different conditional rules, you will have a solid grasp on virtually any Undefined Grouping game.

[PowerSteve]

Thanks for the replies Nikki. They are extremely helpful and very much appreciated. As a quick follow-up to your post regarding game 4: When an undefined grouping game is so conditional in nature, would you stay away from diagramming and instead focus on connecting as many of the conditionals as possible? There isn't really much you'd be able to diagram in this case anyway, other than the fact that L must be in r. If the grouping game were defined, however, I would think it would make sense to consider identifying either templates or even possibilities. Is my thought process correct here?

[Nikki Siclunov]

Absolutely. With Undefined grouping games, it's rather pointless to diagram the two "in/out" groups as part of your main set-up, since their size is unknown. Of course, you can always make a little chart and put L in the (r) group.

When you solve the local questions, however, it makes sense to draw a chart with two columns (s) and (r). Using the local condition, work off your main conditional chain and "plug in" the variables that must go in each group. Any variable that can still go in either group should be kept outside the chart. That way you'll have a much clearer sense of who is in (s), who is in (r), and who can still go in either group. I hope this makes sense

[PowerSteve]

As always, you've been extremely helpful! Thanks again!

[RayMiller]

This grouping game's rules has me stumped. The combination of the third, fourth and fifth rule doesn't make sense to me. They seem likmoste they're contradictory, but I'm sure the problem is most likely my inference.

How I diagrammed the rules and contapositives:
1. J at S K at R
K at S J at R
J at S K at S

2. J at R O at S
O at R J at S
J at R O at R

3. L at S N at R and P at R
N at S or P at S L at R

4. N at R O at R
O at S N at S

5. P at R K at S and O at S
K at R or O at R P at S

My problem: When I linked them beginning with L at S, N and P both are at R. But when this happens N at R puts O at R. But P at R puts O at S. So that mean that L can't be at S? Or have I totally inferred and deduced incorrectly?