Setup and Rule Diagrams

[Administrator]

Setup and Rule Diagram Explanation

This is a Grouping Game: Undefined.

The game scenario establishes that there are seven prospective courses available for a summer program, with the stipulation that at least one is offered. As the group size is undetermined, this game is Undefined.

There are four rules in this game, and each is diagrammed below:

June 06_M12_game#3_L6_explanations_game#4_setup_diagram_1.png

These four rules yield several inferences:

• Rule #2 and Rule #3 Combined
  
  The third rule requires that P be offered but the second rule requires that P not be offered. Thus, the sufficient conditions in each rule cannot occur simultaneously, and we can infer that L and S are never offered together:
  
  
  L S
  
  The presence of this inference actually makes the “but not both” phrase in the first rule superfluous; regardless of that phrase in the first rule, the combination of the second and third rules makes it impossible for L and S to be offered together.
  
  
  Rule #3 and Rule #4 Combined
  
  The fourth rule requires that Z be offered but the third rule requires that Z not be offered. Thus, the sufficient conditions in each rule cannot occur simultaneously, and we can infer that G and S are never offered together:
  
  
  G .....S
  
  
  Note that the presence of S in the two inferences means that when S is offered, neither G nor L is offered, and from the third rule Z cannot be offered either (although from that same rule P would be offered).
  
  Rule #2 and Rule #4 Combined
  
  The second rule contains a sufficient condition of G, and the fourth rule has G as the sufficient. Thus, these two rules can be linked:
  June 06_M12_game#3_L6_explanations_game#4_setup_diagram_2.png
  Of course, we can also add in the contrapositive of the third rule (which could also be described as recycling the first inference) and add the fact that S cannot be offered:
  
  June 06_M12_game#3_L6_explanations_game#4_setup_diagram_3.png
  Note also that the inference regarding S could have been made through Z as well.
  
  Thus, when L is offered, the fate of every course is determined except for that of M. This inference is tested in question #14.

The last point of note is that there are no randoms in this game. Every one of the variables is addressed in at least one of the rules, meaning that every variable can be checked against a rule.

The combination of the information above leads to the final setup for this game:

June 06_M12_game#3_L6_explanations_game#4_setup_diagram_4.png

[mherrera]

i diagrammed the whole sequence but i missed the O inference in spaces 7 and 8. can someone explain to me how i missed that

i got all rules right
[H_J]
[k_ _ L]

J---O
L---O
can you show me where i missed the inference or the connection to making this inference??
thanks

i missed q's 21 22

[Clay Cooper]

Hi mherrera,

Thanks for your question, and welcome to the forum! I hope you find it useful.

The inference that O must be seventh or eighth is not super easy to see. Here's how you get there, though:

The two blocks (the [H _ J] block and the [K _ _ L] block must always occupy at least five spaces. Even if we overlap the two blocks, they will still occupy at least five spaces, because the [K _ _ L] block occupies four spaces itself, and we can't fit the [H _ J] block within the [K _ _ L] block; if we add the smaller one to the bigger one we increase the bigger one's size by at least one blank. That means that these two blocks between them take up at least five spaces.

Also, since the first spot can only be F or G, we won't be able to place either of the blocks in the first spot; the earliest (fastest) they could start is in the second spot. The earliest they could end, therefore, is sixth; in other words, no matter what, the two blocks cannot finish earlier than sixth.

Since we are told that O must come after both J and L, that means that O will, no matter what, come after the end of both blocks. Since they can't finish earlier than sixth, that means O can only ever be seventh or eighth.

Does that help?

[mherrera]

oh ok now i see it thank you, i get it now. i went back and look at it and saw exactly what your talking about. thank you very much mr cooper.

[AspiringLawyer]

I had been having serious trouble with this problem set based on your given set-up. I wonder if the way I hash it out below (inherent chain) is also accurate? Or rather, does it ironically only work for this game?

Rule 1: M L or S but not both
Rule 2: L G not P
Rule 3: S P not Z
Rule 4: G H and Z

Your First Inference: L S
Why? Rules 2 and 3 have P in common. L not P, S P. Need to take contrapositive of one of those rules. Let's do Rule 3: not P not S. New chain is L not P not S. Condense to L not S. Contrapositive of new chain is S P not L. Condense to S not L. Therefore, L S.

Your Third Inference: Combining Rules 2 and 4
Why? They have G in common. I illustrated the outcome slightly differently. L G not P + H + Z

Your Second Inference: G S
Why? Rules 3 and 4 have Z in common. Similar process to first inference. S not Z, G Z. Take contrapositive of Rule 4: not Z not G. New chain is S not Z not G. Condense to S not G. Contrapositive of new chain is G Z not S. Condense to G not S. Therefore, G S.

...bear with me... After attempting all of the questions, my different version of the set-up only got me stuck on Question 17. Because the way I proposed above would make A correct, too [based on Rule 2: L G not P]. Therefore, how should I know to not chain a global rule? I am sorry that my wording is confusing. I guess I want to know why do we have to diagram it as L G + not P instead of L G not P?

Thank you for your patience!

[James Finch]

Hi AL,

It sounds like the issue that you're having with the diagram is in keeping the logical chain intact. The arrows only flow one way, and if you make a mistaken inference, that will cause you to mistakenly eliminate possibilities.

The reason that we diagram L G + P is to allow for the possibility that we don't have M, L or S present, and thus may have both G and P present, as G's inclusion alone does not force P out: only L's inclusion does that. You can chain the rules together, but do so in a manner that visually represents that G and P do not influence the presence of the other, but are influenced by L, which may require branching paths. Here it would look like:

L G H + Z

and

L P S

Let me know if this clears things up.

[TonySteep]

In addition to the two inferences above, can you also assume the following two not-both rules:
1) P L
2) S Z

[Adam Tyson]

Absolutely, Tony! Those two are essentially built right into the diagrams from the beginning, but tracking them off to the side with the additional double-not arrows is a really good idea. Well done!

[T.B.Justin]

Hey,

I am confused.

Can we take the contrapositive of the first rule in this game to work into the conditional diagrams, it appears that may not be warranted in this instance, is that because of the (but not both clause). I ran into this debacle while attempting question #15, and managed to have a sense of the correct answer choice (literature), which I understand the logical chain and why that answer is correct now, however this brings me back to the first rule, can we infer that as a consequence of H not being offered, M cannot be offered, since in my current opinion, if M is offered that starts a chain reaction leading to that H must be offered.

Thank you, I am grateful for your time.

[Jay Donnell]

Hey T.B.!

This game presents one of the nastiest conditional chains in a grouping game to date on the LSAT.

A large part of the complexity, and particularly in the confusion here in number 15, involves the apparent ambiguity of Rule One.

We know that if Mathematics is offered, then at least one of either Literature or Sociology must be taken, but both those courses cannot be taken.

That rule seems especially funky, but by the end of the conditional chains we realize that taking Literature eventually kicks out taking Sociology, and vice versa with the contrapositive.

( L --> ~G --> Z --> ~S) ( S --> ~Z --> ~G --> ~L)

So, the chain(s) beginning with selecting M basically break off into two different possible universes, where M would force us to select L (and its consequences) or S, but we couldn't possibly select both.

The contrapositive of rule one would allow for us to prove that Mathematics cannot be taken, but we would have to know that either BOTH or NEITHER of L and S are selected.

I find a helpful shortcut is helpful for a situation like this. I diagram (when I have the luxury of a pencil rather than a keyboard) this rule as: M --> L/S (with a box around the L/S). That implies to me that we need exactly one from that pair. The contrapositive would then look like ~L/S --> ~M (still with a box around the L/S, and ~ functioning as a slash through the variables) which implies that if it's not exactly one of L/S (so, either 0 or 2), then M cannot be selected.

For 15, the lack of H forces us to know that L is not selected, but without knowing the status of S we cannot force M out of the game.


Hope that helps!