Diagram

[elizv]

I hate this game! Please remind me of the diagram -- thanks!

[noah]

Here's the diagram. Tell me if you have any problems getting to that point, and then if there are any questions that trip you up.



Good luck!

[jrdn_pearl]

i am not sure how you came to some of the conclusions on the diagram...for instance a few of the slots that you needed to bracket off, and i am still having trouble with your "dot" method that we discussed.

[noah]

This game is the one that is taught in the class recording lesson on open assignment, so I'll leave it to those to go through each and every inference. If you have a specific cross out that you don't understand, tell me and I'll explain it here. Most people get confused by the J rule. Remember, if J can't share any with L or M, than it must only have one play (since M must have two for other reasons).

As for the dots, let me try to explain: since there are only 3 types of papers (*, **, and ***). We don't know what each symbol stands for, just that there are 3 types. We'll say that J covers type 1 (one dot). Since J cannot share any with L, we'll give L a different type (* *), and since M has to cover two plays other than the one that J cover, it receives the two other types (** and ***). The confusing piece is that we don't know which type * is. But whatever play it is, that play will not show up in neither L nor M, and M will cover both other plays (and L will cover one of the plays M will cover).

Does that clear that part up?

[ptraye]

Why are 2 out of the 3 slots crossed out in K and 2 out of the 3 slots crossed out in L?

[ptraye]

Noah, i just figured it out. and, thanks for the diagram.

[noah]

Noah, i just figured it out. and, thanks for the diagram.

Nicely done!

[matthew.mainen]

I found frames to be the way to go on this one:



This kind of set up is very common. When you've got an item that can only go in two places, then you have three frames for it. When this item (in this case M) is linked to other items via rules, you can really start to fill the frame out.

[tyra.j.walker]

This Game is super annoying! Not because it is difficult but because the last rule contains a nuance that requires you to think in hardcore LSAT logic, as opposed to understanding the rule using real life assumptions. The last rule is "Exactly two of the students review exactly the same play or plays as each other". In the real world, we might read this as "Exactly two of the students review only the exact same play or plays as each other." Here's where we run into a wall. If we go with matthew.mainen's setup method, we know that it is not possible for exactly two students to review the same PLAY (singular) as each other, because, given the setup, there MUST be at least two SETS of students who review the same play as each other. Thus we move onto the second option: "Exactly two PLAYS". If you read this rule with real world logic you were bound to get stuck because you would have unnecessarily limited your setup to two options instead of three (as matthew correctly has it). Given O and M are the only students who can review multiple plays, you would have split the game so that O and M were ALWAYS together: either reviewing S and T, or reviewing T and U. However, you get to question 21 and realize you cannot find a correct answer - why is that? It's because, in reality, there was a third setup option you missed. O amd M can be together in S and U. Yep, that's right - O can review all three plays. At this point we realize that in LSAT logic, "Exactly two plays" does not mean O and M are ALWAYS together, but rather that they are together exactly twice. This does not preclude the possibility of O reviewing an additional play without M. This game is an example of dirty LSAT tricks because the makers certainly expect test-takers to fall for this "real world logic" pitfall . But if you always remember not to read anything more into the rules than they explicitly say (i.e. leave out real world syntax), then you will be golden .

[cgrosinger]

Is this game similar to another past game? Is it possible for another diagram to solve this particular game?

[ohthatpatrick]

It is possible for more than one diagram to solve any game, but I think there are definitive advantages to making the "choosers" the basis for your column, and the CHOICES (the stuff that everyone has "at least one" of) the flexible pieces.

These are what I would consider equivalent "OPTIONS" games

- Test 43, game 4 (lunch trucks)
- Test 12, game 3 (also about grabbing lunch, strangely)
- Test 35, game 2 (sunroof, leather interior, power windows)
- Test 47, game 3 (folk, jazz, rock, opera)

They all give us a handful of characters that will be choosing from "at least one" or "one or more" of a set of 3-4 options.

They always have rules that compare how many options one column has vs. another

And they always have rules about whether this column can have any matching options with that one

Have fun!

[VendelaG465]

I had a question in regards to Rule #2 "Neither Lopez nor Megregian reviews any play Jiang reviews." I get the inference that none of them can review three but how can we determine anything else numerical distribution wise or without knowing which elements are in either slots? how would I apply this rule in general for opening grouping games I find it to be very confusing

[ohthatpatrick]

When you're doing one of these "Options" games, where some handful of characters all have "one or more" of 3-4 options, you're always keeping track of the min/max for each character.

The game sets the min for each character at 1.

In this game, the max is 3 options for each character (most Options games have three options).

So there are two typical rules with automatic inferences:

1. X has more options than Y
This tells us that X can't be the minimum and that Y can't be the maximum.
Since X can't be the minimum, it's at least two options (draw a box in the 2nd slot of X's column).
Since Y can't be the maximum, it's not allowed to have three options (put a slash in the 3rd slot of Y's column).

2. W and Z have no options in common
This tells us that neither one of them can be the maximum. If W had all three options (we'll call them A, B, and C), then what would we give Z?
Conversely, if Z had all three options, then we wouldn't be able to give any to W.

So when you are "enemies" with another common (you can't have any of the same ingredients), then neither enemy can be the maximum (so put a slash in the 3rd slot of W's and Z's column).

--------------------------

The rule you asked about tells us that L and J are enemies and that M and J are enemies.
Since characters in enemy relationships can't have the max, we'd cross out the 3rd slot in M's, J's, and L's columns.

That's all this rule tells us, by itself. I'm not sure what else you were talking about; it seemed like maybe you were thinking some of the deductions we made combining this rule with others were coming only from this rule.

If you don't understand why characters in enemy relationships can't have the max,
try to do a scenario where M reviews all three movies (put S, T, and U in its column) and see if you can create a legal scenario.

(you can't, because we won't be able to give J or L any of those three plays)

Hope this helps.

[StratosM31]

Hard, but very interesting game The trick is to come to the inference that the number of spots at J, K, L, M can be determined up-front, otherwise it's gonna be very hard.

By the way, I didn't need to apply the last rule at all. Did this happen to anyone else?