Setup and Rule Diagrams

[Administrator]

The second game presents a Defined Grouping scenario (also, once again Unbalanced: Overloaded, as we'll see), where a history project has four students out of a group of six candidates assigned to one of four years: 1921-1924. This is clearly an Unbalanced situation (more people than spots), but the six students--L through Y--can, much like we saw in the first game, be balanced by including an out group:

__ __ __ __ | __ __

1 2 3 4

With this simple setup we've also introduced the Linear aspect of the game, since the 1-4 years (I'm abbreviating 1921-1924 with their units digits) of the assigned group are sequential, and at least one of the rules includes the concept of ordered placement (rule 4, as we'll see).

What is striking to me about this setup, and hopefully got your attention as well, is how small the out group is: only two members! The significance of that really can't be overstated. It means that as soon as we can determine just two people who are unassigned, the entirety of the project's membership is known. And sure enough, several of the questions revolve around that very idea.

The rules are primarily Grouping in nature, and present the following:

Space 3 must be either L or T. Note that this doesn't mean L or T can't go somewhere else (both L and T can both be used, so long as one is on 3). This only prohibits the other four students from going in that space.

M can only go in 1 or 2. Again, pay close attention to a third option for M: out. It's only when M is assigned to the project that she's in the first two years. I showed this simply as a Not Law for M under 4 (M not 3 is implied by the first rule), and a conditional: M --> 1 or 2.

If T is assigned then so is R. This is also conditional: T --> R. But the key, and it's one of the most powerful ideas in this game, is the contrapositive, No R --> No T. Think about what happens when R is out: T must also be out, which fills our two out spots and forces the other four people--L, M, O, and Y--to be in. That would put L on 3, M in 1 or 2, and O/Y in 4. By removing R an entire cascade of consequences occurs, and if you're focused on the heavily restricted unassigned pair you'll catch it.

Lastly, and combining both conditionality and linearity, the fourth rule tells us that if we have R then we get an OR block (in that order). Three things stand out with this rule: first, R becomes very limited in its placement options. For instance, R cannot go in 1 because then there would be no room for O ahead of it. Similarly, R cannot go in 4 because that would force O into 3, which must be either L or T. That means that if R is assigned, it MUST be in 2! That's the only position for R where O could immediately precede it. (Careful here: that doesn't mean that O is similarly limited; O is only tied to this rule when R is in, so until that happens O is free to do as he pleases)

Second, since R can only go in 2, forcing O into 1, then M is left with nowhere to go. Remember, M had to be 1 or 2, so if we fill those spots with the OR block then M is out. Take away? M and R can never be in together; one of the two must always be out! But can they both be out? No! Don't forget that if R is gone then T is gone too, and that fills our out group. So R or M in = the other out. R or M out = the other in.

The third notable aspect of rule 4 is its connection to rule 3, where T --> R. The shared R between these rules is important: T tells us R, which tells us the OR block is in 1 and 2. So if L is out and T is 3, then we'd have OR in 1-2. If T is 4, then L is 3 and OR is 1-2. But what if T is 1, or 2? Then we have a problem: O and R need those spaces if T is in. Inference? T cannot be in 1 or 2.

You're probably getting the sense at this point that there are a lot of Not Laws present in this game. You're right. Take a look:

Game 2.JPG

That's a lot of useful information, and it makes the questions entirely manageable.

[Etsevdos]

Can we make the inference that O must always be in. If o is out, then r is out, which then means T is out. We cannot have 3 in the out group.

[Adam Tyson]

You bet you can! Nice work!

[zah]

When I saw your Not laws for L on 1 and 2, it took me a while to figure out that it was because T would then be on 3, and then the OR block that follows wouldn't fit on 1 and 2.

I am wondering if you have any pointers on spotting these inferences (same with the above commenter's inference that O must always be included, I didn't spot that either). I understand that proper setup before tackling the questions is key, but I keep missing critical inferences in the setup that have me spending way too much time diagramming out answer choices for each question. Other than making sure to include the contrapositives on your setup and making any connections there, any tips on how to recognize inferences?

Also, this is my first post!

[Malila Robinson]

Hi Zah,
Welcome to the Forum!!
In general inferences like the ones you mentioned come with practice. When you are starting out the most important thing is that you can understand where the inferences are coming from when you look at the model diagrams. That understanding will train your brain to look for similar inferences in the future.

But essentially what an inference is is an unstated rule that becomes apparent when you look closely at the interaction of the rules of the game and the diagram that you have created from those rules.

So for now, don't get discouraged when you miss inferences. Just try to understand them and you will likely notice that you are making more and more inferences on your games as you continue your studies!
Hope that helps!
Malila

[hskhader]

I can see that we can derive many inferences by attacking the game using this approach, on the other hand I was wondering why we did not create templates using the first rule (L OR T in 3), as I tend to see that most games with this structure and also containing an either/ or rule like rule 1 tend to split the game into templates.

Thank You

[Brook Miscoski]

Hader,

You are correct that you can use templates/possibilities if you wish. There are a number of them, though, so you do need to use care if you do that.

The big inference is that since L/T blocks the 3rd space, if you use R, you have OR in 1 and 2. If you don't use R, then R and T are both unused, and you must use all of the remaining students. This presents you with two basic possibilities--either you must put OR in 1 and 2, or you must put RT out. If you do write templates, that's where I would start.

There is one basic template if RT is unused.
(M/O/Y) in 1 and 2
L in 3
(O/Y) in 4
R and T unused

There are two basic templates if OR is used.
OR in 1 and 2
(L/T) in 3 and 4
MY unused

or

OR in 1 and 2
(L/T) in 3
Y in 4
(L/T) unused and M unused

[rahimlsat]

Hi,

Can you please clarify whether it's possible for one student to be assigned more than one year? When I read it, I feel like it didn't exclude that possibility. All it said was that "each of the four years will have exactly one student assigned to it." But it did NOT say "no student can be assigned to more than one year". After having done so many questions, I feel like LSAC would have said that if one student could not be assigned to more than one year; but they didn't, so I assumed it was possible.

[Adam Tyson]

The answer is in the first line of the scenario, rahimlsat: "Four students will be assigned" means four students in the group. If one student goes twice, then you would only have three students assigned!

[srr021]

I treated this as a limited possibilities game with the following being my options:

1. In this order: O R T L/Y Out M Y/L
2. In this order ORLT Out MY
3. In this order ORLY Out MT
4. In this order M/Y/O O/Y/M L Y/O Out TR

I was able to answer all of the questions using these possibilities, but I noticed that you guys didn't list them in the setup. Am I missing possibilities that make it such that I should not have treated this game like a limited solution game?

Thanks in advance!