Setup and Rule Diagrams

[Dave Killoran]

Setup and Rule Diagram Explanation

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

This is a deceptively tricky game because it does not at first appear to be controlled by a Numerical Distribution. However, the presence of “singles,” “doubles,” and “triples” lends a natural numerical base to the game, and ultimately controls all of the possible outcomes. In our diagram, we will use the room capacity as the base, with a single represented by “1,” a double represented by “2,” and a triple represented by “3.” With those designations in mind, here is the initial diagram of the game:

d96_Game_#1_setup_diagram 1.png

The third and fourth rules are simple block rules, and can be added to the side of the diagram:

d96_Game_#1_setup_diagram 2.png

The first and second rules create a number of Not Laws, and when combined with the fourth rule, a number of inferences:

• Because no fourth-year student can be assigned to a triple, K and L Not Laws can be placed under the “3.” And, because K and P must share a room, a P Not Law also can be placed under the “3.” With three of the seven students eliminated from sharing a triple, there are only four possible candidates to live in a triple, meaning there is at most one triple in this game.
  
  Because no second-year student can be assigned to a single, S, T, and V Not Laws can be placed under the “1.” And, because K and P must share a room, K and P Not Laws also can be placed under the “1.” Thus, only L or R can be assigned to a single, meaning that there are at most two singles in this game.
  
  Because K and P must share a room, and it cannot be a single or a triple, they must share a double room.

Thus, the game scenario and rules combine to form the following diagram:

d96_Game_#1_setup_diagram 3.png

As mentioned before, the Numerical Distribution is hidden in this game. The singles, doubles, and triples form natural numerical limitations within the game, and the assignment of students-to-rooms further limits the possible distributions. In fact, only three distributions of students-to-rooms exist:

Single/Double/Triple Numerical Distributions:

d96_Game_#1_setup_diagram 4.png

In this distribution, there are two singles, which must be L and R. K and P are assigned to the sole double, and the remaining three students—S, T, and V—are assigned a triple. This is the only distribution where every student is assigned to a specific room.

d96_Game_#1_setup_diagram 5.png

In this distribution, there is one single, which must be L or R. K and P are assigned to one double, and the remaining four students—R/L, S, T, and V—are assigned in pairs to the remaining two doubles.

d96_Game_#1_setup_diagram 6.png

In this distribution, K and P are assigned to one of the doubles. L, a fourth-year student who cannot be assigned to a triple, must be assigned to a double. Because the third rule stipulates that L and R cannot share the same room, R must then be assigned to the triple. Of the three remaining students—S, T, and V—one is assigned to a double and the other two are assigned to the triple.

Most of the questions can be easily answered by using the distributions above.

[ellenb]

dear Powerscore,

I have looked over the explanation for this game in the LSAt logic games, and I understood it. However, the part that I am confused with is the distribution, how did you come up with it? Do we base it on the L/R combination? Since, it seems this is the most limiting option.

Thanks in advance!

Ellen

[Dave Killoran]

Hi Ellen,

Good question. In this case, we distributed the seven students across the rooms, but, since the rooms are divided into predetermined lots of 1, 2, and 3, we used those numbers. It is the most limiting factor in the game, and once you lock on to that idea, it makes the game much easier.

Please let me know if that helps. Thanks!

[ellenb]

Got that! However, the part that I get confused is the assignment of students to rooms. How do you keep track who goes where? L/R is the key, you always have to start with this pair, correct?

I guess, to clarify, I got to the distribution of 1,2,3, on the main diagram and than I was confused with the next step. I could not keep track of who goes where.

So, the first one was 1-1-2-3. Before, for the distribution I remember distributing for example seven cookies into 3 jars. However, it is a bit confusing here. Since, the rooms can repeat.


Hope my question makes sense, basically it is how do you come up with the distribution. I used to count before and than go to see if it is possible. Is it how was done here?


thanks in advance!

Ellen

[Dave Killoran]

Ah, ok! Once you have the room sizes figured out--as in 1-1-2-3--then you use the rules to determine who goes where. For example, KP must share the same room, so they aren't in either one of the singles (the "1s"). But, only third- and fourth-year students (L, R, KP) can have a single, and since KP are never in a single, that means L and R must be the two singles here. K is a fourth year, and can't be assigned to the triple, so the KP block must go into the double ("2"). That leaves STV for the triple ("3"), and that places all the students.

So, we just use the rules about each student to fill in the various solutions.

Does that help?

[ellenb]

But, how do you figure out the room sizes, the distributions in the first place? the different distributions that you have? That is what I am confused with. How do you actually figure out the distribution #1, #2 and #3?(like how would 7ppl fit in 3 one single one double and one tripple room)I hope my question makes sense

Ellen

[Dave Killoran]

Ok, for that, it's like a puzzle where you simply attempt to add 1s, 2s, and 3s, up to make 7.

3+3+3 = 7? No, this one is out.

3+3+2 = 7? No, this one is out.

3 + 2 + 2 = 7? Yes, so that's initially a possibility

3+2+1+1 = 7? Yes, same.

And so on...keeping the rules in mind, which clearly rule out some scenarios (like 1+1+1+1+1+1+1).

Does that help explain it a bit better?

[ellenb]

333 Why did you start specifically with that?

because it is the maximum?


For 3-2-2 and 3-2-1-1 Did you check this scenario by plugging it it and seeing whether it works?

Than it could be, 3-1-1-1-1 right? which we know will not work. And we get to 1-2-2-2.So we stop there since, the options keep repeating .

[Dave Killoran]

Yes, I went with all maximums, but that didn't work, so mentally I would have quickly checked down to 3-2-2.

For 3-2-2 and 3-2-1-1, yes, I knew they worked numerically, but then had to make sure they worked with the rules.

And correct on the rest, too. Good job!

[ellenb]

So, how would you know that you did not miss anything as far as the distribution, how do you know you got all of them and did not miss a distribution?

Btw thanks Dave for your follow-up!