Setup and Rule Diagrams

[Administrator]

Setup and Rule Diagram Explanation

This is an Advanced Linear: Unbalanced: Overloaded, Numerical Distribution game.
This is arguably the toughest game on this test. The game scenario specifies that five talks—F, G, H, I, and L—are held in successive order. Thus, these five talks should be the base of the game. Four employees each then attend two of the talks, leading to the following Linear setup:

D10_game #3_setup_diagram 1.png

With each employee attending two talks, there are eight variables to be placed (which are represented by doubling Q, R, S, and T in the variable list). These eight variables must be placed into the five talks, with the stipulation that no talk can be attended by more than two employees. Thus, we have the classic elements for a Numerical Distribution:

D10_game #3_setup_diagram 2.png

Using this information, two employees-to-talks distributions are possible:

2-2-2-2-0
and
2-2-2-1-1


In the 2-2-2-2-0 distribution, four of the talks are attended by two employees each, and one talk is not attended by any employee.

In the 2-2-2-1-1 distribution, three of the talks are attended by two employees each, and two talks are attended by exactly one employee each.

The drawback to these distributions is that our Linear scenario is somewhat fluid; that is, a talk can have 0, 1, or 2 attendees depending on the distribution, but there is no way to fix the number of attendees for each talk at this point. Let’s examine the rules and hope there is some information that fixes some of the attendees or talks.

The first rule establishes Not Laws for Q on F and H:

D10_game #3_setup_diagram 3.png

Accordingly, Q must attend two of G, I, and L.

The second rule establishes two more Not Laws, this time for R on G and H:

D10_game #3_setup_diagram 4.png

Note that H is particularly restricted at this point, with only S or T as possible attendees. This is not shown as a dual-option because H can have zero attendees. Interestingly (and likely not coincidentally), the next rule establishes that S and T do not attend the same talk:

D10_game #3_setup_diagram 5.png

This is an intriguing rule, because with only five available talks (depending on the distribution), the placement of either S or T has a significant impact on the other variable. For example, in the 2-2-2-2-0 distribution, if you know which two talks S attends, you then know T must attend the other two talks that have attendees.

However, this rule is most interesting when considered in reference to H. Because only S and T can attend H, and S and T cannot both attend the same talk, we can determine that H either has no attendees or that H has exactly one attendee (S or T). That’s a particularly useful inference, because H can be used to determine the proper distribution:

• If the 2-2-2-2-0 distribution is in effect, then H must be the talk with zero attendees. Remember, H can have at most one attendee (S or T), so if another talk were to have no attendees, then H would have to have two attendees. But, since H can never have two attendees, we can infer that no other talk can have zero attendees (and thus, globally, all of the other talks must have at least one attendee). As only H can be the talk with no attendees, in the 2-2-2-2-0 distribution then the other four talks have exactly two attendees each, leading to the following distribution: 2F-2G-2I-2L-0H.
  
  If S or T attends H, then the 2-2-2-1-1 distribution applies. Consequently, F, G, H, I, and L all have at least one attendee, and three of F, G, I, and L have two attendees: 2-2-2-1-1H.

The last two rules are similar in form: both are conditional and each establishes that the first talk attended by a certain employee is also attended by another specific employee (but not necessarily that employees first talk):

D10_game #3_setup_diagram 6.png

The fourth rule can be combined with the first rule to produce a useful inference: T cannot attend F. Because Q cannot attend F, if T were to attend F there is no way to satisfy the fourth rule. Thus, T cannot attend F, which can be shown as a T Not Law on F:

D10_game #3_setup_diagram 7.png

While this is the first inference that follows from these two rules, there are two other powerful inferences that follow from combining these rules with other rules. The two most restricted talks are F and H. H was already discussed above, so the next logical point of examination is F.

As established in the discussion of the distributions above, F (and G, I, and L) must always have at least one attendee. Because Q and T cannot attend F, only R and S are available to attend S. But, when the final rule is considered, a powerful inference results: S must always attend F. Here’s why: if R attends F, then from the final rule S must also attend F. But, if R does not attend F, then only S remains to attend F, and because F must always have at least one attendee, S must then attend F. In either case, S attends F, and we can add that to our diagram:

D10_game #3_setup_diagram 8.png

With F and H both discussed, G is the only remaining restricted talk that has not yet been examined. Interestingly, the inference involving G is even trickier than the inference involving F.

Because R cannot attend G, only Q, S and T are available to attend G. Of course, from the third rule, S and T cannot attend the same talk, so there are only three options for employees to attend G:

• 1. Q and S
  
  Q and S can both attend G under the 2-2-2-2-0 distribution: if S and R attend F, H is empty, and Q and T attend I (T attends L as well).
  
  2. Q and T
  
  If the first talk that T attends is G, then from the fourth rule Q must also attend G.
  
  3. Q alone
  
  Under the 2-2-2-1-1 distribution, Q can attend G alone if R and S attend F, S attends H, and Q and T attend I.

S cannot attend G alone because then S would attend both F and G, and from the fourth rule T would have to attend one of those, which would have to be G. T cannot attend G alone due to the actions of the fourth rule as well. Thus, in each of the three possible scenarios, Q must attend G. This inference can be added to the diagram, leading to the final diagram for the game (the numbers above each talk represent the numerical possibilities):

D10_game #3_setup_diagram 9.png

There is also a case to be made for creating two templates (one for each distribution), but the information in the final diagram is sufficient to attack the questions, especially if you get the inferences that H is the only talk that can have no attendees, and that S attends F, and Q attends G (and regrettably, these are all tough inferences, and it is very hard to get any one of them, let alone all three).

Overall, this is not a fun game. The three key inferences are each difficult to draw, and without them the game is exceptionally difficult. When combined with the second game, the two games form one of the toughest game tandems in LSAT history.

[Brittney]

I was very confused with this game and need help with the setup. I almost got all of the questions wrong on this game because I didn't do the setup right.

[Jon Denning]

Hey Brittney,

Thanks for the question! That's a tricky game for a lot of people, but hopefully I can give you some tips for the setup that'll make things a little easier. Once I walk you through the scenario and rules, take the setup and re-attempt the questions to see if things are more clear.

The first thing to do is to choose a base. Since the talks are listed as having a set order (F G H I L), use them as your base and then assign people to them.

Secondly, you should notice that the numbers are not balanced: There are four employees each attending two talks, for a total of eight attendees, and there are five talks that can have zero, one, or two attendees each (for a total of 8). So we need to start by thinking about the distribution possibilities of 8 people to 5 talks:

2-2-2-2-0 (four of the talks have 2 people each, one talk has 0)

2-2-2-1-1 (three talks have 2 people each, the other two talks have 1 person each)

Those two distributions will play an important role as we move to the rules.

The first two rules are fairly straightforward, and give us Not Laws for Q on F and H, and for R on G and H. That's pretty limiting for H, as now only S or T could attend that talk.

The third rule is specific to S and T: they do not attend any of the same talks. This is a basic Not Block. It is also important for talk H, as S and T were the only two who could attend. Since they cannot attend together, talk H must have either 1 person (S or T) or 0 people attending (cannot have 2). That's useful because it means H can potentially be used to determine which of the two distributions above we've got.

Finally, the fourth and fifth rules tell us that T's first talk includes Q, and R's first talk includes S, respectively. That means that T cannot attend F, since that would be the first talk for T and Q cannot attend F (rule 1).

Note at this point that H and F are the two most restricted talks. H, as mentioned above, has either 0 or 1 attendees (if just 1 it's S or T). Everything else then must have either 1 or 2 attendees. So F must have 1 or 2 attendees, and we should pay attention to whom they could be: only S and R. Of course, from the last rule if R attends F (first talk), then S also attends. And if R does NOT attend F then S must attend since F cannot have 0 attendees (only H can). In either case, S must always attend F, which is a powerful inference.

And we're not done yet! Look at talk 2, which is G. R cannot attend, so only S, T, and Q are available. Of course, S and T cannot attend together (rule 3), so we have these options; Q and S, Q and T, or Q alone. In all instances Q must attend G, which is another great inference to spot early on!

Now we've got several good inferences, including a few placed variables and a number of Not Laws. Take those ideas and go back to the questions. See how you do with the setup I've described and then let me know your thoughts.

Thanks!

[Jon@an]

I'm a little unclear on just one portion of the initial explanation, and I didn't see this touched upon in the later response either. In the first post, when discussing the inferences around who can attend G, I see: "S cannot attend G alone because then S would attend both F and G, and from the fourth rule T would have to attend one of those, which would have to be G." However, I'm not clear on how this actually follows from rule 4 . Can someone please clarify? Is this somehow related to rule 5, which would place R in F?

[Jon@an]

Oh, I think I see the issue now. If S is placed in G alone, that requires the 2-2-2-1-1 distribution, with G and H each containing one empty slot. Under that distribution, the first three slots would look like the below:

F: R and S (because F must have two slots filled, and not laws)
G: S and and an empty slot (per the hypothetical)
H: T and empty slot (the S/T dual option from the initial not laws and the 2-2-2-1-1 distribution results in T being placed here when both of the S's are used up in F and G)

This then presents the issue that T is in violation of the 4th Rule - that the first instance of T places a QT block. To solve this issue, the only other place where you could slot in the first T would be in G, given the not laws in F, which then causes a violation of the 3rd rule (T and S cannot occur together).

[Jeremy Press]

Hi Jon@an,

Yes, you've answered your own question here with your second post. It's extremely tricky to see how S cannot attend G by itself, but the hypothetical you present is accurate and shows the problem with doing that (the distribution won't permit it). Great work on a very difficult game!

Jeremy

[ncolicci11]

Powerscore,

So, if R is placed later in the diagram for the first time in either I or L for example, does that mean S will be reused to satisfy the R1 + S block?

Thank you!

[Jeremy Press]

Hi ncolicci,

Yes, that's exactly right. S's first talk will always be F (for the somewhat complex reasons discussed in the very detailed overview of the setup). If R's first talk is a later talk than F, that first talk will have to be I, and the second S will be on I with it. The second R will then go in L, with some available variable other than S.

Good work!

Jeremy