Grouping Setup Practice Drill #4

[arr0418]

In the Grouping Setup Practice Drill #4 (page 302, 2017 Edition of the Logic Games Bible), the first rule reads "If D is assigned to tester 1, then E is assigned to tester 2." I initially diagrammed this as D1 --> E2 , with the contrapositive as E1 --> D2 . I thought this was correct, given discussion on page 287 and 297 regarding the two-value system. However, it becomes apparent that E can only be placed in the "tester 2" group in this drill.

What makes this scenario different than the scenarios on page 287 and 297?

[Dave Killoran]

Hi Arr,

Thanks for the question! Before reading my answer, I encourage you to go back to pages 287 and 297 as well as this particular drill item, and carefully examine each. Ask yourself, what's happening in this drill that is different from what happens in the other two? What other information is present that might have caused a problem here? I want you to do that because that's what we do when solving games—we look for things that are similar to what we've seen before as well as things that are different. Then we can use that information to start building a setup for the game, and we also then have a sense of what we should be on the lookout for and what rules/conditions might cause us issues.

In the drill, the first indication that something unusual is going on is that we have a unique situation with the variables and the spaces available: there are 6 spaces to be filled, but only 5 variables. That by itself isn't the problem, but it leads to it because the variable that is doubled, D, is also involved in the rule in question. When you put that together, it causes a very unusual effect that ultimately makes it so the contrapositive of the first rule can never be enacted. So, if we didn't have this unusual set of situations, then this rule would work just like the others.

Now, because D appears twice we know that immediately the sufficient condition in the D1 E2 rule is met, and that consequently E must be in the tester 2 group. At that point, E is done because due to the constraints of the game E can only be tested once. So, when we go to consider the contrapositive of the first rule, we have a problem. That contrapositive is E1 D2 and it does exist; the problem is that E never can be in 1, and thus the sufficient condition can't be met. So, it's there but will never be used. If this was a real game and you diagrammed both the first rule and its contrapositive during the setup, you'd just then cross out that contrapositive before moving on to the questions (since the CP can never occur under the rules of the game). Now, when I went to explain this game, I wanted to focus on the unusual result of E being in 2 and thus H being forced into 1, and so I intentionally did not go through the process of diagramming that CP and then later crossing it out since we arrived at that same conclusion via the discussion there about E and H.

Please let me know if that helps. Thanks!

[arr0418]

Makes complete sense! Many thanks.

[leslie7]

Now, because D appears twice we know that immediately the sufficient condition in the D1 E2 rule is met, and that consequently E must be in the tester 2 group. At that point, E is done because due to the constraints of the game E can only be tested once. So, when we go to consider the contrapositive of the first rule, we have a problem. That contrapositive is E1 D2 and it does exist; the problem is that E never can be in 1, and thus the sufficient condition can't be met. So, it's there but will never be used. If this was a real game and you diagrammed both the first rule and its contrapositive during the setup, you'd just then cross out that contrapositive before moving on to the questions (since the CP can never occur under the rules of the game).

Hi Dave, feeling a little poorly at this point because I am getting a sense that my questions are over lapping in terms of their core content even though they are addressed to different questions whereby each has obviously, a unique answer

I didn't understand what you meant when you said that the contrapositive is cancelled out - could you explain that further? Why is it cancelled and why is it only that the CP is cancelled?

I realize this will not sound clear but I did ask a question before on how conditionals are not temporal and how you know when an item is triggered but this question really highlights that question well.

I did get the questions and diagraming perfectly correct but I made a lot of assumptions that I'm not sure are logical and I don't want to carry that thinking forward if it isn't applicable to the LSAT.

I understand D is placed in T1 and T2 - clear

but I only placed E in T2 because D1--> E2 but if I take E1 --> D2 into account E could also be placed in 1 and therefore E and H could be exchangeable but I thought to myself that I would apply the first rule only because the rule came before the E1 rule so it would trigger the conditional before the E1 could trigger the conditional . so additionally when a variable is placed into two slot simultaneously which one triggers what?

But technically I could do

T1 E/H , F/G, D
T2 D, F/G, E/H

I'm not sure why the CP is cancelled but if it is cancelled why is that D1-->E2 is not also cancelled?

Placing D in both T1 and T2 appears to make the entire conditional irrelevant since both E and E and potentially exist together in either T1 or T2

Feeling pretty lost + exhausted.

Ty for you time + patience

[Robert Carroll]

leslie,

It's not that the contrapositive is "canceled", but more like it just doesn't matter. Because D will always be assigned to tester 1, E will always be assigned to tester 2. Because D is the one product assigned to both testers, E cannot be assigned to tester 1. So the sufficient condition of rule 1 is always true, and the sufficient condition of its contrapositive is never true. A conditional with a true sufficient condition is useful to a point - to the point where you use that fact to infer the necessary condition. A conditional with a definitely false sufficient condition is useless - it doesn't constrain the game in any way, because a conditional does something only when its sufficient condition is true. So the contrapositive, by always having its sufficient condition false in this game, is not useful.

This game simply does not have a temporal element to it. The statement "once a tester has tested a product she cannot test that product again" is used in this game only to make sure each tester tests three different things. There is no time order to any of the tests in the scenario or rules, and the question does not get into that either.

As Dave pointed out, E is never tested by 1, so "taking into account" the conditional does not provide any additional information. The rules are not in any temporal order either. All the rules have to be true at the same time. So there's no chronological ordering of rules.

Your hypothetical doesn't work because, as you can see, D is assigned to tester 1 in your situation. The first rule tells us that E should be assigned to tester 2. It's not an option with H - E MUST be assigned to 2.

As I said, the contrapositive is not canceled, but just never gets to do any work. The original conditional itself has an always-true sufficient condition, so the necessary condition is always true. The conditional is not irrelevant - it's and the last rule are what make E have to be in 2.

Robert Carroll