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 Dave Killoran
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#47017
Setup and Rule Diagram Explanation

This is a Linear: Pattern game.

From the nature of the rules, you should deduce that this is a Linear: Pattern game. Pattern games are characterized by rules which control the placement of variables without actually placing the variables. This leads to setups which are largely devoid of concrete information. The setup to this game is typical:
o97_game_#4_setup_diagram 1.png
Because the setup contains no “starting point” for analysis, the best approach is to review the rules in order to ensure a complete understanding of the game. As is often the case in pattern games, the rules are difficult to diagram. However, be sure to symbolize the rules in some way since the focus of the game will be on their application. Fortunately, in this game the rules are relatively simple and thus easy to remember.

The first rule states that “Each candidate must speak either first or second at at least one of the meetings.” Since there are three meetings, it follows that there are six available spaces for the candidates to meet this requirement. Since there are five candidates, each of which must appear once in these six spaces, it can be inferred that only one candidate can appear twice within the first two speaking spaces of all three meetings, and the rest of the candidates can only appear once. This is an unfixed numerical distribution of 2-1-1-1-1 for the six spaces that represent the first and second speaking slots of the three meetings. Essentially, this rule means that if one speaker speaks within the first two slots at two of the meetings, then the remaining slots must be filled with the rest of the speakers. For example, if Q speaks first at meeting 1 and second at meeting 2, then R, S, T, and U each speak once in the remaining first or second positions of the meetings. This inference comes into play on all of the questions, particularly questions #20 and #21.

The second rule states that “Any candidate who speaks fifth at any of the meetings must speak first at at least one of the other meetings.” This is a powerful rule because it establishes a constant connection between the first and fifth spaces. Since the fifth space cannot be filled by the same candidate at all three meetings (that candidate would have to speak first at at least one of the meetings), it follows that there are always two or three different speakers in the fifth slot at all three meetings. If there are three different candidates speaking in the fifth slot, then those same three candidates will also speak in the first slot at a meeting in a different order. If there are two different candidates speaking in the fifth slot, then those same two candidates will speak in the first slot, with either another candidate in the remaining first slot or with one of the two candidates doubling up. Therefore, please note that if two different candidates fill all three of the fifth speaking slots, it is possible for a candidate to speak first at a meeting and not speak fifth. For example, if the fifth speaker at each of the three meetings, is R, R, and T respectively, then the first speaker at each of the three meetings could be T, Q, and R respectively. Although the above explanation is complex, the application of the rule is much easier. Essentially, any candidate placed into the first or fifth slot immediately becomes subject to this rule. Combined with the first rule, slots one, two, and five appear to be the most controlling slots, and thus the most important slots.

Remember, in pattern games there is generally no concrete setup, so you have a greater amount of time to analyze the rules and ascertain their relationship to the pattern of the game. Also, when in doubt,
do the List questions or try a hypothetical to help gain an understanding of the nature of the game.
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 Trackstar21
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#14675
I need help fixing my brain haha. I understand that (D) is the 'best' answer choice, but I'm having difficulty understanding why the answer choice itself is valid. My mind is convinced that the fifth time slot is the ONLY appropriate place that R can be placed. I need help understanding why my reasoning is flawed.

For meeting 3, I understand that because of rule #1, R is restricted from being placed first or second. I'm aware that at this point, because there are no other restrictions on the R variable, that R can be placed third, fourth, and fifth. This is where I'm having problems.

The setup allows each candidate to speak only once per meeting. Because R has already been placed in meetings 1 and 2, he can only be placed once more in meeting 3. Rule #2 states that any candidate that speaks fifth must also speak first at at least one of the meetings. I thought that maybe I was making a mistaken reversal by requiring R to be fifth, but placing R in any other time slot would violate either rule #1 or rule #2.

I know that I am over-thinking this (I told you I need help fixing my brain) but I need help understanding where my reasoning is flawed so I don't make the same mistake on future problems. Any help is greatly appreciated!
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 Dave Killoran
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#14676
Hi Trackstar,

It's not that your brain needs fixing, it's just that this is an extremely tricky game :-D Let's break it down step by step:

First, the question stem tells you that R speaks second at meeting 2 and first at meeting 3, leading to the following basic setup:

  • Meeting 1: ___ ..... ___ ..... ___ ..... ___ ..... ___

    Meeting 2: ___ ..... _R_ ..... ___ ..... ___ ..... ___

    Meeting 3: _R_ ..... ___ ..... ___ ..... ___ ..... ___
Second, as you correctly note, due to the first rule, R is now eliminated from speaking first or second at meeting 1. So, just from considering the first rule, we've eliminated two spaces from meeting 1 (and, in the process of doing that, eliminated answer choices (B), (C), and (E) from contention). Since the difference between (A) and (D) is just whether R can speak third at meeting 1, that then becomes the focus.

Thus far, this looks pretty normal, but it's the second rule--which is a frankly a viciously devious rule--that now seemingly comes into play and causes the confusion. If indeed the rule stated that if you speak first you must also speak fifth (and was thus an MR of what is stated), then only fifth would be possible for R in the meeting 1. But, none of the answers is just "fifth," and so right then you know that the rule doesn't go both ways. This is the first signal that the test makers give that the rule might be a bit unusual in how it works.

So, how can a candidate speak first at a meeting but not fifth at another? The only way this can occur is if another candidate speaks fifth at two different meetings. This is a hard possibility to understand just from glancing at the second rule, but as in all conditional statements, you have to focus on when the sufficient occurs (or when the necessary does not occur). That means that speaking first isn't a guarantee of speaking fifth elsewhere. Here's one scenario that shows how that can work:

  • Meeting 1: _S_ ..... _Q_ ..... _R_ ..... _U_ ..... _T_

    Meeting 2: _T_ ..... _R_ ..... _U_ ..... _Q_ ..... _S_

    Meeting 3: _R_ ..... _U_ ..... _Q_ ..... _S_ ..... _T_
In the above scenario, only S and T speak fifth, and since both also speak first, that means that the second rule is satisfied. Tricky, right?

To get past something of this difficulty, you really have to understand conditional reasoning, and how in this case there are really three sufficient conditions (who speaks fifth at all three meetings) but that two of those three sufficient conditions could be identical (as with T speaking fifth at two different meetings as diagrammed above). This last point is what makes this rule so difficult.

Please let me know if that helps. Thanks!
 Trackstar21
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#14677
Thank you SOOO much Dave!!! Your explanation is so obviously simple but I honestly wouldn't have been able to understand the solution without your help. This question put me through hell. I even gave it a few days rest in between and tried to attack it with a fresh mind. Your books are awesome and coupled with this forum, it's a complete package! Thanks again.
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 Dave Killoran
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#14679
Thanks Trackstar! I'm very glad I could help :-D Please do not hesitate to let me know if you run into any other questions. Thanks again!
 Etsevdos
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#41840
Saw the setup in bible, but curious if one of the inferences that could be made is as follows:

5-->1
CP: not 1--> not 5

Not 1 would mean the item would have to be in 2 due to prior rules , this 2--> 3, or 4 or said differently if 2--> no 5 in other ones. Does this work?
 Adam Tyson
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#42405
Half yes, half no, etsevdos. You are correct that if a candidate is never 1st then it must never be 5th and, further, that it must be 2nd at least once. However, being 2nd at least once does not mean that a candidate is never 1st, because one of the five candidates has to appear in the first two spaces twice (6 spaces but only 5 candidates, so one repeats). In other words, there could be a candidate who appears 1st one time and 2nd another time (or 1st twice, or 2nd twice). That means a candidate who appears 2nd one time could be 5th another time and 1st the third time.
 LSAT2018
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#45567
For the rule, 'Any candidate who speaks fifth at any of the meetings must speak first at at least one of the other meetings' I have a question on the conditional reasoning for this.
Does this mean that if one of the candidates spoke fifth more than once, then the candidate would have to speak first more than once? So for example, if T spoke fifth for meetings 1 and 2, then would it be acceptable that T spoke first in meeting 1 only (since the rule says 'at least one')


And are there any inferences that can be made from the setup?
 Adam Tyson
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#48779
There is no numeric component to that rule other than "at least once", LSAT2018, and we shouldn't infer one. It's possible that a variable goes last twice and then goes first the other time. It would actually be impossible to be first twice and last twice, because there are only three meetings.

No really powerful inferences come immediately from that, but try coming up with a setup where one variable goes first once and last twice, and see what else happens as a result. You may learn something interesting!

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