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Setup and Rule Diagrams

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Setup and Rule Diagram Explanation

This is an Undefined Grouping Game.

The game is Undefined because we do not know how many photographs are displayed in the album, nor do we know exactly how many people are in a photograph. With these elements unspecified, we cannot create a diagram in the traditional sense; that is, we cannot create a group of say, five spaces and then place variables in those spaces. Instead, we must diagram the rules and make inferences, and then proceed to the questions without a defined group in place.

The Rules

Rule #1. One of the easiest mistakes to make in this game is to misinterpret the wording in each of the rules. This rule indicates that whenever S appears in a photograph, then W appears in that same photograph. The correct diagram for that relationship is:


..... ..... ..... ..... ..... ..... S :arrow: W


Many students mistakenly reverse the above diagram. To avoid doing so, consider the rule for a moment: does the rule say that every time W appears in a photograph that S also appears? No. Although the difference is subtle, the wording in the rule is that W appears in every photograph that S appears in. Thus, S is the sufficient condition, and the appearance of S indicates that W will appear in that same photograph.

Rule #2. The wording in this rule is identical to the wording in rule #1, and so it is just as easy to make an error when diagramming this rule. The correct diagram is:


..... ..... ..... ..... ..... ..... U :arrow: S


When we get to the “Inference” section we will discuss how the first and second rules can be linked. Also, note that since both of the first two rules are “positive” (no negative terms involved) we are not diagramming contrapositives since you should know those as an automatic result of seeing any conditional statement.

Rule #3. This is also a tricky rule, not only because of the wording, but also because the conditional relationship between R and Y is easy to misunderstand. First, let us diagram the rule:


..... ..... ..... ..... ..... .....Y :arrow: R


Since this rule contains a negative, let us also diagram the contrapositive:


..... ..... ..... ..... ..... ..... R :arrow: Y


Many students will interpret these two diagrams as reduced to a simple double-not arrow relationship:
Y :dblline: R. This is not the meaning of the rule! Instead, consider the exact relationship between Y and R: when Y is not in the photograph, then R must be in the photograph; and, via the contrapositive, when R is not in the photograph, then Y must be in the photograph. Thus, when one of the two is not in the photograph, the other must be in the photograph, and that relationship can best be expressed by stating that both cannot be absent from the photograph. The correct double-not arrow diagram, then, is:


..... ..... ..... ..... ..... ..... R :dblline: Y


The operating result of this rule is that either Y or R, or both, must appear in each photograph (note that R and Y can appear in a photograph together; the rule does not prohibit this occurrence).

Rule #4. This rule is actually two rules in one: one rule states that T and W do not appear in the same photograph, and the other states that R and W do not appear in the same photograph. Let’s consider each part separately.

As stated directly, when W is in a photograph, T is not in that same photograph:


..... ..... ..... ..... ..... ..... W :arrow: T


This rule can be turned into the a double-not arrow indicating that T and W are never in the same photograph:


..... ..... ..... ..... ..... ..... W :dblline: T


The other part of the rule states that when W is in a photograph, R is not in that same photograph:


..... ..... ..... ..... ..... ..... W :arrow: R


This rule can be turned into the a double-not arrow indicating that R and W are never in the same photograph:


..... ..... ..... ..... ..... ..... W :dblline: R


One note about the rules: every variable is mentioned in the rules except Z. Thus, Z is a random in this game, and since there is not a specified number of spaces, Z is largely powerless in this game. If a question stem includes Z as part of a local condition, there must be other information included in the question stem, and you should focus on the other information first.

In review, the four rules contain five basic grouping relationships that create many different inferences. When you consider the four rules, you can see that the test makers placed a trap for the unwary student. A student who quickly reads through the rules and does not read for meaning can easily mis-diagram one or more of the rules, and of course mis-diagramming during the setup is almost always costly.

After correctly negotiating each of the rule diagrams, the next step is to make inferences by connecting the rules. In this game, there are many inferences, and so the challenge becomes managing the information.


The Inferences

Inference #1. By connecting the first and second rules, we can create the following chain:


..... ..... ..... ..... ..... ..... U :arrow: S :arrow: W


This connection is important because it shows that if U appears in a photograph then two other friends must also appear in that same photograph. Via the contrapositive, the relationship also indicates that if W does not appear in a photograph, then neither S nor U can appear in that photograph.

Inferences #2 and #3. The first and fourth rules can be connected through W:


..... ..... ..... ..... ..... ..... S :arrow: W :dblline: T

..... ..... ..... ..... ..... ..... +

..... ..... ..... ..... ..... ..... S :arrow: W :dblline: R



These two relationships yield the following inferences:


..... ..... ..... ..... ..... ..... S :dblline: T

..... ..... ..... ..... ..... ..... +

..... ..... ..... ..... ..... ..... S :dblline: R


Inferences #4 and #5. The two previous inferences connected S to T and R through S’s relationship with W. Since U also has a relationship with S from the second rule, we can connect the second rule to inferences we just made:


..... ..... ..... ..... ..... ..... U :arrow: S :dblline: T

..... ..... ..... ..... ..... ..... +

..... ..... ..... ..... ..... ..... U :arrow: S :dblline: R


These two relationships yield the following inferences:


..... ..... ..... ..... ..... ..... U :dblline: T

..... ..... ..... ..... ..... ..... +

..... ..... ..... ..... ..... ..... U :dblline: R


Inference #6. Because R appears in the third and fourth rules, we can make a connection using R:


..... ..... ..... ..... ..... ..... Y :arrow: R :arrow: W


This relationship results in the unique inference that:


..... ..... ..... ..... ..... ..... Y :arrow: W



Because both conditions are negative, take the contrapositive:


..... ..... ..... ..... ..... ..... W :arrow: Y


Thus, if W appears in a photograph, then Y must appear in that photograph as well.


Inferences #7 and #8. From the inference we just made we know that when W is in a photograph, then Y must also be in that photograph. We can add this to the chain that appeared in inference #1 (which was the combination of the first two rules):


..... ..... ..... ..... ..... ..... U :arrow: S :arrow: W :arrow: Y


These relationships yield the following two inferences:


..... ..... ..... ..... ..... ..... S :arrow: Y

..... ..... ..... ..... ..... ..... +

..... ..... ..... ..... ..... ..... U :arrow: Y


Thus, the appearance of either S or U will ultimately force Y to appear. This makes U a powerful variable: when U appears in a photograph, then S, W, and Y must also appear, and R and T cannot appear. Thus, the appearance of U allows for only two solutions, depending on whether Z is in the photograph.


There are other ways to arrive at some of these inferences (for example, when S has a relationship with a variable, U has the same relationship because of the second rule), but each inference above is a product of combining the rules or of combining the rules and inferences (do not forget to recycle your inferences!).

Compiling all of the information above, we arrive at the final setup for this game:


pt45_d04_g3_20a.png
pt45_d04_g3_20a.png (8.37 KiB) Viewed 17 times

pt45_d04_g3_20b.png
pt45_d04_g3_20b.png (3.74 KiB) Viewed 17 times



As long as you focus on connecting the variables in the rules and inferences, you can attack the questions and complete each problem with relative ease. Remember, this game is not logically difficult; it is simply an information management game, so do not be intimidated!
eober
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Hi,

I just want to make sure if my inferences are correct in this game.


I made the inferences:
U :dblline: W (from W :dbl: S and S :dbl: U)

R :dblline: T (from T :dblline: W and R :dblline: W)

W :dblline: Y (from R :dblline: W and R not :arrow: Y) because if W is there, R is not, if R is not then Y is.

Y :dblline: R (from Y not :arrow: R)

Does the inferences I made make sense? I get confused when combining a rule with a double arrow. Could you explain as a general rule how we make different inferences from double arrows and double not arrows?

Thank you!
David Boyle
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eober wrote:Hi,

I just want to make sure if my inferences are correct in this game.


I made the inferences:
U :dblline: W (from W :dbl: S and S :dbl: U)

R :dblline: T (from T :dblline: W and R :dblline: W)

W :dblline: Y (from R :dblline: W and R not :arrow: Y) because if W is there, R is not, if R is not then Y is.

Y :dblline: R (from Y not :arrow: R)

Does the inferences I made make sense? I get confused when combining a rule with a double arrow. Could you explain as a general rule how we make different inferences from double arrows and double not arrows?

Thank you!


Hello,

One might do them differently, something like u :arrow: s :arrow: w :arrow: slash t, slash r, with the slash r :arrow: y. (Because we know that slash y :arrow: r)

As for "different inferences from double arrows and double not arrows", the first is about "love", the second is about "hate". Make sure you correctly track any arrows coming in or out of double arrows, e.g., if A :dblline: B, and C :arrow: A, then you can assume that C :dblline: B. But if there were A :arrow: C, you can't automatically assume that C :dblline: B.

David
Dave Killoran
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This morning I received the following question:

I know this is a dumb question but I need some clarification. On Dec 2004 LSAT LG 3: a rule states Raimundo appears in every photograph that Yakira does not appear in. So I diagrammed it as not R then Y. wouldn't the contrapositive be Not Y then R? I reviewed a video lesson and it said that If not R then Y is free variable? which the #1 on the game was B which include R and Y. I just don't understand how the rule effects the game. Help!



First, there are never any dumb questions! I know people say that a lot, but it is still true. One small misunderstanding can sink you on a game, so if you ever are even a bit uncertain, ask! Second, this is a really tricky game, and they intentionally worded the rules in an unusual way in order to get students messed up. Basically, the first two rules look like "B appears when A appears" (A :arrow: B), and the third rule looks like "B appears when A does not appear" (A :arrow: B). So, the necessary condition is being presented first, and the sufficient condition appears later in the sentence (which is very normal and acceptable). Thus, the first two rules would be diagrammed as:


    S :arrow: W

    U :arrow: S

    (and those two can be combined to form the chain diagram: U :arrow: S :arrow: W )


With the third rule, we have a negative on the sufficient condition, leading to the following diagram:


    Y :arrow: R

So, if Y is NOT in a photograph, then R must be in that photograph. The contrapositive both reverses and negates the terms, and appears as follows:


    R :arrow: Y

When those two rules are considered together, the mean that if one of R or Y is absent from the photograph, then the other must be in the photograph. In other words, at least one of R/Y is always in a photograph(and possibly both). Rules like this,w here the sufficient condition is negated, are extremely tricky. I wrote an article about them that I think you would find worthwhile. It's called The Most Dangerous Conditional Rule on the LSAT and it explains exactly how rules like this work.

Ok, so when we look at your diagrams, it looks like you did great! You did it in reverse order from me, but that doesn't matter at all. A statement and its contrapositive are identical in meaning, so whichever is "first" is irrelevant.

As for the video lesson you reviewed, it definitely wasn't one of ours because we wouldn't say that if Y occurs then R is "free." In fact, it's not free at all—if Y, then R must be in that photograph. Could they have meant that if Y was in the photograph, then R was free? Because that would be true.

Hopefully that clears this rule up. It's a tough one, so make sure you are 100% comfortable with it. Please let me know what you think. Thanks!
Dave Killoran
PowerScore Test Preparation
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Zestor
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Hello, I am practicing this game in the LG bible book, and I seem a bit confused regarding the relationship between u>s>w>slash r > y relationship. I am able to infer that w>y, s>y, u>y, and slash y>slash w.

But, wouldn't it be correct to also infer that slash y> slash s, slash y> slash u? I am visually trying to make these inferences so i am looking at the u>s>w>slash r > y relationship, and taking the contrapositive of it to arrive at inferences. I am not sure if my process is icorrect.
James Finch
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Hi Zestor,

The first thing to look for in an undefined grouping game like this is whether we can chain the rules together or not. Here, we definitely can: by taking the contrapositive of rule 3 and the other rules as given, we can create a chain that looks like this:

U :arrow: S :arrow: W :arrow: T and R :arrow: Y,

leaving Z alone as a wildcard. We can also take the contrapositive of that chain and get:

Y :arrow: R, R or T :arrow: W :arrow: S :arrow: U

This diagram gives us numerous inferences, and should be used as the main diagram going forward in the game.

Hope this clears things up!