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#44059
Setup and Rule Diagram Explanation

This is a Defined-Moving, Balanced Grouping game.

At first glance, this Game appears to be Undefined because it is not immediately clear how the seven variables will be divided into the two groups, F and I. However, the third and fourth rules establish conditions that serve to restrict the distribution possibilities to a high degree (as will be discussed momentarily), thus this Game can be considered largely Defined.

Another important point to consider when attacking this Game is the two-value system. Because there are two committees here—Finance and Incentives—and the board members must each be placed into exactly one of the two committees, you can make powerful inferences when you determine both where a person must go, as well as when you determine where a person cannot go. That is, if a member cannot be on the finance committee, you can infer that that member must be on the incentives committee. This idea will also come into play as we take contrapositives of the conditional rules for this Game.

Finally, because each of the five rules is purely conditional, this Game can be described as a Pure Grouping Game. In Pure Grouping games you are generally not able to make definitive inferences about where variables must be placed, but rather you must understand the relationships between certain variables and how they may or may not be grouped with one another. So we will not be able to make a diagram in the traditional sense; 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 set group in place. Some similar games (albeit less defined) are: October 2005 #2 (Light Switches), December 2005 #1 (Electrical Appliances), and June 2006 #3 (Summer Courses).


G H L M U W Z 7

Rule 1:
  • GF :arrow: HI

    HF :arrow: GI
Rule 2 (for space purposes we are going to write out the multiconditionals in this game horizontally):
  • LF :arrow: MI + UI


    MF or UF :arrow: LI
Rule 3:
  • W :dblline: Z
Rule 4:
  • U :dblline: G
Rule 5:
  • ZF :arrow: HF

    HI :arrow: ZI

Inferences:

There are four main inferences (chains of inferences) that result from a combination of two or more of the rules listed previously.

Inference 1:

You know that if G is in F, then H is in I (first rule), and you also know that if H is in I, then Z is in I (last rule). Further, you know that G and U cannot be together, so if G is in F, then U must be in I. This creates:
  • ..... ..... ..... ..... ..... GF :arrow: HI + UI + ZI
Inference 2:

You know that if H is in F, then G is in I (rule 1), and you know that G and U are not together (rule 4) so G in I would place U in F. You also know that if U is in F then L is in I (rule 2), so you have the following:
  • ..... ..... ..... ..... ..... HF :arrow: GI + UF + LI
Inference 3:

You are told that Z in F means that H is in F (rule 5), and H in F indicates something about G, U, and L (Inference 2 above). Further, since Z and W must be separated (rule 3), then Z in F would place W in I. This yields:
  • ..... ..... ..... ..... ..... ZF :arrow: HF + GI + UF + LI + WI
Inference 4:

You know that H in I tells you that Z is also in I (rule 5), and you know that Z and W must be separated. This produces:
  • ..... ..... ..... ..... ..... HI :arrow: ZI + WF
Armed with the rules and these powerful inferences we are now ready to attack the questions.
 amagari
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#34848
Hi,

I still don't understand the initial setup of the conditional chains in this question. How do you combine the Double Not Arrows with the other normal arrows?

I understand how to get the initial individual rules and the contra positives but how to connect them all up and why you can connect certain ones to other ones.

I can combine the G :arrow: /H :arrow: /Z chain and it's contra positive. I also understand the L :arrow: /M and L :arrow: /U but after that I"m lost.

Thanks
 AthenaDalton
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#34892
Hi Amagari!

You're off to a great start on this one.

Our first rule tells us that if G(f) :arrow: H(i)

Rule two gives us the L(f) :arrow: M(i) U(i) inference

You also correctly deduced that:

W :dblline: Z
U :dblline: G

The final rule tells us that if Z(f) :arrow: H (f)

To put these together, I started by considering what happens if G is assigned to committee (f), which will trigger the first rule.

(f) Finance Committee: G
(i) Investment Committee: H

We also need to split up G and U, which gives us:

(f) Finance Committee: G
(i) Investment Committee: H, U

We also know that if Z(f) :arrow: H(f). That can't work here since H(i). That means that Z must also be assigned to the investment committee:

(f) Finance Committee: G
(i) Investment Committee: H, U, Z

Next, we split up Z and W:

(f) Finance Committee: G, W
(i) Investment Committee: H, U, Z

This leaves us with the last rule that if L(f) :arrow: M(i) U(i)

This gives us two possible diagrams, depending on whether L(f) of L(i):

TEMPLATE 1:
(f) Finance Committee: G, W, L
(i) Investment Committee: H, U, Z, M

TEMPLATE 2:
(f) Finance Committee: G, W
(i) Investment Committee: H, U, Z, L, M

So we get a lot of inferences just from assigning G(f).

Next let's consider what happens if G(i):

(f) Finance Committee:
(i) Investment Committee: G

We can start by splitting up G and U:

(f) Finance Committee: U
(i) Investment Committee: G

Then we can consider the L(f) :arrow: M(i) U(i) rule. Here, we know that U is assigned to (f). So L cannot be assigned to (f). This forces L into (i):

(f) Finance Committee: U
(i) Investment Committee: G, L

Then we can split up W and Z, one in either (f) or (i):

(f) Finance Committee: U, W/Z
(i) Investment Committee: G , L, W/Z

At this point, M and H can be assigned to either committee for template 3. Keep in mind that if Z(f) :arrow: F(f):

TEMPLATE 3 (M can be assigned to either committee):
(f) Finance Committee: U, Z, F
(i) Investment Committee: G , L, W

TEMPLATE 4 (M and H can be assigned to either committee):
(f) Finance Committee: U, W
(i) Investment Committee: G , L, Z

Those are the big templates/inferences for this game. Hope this helps!

Athena Dalton
 glasann
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#75943
Hi - for inference #1 that you've come up with, can you also add W(F) to the necessary side if G is in F, considering that'll force Z into Incentives and you can't have Z and W together? thanks!

Inference 1:

You know that if G is in F, then H is in I (first rule), and you also know that if H is in I, then Z is in I (last rule). Further, you know that G and U cannot be together, so if G is in F, then U must be in I. This creates:
  • ..... ..... ..... ..... ..... GF :arrow: HI + UI + ZI
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 KelseyWoods
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#76042
Hi rademaker1!

Yes, you can absolutely add the inference that W must be in F if G is in F. Good job keeping an eye out for all those inferences!

Best,
Kelsey
 emccready24
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#76720
Hello -Could you tell me why you don't make the first and second rule into double not arrows? My rules looked like this:
G <--|--> H
L <--|--> M
L <--|--> U
W <--|--> Z
U <--|--> G
Z(F) ---> H(F)

So then I made inferences that
U <--|--> G <--|--> H ------ so U <--|--> H
L <--|--> U <--|--> G ------ so L <--|--> G

However when I started the game, I began to get confused and I don't think I diagrammed the first and second rules correctly, leading me to make incorrect inferences.
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 KelseyWoods
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#76782
Hi emccready24!

Be careful with those first 2 rules!

Rule #1 says that if G is in F, then H is in I. The contrapositive is that if H is in F, then G is in I. So that means that we can't have G and H together in F. But G and H could still be together in I! When you have a pair of contrapositives, you can't have the 2 sufficient conditions at the same time, but you could have the 2 necessary conditions at the same time. So we can't make a blanket rule that G and F cannot be together. They can be together in I, they just can't be together in F.

The same goes for rule #2. L can't be with M or U in F, but L could be with M or U in I.

Hope this helps!

Best,
Kelsey

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