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#40566
Setup and Rule Diagram Explanation

This is a Grouping: Partially Defined game.

The game scenario establishes that at least four employees will be selected from a group of eight employees:
PT70 -Game_#2_setup_diagram 1.png
With a group being selected, and order not important, this is a Grouping game. Because the research team must contain at least four employees, but could contain more, the game is Partially Defined. Let’s examine the rules, which, unsurprisingly, are all conditional.

The first rule is actually two rules-in-one, and establishes that M cannot be on the team with O or with P. For the sake of clarity, write these relationships out separately:
PT70 -Game_#2_setup_diagram 2.png
Note that this rule automatically eliminates all eight employees from being selected for the team. If M is selected, the maximum number of employees that could be on the team at this point in our analysis is six. If M is not selected, the maximum at this point is seven (all employees except M).

The second rule indicates that if S is selected, then both P and T are selected:
PT70 -Game_#2_setup_diagram 3.png
Because P appears in both the first and second rules, we can link those rules together via the following chain (T is dropped for the moment since it is superfluous to this inference):
PT70 -Game_#2_setup_diagram 4.png
Resulting in the following inference:
PT70 -Game_#2_setup_diagram 5.png
Side note: drawing out the:
PT70 -Game_#2_setup_diagram 6.png
connection is not necessary in order to draw the inference. After diagramming the first two rules, you can visually link the rules and then write out just the:
PT70 -Game_#2_setup_diagram 7.png
inference.

The third rule indicates that if W is selected, then both M and Y are selected:
PT70 -Game_#2_setup_diagram 8.png
This rule can be connected to the other rules to draw several inferences. First, combining the third rule and the first rule creates the following chains:
PT70 -Game_#2_setup_diagram 9.png
The two chains respectively create the following two inferences:
PT70 -Game_#2_setup_diagram 10.png
The inference created by the first and second rules can also be recycled and combined with the third rule via the following chain:
PT70 -Game_#2_setup_diagram 11.png
Which then results in the following inference:
PT70 -Game_#2_setup_diagram 12.png
From these rules and inferences you can see that M and W are both powerful variables. If M is selected, then O, P, and S are eliminated. If W is selected, then M and Y must be selected, and O, P, and S are eliminated. In each instance, then, the maximum number of employees selected for the team is five.

Combining all of the prior information, and the fact that Z is a random, leads to the final diagram for the game:
PT70 -Game_#2_setup_diagram 13.png
Numerically, the research team can only have 4, 5, or 6 members:

1, 2, or 3 members can never occur because 4 is the minimum stipulated in the game scenario.

8 members can never be selected because of the first rule.

7 members can never occur because either M or W would have to be selected in the group of 7, and as detailed earlier they limit the total number of members to 5.

6 could be selected by eliminating both M and W from the team, and choosing all six of the remaining employees.
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 Salvi627
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#16584
I just have a question about setting up the diagram for this problem, At first i thought about setting up 3 different diagrams each depicting if either M S or W were in it... But i found that didn't yield too much information. I ended up getting only one question wrong, because the questions seemed easy to figure out, but I didn't have any original diagram to base them off..
Thanks!
 David Boyle
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#16592
Salvi627 wrote:I just have a question about setting up the diagram for this problem, At first i thought about setting up 3 different diagrams each depicting if either M S or W were in it... But i found that didn't yield too much information. I ended up getting only one question wrong, because the questions seemed easy to figure out, but I didn't have any original diagram to base them off..
Thanks!
Hello Salvi627,

These may help:

slash mo
slash mp
s :arrow: pt
slash p or slash t :arrow: slash s
w :arrow: my
slash m or slash y :arrow: slash w
slash ms (because s leads to pt)
If wmy, then only t or z can fit in, or both of them (not s, o, or p).
If spt, then y, o, or z, or several of them, can fit in (not w or m).
m can only go with w, y, t, or z, or several of them.

Hope that helps,
David
 emilyfoster2013
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#17443
Could I please have some help with the October 2013 Logic Game #2?

Other than diagramming all the rules and their contrapositives I couldn't get much further with inferences. I was able to connect the notP on rule 1 to notS in the contrapositive of rule 2. I also connected the notm in the contrapositive of rule 1 to the notw in the contrapositive of rule 3. I also recognized that I would probably need to use an in group and an out group but could not make any specific assignments to either of these.

Once I moved into the questions it became clear to me that I was probably missing some major inferences. Would it be possible for you to direct me where to proceed from this point?

Thank you so much :)

Emily
 BethRibet
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#17448
Hi Emily,

You caught most of the major inferences (you're off to a good start).

It can be a helpful with a game like this to lay out what's true if certain variables are included. For instance if Wong is included, M & Y are, S, O & P are not, and Z is optional.

W: M,Y ~S, ~O, ~P Z?


Similarly if Schmidt is in:

S: P, T ~M, ~W O? Z?

If you have questions about any specific question, feel free to write back.

Good luck!
Beth
 angelsfan0055
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#86058
I had some difficulty with this game, and I think in part it was because I neglected to use the double not arrows. I was able to get a few of the inferences, that If M is selected, S can't be and that if P and O are selected, W can't be, but I missed some others.
I guess my question is, what in this setup triggered using a double not arrow? I've seen in certain grouping games on here when I've used them that the explanation doesn't include them. Is there anything in particular about this game that made it a good candidate for not arrows?
Thank you!
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 KelseyWoods
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#86140
Hi angelsfan0055!

The Double Not Arrow can be used whenever you have two variables that cannot go together. This frequently happens when you have a rule that says if you have one thing, then you do not have something else.

For example, if I have a rule that states "If A is selected, then B is NOT selected."
The diagram would be: A :arrow: B
The contrapositive would be: B :arrow: A

That's basically telling me that if I have A then I do not have B and if I have B then I do not have A. Thus, A and B can never be together. So instead of writing out the two separate contrapositives, I can make this rule more clear and more efficient by diagramming it with the Double Not Arrow, which essentially just combines the two contrapositives:
A :dblline: B

Here, the first rule tells us "If Myers is on the team, neither Ortega nor Paine can be." That's really just telling us that M cannot be with O and M cannot be with P. So I would diagram that with Double Not Arrows:
M :dblline: O
M :dblline: P

Sometimes in Grouping Games, you might see us use Not Blocks to represent similar rules instead of Double Not Arrows. But it is never wrong to use a Double Not Arrow instead of a Not Block to represent a rule that says you can't have two variables together. And oftentimes it is better to use Double Not Arrows because they can more easily be combined with other rules to create inferences.

Hope this helps!

Best,
Kelsey
 angelsfan0055
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#86171
this does help, thank you Kelsey!
I'll keep that in mind going forward

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