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#27194
Setup and Rule Diagram Explanation

This is a Grouping/Linear Combination Game.

This game requires you to focus on two separate functions: establishing the group and then ordering the group:
June 03_M12_game#2_L7_explanations_game#2_setup_diagram_1.png
Any variable that is in the group of five must then be ordered, and we have shown that relationship with an arrow from the group to the order.

As the first three rules are applied, we can begin to fill in the diagram:
June 03_M12_game#2_L7_explanations_game#2_setup_diagram_2.png
The last three rules are more complex, and must be discussed in detail:

The fourth rule can be diagrammed as follows:
  • GH :arrow: H :longline: G
This rule can be difficult to apply. Some students make the mistake of assuming that the rule implies that G can never be first and that H can never be fifth; this is only true if both G and H are tested. If only G or only H is tested, then the sufficient condition of the rule would not be enacted and the rule would not apply.

If we link the fourth rule with the second rule, we can make the inference that if H is tested then F must be tested first. Here is why: if H is selected, then if G is selected we know from the fourth rule that H would be tested ahead of G; according to the second rule, either F or G must be ranked first, so if H is tested, then G could never be first and we can infer that F would have to be first:
  • H :arrow: F1
Because the fifth rule is identical in structure to the fourth rule, a similar analysis can be applied. The fifth rule can be diagrammed as follows:
  • FK :arrow: K :longline: F
As with the fourth rule, some students make the mistake of assuming that the rule implies that F can never be first and that K can never be fifth; this is only true if both F and K are tested. If only one is tested, then the sufficient condition of the rule would not be enacted and the rule would not apply.

If we combine the fifth rule with the second rule, we can make the inference that if K is tested then G must be tested first. Here is why: if K is selected, then when F is selected (and we will discover below that F must be selected) we know from the fourth rule that K must be tested ahead of F; according to the second rule, either F or G must be ranked first, so if K is tested, then F could never be first and we can infer that G would have to be first:
  • K :arrow: G1
The final rule can be diagrammed as follows:
  • ..... ..... ..... ..... ..... ..... F
    M ..... ..... :arrow: ..... ..... ..... +
    ..... ..... ..... ..... ..... ..... H

This rule is extremely restrictive because the selection of M automatically adds two other members of the group. And, because the group already includes L and I, the selection of M yields only one possible group of five cold medications: M-F-H-L-I. When this group is ordered, F must be ranked first, and, of course, L must be second.

One final inference remains, and this inference is tricky indeed. An examination of the rules reveals that F is a critical variable: F appears in three of the six rules while no other variable appears in more than two of the rules. In fact, F is so critical that F must be one of the five cold medications that are tested. Consider the following: if F is not selected to be tested, then by the contrapositive of the last rule we know that M cannot be selected for testing. This situation forces the remaining five cold medications—G, H, I, K, and L—to comprise the entire testing group. Because both G and H are included in this group, the fourth rule is enacted, and we know H must rank better than G. This causes a violation of the second rule, which requires either F or G to be ranked first. Hence, because an acceptable group cannot be selected when F is not included, we can infer that F must be included in the group of five cold medications tested in the study. The group of cold medications is therefore as follows:
June 03_M12_game#2_L7_explanations_game#2_setup_diagram_3.png
The group above is not given in order; although F can be first, if G is also tested then G could be first instead of F. Please note also that because F must be selected, from the action of the fifth rule we can deduce that K can never be tested last.

Of additional note is that there are no randoms in this game.
 TonySteep
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#49064
Can you also make the inference that the ranked group must have either M or G, but NOT both?

Thanks for the help!
 Jonathan Evans
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#49117
Hi, Tony,

Yes, that is accurate. The setup cannot include both M & G because that would force too many medications to be selected. It is also true that it is not possible for the study to include neither M nor G because in that case it would have to include K and F, which would imply K — F and that F must be 1st, creating a contradiction.

Good observation!
 GGIBA003@FIU.EDU
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#89069
Hi P.S.,
I made some inferences that weren't part of the explanations. Can someone please tell me if they are correct or incorrect.

Inference#1: F1 :dblline: K--- F [G could go 1st. If F is in, it must be 3rd, 4th, or 5th because K must go before it]

From Inference 1 I made a Limited Solution Set of: G, L, K, I , F (M and H are out)
Which lead me to further Inference #2: If K-- F :dblline: M and If K-- F :dblline: H

Inference #3: G1 :dblline: H-- G [ F could go in. If G is in, it must be 3rd, 4th, or 5th because H must go before it]

From Inference #3 I made a Limited Solution set of: K, L, H, I, G (F and M are out)
Which lead me to further inference #4: F :dblline: G. [F or G must be in BUT NOT BOTH because there isn't enough room for the rest of the variables since it's 5 in and 2 out- a total of 7).

Thanks in advance!
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 atierney
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#89196
Hi,

As to the inference #1, if F is first, then yes, K cannot be in the solution set (and therefore not go before F). Likewise, if we know K is tested, then M and H must be out, and we have our five that are tested K, F, G, I, and L, with G, L [I, K F], which I believe is inference #2.

For Inference #3, if G is first, then H and M are necessarily out. If G is first, then H can't be in the solution set, and therefore, M can't be in the solution set either. If G is first, F must appear in the solution set, so your inference #4 is incorrect actually. In general, as indicated in question 8, F must appear in every solution of the game.

In conclusion, #1, #2, and #3, are correct, but your limited solution set from inference #3 is wrong, and I believe that's what led to deriving incorrectly inference #4. Let me know if you have further questions on this, and I hope this helps!
 GGIBA003@FIU.EDU
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#89344
Yes it does help, thank you. I see now that I misinterpreted rule 2 which is why I made incorrect inferences. As a follow up question- when a rule has "EITHER" "OR" in LG game, is it correct to imply/read it "ONLY"? In this case "Either F or G ranks first" means "ONLY F or G can be 1st". And just because 1 of them is first DOESN'T mean the other variable can't be placed in a different ranking, correct? Meaning that, F or G COULD BOTH BE IN as long as 1 of them is first. Am I understanding that correctly?

Thanks in advance!
 Robert Carroll
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#89863
G,

I'll refer here about "only" in this game: viewtopic.php?p=89601#p89601

F and G can both be in. One of them is first; the other may be some position that's not first or second.

Robert Carroll

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