Setup and Rule Diagrams

[Dave Killoran]

Setup and Rule Diagram Explanation

This is a Grouping: Partially Defined game.

This game is classified as Partially Defined because each committee must have at least three members, but can have more. At first glance, the game appears as if it may be difficult. However, as you work through the questions, the game turns out to be fairly reasonable.

This game also serves as a useful reminder that you cannot assume that variables are always placed in just one group. No rules establish that a volunteer is a member of only one committee, but the last two rules explicitly allow for volunteers to be members of both committees. The first rule also allows for the possibility that not every volunteer must be a member of a committee (and the fourth rule suggests this as well).

Accounting for the game scenario and the first rule, the initial setup is as follows:

PT25-Jun 1998 LGE-G1_srd1.png

The remaining four rules are fairly simple, and result in the following diagram:

PT25-Jun 1998 LGE-G1_srd2.png

Three of the seven volunteers—G, H, and L—are randoms, and so your focus must be on the four non-random volunteers.

Also, please note that you cannot make the inference that F and J do not serve on the same committee together. The relationships are not presented in a manner that allows you to make that connection (the arrow between K and J would have to be reversed in order to draw that inference).

[rameday]

Hello,

So I made the inference that F J cannot go together because if F K and k J doesn't that mean F J. Because of this incorrect inference I only got the list question wrong and I was still able to get the other 4 questions right. But I still for the life of me don't understand why that is a wrong inference.


A

[Nikki Siclunov]

Hi rameday,

The following two conditional statements cannot be linked in a conditional chain that produces a valid additive inference:

F K

K J

According to the first statement, F and K cannot be selected together. In other words, F NOT K (and, by the contrapositive, K NOT F). So, if F is selected, K cannot be selected. This has no effect on whether or not J is selected: to conclude that J is not selected would be to make a Mistaken Negation, because J can be selected even if K isn't. Thus, a chain of this type produces no viable inferences: F K J

Notice how in the chain above the straight arrow ( ) goes away from the middle term K. For us to make a valid inference, the arrow must go towards that term. For instance, if the rules stated: A B C, then we can easily conclude that since A requires B, but B cannot be selected with C, it follows that A cannot be selected with C (A C).

Thanks!

[rameday]

ok that makes perfect sense thanks.

[francken]

Could someone please clarify the rule of, if K then J. I understood that to mean that if K is present then J has to be present but if J is present then K does not have to be present. So for question #2, if K is replacing a letter on either committee, then wouldn't F be correct because J doesn't need K's presence to remain in the trails committee but H would remain to be the letter in common and F can not be with K?

Brittany

[Adam Tyson]

Once we put K on a committee, Brittany, then that rule requires that J also be on that same committee. That's why K cannot replace anyone on the Planting committee. If it did, then that committee would need J, and J isn't there! K must replace someone on the Trails committee other than J to avoid violating that rule.

If K replaced F on the Planting committee, then the Planting committee would have K but not J, a clear violation.

Let us know if that clears it up for you!

[francken]

To clarify, the way the question is phrased seemed to throw me off. It is stating that K was already on the trails committee and not where K could go if it had yet to be assigned,correct? Because then K would not be affecting J if it was already there? Please explain if J can be on either committee without K being required first.

[James Finch]

Hi Brittany,

No, the question is saying that the committees are composed as follows:

Planting: F, H, L, M

Trails: G, H, J

With K being the only variable not on either committee.

We're then being asked to replace one of the already-filled committee spots with K, without breaking the rules. Because of the K J, we have to put K on the Trails committee, and our only possibilites for replacement would be G and H. However, there must be one member in common between the two groups, and the only member in common is H, so we can't kick H out. This leaves only G, making (B) the correct answer choice.

Hope this clears things up!

[blade21cn]

Hmm ... I don't think in Question #2 "(C) H" is eliminated because there must be one member in common between the two groups and the only member in common is H and so we cannot kick H out. If H gets replaced by K, then K will be the "new" member that the two groups have in common and therefore there's no violation of the last rule. Rather, H is eliminated as a carryover from the P group, where F is also present. So any variable from group P, except F, cannot be replaced by K, which includes H. F itself cannot be replaced for a different reason, because the group has to have J, due to the first rule.

[Adam Tyson]

But the question is about replacing a member on "a committee," meaning ONE of the committees, Blake, not both! If K replaces H on one committee, then you have H on one and K on the other and nothing in common!