Diagram

[missbernadette]

I believe that this is a binary grouping game...May I see the setup for this one please? I have a setup in mind, but I'm thinking that this game deals more with inferences...

[bbirdwell]

Hey Bernadette!

Since this game involves two groups, and it's a "closed" setup, we can bet that the number of adults vs number of children on each boat will play an important role, so i added this information to the T-chart, which proves quite helpful in games like this.

This diagram can get you everything you need for this game, though there is one more "advanced" inference you could use on the T-chart. Can you figure it out? Hint: it involves only one of the given constraints.

Think about it and let me know.

As for this diagram, here's how I thought it out:
Since there must be one adult on each side, we can infer that the maximum number of children on any side is 3. Since there are 5 children, we can infer that at least 2 kids must be on each side. This leaves one blank slot on each side. One will be filled with a child, one with an adult.

Since X and Y can never go together, there is an x/y option on either side.

How's that?

[missbernadette]

Hello Brian!
Wow, I really need a Dunce hat right about now because I diagrammed the if F then G constraint totally wrong (I diagrammed it as if F is in, G is out, and vice versa).

The only other inference that I could think of was

_ _ x/z v/w l _ _ x/z v/w

Which is if either v/w is in boat 1, then w/v is in boat 2, which would put two children on both sides. But, this cannot work if either v/w is in boat 2 because that is not the contrapositive of the constraints, therefore, we do not know what happens if v/w is boat 2. In fact, both v and w can be in boat 2 and it would not go against the rule (at least thats what I'm thinking).
I also see that if h is in boat 1, g cannot be in boat 1 because that would push f in boat 1 as well, which we cannot do. So if h is in boat 1, it is either in with 3 children or f and two children. G and F can both be in boat 1 along with two children. The opposite goes for boat 2. If h is in boat 2, then f cannot be in boat 2 because it puts g in with it which would not work. h can be in boat 2 with either three children or g and two children.
Am I close to the inference that you were talking about?

[bbirdwell]

In fact, both v and w can be in boat 2 and it would not go against the rule (at least thats what I'm thinking).

Yes! This is it! But the way you symbolized it does not represent this statement. This is the tricky part, but a really helpful thing to know.

It's OK for v and w to be in boat 2 together, but they can NEVER be in boat 1 together. So what does that tell you? No matter what, ONE of them must be in boat 2. Got that? Only one of them can go in boat 1, so at least one of them MUST be in boat 2 (and, as you said, maybe both of them are in boat 2).

So the way to symbolize that is like this:

__ __ __ __ l __ __ __ v/w

If you do it as you have written, we infer that one of them has to go in boat 1, which isn't true. See the difference?

Nice work!

(the rest of what you said about h and g is true, but there's nothing snappy enough there to be deemed a useful "inference," know what i mean?)

[chike_eze]

Can't we also infer that because of rule 1, if F is in (2) then H must be in (1) ?

My Binary grouping diagram was similar to this except for
F in 2 --> H in 1 (and contra)

I think Bernadette pointed this out. I guess it is not as consequential as other inferences?

[ManhattanPrepLSAT1]

Hey Chike! That inference is absolutely correct. You can track that information in the logic chain or you can deal with the implication of the first constraint as you go through your hypotheticals as you work through each question.

But that inference is absolutely correct, great work!

[OliviaB660]

Please help! I realized over the course of this game that I had misinterpreted the v/w and H/G rules (as discussed above), but I really don't understand why my interpretation was wrong. I do not understand why, for example, v & w can be in Boat 2 together. Isn't the contrapositive of [V1 -> W2] [-W2 -> -V1], which would, in this context, mean [W1 -> V2]?

Thank you!

[Misti Duvall]

Sure! I get where you're coming from, but that doesn't mean V and W can't be in group 2 together. For example, if - W2, then that means W1 (cause that's the only other group) --> - V1, then that means V2 (cause that's the only other group).

But what if we start with W2? In that case, nothing is triggered. Same if we start with V2. The condition (ie the trigger for the rule) is only applicable if V1 or - W2 (ie, W1). There's no condition that is triggered if either is in group 2.