Setup and Rule Diagrams

[huhjunn]

I'm running into a brick wall trying to diagram rules 3 and 4.

This is how I have attempted to diagram all the rules:

1. R>P>L

2. T>M>V

3. S>P M>S (Contrapositive: S>M P>S)

4. M>S S>P (Contrapositive: P>S S>M)

From here I create a M>S>P Chain and a P>S>M Chain for rule #4, and then from this chain I diagram the other two possible sequences of S>P>M and S>M>P (I added these two sequences because a similar situation exists in the LGB on pg 382 of the 2013 version).

It is at this point that I am completely lost on what to do.


Could anyone help me figure out what I'm doing wrong?

[Robert Carroll]

h,

You're part of the way there, but look more carefully at the interaction of the third and fourth rules. The fourth rule merely swaps the sufficient and necessary conditions of the third rule, so the combination of these rules is a double arrow:

S > P M > S

So either S > P and M > S or P > S and S > M. Chaining each pair, the possibilities are:

M > S > P

or

P > S > M

These are the only two possibilities. The other two possible sequences you have can't be true. Think about it - in the first, S > P, but then the third rule tells us M > S, so that doesn't work. In the second, S > P again so the third rule tells us M > S. Either way it doesn't work. The two chains you had, and the two I listed above, are the only possibilities.

Once you get to this point, you need two main diagrams, one for each SPM chain. Since the first and second rules are consistent across all possibilities, they don't change. So diagram the first rule, the second rule, and M > S > P, branching as needed and using arrows as needed, just like with any sequencing diagram. Then, on a separate diagram, diagram the first rule, the second rule, and P > S > M, with branching, arrows, and all the normal sequencing elements.

Because of the dual possibilities for that chain, you need two main diagrams rather than one, but for each of the two, you treat it as its own sequencing game diagram. Identify which can be ranked best and which ranked least for each of the two possible situations.

Let's look at a question to see how this impacts your approach. Question 7 says: "If Pastilla ranks second highest, then which one of the following is a complete and accurate list of restaurants any one of which could be ranked fourth highest?"

If P is second, then we must be dealing with the "P > S > M" possibility, as in the other one, P needs too many ranked ahead of it to be second. So for this question, the local condition tells you which of the two diagrams to use.

I hope this has helped!

Robert Carroll

[huhjunn]

Thank you for the explanation it makes a lot more sense now. However, I'm having trouble diagramming the two sequence templates.

This is what I have come up with for sequence template 1 (combining rule 1, 2, and M>S>P):

http://i.imgur.com/QV1a163.png

Is this correct?



Also, how can you tell when a conditional sequencing rule creates more than two chains? For instance the example on pg. 382 (2013 version LGB) shows a sequencing conditional statement that creates 4 chains: "If A's presentation is earlier than B's presentation, then B's presentation is earlier than C's presentation."

This example rule in the LGB is similar to the fourth rule of the game we are currently discussing. This is what confused me and led me to create 4 different chains.

[Lucas Moreau]

Hello, huhjunn,

First, that is indeed a correct diagram for sequence template 1! Well done! The relationship between V, S, and R with respect to M and P is complicated, but you portrayed it correctly. Good job.

As for determining whether you would have two or four chains, you'd just have to look at the specific rule interactions. Here, we have a double arrow between S > P and M > S, so if one is true, the other is also true. That means that there are only two possibilities: either both are true or neither is true. Just one can't be true without the other one being true, so that trims us from four down to two.

As far as the other game, that would be:

If A > B, B > C

The trick here is that it is possible for B > C to be true without A > B also being true. We're not in a "both or neither" situation, one is capable of being true while the other is false!

Hope that helps,
Lucas Moreau

[huhjunn]

That helps very much. Many thanks!