Diagram

[nanagyanewa]

hello,

Could someone please help me with the diagram for this game? It looked easy at first but the last two conditions confused me a little bit. I read the explanation in the text but i still do not get it. Thanks for your help!

[bbirdwell]

So, when you symbolize those two conditional constraints, you get:

S--P --> M--S
and
M--S --> S--P

Hmmm.... This means that either one of these situations (M before S or S before P) is sufficient to create the other. In other words, these two things always happen together.

The efficient way to symbolize this is with a double arrow:

S--P <--> M--S
and the contrapositive means that if either one of these situations does not occur, then neither of them do.
So you can do it one of two ways:
~(S--P) <--> ~(M--S)
or actually negate them by flipping them, since there are no ties.
P--S <--> S--M

At this point you might notice that this relationship between P, S, and M is the point of connection between the first two constraints (R--P--L, T--M--V).

Since there are two distinct roads to follow, setup two possible trees. One tree works if S--P, the other works if P--S. See attached .pdf.

Make sense?

[nanagyanewa]

thanks, that helped.

[BarryM800]

I fully understand the double arrow of S-P ↔︎ M-S when Rules #3 and #4 are combined, which creates the two possibilities, M-S-P and P-S-M. But based on Rule #3 alone, S-P → M-S, thus M-S-P; then, its contrapositive, S-M → P-S, thus P-S-M. So there's that two possibilities already. Wouldn't Rule #4 be redundant? I'm really confused about the co-existance of these two rules. Even though Rule #4 changes Rule #3 from a one-way conditional statement to a bi-directional conditional statement, it does not seem to affect what Rule #3 is already saying. Can someone shed some light on this? Thanks!

[ohthatpatrick]

R - P - L
T - M - V

Yeah, it changes very little to have both rules. As you said, this is rule 3 by itself

S - P --> M - S ..... thus, M - S - P
S - M --> P - S .... thus, P - S - M

But are we guaranteed to have M - S- P or P - S -M on every go?

No, not unless you're guaranteed to always trigger
S - P -->
or
S - M -->

But you're not. S doesn't have to be before either of them. It's possible that S would be after P and after M, and then neither M - S - P nor P - S - M would be triggered.

For example, if we only had the first three rules, then we could write a legal scenario such as
R - P - L - T - M - V - S

However, that scenario would break rule 4. So rule 4 does change the game somewhat.

Hope this helps.