by ohthatpatrick Tue Nov 13, 2018 2:59 pm
From our setup, we knew that
...........R
........./
... [ ? ]
.........\
...........S ---> [ ? ]
If P is the only one transmitting to 2, then P has to be the thing that transmits to R and S.
Updating our diagram, we have
...........R
........./
.... P
.........\
...........S ---> [ ? ]
We also knew that
R/T --> Q
T/U --> P
We'd love to know whether it's T or U that leads to P.
....................R
................../
[ T/U ] --> P
..................\
...................S ---> [ ? ]
I would just draw each of those as frames to figure out what happens next.
...........R
........./
T -> P
........\
...........S ---> [ ? ]
or
...........R
........./
U --> P
.........\
...........S ---> [ ? ]
For frame 1, if T --> P, then T can't lead to Q (only P gets to lead to two things for Q23).
So if T --> P, then R ---> Q.
And since U is the only one left and S needs to lead to someone, that would S --> U.
Updated frame 1:
...........R ---> Q
........./
T -> P
.........\
...........S ---> U
For frame 2, if U --> P, then it could still be T --> Q or R ---> Q.
That's kinda messy, so maybe we see what answers we can kill at this point.
(A) frame 1 is a counterexample
(B) in frame 1, but seemingly not necessary, since frame 2 would work with U ---> P
(C) Maybe? Keep it.
(D) frame 1 is a counterexample
(E) in frame 2, U would lead to P.
This means that (C) is almost definitely our answer. As long as frame 2 is possible, it will rule out (B) and (E).
If we want to fully prove that to ourselves, we can arbitrarily pick whether we're doing T --> Q or R --> Q for frame 2. I'll pick the former.
...........R ---> Q
........./
U -> P
.........\
...........S ---> T
This proves (B) and (E) wrong (as well as D, but we had already killed that from frame 1).
Hope this helps.