Q23

 
ShehryarB30
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Vinny Gambini
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Q23

by ShehryarB30 Fri Nov 09, 2018 2:05 pm

Could you pls explain this question?
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ohthatpatrick
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Atticus Finch
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Re: Q23

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.