by ohthatpatrick Fri Jun 09, 2017 1:41 pm
I would have done all the "IF" questions first, so I'd have several scenarios already written on my page.
So first I would go through those scenarios and find which letters have previously transmitted to two viruses and eliminate those answer choices.
If we're starting from a blank slate, we might start by thinking about who we've been using to do rule 3.
If we had frames, we would look at all the people in our frames who have been "two-pronged".
Otherwise, just start pluggin' and chuggin'.
Can P give to 2?
Let's make it give to R and S
..... R
.../
P
...\
.....S
Let's try to get other people in there
........... R -- Q
........../
T -- P
..........\
.............S
Where could U go?
It actually MUST go with S, since S still needs to give the virus to something.
........... R -- Q
........../
T -- P
..........\
.............S --- U
So we can eliminate (A), since P can give to 2 computers.
Can Q be two-pronged?
........... R
........../
.......Q
..........\
.............S
Let's make it work
........... R
........../
T --- Q
..........\
.............S --- U
P is left.
We could put it after U or after T.
Since Q works, eliminate (B).
Since we could make T two-pronged in this diagram by having T go
Now we think about which parts of this diagram could be interchangeable.
With our freedom to put U pretty much anywhere, can we make anyone else two-pronged?
T could give to U and P. Eliminate (D).
So U and R are left. Could U be the two-pronged that goes to R and S?
........... R
........../
.......U
..........\
.............S
Make it work:
........... R --- Q
........../
.......U
..........\
.............S --- T --- P
It works!
Eliminate (E) and pick (C).
Along the way, with those plug-n-chug experiments, we may have developed a sense of why R won't work.
We never have room to get the 2-prongs to the right of R and S.
S is only allowed to have one letter, and BECAUSE we have to give S a letter, we never end up with two letters to give to R.