P1.
all good:
synchronized
in-synch
deterministically

P2.
3 points:
3-6
all a, all b -> 6
interleave a1 b1 a2 b2 a3 b3 ....

minimum 2 -> +2 points


P3.
3 points deadlock
-1 point if you have something else as a problem

P4.
//Thread A          //Thread B
line a1				line b1
s1.signal			s2.signal
s2.wait				s1.wait
line a2				line b2

deadlock
//Thread A          //Thread B
line a1				line b1
s2.wait				s1.wait
s1.signal			s2.signal
line a2				line b2

//Thread A          //Thread B
line a1				line b1
s2.wait				s2.signal
s1.signal			s1.wait
line a2				line b2

wrong
//Thread A          //Thread B
					s2.signal
line a1				line b1
s2.wait				
s1.signal			s1.wait
line a2				line b2

-1 point for overly synchonizing

extra n threads:
T1        	T2        	T3
s1-2.sig	s2-1.sig	s3-1.signal
s1-3.sig	s2-3.sig	s3-2.signal

s2-1.w    	s1-2.w		s1-3.w
s3-1.w    	s3-2.w		s2-3.w



Another solution:
sem m[n];

Thread i
========
m[i].signal
foreach j except i 
	m[j].wait
	m[j].signal
	

Another solution:
T0
s[1].signal
sem[0].wait

T1
s[1].wait
s[2].signal
sem[1].wait

T2
s[2].wait
s[3].signal
sem[2].wait

....
Tn
s[n].wait
sem[n-1].signal ..... sem[0].signal


Another solution:
sem zzz[n-1]
sem aaa[n-1]

T i=1..N-1
zzz[i].signal
aaa[i].wait


T N
foreach z in zzz
	z.wait
foreach a in aaa
	a.signal


P5.

P6.

P7.
-2 points for counting
-1 point for thing outside of critical section that should be inside

P8.
barrier solution that ALWAYS expects 5 threads to be done
4 points


P8.
2.5 points for each of lightswitch and multiplexer

LS1.wait(m)               	LS2.wait(m)
s4.wait						s6.wait
s4.singal					s6.signal
LS1.signal(m)             	LS2.signal(m)