CS414:	HW #3 Solution 

(This is not the only solution; there can be many correct variations)

1. Gilbert and Shoe

semaphore MUTEX = 1;
semaphore BATHROOM = 1;
Semaphore X, Y = 1;

The Gilbert process:
Repeat
	P(MUTEX)
	P(Y)
	P(X)
	V(MUTEX)
	V(Y) 
	P(BATHROOM)
	V(X)
	V(BATHROOM)
until false

The Shoe process:
Repeat
	P(MUTEX)
	P(X)
	P(Y)
	V(MUTEX)
	V(X)
	V(Y)
	P(BATHROOM)
	V(BATHROOM)
until false

To get magazine X, each process has to get the semaphore MUTEX first, so if
the MUTEX semaphore can't be possesssed by two processes at the same time, we
can guarantee the mutual exclusion of magazine X. In fact, since MUTEX is 
initialized to 1, only one process has the access to it at the same.
	
Suppose these process will deadlock. Then both the Gilbert process and the Shoe
process hold one semaphore and request the other one held by the other process.
However, according to the pseudocode, such situation can never happen.

2. Barbershop

const CHAIRS = 5;
semaphore barber_free = 1;
semaphore mutex = 1;
int WC = 0; //number of waiting customer

Barber process:

while (true)
P(mutex)
if WC !=0
	WC = WC - 1;
	V(mutex);
	snip_snip //cut hair
	V(barber_free);
else
	V(mutex);

Customer process:

P(mutex)
if (WC < CHAIRS)
	WC = WC + 1;
	V(mutex)
	P(barber_free)
	get_my_haircut //get haircut and leave
else 
	V(mutex)