Problem 2: H2O Building Problem
There are two kinds of threads, oxygen and hydrogen. In order to assemble these threads into
water molecules, we have to create a barrier that makes each thread wait until a complete
molecule is ready to proceed. As each thread passes the barrier, it should invoke bond. You must
guarantee that all the threads from one molecule invoke bond before any of the threads from the
next molecule do.
In other words:
• If an oxygen thread arrives at the barrier when no hydrogen threads are present, it has to wait
for two hydrogen threads.
• If a hydrogen thread arrives at the barrier when no other threads are present, it has to wait for
an oxygen thread and another hydrogen thread.
We don’t have to worry about matching the threads up explicitly; that is, the threads do not
necessarily know which other threads they are paired up with. The key is just that threads pass
the barrier in complete sets; thus, if we examine the sequence of threads that invoke bond and
divide them into groups of three, each group should contain one oxygen and two hydrogen
threads.
Puzzle: Write synchronization code for oxygen and hydrogen molecules that enforces these
constraints.
Problem 3: Dining Savages Problem
A tribe of savages eats communal dinners from a large pot that can hold M servings of
stewed missionary. When a savage wants to eat, he helps himself from the pot, unless it is
empty. If the port is empty, the savage wakes up the cook and then waits until the cook has
refilled the pot.
Any number of savage threads run the following code:
Unsynchronized savage code
1. while true:
2.
getServingFromPot()
3.
eat()
And one cook thread runs this code:
Unsynchronized cook code
1. while true:
2.
putServingsInPot(M)
The synchronization constraints are:
 Savages cannot invoke getServingFromPot if the pot is empty.
 The cook can invoke putServingsInPot only if the pot is empty.
Please add code for the savages and the cook that satisfies the synchronization constraints.
Solution1 for the barbershop problem:
customers = 0
mutex = Semaphore(1)
customer = Semaphore(0)
barber = Semaphore(0)
<Customer>
1
mutex.wait()
2
if customers == n+1:
3
mutex.signal()
4
leave()
5
customers += 1
6
mutex.signal()
7
8
customer.signal()
9
barber.wait()
10 getHairCut()
11
12 mutex.wait()
13
customers -= 1
14 mutex.signal()
<Barber>
1
customer.wait()
2
barber.signal()
3
cutHair()
Solution2 for the barbershop problem:
Semaphore Customers = 0
Semaphore Barber = 0
Semaphore accessSeats (mutex) = 1
int NumberOfFreeSeats = N //total number of seats
<Customer>
while(true) { //runs in an infinite loop
SemWait(accessSeats) //tries to get access to the chairs
if ( NumberOfFreeSeats > 0 ) { //if there are any free seats
NumberOfFreeSeats-- //sitting down on a chair
SemSignal(Customers) //notify the barber, who's waiting that there is a customer
SemSignal(accessSeats) //don't need to lock the chairs anymore
SemWait(Barber) //now it's this customer's turn, but wait if the barber is busy
//here the customer is having his hair cut
} else { //there are no free seats
//tough luck
SemSignal(accessSeats) //but don't forget to release the lock on the seats
//customer leaves without a haircut
}
}
<Barber>
while(true) { //runs in an infinite loop
//tries to acquire a customer - if none is available he goes to sleep
SemWait(Customers)
//at this time he has been awakened - want to modify the number of available seats
SemWait(accessSeats)
NumberOfFreeSeats++ //one chair gets free
SemSignal(Barber) //the barber is ready to cut
SemSignal(accessSeats) //we don't need the lock on the chairs anymore
//here the barber is cutting hair
}
Solution for Building H2O Problem:
mutex = Semaphore(1)
oxygen = 0
hydrogen = 0
oxyQueue = Semaphore(0)
hydroQueue = Semaphore(0)
<Oxygen>
1
mutex.wait()
2
oxygen += 1
3
if hydrogen >= 2:
4
hydroQueue.signal()
5
hydroQueue.signal()
6
hydrogen -= 2
7
oxyQueue.signal()
8
oxygen -= 1
9
else
10
mutex.signal()
11
12 oxyQueue.wait()
13 bond()
14 mutex.signal()
<Hydrogen>
1
mutex.wait()
2
hydrogen += 1
3
if hydrogen >= 2 and oxygen >= 1:
4
hydroQueue.signal()
5
hydroQueue.signal()
6
hydrogen -= 2
7
oxyQueue.signal()
8
oxygen -= 1
9
else
10
mutex.signal()
11
12
hydroQueue.wait()
13
bond()
Solution for the dining savages problem:
servings = 0
mutex = Semaphore(1)
emptyPot = Semaphore(0)
fullPot = Semaphore(0)
<Cook>
1
while true:
2
emptyPot.wait()
3
putServingsInPot(M)
4
fullPot.signal()
<Savage>
1
while true:
2
mutex.wait()
3
if servings == 0:
4
emptyPot.signal()
5
fullPot.wait()
6
servings = M
7
servings -= 1
8
getServingFromPot()
9
mutex.signal()
10
11
eat()