Classical IPC Problems:
1. Dining Philosophers Problem
2. The Readers and Writers Problem
3. The Sleeping Barber Problem
1.
Dining
philosophers problems:
There are N philosphers sitting
around a circular table eating spaghetti and discussing philosphy. The problem
is that each philosopher needs 2 forks to eat, and there are only N forks, one
between each 2 philosophers. Design an algorithm that the philosophers can
follow that insures that none starves as long as each philosopher eventually
stops eating, and such that the maximum number of philosophers can eat at once.
· Philosophers
eat/think
· Eating needs 2 forks
· Pick one fork at a time
· How to prevent deadlock
The
problem was designed to illustrate the problem of avoiding deadlock, a system state in which no progress is
possible.
One
idea is to instruct each philosopher to behave as follows:
· think
until the left fork is available; when it is, pick it up
· think until the right fork is available; when
it is, pick it up
· eat
· put the left fork down
· put the right fork down
· repeat from the start.
This solution is
incorrect: it allows the system to reach deadlock. Suppose that all five
philosophers take their left forks simultaneously. None will be able to take their
right forks, and there will be a deadlock.
We could modify the
program so that after taking the left fork, the program checks to see if the
right fork is available. If it is not, the philosopher puts down the left one,
waits for some time, and then repeats the whole process. This proposal too,
fails, although for a different reason. With a little bit of bad luck, all the
philosophers could start the algorithm simultaneously, picking up their left
forks, seeing that their right forks were not available, putting down their
left forks, waiting, picking up their left forks again simultaneously, and so
on, forever. A situation like this, in which all the programs continue to run indefinitely
but fail to make any progress is called starvation.
The solution presented
below is deadlock-free and allows the maximum parallelism for an arbitrary
number of philosophers. It uses an array, state, to keep track of whether a
philosopher is eating, thinking, or hungry (trying to acquire forks). A
philosopher may move into eating state only if neither neighbor is eating.
Philosopher i's neighbors are defined by the macros LEFT and RIGHT. In other
words, if i is 2, LEFT is 1 and RIGHT is 3.
Solution:
#define N 5 /* number of philosophers */
#define LEFT (i+N-1)%N /* number of i's left neighbor */
#define RIGHT (i+1)%N /* number of i's right neighbor */
#define THINKING 0 /* philosopher is thinking */
#define HUNGRY 1 /* philosopher is trying to get
forks */
#define
EATING 2 /* philosopher is eating */
typedef int semaphore; /* semaphores are a special kind
of int */
int state[N]; /* array to keep track of
everyone's state */
semaphore
mutex =1; /* mutual
exclusion for critical regions */
semaphore
s[N]; /* one semaphore
per philosopher */
void philosopher(int i) /* i: philosopher number, from 0 to
N1 */
{
while (TRUE)
{ /* repeat forever */
think(); /* philosopher is thinking */
take_forks(i); /* acquire two forks or block */
eat(); /* yum-yum, spaghetti
*/
put_forks(i); /* put both forks back on table
*/
}
}
void
take_forks(int i) /* i:
philosopher number, from 0 to N1 */
{
down(&mutex); /* enter critical region
*/
state[i]
= HUNGRY; /* record fact that
philosopher i is hungry */
test(i); /* try to acquire 2 forks
*/
up(&mutex); /* exit critical region
*/
down(&s[i]); /* block if forks were not
acquired */
}
void
put_forks(i) /* i:
philosopher number, from 0 to N1 */
{
down(&mutex); /* enter critical region
*/
state[i]
= THINKING; /* philosopher has
finished eating */
test(LEFT); /* see if left neighbor can
now eat */
test(RIGHT); /* see if right neighbor can
now eat */
up(&mutex); /* exit critical region */
}
void
test(i) /* i: philosopher number,
from 0 to N1* /
{
if
(state[i] == HUNGRY && state[LEFT] != EATING && state[RIGHT] !=
EATING)
{
state[i]
= EATING;
up(&s[i]);
}
}
Readers Writer problems:
The
dining philosopher’s problem is useful for modeling processes that are
competing for exclusive access to a limited number of resources, such as I/O
devices. Another famous problem is the readers and writers problem which models
access to a database (Courtois et al., 1971). Imagine, for example, an airline
reservation system, with many competing processes wishing to read and write it.
It is acceptable to have multiple processes reading the database at the same
time, but if one process is updating (writing) the database, no other process
may have access to the database, not even a reader. The question is how do you
program the readers and the writers?
One
solution is shown below.
Solution to Readers Writer problems
typedef
int semaphore; /* use
your imagination */
semaphore
mutex = 1; /* controls
access to 'rc' */
semaphore
db = 1; /* controls
access to the database */
int
rc = 0; /* #
of processes reading or wanting to */
void
reader(void)
{
while
(TRUE)
{ /* repeat forever */
down(&mutex); /* get exclusive access to 'rc'
*/
rc
= rc + 1; /* one reader
more now */
if (rc == 1) down(&db); /* if this is the first reader ...
*/
up(&mutex); /* release exclusive access
to 'rc' */
read_data_base(); /* access the data */
down(&mutex); /* get exclusive access to 'rc'
*/
rc = rc
1; /* one reader
fewer now */
if (rc == 0) up(&db); /* if this is the last reader ...
*/
up(&mutex); /* release exclusive access to
'rc' */
use_data_read(); /* noncritical region */
}
}
void
writer(void)
{
while (TRUE)
{ /* repeat forever */
think_up_data(); /* noncritical region */
down(&db); /* get exclusive access
*/
write_data_base(); /* update the data */
up(&db); /* release exclusive
access */
}
}
In
this solution, the first reader to get access to the data base does a down on
the semaphore db. Subsequent readers merely have to increment a counter, rc. As
readers leave, they decrement the counter and the last one out does an up on
the semaphore, allowing a blocked writer, if there is one, to get in.
No comments:
Post a Comment