Explanation:
Mid-Square Method:-In this method some it takes some digits from the square of a number.This number is a random number.This number can be used as a key for hashing.
This technique is mostly used in hashing so decrease the collision in hash maps.
This method has a limitation that when we square a large number it may go out of range of integer so we have handle those cases.
Answer: The correct answer is A. I just answered this question myself!
Answer: Incomplete question
Explanation: The question attached is incomplete and incoherent, possible solution of this will be difficult to assume. Take adequate time to post complete questions.
Regards.
Answer:
int sumid=0; /* Shared var that contains the sum of the process ids currently accessing the file */
int waiting=0; /* Number of process waiting on the semaphore OkToAccess */
semaphore mutex=1; /* Our good old Semaphore variable ;) */
semaphore OKToAccess=0; /* The synchronization semaphore */
get_access(int id)
{
sem_wait(mutex);
while(sumid+id > n) {
waiting++;
sem_signal(mutex);
sem_wait(OKToAccess);
sem_wait(mutex);
}
sumid += id;
sem_signal(mutex);
}
release_access(int id)
{
int i;
sem_wait(mutex);
sumid -= id;
for (i=0; i < waiting;++i) {
sem_signal(OKToAccess);
}
waiting = 0;
sem_signal(mutex);
}
main()
{
get_access(id);
do_stuff();
release_access(id);
}
Some points to note about the solution:
release_access wakes up all waiting processes. It is NOT ok to wake up the first waiting process in this problem --- that process may have too large a value of id. Also, more than one process may be eligible to go in (if those processes have small ids) when one process releases access.
woken up processes try to see if they can get in. If not, they go back to sleep.
waiting is set to 0 in release_access after the for loop. That avoids unnecessary signals from subsequent release_accesses. Those signals would not make the solution wrong, just less efficient as processes will be woken up unnecessarily.
