You can add row, table, columns or cells
Answer:
Explanation:
Following are the Semaphores:
Customers: Counts waiting customers;
Barbers: Number of idle barbers (0 or 1)
mutex: Used for mutual exclusion.
Cutting: Ensures that the barber won’t cut another customer’s hair before the previous customer leaves
Shared data variable:
count_cust: Counts waiting customers. ------------copy of customers. As value of semaphores can’t access directly.
// shared data
semaphore customers = 0; semaphore barbers = 0; semaphore cutting = 0; semaphore mutex = 1;
int count_cust= 0;
void barber() {
while(true) { //shop is always open
wait(customers); //sleep when there are no waiting customers
wait(mutex); //mutex for accessing customers1
count_cust= count_cust-1; //customer left
signal(barbers);
signal(mutex);
cut_hair();
}
}
void customer() {
wait(mutex); //mutex for accessing count_cust
if (count_cust< n) {
count_cust= count_cust+1; //new customer
signal(customers); signal(mutex);
wait(barbers); //wait for available barbers get_haircut();
}
else { //do nothing (leave) when all chairs are used. signal(mutex);
}
}
cut_hair(){ waiting(cutting);
}
get_haircut(){
get hair cut for some time; signal(cutting);
}
Answer:
e(a) = 0
e(b) = 10
e(c) = 110
e(d) = 1110
Explanation:
The Worst case will happen when f(a) > 2*f(b) ; f(b) > 2*f(c) ; f(c) > 2*f(d) ; f(d) > 2*f(e) and f(e) > 2*f(f).
Where f(x) is frequency of character x.
Lets consider the scenario when
f(a) = 0.555, f(b) = 0.25, f(c) = 0.12, f(d) = 0.05, f(e) = 0.02 and f(f) = 0.005
Please see attachment for image showing the steps of construction of Huffman tree:- see attachment
From the Huffman tree created, we can see that endcoding e() of each character are as follows:-
e(a) = 0
e(b) = 10
e(c) = 110
e(d) = 1110
e(e) = 11110
e(f) = 11111
So we can see that maximum length of encoding is 5 in this case.
