the answer is D. without the right tags the content wont be accurately indexed
Answer:
Explanation:
The system will be deadlock free if the below two conditions holds :
Proof below:
Suppose N = Summation of all Need(i), A = Addition of all Allocation(i), M = Addition of all Max(i). Use contradiction to prove.
Suppose this system isn't deadlock free. If a deadlock state exists, then A = m due to the fact that there's only one kind of resource and resources can be requested and released only one at a time.
Condition B, N + A equals M < m + n. Equals N + m < m + n. And we get N < n. It means that at least one process i that Need(i) = 0.
Condition A, Pi can let out at least 1 resource. So there will be n-1 processes sharing m resources now, Condition a and b still hold. In respect to the argument, No process will wait forever or permanently, so there's no deadlock.
Answer:
The function is as follows:
def divisible_by(listi, n):
mylist = []
for i in listi:
if i%n == 0:
mylist.append(i)
return mylist
pass
Explanation:
This defines the function
def divisible_by(listi, n):
This creates an empty list
mylist = []
This iterates through the list
for i in listi:
This checks if the elements of the list is divisible by n
if i%n == 0:
If yes, the number is appended to the list
mylist.append(i)
This returns the list
return mylist
pass
