Question

One way to find prime numbers less than n is to treat them as array indices. Mark each position as True initially, assuming that all numbers are prime. Then, starting with index 2, mark all multiples of 2 (greater than 2) as not prime (False). Repeat this for multiples of 3, then multples of 4, etc. When the algorithm terminates, only prime indices will still be True; everything else will be False.

The following function takes an integer n as a parameter and returns an array of prime integers less than n. Note that in the final step, we use list comprehension to fetch the indices whose values are True and place those indices as values in a new list. All displayed code is correct. Fill in the missing components, using spaces only where Python requires them (the minimum possible spacing).

def filter(n):
if n<3:
    return []
  primes=[True for i in range(n)]
  primes[0]=False

i=
while i<:
    while i<n and is False:
      i=
    for j in range():
      primes[j]=
    i=
return [i for i in range(2,n) if primes[i] is True]

Answer 1

code:

def filter(n):
if n<3:
return []
primes=[True for i in range(n)]
primes[0]=False
i=2
while i<n:
while i<n and primes[i] is False:
i=i+1
for j in range(2*i,n,i):
primes[j]=False
i=i+1
return [i for i in range(2,n) if primes[i] is True]

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