def prime(n):
    if n==2 or n==3:
        return True
    if n==1:
        return False
    f=True
    for i in range(2,int(n**0.5) 1):
        if n%i==0:
            f=False 
            break
    return f
def n_primes(n):
    primes=[2,3,5,7]
    k=10
    count=4
    itr=[1,3,7,9]
    while count!=n:
        for i in itr:
            if prime(k i):
               primes.append(k i)
               count =1 
        k =10
    return primes
print(len(n_primes(100)))

This program is returning n primes for almost any number that I've tried except 100 for 100 it's returning 101 prime numbers I can easily get 100 primes by slicing the list but I want to know what is wrong with my program I've tried terminating the while loop using count, length of the list but result is not changing

CodePudding user response：

Not only for 100, it would be an issue for other numbers as well. For example:

print(n_primes(5))

[2, 3, 5, 7, 11, 13, 17, 19]

It is because you do the count!=n check only once every 4 iterations (after iterating through the 4 numbers of itr). Also, you should do count<n otherwise you would go into an infinite loop. To avoid infinite loop and get exact number of primes do another check after incrementing count.

def prime(n):
    if n==2 or n==3:
        return True
    if n==1:
        return False
    f=True
    for i in range(2,int(n**0.5) 1):
        if n%i==0:
            f=False 
            break
    return f

def n_primes(n):
    primes=[2,3,5,7]
    k=10
    count=4
    itr=[1,3,7,9]
    while count<n:
        for i in itr:
            if prime(k i):
               primes.append(k i)
               count =1
               if (count>=n):
                  break 
        k =10
    return primes

print(n_primes(5)) # prints [2, 3, 5, 7, 11]
print(len(n_primes(100))) # prints 100