Python：带平方根的素数算法实现

Question

I am trying to solve this problem and despite wrapping my head around it I get stucked over and over and over.

I have a list made of one element [2] and given that and knowing that in order to get a prime number $N$ I have to check that it does not have any prime factor smaller or equal than $\\sqrt{N}$, generate all prime numbers between 3 and 100

My code is the following and I will try to explain how I came up with it:

prime_numbers = [2]

for k in range(3,100):  # k runs on all numbers to check them
    for i in range (0,len(prime_numbers)): # i is the index of the elements in my list
        if prime_numbers[i] <= sqrt(k):    # first I need to check the sqrt condition
            if k % prime_numbers[i] ==0:     # check that element in list divides k
                k+=1                         # if so k++ and break the cicle
                break
            else:
                i +=1                        # otherwise go to the next elem of the list
        else:                              #if the sqrt condition is not met I have found a prime number    
            value = k
        prime_numbers.append(value)

When I run this code it gets into a loop with no exit, and I cannot individuate where the issue is. My best guess would be to correct the most indented else condition, but all the attempts I have done have resulted in failure. Thanks to everyone who is willing to partecipate.

Answer 1

One way to get there is:

from math import sqrt

prime_numbers=[2]

for k in range(3,101):
    prime=True
    for i in range(2, int(sqrt(k))+2):
        if k%i == 0:
            prime=False
            break
    if prime:
            prime_numbers.append(k)

print(prime_numbers)

#Output:
#[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]

Notes:

• It's range(3,101) and not range(3,100) if assuming you want to check all numbers including 100, otherwise your last number checked will be 99.
• Have a boolean variable prime ; start by assuming the number is prime ( prime=True )
• sqrt(k) returns a float, while your range expects integers; as such - int(sqrt(k)) . It's +2 to ensure your range 'to' value is never smaller than your 'from' value of 2.
• if k%i == 0 (if you have a divisor) the number is not prime ( prime=False ) and break the for loop to move to the next k
• This could be my "pythonic" by doing a list comprehension, but it may be easier for you to understand in the current format

Answer 2

This works:

prime_numbers = [2]

for prime_candidate in range(3, 100):
    for known_prime in prime_numbers:
        if known_prime <= sqrt(prime_candidate):
            if prime_candidate % known_prime == 0:
                break
    else:
       prime_numbers.append(prime_candidate)

Your problem mainly was, that i += 1 does not prevent prime_numbers.append to append again.

The for/else can be substituted by a separate flag to trigger all cases without a break happening earlier.

Answer 3

Thanks to everyone who suggested a solution. Meanwhile I came up with one myself.

prime_numbers = [2]


for k in range(3,100):
    for i in range (0,len(prime_numbers)):
        if prime_numbers[i] <= sqrt(k):
            if k % prime_numbers[i] ==0:
                k+=1
                break
            else:
                i +=1
        else:
            value = k
    if(value > max(prime_numbers)):
        prime_numbers.append(value)