求素数

Question

The problem is that you need to find the prime number after the number input, or if the number input is prime, return that. It works fine. It's just not working when the input is print(brute_prime(1000)) . It returns 1001 not 1009. The full code is this:

def brute_prime(n):
    for i in range(2, int(n**(0.5))):
        if n % i == 0:
            n += 1
    else:
        return n

Answer 1

You're not restarting the for i loop when you discover that a number is not prime and go to the next number. This has two problems: you don't check whether the next number is a multiple of any of the factors that you checked earlier, and you also don't increase the end of the range to int(n ** 0.5) with the new value of n .

def brute_prime(n):
    while true:
        prime = true
        for i in range(2, int(n ** 0.5)+1):
            if n % i == 0:
                prime = false
                break
        if prime:
            return n
        n += 1

break will exit the for loop, and while true: will restart it after n has been incremented.

Answer 2

As Barmar suggests, you need to restart the loop each time you increment n. Your range also ends earlier than it should, as a range stops just before the second argument.

def brute_prime(n):
    while True:
        for i in range(2, int(n**(0.5)) + 1):
            if n % i == 0:
                break
        else:
            return n
        n = n+1

Answer 3

remember 2 is a prime number. again you can just check the division by 2 and skip all the even number division

 def brute_prime(n):
        while True:
            if n==2:return n
            elif n%2 ==0 or any(n % i==0 for i in range(3, int(n**(0.5)+1),2)):
                n += 1
            else:
                return n

Answer 4

as mention by Chris Martin the wise solution is define a isPrime function separately and use it to get your desire number.

for example like this

def isPrime(n):
    #put here your favorite primality test

from itertools import count

def nextPrime(n):
    if isPrime(n):
        return n
    n += 1 if n%2==0 else 2
    for x in count(n,2):
        if isPrime(x):
            return x

if the given number is not prime, with n += 1 if n%2==0 else 2 it move to the next odd number and with count check every odd number from that point forward.

for isPrime trial division is fine for small numbers, but if you want to use it with bigger numbers I recommend the Miller-Rabin test (deterministic version) or the Baille-PSW test . You can find a python implementation of both version of the Miller test here: http://rosettacode.org/wiki/Miller%E2%80%93Rabin_primality_test#Python