使用python進行素數計算

Question

我知道，關於素數已經有很多問題了，但我並沒有要求提供代碼。我只是想知道，什麼是錯的我的（我希望意見將幫助你理解，我在做什麼）：

from math import ceil 

def isPrimeNumber (n, out='short'): #Checks, whether n is a prime number 
    answer = 'Yes' 
    for p in range(2, int(ceil(n**0.5))+1): #Checks all numbers lower than 
              #SQRT(n), if they are divisors of n 
     if n%p == 0: #If p is a divisor, n isn't prime 
      answer = 'No' 
      if out == 'verbose': 
       print 'Least divisor is', p 
      return False 
      break 
    if answer == 'Yes': #If there isn't a p, that is a divisor, n is prime 
     if out == 'verbose': 
      print 'No divisors except for 1 and', str(n)+'!' 
     return True 

def primeNumbers (start = 1, stop = 1000, numbers = 0): 
    N = stop 
    if numbers == 0: #Calculates all prime numbers in N numbers in a row 
        #(N=1000 -> calculates all prime numbers in 1000 numbers, 
        #by default from 1 to 997) 
     primes = [] 
     for i in range(start, N+1): 
      if isPrimeNumber(i) == True: 
       primes.append(i) 
    elif numbers == 1: #Calculates N prime numbers in a row 
         #(N=1000 -> calculates 1000 prime numbers) 
     primes = [start] 
     i = len(primes) 
     j = 1 
     while i <= N: #Stops, when we get N prime numbers - doesn't work! 
      n = max(primes) + 1 
      while j != 'stop': 
       if isPrimeNumber(n, out='short') == True: 
        primes.append(n) 
        i = i + 1 
        j = 'stop' #Stops nested cycle, when reached 
           #the first prime number 
       else: 
        n = n + 1 
    else: 
     print 'Wrong input! 3rd number in function call must be either 0 or 1' 
    return primes 

功能isPrimeNumber（）工作正常。當數字= 0時，函數primeNumbers也可以正常工作。但是，如果數量= 1，那麼，因爲它似乎，週期永不停歇的一個，我不能understan爲什麼...

Answer 1

問題是，你的j變量最終被設置爲'停止'，然後再也沒有設置回來，所以while j!='stop'只能在第一次運行。

# don't initialize j here 
    while i <= N: #Stops, when we get N prime numbers - doesn't work! 
     n = max(primes) + 1 
     j = 1 #initialize it here 
     while j != 'stop': 
      if isPrimeNumber(n, out='short') == True: 
       primes.append(n) 
       i = i + 1 
       j = 'stop' #Stops nested cycle, when reached 
          #the first prime number 
      else: 
       n = n + 1 

Answer 2

你有一個無限循環就在這裏：

while i <= N: 
     n = max(primes) + 1 ### resetting `n'! 
     while j != 'stop': 
      if isPrimeNumber(n, out='short') == True: 
       ... 
       j = 'stop' 
      else: 
       n = n + 1 

一旦j是設置爲'stop'，你永遠不會改回它。一旦發生這種情況，內部while實際上成爲無操作，將外部while變成無限循環。

Answer 3

我只是想我會清理有點...

from math import ceil 
def isprime (n,out="short"): 
    answer = True 
    for p in range(2,int(ceil(n ** 0.5)) + 1): 
     if n % p == 0: 
      answer = False 
      if out == "verbose": 
       print "Least Divisor: " + str(p) 
      return False 
    if answer: 
     if out == "verbose": 
      print "No Divisors (Except For 1 & " + str(n) + "!" 
     return True 
def primenumbers (start = 1,stop = 1000,numbers = False): 
    N = stop 
    if numbers: 
     primes = [start] 
     i = len(primes) 
     j = 1 
     while i <= N: 
      n = max(primes) + 1 
      j = true 
      while j: 
       if isPrimeNumber(n): 
        primes.append(n) 
        i = i + 1 
        j = false 
       else: 
        n = n + 1 
    else 
     primes = [] 
     for i in range(start,N + 1): 
      if isPrimeNumber(i): 
       primes.append(i) 
    return primes 

我也稍微改變實際的程序。