CODEWITHSUNDEEP
pythonfor-loop
Nasif Imtiaz Ohi
Nasif Imtiaz Ohi

Reputation: 1713

modifying conditions in for in python

hey this is a code that works fine. i am using in a bigger code and the program gives quick answer for like 999999 or 9999999.

prime=[2,3,5]
f=7
def next_prime(f): #whenever called, measures the next prime of the list and returns it
    j=len(prime)
    while j==len(prime):
        for x in prime:
            if f%x==0:
                f+=2
                break
        else:
            prime.append(f)
    return f

But if i modify the code a little bit (I want add the condition that x should be less than f**.5), the program gives no results.

prime=[2,3,5]
f=7
main=[]
power=[]
def next_prime(f):
    j=len(prime)
    while j==len(prime):
        for x in prime:
            if (x<int(f**.5)+1) and f%x==0:
                f+=2
                break
        else:
            prime.append(f)
    return f

Where is the mistake?

Upvotes: 1

Views: 109

Answers (2)

Luka Rahne
Luka Rahne

Reputation: 10447

First of all this code does not work even numbers. You should write your code somthing like.

def next_prime(n):
   assert(n>0)
   while 1:
      n+=1
      if isprime(n):
          return n

and if you can use any isprime() implementations.

You can write it on your own or use famous miller-rabin algorithm that works fast for any prime as long as you have enough ram to hold computations in it.

http://ideone.com/XO74b

Upvotes: 1

Zozz
Zozz

Reputation: 2025

Both versions work fine in my python interpreter (2.7). Although both your versions will not return any result for an even number, so you probably should add a check for this condition:

f=7
main=[]
power=[]
prime = [2,3,5]
def next_prime(f):

    #Added check:
    if not f%2:
        f += 1

    j=len(prime)
    while j==len(prime):
        for x in prime:
            if (x<int(f**.5)+1) and f%x==0:
                f+=2
                break
        else:
            prime.append(f)
    return f

Upvotes: 2

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