Need Help! Count Primes using python

Question

Count the number of prime numbers less than a non-negative number, n. I have created the following code but the complexity is too high. I would really be grateful if some one give me a better resolution.

import math
class Solution(object):
    def countPrimes(self, n):
        PrimeCount=0
        primelist=[]
        for i in range(2,n):
            if self.primeCheck1(i,primelist)==True:
                primelist.append(i)       #try2 with new logic
                PrimeCount=PrimeCount+1

        return PrimeCount



    def primeCheck1(self,n,primelist):
        flag=False
        if n==2:
            return True
        elif n==3:
            return True
        sqroot=int(math.sqrt(n))

        for j in range(0,sqroot):
            if n%primelist[j]==0:
                flag=True
                break

        if flag!=True:
            return True
        else:
            return False

look here : https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n — gold_cy, Jul 30 '17 at 01:57
This is probably better posted on https://codereview.stackexchange.com — AChampion, Jul 30 '17 at 02:23
Possible duplicate of [Fastest way to list all primes below N](https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n) — obgnaw, Jul 30 '17 at 02:26

score 2 · Accepted Answer · answered Jul 30 '17 at 09:25

I like the way you build a list of primes and use that list to detect larger primes. You are right that your code is more complex than it needs to be, specifically your flag variable is not needed.

You have missed a few tricks in your primeCheck1() method as well that can speed things up. Since my Python is non-existent, this is in Python-like pseudocode. I assume you will be able to convert it.

def primeCheck1(self, n, primelist):

  # Handle even numbers.
  if n % 2 == 0:
        # The only even prime is 2.
        return (n == 2)

  # Handle multiples of 3.
  if n % 3 == 0:
        return (n == 3)

  # Handle remaining numbers.
  sqroot = int(math.sqrt(n))

  for j in range(0, sqroot):
    if n % primelist[j] == 0:
      return False  # Not a prime number.

  # If we get this far then the number is prime.
  return True

end primeCheck1

On a minor point, you have a habit of not spacing out your code: a==b instead of a == b. Using spaces as separators makes code easier to read.

score 1 · Answer 2 · answered Aug 08 '17 at 18:45

A rework of your code that speeds it up 3X counting the number of primes less than a million:

class Solution(object):
    def countPrimes(self, n):
        primelist = []

        for i in range(2, n):
            if self.primeCheck(i, primelist):
                primelist.append(i)

        return len(primelist)

    def primeCheck(self, n, primes):

        if n < 2:
            return False

        if n % 2 == 0:  # 2 is the only even prime
            return n == 2

        for prime in primes:

            if prime * prime > n:  # easier to square many than sqroot one
                return True

            if n % prime == 0:  # divisible by a prime, not a prime
                return False

        return True

solution = Solution()

print(solution.countPrimes(1000000))

This includes some of the tricks @rossum pointed out to you.

However, we can do better. Here's a reimplementation using @ClockSlave's prime sieve code that's 22 times faster than your original when counting the number of primes less than a million:

class Solution(object):

    def __init__(self):
        self.sieve = None

    def countPrimes(self, n):

        self.sieve = [True] * n

        if n > 0:
            self.sieve[0] = False
            if n > 1:
                self.sieve[1] = False

        def mark_sieve(prime):
            for index in range(prime + prime, n, prime):
                self.sieve[index] = False

        for number in range(2, int(n ** 0.5) + 1):
            if self.sieve[number]:
                mark_sieve(number)

        return sum(self.sieve)

solution = Solution()

print(solution.countPrimes(1000000))

The win here is eliminating all those expensive divisions.

whackamadoodle3000 · Answer 3 · 2017-07-30T04:27:26.833

0

First install the sympy package with pip install sympy. Then try this:

import sympy
PrimeCount=0
n=int(input("Number?"))
for elem in range(n):
   if sympy.isPrime(elem):
      PrimeCount+=1
print(PrimeCount)

edited Jul 30 '17 at 04:27

answered Jul 30 '17 at 01:54

whackamadoodle3000

6,185
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3

I'm pretty sure the OP has to do this for school. So using `sympy` would defeat the whole purpose of the assignment. – Christian Dean Jul 30 '17 at 01:56
1

Also, there is not `++` operator in Python. You need to use `+= 1` instead. – Christian Dean Jul 30 '17 at 02:06
oops, was using java for a while. Fixed it – whackamadoodle3000 Jul 30 '17 at 04:28

Shaon shaonty · Answer 4 · 2017-08-09T06:06:28.727

0

You can use bitwise sieve algorithm.

SIZE = 1000000
LIMIT=int(math.sqrt(SIZE)+1)

prime = []

def sieve():
    for i in range(SIZE/32):
        prime[i]=0xffff
        prime[0]&=~(1<<(1%32))

    for i in range(LIMIT+1):
        if(prime[i/32]&(1<<(i%32))):
            for j in range(2*i, SIZE, j=j+i):
                prime[j/32]&=~(1<<(j%32))

def isPrime(n):
        return prime[n/32]&(1<<(n%32))

edited Aug 09 '17 at 06:06

answered Jul 30 '17 at 02:06

Shaon shaonty

1,097
1
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Did you, or the person who up voted this, actually try to run it? First, it doesn't appear to count the number of primes, just generate primes. Second, it doesn't appear to be runnable Python, e.g. `def isPrime(int n):`, `range(2*i, SIZE, j=j+i)`, etc. – cdlane Aug 08 '17 at 17:45
I just show him a example of code which reduce time complexity – Shaon shaonty Aug 09 '17 at 06:08

score 0 · Answer 5 · edited Aug 20 '19 at 16:42

0

def count_primes(prime):

    check=2
    main_res=0

    while check <= prime:
        result=0
        control=1
        while control <= check: 
            if check%control==0:
                result+=1
                control+=1
            else:
                control+=1
        if result==2:
            main_res+=1


        check+=1

    return main_res


# Test for COUNT PRIME
#print(count_primes(100))

edited Aug 20 '19 at 16:42

N Alex

986
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answered Aug 20 '19 at 16:04

Mastan Abdulkhaligli

1
2

The question was about finding the prime numbers in a better complexity. The code you posted although it works, it's not in better complexity. – N Alex Aug 20 '19 at 16:19

Need Help! Count Primes using python

5 Answers5