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Add doctest to maths/sieve_of_eratosthenes.py and remove other/finding_primes.py (#1078)
Both of the two files implemented sieve of eratosthenes. However, there was a bug in other/finding_primes.py, and the time complexity was larger than the other. Therefore, remove other/finding_primes.py and add doctest tomaths/sieve_of_eratosthenes.py.
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"""Sieve of Eratosthones."""
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# -*- coding: utf-8 -*-
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"""
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Sieve of Eratosthones
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The sieve of Eratosthenes is an algorithm used to find prime numbers, less than or equal to a given value.
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Illustration: https://upload.wikimedia.org/wikipedia/commons/b/b9/Sieve_of_Eratosthenes_animation.gif
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Reference: https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
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doctest provider: Bruno Simas Hadlich (https://github.com/brunohadlich)
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Also thanks Dmitry (https://github.com/LizardWizzard) for finding the problem
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"""
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import math
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import math
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def sieve(n):
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def sieve(n):
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"""Sieve of Eratosthones."""
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"""
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Returns a list with all prime numbers up to n.
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>>> sieve(50)
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[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
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>>> sieve(25)
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[2, 3, 5, 7, 11, 13, 17, 19, 23]
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>>> sieve(10)
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[2, 3, 5, 7]
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>>> sieve(9)
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[2, 3, 5, 7]
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>>> sieve(2)
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[2]
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>>> sieve(1)
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[]
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"""
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l = [True] * (n + 1)
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l = [True] * (n + 1)
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prime = []
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prime = []
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start = 2
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start = 2
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end = int(math.sqrt(n))
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end = int(math.sqrt(n))
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while start <= end:
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while start <= end:
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# If start is a prime
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if l[start] is True:
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if l[start] is True:
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prime.append(start)
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prime.append(start)
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# Set multiples of start be False
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for i in range(start * start, n + 1, start):
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for i in range(start * start, n + 1, start):
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if l[i] is True:
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if l[i] is True:
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l[i] = False
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l[i] = False
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start += 1
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start += 1
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for j in range(end + 1, n + 1):
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for j in range(end + 1, n + 1):
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@ -1,21 +0,0 @@
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'''
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-The sieve of Eratosthenes is an algorithm used to find prime numbers, less than or equal to a given value.
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-Illustration: https://upload.wikimedia.org/wikipedia/commons/b/b9/Sieve_of_Eratosthenes_animation.gif
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'''
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from __future__ import print_function
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from math import sqrt
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def SOE(n):
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check = round(sqrt(n)) #Need not check for multiples past the square root of n
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sieve = [False if i <2 else True for i in range(n+1)] #Set every index to False except for index 0 and 1
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for i in range(2, check):
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if(sieve[i] == True): #If i is a prime
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for j in range(i+i, n+1, i): #Step through the list in increments of i(the multiples of the prime)
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sieve[j] = False #Sets every multiple of i to False
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for i in range(n+1):
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if(sieve[i] == True):
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print(i, end=" ")
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