质数小于或等于n

Question

The statement says:

Write a function era1() that ask the user for a number n and then use this algorithm to print all prime numbers less than or equal to n .

Algorithm:

• Write a list with numbers from 2 to largest integer N you want to calculate.
• The first number in the list is a prime number. Write this number a list of primes, B .
• Remove from the list A , the first element and its multiples.
• If the first number in the list A is less than the square root N, come back to second point.
• The numbers in the B list and those left in List A are all primes searched.

Now, I put this code:

import math

def primo(num):
    if num < 2:
        return False

    i = 2
    for i in range(2, int(math.sqrt(num) + 1)):
        if (num % i == 0):
            return False

    return True

def main():
    n = input("Introdueix un nombre: ")
    B = range(2, n)
    for i in B:
        if primo(i):
            print i        

main()

def era1():
    n = input("Introdueix un nombre: ")
    A = range(2, n + 1)
    B = [A[0]]

    for i in A:
        if i % 2 == 0:
            A.remove(i)

    if A[0] < math.sqrt(n):
        print B + A

era1()

The result is incorrect because if I remove one of the input, appears error and I have to put only one time the input. Also the result is incorrect because the A + B , the list B is not the list of function main and the final result is only the numbers not multiples of 2 and 2. How can I put only one input and then the final result will be correct?

Answer 1

This Algorithm is called Sieve of Eratosthenes .

It is a simple algorithm for finding all prime numbers up to a specified integer. It was created in the 3rd century BC by Eratosthenes, an ancient Greek mathematician.

In order to develop this algorithm, we'll go through the different above-mentioned steps.

• First, we generate a list with numbers from 2 to largest integer N you want to calculate.

 A = range(2, n + 1) 

• We use another list C, as we might use A later to print the initial list.

• We go through C, processing all the numbers less than the square root N.

• We initialize an empty list B, and add every time a prime number (Which is first element of the list).
• We use list comprehension to filter multiplies using : (x%firstElement!=0) .

C= [x for x in C if x%firstElement!=0]

• B is the union of the rest of numbers (prime numbers that are larger than the square root N), and the prime numbers that we detected already.

Your code should look like:

def era1():
    n = input("Introduce a nombre: ")
    #n=120 #To test the
    A = range(2, n + 1) 
    B, C= [],A
    while C[0]< math.sqrt(n): #Condition
        firstElement= C[0]
        B+= [firstElement] #The first number in the list is a prime number. Write this number a list of primes, B.
        C= [x for x in C if x%firstElement!=0] #We use comprehension List to filter multiplies using
    return B+C #The numbers in the B list and those left in List A are all primes searched.

print era1()

Output in case of n=120: [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113]

This picture visualize the Algorithm, Source of picture .

Answer 2

Removing items from a list while iterating over it will have unexpected results, it interferes with the indexing.

a = [1,2,3,4,5,6,7,8,9]
for thing in a:
    a.remove(thing)


>>> a
[2, 4, 6, 8]
>>> 

You will need to figure out another way to accomplish that - perhaps create a new list with the items you want to keep.