Python Program for Sieve of Eratosthenes (original) (raw)

Given a number n, the task is to find and print all the prime numbers smaller than or equal to n. A prime number is a number greater than 1 that is divisible only by 1 and itself. For example:

**Input: n = 20
**Output: 2, 3, 5, 7, 11, 13, 17, 19

Let’s explore different methods to find all prime numbers up to a given number in Python.

Using Sieve of Eratosthenes

In this method, we start by assuming all numbers are prime and progressively mark the multiples of each prime as non-prime. This efficiently eliminates composites using a boolean list.

Python `

import math n = 30 prime = [True for _ in range(n + 1)] prime[0], prime[1] = False, False

for p in range(2, int(math.sqrt(n)) + 1): if prime[p]: for i in range(p * p, n + 1, p): prime[i] = False

for i in range(2, n + 1): if prime[i]: print(i, end=" ")

`

Output

2 3 5 7 11 13 17 19 23 29

**Explanation:

• math.sqrt(n) reduces the upper limit for factor checks, improving efficiency.
• Multiples of each prime are marked False in the list.
• Finally, numbers that remain True are printed as primes.

Using NumPy for Vectorized Sieve

Here, we use NumPy arrays to speed up marking non-prime numbers with vectorized operations. It’s faster for large n due to optimized array handling.

Python `

import numpy as np import math

n = 30 prime = np.ones(n + 1, dtype=bool) prime[0:2] = False

for p in range(2, int(math.sqrt(n)) + 1): if prime[p]: prime[p*p:n+1:p] = False

print(np.nonzero(prime)[0].tolist())

`

Output

[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]

**Explanation:

• np.ones() initializes a boolean array to mark potential primes.
• Vectorized slicing marks multiples of each prime as False.
• np.nonzero(prime)[0] returns indices of all remaining True values, i.e., the primes.

Using List Comprehension

In this concise method, list comprehension checks each number’s divisibility by all numbers up to its square root and keeps only primes. It’s compact but slower than the sieve methods.

Python `

import math n = 30 p = [x for x in range(2, n + 1) if all(x % i != 0 for i in range(2, int(math.sqrt(x)) + 1))] print(p)

`

Output

[2, 3, 5, 7, 11, 13, 17, 19, 23, 29]

**Explanation:

• For each number, divisibility is checked up to its square root.
• all() ensures no divisor divides x evenly.
• The result is a list of prime numbers up to n.

Simple Iterative Check

This method checks divisibility for each number up to n. Though simple and easy to understand, it’s computationally expensive for large numbers.

Python `

n = 30 for num in range(2, n + 1): for i in range(2, num): if num % i == 0: break else: print(num, end=" ")

`

Output

2 3 5 7 11 13 17 19 23 29

**Explanation:

• Each number is tested for divisibility by all smaller numbers.
• If no divisor divides it evenly, it’s identified as a prime and printed.

Please refer complete article on Sieve of Eratosthenes for more details!