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Learn Python Coding

Learn Python Coding

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Learn Python through simple, practical examples and real coding ideas. Clear explanations, useful snippets, and hands-on learning for anyone starting or improving their programming skills. Admin: @HusseinSheikho || @Hussein_Sheikho

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✨ How to Create a Django Project ✨ 📖 Learn how to create a Django project and app in clear, guided steps. Use it as a refere
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Accelerating the Sieve of Eratosthenes 1. Quickly recall the algorithm Classic implementation:
def eratosthenes(n):
    is_prime = [True] * (n + 1)
    is_prime[0] = is_prime[1] = False

    for i in range(2, int(n ** 0.5) + 1):
        if is_prime[i]:
            for j in range(i * i, n + 1, i):
                is_prime[j] = False

    return is_prime
Time — O(N log log N). We're not interested in the asymptotics, but in how much we can speed up the implementation itself. 2. Optimization #1 — don't bother with even numbers The idea is simple: * all even numbers except 2 are composite * if we only work with odd numbers, we reduce the array size and the number of iterations by about half Implementation:
def eratosthenes_odd(n):
    if n < 2:
        return []

    size = (n + 1) // 2
    is_prime = [True] * size
    is_prime[0] = False

    limit = int(n ** 0.5) // 2
    for i in range(1, limit + 1):
        if is_prime[i]:
            p = 2 * i + 1
            start = (p * p) // 2
            for j in range(start, size, p):
                is_prime[j] = False

    return is_prime
3. Optimization #2 — use bytearray instead of list[bool] Thought: * bool in Python is an object * bytearray is a tightly packed buffer * less overhead and better fits into the CPU cache Example:
def eratosthenes_bytearray(n):
    is_prime = bytearray(b"\x01") * (n + 1)
    is_prime[0:2] = b"\x00\x00"

    for i in range(2, int(n ** 0.5) + 1):
        if is_prime[i]:
            for j in range(i * i, n + 1, i):
                is_prime[j] = 0

    return is_prime
4. Optimization #3 — a hybrid of the two approaches
def eratosthenes_fast(n):
    if n < 2:
        return []

    size = (n + 1) // 2
    is_prime = bytearray(b"\x01") * size
    is_prime[0] = 0

    limit = int(n ** 0.5) // 2
    for i in range(1, limit + 1):
        if is_prime[i]:
            p = 2 * i + 1
            start = (p * p) // 2
            is_prime[start::p] = b"\x00" * (((size - start - 1) // p) + 1)

    return is_prime
5. Time comparison Test with n = 10_000_000: >>> eratosthenes.py real  0.634s >>> eratosthenes_odd.py real  0.245s >>> eratosthenes_bytearray.py real  0.801s >>> eratosthenes_fast.py real  0.028s Conclusions: * skipping even numbers (#1) gives ~2.6× speedup * bytearray itself doesn't speed up — it's more about memory * the hybrid (#3) gives ~22.6× speedup Key trick in #3:
is_prime[start::p] = b"\x00" * (((size - start - 1) // p) + 1)
There's no Python loop here — everything is done by a C-level operation on the slice. On such tasks, this makes a huge difference. General idea: in Python, we often speed up not the asymptotics, but the memory model and the number of passes over the data. Loops + memory → the main factors. 👉 @DataScience4

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