To solve this problem, we need to compute the sum of all prime numbers up to a given number ( n ). A prime number is a number greater than 1 that has no divrs other than 1 and itself. The most efficient way to find all primes up to ( n ) is using the Sieve of Eratosthenes algorithm, which allows us to mark non-prime numbers in linear time relative to ( n ).

Approach

The Sieve of Eratosthenes works as follows:

1. Initialize: Create a boolean array (or list) of size ( n+1 ) where each element is set to True (indicating the number is initially considered prime), except indices 0 and 1 which are set to False (since they are not primes).
2. Mark Non-Primes: For each number ( i ) starting from 2 up to the square root of ( n ):
   • If ( i ) is still marked as prime (i.e., sieve[i] is True), mark all multiples of ( i ) starting from ( i^2 ) as non-prime (set to False). This is because smaller multiples of ( i ) would already have been marked by smaller primes.
3. Sum Primes: Sum all indices in the sieve array that are still marked as True (these are the primes).

Solution Code

def sum_of_primes(n):
    if n < 2:
        return 0
    # Initialize sieve: sieve[i] is True if i is prime
    sieve = [True] * (n + 1)
    sieve[0] = sieve[1] = False
    # Mark non-primes
    for i in range(2, int(n ** 0.5) + 1):
        if sieve[i]:
            # Mark multiples of i starting from i*i
            sieve[i*i : n+1 : i] = [False] * len(sieve[i*i : n+1 : i])
    # Sum all primes
    return sum(i for i, is_prime in enumerate(sieve) if is_prime)

Explanation

• Edge Case Handling: If ( n ) is less than 2, the sum of primes is 0 since there are no primes below 2.
• Sieve Initialization: The sieve list is initialized to True for all numbers from 0 to ( n ), except 0 and 1 which are non-prime.
• Marking Non-Primes: For each prime ( i ), we mark all multiples of ( i ) starting from ( i^2 ) as non-prime. This is efficient because we skip smaller multiples that have already been marked by previous primes.
• Summing Primes: We iterate over the sieve list and sum all indices where the value is True (primes).

This approach has a time complexity of ( O(n \log \log n) ), which is optimal for finding primes up to ( n ). It is much faster than checking each number individually for primality, especially for large values of ( n ).

Example: For ( n = 10 ), the primes are 2,3,5,7, so the sum is ( 2+3+5+7 =17 ). The function returns 17 as expected. For ( n=2 ), it returns 2, and for ( n=1 ), it returns 0.

This solution efficiently computes the sum of primes up to any given ( n ) using the Sieve of Eratosthenes algorithm.

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