**π‘ Problem Formulation:** In computational number theory, a semi-prime is a natural number that is the product of two prime numbers. This problem involves determining whether a given integer `n`

can be decomposed into the sum of two semi-primes. The input is a single integer; for example, if `n = 30`

, the output should confirm that 30 can be expressed as the sum of two semi-primes (15 + 15).

## Method 1: Brute Force Approach

This brute force method involves iterating through all possible pairs of numbers to check if they are semi-primes and, if so, whether their sum equals the target integer. This is done using a function that maps an integer to its prime factors, and then to identify semi-primes among the factors.

Here’s an example:

def is_prime(num): if num < 2: return False for i in range(2, int(num**0.5) + 1): if num % i == 0: return False return True def is_semi_prime(num): primes = [i for i in range(2, num // 2 + 1) if is_prime(i)] for prime in primes: if num % prime == 0 and is_prime(num // prime): return True return False def can_be_expressed_as_sum_of_two_semi_primes(n): for i in range(2, n): if is_semi_prime(i) and is_semi_prime(n - i): return True return False n = 30 print(can_be_expressed_as_sum_of_two_semi_primes(n))

The output of this code snippet will be:

True

This code snippet defines a method to check if an integer is prime or a semi-prime. The last function checks for every pair of integers if both are semi-primes and their sum equals the given number. If such a pair is found, it returns True, indicating that the integer can be expressed as a sum of two semi-primes.

## Method 2: Optimized Semi-Prime Identification

For a more efficient approach, the semi-prime checking function is enhanced by reducing the search space and using memoization to store previous prime checks. This reduces the number of calculations and speeds up the process significantly for larger numbers.

Here’s an example:

def is_prime(num, memo={}): if num in memo: return memo[num] if num < 2: return False for i in range(2, int(num**0.5) + 1): if num % i == 0: memo[num] = False return False memo[num] = True return True def can_be_expressed_as_sum_of_two_semi_primes(n): primes = [i for i in range(2, n) if is_prime(i)] semi_primes = [i * j for i in primes for j in primes if i * j < n] for semi_prime in semi_primes: if (n - semi_prime) in semi_primes: return True return False n = 30 print(can_be_expressed_as_sum_of_two_semi_primes(n))

The output of this code snippet will be:

True

In this snippet, the `is_prime`

function uses a memoization dictionary to avoid recalculating the primality of numbers. The `can_be_expressed_as_sum_of_two_semi_primes`

function generates all possible semi-primes less than `n`

and checks if any two can sum up to `n`

.

## Method 3: Sieve of Eratosthenes with Semi-Prime Generation

Building on classic algorithms, this method involves using the Sieve of Eratosthenes to generate a list of prime numbers efficiently. Then, semi-primes are identified more swiftly by using these precomputed primes. This is especially effective for higher ranges of integers.

Here’s an example:

def sieve_of_eratosthenes(limit): is_prime = [True] * (limit + 1) for num in range(2, int(limit**0.5) + 1): if is_prime[num]: for i in range(num*num, limit + 1, num): is_prime[i] = False return [num for num in range(2, limit) if is_prime[num]] def can_be_expressed_as_sum_of_two_semi_primes(n): primes = sieve_of_eratosthenes(n) semi_primes = set(i * j for i in primes for j in primes if i * j < n) for semi_prime in semi_primes: if (n - semi_prime) in semi_primes: return True return False n = 30 print(can_be_expressed_as_sum_of_two_semi_primes(n))

The output of this code snippet will be:

True

The sieve produces a list of primes up to `n`

in an optimized manner, which is then used to compute semi-primes. The `can_be_expressed_as_sum_of_two_semi_primes`

function iterates through the set of semi-primes, returning True when a valid pair is found.

## Method 4: Factorization with Wheel Factorization

Wheel factorization is an advanced technique for identifying prime numbers by skipping multiples of first few primes in a cyclic manner. One can generate semi-primes quicker with wheel factorization and then check if their sum can represent the integer in question.

Here’s an example:

from itertools import cycle def wheel_factorization(): yield 2; yield 3; yield 5 wheel = cycle([4, 2, 4, 2, 4, 6, 2, 6]) wheels = [7, 11, 13, 17, 19, 23, 29, 31] for i in wheel: yield wheels[-1] + i wheels.append(wheels[-1] + i) def can_be_expressed_as_sum_of_two_semi_primes(n): primes = [] for prime in wheel_factorization(): if prime * prime > n: break primes.append(prime) semi_primes = set(i * j for i in primes for j in primes if i * j < n) for semi_prime in semi_primes: if (n - semi_prime) in semi_primes: return True return False n = 30 print(can_be_expressed_as_sum_of_two_semi_primes(n))

The output of this code snippet will be:

True

This code demonstrates wheel factorization to efficiently generate prime numbers and then form semi-primes from these primes. It concludes with verification if a given integer `n`

can be expressed as a sum of two semi-primes.

## Bonus One-Liner Method 5: Use of Python Libraries

Utilizing a Python library such as ‘sympy’ which specializes in symbolic mathematics, one can quickly identify prime numbers and thus semi-primes, outperforming manual implementations for checking if an integer is the sum of two semi-primes.

Here’s an example:

from sympy import primerange def can_be_expressed_as_sum_of_two_semi_primes(n): primes = list(primerange(2, n)) semi_primes = {i * j for i in primes for j in primes if i * j < n} return any((n - sp) in semi_primes for sp in semi_primes) n = 30 print(can_be_expressed_as_sum_of_two_semi_primes(n))

The output of this code snippet will be:

True

This one-liner approach uses the ‘sympy’ library to generate prime numbers and then checks for semi-prime pairs in a set comprehension, providing an elegant and efficient solution to the problem.

## Summary/Discussion

**Method 1: Brute Force Approach.**Simple to understand and implement. Not efficient for large numbers.**Method 2: Optimized Semi-Prime Identification.**Improved by memoization. More efficient than brute force but still slow for very large numbers.**Method 3: Sieve of Eratosthenes with Semi-Prime Generation.**Significantly faster for larger ranges. Requires more memory to store primes.**Method 4: Factorization with Wheel Factorization.**Provides an advanced approach for prime generation. More complex but efficient for huge numbers.**Method 5: Use of Python Libraries.**Utilizes existing efficient implementations. Fastest and easiest but requires an external dependency.