feat(project_euler): add solution to problem 60

Author: AksharGoyal

project_euler/problem_060/__init__.py

@@ -0,0 +1,241 @@
+"""
+Project Euler Problem 60: https://projecteuler.net/problem=60
+
+# Problem Statement:
+
+The primes 3, 7, 109, and 673 are quite remarkable. By taking any two primes
+and concatenating them in any order the result will always be prime.
+For example, taking 7 and 109, both 7109 and 1097 are prime.
+The sum of these four primes, 792, represents the lowest sum for a set of four primes
+with this property.
+Find the lowest sum for a set of five primes for which any two primes concatenate
+to produce another prime.
+
+# Solution Explanation:
+
+The brute force approach would be to check all combinations of 5 primes and check
+if they satisfy the concatenation property. However, this is computationally
+expensive. Instead, we can use a backtracking approach to build sets of primes
+that satisfy the concatenation property. We can further optimize by using property
+of divisibility by 3 to eliminate certain candidates and memoization to avoid
+redundant prime checks.
+Throughout the code, we have used a parameter flag to indicate whether
+we are working with primes that are congruent to 1 or 2 modulo 3.
+This helps in reducing the search space.
+
+## Eliminating candidates using divisibility by 3:
+Consider any 2 primes p1 and p2 that are not divisible by 3. If p1 divided by 3
+gives a remainder of 1 and p2 divided by 3 gives a remainder of 2, then
+the concatenated number p1p2 will be divisible by 3 and hence not prime.
+This can be easily proven using the property of modular arithmetic.
+    Consider p1 ≡ 1 (mod 3) and p2 ≡ 2 (mod 3). Define a1 = p1, b1 = 1, a2 = p2, b2 = 2.
+    concat(p1, p2) = (p1 * 10^k + p2) where k is the number of digits in p2.
+    Now, (p1 * 10^k + p2) mod 3 = ((p1 * 10^k) + p2) mod 3
+    As 10^k mod 3 = 1, we have (p1 * 1 + p2) mod 3 (ka mod 3 = kb mod 3)
+    Which implies (p1 + p2) mod 3 = (1 + 2) mod 3 = 0 (a1 + a2 mod 3 = b1 + b2 mod 3)
+
+Thus, we can eliminate such pairs from our search space and reach the solution faster.
+The solution uses this property to divide the primes into two lists based on their
+remainder when divided by 3. This way, we only need to check combinations within
+either list, reducing the number of checks significantly.
+
+## Memoization:
+We can use a dictionary to store the results of prime checks for concatenated numbers.
+This way, if we encounter the same concatenated number again, we can simply look up
+the result instead of recalculating it.
+
+## Backtracking:
+We can use a recursive function to build sets of primes. Starting with an empty set,
+we can add primes one by one, checking at each step if the current set satisfies
+the concatenation property. If it does, we can continue adding more primes.
+If we reach a set of 5 primes, we can check if their sum is the lowest
+
+References:
+- [Modular Arithmetic Explanation](https://en.wikipedia.org/wiki/Modular_arithmetic)
+- [Project Euler Forum Discussion](https://projecteuler.net/problem=60)
+- [Prime Checking Optimization](https://en.wikipedia.org/wiki/Primality_test)
+- [Backtracking Algorithm](https://en.wikipedia.org/wiki/Backtracking)
+"""
+
+from functools import cache
+
+prime_mod_3_is_1_list: list[int] = [3, 7, 13, 19]
+prime_mod_3_is_2_list: list[int] = [3, 5, 11, 17]
+
+prime_pairs: dict[tuple, bool] = {}
+
+
+@cache
+def is_prime(num: int) -> bool:
+    """
+    Efficient primality check using 6k ± 1 optimization.
+
+    >>> is_prime(0)
+    False
+    >>> is_prime(1)
+    False
+    >>> is_prime(2)
+    True
+    >>> is_prime(3)
+    True
+    >>> is_prime(77)
+    False
+    >>> is_prime(673)
+    True
+    >>> is_prime(1097)
+    True
+    >>> is_prime(7109)
+    True
+    """
+
+    if num < 2:
+        return False
+    if num in (2, 3):
+        return True
+    if num % 2 == 0 or num % 3 == 0:
+        return False
+    # Check divisibility up to sqrt(num)
+    n_sqrt = int(num**0.5)
+    for i in range(5, n_sqrt + 1, 6):
+        if num % i == 0 or num % (i + 2) == 0:
+            return False
+    return True
+
+
+def sum_digits(num: int) -> int:
+    """
+    Returns the sum of digits of num. If the sum is greater than 10,
+    it recursively sums the digits of the result until a single digit is obtained.
+
+    >>> sum_digits(-18)
+    Traceback (most recent call last):
+        ...
+    ValueError: num must be non-negative
+    >>> sum_digits(0)
+    0
+    >>> sum_digits(5)
+    5
+    >>> sum_digits(79)
+    7
+    >>> sum_digits(999)
+    9
+    """
+    if num < 0:
+        raise ValueError("num must be non-negative")
+    if num < 10:
+        return num
+    return sum_digits(sum(map(int, str(num))))
+
+
+def is_concat(num1: int, num2: int) -> bool:
+    """
+    Check if concatenations of num1+num2 and num2+num1 are both prime.
+    Uses memoization to store previously computed results in prime_pairs dictionary.
+    Effects: Updates the prime_pairs dictionary with the result.
+             Only stores (min(num1, num2), max(num1, num2)) as key to avoid duplicates.
+
+    >>> is_concat(3, 7)
+    True
+    >>> is_concat(1, 6)
+    False
+    >>> is_concat(7, 109)
+    True
+    >>> is_concat(13, 31)
+    False
+    """
+    if num1 > num2:
+        num1, num2 = num2, num1
+    key = (num1, num2)
+    if key in prime_pairs:
+        return prime_pairs[key]
+    concat1 = int(f"{num1}{num2}")
+    concat2 = int(f"{num2}{num1}")
+    result = is_prime(concat1) and is_prime(concat2)
+    prime_pairs[key] = result
+    return result
+
+
+def add_prime(primes: list[int]) -> list[int]:
+    """
+    Add a new prime number to the input list of primes based on its modulo 3 value.
+    Effects: Modifies the input list by appending a new prime number.
+
+    >>> add_prime([3, 7, 13, 19])
+    [3, 7, 13, 19, 31]
+    >>> add_prime([3, 5, 11, 17])
+    [3, 5, 11, 17, 23]
+    >>> add_prime([3, 7, 13, 19, 31])
+    [3, 7, 13, 19, 31, 37]
+    """
+
+    next_num = primes[-1] + 3  # using modular arithmetic to get similar primes
+    while not is_prime(next_num):
+        next_num += 3
+    primes.append(next_num)
+    return primes
+
+
+def generate_primes(n: int, flag: int = 1) -> list[int]:
+    """
+    Ensure we have at least n primes in the selected list.
+
+    >>> generate_primes(5, 1)
+    [3, 7, 13, 19, 31]
+    >>> generate_primes(5, 2)
+    [3, 5, 11, 17, 23]
+    """
+    primes = prime_mod_3_is_1_list if flag == 1 else prime_mod_3_is_2_list
+    while len(primes) < n:
+        primes = add_prime(primes)
+    return primes
+
+
+def solution(
+    target_size: int = 5, prime_limit: int = 1000, flag: int = 1
+) -> int | None:
+    """
+    Search for a set of primes with the concat-prime property.
+    Returns the sum of the lowest such set found else returns None.
+
+    >>> solution(3, 100, None)
+    Traceback (most recent call last):
+        ...
+    ValueError: flag must be either 1 or 2
+    >>> solution(4, 100, 1)
+    792
+    >>> solution(3, 100, 2)
+    715
+    >>> solution(5, 1000, 1)
+    26033
+    """
+    if flag not in (1, 2):
+        raise ValueError("flag must be either 1 or 2")
+    primes = generate_primes(prime_limit, flag)
+
+    def search(chain):
+        """
+        Recursive backtracking search to find a valid set of primes.
+        A threshold is used to ensure we don't exceed the smallest sum.
+        Returns the valid set if found, else None.
+        """
+        if len(chain) == target_size:
+            return chain
+        for p in primes:
+            if p <= chain[-1]:
+                continue
+            if all(is_concat(p, c) for c in chain):
+                result = search((*chain, p))
+                if result:
+                    return result
+        return None
+
+    for _, p in enumerate(primes):
+        result = search((p,))
+        if result and len(result) == target_size:
+            return sum(result)
+
+    return None  # No valid set found
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")