feat: add algorithm 95 to find the longest amicable chain

Author: gaye-lamine

project_euler/problem_095/sol1.py

@@ -0,0 +1,65 @@
+"""
+Project Euler Problem: https://projecteuler.net/problem=95
+
+An amicable chain is a sequence of numbers where each number is the sum of the proper divisors of the previous one, and the chain eventually returns to the starting number.
+
+The problem is to find the smallest member of the longest amicable chain under a given limit.
+
+In this implementation, we aim to identify all amicable chains and find the one with the maximum length, while also returning the smallest member of that chain.
+"""
+
+def sum_of_proper_divisors(n):
+    """Calculate the sum of proper divisors of n."""
+    if n < 2:
+        return 0  # Proper divisors of 0 and 1 are none.
+    total = 1  # Start with 1, since it is a proper divisor of any n > 1
+    sqrt_n = int(n**0.5)  # Calculate the integer square root of n.
+    
+    # Loop through possible divisors from 2 to the square root of n
+    for i in range(2, sqrt_n + 1):
+        if n % i == 0:  # Check if i is a divisor of n
+            total += i  # Add the divisor
+            if i != n // i:  # Avoid adding the square root twice
+                total += n // i  # Add the corresponding divisor (n/i)
+    
+    return total
+
+def find_longest_amicable_chain(limit):
+    """Find the smallest member of the longest amicable chain under a given limit."""
+    sum_divisors = {}  # Dictionary to store the sum of proper divisors for each number
+    for i in range(1, limit + 1):
+        sum_divisors[i] = sum_of_proper_divisors(i)  # Calculate and store sum of proper divisors
+
+    longest_chain = []  # To store the longest amicable chain found
+    seen = {}  # Dictionary to track numbers already processed
+
+    # Iterate through each number to find amicable chains
+    for start in range(1, limit + 1):
+        if start in seen:  # Skip if this number is already processed
+            continue
+
+        chain = []  # Initialize the current chain
+        current = start  # Start with the current number
+        while current <= limit and current not in chain:
+            chain.append(current)  # Add the current number to the chain
+            seen[current] = True  # Mark this number as seen
+            current = sum_divisors.get(current, 0)  # Move to the next number in the chain
+
+        # Check if we form a cycle and validate the chain
+        if current in chain and current != start:
+            cycle_start_index = chain.index(current)  # Find where the cycle starts
+            if current in sum_divisors and sum_divisors[current] in chain:
+                # This means we have a valid amicable chain
+                chain = chain[cycle_start_index:]  # Take only the cycle part
+                if len(chain) > len(longest_chain):
+                    longest_chain = chain  # Update longest chain if this one is longer
+
+    return min(longest_chain) if longest_chain else None  # Return the smallest member of the longest chain
+
+def solution():
+    """Return the smallest member of the longest amicable chain under one million."""
+    return find_longest_amicable_chain(10**6)
+
+if __name__ == "__main__":
+    smallest_member = solution()  # Call the solution function
+    print(smallest_member)  # Output the smallest member of the longest amicable chain