feat: add algorithm 95 to find the longest amicable chain

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