Symbol(clack:cancel)

solutions/0000_0029/0014.CollatzSeq/eular_14.py

@@ -0,0 +1,77 @@
+"""
+The following iterative sequence is defined for the set of positive integers:
+
+n -> n/2 (is even)
+n -> 3n + 1 (is odd)
+
+Using the rule above and starting with 13, we generate the following sequence:
+    13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
+
+It can be seen that this sequence (starting at 13 and finishing at 1 ) contains 10 terms.
+Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
+
+Which starting number, under one million, produces the longest chain?
+"""
+
+import time
+from functools import cache
+
+
+def timer(func):
+    def wrapper(*args, **kwargs):
+        start_time = time.time()
+        result = func(*args, **kwargs)
+        end_time = time.time()
+        print(f"{func.__name__} took {end_time - start_time:.6f} seconds")
+        return result
+
+    return wrapper
+
+
+def is_even(n: int) -> bool:
+    return n & 1 == 0
+
+
+def collatz_sequence(n: int) -> list[int]:
+    sequence = [n]
+    while sequence[-1] != 1:
+        if is_even(n):
+            n = n // 2
+        else:
+            n = 3 * n + 1
+        sequence.append(n)
+    return sequence
+
+
+def longest_collatz_sequence(limit: int) -> int:
+    longest_length = 0
+    longest_start = 0
+    for i in range(1, limit + 1):
+        length = len(collatz_sequence(i))
+        if length > longest_length:
+            longest_length = length
+            longest_start = i
+    return longest_start
+
+
+@cache
+def chain_length(n):
+    if n == 1:
+        return 0
+    next_term = 3 * n + 1 if n % 2 else n // 2
+    return chain_length(next_term) + 1
+
+
+@timer
+def main_do() -> None:
+    print(longest_collatz_sequence(1000000))
+
+
+@timer
+def main_cache() -> None:
+    print(max(range(1, 1000001), key=chain_length))
+
+
+if __name__ == "__main__":
+    main_do()
+    main_cache()