added a faster solution to project euler problem 14

Author: crazazy

Diff for: project_euler/problem_14/sol3.py

@@ -0,0 +1,79 @@
+#!/usr/bin/env python
+
+"""
+Collatz conjecture: start with any positive integer n. Next term obtained from
+the previous term as follows:
+If the previous term is even, the next term is one half the previous term.
+If the previous term is odd, the next term is 3 times the previous term plus 1.
+The conjecture states the sequence will always reach 1 regardless of starting
+n.
+Problem Statement:
+The following iterative sequence is defined for the set of positive integers:
+    n → n/2 (n is even)
+    n → 3n + 1 (n 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?
+"""
+
+
+def solution(m):
+    """ Returns the number under n that generates the longest Collatz sequence.
+
+    >>> solution(1000000)
+    {'counter': 525, 'largest_number': 837799}
+    >>> solution(200)
+    {'counter': 125, 'largest_number': 171}
+    >>> solution(5000)
+    {'counter': 238, 'largest_number': 3711}
+    >>> solution(15000)
+    {'counter': 276, 'largest_number': 13255}
+    """
+    # we are going to avoid repeat computations by creating a knowledge base
+    # where we store the length of all collatz chains we calculated so far
+
+    knowledge = {1: 1}
+
+    # a single step in a collatz chain
+    def collatz(n):
+        if n % 2 == 0:
+            return n // 2
+        else:
+            return 3 * n + 1
+
+    # calculates a collatz chain of a certain number this calculation is halted
+    # whenever we find a key with a know collatz chain in our knowledge base
+    def calculateChain(n):
+        entries = []
+        while n not in knowledge:
+            entries.append(n)
+            n = collatz(n)
+        chainSize = knowledge[n]
+        for i in entries[::-1]:
+            chainSize += 1
+            knowledge[i] = chainSize
+
+    maxChain = {
+        "counter": 1,
+        "largest_number": 1
+    }
+    for i in range(1, m + 1):
+        calculateChain(i)
+        if maxChain["counter"] < knowledge[i]:
+            maxChain = {
+                "counter": knowledge[i],
+                "largest_number": i
+            }
+
+    return maxChain
+
+
+if __name__ == "__main__":
+    print("calculate the number with the longest collatz chain in the range between 1 and the following number:")
+    inputNumber = int(input().strip())
+    number, chainLength = solution(inputNumber)
+    print(f"the maximum collatz chain of all numbers between 1 and {inputNumber} is {chainLength}, starting with the number {number}")
+