Solution for the Euler Project Problem 122

project_euler/problem_122/sol1.py

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+"""
+Project Euler Problem 122: https://projecteuler.net/problem=122
+
+Efficient Exponentiation
+
+The most naive way of computing n^15 requires fourteen multiplications:
+
+                                               n x n x ... x n = n^15.
+
+But using a "binary" method you can compute it in six multiplications:
+
+                                                         n x n = n^2
+                                                     n^2 x n^2 = n^4
+                                                     n^4 x n^4 = n^8
+                                                     n^8 x n^4 = n^12
+                                                    n^12 x n^2 = n^14
+                                                      n^14 x n = n^15
+
+However it is yet possible to compute it in only five multiplications:
+
+                                                                n x n = n^2
+                                                              n^2 x n = n^3
+                                                            n^3 x n^3 = n^6
+                                                            n^6 x n^6 = n^12
+                                                           n^12 x n^3 = n^15
+
+We shall define m(k) to be the minimum number of multiplications to compute n^k;
+for example m(15) = 5.
+
+Find sum_{k = 1}^200 m(k).
+
+It uses the fact that for rather small n, applicable for this problem, the solution
+for each number can be formed by increasing the largest element.
+
+References:
+- https://en.wikipedia.org/wiki/Addition_chain
+"""
+
+
+def solve(nums: list[int], goal: int, depth: int) -> bool:
+    """
+    Checks if nums can have a sum equal to goal, given that length of nums does
+    not exceed depth.
+
+    >>> solve([1], 2, 2)
+    True
+    >>> solve([1], 2, 0)
+    False
+    """
+    if len(nums) > depth:
+        return False
+    for el in nums:
+        if el + nums[-1] == goal:
+            return True
+        nums.append(el + nums[-1])
+        if solve(nums=nums, goal=goal, depth=depth):
+            return True
+        del nums[-1]
+    return False
+
+
+def solution(n: int = 200) -> int:
+    """
+    Calculates sum of smallest number of multiplactions for each number up to
+    and including n.
+
+    >>> solution(1)
+    0
+    >>> solution(2)
+    1
+    >>> solution(14)
+    45
+    >>> solution(15)
+    50
+    """
+    total = 0
+    for i in range(2, n + 1):
+        max_length = 0
+        while True:
+            nums = [1]
+            max_length += 1
+            if solve(nums=nums, goal=i, depth=max_length):
+                break
+        total += max_length
+    return total
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")