Added solution to Project Euler problem 301

project_euler/problem_301/sol1.py

@@ -0,0 +1,75 @@
+"""
+Project Euler Problem 301: https://projecteuler.net/problem=301
+
+Problem Statement:
+Nim is a game played with heaps of stones, where two players take
+it in turn to remove any number of stones from any heap until no stones remain.
+
+We'll consider the three-heap normal-play version of
+Nim, which works as follows:
+- At the start of the game there are three heaps of stones.
+- On each player's turn, the player may remove any positive
+  number of stones from any single heap.
+- The first player unable to move (because no stones remain) loses.
+
+If (n1, n2, n3) indicates a Nim position consisting of heaps of size
+n1, n2, and n3, then there is a simple function, which you may look up
+or attempt to deduce for yourself, X(n1, n2, n3) that returns:
+- zero if, with perfect strategy, the player about to
+  move will eventually lose; or
+- non-zero if, with perfect strategy, the player about
+  to move will eventually win.
+
+For example X(1,2,3) = 0 because, no matter what the current player does,
+the opponent can respond with a move that leaves two heaps of equal size,
+at which point every move by the current player can be mirrored by the
+opponent until no stones remain; so the current player loses. To illustrate:
+- current player moves to (1,2,1)
+- opponent moves to (1,0,1)
+- current player moves to (0,0,1)
+- opponent moves to (0,0,0), and so wins.
+
+For how many positive integers n <= 2^30 does X(n,2n,3n) = 0?
+"""
+
+
+def x(n: int, n2: int, n3: int) -> int:
+    """
+    Returns:
+    - zero if, with perfect strategy, the player about to
+      move will eventually lose; or
+    - non-zero if, with perfect strategy, the player about
+      to move will eventually win.
+
+    >>> x(1, 2, 3)
+    0
+    >>> x(3, 6, 9)
+    12
+    >>> x(8, 16, 24)
+    0
+    >>> x(11, 22, 33)
+    60
+    >>> x(1000, 2000, 3000)
+    3968
+    """
+    return n ^ n2 ^ n3
+
+
+def solution(n: int = 2 ** 30) -> int:
+    """
+    For a given integer n <= 2^30, returns how many Nim games are lost.
+    >>> solution(2)
+    2
+    >>> solution(2 ** 10)
+    144
+    """
+    loss_count = 0
+    for i in range(1, n + 1):
+        if x(i, 2 * i, 3 * i) == 0:
+            loss_count += 1
+
+    return loss_count
+
+
+if __name__ == "__main__":
+    print(solution())