Solve problem 301.

python/e301.py

@@ -0,0 +1,70 @@
+from functools import lru_cache
+
+
+@lru_cache
+def x_naiv(s):
+    zcount = s.count(0)
+    if zcount == 3:
+        return 0
+    elif zcount == 2:
+        return 1
+    elif zcount == 1:
+        sl = list(s)
+        sl.remove(0)
+        if sl[0] == sl[1]:
+            return 0
+
+    s = list(s)
+    for i in range(len(s)):
+        for v in range(1, s[i] + 1):
+            ns = list(s)
+            ns[i] -= v
+            if x_naiv(tuple(sorted(ns))) == 0: 
+                # Other player loses.
+                return 1
+    return 0
+
+
+def x_nim_sum(s):
+    """ I wasn't able to deduce this myself quickly and looked it up instead. """
+    return 0 if (s[0] ^ s[1] ^ s[2]) == 0 else 1
+
+
+def test_nim_functions():
+    assert(x_naiv(tuple([0, 0, 0])) == 0)
+    assert(x_naiv(tuple([1, 0, 0])) == 1)
+    assert(x_naiv(tuple([1, 1, 0])) == 0)
+    assert(x_naiv(tuple([1, 1, 1])) == 1)
+    assert(x_naiv(tuple([1, 2, 3])) == 0)
+
+    for n1 in range(0, 8):
+        for n2 in range(1, n1 + 1):
+            for n3 in range(1, n2 + 1):
+                s = tuple([n1, n2, n3])
+                assert x_naiv(s) == x_nim_sum(s)
+
+
+def euler_301():
+    test_nim_functions()
+
+    # log(2**30, 10) ~= 9 -> 1 billion easy brute force
+    # Probably a more efficient computation is possible by counting the permutations
+    # where all binary digits are zero.
+    # 0 0 0 -> 0
+    # 0 0 1 -> 0
+    # 0 1 1 -> 0
+
+    r = 0
+    for n in range(1, 2**30 + 1):
+        s = tuple([n, 2 * n, 3 * n])
+        xr = x_nim_sum(s)
+        if xr == 0:
+            r += 1
+    return r
+
+
+if __name__ == "__main__":
+    solution = euler_301()
+    print("e301.py: " + str(solution))
+    # assert(solution == 0)
+