Permalink
Switch branches/tags
Nothing to show
Find file Copy path
Fetching contributors…
Cannot retrieve contributors at this time
54 lines (49 sloc) 2.55 KB
#
# Solution to Project Euler problem 301
# Copyright (c) Project Nayuki. All rights reserved.
#
# https://www.nayuki.io/page/project-euler-solutions
# https://github.com/nayuki/Project-Euler-solutions
#
# In a game of Nim where both players play with perfect strategy, if the current state is a collection (multiset) of piles
# with sizes {n1, n2, ..., n_m}, then the current player will lose if and only if n1 XOR n2 XOR ... XOR n_m = 0.
# In this problem, we specialize the condition to just n XOR 2n XOR 3n = 0.
#
# Facts:
# 3n = 2n + n.
# a ^ b = 0 iff a = b. (from digital logic)
# a + b = (a ^ b) + ((a & b) << 1). (from digital logic)
# Hence:
# n ^ 2n ^ 3n = 0 (what we want)
# iff n ^ 2n = 3n (from the second fact)
# iff n ^ 2n = (n ^ 2n) + ((n & 2n) << 1) (from the third fact)
# iff (n & 2n) << 1 = 0 (by cancelling on both sides)
# iff n & 2n = 0 (left-shifting doesn't change zeroness)
# iff the binary representation of n does not have consecutive '1' bits.
#
# How many binary strings of length i have no consecutive 1's?
# First partition the set into strings that begin with a 0 and strings that begin with a 1.
# For those that begin with a 0, the rest of the string can be any string of length i-1 that doesn't have consecutive 1's.
# For those that begin with a 1, the rest of the string can be any string of length i-1 that begins with a 0 and doesn't have consecutive 1's.
# Let numStrings(i, j) be the number of bit strings of length i that begin with the bit j and have no consecutive 1's. Then:
# numStrings(1, 0) = 1. (base case)
# numStrings(1, 1) = 1. (base case)
# numStrings(i, 0) = numStrings(i-1, 0) + numStrings(i-1, 1). (for i >= 2)
# numStrings(i, 1) = numStrings(i-1, 0). (for i >= 2)
# This corresponds to a shifted Fibonacci sequence, because:
# numStrings(i, 0) = numStrings(i-1, 0) + numStrings(i-2, 0). (substitute)
# numStrings(1, 0) = 1. (base case)
# numStrings(2, 0) = 2. (derived)
# So numStrings(i, 0) = fibonacci(i + 1).
# What we want is numStrings(30, 0) + numStrings(30, 1) = numStrings(31, 0) = fibonacci(32).
#
# Actually, that answer considers numbers in the range [0, 2^30), which is not exactly what we want.
# According to the problem statement, we need to exclude 0 and include 2^30. But both are losing positions, so the adjustments cancel out.
def compute():
a = 0
b = 1
for i in range(32):
a, b = b, a + b
return str(a)
if __name__ == "__main__":
print(compute())