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#
# Solution to Project Euler problem 100
# Copyright (c) Project Nayuki. All rights reserved.
#
# https://www.nayuki.io/page/project-euler-solutions
# https://github.com/nayuki/Project-Euler-solutions
#
import eulerlib
# Suppose the box has b blue discs and r red discs.
# The probability of taking 2 blue discs is [b / (b + r)] * [(b - 1) / (b + r - 1)],
# which we want to be equal to 1/2. Rearrange the equation:
# [b(b - 1)] / [(b + r)(b + r - 1)] = 1 / 2.
# 2b(b - 1) = (b + r)(b + r - 1).
# 2b^2 - 2b = b^2 + br - b + br + r^2 - r.
# b^2 - b = r^2 + 2br - r.
# b^2 - (2r + 1)b + (r - r^2) = 0.
# Apply the quadratic equation to solve for b:
# b = [(2r + 1) +/- sqrt((2r + 1)^2 - 4(r - r^2))] / 2
# = r + [1 +/- sqrt(8r^2 + 1)]/2
# = r + [sqrt(8r^2 + 1) + 1]/2. (Discard the minus solution because it would make b < r)
#
# For b to be an integer, we need sqrt(8r^2 + 1) to be odd, and also 8r^2 + 1 be a perfect square.
# Assume 8y^2 + 1 = x^2 for some integer x > 0.
# We can see this is in fact a Pell's equation: x^2 - 8y^2 = 1.
#
# Suppose we have the solution (x0, y0) such that x0 > 0 and x0 is as small as possible.
# This is called the fundamental solution, and all other solutions be derived from it (proven elsewhere).
# Suppose (x0, y0) and (x1, y1) are solutions. Then we have:
# x0^2 - 8*y0^2 = 1.
# (x0 + y0*sqrt(8))(x0 - y0*sqrt(8)) = 1.
# (x1 + y1*sqrt(8))(x1 - y1*sqrt(8)) = 1. (Similarly)
# Multiply them together:
# [(x0 + y0*sqrt(8))(x0 - y0*sqrt(8))][(x1 + y1*sqrt(8))(x1 - y1*sqrt(8))] = 1 * 1.
# [(x0 + y0*sqrt(8))(x1 + y1*sqrt(8))][(x0 - y0*sqrt(8))(x1 - y1*sqrt(8))] = 1.
# [x0*x1 + x0*y1*sqrt(8) + x1*y0*sqrt(8) + 8y0*y1][x0*x1 - x0*y1*sqrt(8) - x1*y0*sqrt(8) + 8y0*y1] = 1.
# [(x0*x1 + 8y0*y1) + (x0*y1 + x1*y0)*sqrt(8)][(x0*x1 + 8y0*y1) - (x0*y1 + x1*y0)*sqrt(8)] = 1.
# (x0*x1 + 8y0*y1)^2 - 8*(x0*y1 + x1*y0)^2 = 1.
# Therefore (x0*x1 + 8y0*y1, x0*y1 + x1*y0) is also a solution.
# By inspection, the fundamental solution is (3, 1).
def compute():
# Fundamental solution
x0 = 3
y0 = 1
# Current solution
x = x0
y = y0 # An alias for the number of red discs
while True:
# Check if this solution is acceptable
sqrt = eulerlib.sqrt(y**2 * 8 + 1)
if sqrt % 2 == 1: # Is odd
blue = (sqrt + 1) // 2 + y
if blue + y > 10**12:
return str(blue)
# Create the next bigger solution
nextx = x * x0 + y * y0 * 8
nexty = x * y0 + y * x0
x = nextx
y = nexty
if __name__ == "__main__":
print(compute())