# nayuki/Project-Euler-solutions

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 # # Solution to Project Euler problem 73 # Copyright (c) Project Nayuki. All rights reserved. # # https://www.nayuki.io/page/project-euler-solutions # https://github.com/nayuki/Project-Euler-solutions # # The Stern-Brocot tree is an infinite binary search tree of all positive rational numbers, # where each number appears only once and is in lowest terms. # It is formed by starting with the two sentinels 0/1 and 1/1. Iterating infinitely in any order, # between any two currently adjacent fractions Ln/Ld and Rn/Rd, insert a new fraction (Ln+Rn)/(Ld+Rd). # See MathWorld for a visualization: http://mathworld.wolfram.com/Stern-BrocotTree.html # # The natural algorithm is as follows: # # Counts the number of reduced fractions n/d such that leftN/leftD < n/d < rightN/rightD and d <= 12000. # # leftN/leftD and rightN/rightD must be adjacent in the Stern-Brocot tree at some point in the generation process. # def stern_brocot_count(leftn, leftd, rightn, rightd): # d = leftd + rightd # if d > 12000: # return 0 # else: # n = leftn + rightn # return 1 + stern_brocot_count(leftn, leftd, n, d) + stern_brocot_count(n, d, rightn, rightd) # But instead we use depth-first search on an explicit stack, because having # a large number of stack frames seems to be supported on Linux but not on Windows. def compute(): ans = 0 stack = [(1, 3, 1, 2)] while len(stack) > 0: leftn, leftd, rightn, rightd = stack.pop() d = leftd + rightd if d <= 12000: n = leftn + rightn ans += 1 stack.append((n, d, rightn, rightd)) stack.append((leftn, leftd, n, d)) return str(ans) if __name__ == "__main__": print(compute())