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 - - This algorithm uses a lot of stack space (about 12000 frames). You probably need to use a JVM option like "-Xss4M". -} main = putStrLn (show ans) ans = sternBrocotCount 1 3 1 2 -- 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. sternBrocotCount :: Int -> Int -> Int -> Int -> Integer sternBrocotCount leftN leftD rightN rightD = let n = leftN + rightN d = leftD + rightD in if (d > 12000) then 0 else 1 + (sternBrocotCount leftN leftD n d) + (sternBrocotCount n d rightN rightD)