/* 
 * 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
 */


public final class p073 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p073().run());
	}
	
	
	/* 
	 * 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".
	 */
	public String run() {
		return Integer.toString(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.
	private static int sternBrocotCount(int leftN, int leftD, int rightN, int rightD) {
		int n = leftN + rightN;
		int d = leftD + rightD;
		if (d > 12000)
			return 0;
		else
			return 1 + sternBrocotCount(leftN, leftD, n, d) + sternBrocotCount(n, d, rightN, rightD);
	}
	
}