Solution to Project Euler Problem 57: Square root convergents - It is possible to show that the square root of two can be expressed as an infinite continued fraction. √2 = 1 + (1/(2 + 1/(2 + 1/(2 + ...)))). By expanding this for the first four iterations, we get: 1 + 1/2 = 3/2 = 1.5, 1 + 1/(1 + 1/2) = 7/5 = 1.4, 1 + (1/(2 + 1/(2 + 1/2))) = 1.41666... 1 + (1/(2 + 1/(2 + 1/(2 + 1/2)))) = 1.41379... The next three expansions are 99/70, 239/269, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator. In the first one-thousand expansions, how many fractions contain a numerator with more digits than the denominator?

• Java Code

package sikademy;
/**
 *
 * @author Archangel Macsika
 * Copyright (c) Sikademy. All rights reserved
 */

import java.math.BigInteger;

public class SikademyEulerSolution {
	
	private static final int LIMIT = 1000;
	
	public String run() {
		BigInteger n = BigInteger.ZERO;
		BigInteger d = BigInteger.ONE;
		int count = 0;
		for (int i = 0; i < LIMIT; i++) {
			BigInteger temp = d.multiply(BigInteger.valueOf(2)).add(n);
			n = d;
			d = temp;
			if (n.add(d).toString().length() > d.toString().length())
				count++;
		}
		return Integer.toString(count);
	}

	public static void main(String[] args) {
                SikademyEulerSolution solution = new SikademyEulerSolution();
		System.out.println(solution.run());
	}	
}

• Python Code

# 
# @author Archangel Macsika
# Copyright (c) Sikademy. All rights reserved.
# 

def compute():
	LIMIT = 1000
	ans = 0
	numer = 0
	denom = 1
	for _ in range(LIMIT):
		numer, denom = denom, denom * 2 + numer
		# Now numer/denom is the i'th (0-based) continued fraction approximation of sqrt(2) - 1
		if len(str(numer + denom)) > len(str(denom)):
			ans += 1
	return str(ans)


if __name__ == "__main__":
	print(compute())