Project Euler/57
From charlesreid1
Problem Statement
Continued fractions problem - this problem asks about the continued fraction representation of . In particular, we are to examine the first 1,000 continued fraction terms, determine what they are, and count how many times the numerator has more digits than the denominator.
Link: https://projecteuler.net/problem=57
Solution Technique
Also see blog post: Computing square roots using continued fractions: https://charlesreid1.github.io/computing-square-roots-part-2-using-continued-fractions.html
This utilizes a recurrence relation for the numerator and denominator, so we can start with the first few terms and obtain as many terms as we wish.
The recurrence relation is:
For , we have
Code
import java.math.BigInteger;
/**
* Find the number of terms in the first 1,000 iterations
* of the continued fraction of sqrt(2) whose denominator
* has more digits than its numerator.
*
* This utilizes the recurrence relation for the nth iteration,
* a(1) = 3, b(1) = 2
*
* a_i = a_{i-1} + 2 b_{i-1}
*
* b_i = a_{i-1} + b_{i-1}
*
* I actually implemented this in Excel, to begin with, just cuz,
* but these numbers get REALLY huge, REALLY fast.
*/
public class ContinuedFraction {
public static final BigInteger TWO = new BigInteger("2");
public static void main(String[] args) {
BigInteger aim1 = new BigInteger("3");
BigInteger bim1 = new BigInteger("2");
BigInteger ai, bi;
int nterms = 1000;
int noverflows = 0;
for(int i=0; i<nterms; i++) {
ai = aim1.add( bim1.multiply(TWO) );
bi = aim1.add( bim1 );
if( ai.toString().length() > bi.toString().length() ) {
noverflows++;
}
aim1 = ai;
bim1 = bi;
}
System.out.println(noverflows);
}
}
Link: https://git.charlesreid1.com/cs/euler/src/master/scratch/Round2_050-070/057/ContinuedFraction.java
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