# Stern-Brocot Tree

Stern-Brocot tree is a tree data structure whose vertices correspond to the set of non-negative rational numbers. Thus, this tree provides a very elegant way for constructing the set of fractions m/n, where m and n are relatively prime. To construct the tree, the basic idea is to start with two fractions (0/1, 1/0) and then repeat the following operation:

Insert (m+m’)/(n+n’) between two adjacent fractions m/n and m’/n’

The first step gives us the entry 1/1 between 0/1 and 1/0. Similarly, the 2nd step gives us two more: 0/1,1/2,1/1,2/1,1/0

Continuing on like this results in an infinite binary search tree which preserves the usual ordering of rational numbers.

The figure below shows the 1st 4 levels of the Stern-Brocot tree.

### Finding the Path to k in Stern-Brocot Tree

The path from the root of the tree to a number k in the Stern-Brocot tree can be found using binary search. At each node, k will either be in the left half of the tree, or the right half. We continue down the left or right subtree until we finally find k.

• Initialize the left fraction L to 0/1 and right fraction R to 1/0
• Repeat the following until k is found:
• Compute the mediant M (which is (m+m’)/(n+n’) )
• If (k>M), then k is in the right half of the tree. L:=M and continue.
• Else If (M>k), then k is in the left half of the tree. R:=M and continue.
• Else k=M, terminate search.

Implementation

There’s a couple of things to tackle in our implementation. First, I need an easy way to represent fractions, so I create my own SternBrocotFraction class. I deliberately chose to make it very specific to this algorithm because I needed a special way to handle the fraction 1/0 (which by definition is greater than all other rationals).

Secondly, I needed a good way to represent the path from the root of the tree to k. I do this by using a StringBuilder, and at each step I append either the letter ‘L’ or ‘R’ depending on which sub-tree we take. When the search is finished, this gives us a String representation of the path from the root of the tree to the number k. This approach is similar to the approach advocated by ACM Programming Competitions for the “Stern-Brocot Number System” problem.

Here’s the code to find path to a number k:

```package com.umairsaeed.algorithm;

public class SternBrocotPath {
private static final char LEFT_SUB = 'L';
private static final char RIGHT_SUB = 'R';

public String findPathTo(SternBrocotFraction f) {
SternBrocotFraction L = new SternBrocotFraction(0, 1);
SternBrocotFraction R = new SternBrocotFraction(1, 0);

StringBuilder results = new StringBuilder();
SternBrocotPath.find(f, L, R, results);
return results.toString();
}

public static void find(SternBrocotFraction f,
SternBrocotFraction L,
SternBrocotFraction R,
StringBuilder results)
{

if (M.compareTo(f) < 0) {
L = M;
results.append(RIGHT_SUB);
SternBrocotPath.find(f, L, R, results);
} else if (M.compareTo(f) > 0) {
R = M;
results.append(LEFT_SUB);
SternBrocotPath.find(f, L, R, results);
}
return;
}
}
```

The special SternBrocotFraction class is:

```package com.umairsaeed.algorithm;

public class SternBrocotFraction implements
Comparable<SternBrocotFraction> {
private int numerator;
private int denominator;

public SternBrocotFraction(int numerator, int denominator) {
if (denominator < 0) {
numerator *= -1;
denominator *= -1;
}

this.numerator = numerator;
this.denominator = denominator;
}

public double doubleValue() {
if (this.denominator == 0) {
return Double.MAX_VALUE;
} else {
return (double) this.numerator /
(double) this.denominator;
}
}

return new SternBrocotFraction(
this.numerator + other.numerator,
this.denominator + other.denominator);
}

public int compareTo(SternBrocotFraction other) {
if (this.doubleValue() < other.doubleValue()) {
return -1;
} else if (this.doubleValue() > other.doubleValue()) {
return 1;
}
return 0;
}
}
```

Finally, some test code to exercise my class:

```package com.umairsaeed.algorithm;

public class SternBrocotTester {
public static void main(String[] args) {
testSternBrocotPath();
}

public static void testSternBrocotPath() {
SternBrocotPath t = new SternBrocotPath();

SternBrocotFraction f = new SternBrocotFraction(5, 7);
System.out.println(t.findPathTo(f));

f = new SternBrocotFraction(19, 101);
System.out.println(t.findPathTo(f));

f = new SternBrocotFraction(977, 331);
System.out.println(t.findPathTo(f));

f = new SternBrocotFraction(1049, 7901);
System.out.println(t.findPathTo(f));
}
}
```

Filed under Algorithms

### 2 Responses to Stern-Brocot Tree

1. pedant

Else q=M, terminate search.
typo?

“k=M”?

• Umair

Ah, good catch. Thank you. I’ve fixed it now.