Binary Search is a simple algorithm to find an item in an sorted array, and it’s usually referenced as a code sample to study when learning a new programming language.

Efficency

It’s very efficient:

  • Time: O(log n), it’s at worst logaritmic
  • Space: 0(1), takes constant time

Theory

Given a sorted array, we pick the item Y in the middle and we compare it to the target value X.

If Y matches X, we return the Y position and we exit.

We determine if X < Y, in this case we discard the right side, and we go in the middle of the left side, and we repeat the same operation as above.

The search ends when Y finally matches X. If this does not happen, we return the position that X would have taken if it was in the array.

Implementation

We’re lucky, the Go standard library provides a binary tree implementation in its sort package, in sort.Search().

Let’s see the usage of the API, as taken from the package documentation, so we know what sort.Search should return:

package main

import (
    "fmt"
    "sort"
)

func main() {
    a := []int{1, 3, 6, 10, 15, 21, 28, 36, 45, 55}
    x := 6

    i := sort.Search(len(a), func(i int) bool { return a[i] >= x })
    if i < len(a) && a[i] == x {
        fmt.Printf("found %d at index %d in %v\n", x, i, a)
    } else {
        fmt.Printf("%d not found in %v\n", x, a)
    }
}

sort.Search returns an index i, and we just need to make sure that that index actually contains x.

Of course we want to know how this is implemented internally. Since the standard library is written in Go, and open source, it’s really easy to see how sort.Search is implemented:

func Search(n int, f func(int) bool) int {
    // Define f(-1) == false and f(n) == true.
    // Invariant: f(i-1) == false, f(j) == true.
    i, j := 0, n
    for i < j {
        h := i + (j-i)/2 // avoid overflow when computing h
        // i ≤ h < j
        if !f(h) {
            i = h + 1 // preserves f(i-1) == false
        } else {
            j = h // preserves f(j) == true
        }
    }
    // i == j, f(i-1) == false, and f(j) (= f(i)) == true  =>  answer is i.
    return i
}

Let’s break it down:

Given an array with length n, and a comparison function f (that internally evaluates x >= a[h]), we start iterating the array a.

Let’s use the actual values we use in the example, it’s easier to show what’s happening.

Data:

a := []int{1, 3, 6, 10, 15, 21, 28, 36, 45, 55}
x := 6

Iteration 1:

  • i is 0, j is 9.
  • h is calculated as (0 + (9-0) / 2) = 4
  • a[h] is 15. This means we execute j = h

Iteration 2:

  • i is 0, j is 4.
  • h is calculated as (0 + (4-0) / 2) = 2
  • a[h] is 6. We found the position of x, this means we return h = 2

Searching reverse order

Since we can pass a function to sort.Search, it’s easy to search on an array sorted in the reverse order, like

a := []int{55, 45, 36, 28, 21, 15, 10, 6, 3, 1}
x := 6

by passing a function that compares a[i] <= x instead of a[i] >= x.

i := sort.Search(len(a), func(i int) bool { return a[i] <= x })

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