ðŸŽ‰ Exercism Research is now launched. Help Exercism, help science and have some fun at research.exercism.io ðŸŽ‰

Published at Sep 23 2019
·
0 comments

Instructions

Test suite

Solution

Implement a binary search algorithm.

Searching a sorted collection is a common task. A dictionary is a sorted list of word definitions. Given a word, one can find its definition. A telephone book is a sorted list of people's names, addresses, and telephone numbers. Knowing someone's name allows one to quickly find their telephone number and address.

If the list to be searched contains more than a few items (a dozen, say) a binary search will require far fewer comparisons than a linear search, but it imposes the requirement that the list be sorted.

In computer science, a binary search or half-interval search algorithm finds the position of a specified input value (the search "key") within an array sorted by key value.

In each step, the algorithm compares the search key value with the key value of the middle element of the array.

If the keys match, then a matching element has been found and its index, or position, is returned.

Otherwise, if the search key is less than the middle element's key, then the algorithm repeats its action on the sub-array to the left of the middle element or, if the search key is greater, on the sub-array to the right.

If the remaining array to be searched is empty, then the key cannot be found in the array and a special "not found" indication is returned.

A binary search halves the number of items to check with each iteration, so locating an item (or determining its absence) takes logarithmic time. A binary search is a dichotomic divide and conquer search algorithm.

To run the tests, run the command `busted`

from within the exercise directory.

For more detailed information about the Lua track, including how to get help if you're having trouble, please visit the exercism.io Lua language page.

Wikipedia http://en.wikipedia.org/wiki/Binary_search_algorithm

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

```
local find = require('binary-search')
local TracedArray = require('TracedArray')
require('assertions')
describe('binary-search', function()
it('should return -1 when an empty array is searched', function()
local array = TracedArray{}
assert.equal(-1, find(array, 6))
end)
it('should be able to find a value in a single element array with one access', function()
local array = TracedArray{ 6 }
assert.equal(1, find(array, 6))
assert.equal(1, array.access_count)
end)
it('should return -1 if a value is less than the element in a single element array', function()
local array = TracedArray{ 94 }
assert.equal(-1, find(array, 6))
assert.equal(1, array.access_count)
end)
it('should return -1 if a value is greater than the element in a single element array', function()
local array = TracedArray{ 94 }
assert.equal(-1, find(array, 602))
assert.equal(1, array.access_count)
end)
it('should find an element in a longer array in less than log(n) accesses', function()
local array = TracedArray{ 6, 67, 123, 345, 456, 457, 490, 2002, 54321, 54322 }
assert.equal(8, find(array, 2002))
assert.lteq(array.access_count, 4)
end)
it('should find elements at the beginning of an array in less than log(n) accesses', function()
local array = TracedArray{ 6, 67, 123, 345, 456, 457, 490, 2002, 54321, 54322 }
assert.equal(1, find(array, 6))
assert.lteq(array.access_count, 4)
end)
it('should find elements at the end of an array in less than log(n) accesses', function()
local array = TracedArray{ 6, 67, 123, 345, 456, 457, 490, 2002, 54321, 54322 }
assert.equal(10, find(array, 54322))
assert.lteq(array.access_count, 4)
end)
it('should return -1 if a value is less than all elements in a long array', function()
local array = TracedArray{ 6, 67, 123, 345, 456, 457, 490, 2002, 54321, 54322 }
assert.equal(-1, find(array, 2))
assert.lteq(array.access_count, 4)
end)
it('should return -1 if a value is greater than all elements in a long array', function()
local array = TracedArray{ 6, 67, 123, 345, 456, 457, 490, 2002, 54321, 54322 }
assert.equal(-1, find(array, 54323))
assert.lteq(array.access_count, 4)
end)
end)
```

```
return function (array, target)
function search(start, finish)
if start >= finish then return array[start] == target and start or -1 end
local mid = math.floor((start + finish)/2)
if target == array[mid] then
return mid
elseif target < array[mid] then
return search(start, mid - 1)
else
return search(mid + 1, finish)
end
end
return search(1, #array)
end
```

A huge amount can be learned from reading other peopleâ€™s code. This is why we wanted to give exercism users the option of making their solutions public.

Here are some questions to help you reflect on this solution and learn the most from it.

- What compromises have been made?
- Are there new concepts here that you could read more about to improve your understanding?

Level up your programming skills with 3,394 exercises across 50 languages, and insightful discussion with our volunteer team of welcoming mentors.
Exercism is
**100% free forever**.

## Community comments