Avatar of ajoshguy

ajoshguy's solution

to Binary Search in the Lua Track

Published at Sep 11 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.

Running the tests

To run the tests, run the command busted from within the exercise directory.

Further information

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.

Source

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

Submitting Incomplete Solutions

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

binary-search_spec.lua

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)
local function middle(left, right)
    return math.floor((left+right)/2)
end

return function(array, target)
    local left = 1
    local right = #array

    if right == 0 then
        return -1
    end

    repeat
        local middle = middle(left,right)
        local value = array[middle]
        if value < target then
            left = middle + 1
        elseif value > target then
            right = middle - 1
        elseif value == target then
            return middle
        end
    until left > right

    return -1
end

Community comments

Find this solution interesting? Ask the author a question to learn more.

What can you learn from this solution?

A huge amount can be learned from reading other people’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?