Heapsort

Heapsort is a pyramid sort. The time complexity of the algorithm is O(n log n), fast, right? I would call this sorting – sorting of falling stones. It seems to me that the easiest way to explain it is visually.

The input is a list of numbers, for example:
5, 0, 7, 2, 3, 9, 4

From left to right, a data structure is made – a binary tree, or as I call it – a pyramid. Pyramid elements can have a maximum of two child elements, but only one top element.

Let’s make a binary tree:
⠀⠀5
⠀0⠀7
2 3 9 4

If you look at the pyramid for a long time, you can see that it is simply numbers from an array, following one after another, the number of elements on each floor is multiplied by two.

Next comes the most interesting part, we sort the pyramid from the bottom up, using the falling pebbles method (heapify). Sorting could start from the last floor (2 3 9 4), but there is no point because there is no floor below to fall to.

Therefore, we begin to drop elements from the penultimate floor (0 7)
⠀⠀5
⠀0⠀7
2 3 9 4

The first element to fall is chosen from the right, in our case it is 7, then we look at what is under it, and under it is 9 and 4, nine is greater than four, and nine is also greater than seven! We drop 7 on 9, and lift 9 to the place of 7.
⠀⠀5
⠀0⠀9
2 3 7 4

Next, we understand that the seven has nowhere to fall lower, we move on to the number 0, which is on the penultimate floor on the left:
⠀⠀5
⠀0⠀9
2 3 7 4

We look at what’s underneath it – 2 and 3, two is less than three, three is greater than zero, so we swap zero and three:
⠀⠀5
⠀3⠀9
2 0 7 4

When you reach the end of the floor, go to the floor above and drop everything there if you can.
The result will be a data structure – a heap, namely a max heap, since the largest element is at the top:
⠀⠀9
⠀3⠀7
2 0 5 4

If you return to the array representation, you will get a list:
[9, 3, 7, 2, 0, 5, 4]

From this we can conclude that by swapping the first and last elements, we will get the first number in the final sorted position, namely 9 should be at the end of the sorted list, swap them:
[4, 3, 7, 2, 0, 5, 9]

Let’s look at a binary tree:
⠀⠀4
⠀3⠀7
2 0 5 9

We have a situation where the bottom of the tree is sorted, we just need to drop 4 to the correct position, we repeat the algorithm, but we do not take into account the already sorted numbers, namely 9:
⠀⠀4
⠀3⠀7
2 0 5 9

⠀⠀7
⠀3⠀4
2 0 5 9

⠀⠀7
⠀3⠀5
2 0 4 9

It turned out that, having dropped 4, we raised the next largest number after 9 – 7. We swap the last unsorted number (4) and the largest number (7)
⠀⠀4
⠀3⠀5
2 0 7 9

It turns out that now we have two numbers in the correct final position:
4, 3, 5, 2, 0, 7, 9

Then we repeat the sorting algorithm, ignoring those already sorted, and as a result we get a heap of the following type:
⠀⠀0
⠀2⠀3
4 5 7 9

Or as a list:
0, 2, 3, 4, 5, 7, 9

Implementation

The algorithm is usually divided into three functions:

Creating a heap
Heapify Algorithm
Replace the last unsorted element with the first

The heap is created by going through the penultimate row of the binary tree using the heapify function, from right to left until the end of the array. Then the first number swap is made in the loop, after which the first element falls/stays in place, as a result of which the largest element gets to the first place, the loop is repeated with the participants decreasing by one, since after each pass at the end of the list there are sorted numbers.

Heapsort example in Ruby:






module Colors



    BLUE = "\033[94m"



    RED = "\033[31m"



    STOP = "\033[0m"



end







def heapsort(rawNumbers)



    numbers = rawNumbers.dup







    def swap(numbers, from, to)



        temp = numbers[from]



        numbers[from] = numbers[to]



        numbers[to] = temp



    end







    def heapify(numbers)



        count = numbers.length()



        lastParentNode = (count - 2) / 2







        for start in lastParentNode.downto(0)



            siftDown(numbers, start, count - 1)



            start -= 1 



        end







        if DEMO



            puts "--- heapify ends ---"



        end



    end







    def siftDown(numbers, start, rightBound)      



        cursor = start



        printBinaryHeap(numbers, cursor, rightBound)







        def calculateLhsChildIndex(cursor)



            return cursor * 2 + 1



        end







        def calculateRhsChildIndex(cursor)



            return cursor * 2 + 2



        end            







        while calculateLhsChildIndex(cursor) <= rightBound



            lhsChildIndex = calculateLhsChildIndex(cursor)



            rhsChildIndex = calculateRhsChildIndex(cursor)







            lhsNumber = numbers[lhsChildIndex]



            biggerChildIndex = lhsChildIndex







            if rhsChildIndex <= rightBound



                rhsNumber = numbers[rhsChildIndex]



                if lhsNumber < rhsNumber



                    biggerChildIndex = rhsChildIndex



                end



            end







            if numbers[cursor] < numbers[biggerChildIndex]



                swap(numbers, cursor, biggerChildIndex)



                cursor = biggerChildIndex



            else



                break



            end



            printBinaryHeap(numbers, cursor, rightBound)



        end



        printBinaryHeap(numbers, cursor, rightBound)



    end







    def printBinaryHeap(numbers, nodeIndex = -1, rightBound = -1)



        if DEMO == false



            return



        end



        perLineWidth = (numbers.length() * 4).to_i



        linesCount = Math.log2(numbers.length()).ceil()



        xPrinterCount = 1



        cursor = 0



        spacing = 3



        for y in (0..linesCount)



            line = perLineWidth.times.map { " " }



            spacing = spacing == 3 ? 4 : 3



            printIndex = (perLineWidth / 2) - (spacing * xPrinterCount) / 2



            for x in (0..xPrinterCount - 1)



                if cursor >= numbers.length



                    break



                end



                if nodeIndex != -1 && cursor == nodeIndex



                    line[printIndex] = "%s%s%s" % [Colors::RED, numbers[cursor].to_s, Colors::STOP]



                elsif rightBound != -1 && cursor > rightBound



                    line[printIndex] = "%s%s%s" % [Colors::BLUE, numbers[cursor].to_s, Colors::STOP]



                else



                    line[printIndex] = numbers[cursor].to_s



                end



                cursor += 1



                printIndex += spacing



            end



            print line.join()



            xPrinterCount *= 2           



            print "\n"            



        end



    end







    heapify(numbers)



    rightBound = numbers.length() - 1







    while rightBound > 0



        swap(numbers, 0, rightBound)   



        rightBound -= 1



        siftDown(numbers, 0, rightBound)     



    end







    return numbers



end







numbersCount = 14



maximalNumber = 10



numbers = numbersCount.times.map { Random.rand(maximalNumber) }



print numbers



print "\n---\n"







start = Time.now



sortedNumbers = heapsort(numbers)



finish = Time.now



heapSortTime = start - finish







start = Time.now



referenceSortedNumbers = numbers.sort()



finish = Time.now



referenceSortTime = start - finish







print "Reference sort: "



print referenceSortedNumbers



print "\n"



print "Reference sort time: %f\n" % referenceSortTime



print "Heap sort:      "



print sortedNumbers



print "\n"



if DEMO == false



    print "Heap sort time:      %f\n" % heapSortTime



else



    print "Disable DEMO for performance measure\n"



end







if sortedNumbers != referenceSortedNumbers



    puts "Validation failed"



    exit 1



else



    puts "Validation success"



    exit 0



end

Without visualization, this algorithm is not easy to understand, so the first thing I recommend is to write a function that will print the current form of the binary tree.