BUBBLEBUBBLESORTSORT

tobiasstraub.com

Bubble Sortis a very simple sort algorithm

tobiasstraub.com

Ah, okayand how does the process work?

tobiasstraub.com

Let us seewe want to sort the following data sequence

seqA = 6 | 8 | 10 | 3

tobiasstraub.com

Let's lookat the first two numbers

seqA = 6 | 8 | 10 | 3

6 8

Okay! Now the question isIs the second number (8) smaller thenthe first (6)? Answer: NO, it is not

tobiasstraub.com

Perfect,then we need to do nothing in this step

tobiasstraub.com

Okay,our seqA has not changed.The next step is to perform the samecheck again, but this time we move ourpointers around a number to the right

tobiasstraub.com

A numberto the right

seqA = 6 | 8 | 10 | 3

8 10

Okay! Now the question is againIs the second number (10) smaller thenthe first (8)? Answer: NO, it is not

tobiasstraub.com

Perfect,then we need to do nothing in this stepagain

tobiasstraub.com

Good,then let us again move a number to theright

tobiasstraub.com

A numberto the right

seqA = 6 | 8 | 10 | 3

10 3

Okay! Now the question is againIs the second number (3) smaller thenthe first (10)? Answer: YES, it is

tobiasstraub.com

Now,we have to swap the numbers

seqA = 6 | 8 | 10 | 3

10 3

seqA = 6 | 8 | 3 | 10

tobiasstraub.com

Awesome!So this step is completed

tobiasstraub.com

What have we achieved?The data sequence is now completelypass through, so that the largest elementis at the end of the data sequence

tobiasstraub.com

Now what?Now we pass through the sequence ofdata once again, so that the secondlargest number (8) is on the second lastposition

At the last position of our data sequencewe have already the largest number (10)

tobiasstraub.com

How do we do that?It's easy! - We take our previously sorteddata sequence and complete all the stepsagain

tobiasstraub.com

Let's lookat the first two numbers

seqA = 6 | 8 | 3 | 10

6 8

Okay! Now the question isIs the second number (8) smaller thenthe first (6)? Answer: NO, it is not

tobiasstraub.com

Perfect,then we need to do nothing in this step

tobiasstraub.com

Good,then let us move a number to theright

tobiasstraub.com

A numberto the right

seqA = 6 | 8 | 3 | 10

8 3

Okay! Now the question is againIs the second number (3) smaller thenthe first (8)? Answer: YES, it is

tobiasstraub.com

Now,we have to swap the numbers

seqA = 6 | 8 | 3 | 10

8 3

seqA = 6 | 3 | 8 | 10

tobiasstraub.com

Good,the last number we may exclude from thecomparison.

We remember: In the first pass we havealready promoted the largest number tothe last position

tobiasstraub.com

What have we achieved?The data sequence has now been reruncompletely, so that the second largestnumber is at the second last positionof the data sequence

tobiasstraub.com

Okay, but what's next?Guess what!

tobiasstraub.com

Correctly!We take our previously sorteddata sequence and complete all the stepsagain

tobiasstraub.com

Let's lookat the first two numbers

seqA = 6 | 3 | 8 | 10

6 3

Okay! Now the question isIs the second number (3) smaller thenthe first (6)? Answer: YES, it is

tobiasstraub.com

Now,we have to swap the numbers

seqA = 6 | 3 | 8 | 10

6 3

seqA = 3 | 6 | 8 | 10

tobiasstraub.com

Very well,the second last and the last element wemay exclude from the comparison.

We remember: In the first and the secondpass we have already promoted thelargest number to the last position and thesecond last number to the second lastposition

tobiasstraub.com

Yay!That was all.We have reached the end of thesequence

tobiasstraub.com

You could see,that there is a very simple algorithm forfor sorting elements

tobiasstraub.com

And you know what?This is even the optimized version ofbubblesort

tobiasstraub.com

In the traditionalversion of the algorithm, all comparisonsare made for each run. It does not matterwhether the elements are already in thecorrect position

tobiasstraub.com

Let's talk about complexityLet us consider in terms of complexityat first the traditional algorithm

tobiasstraub.com

worst case O(n²)If we want to bring a number to thedesired position, we need in the worstcase n-1 comparisons

tobiasstraub.com

worst case O(n²)In our example, we had a data sequencewith four elements

seqA = 6 | 8 | 10 | 3

Do you remember?

tobiasstraub.com

worst case O(n²)Okay, in the worst case, we need toperform three comparisons, becauseseqA has four elements 4 – 1 = 3 (n-1)

seqA = 6 | 8 | 10 | 3

1 : Compare 6 with 82 : Compare 8 with 103 : Compare 10 with 3

tobiasstraub.com

worst case O(n²)So, now we have one number on thedesired position

The question we must ask ourselvesnow is

How many times must we repeatthis procedure in the worst case, so thatall our numbers are in the correctposition?

tobiasstraub.com

worst case O(n²)It's logical! - Until all the numbers havereached the correct position

In the worst case, the n-1 passes

tobiasstraub.com

worst case O(n²)The O-notation for the worst case,given by the number of passesmultiplied by the number of comparisons

(n-1) * (n-1) = O(n²)

Thus we have a quadratic term, whichmakes the bubblesort algorithm withincreasing number of data extremelyinefficient

tobiasstraub.com

best case O(n)The best case would be if the data isalready completely stored

tobiasstraub.com

For exampleWe have the following data sequence

seqB = 3 | 6 | 8 | 10

We pass through the data sequence.At the end of the pass we would noticethat there was no swapping. Thus, thedata sequence is already stored.

tobiasstraub.com

The big OThe O-notation for the best case,given by the number of passesmultiplied by the number of comparisons

1 * (n-1)

tobiasstraub.com

Okay, let's see nowhow the complexity behaves in Bubblesortoptimized version

tobiasstraub.com

As we know,we can at eatch iteration of the datasequence save one comparison(namely comparison with the alreadysorted number from the last run)

tobiasstraub.com

The number of comparisons..

seqA = 6 | 8 | 10 | 3Pass 1: 6 | 8 | 3 | 10 (3 comparisons: n-1)Pass 2: 6 | 3 | 8 | 10 (2 comparisons: n-2)Pass 3: 3 | 6 | 8 | 10 (1 comparisons: n-3)Pass 4: 3 | 6 | 8 | 10 (0 comparisons: 1)

..can thus be described as follows:(n-1) + (n-2) + … + 1 also n/2(linear O-notation)

tobiasstraub.com

The number of passeswill not change

tobiasstraub.com

The O-notationfor the optimized algorithm,thus obtained again by the number ofpasses multiplied by the number ofcomparisons

(n-1) * (n/2) = O(n²)

Thus we have again a quadratic term,but in terms of the linear portion(comparisons) much faster on the run time

tobiasstraub.com

Okay, nowa few facts about Bubblesort

tobiasstraub.com

Bubblesort....is hardly used in practice...has a bad runtime behavior...is very easy to understand.

tobiasstraub.com

Code? Now!You can download the code discussedhere (Java)

Optimized Versionhttp://tobiasstraub.com/bubblesort-ex1

Traditional Versionhttp://tobiasstraub.com/bubblesort-ex2

tobiasstraub.com

About the Author

Tobias Straub is a Web and Mobile Developer from Germany. Write an email to talk with him or follow him on LinkedIn.

tobiasstraub.com/linkedin

hello@tobiasstraub.com

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/3.0/.

Image sourcesPage 1: Keith, http://www.flickr.com/photos/outofideas/

License