Bubblesort Algorithm

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Bubble Sort
is a very simple sort algorithm

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Ah, okay
and how does the process work?

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Let us see
we want to sort the following data
sequence
seqA = 6 | 8 | 10 | 3

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Let's look
at the first two numbers
seqA = 6 | 8 | 10 | 3
6 8
Okay! Now the question is
Is the second number (8) smaller then
the first (6)? Answer: NO, it is not

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Perfect,
then we need to do nothing in this step

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

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A number
to the right
seqA = 6 | 8 | 10 | 3
8 10
Okay! Now the question is again

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Perfect,
then we need to do nothing in this step
again

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Good,
then let us again move a number to the
right

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A number
to the right
seqA = 6 | 8 | 10 | 3
10 3
the first (10)? Answer: YES, it is

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Now,
we have to swap the numbers
seqA = 6 | 8 | 10 | 3
10 3
seqA = 6 | 8 | 3 | 10

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Awesome!
So this step is completed

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What have we achieved?
The data sequence is now completely
pass through, so that the largest element
is at the end of the data sequence

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Now what?
Now we pass through the sequence of
data once again, so that the second
largest number (8) is on the second last
position
At the last position of our data sequence
we have already the largest number (10)

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How do we do that?
It's easy! - We take our previously sorted
data sequence and complete all the steps
again

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Let's look
seqA = 6 | 8 | 3 | 10
6 8

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Good,
then let us move a number to the
right

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A number
to the right
seqA = 6 | 8 | 3 | 10
8 3

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Now,
seqA = 6 | 8 | 3 | 10
8 3
seqA = 6 | 3 | 8 | 10

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Good,
the last number we may exclude from the
comparison.
We remember: In the first pass we have
already promoted the largest number to
the last position

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What have we achieved?
The data sequence has now been rerun
completely, so that the second largest
number is at the second last position
of the data sequence

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Okay, but what's next?
Guess what!

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Correctly!
We take our previously sorted
data sequence and complete all the steps
again

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Let's look
seqA = 6 | 3 | 8 | 10
6 3

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Now,
seqA = 6 | 3 | 8 | 10
6 3
seqA = 3 | 6 | 8 | 10

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Very well,
the second last and the last element we
may exclude from the comparison.
We remember: In the first and the second
pass we have already promoted the
largest number to the last position and the
second last number to the second last
position

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

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

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And you know what?
This is even the optimized version of
bubblesort

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In the traditional
version of the algorithm, all comparisons
are made for each run. It does not matter
whether the elements are already in the
correct position

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Let's talk about complexity
Let us consider in terms of complexity
at first the traditional algorithm

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

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worst case O(n²)
In our example, we had a data sequence
with four elements
seqA = 6 | 8 | 10 | 3
Do you remember?

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worst case O(n²)
Okay, in the worst case, we need to
perform three comparisons, because
seqA has four elements 4 – 1 = 3 (n-1)
seqA = 6 | 8 | 10 | 3
1 : Compare 6 with 8

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worst case O(n²)
So, now we have one number on the
desired position
The question we must ask ourselves
now is
How many times must we repeat
this procedure in the worst case, so that
all our numbers are in the correct
position?

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worst case O(n²)
It's logical! - Until all the numbers have
reached the correct position
In the worst case, the n-1 passes

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worst case O(n²)
The O-notation for the worst case,
given by the number of passes
multiplied by the number of comparisons
(n-1) * (n-1) = O(n²)
Thus we have a quadratic term, which
makes the bubblesort algorithm with
increasing number of data extremely
inefficient

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

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For example
We have the following data sequence
seqB = 3 | 6 | 8 | 10
We pass through the data sequence.
At the end of the pass we would notice
that there was no swapping. Thus, the
data sequence is already stored.

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The big O
The O-notation for the best case,
given by the number of passes
multiplied by the number of comparisons
1 * (n-1)

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Okay, let's see now
how the complexity behaves in Bubblesort
optimized version

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As we know,
we can at eatch iteration of the data
sequence save one comparison
(namely comparison with the already
sorted number from the last run)

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The number of comparisons..
seqA = 6 | 8 | 10 | 3
Pass 1: 6 | 8 | 3 | 10 (3 comparisons: n-1)
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)

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The number of passes
will not change

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The O-notation
for the optimized algorithm,
thus obtained again by the number of
passes multiplied by the number of
comparisons
(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

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Okay, now
a few facts about Bubblesort

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

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Code? Now!
You can download the code discussed
here (Java)
Optimized Version
http://tobiasstraub.com/bubblesort-ex1
Traditional Version
http://tobiasstraub.com/bubblesort-ex2

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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 sources
Page 1: Keith, http://www.flickr.com/photos/outofideas/
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Bubblesort Algorithm

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