Quicksort with first element as pivot example

Question

I am currently studying quicksort and would like to know how it works when the first (or last) element is chosen as the pivot point.

Say for example I have the following array:

{15, 19, 34, 41, 27, 13, 9, 11, 44}

This is what I think happens:

{15, 19, 34, 41, 27, 13, 9, 11, 44}
 ^
pivot

{15, 19, 34, 41, 27, 13, 9, 11, 44}
 ^                              ^
compare these two, they are good

{15, 19, 34, 41, 27, 13, 9, 11, 44}
 ^                          ^
compare these two and swap

{11, 19, 34, 41, 27, 13, 9, 15, 44}
 ^                       ^
compare these two and swap

{9, 19, 34, 41, 27, 13, 11, 15, 44}
 ^                  ^
compare these two, they are good

{9, 19, 34, 41, 27, 13, 11, 15, 44}
 ^              ^
 compare these two, they are good

{9, 19, 34, 41, 27, 13, 11, 15, 44}
 ^          ^
compare these two, they are good

{9, 19, 34, 41, 27, 13, 11, 15, 44}
 ^      ^
 compare these two, they are good

{9, 19, 34, 41, 27, 13, 11, 15, 44}
 ^  ^
 compare these two, they are good

{9, 19, 34, 41, 27, 13, 11, 15, 44}

End of first partition

Is this how it works? If so, would 19 be the new pivot point, or do you divide the array in half to find it (so that it would be 27/13), or does it depend on the implementation of the quicksort? Thanks for your time!

Answer 1

Check wikipedia, there is a little example with a bit smaller list of inplace quicksort http://en.wikipedia.org/wiki/Quicksort

With your example the idea is to partition

{15, 19, 34, 41, 27, 13, 9, 11, 44}

into

{13, 9, 11 -- 15 -- 19, 34, 41, 27, 44}

So first we move pivot to the end

Swap 44, and 15
{44, 19, 34, 41, 27, 13, 9, 11, 15}
 ^                          ^

Than check 44, its larger than pivot, so swap with one one before last...

{11, 19, 34, 41, 27, 13, 9, 44, 15}
 ^                       ^

than check element at some position as last one was larger than pivot.
9 < 15, so proceed to the next, 19 > 15 => swap

{11, 9, 34, 41, 27, 13, 19, 44, 15}
        ^            ^

swap again
{11, 9, 13, 41, 27, 34, 19, 44, 15}
        ^       ^

next
{11, 9, 13, 41, 27, 34, 19, 44, 15}
            ^   ^

and second last swap

{11, 9, 13, 27, 41, 34, 19, 44, 15}
            ^    

Now as forward and backward indices reached each other,
we swap pivot into right position

{11, 9, 13, 15, 41, 34, 19, 44, 27}

And we got partitioned set. Items less than 15 at the beginning, than pivot = 15, and then greater elements.

EDIT: algorithm described in wikipedia article is a bit different:

Legend:
^ = storeindex
# = i

{44, 19, 34, 41, 27, 13, 9, 11, 15}
 ^#

{44, 19, 34, 41, 27, 13, 9, 11, 15}
 ^   #

... until ...   

{44, 19, 34, 41, 27, 13, 9, 11, 15}
 ^                   #

{13, 19, 34, 41, 27, 44, 9, 11, 15}
     ^                   #

{13, 9, 34, 41, 27, 44, 19, 11, 15}
        ^                   #

{13, 9, 11, 41, 27, 44, 19, 34, 15}
            ^                   #

{13, 9, 11, 15, 27, 44, 19, 34, 41}
            ^- pivot