Tutorial 4 The Quicksort Algorithm

QuickSort  Divide: Choose a pivot, P Form a subarray with all elements ≤ P Form a subarray with all elements > P  Conquer Call “ divide ” recursive call until there is all the subarraies have one element only.  Combine Nothing to do because all elements have been sorted already after the “ conquer ” step.

Quicksort  Complexity of the Partition step: O(n). How to obtain?  Worst case partition has complexity O(n 2 ). How to obtain?  Best case partition has complexity O(nlogn). How to obtain?  How about average case?  Why we need the randomized version Quicksort?

Quicksort  Zij={z i, z i+1,z i+2, ……,z j }  Each pair of elements are compared either 0 time or 1 time only. Why?  Xij=I{z i compares with z j }, it can be 1 or 0.  Total No. of comparison X

Expectation of X: Average number of comparisons for randomized Quicksort Using the lemma for the Indicator function, it equals P(zi compares with zj).

QuickSort  P(X ij )=Prob of {(zi is pivot) U (zj is pivot)}  P(X ij )=P(zi is pivot)+P(zj is pivot), why?  P(X ij )=1/(j-i+1)+1/(j-i+1)

Quicksort  What happens if all elements have the same values?  Good news or Bad news?  What happens if all elements have been sorted?  Any ways to modify?

Original Quicksort Algorithm: Hoare-Partition Hoare-Partition(A,p,r) x ← A[p]; i ← p-1; j ← r+1 While true do repeat j ← j-1 until A[j] ≤ x do repeat i ← i+1 until A[i]  x if i < j then exchange A[i], A[j] else return j

True or False?  Q1: The indices i and j have values such that elements outside A[p..r] are not accessed.  Q2: When the procedure terminates, p ≤ j < r.  Q3: elements of A[p..j] is less than or equal to elements of A[j+1..r].