Quick sort Algorithm Discussion And Analysis

Algorithms Analysis
Lecture 6
Quicksort

Quick Sort
88
14
9825
62
52
79
30
23
31
Divide and Conquer

Quick Sort
88
14
9825
62
52
79
30
23
31
Partition set into two using
randomly chosen pivot
14
25
30
2331
88
98
62
79
≤ 52 ≤

Quick Sort
14
25
30
2331
88
98
62
79
≤ 52 ≤
14,23,25,30,31
sort the first half.
62,79,98,88
sort the second half.

Quick Sort
14,23,25,30,31
62,79,88,98
52
Glue pieces together.
14,23,25,30,31,52,62,79,88,98

Quicksort
• Quicksort pros [advantage]:
– Sorts in place
– Sorts O(n lg n) in the average case
– Very efficient in practice , it’s quick
• Quicksort cons [disadvantage]:
– Sorts O(n2
) in the worst case
– And the worst case doesn’t happen often … sorted

Quicksort
• Another divide-and-conquer algorithm:
• Divide: A[p…r] is partitioned (rearranged) into two
nonempty subarrays A[p…q-1] and A[q+1…r] s.t.
each element of A[p…q-1] is less than or equal to
each element of A[q+1…r]. Index q is computed here,
called pivot.
• Conquer: two subarrays are sorted by recursive calls
to quicksort.
• Combine: unlike merge sort, no work needed since
the subarrays are sorted in place already.

Quicksort
• The basic algorithm to sort an array A consists of the following four
easy steps:
– If the number of elements in A is 0 or 1, then return
– Pick any element v in A. This is called the pivot
– Partition A-{v} (the remaining elements in A) into two disjoint
groups:
• A1 = {x ∈ A-{v} | x ≤ v}, and
• A2 = {x ∈ A-{v} | x ≥ v}
– return
• { quicksort(A1) followed by v followed by
quicksort(A2)}

Quicksort
• Small instance has n ≤ 1
– Every small instance is a sorted instance
• To sort a large instance:
– select a pivot element from out of the n elements
• Partition the n elements into 3 groups left, middle and
right
– The middle group contains only the pivot element
– All elements in the left group are ≤ pivot
– All elements in the right group are ≥ pivot
• Sort left and right groups recursively
• Answer is sorted left group, followed by middle group
followed by sorted right group

Quicksort Code
P: first element
r: last element
Quicksort(A, p, r)
{
if (p < r)
{
q = Partition(A, p, r)
Quicksort(A, p , q-1)
Quicksort(A, q+1 , r)
}
}
• Initial call is Quicksort(A, 1, n), where n in the length of A

Partition
• Clearly, all the action takes place in the
partition() function
– Rearranges the subarray in place
– End result:
• Two subarrays
• All values in first subarray ≤ all values in second
– Returns the index of the “pivot” element
separating the two subarrays

Partition Code
Partition(A, p, r)
{
x = A[r] // x is pivot
i = p - 1
for j = p to r – 1
{
do if A[j] <= x
then
{
i = i + 1
exchange A[i] ↔ A[j]
}
}
exchange A[i+1] ↔ A[r]
return i+1
}
partition() runs in O(n) time

Partition Example
A = {2, 8, 7, 1, 3, 5, 6, 4{
2 8 7 1 3 5 62 8 7 1 3 5 6 44
rp ji
2 8 7 1 3 5 62 8 7 1 3 5 6 44
rp i j
rp i j
2 8 7 1 3 5 62 8 7 1 3 5 6 44
rp i j
8822 77 11 33 55 66 44
rp j
1122 77 88 33 55 66 44
i rp j
1122 33 88 77 55 66 44
i
rp j
1122 33 88 77 55 66 44
i rp
1122 33 88 77 55 66 44
i
rp
1122 33 44 77 55 66 88
i
22
22
22 22
22 22
22
11
11 33
33 11 33
11 33

Partition Example Explanation
• Red shaded elements are in the first partition
with values ≤ x (pivot)
• Gray shaded elements are in the second
partition with values ≥ x (pivot)
• The unshaded elements have no yet been put in
one of the first two partitions
• The final white element is the pivot

Choice Of Pivot
Three ways to choose the pivot:
• Pivot is rightmost element in list that is to be sorted
– When sorting A[6:20], use A[20] as the pivot
– Textbook implementation does this
• Randomly select one of the elements to be sorted as
the pivot
– When sorting A[6:20], generate a random number r in
the range [6, 20]
– Use A[r] as the pivot

Choice Of Pivot
• Median-of-Three rule - from the leftmost, middle, and rightmost
elements of the list to be sorted, select the one with median key as
the pivot
– When sorting A[6:20], examine A[6], A[13] ((6+20)/2), and A[20]
– Select the element with median (i.e., middle) key
– If A[6].key = 30, A[13].key = 2, and A[20].key = 10, A[20]
becomes the pivot
– If A[6].key = 3, A[13].key = 2, and A[20].key = 10, A[6] becomes
the pivot

Worst Case Partitioning
• The running time of quicksort depends on whether the partitioning is
balanced or not.
∀ Θ(n) time to partition an array of n elements
• Let T(n) be the time needed to sort n elements
• T(0) = T(1) = c, where c is a constant
• When n > 1,
– T(n) = T(|left|) + T(|right|) + Θ(n)
• T(n) is maximum (worst-case) when either |left| = 0 or |right| = 0
following each partitioning

Worst Case Partitioning
• Worst-Case Performance (unbalanced):
– T(n) = T(1) + T(n-1) + Θ(n)
• partitioning takes Θ(n)
= [2 + 3 + 4 + …+ n-1 + n ]+ n =
= [∑k = 2 to n k ]+ n = Θ(n2
)
• This occurs when
– the input is completely sorted
• or when
– the pivot is always the smallest (largest) element
)(2/)1(...21 2
1
nnnnk
n
k
Θ=+=+++=∑=

Best Case Partition
• When the partitioning procedure produces two regions of
size n/2, we get the a balanced partition with best case
performance:
– T(n) = 2T(n/2) + Θ(n) = Θ(n lg n)
• Average complexity is also Θ(n lg n)

Average Case
• Assuming random input, average-case running time is
much closer to Θ(n lg n) than Θ(n2
)
• First, a more intuitive explanation/example:
– Suppose that partition() always produces a 9-to-1
proportional split. This looks quite unbalanced!
– The recurrence is thus:
T(n) = T(9n/10) + T(n/10) + Θ(n) = Θ(nlg n)?
[Using recursion tree method to solve]

Average Case
( ) ( /10) (9 /10) ( ) ( log )!T n T n T n n n n= + + Θ = Θ
log2n = log10n/log102

Average Case
• Every level of the tree has cost cn, until a boundary condition
is reached at depth log10 n = Θ( lgn), and then the levels have
cost at most cn.
• The recursion terminates at depth log10/9 n= Θ(lg n).
• The total cost of quicksort is therefore O(n lg n).

Average Case
• What happens if we bad-split root node, then good-split
the resulting size (n-1) node?
– We end up with three subarrays, size
• 1, (n-1)/2, (n-1)/2
– Combined cost of splits = n + n-1 = 2n -1 = Θ(n)
n
1 n-1
(n-1)/2 (n-1)/2
( )nΘ
(n-1)/2 (n-1)/2
n ( )nΘ
)1( −Θ n

Intuition for the Average Case
• Suppose, we alternate lucky and unlucky cases to get
an average behavior
( ) 2 ( / 2) ( ) lucky
( ) ( 1) ( ) unlucky
we consequently get
( ) 2( ( / 2 1) ( / 2)) ( )
2 ( /2 1) ( )
( log )
L n U n n
U n L n n
L n L n n n
L n n
n n
= + Θ
= − + Θ
= − + Θ + Θ
= − + Θ
= Θ
The combination of good and bad splits would result in
T(n) = O (n lg n), but with slightly larger constant hidden by
the O-notation.

Randomized Quicksort
• An algorithm is randomized if its behavior is determined
not only by the input but also by values produced by a
random-number generator.
• Exchange A[r] with an element chosen at random from
A[p…r] in Partition.
• This ensures that the pivot element is equally likely to be
any of input elements.
• We can sometimes add randomization to an algorithm in
order to obtain good average-case performance over all
inputs.

Randomized Quicksort
Randomized-Partition(A, p, r)
1. i ← Random(p, r)
2. exchange A[r] ↔ A[i]
3. return Partition(A, p, r)
Randomized-Quicksort(A, p, r)
1. if p < r
2. then q ← Randomized-Partition(A, p, r)
3. Randomized-Quicksort(A, p , q-1)
4. Randomized-Quicksort(A, q+1, r)
pivot
swap

Review: Analyzing Quicksort
• What will be the worst case for the algorithm?
– Partition is always unbalanced
• What will be the best case for the algorithm?
– Partition is balanced

Summary: Quicksort
• In worst-case, efficiency is Θ(n2
)
– But easy to avoid the worst-case
• On average, efficiency is Θ(n lg n)
• Better space-complexity than mergesort.
• In practice, runs fast and widely used

Quick sort Algorithm Discussion And Analysis

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