Ch. 7 - QuickSort
Quick but not Guaranteed
Ch.7 - QuickSort
Another Divide-and-Conquer sorting algorithm…
As it turns out, MERGESORT and HEAPSORT, although O(n
lg n) in their time complexity, have fairly large constants
and tend to move data around more than desirable (e.g.,
equal-key items may not maintain their relative position
from input to output).
We introduce another algorithm with better constants, but a
flaw: its worst case in O(n2). Fortunately, the worst case
is “rare enough” so that the speed advantages work an
overwhelming amount of the time… and it is O(n lg n) on
average.
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Ch.7 - QuickSort
Like in MERGESORT, we use Divide-and-Conquer:
1. Divide: partition A[p..r] into two subarrays A[p..q-1] and
A[q+1..r] such that each element of A[p..q-1] is ≤ A[q],
and each element of A[q+1..r] is ≥ A[q]. Compute q as
part of this partitioning.
2. Conquer: sort the subarrays A[p..q-1] and A[q+1..r] by
recursive calls to QUICKSORT.
3. Combine: the partitioning and recursive sorting leave us
with a sorted A[p..r] – no work needed here.
An obvious difference is that we do most of the work in the
divide stage, with no work at the combine one.
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Ch.7 - QuickSort
The Pseudo-Code
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Ch.7 - QuickSort
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Ch.7 - QuickSort
Proof of Correctness: PARTITION
We look for a loop invariant and we observe that at the
beginning of each iteration of the loop (l.3-6) for any
array index k:
1. If p ≤ k ≤ i, then A[k] ≤ x;
2. If i+1 ≤ k ≤ j-1, then A[k] > x;
3. If k = r, then A[k] = x.
4. If j ≤ k ≤ r-1, then we don’t know anything about A[k].
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Ch.7 - QuickSort
The Invariant
• Initialization. Before the first iteration: i=p-1, j=p. No values between
p and i; no values between i+1 and j-1. The first two conditions are
trivially satisfied; the initial assignment satisfies 3.
• Maintenance. Two cases
– 1. A[j] > x.
– 2. A[j] ≥ x.
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Ch.7 - QuickSort
The Invariant
• Termination. j=r. Every entry in the array is in one of the three sets
described by the invariant. We have partitioned the values in the
array into three sets: less than or equal to x, greater than x, and a
singleton containing x.
Running time of PARTITION on A[p..r] is Q(n), where n = r – p + 1.
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Ch.7 - QuickSort
QUICKSORT: Performance – a quick look.
• We first look at (apparent) worst-case partitioning:
T(n) = T(n-1) + T(0) + Q(n) = T(n-1) + Q(n).
It is easy to show – using substitution - that T(n) = Q(n2).
• We next look at (apparent) best-case partitioning:
T(n) = 2T(n/2) + Q(n).
It is also easy to show (case 2 of the Master Theorem)
that T(n) = Q(n lg n).
• Since the disparity between the two is substantial, we
need to look further…
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Ch.7 - QuickSort
QUICKSORT: Performance – Balanced Partitioning
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Ch.7 - QuickSort
QUICKSORT: Performance – the Average Case
As long as the number of “good splits” is bounded below as
a fixed percentage of all the splits, we maintain
logarithmic depth and so O(n lg n) time complexity.
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Ch.7 - QuickSort
QUICKSORT: Performance – Randomized QUICKSORT
We would like to ensure that the choice of pivot does not
critically impair the performance of the sorting algorithm
– the discussion to this point would indicate that
randomizing the choice of the pivot should provide us
with good behavior (if at all possible with the data-set we
are trying to sort). We introduce
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Ch.7 - QuickSort
QUICKSORT: Performance – Randomized QUICKSORT
And the recursive procedure becomes:
Every call to RANDOMIZED-PARTITION has introduced
the (constant) extra overhead of a call to RANDOM.
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Ch.7 - QuickSort
QUICKSORT: Performance – Rigorous Worst Case
Analysis
Since we do not, a priori, have any idea of what the splits of
the subarrays will be, we have to represent a possible
“worst case” (we already have an O(n2) bound from the
“bad split” example – so it could be worse… although we
hope not). The worst case leads to the recurrence
T(n) = max0≤q≤n-1(T(q) + T(n – q - 1)) + Q(n),
where we remember that the pivot does not appear at the
next level (down) of the recursion.
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Ch.7 - QuickSort
QUICKSORT: Performance – Rigorous Worst Case
Analysis
We have to come up with a “guess” and the basis for the
guess is our likely “bad split case”: it tells us we cannot
hope for any better than W(n2). So we just hope it is no
worse… Guess T(n) ≤ cn2 for some c > 0 and start doing
algebra for the induction:
T(n) ≤ max0≤q≤n-1(T(q) + T(n – q - 1)) + Q(n)
≤ max0≤q≤n-1(cq2 + c(n – q - 1)2) + Q(n).
Differentiate cq2 + c(n – q - 1)2 twice with respect to q, to
obtain 4c > 0 for all values of q.
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Ch.7 - QuickSort
QUICKSORT: Performance – Rigorous Worst Case
Analysis
Since the expression represents a quadratic curve,
concave up, it reaches it maximum at one of the
endpoints q = 0 and q = n – 1. As we evaluate, we find
max0≤q≤n-1(cq2 + c(n – q - 1)2) + Q(n) ≤
c max0≤q≤n-1(q2 + (n – q - 1)2) + Q(n) ≤
c (n – 1)2 + Q(n) = cn2 – 2cn + 1 + Q(n) ≤ cn2
by choosing c large enough to overcome the positive
constant in Q(n).
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Ch.7 - QuickSort
QUICKSORT: Performance – Expected RunTime
Understanding partitioning.
1. Each time PARTITION is called, it selects a pivot element
and this pivot element is never included in successive
calls: the total number of calls to PARTITION is n.
2. Each call to PARTITION costs O(1) plus an amount of
time proportional to the number of iterations of the for
loop.
3. Each iteration of the for loop (in line 4) performs a
comparison , comparing the pivot to another element in
A.
4. We need to count the number of times l. 4 is executed.
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Ch.7 - QuickSort
QUICKSORT: Performance – Expected RunTime
Lemma 7.1. Let X be the number of comparisons
performed in l. 4 of PARTITION over the entire execution
of QUICKSORT on an n-element array. Then the running
time of QUICKSORT is O(n + X).
Proof: the observations on the previous slide.
We need to find X, the total number of comparisons
performed over all calls to PARTITION.
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Ch.7 - QuickSort
QUICKSORT: Performance – Expected RunTime
1. Rename the elements of A as z1, z2, …, zn, so that zi is the
ith smallest element of A.
2. Define the set Zij = {zi, zi+1,…, zj}.
3. Question: when does the algorithm compare zi and zj?
4. Answer: at most once – notice that all elements in every
(sub)array are compared to the pivot once, and will
never be compared to the pivot again (since the pivot is
removed from the recursion).
5. Define Xij = I{zi is compared to zj}, the indicator variable of
this event. Comparisons are over the full run of the
algorithm.
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Ch.7 - QuickSort
QUICKSORT: Performance – Expected RunTime
6. Since each pair is compared at most once, we can write
n 1
n
 X
X 
ij
.
i 1 j  i 1
7. Taking expectations of both sides:
n 1 n

E  X   E   X ij  


i 1 

j  i 1
n 1
n 1
n
n
  E X     Pr z
ij
i 1 j  i 1
i
is compared to

zj .
i 1 j  i 1
8. We need to compute Pr{zi is compared to zj}.
 9. We will assume all zi and zj are distinct.
10.For any pair zi, zj, once a pivot x is chosen so that zi < x <
zj, zi and zj will never be compared again (why?).
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Ch.7 - QuickSort
QUICKSORT: Performance – Expected RunTime
11.If zi is chosen as a pivot before any other item in Zij, then
zi will be compared to every other item in Zij.
12.Same for zj.
13. zi and zj are compared if and only if the first element to
be chosen as a pivot from Zij is either zi or zj.
14.What is that probability? Until a point of Zij is chosen as
a pivot, the whole of Zij is in the same partition, so every
element of Zij is equally likely to be the first one chosen
as a pivot.
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Ch.7 - QuickSort
QUICKSORT: Performance – Expected RunTime
15.Because Zij has j – i + 1 elements, and because pivots
are chosen randomly and independently, the probability
that any given element is the first one chosen as a pivot
is 1/(j-i+1). It follows that:
16. Pr{zi is compared to zj}
= Pr{zi or zj is first pivot chosen from Zij}
= Pr{zi is first pivot chosen from Zij}+
Pr{ zj is first pivot chosen from Zij}
= 1/(j-i+1) + 1/(j-i+1) = 2/(j-i+1).
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Ch.7 - QuickSort
QUICKSORT: Performance – Expected RunTime
17.Replacing the right-hand-side in 7, and grinding through
some algebra:
n 1
E X  
n

i 1 j  i 1
2
j  i 1
n 1 n  i

2
n 1
n
2
n 1
  k  1    k   2H
i 1 k 1
i 1 k 1
i 1
n 1
n

 O lg n   O( n lg n ).
i 1
And the result follows.

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