Analyzing the Average Case of Quicksort Algorithm, Slides of Computer Science

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Algorithms

Quicksort

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

 Which is more likely?

 The latter, by far, except...

 Will any particular input elicit the worst case?

 Yes: Already-sorted input

Review: Analyzing Quicksort

 In the best case:

T(n) = 2T(n/2) + Θ(n)

 What does this work out to?

T(n) = Θ(n lg n)

Review: Analyzing Quicksort

(Average Case)

 Intuitively, a real-life run of quicksort will

produce a mix of “bad” and “good” splits

 Randomly distributed among the recursion tree

 Pretend for intuition that they alternate between best-case (n/2 : n/2) and worst-case (n-1 : 1)

 What happens if we bad-split root node, then good-split the resulting size (n-1) node?

Review: Analyzing Quicksort

(Average Case)

 Intuitively, the O(n) cost of a bad split

(or 2 or 3 bad splits) can be absorbed

into the O(n) cost of each good split

 Thus running time of alternating bad and good

splits is still O(n lg n), with slightly higher

constants

 How can we be more rigorous?

Analyzing Quicksort: Average Case

 For simplicity, assume:

 All inputs distinct (no repeats)

 Slightly different partition() procedure

 partition around a random element, which is not included in subarrays

 all splits (0:n-1, 1:n-2, 2:n-3, … , n-1:0) equally likely

 What is the probability of a particular split

happening?

 Answer: 1/n

Analyzing Quicksort: Average Case

 So…

( ) [ ( ) ( )] ( )

∑^ ( )^ ( )

−

=

−

=

1

0

1

0

n

k

n

k

T k n n

T k T n k n n

T n

Write it on the board

Analyzing Quicksort: Average Case

 We can solve this recurrence using the dreaded

substitution method

 Guess the answer

 Assume that the inductive hypothesis holds

 Substitute it in for some value < n

 Prove that it follows for n

Analyzing Quicksort: Average Case

 We can solve this recurrence using the dreaded

substitution method

 Guess the answer

 T(n) = O(n lg n)

 Assume that the inductive hypothesis holds

 Substitute it in for some value < n

 Prove that it follows for n

Analyzing Quicksort: Average Case

 We can solve this recurrence using the dreaded

substitution method

 Guess the answer

 T(n) = O(n lg n)

 Assume that the inductive hypothesis holds

 What’s the inductive hypothesis?

 Substitute it in for some value < n

 Prove that it follows for n

Analyzing Quicksort: Average Case

 We can solve this recurrence using the dreaded

substitution method

 Guess the answer

 T( n ) = O( n lg n )

 Assume that the inductive hypothesis holds

 T( n ) ≤ an lg n + b for some constants a and b

 Substitute it in for some value < n

 What value?

 Prove that it follows for n

Analyzing Quicksort: Average Case

 We can solve this recurrence using the dreaded

substitution method

 Guess the answer

 T( n ) = O( n lg n )

 Assume that the inductive hypothesis holds

 T( n ) ≤ an lg n + b for some constants a and b

 Substitute it in for some value < n

 The value k in the recurrence

 Prove that it follows for n

Note: leaving the same recurrence as the book

What are we doing here?

Analyzing Quicksort: Average Case

∑^ (^ )^ ( )

−

=

−

=

−

=

−

=

−

=

1

1

1

1

1

1

1

0

1

0

lg

lg

lg

lg

n

k

n

k

n

k

n

k

n

k

ak k b n n

n n

b ak k b n

b ak k b n n

ak k b n n

T k n n

T n The recurrence to be solved

What are we doing here?

What are we doing here?

Plug in inductive hypothesis

Expand out the k=0 case

2b/n is just a constant, so fold it into Θ (n)

What are we doing here?

What are we doing here?

Evaluate the summation: b+b+…+b = b (n-1)

The recurrence to be solved

Since n-1<n, 2b(n-1)/n < 2b

Analyzing Quicksort: Average Case

k k b ( ) n

n

a

n n n

b k k n

a

b n n

ak k n

ak k b n n

T n

n

k

n

k

n

k

n

k

n

k

−

=

−

=

−

=

−

=

−

=

lg 2

lg

lg

lg

1

1

1

1

1

1

1

1

1

1

Distribute the summation What are we doing here?

This summation gets its own set of slides later