CSE 431/531: Analysis of Algorithms
Lectures 9-13: Divide-and-Conquer
Lecturer: Shi Li
Department of Computer Science and Engineering
University at Buffalo
Spring 2016
MoWeFr 3:00-3:50pm
Knox 110
Outline
1
Divide-and-Conquer
2
Solving Recurrences
3
Counting Inversions
4
Polynomial Multiplication
5
Strassen’s Algorithm for Matrix Multiplication
6
Quicksort
Randomized Quicksort
Average-Case Analysis
7
Lower Bound for Comparison-Based Sorting Algorithms
8
Selection Problem
Divide-and-Conquer
Divide: Divide instance into many smaller instances
Conquer: Solve each of smaller instances recursively and
separately
Combine: Combine solutions to small instances to obtain a
solution for original big instance
merge-sort(A, n)
1
2
3
4
5
6
if n = 1 then
return A
else
B ← merge-sort A 1..bn/2c , bn/2c
C ← merge-sort A bn/2c + 1..n , dn/2e
return merge(B, C, bn/2c, dn/2e)
merge-sort(A, n)
1
2
3
4
5
6
if n = 1 then
return A
else
B ← merge-sort A 1..bn/2c , bn/2c
C ← merge-sort A bn/2c + 1..n , dn/2e
return merge(B, C, bn/2c, dn/2e)
Divide: trivial
Conquer: 4 , 5
Combine: 6
Merge Two Sorted Arrays
3
8 12 20 32 48
5
7
9 25 29
Merge Two Sorted Arrays
3
8 12 20 32 48
5
7
9 25 29
Merge Two Sorted Arrays
3
8 12 20 32 48
5
7
3
9 25 29
Merge Two Sorted Arrays
3
8 12 20 32 48
5
7
3
9 25 29
Merge Two Sorted Arrays
3
8 12 20 32 48
5
7
3
5
9 25 29
Merge Two Sorted Arrays
3
8 12 20 32 48
5
7
3
5
9 25 29
Merge Two Sorted Arrays
3
8 12 20 32 48
5
7
9 25 29
3
5
7
Merge Two Sorted Arrays
3
8 12 20 32 48
5
7
9 25 29
3
5
7
Merge Two Sorted Arrays
3
8 12 20 32 48
5
7
9 25 29
3
5
7
8
Merge Two Sorted Arrays
3
8 12 20 32 48
5
7
9 25 29
3
5
7
8
Merge Two Sorted Arrays
3
8 12 20 32 48
5
7
9 25 29
3
5
7
8
9 12 20 25 29
Merge Two Sorted Arrays
3
8 12 20 32 48
5
7
9 25 29
3
5
7
8
9 12 20 25 29 32 48
Merge Two Sorted Arrays
Merge(B, C, n1 , n2 )
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2
3
4
5
6
7
8
9
A ← []; i ← 1; j ← 1
while i ≤ n1 and j ≤ n2
if (B[i] ≤ C[j]) then
append B[i] to A; i ← i + 1;
else
append C[j] to A; j ← j + 1
if i ≤ n1 then append B[i..n1 ] to A
if j ≤ n2 then append C[j..n2 ] to A
return A
Recurrence for Running Time for Merge-Sort
T (n): running time for sorting an array of size n
(
c
if n = 1
T (n) =
T (bn/2c) + T (dn/2e) + cn if n ≥ 2
Recurrence for Running Time for Merge-Sort
T (n): running time for sorting an array of size n
(
c
if n = 1
T (n) =
2T (n/2) + cn if n ≥ 2
Recurrence for Running Time for Merge-Sort
T (n): running time for sorting an array of size n
T (n) = 2T (n/2) + O(n)
Outline
1
Divide-and-Conquer
2
Solving Recurrences
3
Counting Inversions
4
Polynomial Multiplication
5
Strassen’s Algorithm for Matrix Multiplication
6
Quicksort
Randomized Quicksort
Average-Case Analysis
7
Lower Bound for Comparison-Based Sorting Algorithms
8
Selection Problem
Methods for Solving Recurrences
The recursion-tree method
The substitution method
The master method
Recursion-Tree Method
T (n) = 2T (n/2) + O(n)
Recursion-Tree Method
T (n) = 2T (n/2) + O(n)
n
n/2
n/2
n/4
n/4
n/4
n/4
n/8
n/8
n/8
n/8
n/8
n/8
n/8
n/8
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Recursion-Tree Method
T (n) = 2T (n/2) + O(n)
n
n/2
n/2
n/4
n/4
n/4
n/4
n/8
n/8
n/8
n/8
n/8
n/8
n/8
n/8
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Each level takes running time O(n)
Recursion-Tree Method
T (n) = 2T (n/2) + O(n)
n
n/2
n/2
n/4
n/4
n/4
n/4
n/8
n/8
n/8
n/8
n/8
n/8
n/8
n/8
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Each level takes running time O(n)
There are O(lg n) levels
Recursion-Tree Method
T (n) = 2T (n/2) + O(n)
n
n/2
n/2
n/4
n/4
n/4
n/4
n/8
n/8
n/8
n/8
n/8
n/8
n/8
n/8
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Each level takes running time O(n)
There are O(lg n) levels
Running time = O(n lg n)
Recursion-Tree Method
T (n) = 4T (n/2) + O(n)
Recursion-Tree Method
T (n) = 4T (n/2) + O(n)
n
Recursion-Tree Method
T (n) = 4T (n/2) + O(n)
n
n/2
n/2
n/2
n/2
Recursion-Tree Method
T (n) = 4T (n/2) + O(n)
n
n/2
n/4
n/4
n/4
n/2
n/4
···
n/2
···
···
n/2
n/4
n/4
n/4
n/4
Recursion-Tree Method
T (n) = 4T (n/2) + O(n)
n
n/2
n/4
n
8
n/4
n
8
n
8
n/4
n
8
n/2
n/4
n/2
···
···
···
···
···
···
n/2
n/4
n/4
n
8
n/4
n
8
n/4
n
8
n
8
Recursion-Tree Method
T (n) = 4T (n/2) + O(n)
n
n/2
n/4
n
8
n/4
n
8
n
8
n/4
n
8
n/2
n/4
n/2
···
···
···
···
···
···
Total running time at level i?
n/2
n/4
n/4
n
8
n/4
n
8
n/4
n
8
n
8
Recursion-Tree Method
T (n) = 4T (n/2) + O(n)
n
n/2
n/4
n
8
n/4
n
8
n
8
n/4
n
8
n/2
n/4
n/2
···
···
···
···
···
···
Total running time at level i?
n
2i
n/2
n/4
n/4
n
8
× 4i = 2i n
n/4
n
8
n/4
n
8
n
8
Recursion-Tree Method
T (n) = 4T (n/2) + O(n)
n
n/2
n/4
n
8
n/4
n
8
n
8
n/4
n
8
n/2
n/4
n/2
···
···
···
···
···
···
Total running time at level i?
Total number of levels?
n
2i
n/2
n/4
n/4
n
8
× 4i = 2i n
n/4
n
8
n/4
n
8
n
8
Recursion-Tree Method
T (n) = 4T (n/2) + O(n)
n
n/2
n/4
n
8
n/4
n
8
n
8
n/4
n
8
n/2
n/4
n/2
···
···
···
···
···
···
Total running time at level i?
Total number of levels? lg2 n
n
2i
n/2
n/4
n/4
n
8
× 4i = 2i n
n/4
n
8
n/4
n
8
n
8
Recursion-Tree Method
T (n) = 4T (n/2) + O(n)
n
n/2
n/4
n
8
n/4
n
8
n
8
n/4
n
8
n/2
n/4
n/2
···
···
···
···
···
···
Total running time at level i?
Total number of levels? lg2 n
Total running time?
n
2i
n/2
n/4
n/4
n
8
× 4i = 2i n
n/4
n
8
n/4
n
8
n
8
Recursion-Tree Method
T (n) = 4T (n/2) + O(n)
n
n/2
n/4
n
8
n/4
n
8
n
8
n/4
n
8
n/2
n/4
n/2
···
···
···
···
···
···
Total running time at level i?
Total number of levels? lg2 n
Total running time?
n
2i
n/2
n/4
n/4
n
8
n/4
n
8
n/4
n
8
n
8
× 4i = 2i n
lg2 n
X
i=1
2i n = n(2lg2 n+1 − 1) = n(2n − 1) = O(n2 ).
Substitution Method
1. Guess the form of the solution
2. Use mathematical induction to find the constants
Substitution Method
1. Guess the form of the solution
2. Use mathematical induction to find the constants
Ex: T (n) = 2T (n/2) + O(n)
Substitution Method
1. Guess the form of the solution
2. Use mathematical induction to find the constants
Ex: T (n) = 2T (n/2) + O(n)
Ex: T (n) = 4T (n/2) + O(n3)
Substitution Method
1. Guess the form of the solution
2. Use mathematical induction to find the constants
Ex: T (n) = 2T (n/2) + O(n)
Ex: T (n) = 4T (n/2) + O(n3)
Ex: T (n) = T (n/5) + T (7n/10) + O(n)
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,
T (n) =


??


if c < lgb a
if c = lgb a
if c > lgb a
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,
T (n) =

lg a

O(n b )


if c < lgb a
if c = lgb a
if c > lgb a
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,
T (n) =

lg a

O(n b )


??
if c < lgb a
if c = lgb a
if c > lgb a
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,
T (n) =

lg a

O(n b )


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,

lg a

O(n b )
T (n) = ??


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,

lg a

O(n b )
T (n) = O(nc lg n)


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,

lg a

O(n b )
T (n) = O(nc lg n)


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
Ex: T (n) = 4T (n/2) + O(n2 ). Which Case?
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,

lg a

O(n b )
T (n) = O(nc lg n)


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
Ex: T (n) = 4T (n/2) + O(n2 ). Case 2.
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,

lg a

O(n b )
T (n) = O(nc lg n)


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
Ex: T (n) = 4T (n/2) + O(n2 ). Case 2. T (n) = O(n2 lg n)
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,

lg a

O(n b )
T (n) = O(nc lg n)


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
Ex: T (n) = 4T (n/2) + O(n2 ). Case 2. T (n) = O(n2 lg n)
Ex: T (n) = 3T (n/2) + O(n). Which Case?
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,

lg a

O(n b )
T (n) = O(nc lg n)


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
Ex: T (n) = 4T (n/2) + O(n2 ). Case 2. T (n) = O(n2 lg n)
Ex: T (n) = 3T (n/2) + O(n). Case 1.
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,

lg a

O(n b )
T (n) = O(nc lg n)


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
Ex: T (n) = 4T (n/2) + O(n2 ). Case 2. T (n) = O(n2 lg n)
Ex: T (n) = 3T (n/2) + O(n). Case 1. T (n) = O(nlg2 3 )
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,

lg a

O(n b )
T (n) = O(nc lg n)


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
Ex: T (n) = 4T (n/2) + O(n2 ). Case 2. T (n) = O(n2 lg n)
Ex: T (n) = 3T (n/2) + O(n). Case 1. T (n) = O(nlg2 3 )
Ex: T (n) = T (n/2) + O(1). Which Case?
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,

lg a

O(n b )
T (n) = O(nc lg n)


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
Ex: T (n) = 4T (n/2) + O(n2 ). Case 2. T (n) = O(n2 lg n)
Ex: T (n) = 3T (n/2) + O(n). Case 1. T (n) = O(nlg2 3 )
Ex: T (n) = T (n/2) + O(1). Case 2.
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,

lg a

O(n b )
T (n) = O(nc lg n)


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
Ex: T (n) = 4T (n/2) + O(n2 ). Case 2. T (n) = O(n2 lg n)
Ex: T (n) = 3T (n/2) + O(n). Case 1. T (n) = O(nlg2 3 )
Ex: T (n) = T (n/2) + O(1). Case 2. T (n) = O(lg n)
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,
Ex:
Ex:
Ex:
Ex:

lg a

O(n b )
T (n) = O(nc lg n)


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
T (n) = 4T (n/2) + O(n2 ). Case 2. T (n) = O(n2 lg n)
T (n) = 3T (n/2) + O(n). Case 1. T (n) = O(nlg2 3 )
T (n) = T (n/2) + O(1). Case 2. T (n) = O(lg n)
T (n) = 2T (n/2) + O(n2 ). Which Case?
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,
Ex:
Ex:
Ex:
Ex:

lg a

O(n b )
T (n) = O(nc lg n)


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
T (n) = 4T (n/2) + O(n2 ). Case 2. T (n) = O(n2 lg n)
T (n) = 3T (n/2) + O(n). Case 1. T (n) = O(nlg2 3 )
T (n) = T (n/2) + O(1). Case 2. T (n) = O(lg n)
T (n) = 2T (n/2) + O(n2 ). Case 3.
Master Theorem
a ≥ 1, b > 1, c ≥ 0: constants
T (n) = aT (n/b) + O(nc )
Then,
Ex:
Ex:
Ex:
Ex:

lg a

O(n b )
T (n) = O(nc lg n)


O(nc )
if c < lgb a
if c = lgb a
if c > lgb a
T (n) = 4T (n/2) + O(n2 ). Case 2. T (n) = O(n2 lg n)
T (n) = 3T (n/2) + O(n). Case 1. T (n) = O(nlg2 3 )
T (n) = T (n/2) + O(1). Case 2. T (n) = O(lg n)
T (n) = 2T (n/2) + O(n2 ). Case 3. T (n) = O(n2 )
Recurssion Tree for Master Theorem
nc
1 node
(n/b)c
a nodes
(n/b2)c
a2 nodes
a3 nodes
n c
b3
.
.
.
n c
b3
.
.
.
(n/b)c
(n/b2)c
n c
b3
.
.
.
n c
b3
.
.
.
(n/b2)c
n c
b3
.
.
.
(n/b2)c
n c
b3
.
.
.
n c
b3
.
.
.
n c
b3
.
.
.
Recurssion Tree for Master Theorem
nc
1 node
(n/b)c
a nodes
(n/b2)c
a2 nodes
a3 nodes
nc
n c
b3
.
.
.
n c
b3
.
.
.
(n/b2)c
n c
b3
.
.
.
a c
bc n
(n/b)c
n c
b3
.
.
.
(n/b2)c
n c
b3
.
.
.
(n/b2)c
n c
b3
.
.
.
n c
b3
.
.
.
n c
b3
.
.
.
a 2 c
n
bc
a 3 c
n
bc
Recurssion Tree for Master Theorem
nc
1 node
(n/b)c
a nodes
(n/b2)c
a2 nodes
a3 nodes
nc
n c
b3
.
.
.
n c
b3
.
.
.
(n/b2)c
n c
b3
.
.
.
a c
bc n
(n/b)c
n c
b3
.
.
.
(n/b2)c
n c
b3
.
.
.
c < lgb a : bottom-level dominates:
(n/b2)c
n c
b3
.
.
.
n c
b3
.
.
.
a lgb n
bc
n c
b3
a 2 c
n
bc
a 3 c
n
bc
.
.
.
nc = nlgb a
Recurssion Tree for Master Theorem
nc
1 node
(n/b)c
a nodes
(n/b2)c
a2 nodes
a3 nodes
nc
n c
b3
.
.
.
n c
b3
.
.
.
(n/b2)c
n c
b3
.
.
.
a c
bc n
(n/b)c
n c
b3
.
.
.
(n/b2)c
n c
b3
.
.
.
(n/b2)c
n c
b3
.
.
.
n c
b3
.
.
.
n c
b3
a 2 c
n
bc
a 3 c
n
bc
.
.
.
lg n
c < lgb a : bottom-level dominates: bac b nc = nlgb a
c = lgb a : all levels are the same: nc lgb n = O(nc lg n)
Recurssion Tree for Master Theorem
nc
1 node
(n/b)c
a nodes
(n/b2)c
a2 nodes
a3 nodes
nc
n c
b3
.
.
.
n c
b3
.
.
.
(n/b2)c
n c
b3
.
.
.
a c
bc n
(n/b)c
n c
b3
.
.
.
(n/b2)c
n c
b3
.
.
.
(n/b2)c
n c
b3
.
.
.
n c
b3
.
.
.
n c
b3
a 2 c
n
bc
a 3 c
n
bc
.
.
.
lg n
c < lgb a : bottom-level dominates: bac b nc = nlgb a
c = lgb a : all levels are the same: nc lgb n = O(nc lg n)
c > lgb a : top-level dominates: O(nc )
Outline
1
Divide-and-Conquer
2
Solving Recurrences
3
Counting Inversions
4
Polynomial Multiplication
5
Strassen’s Algorithm for Matrix Multiplication
6
Quicksort
Randomized Quicksort
Average-Case Analysis
7
Lower Bound for Comparison-Based Sorting Algorithms
8
Selection Problem
Def. Given an array A of n integers, an inversion of A is a pair
(i, j) of indices such that i < j and ai > aj .
Def. Given an array A of n integers, an inversion of A is a pair
(i, j) of indices such that i < j and ai > aj .
Counting Inversions
Input: a sequence a1 , a2 , · · · , an of n distinct numbers
Output: number of inversions
Def. Given an array A of n integers, an inversion of A is a pair
(i, j) of indices such that i < j and ai > aj .
Counting Inversions
Input: a sequence a1 , a2 , · · · , an of n distinct numbers
Output: number of inversions
Example:
10
8
15
9
12
Def. Given an array A of n integers, an inversion of A is a pair
(i, j) of indices such that i < j and ai > aj .
Counting Inversions
Input: a sequence a1 , a2 , · · · , an of n distinct numbers
Output: number of inversions
Example:
10
8
15
9
12
8
9
10
12
15
Def. Given an array A of n integers, an inversion of A is a pair
(i, j) of indices such that i < j and ai > aj .
Counting Inversions
Input: a sequence a1 , a2 , · · · , an of n distinct numbers
Output: number of inversions
Example:
10
8
15
9
12
8
9
10
12
15
Def. Given an array A of n integers, an inversion of A is a pair
(i, j) of indices such that i < j and ai > aj .
Counting Inversions
Input: a sequence a1 , a2 , · · · , an of n distinct numbers
Output: number of inversions
Example:
10
8
15
9
12
8
9
10
12
15
4 inversions (for convenience, using numbers, not indices):
(10, 8), (10, 9), (15, 9), (15, 12)
Naive Algorithm for Counting Inversions
count-inversions(A, n)
1
2
3
4
5
c←0
for every i ← 1 to n − 1
for every j ← i + 1 to n
if A[i] > A[j] then c ← c + 1
return c
Divide-and-Conquer
A:
B
C
Recursion:
inversions(A) = inversions(B) + inversions(C) + m
m = {(i, j) : 1 ≤ i ≤ bn/2c, bn/2c < j ≤ n, Ai > Aj }
Computing m in O(n2 ) time: T (n) = 2T (n/2) + O(n2 )
T (n) = O(n2 ): bad
Computing m in O(n) time: T (n) = 2T (n/2) + O(n)
T (n) = O(n lg n): good
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
total= 0
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
total= 0
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
3
9 25 29
total= 0
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
3
9 25 29
total= 0
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
3
5
9 25 29
total= 0
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
3
5
9 25 29
total= 0
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
total= 0
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
total= 0
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
total= 022
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
total= 022
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
9
total= 022
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
9
total= 022
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
9 12
total= 0225
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
9 12
total= 0225
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
9 12 20
total= 02258
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
9 12 20
total= 02258
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
total= 02258
9 12 20 25
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
total= 02258
9 12 20 25
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
total= 02258
9 12 20 25 29
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
total= 02258
9 12 20 25 29
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
02258
total= 13
9 12 20 25 29 32
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
02258
total= 13
9 12 20 25 29 32
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
13
02258
total= 18
9 12 20 25 29 32 48
Counting Inversions between B and C
Count pairs i, j such that B[i] > C[j]:
B:
3
8 12 20 32 48
C:
5
7
9 25 29
3
5
7
8
13
02258
total= 18
9 12 20 25 29 32 48
Count Inversions between B and C
inversions-between(B, C, n1 , n2 )
1
2
3
4
5
6
7
8
9
count ← 0;
A ← []; i ← 1; j ← 1
while i ≤ n1 or j ≤ n2
j > n2 or (i ≤ n1 and B[i] ≤ C[j]) then
append B[i] to A; i ← i + 1
count ← count + (j − 1)
else
append C[j] to A; j ← j + 1
return (A, count)
Count Inversions in A
inversions(A, n)
1
2
3
4
5
6
7
if n = 1 then
return (A, 0)
else
(B, m1 ) ← inversions A 1..bn/2c , bn/2c
(C, m2 ) ← inversions A bn/2c + 1..n , dn/2e
(A, m3 ) ← inversions-between(B, C, bn/2c, dn/2e)
return (A, m1 + m2 + m3 )
Count Inversions in A
inversions(A, n)
1
2
3
4
5
6
7
Divide: trivial
Conquer: 5 , 6
Combine: 7
if n = 1 then
return (A, 0)
else
(B, m1 ) ← inversions A 1..bn/2c , bn/2c
(C, m2 ) ← inversions A bn/2c + 1..n , dn/2e
(A, m3 ) ← inversions-between(B, C, bn/2c, dn/2e)
return (A, m1 + m2 + m3 )
Outline
1
Divide-and-Conquer
2
Solving Recurrences
3
Counting Inversions
4
Polynomial Multiplication
5
Strassen’s Algorithm for Matrix Multiplication
6
Quicksort
Randomized Quicksort
Average-Case Analysis
7
Lower Bound for Comparison-Based Sorting Algorithms
8
Selection Problem
Polynomial Multiplication
Input: two polynomials of degree n − 1
Output: product of two polynomials
Polynomial Multiplication
Input: two polynomials of degree n − 1
Output: product of two polynomials
Example:
(3x3 + 2x2 − 5x + 4) × (2x3 − 3x2 + 6x − 5)
Polynomial Multiplication
Input: two polynomials of degree n − 1
Output: product of two polynomials
Example:
(3x3 + 2x2 − 5x + 4) × (2x3 − 3x2 + 6x − 5)
= 6x6 − 9x5 + 18x4 − 15x3
+ 4x5 − 6x4 + 12x3 − 10x2
− 10x4 + 15x3 − 30x2 + 25x
+ 8x3 − 12x2 + 24x − 20
= 6x6 − 5x5 + 2x4 + 20x3 − 52x2 + 49x − 20
Polynomial Multiplication
Input: two polynomials of degree n − 1
Output: product of two polynomials
Example:
(3x3 + 2x2 − 5x + 4) × (2x3 − 3x2 + 6x − 5)
= 6x6 − 9x5 + 18x4 − 15x3
+ 4x5 − 6x4 + 12x3 − 10x2
− 10x4 + 15x3 − 30x2 + 25x
+ 8x3 − 12x2 + 24x − 20
= 6x6 − 5x5 + 2x4 + 20x3 − 52x2 + 49x − 20
Input: (4, −5, 2, 3), (−5, 6, −3, 2)
Output: (−20, 49, −52, 20, 2, −5, 6)
Naı̈ve Algorithm
polynomial-multiplication(A, B, n)
1
2
3
4
5
let C[k] = 0 for every k = 0, 1, 2, · · · , 2n − 2
for i ← 0 to n − 1
for j ← 0 to n − 1
C[i + j] ← C[i + j] + A[i] × B[j]
return C
Naı̈ve Algorithm
polynomial-multiplication(A, B, n)
1
2
3
4
5
let C[k] = 0 for every k = 0, 1, 2, · · · , 2n − 2
for i ← 0 to n − 1
for j ← 0 to n − 1
C[i + j] ← C[i + j] + A[i] × B[j]
return C
Running time: O(n2 )
Divide-and-Conquer for Polynomial Multiplication
p(x) = 3x3 + 2x2 − 5x + 4 = (3x + 2)x2 + (−5x + 4)
q(x) = 2x3 − 3x2 + 6x − 5 = (2x − 3)x2 + (6x − 5)
Divide-and-Conquer for Polynomial Multiplication
p(x) = 3x3 + 2x2 − 5x + 4 = (3x + 2)x2 + (−5x + 4)
q(x) = 2x3 − 3x2 + 6x − 5 = (2x − 3)x2 + (6x − 5)
p(x): degree of n − 1 (assume n is even)
p(x) = pH (x)xn/2 + pL (x),
pH (x), pL (x): polynomials of degree n/2 − 1.
Divide-and-Conquer for Polynomial Multiplication
p(x) = 3x3 + 2x2 − 5x + 4 = (3x + 2)x2 + (−5x + 4)
q(x) = 2x3 − 3x2 + 6x − 5 = (2x − 3)x2 + (6x − 5)
p(x): degree of n − 1 (assume n is even)
p(x) = pH (x)xn/2 + pL (x),
pH (x), pL (x): polynomials of degree n/2 − 1.
pq = pH xn/2 + pL qH xn/2 + qL
Divide-and-Conquer for Polynomial Multiplication
p(x) = 3x3 + 2x2 − 5x + 4 = (3x + 2)x2 + (−5x + 4)
q(x) = 2x3 − 3x2 + 6x − 5 = (2x − 3)x2 + (6x − 5)
p(x): degree of n − 1 (assume n is even)
p(x) = pH (x)xn/2 + pL (x),
pH (x), pL (x): polynomials of degree n/2 − 1.
pq = pH xn/2 + pL qH xn/2 + qL
= pH qH xn + pH qL + pL qH xn/2 + pL qL
Divide-and-Conquer for Polynomial Multiplication
pq = pH xn/2 + pL qH xn/2 + qL
= pH qH xn + pH qL + pL qH xn/2 + pL qL
Divide-and-Conquer for Polynomial Multiplication
pq = pH xn/2 + pL qH xn/2 + qL
= pH qH xn + pH qL + pL qH xn/2 + pL qL
multiply(p, q) = multiply(pH , qH ) × xn
+ multiply(pH , qL ) + multiply(pL , qH ) × xn/2
+ multiply(pL , qL )
Divide-and-Conquer for Polynomial Multiplication
pq = pH xn/2 + pL qH xn/2 + qL
= pH qH xn + pH qL + pL qH xn/2 + pL qL
multiply(p, q) = multiply(pH , qH ) × xn
+ multiply(pH , qL ) + multiply(pL , qH ) × xn/2
+ multiply(pL , qL )
Recurrence: T (n) = 4T (n/2) + O(n)
Divide-and-Conquer for Polynomial Multiplication
pq = pH xn/2 + pL qH xn/2 + qL
= pH qH xn + pH qL + pL qH xn/2 + pL qL
multiply(p, q) = multiply(pH , qH ) × xn
+ multiply(pH , qL ) + multiply(pL , qH ) × xn/2
+ multiply(pL , qL )
Recurrence: T (n) = 4T (n/2) + O(n)
T (n) = O(n2 )
Reduce Number from 4 to 3
Reduce Number from 4 to 3
pq = pH xn/2 + pL qH xn/2 + qL
= pH qH xn + pH qL + pL qH xn/2 + pL qL
Reduce Number from 4 to 3
pq = pH xn/2 + pL qH xn/2 + qL
= pH qH xn + pH qL + pL qH xn/2 + pL qL
pH qL + pL qH = (pH + pL )(qH + qL ) − pH qH − pL qL
Divide-and-Conquer for Polynomial Multiplication
rH = multiply(pH , qH )
rL = multiply(pL , qL )
multiply(p, q) = rH × xn
+ multiply(pH + pL , qH + qL ) − rH − rL × xn/2
+ rL
Divide-and-Conquer for Polynomial Multiplication
rH = multiply(pH , qH )
rL = multiply(pL , qL )
multiply(p, q) = rH × xn
+ multiply(pH + pL , qH + qL ) − rH − rL × xn/2
+ rL
Recurrence: T (n) = 3T (n/2) + O(n)
T (n) = O(nlg2 3 ) = O(n1.585 )
multiply(A, B, n)
1
2
3
4
5
6
7
8
9
10
11
12
13
\\ assume n is power of 2
if n = 1 then return (A[0]B[0])
AL ← A[0 .. n/2 − 1], AH ← A[n/2 .. n − 1]
BL ← B[0 .. n/2 − 1], BH ← B[n/2 .. n − 1]
CL ← multiply(AL , BL , n/2)
CH ← multiply(AH , BH , n/2)
CM ← multiply(AL + AH , BL + BH , n/2)
for i ← 0 to 2n − 2
C[i] ← 0
for i ← 0 to n − 2
C[i] ← C[i] + CL [i]
C[i + n] ← C[i + n] + CH [i]
C[i + n/2] ← C[i + n/2] + CM [i] − CL [i] − CH [i]
return C
Outline
1
Divide-and-Conquer
2
Solving Recurrences
3
Counting Inversions
4
Polynomial Multiplication
5
Strassen’s Algorithm for Matrix Multiplication
6
Quicksort
Randomized Quicksort
Average-Case Analysis
7
Lower Bound for Comparison-Based Sorting Algorithms
8
Selection Problem
Strassen’s Algorithm for Matrix Multiplication
Matrix Multiplication
Input: two n × n matrices A and B
Output: C = AB
Running time: O(n3 )
Strassen’s Algorithm for Matrix Multiplication
Matrix Multiplication
Input: two n × n matrices A and B
Output: C = AB
Naive Algorithm: matrix-multiplication(A, B, n)
1
2
3
4
5
6
for i ← 1 to n
for j ← 1 to n
C[i, j] ← 0
for k ← 1 to n
C[i, j] ← C[i, j] + A[i, k] × B[k, j]
return C
Running time: O(n3 )
Try to Use Divide-and-Conquer
n/2
A11 A12
A=
B11 B12
n/2
n/2
B=
A21 A22
n/2
B21 B22
A11 B11 + A12 B21 A11 B12 + A12 B22
C=
A21 B11 + A22 B21 A21 B12 + A22 B22
matrix multiplication(A, B) recursively calls
matrix multiplication(A11 , B11 ),
matrix multiplication(A12 , B21 ),
···
Try to Use Divide-and-Conquer
n/2
A11 A12
A=
B11 B12
n/2
n/2
B=
A21 A22
n/2
B21 B22
A11 B11 + A12 B21 A11 B12 + A12 B22
C=
A21 B11 + A22 B21 A21 B12 + A22 B22
matrix multiplication(A, B) recursively calls
matrix multiplication(A11 , B11 ),
matrix multiplication(A12 , B21 ),
···
Recurrence for running time: T (n) = 8T (n/2) + O(n2 )
T (n) = O(n3 )
Strassen’s Algorithm
T (n) = 8T (n/2) + O(n2 )
Strassen’s Algorithm: improve the number of multiplications
from 8 to 7!
New recurrence: T (n) = 7T (n/2) + O(n2 )
New running time?
Strassen’s Algorithm
T (n) = 8T (n/2) + O(n2 )
Strassen’s Algorithm: improve the number of multiplications
from 8 to 7!
New recurrence: T (n) = 7T (n/2) + O(n2 )
New running time?
T (n) = O(nlog2 7 ) = O(n2.808 )
Strassen’s Algorithm: Reduce 8 to 7
M1
M2
M3
M4
M5
M6
M7
:= (A11 + A22 )(B11 + B22 )
:= (A21 + A22 )B11
:= A11 (B12 − B22 )
:= A22 (B21 − B11 )
:= (A11 + A12 )B22
:= (A21 − A11 )(B11 + B12 )
:= (A12 − A22 )(B21 + B22 )
= M1 + M4 − M5 + M7
= M3 + M5
= M2 + M4
= M1 − M2 + M3 + M6
C11 C12
C=
.
C21 C22
C11
C12
C21
C22
Strassen’s Algorithm: Reduce 8 to 7
M1
M2
M3
M4
M5
M6
M7
:= (A11 + A22 )(B11 + B22 )
:= (A21 + A22 )B11
:= A11 (B12 − B22 )
:= A22 (B21 − B11 )
:= (A11 + A12 )B22
:= (A21 − A11 )(B11 + B12 )
:= (A12 − A22 )(B21 + B22 )
= M1 + M4 − M5 + M7
= M3 + M5
= M2 + M4
= M1 − M2 + M3 + M6
C11 C12
C=
.
C21 C22
C11
C12
C21
C22
Current best running time : O(n2.373 )
Outline
1
Divide-and-Conquer
2
Solving Recurrences
3
Counting Inversions
4
Polynomial Multiplication
5
Strassen’s Algorithm for Matrix Multiplication
6
Quicksort
Randomized Quicksort
Average-Case Analysis
7
Lower Bound for Comparison-Based Sorting Algorithms
8
Selection Problem
Quicksort Example
Assumption: we can choose median of an array in linear time.
29 82 75 64 38 45 94 69 25 76 15 92 37 17 85
Quicksort Example
Assumption: we can choose median of an array in linear time.
29 82 75 64 38 45 94 69 25 76 15 92 37 17 85
Quicksort Example
Assumption: we can choose median of an array in linear time.
29 82 75 64 38 45 94 69 25 76 15 92 37 17 85
29 38 45 25 15 37 17 64 82 75 94 92 69 76 85
Quicksort Example
Assumption: we can choose median of an array in linear time.
29 82 75 64 38 45 94 69 25 76 15 92 37 17 85
29 38 45 25 15 37 17 64 82 75 94 92 69 76 85
Quicksort Example
Assumption: we can choose median of an array in linear time.
29 82 75 64 38 45 94 69 25 76 15 92 37 17 85
29 38 45 25 15 37 17 64 82 75 94 92 69 76 85
25 15 17 29 38 45 37 64 82 75 94 92 69 76 85
Quicksort
quicksort(A, n)
1
2
3
4
5
6
7
if n = 1 return A;
x ← median of A;
AL ← elements in A that are less than x;
AR ← elements in A that are greater than x;
BL ← quicksort(AL , AL .size)
BR ← quicksort(AR , AR .size)
return BL concatenating x concatenating BR
\\ Divide
\\ Divide
\\ Conquer
\\ Conquer
\\ Combine
Quicksort
quicksort(A, n)
1
2
3
4
5
6
7
if n = 1 return A;
x ← median of A;
AL ← elements in A that are less than x;
AR ← elements in A that are greater than x;
BL ← quicksort(AL , AL .size)
BR ← quicksort(AR , AR .size)
return BL concatenating x concatenating BR
how?
\\ Divide
\\ Divide
\\ Conquer
\\ Conquer
\\ Combine
Quicksort
quicksort(A, n)
1
2
3
4
5
6
7
if n = 1 return A;
x ← median of A;
AL ← elements in A that are less than x;
AR ← elements in A that are greater than x;
BL ← quicksort(AL , AL .size)
BR ← quicksort(AR , AR .size)
return BL concatenating x concatenating BR
Recurrence T (n) = 2T (n/2) + O(n)
how?
\\ Divide
\\ Divide
\\ Conquer
\\ Conquer
\\ Combine
Quicksort
quicksort(A, n)
1
2
3
4
5
6
7
if n = 1 return A;
x ← median of A;
AL ← elements in A that are less than x;
AR ← elements in A that are greater than x;
BL ← quicksort(AL , AL .size)
BR ← quicksort(AR , AR .size)
return BL concatenating x concatenating BR
Recurrence T (n) = 2T (n/2) + O(n)
Running time = O(n lg n)
how?
\\ Divide
\\ Divide
\\ Conquer
\\ Conquer
\\ Combine
Quicksort in Practice
quicksort(A, n)
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7
if n = 1 return A;
x ← pivot of A;
AL ← elements in A that are less than x;
AR ← elements in A that are greater than x;
BL ← quicksort(AL , AL .size)
BR ← quicksort(AR , AR .size)
return BL concatenating x concatenating BR
\\ Divide
\\ Divide
\\ Conquer
\\ Conquer
\\ Combine
Quicksort in Practice
quicksort(A, n)
1
2
3
4
5
6
7
if n = 1 return A;
x ← pivot of A;
AL ← elements in A that are less than x;
AR ← elements in A that are greater than x;
BL ← quicksort(AL , AL .size)
BR ← quicksort(AR , AR .size)
return BL concatenating x concatenating BR
Ways to chose pivot:
(a) first element of A
(b) middle element of A
\\ Divide
\\ Divide
\\ Conquer
\\ Conquer
\\ Combine
worst case running time = O(n2 )
worst case running time=
Quicksort in Practice
quicksort(A, n)
1
2
3
4
5
6
7
if n = 1 return A;
x ← pivot of A;
AL ← elements in A that are less than x;
AR ← elements in A that are greater than x;
BL ← quicksort(AL , AL .size)
BR ← quicksort(AR , AR .size)
return BL concatenating x concatenating BR
Ways to chose pivot:
(a) first element of A
(b) middle element of A
\\ Divide
\\ Divide
\\ Conquer
\\ Conquer
\\ Combine
worst case running time = O(n2 )
worst case running time= O(n2 )
Ways to chose pivot:
(a) first element of A
(b) middle element of A
worst case running time = Θ(n2 )
worst case running time= Θ(n2 )
Any deterministic way of choosing pivot has worst-case
running time Θ(n2 )
Ways to chose pivot:
(a) first element of A
(b) middle element of A
worst case running time = Θ(n2 )
worst case running time= Θ(n2 )
Any deterministic way of choosing pivot has worst-case
running time Θ(n2 )
Idea: using a random pivot
Outline
1
Divide-and-Conquer
2
Solving Recurrences
3
Counting Inversions
4
Polynomial Multiplication
5
Strassen’s Algorithm for Matrix Multiplication
6
Quicksort
Randomized Quicksort
Average-Case Analysis
7
Lower Bound for Comparison-Based Sorting Algorithms
8
Selection Problem
Randomized Algorithm Model
Assumption: we can choose a random integer in
{a, a + 1, · · · , b}
Randomized Algorithm Model
Assumption: we can choose a random integer in
{a, a + 1, · · · , b}
Can computers really produce random numbers?
Randomized Algorithm Model
Assumption: we can choose a random integer in
{a, a + 1, · · · , b}
Can computers really produce random numbers?
No! Computer programs are deterministic!
Randomized Algorithm Model
Assumption: we can choose a random integer in
{a, a + 1, · · · , b}
Can computers really produce random numbers?
No! Computer programs are deterministic!
In practice: use pseudo-random-generator, a deterministic
algorithm returning numbers that “look” random
Randomized Algorithm Model
Assumption: we can choose a random integer in
{a, a + 1, · · · , b}
Can computers really produce random numbers?
No! Computer programs are deterministic!
In practice: use pseudo-random-generator, a deterministic
algorithm returning numbers that “look” random
In theory: assuming the answer is yes.
Randomized Quick-Sort
Quicksort(A, n)
1
2
3
4
5
6
7
if n = 1 return A;
x ← random element of A;
AL ← elements in A that are less than x;
AR ← elements in A that are greater than x;
BL ← Quicksort(AL , AL .size)
BR ← Quicksort(AR , AR .size)
return BL concatenating x concatenating BR
\\ Divide
\\ Divide
\\ Conquer
\\ Conquer
\\ Combine
Randomized Quick-Sort
Quicksort(A, n)
1
2
3
4
5
6
7
if n = 1 return A;
x ← random element of A;
AL ← elements in A that are less than x;
AR ← elements in A that are greater than x;
BL ← Quicksort(AL , AL .size)
BR ← Quicksort(AR , AR .size)
return BL concatenating x concatenating BR
\\ Divide
\\ Divide
\\ Conquer
\\ Conquer
\\ Combine
T (n) = expected running time of sorting n elements
Randomized Quick-Sort
Quicksort(A, n)
1
2
3
4
5
6
7
if n = 1 return A;
x ← random element of A;
AL ← elements in A that are less than x;
AR ← elements in A that are greater than x;
BL ← Quicksort(AL , AL .size)
BR ← Quicksort(AR , AR .size)
return BL concatenating x concatenating BR
\\ Divide
\\ Divide
\\ Conquer
\\ Conquer
\\ Combine
T (n) = expected running time of sorting n elements
P
T (n) = n1 ni=1 T (i − 1) + T (n − i) + O(n)
Randomized Quick-Sort
Quicksort(A, n)
1
2
3
4
5
6
7
if n = 1 return A;
x ← random element of A;
AL ← elements in A that are less than x;
AR ← elements in A that are greater than x;
BL ← Quicksort(AL , AL .size)
BR ← Quicksort(AR , AR .size)
return BL concatenating x concatenating BR
\\ Divide
\\ Divide
\\ Conquer
\\ Conquer
\\ Combine
T (n) = expected running time of sorting n elements
P
T (n) = n1 ni=1 T (i − 1) + T (n − i) + O(n)
T (n) = O(n lg n)
Outline
1
Divide-and-Conquer
2
Solving Recurrences
3
Counting Inversions
4
Polynomial Multiplication
5
Strassen’s Algorithm for Matrix Multiplication
6
Quicksort
Randomized Quicksort
Average-Case Analysis
7
Lower Bound for Comparison-Based Sorting Algorithms
8
Selection Problem
Quicksort: Average-Case Analysis
Deterministic ways to choose pivot:
(a) first element of A
worst-case running time = Θ(n2 )
(b) the middle element of A worst-case running time = Θ(n2 )
Quicksort: Average-Case Analysis
Deterministic ways to choose pivot:
(a) first element of A
worst-case running time = Θ(n2 )
(b) the middle element of A worst-case running time = Θ(n2 )
what if the input array is already randomly perturbed?
i.e, all n! permutations are equally likely to happen
Quicksort: Average-Case Analysis
Deterministic ways to choose pivot:
(a) first element of A
worst-case running time = Θ(n2 )
(b) the middle element of A worst-case running time = Θ(n2 )
what if the input array is already randomly perturbed?
i.e, all n! permutations are equally likely to happen
O(n lg n) expected time: equivalent to choose pivot
randomly
Quicksort: Average-Case Analysis
Deterministic ways to choose pivot:
(a) first element of A
worst-case running time = Θ(n2 )
(b) the middle element of A worst-case running time = Θ(n2 )
what if the input array is already randomly perturbed?
i.e, all n! permutations are equally likely to happen
O(n lg n) expected time: equivalent to choose pivot
randomly
average-case running time of quicksort using (a) or (b) is
O(n lg n)
Practical Issue
Deterministic ways to choose pivot:
(a) first element of A
(b) the middle element of A
average-case running time of quicksort using (a) or (b) is
O(n lg n)
Practical Issue
Deterministic ways to choose pivot:
Bad in Practice
(a) first element of A
Good in Practice
(b) the middle element of A
average-case running time of quicksort using (a) or (b) is
O(n lg n)
Sort “in-Place”
Quick-sort can be implemented as an “in-place” algorithm;
Merge-sort can not
3
8 12 20 32 48
5
7
9 25 29
Sort “in-Place”
Quick-sort can be implemented as an “in-place” algorithm;
Merge-sort can not
3
8 12 20 32 48
5
7
9 25 29
Sort “in-Place”
Quick-sort can be implemented as an “in-place” algorithm;
Merge-sort can not
3
8 12 20 32 48
5
7
3
9 25 29
Sort “in-Place”
Quick-sort can be implemented as an “in-place” algorithm;
Merge-sort can not
3
8 12 20 32 48
5
7
3
9 25 29
Sort “in-Place”
Quick-sort can be implemented as an “in-place” algorithm;
Merge-sort can not
3
8 12 20 32 48
5
7
3
5
9 25 29
Sort “in-Place”
Quick-sort can be implemented as an “in-place” algorithm;
Merge-sort can not
3
8 12 20 32 48
5
7
3
5
9 25 29
Sort “in-Place”
Quick-sort can be implemented as an “in-place” algorithm;
Merge-sort can not
3
8 12 20 32 48
5
7
9 25 29
3
5
7
Sort “in-Place”
Quick-sort can be implemented as an “in-place” algorithm;
Merge-sort can not
3
8 12 20 32 48
5
7
9 25 29
3
5
7
Sort “in-Place”
Quick-sort can be implemented as an “in-place” algorithm;
Merge-sort can not
3
8 12 20 32 48
5
7
9 25 29
3
5
7
8
Sort “in-Place”
Quick-sort can be implemented as an “in-place” algorithm;
Merge-sort can not
3
8 12 20 32 48
5
7
9 25 29
3
5
7
8
Sort “in-Place”
Quick-sort can be implemented as an “in-place” algorithm;
Merge-sort can not
3
8 12 20 32 48
5
7
9 25 29
3
5
7
8
9 12 20 25 29
Sort “in-Place”
Quick-sort can be implemented as an “in-place” algorithm;
Merge-sort can not
3
8 12 20 32 48
5
7
9 25 29
3
5
7
8
9 12 20 25 29 32 48
Sort “in-Place”
29 82 75 64 38 45 94 69 25 76 15 92 37 17 85
Sort “in-Place”
64 82 75 29 38 45 94 69 25 76 15 92 37 17 85
Sort “in-Place”
i
j
64 82 75 29 38 45 94 69 25 76 15 92 37 17 85
Sort “in-Place”
i
j
64 82 75 29 38 45 94 69 25 76 15 92 37 17 85
Sort “in-Place”
i
j
64 82 75 29 38 45 94 69 25 76 15 92 37 17
17
64 85
Sort “in-Place”
i
j
64 82 75 29 38 45 94 69 25 76 15 92 37 17
17
64 85
Sort “in-Place”
i
j
64 82
17
64 75 29 38 45 94 69 25 76 15 92 37 82
64 85
17
Sort “in-Place”
i
j
64 82
17
64 75 29 38 45 94 69 25 76 15 92 37 17
82 85
64
Sort “in-Place”
i
j
64 82
17
64 75 29 38 45 94 69 25 76 15 92 64
37
37 17
82 85
64
Sort “in-Place”
i
j
64 82
17
37 75 29 38 45 94 69 25 76 15 92 64
64
37 17
82 85
64
Sort “in-Place”
i
j
64 82
17
37 64
64
75 29 38 45 94 69 25 76 15 92 75
37 17
64
82 85
64
Sort “in-Place”
i
j
64 82
17
37 64
64
75 29 38 45 94 69 25 76 15 92 64
37 17
75
82 85
64
Sort “in-Place”
i
j
64 82
17
37 15
64
75 29 38 45 94 69 25 76 64
64
15 92 64
37 17
75
82 85
64
Sort “in-Place”
i
j
64 82
17
37 64
64
75 29 38 45 94 69 25 76 64
15
15 92 64
37 17
75
82 85
64
Sort “in-Place”
i
j
64 82
17
37 64
64
75 29 38 45 64
15
94 69 25 76 94
15 92 64
64
37 17
75
82 85
64
Sort “in-Place”
i
j
64 82
17
37 64
64
75 29 38 45 64
15
94 69 25 76 64
15 92 64
94
37 17
75
82 85
64
Sort “in-Place”
i
j
64 82
17
37 64
64
75 29 38 45 25
15
94 69 64
64
25 76 64
15 92 64
94
37 17
75
82 85
64
Sort “in-Place”
i
j
64 82
17
37 64
64
75 29 38 45 64
15
94 69 64
25
25 76 64
15 92 64
94
37 17
75
82 85
64
Sort “in-Place”
i
j
64 82
17
37 64
64
75 29 38 45 64
15
94 64
25
69 69
25 76 64
64
15 92 64
94
37 17
75
82 85
64
Sort “in-Place”
ij
64 82
17
37 64
64
75 29 38 45 64
15
94 64
25
69 69
25 76 64
64
15 92 64
94
37 17
75
82 85
64
Sort “in-Place”
partition(A, `, r)
1
2
3
4
5
6
7
8
9
p ← random integer between ` and r
swap A[p] and A[`]
i ← `, j ← r
while i < j do
while i < j and A[i] ≤ A[j] do j ← j − 1
swap A[i] and A[j]
while i < j and A[i] ≤ A[j] do i ← i + 1
swap A[i] and A[j]
return i
quicksort(A, `, r)
1
2
3
4
if ` ≥ r return
p = Patition(`, r)
quicksort(A, `, p − 1)
quicksort(A, p + 1, r)
Outline
1
Divide-and-Conquer
2
Solving Recurrences
3
Counting Inversions
4
Polynomial Multiplication
5
Strassen’s Algorithm for Matrix Multiplication
6
Quicksort
Randomized Quicksort
Average-Case Analysis
7
Lower Bound for Comparison-Based Sorting Algorithms
8
Selection Problem
Comparison-Based Sorting Algorithms
Q: Can we do better than O(n log n) for sorting?
Comparison-Based Sorting Algorithms
Q: Can we do better than O(n log n) for sorting?
A: No, for comparison-based sorting algorithms.
Comparison-Based Sorting Algorithms
Q: Can we do better than O(n log n) for sorting?
A: No, for comparison-based sorting algorithms.
Bob has one number x in his hand, x ∈ {1, 2, 3, · · · , N }.
Comparison-Based Sorting Algorithms
Q: Can we do better than O(n log n) for sorting?
A: No, for comparison-based sorting algorithms.
Bob has one number x in his hand, x ∈ {1, 2, 3, · · · , N }.
You can ask Bob “yes/no” questions about x.
Comparison-Based Sorting Algorithms
Q: Can we do better than O(n log n) for sorting?
A: No, for comparison-based sorting algorithms.
Bob has one number x in his hand, x ∈ {1, 2, 3, · · · , N }.
You can ask Bob “yes/no” questions about x.
Q: How many questions do you need to ask in order to get the
value of x?
Comparison-Based Sorting Algorithms
Q: Can we do better than O(n log n) for sorting?
A: No, for comparison-based sorting algorithms.
Bob has one number x in his hand, x ∈ {1, 2, 3, · · · , N }.
You can ask Bob “yes/no” questions about x.
Q: How many questions do you need to ask in order to get the
value of x?
A: dlog2 N e.
Comparison-Based Sorting Algorithms
x ≤ 2?
x = 3?
x = 1?
1
2
3
4
Comparison-Based Sorting Algorithms
Q: Can we do better than O(n log n) for sorting?
A: No, for comparison-based sorting algorithms.
Bob has a permutation π over {1, 2, 3, · · · , n} in his hand.
You can ask Bob “yes/no” questions about π.
Comparison-Based Sorting Algorithms
Q: Can we do better than O(n log n) for sorting?
A: No, for comparison-based sorting algorithms.
Bob has a permutation π over {1, 2, 3, · · · , n} in his hand.
You can ask Bob “yes/no” questions about π.
Q: How many questions do you need to ask in order to get the
permutation π?
Comparison-Based Sorting Algorithms
Q: Can we do better than O(n log n) for sorting?
A: No, for comparison-based sorting algorithms.
Bob has a permutation π over {1, 2, 3, · · · , n} in his hand.
You can ask Bob “yes/no” questions about π.
Q: How many questions do you need to ask in order to get the
permutation π?
A: log2 n! = Θ(n lg n)
Comparison-Based Sorting Algorithms
Q: Can we do better than O(n log n) for sorting?
A: No, for comparison-based sorting algorithms.
Bob has a permutation π over {1, 2, 3, · · · , n} in his hand.
You can ask Bob questions of the form “does i appear before
j in π?”
Comparison-Based Sorting Algorithms
Q: Can we do better than O(n log n) for sorting?
A: No, for comparison-based sorting algorithms.
Bob has a permutation π over {1, 2, 3, · · · , n} in his hand.
You can ask Bob questions of the form “does i appear before
j in π?”
Q: How many questions do you need to ask in order to get the
permutation π?
Comparison-Based Sorting Algorithms
Q: Can we do better than O(n log n) for sorting?
A: No, for comparison-based sorting algorithms.
Bob has a permutation π over {1, 2, 3, · · · , n} in his hand.
You can ask Bob questions of the form “does i appear before
j in π?”
Q: How many questions do you need to ask in order to get the
permutation π?
A: log2 n! = Θ(n lg n)
Outline
1
Divide-and-Conquer
2
Solving Recurrences
3
Counting Inversions
4
Polynomial Multiplication
5
Strassen’s Algorithm for Matrix Multiplication
6
Quicksort
Randomized Quicksort
Average-Case Analysis
7
Lower Bound for Comparison-Based Sorting Algorithms
8
Selection Problem
Selection Problem
Input: a set A of n (distinct) numbers, and 1 ≤ i ≤ n
Output: the i-th smallest number in A
Selection Problem
Input: a set A of n (distinct) numbers, and 1 ≤ i ≤ n
Output: the i-th smallest number in A
Sorting solves the problem in time O(n lg n).
Selection Problem
Input: a set A of n (distinct) numbers, and 1 ≤ i ≤ n
Output: the i-th smallest number in A
Sorting solves the problem in time O(n lg n).
Our goal: O(n) running time
Recall: Randomized Quick-Sort
quicksort(A, n)
1
2
3
4
5
6
7
if n = 1 return A;
x ← random element of A;
AL ← elements in A that are less than x;
AR ← elements in A that are greater than x;
BL ← Quicksort(AL , AL .size)
BR ← Quicksort(AR , AR .size)
return BL concatenating x concatenating BR
\\ Divide
\\ Divide
\\ Conquer
\\ Conquer
\\ Combine
Randomized Algorithm for Selection
select(A, n, i)
1
2
3
4
5
6
7
8
9
10
if n = 1 return A[1];
x ← random element of A;
AL ← elements in A that are less than x;
AR ← elements in A that are greater than x;
if i ≤ AL .size then
return select(AL , AL .size, i)
else if i = AL .size + 1 then
return x
else
return select(AR , AR .size, i − AL .size − 1)
Deterministic Algorithm for Selection
Deterministic Algorithm for Selection
Deterministic Algorithm for Selection
Deterministic Algorithm for Selection
select(A, n, i)
1
2
3
4
5
6
7
8
9
10
11
12
if n is small enough, run the simple algorithm and return
partition A into groups of size 3
for each group, pick the median of the group
call select to find the median x of the dn/3e picked medians
AL = numbers in A that are less than x
AR = numbers in A that are greater than x
if i ≤ AL .size then
return select(AL , AL .size, i)
else if i = AL .size + 1 then
return x
else
return select(AR , AR .size, i − AL .size − 1)
Running Time
T (n) ≤ T (n/3) + T (2n/3) + O(n)
Running Time
T (n) ≤ T (n/3) + T (2n/3) + O(n)
T (n) = O(n lg n) (not good!)
Running Time
T (n) ≤ T (n/3) + T (2n/3) + O(n)
T (n) = O(n lg n) (not good!)
Groups of size 5 instead of 3
Running Time
T (n) ≤ T (n/3) + T (2n/3) + O(n)
T (n) = O(n lg n) (not good!)
Groups of size 5 instead of 3
T (n) ≤ T (n/5) + T (7n/10) + O(n)
Running Time
T (n) ≤ T (n/3) + T (2n/3) + O(n)
T (n) = O(n lg n) (not good!)
Groups of size 5 instead of 3
T (n) ≤ T (n/5) + T (7n/10) + O(n)
T (n) = O(n) (good!)