CS 321. Algorithm Analysis & Design Lecture 8

Divide and ConquerPart III - Multiplication and Selection

T (n) aT (n/b) +O(nd)

T (n) = O(ndlog n)

T (n) = O(nd)

T (n) = O(nlogb a)

if a = bd

if a < bd

if a > bd

Multiplying two n-bit integers

1 1 0 1

1 0 1 1

1 1 0 1

1 0 1 1

1 1 0 1

1 0 1 1

1 1 0 1

1 1 0 1

1 0 1 1

1 1 0 1

1 1 0 1

1 0 1 1

1 1 0 1

1 1 0 1

1 1 0 1

1 0 1 1

1 1 0 1

1 1 0 1

0 0 0 0

1 1 0 1

1 0 0 0 1 1 1 1

1 1 0 1

1 0 1 1

1 1 0 1

1 1 0 1

0 0 0 0

1 1 0 1

1 0 0 0 1 1 1 1

O(n2) operations.

x

y

x

y

xy = (2(n/2)xL + xR)(2(n/2)yL + yR)

xy = 2nxLyL + 2(n/2)(xLyR +yLxR) + xRyR

x

y

xL xR

yL yR

xy = (2(n/2)xL + xR)(2(n/2)yL + yR)

xy = 2nxLyL + 2(n/2)(xLyR +yLxR) + xRyR

x

y

xL xR

yL yR

xy = 2nxLyL + 2(n/2)(xLyR +yLxR) + xRyR

x

y

xL xR

yL yR

T(n) = 4T(n/2) + O(n)

xy = 2nxLyL + 2(n/2)(xLyR +yLxR) + xRyR

x

y

xL xR

yL yR

T(n) = 4T(n/2) + O(n)

xy = 2nxLyL + 2(n/2)(xLyR +yLxR) + xRyR

bd = 2 and a = 4

xy = 2nxLyL + 2(n/2)(xLyR +yLxR) + xRyR

xy = 2nxLyL + 2(n/2)(xLyR +yLxR) + xRyR

(xL + xR)(yL + yR) = (xLyL + xLyR + yLxR + xRyR)

xy = 2nxLyL + 2(n/2)(xLyR +yLxR) + xRyR

(xL + xR)(yL + yR) = (xLyL + xLyR + yLxR + xRyR)

xy = 2nxLyL + 2(n/2)(xLyR +yLxR) + xRyR

(xL + xR)(yL + yR) = (xLyL + xLyR + yLxR + xRyR)

xy = 2nxLyL + 2(n/2)(xLyR +yLxR) + xRyR

(xL + xR)(yL + yR) = (xLyL + xLyR + yLxR + xRyR)

(xL + xR)(yL + yR) - (xLyL + xRyR)

xy = 2nxLyL + 2(n/2)(xLyR +yLxR) + xRyR

(xL + xR)(yL + yR) = (xLyL + xLyR + yLxR + xRyR)

(xL + xR)(yL + yR) - (xLyL + xRyR)

T(n) = 3T(n/2) + O(n)

xy = 2nxLyL + 2(n/2)(xLyR +yLxR) + xRyR

(xL + xR)(yL + yR) = (xLyL + xLyR + yLxR + xRyR)

(xL + xR)(yL + yR) - (xLyL + xRyR)

T(n) = 3T(n/2) + O(n)

≤ n(log3) = n1.59

Finding the kth smallest element

Pick a “pivot”.

Pick a “pivot”.

p elements on the left and (n-p) on the right.

p elements on the left and (n-p) on the right.

if p is larger than k:

p elements on the left and (n-p) on the right.

if p is larger than k:

p elements on the left and (n-p) on the right.

p elements on the left and (n-p) on the right.

if p is smaller than k:

p elements on the left and (n-p) on the right.

if p is smaller than k:

p elements on the left and (n-p) on the right.

if p is smaller than k:

(k-p)th smallest element

Group the n elements into (n/5) buckets of size five each.

Find the median in each bucket.

Identify the Median of these Medians (MOM).

Use the MOM as the pivot.

Use the MOM as the pivot.

4 4 2 1 2

5

1 2 2 4 7

9

6 4 5 5 9

4

7 4 5 9 7

3 8

4 4 2 1 2

5

1 2 2 4 7

9

6 4 5 5 9

4

7 4 5 9 7

3 8

2

4 4 2 1 2

5

1 2 2 4 7

9

6 4 5 5 9

4

7 4 5 9 7

3 8

2

2

4 4 2 1 2

5

1 2 2 4 7

9

6 4 5 5 9

4

7 4 5 9 7

3 8

2

2

5

4 4 2 1 2

5

1 2 2 4 7

9

6 4 5 5 9

4

7 4 5 9 7

3 8

2

2

5

7

4 4 2 1 2

5

1 2 2 4 7

9

6 4 5 5 9

4

7 4 5 9 7

3 8

2

2

5

7

5

4 4 2 1 2

5

1 2 2 4 7

9

6 4 5 5 9

4

7 4 5 9 7

3 8

2

2

5

7

5

4 4 2 1 2

5

1 2 2 4 7

9

6 4 5 5 9

4

7 4 5 9 7

3 8

2

2

5

7

5

5

4 4 2 1 2

5

1 2 2 4 7

9

6 4 5 5 9

4

7 4 5 9 7

3 8

2

2

5

7

5

5

At least 3n/10 elements are smaller than the pivot, and at least 3n/10 elements are larger.

At least 3n/10 elements are smaller than the pivot, and at least 3n/10 elements are larger.

Dealing with at most 7n/10 elements in the next level of recursion.

At least 3n/10 elements are smaller than the pivot, and at least 3n/10 elements are larger.

Dealing with at most 7n/10 elements in the next level of recursion.

T(n) ≤ T(n/5) + T(7n/10) + O(n)

At least 3n/10 elements are smaller than the pivot, and at least 3n/10 elements are larger.

Dealing with at most 7n/10 elements in the next level of recursion.

T(n) ≤ T(n/5) + T(7n/10) + O(n)

Exercise: T(n) = O(n)