Chapter 6

Recurrence and Solution

6.2Recurrence Relation

6.3Solve Homogeneous Recurrence

6.4Solve Nonhomogeneous Recurrence

1 2 3

1 2 3

1 2 3

6.2Recurrence Relation

1 2 3

1 2 3

1 2 3

1 2

( ) 2 ( 1) 1

2[ 2 ( 2) 1] 1

2 2 1

2 1

n n

n

T n T n

T n

Example 2 Region partition for n lines.

( ) ( 1) , 0R n R n n n

( ) ( 2) 1

(0) 1 2 1

( 1) 1

2

R n R n n n

R n n

n n

6.3Solve Homogeneous Recurrence

Try . Is that O.K.?

Yes!

0 1 1 2( ) ( )n nnF C r C r

1 1 2 20 1 1 2 0 1 1 2 0 1 1 2

1 2 1 20 1 1 1 1 2 2 2

2 2 2 20 1 1 1 1 2 2 2

( ) ( ) [ ( ) ( ) ] [ ( ) ( ) ]

[ ] [ ]

[ 1] [ 1]

0

n n n n n n

n n n n n n

n n

C r C r C r C r C r C r

C r r r C r r r

C r r r C r r r

Definition:

Fibonacci Sequence: 0,1,1,2,3,5,8,…

Fibonacci Sequence: 021 nnn FFF

Characteristic Equation:

Using boundaries: and

012 rr

2

511

r

2

512

r

0 11 5 1 5

2 2

n n

nF C C

00 F 11 F

We have 0100 CCF

1 0 11 5 1 5 1

2 2F C C

5

10 C

5

11

C

1 11 5 1 55 52 2

n n

nF

e.g. 3.3 (NTU) Solve with and1 26 9n n nT T T 0 5T 1 6T

1 2 3r r

nnn nCCT 33 10

0 0

1 1 1

5

5 3 1 3 15 3 6

T C

T C C

nnn nT 3335

6.4Solve Nonhomogeneous Recurrence

Example 1 Solve with and 1 22 4n n nT T T 0 2T 1 3T

( ) ( )p h

n n nT T T ( )pnT C

)()( hn

pnn TTT

( ) ( ) ( ) ( )1 1 2 22 4h p h p

n n n nT T T T

Uisng

We have

Let , we have

It yields , and then we have

( )( )1hh

n nT T ( )22 0h

nT

( ) ( )( )1 22 4p pp

n n nT T T

CT pn )( 42 CCC

2C

0 1 2 ( 1) 2

p hn n n

n n

T T T

C C

Finally, we have

0 0 1 2 2T C C

1 0 12 2 3T C C

From

3 2 ( 1) 2n nnT