chapter 6 recurrence and solution. 6.2 recurrence relation 6.3 solve homogeneous recurrence 6.4...
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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