4.Recurrences

Hsu, Lih-Hsing

Chapter 4 P.2

Computer Theory Lab.

Recurrences -- Substitution method Recursion-tree method Master method

T n aT n b f n( ) ( / ) ( )

Chapter 4 P.3

Computer Theory Lab.

Technicalities We neglect certain technical

details when we state and solve recurrences. A good example of a detail that is often glossed over is the assumption of integer arguments to functions. Boundary conditions is ignored. Omit floors, ceilings.

Chapter 4 P.4

Computer Theory Lab.

4.1 The substitution method : Mathematical induction

The substitution method for solving recurrence entails two steps:

1. Guess the form of the solution. 2. Use mathematical induction to find

the constants and show that the solution works.

Chapter 4 P.5

Computer Theory Lab.

Example

(We may omit the initial condition later.)

Guess  Assume

T n T n n

T

( ) ( / )

( )

2 2

1 1

)log()( nnOnT

T n c n n( / ) / log /2 2 2

Chapter 4 P.6

Computer Theory Lab.

Initial condition However,

T n c n n n cnn

n

cn n cn n cn n c .

( ) / log / ) log

log log log )

2( 2 22

2 1 (if

)(01log)1(1 cnT

4 2 2 4 T cn c( ) log ( )if

Chapter 4 P.7

Computer Theory Lab.

Making a good guess

We guess

Making guess provides loose upper bound and lower bound. Then improve the gap.

T n T n n( ) ( / ) 2 2 17

T n O n n( ) ( log )

Chapter 4 P.8

Computer Theory Lab.

Subtleties

Guess Assume

 However, assume

T n T n T n( ) ( / ) ( / ) 2 2 1

T n O n( ) ( )

T n cn( )

T n c n c n cn cn( ) / / 2 2 1 1

T n cn b( )

T n c n b c n b

cn b cn b b

( ) ( / ) ( / )

( )

2 2 1

2 1 1Choose

Chapter 4 P.9

Computer Theory Lab.

Show that the solution to T(n) = 2T(n2 + 17) + n is O(n lg n)

Solution:

assume a > 0, b > 0, c > 0 and T(n) ≦ an lg n – blg n - c

T(n) ≦ 2[(n2 + 17)lg(

n2 +17) - blg(

n2 + 17)-c]+n

≦ (an + 34a)lg(n2 +17) - 2blg(

n2 +17) - 2c + n

≦ anlg(n2 +17) + anlg21/a + (34a-2b)lg(

n2 +17) - 2c

≦ anlg(n) 21/a+ (34a-2b)lg(n) - 2c

Chapter 4 P.10

Computer Theory Lab.

anlg(n) 21/a+ (34a-2b)lg(n) - 2c

n≧ n2 +17，n≧ 34

n≧ (n2 +17) 21/a ∵， 21/2≦ 1.5∴ n ≧ 12

34a-2b≦ -b，b ≧ 34a

c > 0，-c > -2c

T(n) ≦ anlgn - blgn - c，T(n) ≦ anlgn

T(n) = O(nlgn)

Chapter 4 P.11

Computer Theory Lab.

Avoiding pitfalls

Assume Hence  

(WRONG!) You cannot find such a c.

T n T n n

T

( ) ( / )

( )

2 2

1 1

T n O n( ) ( )

T n cn( )

constant) a is Since(

)()2/(2)(

c

nOncnnncnT

Chapter 4 P.12

Computer Theory Lab.

Changing variables T n T n n( ) ( ) lg 2

Let m nlg .

T T mm m( ) ( )/2 2 2 2

Then S m S m m( ) ( / ) 2 2 .

S m O m m

T n T S m O m m

O n n

m

( ) ( lg )

( ) ( ) ( ) ( lg )

(lg lg lg )

2

Chapter 4 P.13

Computer Theory Lab.

4.2 the Recursion-tree method

)()4/(3)( 2nnTnT

Chapter 4 P.14

Computer Theory Lab.

Chapter 4 P.15

Computer Theory Lab.

).(1)16/3(

1)16/3(

16

3

)(16

3...

16

3

16

3)(

3log2log

1log

0

3log2

3log21log

22

22

4

4

4

4

4

4

ncn

ncn

ncncncncnnT

n

n

i

i

n

The cost of the entire tree

Chapter 4 P.16

Computer Theory Lab.

)(

)(13

16

)16/3(1

1

16

3

16

3)(

2

3log2

3log2

3log2

0

1log

0

3log2

4

4

4

4

4

nO

ncn

ncn

ncn

ncnnT

i

i

n

i

i

Chapter 4 P.17

Computer Theory Lab.

substitution methodWe want to Show that T(n) ≤ dn2 for some consta

nt d > 0. using the same constant c > 0 as before, we have

Where the last step holds as long as d (16/13)c.

,

16

3

)4/(3

4/3

)4/(3)(

2

22

22

22

2

dn　 　 　

cndn　 　 　

cnnd　 　 　

cnnd　 　 　

cnnT　nT

Chapter 4 P.18

Computer Theory Lab.

cnnTnTnT )3/2()3/()(

Chapter 4 P.19

Computer Theory Lab.

substitution method

,lg

)3/23(lglg

2lg)3/2(3lg)3/2(3lg)3/((lg

))2/3lg()3/2(3lg)3/((lg

))2/3lg()3/2(lg)3/2(()3lg)3/(lg)3/((

)3/2lg()3/2()3/lg()3/(

)3/2()3/()(

ndn　 　 　

cndnndn　 　 　

cnnnndndn　 　 　

cnnndndn　 　 　

cnndnndndnnd　 　 　

cnnndnnd　 　 　

cnnTnTnT

As long as d c/lg3 – (2/3)).

Chapter 4 P.20

Computer Theory Lab.

4.3 The master method Theorem 4.1 (((MMMaaasssttteeerrr ttthhheeeooorrreeemmm)))

Let a 1 and b 1 be constants, let f n( ) be a function, and T n( )

be defined on the nonnegative integers by the recurrence T n aT n b f n( ) ( / ) ( )

where we interpret n b/ mean either n b/ or n b/ .

1. If f n O n b a( ) ( )log for some constant 0 , then T n n b a( ) ( )log .

2. If f n O n b a( ) ( )log then T n n nb a( ) ( log )log .

3. If f n O n b a( ) ( )log for some constant 0 and if af n b cf n( / ) ( )

for some constant c 1 and all sufficiently large　n, then T n f n( ) ( ( )) .

Proof. (In section 4.4 by recursive tree)

Chapter 4 P.21

Computer Theory Lab.

T n T n n( ) ( / ) 9 3

a b f n n

n n f n O n

9 3

3 39 2 9 1

, , ( )

, ( ) ( )log log

Case T n n1 2 ( ) ( )

T n T n( ) ( / ) 2 3 1

a b f n

n n f n

1, 3 2 1

13 21 0

/ , ( )

( ),log /

Case T n n2 ( ) (log )

Chapter 4 P.22

Computer Theory Lab.

T n T n n n( ) ( / ) log 3 4

a b f n n n

n n f n O n

3 4

4 43 0 793 3

, , ( ) log

, ( ) ( )log . log

Case

af(n/b)= (n) (

n)

nn=cf(n)

c= n

T n n n

3

34 4

34

34

Check

for and sufficiently large

log log

,

( ) ( log )

Chapter 4 P.23

Computer Theory Lab.

The master method does not apply to the recurrence

even though it has the proper form: a = 2, b=2, f(n)= n lgn, and It might seem that case 3 should apply, since f(n)= n lgn is asymptotically larger than

The problem is that it is not polynomially larger.

,lg)2/(2)( nnnTnT

.log nn ab

.log nn ab