The master theorem method
Chapter 4. Recurrences
한양대학교 정보보호 및 알고리즘 연구실
2008. 2. 26
이재준
담당교수님 : 박희진 교수님
1
2. Recurrence solution methods
• Master method
- Definition
determining asymptotic tight bounds
for many simple recurrences of the form
T (n)  aT (n / b)  f (n)
T (n)
( a ≥ 1, b ≥ 1, and f(n) is given function )
f(n)
a
T (n / b) ……. T (n / b) ……. T (n / b)
2
2. Recurrence solution methods
• Master theorem Intuition
log a
Comparing the function f (n) with the function n b
Case 1 :
Case 2 :
Case 3 :
f (n)  n logb a
f (n)  nlogb a
f (n)  nlogb a
,then the solution is
,then the solution is
,then the solution is
T (n)  (n logb a )
T (n)  (nlogb a lg n)
T (n)  ( f (n))
3
2. Recurrence solution methods
• Master theorem Technicalities
Case 1 : f (n)  n
logb a
, and f(n) must be polynomially smaller than
n logb a
Case 2 : f (n)  n logb a
Case 3 : f (n)  n
logb a
, and f(n) must be polynomially lager than
polynomially smaller
f (n)  o(n logb a )
≈ f ( n) n   n log b a
f (n)  n logb a n 
f ( n)


n
n logb a
f (n)  o(n logb a  )
n logb a
polynomially lager
f (n)   (n logb a )
≈ f ( n)  n log b a n 
f ( n)


n
n logb a
f (n)   (n logb a  )
4
2. Recurrence solution methods
• Master theorem Exceptionality
Gap between cases 1 and 2 : f(n) is asymptotically smaller but, not polynomially smaller
f ( n)  n
logb a
, but not
f ( n)


n
n logb a
( for some positive constant ε )
Gap between cases 2 and 3 : f(n) is asymptotically larger but, not polynomially larger
f ( n)  n
logb a
, but not
f ( n)


n
n logb a
( for some positive constant ε )
or regularity condition fail to hold
5
2. Recurrence solution methods
• Master theorem Example
The recurrence is
T (n) = 2T (n/2) + n lg n
Use the master theorem to making to conclude the solution.
6
2. Recurrence solution methods
• Master theorem Solution to the example
T (n) = 2T (n/2) + n lg n ( a=2, b=2, f(n) = n lg n )
f ( n)

Since f (n)  n lg n
, but not log a  n
n b
a
 n logb
n
f (n) n lg n
 log2 2
logb a
n
n
 lg n
 n
( for any positive constant ε , and some
)
f(n) is asymptotically larger ,but not polynomially larger
Thus, the recurrence falls into the gap between case 2 and 3
7
2. Recurrence solution methods
• Master theorem T (n)  aT (n / b)  f (n)
1. If f (n)  O(nlog a  ), for some constant ε > 0, then T (n)  (nlog a )
b
b
2. If f (n)  (nlog a ), then T (n)  (nlog a lg. n)
b
b
3. If f (n)  (nlog a  ) for some constant ε > 0,
and if af (n/b) ≤ cf (n) for some constant c < 1 and all
sufficiently large n, then T (n)  ( f (.n))
b
Q : Why o/ω-notation change to O/Ω-notation bound ?
8
2. Recurrence solution methods
• Master theorem Answer to the question
T (n) = 2T (n/2) + n lg n
Since f (n)  n lg n , but not
n
n
logba
f ( n)


n
n logb a
f (n) n lg n
 log2 2
logb a
n
n
 lg n
 n
( for any positive constant ε )
To prevent something happen like above, we use O-notation and Ω-notation
for define master theorem.
9
2. Recurrence solution methods
• Master theorem Example
The recurrence is
T (n) = 3T (n/4) + n lg n
Use the master theorem to making to conclude the solution.
10
2. Recurrence solution methods
• Master theorem Solution to the example
T (n) = 3T (n/4) + n lg n ( a=3, b=4, f(n) = n lg n )
Since f (n)  n lg n
n
logba
n
log34
 O(n 0.793 )
→ n lg n  O(n0.793 )
f (n)  n
logba
, and
f ( n)


n
n logb a
f (n) n lg n
 0.793
logb a
n
n
 n 0.207 lg n
 n
( where ε=0.2 )
So, f(n) is asymptotically larger and polynomially larger than n
logba
11
2. Recurrence solution methods
• Master theorem Solution to the example
T (n) = 3T (n/4) + n lg n ( a=3, b=4, f(n) = n lg n )
Regularity condition
af(n/b)≤ cf(n)
a f(n/b)=3(n/4)lg(n/4)
≤ cf(n)
=(3/4)n lgn ( for c=3/4 )
Thus, the solution is T (n)  ( f (n))  (n lg n)
12
2. Recurrence solution methods
• Master theorem Proof of the master theorem
PART1. Analyzes with assumption that T(n) is
defined only on exact powers of b>1
PART2. Shows how the analysis can be
extended to all positive integers n.
13
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Abuse asymptotic notation when proof for exact powers
T(n)=
n
n2
if n = 1,2,4,8
otherwise
→ T(n)=O( n 2 )
Making it absolutely clear from the context that we are doing so
When we use asymptotic notation over a limited domain
14
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma for proof
Lemma 1. Make an expression that contains a summation
Lemma 2. Determine asymptotic bounds on this summation
Lemma 3. Use Lemma1 and Lemma2 prove master theorem (n is exact power of b)
15
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 1. Make an expression that contains a summation
Define T(n) on exact power of b by the recurrence
T(n)=
{
(1)
if n = 1,
aT (n / b)  f (n) if n = b
i
( a≥1 and b≥1, f(n) defined on exact power of b )
Where i is a positive integer. Then,
T ( n )  ( n
logba
)
logbn1
a
j
f (n / b j )
j 0
16
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 1. Make an expression that contains a summation Proof
f(n)
a
f ( n / b)
a
logbn
… f ( n / b) →
a
a
…
…
f ( n / b)
…
af (n / b)
logbn 1
f (n / b 2 ) f (n / b 2 ) f (n / b 2 ) f (n / b 2 ) f (n / b 2 ) f (n / b 2 ) f (n / b 2 ) f (n / b 2 ) f (n / b 2 ) →
…
… … … … … …
(1) (1) (1) (1)(1) (1) (1) (1)
n
logb a
…
… …→
(1) (1) (1) (1)
a 2 f (n / b 2 )
→
j
j
a
f
(
n
/
b
)

j 0
a j f (n / b j )
→ (a logb )
T ( n )  ( n
n
logba
)
logbn1
a
j 0
j
f (n / b j )
17
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 2. Determine asymptotic bounds on this summation
g ( n) 
logbn 1
a
j
f (n / b j )
j 0
Case 1. If f (n)  O(n
logb a 
) , for some constant ε > 0, then g (n)  O(n logb a )
log a
Case 2. If f (n)  (n b ) , then g (n)  (n
logb a
lg n)
Case 3. If , af (n/b) ≤ cf (n) for some constant c < 1 and for all n≥b, then g (n)  ( f (n))
18
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 2. Determine asymptotic bounds on this summation Proof for Case1
g ( n) 
logbn 1
a
j
f (n / b j )
j 0
Case 1. If f (n)  O(n
Assumed fact :
Substitution :
logb a 
) , for some constant ε > 0, then g (n)  O(n logb a )
f (n)  O(n logb a  ) → f (n / b j )  O(( n / b j )logb a  )
g ( n)  O (
logbn 1
a
j
j logba 
(n / b )
)
j 0
Let‘s factoring out terms and simplifying the summation which within the O-notation.
19
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 2. Determine asymptotic bounds on this summation Proof for Case1
Substitute into g(n) : g ( n)  O (
logbn 1

j
j logba 
logbn 1
ab 
a (n / b )
)
j 0
→
logbn 1

j
j
a (n / b )
logba

n
logba


j 0
(
j 0
n
logba

a
b logb
)j
logbn 1
 (b )
j
j 0
 logbn
1
)

b 1

logba  n  1
n
( 
)
b 1
 n logb  (
a
b
20
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 2. Determine asymptotic bounds on this summation Proof for Case1
Substitute into g(n) : g ( n)  O (
logbn 1

j
j logba 
a (n / b )
)
j 0
g ( n)  O ( n
 O(n
logba 
 O(n
Case 1. If f (n)  O(n
logb a 
logba 
logba
n  1
( 
))
b 1
n )
) (Now we proved case1)
) , for some constant ε > 0, then g (n)  O(n logb a )
21
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 2. Determine asymptotic bounds on this summation Proof for Case2
g ( n) 
logbn 1
a
j
f (n / b j )
j 0
log a
log a
Case 2. If f (n)  (n b ) , then g (n)  (n b lg n)
Assumed fact :
f (n)  (n
Substitute into g(n) :
logb a
g ( n )  (
j logba
) → f (n / b )  ((n / b )
j
logbn 1
a
j
j logba
(n / b )
)
)
j 0
22
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 2. Determine asymptotic bounds on this summation Proof for Case2
Substitute into g(n) : g (n)  (
logbn 1

j logba
j
a (n / b )
)
j 0
logbn 1
→

a j (n / b j )
logba
n
logba
logbn 1

j 0
j 0
n
logba
a
(
b
logba
)j
logbn 1
 (1)
j 0
a
 n logb log bn
23
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 2. Determine asymptotic bounds on this summation Proof for Case2
Substitute into g(n) : g (n)  (
logbn 1

j logba
j
a (n / b )
)
j 0
→
logbn 1

j
j logba
a (n / b )
n
j 0
logba
log bn
a
logb
log bn )
→ g ( n )  ( n
a
 (n logb lg n) (Now we proved case2)
log a
log a
Case 2. If f (n)  (n b ) , then g (n)  (n b lg n)
24
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 2. Determine asymptotic bounds on this summation Proof for Case3
g ( n) 
logbn 1
j
j
a
f
(
n
/
b
)

j 0
Case 3. If , af (n/b) ≤ cf (n) for some constant c < 1 and for all n≥b, then g (n)  ( f (n))
25
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 2. Determine asymptotic bounds on this summation Proof for Case3
g ( n) 
logbn 1
a
j
f (n / b j )
j 0
→ g (n)  f (n)  af (n / b)  ...  a
logbn 1
f (n / b
logbn 1
)
 f (n)
g(n) include f(n) when j=0
it means g(n) ≥ f(n)
so, we can conclude g(n) =Ω(f(n))
26
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 2. Determine asymptotic bounds on this summation Proof for Case3
g ( n) 
logbn 1
a
f (n / b j )
a
f ( n / b)  f ( n)
c
substitute
2
a
f ( n / b 2 )  f ( n)
2
c
substitute
3
a
f ( n / b 3 )  f (n)
3
c
substitute
a4
4
f ( n / b )  f ( n)
4
c
→
j 0
j
→
→
a
n
n
f( 2) f( )
c b
b
a
n
n
f ( 3) f ( 2)
c b
b
a
n
n
f( ) f( 4)
c b3
b
→…→
aj
j
j
j
j
f
(
n
/
b
)

f
(
n
)
a
f
(
n
/
b
)

c
f ( n)
→
j
c
27
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 2. Determine asymptotic bounds on this summation Proof for Case3
Case 3. If , af (n/b) ≤ cf (n) for some constant c < 1 and for all n≥b, then g (n)  ( f (n))
g ( n) 
logbn 1
a
j
f (n / b j )
j 0

logbn 1
c
j
f (n) ( because a j f (n / b j )  c j f (n) )
j 0

 f ( n) c j
j 0
1
)
1 c
 O ( f ( n)) ( Since c is constant )
 f (n)(
28
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 2. Determine asymptotic bounds on this summation Proof for Case3
g(n) =Ω(f(n)) and g(n)=O(f(n))
so g(n)= θ(g(n)) ( for extra powers of b ) (Now we proved case3)
Case 3. If , af (n/b) ≤ cf (n) for some constant c < 1 and for all n≥b, then g (n)  ( f (n))
29
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 3. Prove master theorem (n is exact power of b)
T(n)=
{
(1)
if n = 1,
aT (n / b)  f (n) if n = b ( a≥1 and b≥1, f(n) defined on exact power of b )
i
1. If f (n)  O(nlog a  ), for some constant ε > 0, then T (n)  (nlog a )
b
b
2. If f (n)  (nlog a ), then T (n)  (nlog a lg. n)
b
b
3. If f (n)  (nlog a  ) for some constant ε > 0,
and if af (n/b) ≤ cf (n) for some constant c < 1 and all
sufficiently large n, then T (n)  ( f (n)) .
b
30
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 3. Prove master theorem (n is exact power of b) Proof for Case1
(1)
if n = 1,
T(n)=
i
aT (n / b)  f (n) if n = b ( a≥1 and b≥1, f(n) defined on exact power of b )
{
Lemma1)
T ( n )  ( n
logba
)
logbn1
a
j
f (n / b j )
j 0
T ( n )  ( n
logba
)
logbn1
a
j
f (n / b j ) ( From Lemma1 )
j 0
31
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 3. Prove master theorem (n is exact power of b) Proof for Case1
T ( n )  ( n
logba
)
logbn1
a
j
f (n / b j )
j 0
log a 
Case1. If f (n)  O(n b ) , for some constant ε > 0, thenT (n)  (n
logb a
)
Lemma2) case1
Case 1. If f (n)  O(n
logb a 
) , for some constant ε > 0, then g (n)  O(n logb a )
T (n)  (nlogb a )  O(nlogb a )
( From case1 in lemma2 )
32
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 3. Prove master theorem (n is exact power of b) Proof for Case2
T ( n )  ( n
logba
)
logbn1
a
j
f (n / b j )
j 0
log a
log a
Case 2. If f (n)  (n b ) , then T (n)  (n b lg n)
Lemma2) case2
Case 2. If f (n)  (n logb a ) , then g (n)  (n
logb a
lg n)
T (n)  (n logb a )  (n logb a lg n)
 (n logb a lg n) ( From case2 in lemma2 )
33
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 3. Prove master theorem (n is exact power of b) Proof for Case3
T ( n )  ( n
logba
)
logbn1
a
j
f (n / b j )
j 0
Case3. If f (n)  (nlogb a  ) for some constant ε > 0,
and if af (n/b) ≤ cf (n) for some constant c < 1
and all sufficiently large n, then T (n)  ( f (n)) .
Lemma2) case3
Case 3. If , af (n/b) ≤ cf (n) for some constant c < 1 and for all n≥b, then g (n)  ( f (n))
T (n)  (n logb a )  ( f (n)) ( From case3 in lemma2 )
 ( f (n)) ( Because f (n)  (nlog a  ) )
b
34
2. Recurrence solution methods
• Proof of the master theorem n is exact power of b
Lemma 3. Prove master theorem (n is exact power of b)
T(n)=
{
(1)
if n = 1,
aT (n / b)  f (n) if n = b ( a≥1 and b≥1, f(n) defined on exact power of b )
i
log a 
Case1. If f (n)  O(n b ) , for some constant ε > 0, thenT (n)  (n
logb a
)
log a
log a
Case 2. If f (n)  (n b ) , then T (n)  (n b lg n)
Case3. If f (n)  (nlogb a  ) for some constant ε > 0,
and if af (n/b) ≤ cf (n) for some constant c < 1
and all sufficiently large n, then T (n)  ( f (n)) .
( Now we proved master theorem when n is exact power of b )
35
2. Recurrence solution methods
• Proof of the master theorem n is all integers
The recurrence defined for all integers
Upper bounding )
T (n)  aT ( n / b)  f (n)
Lower bounding )
T (n)  aT ( n / b)  f (n)
36
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n)
f(n)
a
f (n1 )
a
logbn
f (n)
a
…
af (n / b)
… f (n1 )
f (n1 )
a
…
…
f (n2 ) f (n2 ) f (n2 ) f (n2 ) f (n2 ) f (n2 ) f (n2 ) f (n2 ) f (n2 )
a 2 f ( n / b  / b,)
.....…
… … … … … …… … …
…
(1) (1) (1) (1)(1) (1) (1) (1)
(n
logb a
)
(1) (1) (1) (1)
(nlogb a )
T ( n )  ( n
logba
logbn1 
)   a j f (n / b j )
j 0
37
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n)
n
n / b ,
n / b  / b,,
n / b  / b, / b ,
nj =
n
{ n
if j=0
j 1/ b

if j>0
………
constant
Goal : determine the depth k
such that nk is a constant
38
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n)
Goal : determine the depth k such that nk is a constant
nj =
n0  n,
n
{ n
if j=0
j 1/ b

if j>0
n
n1  n0 / b    1,
b
n  n  1
n n
n2  n1 / b   2   1,
b b
n n n
n3  n2 / b   3  2   1,
b b b
....
39
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n)
Goal : determine the depth k such that nk is a constant
n0  n,
n
n1  n0 / b    1,
b
n n
n2  n1 / b   2   1,
b
b
n
n n
n3  n2 / b   3  2   1,
b
b
b
n j 1 1
nj  j   i
b
i 0 b

n
1
 j  i
b
i 0 b
n
1
 j 
b
b 1
....
n j 1 1
n j  n j 1 / b  j   i
b
i 0 b
40
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n)
Goal : determine the depth k such that nk is a constant
n
1
nj  j 
b
b 1
( Letting j  log
Thus, at depth
n
b
)
1
n
b logb  b  1
n
1
 logn1 
b  1 ( Because,
b b
n
1


n / b b 1
1
b
b 1
 O (1)
nlogn  
b
n

A  A  1 )
log bn  , the problem size is at most a constant
41
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n)
T ( n )  ( n
logba
)
logbn 1
Except that n is an arbitrary integer
j
j
a
f
(
n
/
b
)

and to be all integer
j 0
logbn 1
g ( n)   a j f ( n / b j )
j 0
Case 1. If f (n j )  O(nlogb a ), for some constant ε > 0, then g (n)  O(n logb a )
log a
Case 2. If f (n)  (n b ) , then g (n)  (n
logb a
lg n)
Case 3. If af ( n / b)  cf (n) , for some constant c < 1
and for all n≥b+b/(b-1), then g (n)  ( f (n))
42
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n) Proof for Case3
logbn 1
g ( n)   a j f ( n / b j )
j 0
Case 3. If af ( n / b)  cf (n) , for some constant c < 1
and for all n≥b, then g (n)  ( f (n))
43
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n) Proof for Case3
logbn 1
g ( n) 
a j f ( n / b j ) → a f ( n / b )  f ( n)
 
j 0
c
substitute
2
a
f ( n / b 2 )  f (n)
2
c
substitute
3
a
f ( n / b 3 )  f (n)
3
c
substitute
a4
4
f ( n / b )  f (n)
4
c
→

a n
n
f ( 2 )  f ( )
c b 
b 
→
a n
n
f ( 3 )  f ( 4 )
c b 
b 
→…→
→
a n
n
f ( 3 )  f ( 2 )
c b 
b 
→ a j f ( n / b j )  c j f ( n)
aj
j
f
(
n
/
b
)  f (n) → a j f (n )  c j f (n)


j
j
c
44
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n) Proof for Case3
Case 3. If af ( n / b)  cf (n) , for some constant c < 1
and for all n≥b, then g (n)  ( f (n))
logbn 1
g ( n)   a j f ( n / b j )
j 0
logbn 1
  c j f ( n) ( because a j f (n j )  c j f (n) )
j 0

 f ( n) c j
j 0
1
)
1 c
 O ( f ( n)) ( Since c is constant )
 f ( n)(
45
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n) Proof for Case3
g(n) =Ω(f(n)) and g(n)=O(f(n))
so g(n)= θ(g(n)) ( for extra powers of b ) (Now we proved case3)
Case 3. If af ( n / b)  cf (n) , for some constant c < 1
and for all n≥b, then g (n)  ( f (n))
46
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n) Proof for Case2
logbn 1
g ( n)   a j f ( n / b j )
j 0
log a
log a
Case 2. If f (n)  (n b ) , then g (n)  (n b lg n)
47
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n) Proof for Case2
a j f (n j )  c j f (n)
cj
→ f (n j ) 
f ( n)
j
a
c j logba
f (n j )  j n
a
a
f (n)  O(nlogb )
a
→ f (n j )  O(( n / b j ) logb )
log a
(because f (n)  (n b )
logba
n
 O( j )
a
)
logba
If we can show
n
j logba
O( j )  O(( n / b ) ) then
a
proof for case2
48
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n) Proof for Case2
j  log bn 
j
→b n
→ bj / n 1
49
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n) Proof for Case2
logba
n
j logba
O( j )  O(( n / b ) )
a
show
f ( n)  O ( n
logba
f ( n j )  O( n j
n
b logba

)
j
b
b 1
n
bj b
logba
 c( j (1 
))
b
n b 1
→ f (n j )  c(
)
logba
then proof for case2
)
substitute
n
1
nj  j 
b
b 1
logba
bj b
n
logba
 c( j )(1  (
))
n b 1
a
logba
n
b logba
 c( j )(1 
)
a
b 1
 O(
logba
n
)
aj
(Now we proved case2)
50
2. Recurrence solution methods
• Proof of the master theorem n is all integers
Proving upper bounding in recurrence T (n)  aT ( n / b)  f (n) Proof for Case2
Substitute into g(n) : g (n)  (
logbn 1

j logba
j
a (n / b )
)
j 0
→
logbn 1

j
j logba
a (n / b )
n
j 0
logba
log bn
a
logb
log bn )
→ g ( n )  ( n
a
 (n logb lg n) (Now we proved case2)
log a
log a
Case 2. If f (n)  (n b ) , then g (n)  (n b lg n)
51