c
n 1

 n
T (n)  aT
   cn n  1

 b

T(n) =
=
=
=
=
=
=
=
=
aT(n/b) + cn
a(aT(n/b/b) + cn/b) + cn
a2T(n/b2) + cna/b + cn
a2T(n /b2) + cn(a/b + 1)
a2(aT(n/b2/b) + cn/b2) + cn(a/b + 1)
a3T(n /b3) + cn(a2/b2) + cn(a/b + 1)
a3T(n/b3) + cn(a2/b2 + a/b + 1)
…
akT(n/bk) + cn(ak-1/bk-1 + ak-2/bk-2 + … +
a2/b2 + a/b + 1)
1
2
3
k
1
c
n 1

 n
T (n)  aT
   cn n  1

 b

So we have


T(n) = akT(n/bk) + cn(ak-1/bk-1 + ... + a2/b2 +a/b+1)
To stop the recursion, we should have


n/bk = 1
 n = bk  k = logb n
T(n) = akT(1) + cn(ak-1/bk-1 +...+ a2/b2 + a/b+1)
= akc + cn(ak-1/bk-1 + ... + a2/b2 + a/b + 1)
= cak + cn(ak-1/bk-1 + ... + a2/b2 + a/b + 1)
= cnak/bk + cn(ak-1/bk-1 + ... +a2/b2 +a/b+1)
= cn(ak/bk + ... + a2/b2 + a/b + 1)
2

c
n 1

 n
T (n)  aT
   cn n  1

 b
So with k = logb n


T(n) = cn(ak/bk + ... + a2/b2 + a/b + 1)
What if a = b?

T(n) = cn(1+…+1+1+1)
//k+1 times
= cn(k + 1)
= cn(logb n + 1)
= (n logb n)
3

c
n 1

 n
T (n)  aT
   cn n  1

 b
So with k = logb n


T(n) = cn(ak/bk + ... + a2/b2 + a/b + 1)
What if a < b?

Recall that

k

(xk + xk-1 + … + x + 1) = (xk+1 -1)/(x-1)
So:
k 1
a a
a
 k 1     1 
k
b
b b

a b k 1  1
a b   1
1  a b 

1  a b 
k 1
1

1 a b
T(n) = cn ·(1) = (n)
4
c
n 1

 n
T (n)  aT
   cn n  1

 b

So with k = logb n


T(n) = cn(ak/bk + ... + a2/b2 + a/b + 1)
What if a  b?
a k a k 1
a




1 
k
k 1
b
b b

a b k 1  1
a b   1

  a b 
k

T(n) = cn · (ak / bk)
= cn · (alogbn / blogbn) = cn · (alogbn / n)
recall logarithm fact: alogbn = nlogba
= cn · (nlogba / n) = (cn · nlogba / n)
= (nlogba)
5
c
n 1

 n
T (n)  aT
   cn n  1

 b

So…
 n 

T (n)  n log b n 
  n logb a



ab
ab
ab
6
The Master Method

Based on the Master theorem.

“Cookbook” approach for solving recurrences
of the form
T(n) = aT(n/b) + f(n)



a  1, b > 1 are constants.
f(n) is asymptotically positive.
Requires memorization of three cases.
7
The Master Theorem
Let a  1 and b > 1 be constants, let f(n) be a function, and
Let T(n) be defined on nonnegative integers by the recurrence
T(n) = aT(n/b) + f(n), where we can replace n/b by n/b or
n/b.
T(n) can be bounded asymptotically in three cases:
1. If f(n) = O(nlogba–) for some constant  > 0, then T(n) =
(nlogba).
2. If f(n) = (nlogba), then T(n) = (nlogbalg n).
3. If f(n) = (nlogba+) for some constant  > 0,
and if, for some constant c < 1 and all sufficiently large n,
we have a·f(n/b)  c f(n), then T(n) = (f(n)).
8
The Master Theorem


Given: a divide and conquer algorithm

An algorithm that divides the problem of size n into a
subproblems, each of size n/b

Let the cost of each stage (i.e., the work to divide the
problem + combine solved subproblems) be
described by the function f(n)
Then, the Master Theorem gives us a cookbook for
the algorithm’s running time:
9
The Master Theorem


if T(n) = aT(n/b) + f(n) where a ≥ 1 & b > 1
then 

 logb a
logb a 

f ( n)  O n
 n



 logb a
  0
logb a
T (n)   n
lg n f (n)   n


 c 1


logb a 
 f (n) 

f ( n)   n
&


af (n / b)  cf (n) for large n











10
Understanding Master Theorem

In each of the three cases, we are comparing f(n)
with nlogba , the solution to the recurrence is
determined by the larger of the two functions.

In case 1, if the function nlogba is the larger, then the
solution T(n) = Θ(nlogba).

In case 3, if the function f(n) is the larger, then the
solution is T(n) = (f(n)).

In case 2, if the two functions are the same size,
then the solution is T(n) = Θ(nlogba lg n) = Θ(f(n) lg
n).
11
Understanding Master Theorem

In case 1, not only must f(n) be smaller than
nlogba, it must be polynomially smaller. That is
f(n) must be asymptotically smaller than
nlogba by a factor of nε for some constant ε >
0.

In case 3, not only must f(n) be larger than
nlogba , it must be polynomially larger and in
addition satisfy the “regularity” condition that:
a f(n/b) ≤ c f(n).
12
Understanding Master Theorem

It is important to realize that the three cases do not
cover all the possibilities for f(n).

There is a gap between cases 1 and 2 when f(n) is
smaller than nlogba but not polynomially smaller.

There is a gap between cases 2 and 3 when f(n) is
larger than nlogba but not polynomially larger.

If f(n) falls into one of these gaps, or if the regularity
condition in case 3 fails to hold, the master method
cannot be used to solve the recurrence.
13