5. Recursive.ppt

Prof. Amr Goneid, AUC 1
Analysis & Design of
Algorithms
(CSCE 321)
Prof. Amr Goneid
Department of Computer Science, AUC
Part 5. Recursive Algorithms

Recursive Algorithms

Recursive Algorithms
 Modeling Recursive Algorithms
 Examples of Recurrences
 Exclude & Conquer
 General Solution of 1st Order Linear
Recurrences
 2nd Order Linear Recurrence

1. Modeling Recursive Algorithms
Recursive Algorithm
Model as a
Recurrence Relation
T(n) in terms of T(n-1)
Solve to obtain T(n) as
a function of n

Recurrence Relations
 A recurrence relation is an equation describing a
function in terms of its value for smaller instances.
 Many algorithms, particularly exclude & conquer
and divide and conquer algorithms, have time
complexities which are naturally modeled by
recurrence relations.
 Many counting problems can be modeled as
recurrences. Solving the recurrence relation
solves the problem.

 Many natural functions are expressed as
recurrences:
)
(Factorial
1
with
al)
(Exponenti
2
1
with
2
l)
(Polynomia
1
with
1
1
1
1
1
1
1
1
!
n
a
a
na
a
a
a
a
a
n
a
a
a
a
n
n
n
n
n
n
n
n
n
n


















To derive the recurrence relation:
 Identify the problem size (n).
 Find the base case size n0
(usually n0 = 0 ,1 or 2) and find T(n0)
 The general case consists of “a” recursive
calls to do smaller sub-problems (usually of
size n-1) plus a cost “b” for all the work
needed other than in the recursive calls.

Hence, the recurrence relation will have the
form:











0
0
0
)
1
(
)
(
)
(
n
n
for
b
n
aT
n
n
for
n
T
n
T
base case
cost of base case
no. of
recursive
calls
cost of a
recursive
call
cost of
other
operations

T(0) = 0
 A Recursive Factorial Function:
factorial (n)
{
if (n == 0) return 1;
else return ( factorial (n-1) * n);
}
Find T(n) = number of integer multiplications
2. Examples of Recurrences
T(n)
T(n-1) 1

Examples
 The base case with n = 0 does zero
multiplications. Hence T(0) = 0.
 The general case will cost T(n-1) plus one
multiplication.
 Recurrence Relation:
0
T(0)
0 with
n
for
1
1)
-
T(n
T(n) 




Examples
Recurrence relations of this kind can be solved by
successive substitution:
 Substitute several times (m) to show a pattern.
 Choose (m) so that each T( ) in which (m) appears
reduces to T(n0)
 Substitute for T(n0)
 Solve the resulting equation

Examples
For example, for the recurrence relation of the
recursive factorial:
)
n
(
)
n
(
T
Hence
n
n
)
(
T
)
n
(
T
then
n
m
With
m
)
m
n
(
T
....
)
n
(
T
)
n
(
T
)
n
(
T
)
n
(
T
)
(
T
with
)
n
(
T
)
n
(
T
























0
3
3
2
2
1
1
2
0
0
1
1

 A Recursive Function to return xn:
power (x, n)
{
if (n == 0 ) return 1.0;
else return ( power(x,n-1)*x );
}
Find T(n) = number of arithmetic operations
T(n) = T(n-1) + 1 for n > 0 with T(0) = 0
Hence T(n) = n = (n)
Examples

3. Exclude and Conquer
n
1 n-1
Base
1
1

// Assume the data are in locations s….e of array a[0..n-1]
// A problem of size (n) arises when the function is invoked as
// selectsort(a,0,n-1), i.e. with s = 0 and e = n-1
Selectsort (a[ ], s, e)
{
if (s < e) {
m = minimum (a , s , e) ;
swap (a[s] , a[m]);
Selectsort (a,s+1,e); }
}
Example: Recursive Selection Sort
T(s,e)
T(s+1,e)
comp
s
e )
( 
swap
1

Selection Sort (continued)
Recurrence Relation: ( problem of size n when s = 0 , e = n-1):
)
1
n
(
)
n
(
T
that
So
0
)
1
(
T
with
1
n
for
1
)
1
n
(
T
)
n
(
T
Also
2
)
1
n
(
n
i
)
i
n
(
)
1
(
T
)
n
(
T
,
1
n
m
Set
)
i
n
(
)
m
n
(
T
)
1
n
(
)
2
n
(
)
2
n
(
T
)
n
(
T
Hence
0
)
1
(
T
with
1
n
for
)
1
n
(
)
1
n
(
T
)
n
(
T
1
n
e
,
0
s
Let
)
s
e
(
)
e
,
1
s
(
T
)
e
,
s
(
T
swap
swap
swap
1
n
1
i
1
n
1
i
comp
m
1
i
comp
comp
comp
comp











 




 




























4. General Solution of 1st Order
Linear Recurrences
Many recursive algorithms are modeled to the
following general 1st order linear recurrence
relation:
e.g. the factorial algorithm gives an = 1 and
bn = 1, T(0) given.
Select sort gives an = 1 and bn = (n-1),
T(1) given.
T(1)
or
T(0)
given
,
b
1)
-
T(n
a
T(n) n
n 


General Solution of 1st Order
Linear Recurrences
Such recurrence can be solved by successive
substitution. e.g. for T(0) given:
n
n
i
j
j
n
i
i
n
i
i
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
b
a
b
a
)
(
T
...
b
b
a
b
a
a
)
n
(
T
a
a
a
........
b
b
a
)
n
(
T
a
a
b
}
b
)
n
(
T
a
{
a
b
)
n
(
T
a
)
n
(
T






































1
1
1
1
1
2
1
2
1
1
1
1
1
0
3
2
2
1

General Solution of 1st Order
Linear Recurrences
Successive substitution gives for the cases of
T(0) given and T(1) given:
n
n
i
j
j
n
i
i
n
i
i
n
n
i
j
j
n
i
i
n
i
i
b
a
b
a
)
(
T
)
n
(
T
b
a
b
a
)
(
T
)
n
(
T






















1
1
2
2
1
1
1
1
1
0

Examples
 For the recursive factorial function, an=1 so that ai=1 and
similarly bi=1 with T(0)=0. Substitution in the general
solution gives:
hence T(n) = n as before.
 For the selection sort algorithm, ai=1,
bi= (i-1) with T(1) = 0 so that
n
)
n
(
T
n
i
n
i
n
i
j




 



 
 1
1
1 1
1
1
1
1


n
comp
i
comp
T ( n ) ( i ) ... ( n )
T ( n ) n( n ) /

        
 

2
1 1 2 3 4 1
1 2

Examples
 T(n) = 2 T(n-1) + 1 with T(0) = 0
 T(n) = n T(n-1) with T(0) = 1
)
2
(
1
2
2
1
2
1
)
(
1
0
1
1 1
n
n
n
k
k
n
i
n
i
j
O
n
T 




 
 



 

)
!
(
!
)
0
(
)
(
1
1
n
O
n
i
a
T
n
T
n
i
n
i
i 


 
 


Exercise(1): Exclude & Conquer
with a Linear Process
int FOO (int a[ ], int n)
{
if (n > 0)
{
int m = 5 * Process (a, n);
return m * FOO(a,n-1);
}
}
Show that the number of integer multiplications if Process
makes 2n such operations is:
)
n
(
O
n
n
)
n
(
T 2
2
3 



Solution
)
n
(
O
n
n
}
n
/
)
n
(
n
{
)
i
(
b
b
b
)
n
(
T
)
i
(
b
,
a
)
(
T
with
)
n
(
)
n
(
T
)
n
(
T
n
i
n
i
i
n
n
i
i
i
i
2
2
1
1
1
1
3
2
1
2
1
2
1
2
1
0
0
1
2
1






 



















Exercise(2): History Problem
double Eval ( int n )
{ int k ; double sum = 0.0;
if (n == 0) return 1.0 ;
else
{ for (k = 0; k < n; k++) sum += Eval (k);
return 2.0 * sum / n + n ; }
}
Show that the number of double arithmetic
operations performed as a function of (n) is:
)
(
O
)
n
(
T n
n
2
1
2 



Solution
1
1
2
1
1
1
2
1
3
0
0
3
1
2
0
1
0
1
0




























)
n
(
T
)
n
(
T
)
n
(
T
)
n
(
T
)
n
(
T
)
n
(
)
k
(
T
)
n
(
T
)
n
(
)
k
(
T
)
n
(
T
)
(
T
with
}
)
k
(
T
{
)
n
(
T
n
k
n
k
n
k

Solution
)
(
O
b
a
b
)
n
(
T
)
(
T
,
b
,
a
n
n
n
i
i
n
i
i
n
n
n
i
j
j
n
i
i
i
i
2
1
2
2
1
2
0
0
1
2
1
0
1
1
1
1
1

























Exercise (3): Hanoi Towers Game
Find the number of moves required for a Tower of
Hanoi with n discs. An animation is available at:
http://www.cosc.canterbury.ac.nz/people/mukundan/dsal
/ToHdb.html

5. 2nd Order Linear Recurrence
Example: Fibonacci Sequence
F(n) = F(n-1) + F(n-2) , F(0) = 0 F(1) = 1
Let to obtain the characteristic equation
Solutions:
n
n n
n n
F(n) r r r
,
F( n )
With F( ) and F( ) then and
F( n ) , Golden Ratio .
  
 
  
 

   
 
   
 
    

  
2
1 0
1 5 1 5
1
2 2
1
0 0 1 1
5
1 618034
5

5. Recursive.ppt

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5. Recursive.ppt