Recursion
Recursion is a concept of defining a
method that makes a call to itself.
Trees
1
Recursion
Recursion is a concept of
defining a method that makes a
call to itself.
Trees
2
Factorial
Example:
f(n)=n!=n×(n-1)×(n-2)×…×2×1
Initialization: f(0)=1
Recursive Call: f(n)=n×f(n-1) and.
Java code:
public static int recursiveFactorial(int n) {
if (n==0) return 1;
else return n*recursiveFactorial(n-1);
}
Trees
3
Fibonacci sequence
Fibonacci sequence:
{fn } = 0,1,1,2,3,5,8,13,21,34,55,…
Initialization: f0 = 0, f1 = 1
Recursive Call: fn = fn-1+fn-2 for n > 1.
Java code:
public static int recursiveFibonacci(int n) {
if (n==0) return 0;
if (n==1) return 1;
else return recursiveFibonacci(n-1)+recursiveFibonacci (n-2);
}
L16
4
A={4,3,6,2,5}
LinearSum
return 15+A[4]=20
Algorithm LinearSum(A, n)
LinearSum(A,5)
return 13+A[3]=15
Input: an integer array A of n elements
Output: The sum of the n elements
LinearSum(A,4)
return 7+A[2]=13
if n=1 then
LinearSum(A,3)
return A[0]
return 4+A[1]=7
return LinearSum(A, n-1)+A[n-1]
LinearSum(A,2)
• The recursive method should always
possess—the method terminates.
return A[0]=4
•We did it by setting :
LinearSum(A,1)
•” if n=1 then return A[0] ”
f(n)=A[n-1]+f(n-1) for n>0 and f(1)=A[0]
Trees
The compiler of any high level computer
language uses a stack to handle
recursive calls.
5
n=4
Factorial
return f(4)=4*f(3)=24
public static int recursiveFactorial(int n)
if (n==0)
return 1;
return n*recursiveFactorial(n-1);}
recursiveFactorial(4)
return f(3)=3*f(2)=6
recursiveFactorial(3)
The recursive method should always
possess—the method terminates.
return f(2)=2*f(1)=2
recursiveFactorial(2)
•We did it by setting:
return f(1)=1*1=1
•” if n=0 then return 1 ”
recursiveFactorial(1)
return f(0)=1
f(n)=n*f(n-1) for n>0
recursiveFactorial (0)
f(0)=1.
Trees
6
Fibonacci sequence
public static int recursiveFibonacci(int n)
{
if (n==0)
return 0;
if (n==1)
return 1;
return recursiveFibonacci(n-1)
+recursiveFibonacci (n-2);
}
L16
7
ReverseArray
Algorithm ReverseArray(A, i, j):
input: An array A and nonnegative integer indices i and j
output: The reversal of the elements in A starting at index i
and ending at j
if i<j then
A={1, 2, 3, 4}.
{
ReverseArray(A, 0, 3)
swap A[i] and A[j]
ReverseArray(A, i+1, j-1)}
A={4,2,3,1}
}
What is the base case?
ReverseArray(A, 1 2)
A=(4,3,2,1}
Trees
8
FindMax
Algorithm FindMax(A, i, j):
input: Array A , indices i and j, i≤j
output: The maximum element starting i
and ending at j
if i<j then
1
{
a←FindMax(A, i, (i+j)/2)
T(n/2)+1
b←FindMax(A, (i+j)/2+1, j) T(n/2)+1
return max(a, b)
1
}
return A[i]
1
Trees
Running time:
T(n)=2T(n/2)+c1
T(1)=c2
where c1 and c2 are some
constants.
T(n)=2T(n/2)+c1
=2[2T(n/4)+c1]+c1
=4T(n/4)+3c1
=…
=2kT(1)+(1+2+4+…2k)c1
=nT(1)+2k+1 c1
=O(n)
9
Binary Search
Algorithm BinarySearch(A, i, j, key):
input: Sorted Array A , indices i and j, i≤j,
and key
output: If key appears between elements
from i to j, inclusively
if i≤j
mid  (i + j) / 2
if A[mid] = key
return mid
if A[mid] < key
return BinarySearch(A, mid+1, j, key)
else
return BinarySearch(A, i, mid-1, key)
return -1
Trees
Running time:
T(n)=T(n/2)+c1
T(1)=c2
where c1 and c2 are some
constants.
T(n)=T(n/2)+c1
=[T(n/4)+c1]+c1
=T(n/4)+2c1
=…
=T(1) + kc1
=?
10