Open In App
Related Articles

Master Theorem For Subtract and Conquer Recurrences

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Report issue
Report

Master theorem is used to determine the Big – O upper bound on functions which possess recurrence, i.e which can be broken into sub problems. 
Master Theorem For Subtract and Conquer Recurrences
Let T(n) be a function defined on positive n as shown below: 
 

Screenshot from 2017-07-12 14-14-44


for some constants c, a>0, b>0, k>=0 and function f(n). If f(n) is O(nk), then
1. If a<1 then T(n) = O(nk
2. If a=1 then T(n) = O(nk+1
3. if a>1 then T(n) = O(nkan/b)
Proof of above theorem( By substitution method ):
From above function, we have: 
T(n) = aT(n-b) + f(n) 
T(n-b) = aT(n-2b) + f(n-b) 
T(n-2b) = aT(n-3b) + f(n-2b)
Now, 
T(n-b) = a2T(n-3b) + af(n-2b) + f(n-b) 
T(n) = a3T(n-3b) + a2f(n-2b) + af(n-b) + f(n) 
T(n) = ?i=0 to n ai f(n-ib) + constant, where f(n-ib) is O(n-ib) 
T(n) = O(nk ?i=0 to n/b ai )

 
Where, 
If a<1 then ?i=0 to n/b ai = O(1), T(n) = O(nk)
If a=1 then ?i=0 to n/b ai = O(n), T(n) = O(nk+1
If a>1 then ?i=0 to n/b ai = O(an/b), T(n) = O(nkan/b)
Consider the following program for nth fibonacci number
 

C++

#include<stdio.h>
int fib(int n)
{
   if (n <= 1)
      return n;
   return fib(n-1) + fib(n-2);
}
  
int main ()
{
  int n = 9;
  printf("%d", fib(n));
  getchar();
  return 0;
}

                    

Python3

# Python3 code for the above approach
def fib(n):
    if (n <= 1):
        return n
    return fib(n - 1) + fib(n - 2)
 
# Driver code
n = 9
print(fib(n))
 
# This code is contributed
# by sahishelangia

                    

Java

//Java code for above the approach.
class clg
{
 static int fib(int n)
{
if (n <= 1)
    return n;
return fib(n-1) + fib(n-2);
}
// Driver Code
public static void main (String[] args)
{
int n = 9;
System.out.println( fib(n));
}
}
// This code is contributed by Mukul Singh.

                    

C#

// C# code for above the approach.
using System;
     
class GFG
{
    static int fib(int n)
    {
        if (n <= 1)
            return n;
        return fib(n - 1) + fib(n - 2);
    }
     
    // Driver Code
    public static void Main(String[] args)
    {
        int n = 9;
        Console.WriteLine(fib(n));
    }
}
 
// This code has been contributed
// by Rajput-Ji

                    

PHP

<?php
// PHP code for the above approach
function fib($n)
{
    if ($n <= 1)
        return $n;
    return fib($n - 1) +
           fib($n - 2);
}
 
// Driver Code
$n = 9;
echo fib($n);
 
// This code is contributed
// by Akanksha Rai
?>

                    

Javascript

<script>
    // Javascript code for above the approach.
     
    function fib(n)
    {
        if (n <= 1)
            return n;
        return fib(n - 1) + fib(n - 2);
    }
     
    let n = 9;
      document.write(fib(n));
     
</script>

                    

Output 
 

34


Time complexity Analysis: 
The recursive function can be defined as, T(n) = T(n-1) + T(n-2) 
 

  • For Worst Case, Let T(n-1) ? T(n-2) 
    T(n) = 2T(n-1) + c 
    where,f(n) = O(1) 
    ? k=0, a=2, b=1;
    T(n) = O(n02n/1
    = O(2n
     
  • For Best Case, Let T(n-2) ? T(n-1) 
    T(n) = 2T(n-2) + c 
    where,f(n) = O(1) 
    ? k=0, a=2, b=2;
    T(n) = O(n02n/2
    = O(2n/2
     


 

More Examples:


 


  • Example-1
    T(n) = 3T(n-1), n>0 
         = c, n<=0
    Sol:a=3, b=1, f(n)=0 so k=0;
    Since a>0, T(n) = O(nkan/b
    T(n)= O(n03n/1
    T(n)= 3n 
     

  • Example-2
    T(n) = T(n-1) + n(n-1), if n>=2 
         = 1, if n=1
    Sol:a=1, b=1, f(n)=n(n-1) so k=2;
    Since a=1, T(n) = O(nk+1
    T(n)= O(n2+1
    T(n)= O(n3
     

  • Example-3
    T(n) = 2T(n-1) – 1, if n>0 
         = 1, if n<=0
    Sol: This recurrence can’t be solved using above method 
    since function is not of form T(n) = aT(n-b) + f(n) 
     




 



Last Updated : 31 May, 2021
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads