Master Theorem For Subtract and Conquer Recurrences

31 May, 2021
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:  
for some constants c, a>0, b>0, k>=0 and function f(n). If f(n) is O(nk), then1. 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) = Σi=0 to n ai f(n-ib) + constant, where f(n-ib) is O(n-ib) T(n) = O(nk Σi=0 to n/b ai )
 Where, If a<1 then Σi=0 to n/b ai = O(1), T(n) = O(nk)If a=1 then Σi=0 to n/b ai = O(n), T(n) = O(nk+1) If a>1 then Σi=0 to n/b ai = O(an/b), T(n) = O(nkan/b)Consider the following program for nth fibonacci number:  
C++
Python3
Java
C#
PHP
Javascript
#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 code for the above approachdef fib(n):    if (n <= 1):        return n    return fib(n - 1) + fib(n - 2) # Driver coden = 9print(fib(n)) # This code is contributed# by sahishelangia
//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 Codepublic static void main (String[] args){int n = 9;System.out.println( fib(n));}}// This code is contributed by Mukul Singh.
// 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 code for the above approachfunction 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?>
<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<=0Sol: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=1Sol: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<=0Sol: This recurrence can’t be solved using above method since function is not of form T(n) = aT(n-b) + f(n)  
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