Master Theorem For Subtract and Conquer Recurrences

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), 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) = 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++

 #include 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



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)

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