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:
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)

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 )

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:

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));
  return 0;



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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