Improvised Master's theorem in Design and Analysis of Algorithm .Introduction to Master's Theorem .Master's Theorem Cases .Problem Statement .Solution .Conclusion

1. Improvised Master’s Theorem
Presented By:
Laiba Younas

2. INTRODUCTION:
• Master’s Theorem is a Popular method for solving the recurrence
relations.
• It is used to compute the complexity of a given divide and conquer
problem in asymptotic time.
• In divide and conquer problem we have three steps:
Divide a large problem into smaller subproblems;
Recursively solve each subproblem, then
Combine solutions of the subproblems to solve the original
problem.
2

3. The form of the recurrence relation of the master’s theorem is:
T (n) = aT (n/b) + f(n)
where a ≥ 1 and b > 1 are constants and f(n) is an asymptotically
positive function.
In above equation:
 n = problem size
 a = number of sub problems
 n/b = size of each sub problem
 f (n) = cost of the work done outside the recursive calls
3

4. Master’s Theorem cases:
• If T (n) = aT (n/b) + O(nd) where d>=0, then
Example:
T(n) = 4T(n/2) + n
a = 4, b=2, f(n) = n1, so d = 1
logb a ----> log2 4 = 2
d < logba
1<2
so,
n = O (nlog2 4)
= O (n2)
4

5. PROBLEM STATEMENT:
The earlier used theorem has some limitations. Some equations could not be solved
using the original master’s theorem.
Here are some examples:
T(n)= 0.5T(n/2) + n
In this equation a<1. The number of subproblems is 0.5 whereas there needs to be
at least 1 subproblem.
T(n)= 2T(n/2) + n/log n
Master theorem apply only for polynomial, the difference here is not polynomial but
logarithmic.
T(n)= 2n T(n/2) + nn
Now in this case we can see that ‘a’ is not a constant therefore master’s theorem
won’t be applied.
So, to resolve such problems the master’s theorem has been revised.
5

6. SOLUTION:
T(n)=aT(n/b) + θ (nd logp n)
(a>=1, b>=1, d>=0 and p is a real number)
• Case 1:
If d < logb a, then T(n) = θ (nlogb a)
• Case 2:
If d = logb a
 If p > -1 then, T(n) = θ (nd logp+1 n)
 If p = -1 then, T(n) = θ (nd log log n)
 If p < -1 then, T(n) = θ (nd)
• Case 3:
If d > logb a
 If p >= 0 then, T(n) = θ (nd logp n)
 If p < 0 then, T(n) = O (nd)
6

7. The proposed solution is used to solve the problem like:
Example:
T(n) = 64T (n/8) + n2 log n
Now we will compare the above equation with proposed format we got.
a = 64, b = 8, d = 2, p = 1
Let’s compute logba: log8 64 = 2
2 = 2
logba = d
So, it satisfies the 2nd case i.e T(n) = θ (nd logp+1 n)
Therefore, the solution is θ (n2 log2 n)
7

8. Here’s another example:
T(n)= 2T(n/2) + n/log n
Now we will compare the above equation with proposed format we got.
a = 2, b = 2, d = 1, p = -1;
Let’s compute logba: log2 2 = 1
1 = 1
logba = d
So, it satisfies the 2nd case i.e T(n) = θ (nd loglog n)
Therefore the solution is θ (n loglog n)
8

9. CONCLUSION:
• Those equations that could not be solved using the original master’s
method but after the relaxation of the constraints they can be solved
using the new approach.
• It also helps us to reduce the time complexity in comparison if the
relation is solved using iterative and substitution method.
• The proposed method is able to relax some of the constraints of the
original method.
9

10. THANK YOU!
10