The document discusses an improvised master's theorem for solving recurrence relations, addressing the limitations of the original theorem. It presents a revised formulation that includes constraints to solve previously unsolvable problems in divide and conquer algorithms. The conclusion emphasizes the efficiency gained through this new approach compared to traditional methods.

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.
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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
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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)
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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.
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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)
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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)
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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)
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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.
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10. THANK YOU!
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