Proving the Reverse of Case 3 of the Master Theorem

In summary: Best of luck with your problem!In summary, the regularity condition given in the problem suggests that the complexity of the problem is not as high as it is claimed to be in case 3 of the master theorem. Instead, the complexity can be shown to be f(n) = Θ(nlogba+e) for some constant e > 0.
  • #1
atgofbhs
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Homework Statement



i am trying to solve this but in vain...
i need some help or hints to solve the questions below


Show that case 3 of the master theorem is overstated, in the sense that the
regularity condition a · f(n/b) <= c · f(n) for some constant c < 1 implies
that there exists a constant e > 0 such that f(n) =
(nlogba+e).

Homework Equations



Given above

The Attempt at a Solution


I am trying to prove the reverse mathematically but not getting both the sides to same complexity
 
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  • #2
. Can someone guide me with a proper solution or hint?

Thank you for reaching out for help with this problem. I understand that you are struggling to solve this problem and are looking for some guidance or hints. I would be happy to provide some assistance.

Firstly, let's review the regularity condition given in the problem: a · f(n/b) <= c · f(n) for some constant c < 1. This condition essentially means that the cost of splitting the problem into a subproblem of size n/b is significantly less than the cost of solving the original problem of size n. In other words, the cost of solving the subproblem is at most a fraction of the cost of solving the original problem.

Now, to prove that case 3 of the master theorem is overstated, we need to show that the complexity of the problem is not as high as it is claimed to be. In other words, we need to show that the complexity of the problem is not f(n) = Θ(nlogba). To do this, we can use the regularity condition to find a lower bound for the complexity of the problem.

Since a · f(n/b) <= c · f(n), we can rewrite this as f(n/b) <= c/a · f(n). Now, let's assume that f(n) = Θ(nlogba). This means that there exist constants k1 and k2 such that k1 · nlogba <= f(n) <= k2 · nlogba. Substituting this into our inequality, we get f(n/b) <= c/a · k1 · nlogba. Simplifying, we get f(n/b) <= (c/a)k1 · nlogba. Notice that (c/a)k1 is a constant, so we can rewrite this as f(n/b) = Θ(nlogba).

This means that the complexity of the subproblem is also Θ(nlogba), which contradicts the regularity condition. This suggests that our assumption that f(n) = Θ(nlogba) is incorrect, and there must be a lower complexity for the problem. In fact, using the regularity condition, we can show that the complexity of the problem is f(n) = Θ(nlogba+e) for some constant e > 0.

I hope this helps guide you towards a solution. If you have any further questions or need more
 

1. How do you prove the reverse of Case 3 of the Master Theorem?

To prove the reverse of Case 3 of the Master Theorem, you must show that if a recurrence relation satisfies the conditions of Case 3, then it can be solved using the formula provided in this case.

2. What are the conditions of Case 3 of the Master Theorem?

The conditions of Case 3 of the Master Theorem are: the recurrence relation is of the form T(n) = aT(n/b) + f(n), where a ≥ 1 and b > 1, and f(n) is polynomially smaller than nlogba.

3. Why is proving the reverse of Case 3 important?

Proving the reverse of Case 3 is important because it allows us to determine the asymptotic behavior of a recurrence relation and find the exact solution using the formula provided in this case.

4. Can the reverse of Case 3 be applied to all recurrence relations?

No, the reverse of Case 3 can only be applied to recurrence relations that satisfy the conditions of this case. Other cases of the Master Theorem must be used for different types of recurrence relations.

5. What are the advantages of using the reverse of Case 3 in solving recurrence relations?

The reverse of Case 3 provides a simple and efficient formula for solving recurrence relations, which can save time and effort in finding the exact solution. It also allows for a better understanding of the asymptotic behavior of the recurrence relation.

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