Math Problem Statement

Solve T(n) = 9T(n/6) + n log^2 n, by master method. (10 marks)

Solution

Let's solve the recurrence relation T(n)=9T(n6)+nlog2nT(n) = 9T\left(\frac{n}{6}\right) + n \log^2 n using the Master Theorem for divide-and-conquer recurrences.

The recurrence is of the form: T(n)=aT(nb)+f(n)T(n) = aT\left(\frac{n}{b}\right) + f(n) Where:

  • a=9a = 9
  • b=6b = 6
  • f(n)=nlog2nf(n) = n \log^2 n

To apply the Master Theorem, we compare f(n)f(n) with nlogban^{\log_b a}.

Step 1: Calculate logba\log_b a

We need to compute logba=log69\log_b a = \log_6 9. Using the change of base formula: log69=log9log6\log_6 9 = \frac{\log 9}{\log 6} We approximate: log90.9542,log60.7782\log 9 \approx 0.9542, \quad \log 6 \approx 0.7782 Thus: log690.95420.77821.226\log_6 9 \approx \frac{0.9542}{0.7782} \approx 1.226 So, nlog69n1.226n^{\log_6 9} \approx n^{1.226}.

Step 2: Compare f(n)=nlog2nf(n) = n \log^2 n with nlogban^{\log_b a}

We now compare the asymptotic behavior of f(n)=nlog2nf(n) = n \log^2 n with nlog69n^{\log_6 9}:

  • f(n)=nlog2nf(n) = n \log^2 n grows slightly slower than n1.226n^{1.226}, because nlog2nn \log^2 n is nn times a polylogarithmic term, while n1.226n^{1.226} has a strictly larger exponent of nn.

Step 3: Apply Master Theorem

The Master Theorem cases are:

  1. Case 1: If f(n)=O(nlogbaϵ)f(n) = O(n^{\log_b a - \epsilon}) for some ϵ>0\epsilon > 0, then T(n)=Θ(nlogba)T(n) = \Theta(n^{\log_b a}).
  2. Case 2: If f(n)=Θ(nlogbalogkn)f(n) = \Theta(n^{\log_b a} \log^k n) for some k0k \geq 0, then T(n)=Θ(nlogbalogk+1n)T(n) = \Theta(n^{\log_b a} \log^{k+1} n).
  3. Case 3: If f(n)=Ω(nlogba+ϵ)f(n) = \Omega(n^{\log_b a + \epsilon}) for some ϵ>0\epsilon > 0 and if af(nb)cf(n)a f\left(\frac{n}{b}\right) \leq c f(n) for some constant c<1c < 1 and sufficiently large nn, then T(n)=Θ(f(n))T(n) = \Theta(f(n)).

In this case, f(n)=nlog2nf(n) = n \log^2 n is asymptotically smaller than nlog69n^{\log_6 9} but not polynomially smaller. Therefore, we are in Case 2 of the Master Theorem. Hence, the solution to the recurrence is:

T(n)=Θ(nlog69log3n)T(n) = \Theta(n^{\log_6 9} \log^3 n)

Final Answer:

The solution to the recurrence T(n)=9T(n/6)+nlog2nT(n) = 9T(n/6) + n \log^2 n by the Master Method is: T(n)=Θ(n1.226log3n)T(n) = \Theta(n^{1.226} \log^3 n)

Would you like any additional details or clarifications?


Here are 5 related questions:

  1. How does the Master Theorem handle logarithmic factors in f(n)f(n)?
  2. What conditions are needed for each case of the Master Theorem to apply?
  3. How do we handle non-standard functions like log2n\log^2 n in recurrence relations?
  4. Can the Master Theorem be applied to all recurrences?
  5. How does the change of base formula help in solving recurrences with non-integer bases?

Tip: When solving recurrence relations, carefully compare the growth rates of the terms to choose the correct case of the Master Theorem.

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Math Problem Analysis

Mathematical Concepts

Recurrence Relations
Master Theorem
Asymptotic Analysis

Formulas

Master Theorem for Divide-and-Conquer Recurrences

Theorems

Master Theorem

Suitable Grade Level

Advanced Undergraduate