Math Problem Statement

Master Method Compute the complexity of the recursive algorithms based on the recursive equation and stop condition. Show your work. T(n) = 2T(n/4) + n^2 and T(0) = 1 Note that n^2 is the square root of n.

Solution

To compute the time complexity of the recursive algorithm described by the recurrence relation:

T(n)=2T(n4)+n2withT(0)=1T(n) = 2T\left(\frac{n}{4}\right) + n^2 \quad \text{with} \quad T(0) = 1

we can apply the Master Theorem. The general form of a recurrence that we can apply the Master Theorem to is:

T(n)=aT(nb)+f(n)T(n) = aT\left(\frac{n}{b}\right) + f(n)

Where:

  • aa is the number of subproblems,
  • bb is the factor by which the problem size is divided,
  • f(n)f(n) is the additional work done outside the recursive calls.

In our case, we can match the recurrence to the Master Theorem by identifying:

  • a=2a = 2 (since there are 2 subproblems),
  • b=4b = 4 (because the problem size is divided by 4),
  • f(n)=n2f(n) = n^2 (the additional work done at each level is proportional to n2n^2).

Step 1: Calculate the Critical Exponent

The critical exponent is given by:

d=logba=log42d = \log_b a = \log_4 2

Using the change of base formula:

log42=log2log4=log22log2=12\log_4 2 = \frac{\log 2}{\log 4} = \frac{\log 2}{2 \log 2} = \frac{1}{2}

Thus, d=12d = \frac{1}{2}.

Step 2: Compare the Growth of f(n)f(n) with ndn^d

Now, we compare the growth rate of f(n)=n2f(n) = n^2 with nd=n12n^d = n^{\frac{1}{2}}.

  • f(n)=n2f(n) = n^2 grows faster than n12n^{\frac{1}{2}} because 2>122 > \frac{1}{2}.

Step 3: Apply the Master Theorem

The Master Theorem tells us that the complexity depends on the comparison between f(n)f(n) and ndn^d. There are three cases to consider:

  • Case 1: If f(n)=O(nd)f(n) = O(n^d), the complexity is T(n)=O(nd)T(n) = O(n^d).
  • Case 2: If f(n)=Θ(nd)f(n) = \Theta(n^d), the complexity is T(n)=O(ndlogn)T(n) = O(n^d \log n).
  • Case 3: If f(n)=Ω(nd)f(n) = \Omega(n^d) and f(n)f(n) is polynomially larger, the complexity is T(n)=O(f(n))T(n) = O(f(n)).

Since f(n)=n2f(n) = n^2 grows faster than n12n^{\frac{1}{2}}, we are in Case 3. Therefore, the complexity is dominated by f(n)f(n), and we have:

T(n)=O(n2)T(n) = O(n^2)

Conclusion:

The time complexity of the recurrence relation T(n)=2T(n/4)+n2T(n) = 2T(n/4) + n^2 is O(n2)O(n^2).


Would you like further details or clarification on any step? Here are some related questions to explore:

  1. What happens to the time complexity if the recurrence was T(n)=3T(n/4)+n2T(n) = 3T(n/4) + n^2?
  2. How do you solve recurrences that do not fit the Master Theorem?
  3. How can we use the recursion tree method to derive the time complexity?
  4. What is the impact of the base case T(0)=1T(0) = 1 on the overall time complexity?
  5. What happens to the complexity if f(n)f(n) was linear, i.e., f(n)=nf(n) = n?

Tip: If you encounter a recurrence that doesn’t fit neatly into the Master Theorem, you can often solve it by using the recursion tree method or iteration, expanding the recurrence to find a pattern.

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Math Problem Analysis

Mathematical Concepts

Recursion
Master Theorem
Time Complexity
Logarithmic Functions

Formulas

T(n) = 2T(n/4) + n^2
d = log_b a
f(n) = n^2

Theorems

Master Theorem

Suitable Grade Level

College-Level