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3 votes
3 votes

The master theorem

  1. assumes the subproblems are unequal sizes
  2. can be used if the subproblems are of equal size
  3. cannot be used for divide and conquer algorithms
  4. cannot be used for asymptotic complexity analysis
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2 Answers

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$\underline{\textbf{Answer:}\Rightarrow}\;\mathbf{b.}$

It divides the subproblem into each of size $\mathbf{\dfrac{n}{b}}$.
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Answer might be B as master theorm never give upper bound it gives tightest bound
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Yes right, so this was also a negative for me. :(
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Why not (D) ?
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1 vote
1 vote
Answer will be (b) because

According to master's theorem we can only find the time complexity if the recurrence relation is of form

$T(n) = a\times T \left( \frac{n}{b} \right) +n^k *log^p n$

it assumes the subproblems to be of size $\left( \frac{n}{b} \right)$
Answer:

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