I'm attempting to solve the recurrence relation:

$$ T(n) = 4T\left(\frac{n}{2}\right) + \frac{n}{\lg n} $$

in terms of its asymptotic behavior ($\Theta$), specifically using the first case of the Master's Theorem. We know that if $O(n^{2-\epsilon}) = \frac{n}{\lg n}$ is correct, then $T(n) = \Theta(n^2)$. However, is this correct? And how can I demonstrate it?


1 Answer 1


If you want to see it explicitly here is the calculation:

$T(n)=4T\left(\frac{n}{2}\right) + \frac{n}{\log n}$ implies that the total number of recursions we do is $\log_2 n$. At the pass of the recursion $i$, we have $4^iT\left(\frac{n}{2^i}\right)$ plus all the other terms deriving by the non-recursive term given by $\sum_{j=1}^i 4^{j-1}\frac{n/2^{j-1}}{\log(n/2^{j-1})}$.

Putting all together and assuming that the recursion for $n=1$ is $T(1)=c=O(1)$, we have:

$$T(n)=c\cdot 4^{\log_2 n} + \sum_{i=0}^{(\log_2 n) -1} 4^i \cdot \frac{n/2^{i}}{\log(n/2^{i})} \ .$$ Simplifying $4^{\log_2 n}$ as $n^2$ and the fractions inside the summation, we obtain:

$$T(n) = c \, n^2 + n \cdot \sum_{i=0}^{(\log_2 n) -1} \frac{2^i}{\log(n/2^{i})} \ .$$

We can now conclude with a double inequality:

  • From one side, it is obvious that $T(n)\geq n^2$, so we have $T(n)=\Omega(n^2)$.
  • On the other hand, we have, from the formula above, that: $$T(n) \ \leq \ c \, n^2 + n\cdot \sum_{i=0}^{(\log_2 n) -1} 2^i \ = \ c \, n^2 + n\cdot\bigl( 2^{\log_2 n} -1 \bigr) \\ \ = \ c \, n^2 + n(n-1) \ \leq \ c \, n^2 + n^2 = (c+1)n^2 \ .$$ With this, we can conclude also that $T(n)=O(n^2)$.

Finally, we have $T(n)=\Theta(n^2)$.


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