I already saw this answer. But I didn't understand totally.

So I found the solution to this equation $T(n)=4T(n/2+2)+n$.

enter image description here

Here is the recursion tree. But I can't compute the height as $lgn$ from this tree.

How to compute height? (Image is from solution)

  • $\begingroup$ The height is very close to $\log_2 n$. $\endgroup$ – Yuval Filmus Apr 16 '18 at 7:51

Suppose that $n = 2^m+4$. Then $$ \frac{n}{2} + 2 = \frac{2^m+4}{2} + 2 = (2^{m-1} + 2) + 2 = 2^{m-1} + 4. $$ We conclude that if $n = 2^m+4$ then the height is $m$ (or $m+1$, depending on how you count), with $T(5)$ in the leaves. This shows that for this type of $n$, the height of the recursion is $\log_2(n-4)$ (or $\log_2(n-4)+1$, depending on how you count).

For general $n$, the exact answer depends on the interpretation of $n/2$ — whether you take floor, ceiling, or anything else — but the height will still be very close to $\log_2 n$.

  • 1
    $\begingroup$ You mean just ignore a small constant or coefficient? $\endgroup$ – molamola Apr 16 '18 at 8:13
  • $\begingroup$ The height is very close to $\log_2 n$. $\endgroup$ – Yuval Filmus Apr 16 '18 at 8:15

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