Average number of comparisons made by the Binary search (summation computation)

Asked Sep 29 '20 at 12:59

Active Sep 29 '20 at 13:24

Viewed 249 times

I am trying to find out the average number of comparisons made by the Binary search. In that, I'm confronted with following summation :-

Avg = $(1/n)\sum_{i=1}^{\log_2 n} i*2^i$

In this summation, $i$ are the number of comparisons made at each node in $i^{th}$ level of the Binary tree and $2^i$ is the number of nodes in the tree at level $i$.

Please help me solve this summation. If not completely solve the summation, at least please help me get started.

edited Sep 29 '20 at 13:24

asked Sep 29 '20 at 12:59

Onkar N Mahajan

1

Does this answer your question? How to find the partial sum of $n/2^n$? Also this one This search finds them and others – Ross Millikan Sep 29 '20 at 13:43
Remember, that $\log_2 n$ is not necessarily integer. – zkutch Sep 29 '20 at 16:28

Average number of comparisons made by the Binary search (summation computation)

0 Answers0