6
$\begingroup$

I am reading section 6.4 on Heapsort algorithm in CLRS, page 160.

HEAPSORT(A)  
1 BUILD-MAX-HEAP(A)  
2 for i to A.length downto 2  
3   exchange A[i] with A[i]
4   A.heap-size = A.heap-size-1  
5   MAX-HEAPIFY(A,1)

Why is the running time, according to the book is $\Theta (n\lg{n})$ rather than $\Theta (n^2\lg{n})$ ? BUILD-MAX-HEAP(A) takes $\Theta(n)$, MAX-HEAPIFY(A,1) takes $\Theta(\lg{n})$ and repeated $n-1$ times (line 3).

$\endgroup$
  • $\begingroup$ Where does the n^2 lg n come from? Your numbers don't add up to n^2 log n but to n lg n. $\endgroup$ – gnasher729 Oct 5 '16 at 10:39
11
$\begingroup$

Let us count operations line by line. You construct the heap in linear time. Then, you execute the loop and perform a logarithmic time operation $n-1$ times. Other operations take constant time. Hence, your running time is

$\qquad \begin{align} & n + (n-1) \log n + O(1) \\ &= n + n \log n - \log n + O(1) \\ &= \Theta(n \log n). \end{align}$

In other words, as $n$ grows the $n \log n$ term dominates. That is, the cost of building the heap on line 1 is negligible compared to the cost of executing the loop.

$\endgroup$
  • 1
    $\begingroup$ Oh, I messed up. I had in mind $n*(n-1)*\log{n}$, rather than $n + ....$ The For loop is $(n-1)\log{n}$ and BUILD-MAX-HEAP(A)should have been just added. Thanks for the answer ! $\endgroup$ – newprint Sep 16 '12 at 22:38
0
$\begingroup$

I had a different view altogether on this. I could not ignore the fact that in the MAX_HEAPIFY step in the sort (which has a order of $\log n$, where $n$ is the heap size) receives a diminished heap size in each iteration. So to me the Summation looked more or less like $\sum_{k=1}^{n-1}\log k\approx \log (n!)$. Now I know that $n! =\Theta(n^n)$ that way $\log(n!)=\Theta(n\log n)$, but it was never convincing.

$\endgroup$
  • 3
    $\begingroup$ I'm not sure what you're not convinced about (or why you're posting an answer that you're not convinced is true). $\endgroup$ – David Richerby Oct 4 '16 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.