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<b>Data Structures and Algorithms
with Object-Oriented Design Patterns in C++</b><br>
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<H3><A NAME="SECTION0016531000000000000000">Running Time Analysis</A></H3>
<P>
The <tt>BuildHeap</tt> routine does exactly <IMG WIDTH=36 HEIGHT=26 ALIGN=MIDDLE ALT="tex2html_wrap_inline58823" SRC="img180.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img180.gif" >
<tt>PercolateDown</tt> operations.
As discussed above, the running time for <tt>PercolateDown</tt> is <IMG WIDTH=37 HEIGHT=24 ALIGN=MIDDLE ALT="tex2html_wrap_inline70367" SRC="img2192.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img2192.gif" >,
where <IMG WIDTH=12 HEIGHT=24 ALIGN=MIDDLE ALT="tex2html_wrap_inline62992" SRC="img1023.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img1023.gif" > is the height in the tree of the node at array position <I>i</I>.
The highest node in the tree is the root
and its height is <IMG WIDTH=56 HEIGHT=24 ALIGN=MIDDLE ALT="tex2html_wrap_inline59891" SRC="img403.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img403.gif" >.
If we make the simplifying assumption that the running time for
<tt>PercolateDown</tt> is <IMG WIDTH=56 HEIGHT=24 ALIGN=MIDDLE ALT="tex2html_wrap_inline59891" SRC="img403.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img403.gif" > for every value of <I>i</I>,
we get that the total running time for <tt>BuildHeap</tt> is <IMG WIDTH=68 HEIGHT=24 ALIGN=MIDDLE ALT="tex2html_wrap_inline59897" SRC="img405.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img405.gif" >.
<P>
However, <IMG WIDTH=44 HEIGHT=24 ALIGN=MIDDLE ALT="tex2html_wrap_inline60803" SRC="img558.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img558.gif" > is not a tight bound.
The maximum number of iterations of the <tt>PercolateDown</tt>
loop done during the entire process of building the heap
is equal to the sum of the heights of all of the nodes in the tree!
The following theorem shows that this is <I>O</I>(<I>n</I>).
<P>
<BLOCKQUOTE> <b>Theorem</b><A NAME="theoremsortingii"> </A>
Consider a <em>perfect</em> binary tree <I>T</I> of height <I>h</I>
having <IMG WIDTH=90 HEIGHT=24 ALIGN=MIDDLE ALT="tex2html_wrap_inline66162" SRC="img1433.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img1433.gif" > nodes.
The sum of the heights of the nodes in <I>T</I>
is <IMG WIDTH=254 HEIGHT=27 ALIGN=MIDDLE ALT="tex2html_wrap_inline70393" SRC="img2193.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img2193.gif" >.
</BLOCKQUOTE>
<P>
extbfProof
A perfect binary tree has 1 node at height <I>h</I>,
2 nodes at height <I>h</I>-1, 4 nodes at height <I>h</I>-2 and so on.
In general, there are <IMG WIDTH=11 HEIGHT=13 ALIGN=BOTTOM ALT="tex2html_wrap_inline68283" SRC="img1789.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img1789.gif" > nodes at height <I>h</I>-<I>i</I>.
Therefore, the sum of the heights of the nodes is <IMG WIDTH=94 HEIGHT=33 ALIGN=MIDDLE ALT="tex2html_wrap_inline70405" SRC="img2194.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img2194.gif" >.
<P>
The summation can be solved as follows:
First, we make the simple variable substitution <I>i</I>=<I>j</I>-1:
<P><A NAME="eqnsortingviii"> </A> <IMG WIDTH=500 HEIGHT=283 ALIGN=BOTTOM ALT="eqnarray41061" SRC="img2195.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img2195.gif" ><P>
Note that the summation which appears on the right hand side
is identical to that on the left.
Rearranging Equation <A HREF="page507.html#eqnsortingviii" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/page507.html#eqnsortingviii"><IMG ALIGN=BOTTOM ALT="gif" SRC="cross_ref_motif.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/icons/cross_ref_motif.gif"></A> and simplifying gives:
<P> <IMG WIDTH=500 HEIGHT=71 ALIGN=BOTTOM ALT="eqnarray41089" SRC="img2196.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/img2196.gif" ><P>
<P>
It follows directly from Theorem <A HREF="page507.html#theoremsortingii" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/page507.html#theoremsortingii"><IMG ALIGN=BOTTOM ALT="gif" SRC="cross_ref_motif.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/icons/cross_ref_motif.gif"></A>
that the sum of the heights of a perfect binary tree is <I>O</I>(<I>n</I>).
But a heap is not a <em>perfect</em> tree--it is a <em>complete</em> tree.
Nevertheless, it is easy to show that the same bound applies to
a complete tree.
The proof is left as an exercise for the reader (Exercise <A HREF="page521.html#exercisesortingcomplete" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/html/page521.html#exercisesortingcomplete"><IMG ALIGN=BOTTOM ALT="gif" SRC="cross_ref_motif.gif" tppabs="http://dictator.uwaterloo.ca/Bruno.Preiss/books/opus4/icons/cross_ref_motif.gif"></A>).
Therefore, the running time for the <tt>BuildHeap</tt> routine is <I>O</I>(<I>n</I>),
were <I>n</I> is the length of the array to be heapified.
<P>
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