COMP 250 Winter 2010
23  building a heap (slow)
March 10, 2010
Worst case for building a heap (using slow algorithm)
The algorithm I presented last lecture for building a heap is relatively inefficient in the worst case.
(We will see a faster algorithm next lecture.) Let’s see why it is inefficient. If we label the nodes
from
i
= 1 to
i
=
n
, then if node
i
is at level
l
(the root is at level
l
= 0), we can see by inspection
(see lecture slides) that
2
l
≤
i
≤
2
l
+1
−
1
and so
⌊
log
i
⌋
=
l
. Thus, when we add node
i
to the heap, we need to do at most
⌊
log
i
⌋
swaps up
the tree to bring the new element
i
to a position where it is greater than its parent. In the worst
case, the element at
i
is smaller than all its ancesters and we need to swap it
⌊
log
i
⌋
times, which
brings it to the root.
Since we are adding
n
nodes in total, the worst case number of swaps is:
t
(
n
) =
n
summationdisplay
i
=1
⌊
log
i
⌋
To understand this sum, we ignore the floor and we plot the log
i
as a function of
i
. See
thick
solid curves. We give two plots: from
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 Spring '08
 BLANCHETTE
 Computer Science, Binary relation, worst case, 2L

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