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Heapsort

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Based off of a 'heap', which has several uses ... Max-heaps are used for sorting. Min-heaps are used for priority queues (later) ... log n time? ... – PowerPoint PPT presentation

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Title: Heapsort

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Heapsort

A minimalist's approach

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Heapsort

Like MERGESORT, it runs in O(n lg n)
Unlike MERGESORT, it sorts in place
Based off of a heap, which has several uses
The word heap doesnt refer to memory management

3
The Heap

A binary heap is a nearly complete binary tree
Implemented as an array A
Two similar attributes
lengthA is the size (number of slots) in A
heap-sizeA is the number of elements in A
Thus, heap-sizeA ? lengthA
Also, no element past Aheap-sizeA is an
element

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The Heap

Can be a min-heap or a max-heap

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Simple Functions

PARENT(i)
return (i/2)
LEFT(i)
return (2i)
RIGHT(i)
return (2i 1)

6
Properties

Max-heap property
APARENT(i) ? Ai
Min-heap property
APARENT(i) ? Ai
Max-heaps are used for sorting
Min-heaps are used for priority queues (later)
We define the height of a node to be the longest
path from the node to a leaf.
The height of the tree is ?(lg n)

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MAX-HEAPIFY

This is the heart of the algorithm
Determines if an individual node is smaller than
its children
Parent swaps with largest child if that child is
larger
Calls itself recursively
Runs in O(lg n) or O(h)

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HEAPIFY

MAX-HEAPIFY (A, i)
l ? LEFT (i)
r ? RIGHT(i)
if l heap-sizeA and Al gt Ai
then largest ? l
else largest ? i
if r heap-sizeA and ArgtAlargest
then largest ? r
if largest ? i
then exchange Ai with Alargest
MAX-HEAPIFY (A, largest)

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Of Note

The childrens subtrees each have size at most
2n/3 when the last row is exactly ½ full
Therefore, the running time is
T (n) T(2n/3) ?(1) O(lg n)

13
BUILD-HEAP

Use MAX-HEAPIFY in bottom up manner
Why does the loop start at lengthA/2?
At the start of each loop, each node i is the
root of a max-heap!
BUILD-HEAP (A)
heap-sizeA ? lengthA
for i ? lengthA/2 downto 1
do MAX-HEAPIFY(A, i)

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Analysis of Building a Heap

Since each call to MAX-HEAPIFY costs O(lg n) and
there are O(n) calls, this is O(n lg n)...
Can derive a tighter bound do all nodes take
log n time?
Has at most n/2h1 nodes at any height (the more
the height, the less nodes there are)
It takes O(h) time to insert a node of height h.

Thus, the running time is 2n O(n)

The number of nodes at height h
Multiplied by their height
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HEAPSORT

HEAPSORT (A)
BUILD-HEAP(A)
for i ? lengthA downto 2
do exchange A1 with Ai
heap-sizeA ? heap-sizeA - 1
MAX-HEAPIFY(A, 1)

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Swap A1 ? Ai
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heap-size ? heap-size 1 MAX-HEAPIFY(A, 1)
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Swap A1 ? Ai
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heap-size ? heap-size 1 MAX-HEAPIFY(A, 1)
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Swap A1 ? Ai
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heap-size ? heap-size 1 MAX-HEAPIFY(A, 1)
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Swap A1 ? Ai
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heap-size ? heap-size 1 MAX-HEAPIFY(A, 1)
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Swap A1 ? Ai
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heap-size ? heap-size 1 MAX-HEAPIFY(A, 1)
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Swap A1 ? Ai
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heap-size ? heap-size 1 MAX-HEAPIFY(A, 1)
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Swap A1 ? Ai
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heap-size ? heap-size 1 MAX-HEAPIFY(A, 1)
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Swap A1 ? Ai
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heap-size ? heap-size 1 MAX-HEAPIFY(A, 1)
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Swap A1 ? Ai
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heap-size ? heap-size 1 MAX-HEAPIFY(A, 1)
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Priority Queues