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Heaps
Heaps
Properties
Deletion, Insertion, Construction
Implementation of the Heap
Implementation of Priority Queue
using a Heap
An application: HeapSort
2
Heaps (Min-heap)Complete binary tree that stores a collection of
keys
(or key-element pairs) at its internal nodes and that
satisfies
the additional property:
key(parent) key(child)4
6
207
811
5
9
1214
15
2516
REMEMBER:
complete binary tree
all levels are full, except the
last one, which is left-filled
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Max-heap
key(parent) key(child)
40
26
2017
811
35
19
1214
15
131
4
We store the keys in the internal nodes only
5
915
16
After adding the leaves the resulting tree is full
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Height of a Heap
Theorem: A heap storing nkeys has height O(log n)
Proof:
Let hbe the height of a heap storing nkeys
Since there are 2ikeys at depth i= 0, , h- 2 and at least
one
key at depth h- 1, we have n 1 + 2 + 4 + + 2h-2 + 1
Thus, n 2h-1 , i.e., h log n+ 1
1
2
2h-2
at least 1
keys
0
1
h-2
h-1
depth
h6
We could use a heap to implement a priority queue
We store a (key, element) item at each internal
node
(2, Sue)
(6, Mark)(5, Pat)
(9, Jeff) (7, Anna)
removeMin():
Remove the root
Re-arrange the heap!
Notice that .
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Removal From a Heap
The removal of thetop key leaves a hole
We need to fix theheap
First, replace the holewith the last key inthe heap
Then, begin Downheap
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Downheap
Downheap compares the parent with the smallest
child. If the child is smaller, it switches the two.
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Downheap Continues
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Downheap Continues
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End of Downheap
Downheap terminates when the key is greater than thekeys of both
its children or the bottom of the heap isreached.
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Heap Insertion
The key to insert is 6
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Heap Insertion
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Upheap
Swap parent-child keys out of order
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Upheap Continues
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End of Upheap
Upheap terminates when new key is greater than the keyof its
parent or the top of the heap is reached
total #swa s h - 1 , which is O lo n
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Heap Construction
We could insert the Items one at the time with
a sequence of Heap Insertions:
log k = O(n log n)k=1
n
But we can do better .
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We can construct a heap
storing ngiven keys using
a bottom-up construction
Bottom-up Heap
Construction
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Construction of a Heap
Idea: Recursively re-arrange each sub-
tree in the heap starting with the leaves
begin here
HEAP
HEAPHEAP
1st2nd3rd
4th
5th6th
begin here
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Example 1 (Max-Heap)
3 6
2
4 5
10 87 3
1 9 6
6
3
2
4 5
10 87 6
1 9 3
2
4 5
10 89 6
1 7 3
7 9 5 10
I am now drawing theleaves anymore here
2
4 10
5 89 6
1 7 3
--- keys already in the tree ---
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Let j = h-i, then i = h-j and
(h i)2i = j2h-j = 2h j2-j
Consider j2-j : j2-j = 1/2 + 2 1/4 + 3 1/8 + 4 1/16 +
= 1/2 + 1/4 + 1/8 + 1/16 +
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2i > nifTRUELeaf? T[i]
T 0ifT[1]The Root
i > 1ifT[i div 2]Parent of T[i]
2i + 1 nifT[2i+1]Right child of
T[i]
2i nifT[2i]Left child of T[i]
n = 111
2
4 5 6 7
8 9 10 11
I
3
Reminder ..
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Implementation of a Priority
Queue with a Heap
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O(log n)
O(log n)O(log 1)
(remove root + downheap)
(upheap)
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Algorithm PriorityQueueSort(S, P):
Input: A sequence S storing n elements, on which a
total order relation is defined, and a Priority
Queue P that compares keys with the same relation
Output: The Sequence S sorted by the total
order relation
while S.isEmpty() do
e S.removeFirst()
P.insertItem(e, e)
while P.isEmpty() doe P.removeMin()
S.insertLast(e)
Application: Sorting Heap Sort
PriorityQueueSort where the PQ is implemented with a HEAP
Build Heap
Remove from heap
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Application: Sorting
Heap Sort
Construct initial heap O(n)
remove root O(1)
re-arrange O(log n)
remove root O(1)
re-arrange O(log (n-1))
L M
L
n
times
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When there are i nodes left in the PQ: log i
TOT =i=1
n
log i
= O(n log n)
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HeapSort in Place
Example on the board .
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How about implementing a double PQ ??
removeMin() AND removeMax()
With a Min-heap:
removeMin() is O(log n)
removeMax() is ?
With a Max-Heap:
removeMin() is ?
removeMax() is O(log n)
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