Chapter 6: Priority Queues SCSC 321. Priority Queues The model Implementations Binary heaps ...
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Transcript of Chapter 6: Priority Queues SCSC 321. Priority Queues The model Implementations Binary heaps ...
Chapter 6: Chapter 6: Priority QueuesPriority Queues
SCSC 321
Priority Queues
The modelImplementationsBinary heapsBasic operations
2
The ModelA priority queue is a queue where:
Requests are inserted in the order of arrival
The request with highest priority is processed first (deleted from the queue)
The priority is indicated by a number, the lower the number - the higher the priority.
3
Implementations
4
Linked list:
- Insert at the beginning - O(1)
- Find the minimum - O(N)
Binary search tree (BST):
- Insert O(logN)
- Find minimum - O(logN)
Implementations
5
Binary heap
Better than BST because it does not support links
- Insert O(logN)
- Find minimum O(logN)
Deleting the minimal element takes a constant time, however after that the heap structure has to be adjusted, and this requires O(logN) time.
Binary Heap
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Heap-Structure Property:
Complete Binary Tree - Each node has two children, except for the last two levels.
The nodes at the last level do not have children. New nodes are inserted at the last level from left to right.
Heap-Order Property:
Each node has a higher priority than its children
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Binary Heap
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1012
15 17 18 23
20 19 34
Next node to be inserted - right child of the yellow node
Binary heap implementation with an array
8
Root - A(1)
Left Child of A(i) - A(2i)
Right child of A(i) - A(2i+1)
Parent of A(I) - A([i/2]).
The smallest element is always at the root, the access time to the element with highest priority is constant O(1).
Example
9
6 10 12 15 17 18 23 20 19 34
Consider 17:
position in the array - 5.
parent 10 is at position [5/2] = 2
left child is at position 5*2 = 10 (this is 34)
right child - position 2*5 + 1 = 11 (empty.)
Problems
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Problem 1:
2 8 10 16 17 18 23 20 21 30
Reconstruct the binary heap
Problems
11
Problem 2: Give the array representation for
3
1016
13 12 18 23
15 19 30 14
Basic Operations
Insert a node – Percolate UpDelete a node – Percolate DownDecrease key, Increase key, Remove keyBuild the heap
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Percolate Up – Insert a Node
13
A hole is created at the bottom of the tree, in the next available position.
6
1012
15 17 18 23
20 19 34
Percolate Up
14
Insert 206
1012
15 17 18 23
21 19 34 20
Percolate Up
15
Insert 16
6
1012
15 18 23
21 19 34 17
Percolate Up
16
6
1012
15 18 23
21 19 34 17
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Complexity of insertion: O(logN)
Percolate Down – Delete a Node
17
1016
13 12 18 23
15 19 30 34
Percolate Down – the wrong way
18
10
16
13 12 18 23
15 19 30 34
Percolate Down – the wrong way
19
16
13 18 23
15 19 34
10
12
30
The empty hole violates the heap-structure property
Percolate Down
20
Last element - 20. The hole at the root.
1016
12 13 18 23
15 19 30 20
We try to insert 20 in the hole by percolating the hole down
Percolate Down
21
16
12 13 18 23
15 19 3020
10
Percolate Down
22
16
13 18 23
15 19 3020
10
12
Percolate Down
23
16
13 18 23
20 19 30
10
12
15
Complexity of deletion: O(logN)
Other Heap Operations
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1. DecreaseKey(p,d)
increase the priority of element p in the heap with a positive value d. percolate up.
2. IncreaseKey(p,d)
decrease the priority of element p in the heap with a positive value d. percolate down.
Other Heap Operations
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3. Remove(p) a. Assigning the highest priority to p - percolate p up to the root.
b. Deleting the element in the root and filling the hole by percolating down and trying to insert the last element in the queue.
4. BuildHeap input N elements
place them into an empty heap through successive inserts. The worst case running time is O(NlogN).
Build Heap - O(N)
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Given an array of elements to be inserted in the heap,
treat the array as a heap with order property violated,
and then do operations to fix the order property.
Example:150 80 40 30 10 70 110 100 20 90 60 50 120 140 130
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150
80 40
30 1070 110
100 20 90 60 50 120 140 130
Example (cont)
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150
80 40
20 1050 110
100 30 90 60 70 120 140 130
After processing height 1
Example (cont)
29
150
10 40
20 6050 110
100 30 90 80 70 120 140 130
After processing height 2
Example (cont)
30
10
20 40
30 6050 110
100 150 90 80 70 120 140 130
After processing height 3
Theorem
31
For a perfect binary tree of height h containing N = 2 h+1 - 1 nodes,
the sum S of the heights of the nodes is S = 2 h+1 - 1 - (h + 1) = O(N)
Proof
32
The tree has 1 node at height h, 2 nodes at height h-1, 4 nodes at height h -2, etc
1 (20) h2 (21) h - 14 (22) h - 28 (23) h - 3…….2 h 0
S = 2i (h - i), i = 0 to h
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S = h + 2(h - 1) + 4(h - 2) + 8 (h -
3) + …. 2(h-2).2 + 2(h-1).1 (1)
2S = 2h + 4(h - 1) + 8(h - 2) + 16 (h - 3)
+ …. 2(h-1).2 + 2(h).1 (2)
Subtract (1) from (2):
S = h -2h + 2 + 4 + 8 + …. 2 h = - h - 1 + (2 h+1 - 1) = (2 h+1 - 1) - (h + 1)
Proof (cont.)
34
Note: 1 + 2 + 4 + 8 + …. 2 h = (2 h+1 - 1)
Hence S = (2 h+1 - 1) - (h + 1)
Hence the complexity of building a heap with N nodes is linear O(N)