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Algorithms and data structures
24.5.2017.
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Algorithms and data structures, FER
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Heap
Heap
Heap sort
24.5.2017.
Basic terms

Priority queue is a data structure resembling in some aspects to the
ordinary queue
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The priority queue can be presented in different ways:


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Similarity to the ordinary queue - enqueue and dequeue operations are also
defined
Difference is in the fact that dequeued is not the element that was first
enqueued, but the one with the extreme value of some attribute (i.e. highest
priority)
As a sorted linked list
As a sorted binary tree
The most natural and most efficient way to present it is using the
structure heap
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Heap

Heap is a complete binary tree where the nodes can be mutually
compared using some ordering relation (e.g. >=) and where any
node satisfies this relation toward its children (if existent).
75
46
13
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Heap structure - implementation

n elements to form a heap are stored like a complete binary tree.
The simplest storage and access is achieved if an array is used. For
semantic reasons the node numbering shall proceed from 1 to n.
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It is achieved by starting with an empty tree, satisfying the heap properties,
and insertion of single members, while maintaining the heap properties
A new member is added to the “bottom” (leaf) of the heap and then it is
compared and swapped with its parent, grandparent and so on until its
compared attribute is found to be less or equal to the correposnding value
of its predecessor
 GomiluStvori (CreateHeap)
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Worst case analysis

For the worst case scenario let us take n elements
On the i-th level of a complete binary tree there are at most 2i-1 nodes
 On all the lower levels until then there are a total of 2i-1 - 1 nodes, for i > 1
 A tree with k levels contains at most 2k-1 nodes
 A tree with k-1 levels contains at most 2k-1-1 nodes

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If the tree is complete, the last level has started if
2k-1 - 1 < n ≤ 2k - 1

If follows:
2k-1 < n + 1  (k – 1) log 2 < log (n + 1)  k < log2 (n + 1) + 1
n + 1 ≤ 2k  log (n+1) ≤ k log 2  log2 (n+1) ≤ k
log2 (n+1) ≤ k < log2 (n + 1) + 1 or k = log2(n+1)
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Example
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For n = 14 there are required log2 15 = ln 15/ln 2 =
2.70805/0.693147 = 3.9 = 4 levels

For n = 15 there are required log2 16 = 4 = 4 levels
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For n = 16 there are required log2 17 = 4.087 = 5 levels
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Acceleration of the algorithm - I
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In the worst case, the while loop is repeated in proportion to the
number of levels in the heap
The worst case is caused by an ascending sequence of input data.
 In that case, every new element inserted using the function insert,
becomes the root, so that swapping is performed through k levels
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The execution time is then O(n log2 n)
For average data, the execution time to form a heap is O(n), what is
better for an order of magnitude
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Acceleration of the algorithm - II
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To accelerate the forming of heap, an algorithm was devised
starting from the leafs towards the root, level by level.
Only the root can violate the heap property, while the subtrees
preserve it.
Therefore, it is necessary only to correct this single violation and the
heap is restored.
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In the example, the function adjust performs this
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The heap property is satisfied for leafs; restoration of the heap property,
done in the function create_heap , is necessary only for the root
 GomiluPodesi (HeapAdjust)
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Example - heap creation
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Creation of the heap for the input sequence: 10,63,18,71,7,51,32
10
10
71
7
51
51
63
18
63
71
32
7
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51
63
51
71
7
32
71
10
63
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10
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Acceleration of the algorithm
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For n data, 2k-1  n < 2k , the number of levels is k = log2(n+1)
For the worst case, the number of iterations in adjust is k-i for a node at
level i where can be at most 2i-1 nodes
 Execution time for create_heap is:
k
The exponent changes from 0 to k -1,
while the factor changes from k -1 to 0
 2 k  i 
i 1
i 1

If the factor 0 is omitted and the summation order reversed, it follows:
k 1
i 2
i 1
k i 1
k 1
 2
k 1
i 1
i
 i  2  n  i  2n  O ( n)
i 1 2
i
The series sum
tends towards 2
2k-1  n
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k 1
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Acceleration of the algorithm


In the worst case, the execution time of the algorithm
create_heap is O(n), what is for an order of magnitude
better than O(n log2 n) for subsequent calls of insert
The function create_heap requires all the elements to be
known at creation time, while insert can introduce a new
element at any time


The reason why heap has been designed are the functions that heap
performs quickly: the insertion of new elements and removal of the largest
one from the data set.
Removal of the largest element is performed by deleting of the root and by
calling of the function adjust, while insertion of new elements is
performed with the function insert
– In this way both functions are performed in time O(log2 n)
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Variation of the algorithm
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A heap can be formed for different relationships among
their elements:
 The heap with relationship larger than is called
max heap
 The heap with relationship smaller than is called
min heap
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Heap sort
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Heap sort:
 Element from the heap root is swapped with the last element in
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the array
 The array length is decreased by 1, it is adjusted and becomes a
heap again
Adjustment complexity is O(log2n)
It is repeated n times so that the sort complexity is O(nlog2n)
Difference in execution times for different orders of magnitude in
complexity becomes significant for a large n
 Sortovi (Sorting)
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Exercises
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A sequence of input data is given:
12, 15, 5, 3, 7, 2 18, 11, 4, 10
Make a sketch of a complete binary tree created by consecutive
input of the given sequence of numbers
b) Make a sketch of the data structure heap containing the above
input data
c) What is the a priori execution time for transformation of a
complete binary tree into the structure heap?
a)
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Exercises

A sequence of input data of type int is given:
12, 5, 4, 10, 7, 8, 11
Make a sketch of the tree representing the structure heap, such
to enable the solution of the problem under b)
b) Illustrate the functioning of descending sort using the structure
heap (heap sort)
c) What is the a priori execution time for sorting of n records?
a)
Algorithms and data structures, FER
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Exercises

Write a program to find the k-th largest array member in an integer
array of n members.
Sort the input array in descending order and print the member
with index k-1 (or k if heap sort was used).
b) Input k array members, sort them in descending order. Start
reading the remaining array members. If some member is
smaller than the one with index k-1 (or k), ignore it, if it is
larger, insert it at the right place and throw out the member that
would now have the index k (or k+1).
a)
c)
Use different sorting methods and determine the corresponding
a priori execution times and measure the a posteriori execution
times.
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