ds and algorithms oct 25th and 26th classes

7/30/2019 DS and Algorithms Oct 25th and 26th Classes

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Data Structures

&

Algorithms

Prof. Ravi Prakash Gorthi

Aug-Dec, 2010


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Data Structures & Algorithms

Topics

Why Study DS & Algorithms?

Abstract Data Types

Arrays and Linked Lists

Stacks & Queues

Trees & Graphs

Sorting and Searching

Complexity of Algorithms


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Heapsort

Reference:

www.cis.upenn.edu/~matuszek/cit594-2002/Slides/heapsort.ppt


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A binary tree, which is, Balanced,

Left-justified and

Heapified

is useful to achieve,

Sorting and

Priority Queues


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Binary tree

Every node has at most two sons!

Binary

tree

Binary

tree

Not a binary

tree


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Balanced binary trees Depth of a Tree

The depth of a node is its distance from the root

The depth of a tree, n, is the depth of the deepest node

A binary tree of depth n (n > = 2) is balanced iff allthe nodes at depths 0, 1, 2 through n-2 have twochildren

Balanced Balanced Not balanced

n-2n-1n


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Balanced binary trees

In other words, you can add a son-node to a

node at level (n-1), iff, all nodes at the

previous level, (n2), have two sons

Balanced Balanced Not balanced

n-2n-1n


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A Complete Binary trees

A balanced binary tree of depth n iscomplete, iff, all nodes at level (n-1) have two

sons

A Complete

Binary Tree

Not a Complete

Binary Tree

n-2n-1n


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Left-justified balanced binary trees

A balanced binary tree of depth, n, is left-justified if: It is a Complete Binary Tree, OR

a node at level (n-1),

Has a left-son iff all nodes to its left at level (n-1) havetwo sons

Has a right-son iff it already has a left-son

Left-justified Not left-justified


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Let us construct a few Left-justified, balancedbinary trees:

0 0

1

0

1


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0 0

1

2

0

1 2

0

1


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0 0

1 2

0

1 2

0

1

3

2

0

1

3

2

0

1

33


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0 0

1 2

0

1 2

0

1

3

2

0

1

3 4

2

0

1

3 4


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0 0

1 2

0

1 2

0

1

3

2

0

1

3 4

2

0

1

3 4 5


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0 0

1 2

0

1 2

0

1

3

2

0

1

3 4

2

0

1

3 4 5

2

0

1

3 4 5 6


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Incrementally construct a left-justifiedbalanced binary tree using the following

sequence of integers:

31, 76, 29, 64, 52, 48, 13, 85, 90


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31, 76, 29, 64, 52, 48, 13, 85, 9031


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31, 76, 29, 64, 52, 48, 13, 85, 9031

76


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31, 76, 29, 64, 52, 48, 13, 85, 90

29

31

76


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31, 76, 29, 64, 52, 48, 13, 85, 90

29

31

76

64


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31, 76, 29, 64, 52, 48, 13, 85, 90

29

31

76

64 52


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31, 76, 29, 64, 52, 48, 13, 85, 90

29

31

76

64 52 48


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31, 76, 29, 64, 52, 48, 13, 85, 90

29

31

76

64 52 48 13


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31, 76, 29, 64, 52, 48, 13, 85, 90

29

31

76

64 52 48 13

85


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31, 76, 29, 64, 52, 48, 13, 85, 90

29

31

76

64 52 48 13

85 90


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Left-justified balanced binary trees Here is a very interesting coincidence:

29

31

76

64 52 48 13

85 90

2

0

1

3 4 5 6

7 8

0 1 2 3 4 5 6 7 8

31 76 29 64 52 48 13 85 90

Array Index

Unsorted List

Node Values


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The heap property

A node has the max heap property if the value inthe node is as large as or larger than the values in

its children

All leaf nodes automatically have the heap property

A binary tree is a heap ifallnodes in it have the

heap property

12

8 3

Blue node has

heap property

12

8 12

Blue node has

heap property

12

8 14

Blue node does not

have heap property


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siftUp

Given a node that does not have the heapproperty, you can give it the heap property by

exchanging its value with the value of the largest

child

This is sometimes called sifting up

14

8 12

Blue node hasheap property

12

8 14

Blue node does nothave heap property


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Constructing a heap I

A tree consisting of a single node is automatically

a heap

We construct a heap by adding nodes one at a

time:

Add nodes such that the binary tree remainsbalanced and left-justified

Examples:

Add a new

node here

Add a new

node here


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Constructing a heap II

Each time we add a node, we may destroy theheap property of its parent node;

To fix this, we sift up;

We repeat the sifting up process, moving up inthe tree, until either:

We reach nodes whose values dont need to

be swapped (because the parent is stilllarger than both children), or

We reach the root


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Constructing a heap III

8 8

10

10

8

10

8 5

10

8 5

12

10

12 5

8

12

10 5

8

1 2 3

4


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Other children are not affected

The node containing 8 is not affected because its parent gets

larger, not smaller

The node containing 5 is not affected because its parent getslarger, not smaller

12

10 5

8 14

12

14 5

8 10

14

12 5

8 10


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A sample heap

Heres a sample binary tree after it has beenheapified

Notice that heapified does notmean sorted

Heapifying does notchange the shape of the

binary tree; this binary tree is balanced and left-

justified because it started out that way

19

1418

22

321

14

119

15

25

1722


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Mapping into an array

Notice:

The left child of index i is at index 2*i+1

The right child of index i is at index 2*i+2

Example: the children of node 3 (19) are 7 (18) and 8 (14)

19

1418

22

321

14

119

15

25

1722

25 22 17 19 22 14 15 18 14 21 3 9 11

0 1 2 3 4 5 6 7 8 9 10 11 12


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Notice:




19

1418

22

321

14

119

15

25

1722

25 22 17 19 22 14 15 18 14 21 3 9 11

0 1 2 3 4 5 6 7 8 9 10 11 12


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Notice:




19

1418

22

321

14

119

15

25

1722

25 22 17 19 22 14 15 18 14 21 3 9 11

0 1 2 3 4 5 6 7 8 9 10 11 12


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Notice:




19

1418

22

321

14

119

15

25

1722

25 22 17 19 22 14 15 18 14 21 3 9 11

0 1 2 3 4 5 6 7 8 9 10 11 12


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Removing the root

Notice that the largest number is now in the root

Suppose we discardthe root:

How can we fix the binary tree so it is once again

balanced and left-justified?

Solution: remove the rightmost leaf at the

deepest level and use it for the new root

19

1418

22

321

14

119

15

1722

11

Th eHeap th d I


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The reHeap method I

Our tree is balanced and left-justified, but no longer a

heap However, only the rootlacks the heap property

We can siftUp() the root

After doing this, one and only one of its children may have

lost the heap property

19

1418

22

321

14

9

15

1722

11


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The reHeap method II

Now the left child of the root (still the number 11) lacksthe heap property

We can siftUp() this node


lost the heap property

19

1418

22

321

14

9

15

1711

22


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The reHeap method III Now the right child of the left child of the root (still the

number 11) lacks the heap property:



lost the heap property but it doesnt, because its a leaf

19

1418

11

321

14

9

15

1722

22


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The reHeap method III Now the right child of the left child of the root (still the

number 11) lacks the heap property:



lost the heap property but it doesnt, because its a leaf

19

1418

21

311

14

9

15

1722

22


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Removing and replacing the root

The root is the first element in the array

The rightmost node at the deepest level is the lastelement in the array

Swap them...

...And pretend that the last element in the array no longer

existsthat is, the last index is 11 (9)

25 22 17 19 22 14 15 18 14 21 3 9 11

0 1 2 3 4 5 6 7 8 9 10 11 12

11 22 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12


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Heapifying the Tree Again . . .

11 22 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 11 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12


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11 22 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 11 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 22 17 19 11 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

f h


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11 22 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 11 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 22 17 19 11 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

f h


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11 22 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 11 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 22 17 19 11 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 22 17 19 21 14 15 18 14 11 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

h d


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Reheap and repeat

Reheap the root node (index 0, containing 11)...

...And again, remove and replace the root node

Remember, though, that the last array index is changed

Repeat until the last becomes first, and the array is sorted!

22 22 17 19 21 14 15 18 14 11 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

11 22 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

9 22 17 19 21 14 15 18 14 11 3 22 25

0 1 2 3 4 5 6 7 8 9 10 11 12

R h d


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Reheap and repeat

Finally . . .

3 9 11 14 14 15 17 18 19 21 22 22 25

0 1 2 3 4 5 6 7 8 9 10 11 12

H S Al i h


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Heap Sort Algorithm

Input: Unsorted array of n integers:

Construct a balanced, left-justified, heapified binary tree from

the given array, by taking one node-value at a time, starting

from index 0 till the last index

18 22 15 22 21 11 17 19 14 25 3 9 14

0 1 2 3 4 5 6 7 8 9 10 11 12

25 22 17 19 22 14 15 18 14 21 3 9 11

0 1 2 3 4 5 6 7 8 9 10 11 12

H S Al i h


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Heap Sort Algorithm

Interchange node values at 0 and 12

25 22 17 19 22 14 15 18 14 21 3 9 11

0 1 2 3 4 5 6 7 8 9 10 11 12

11 22 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

Consider that the current tree has only (n-1) nodes and re-

heapify it by sifting down the node value 11 to the extent

required;

22 22 17 19 21 14 15 18 14 11 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

H if i th T A i


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11 22 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 11 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

H if i th T A i


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11 22 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 11 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 22 17 19 11 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

H if i th T A i


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11 22 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 11 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 22 17 19 11 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

H if i th T A i


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11 22 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 11 17 19 22 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 22 17 19 11 14 15 18 14 21 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 22 17 19 21 14 15 18 14 11 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

H S t Al ith


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Heap Sort Algorithm

Interchange node values at 0 and 11

Consider that the current tree has only (n-2) nodes and re-

heapify it by sifting down the node value 9 to the extent

required;

22 21 17 19 11 14 15 18 14 9 3 22 25

0 1 2 3 4 5 6 7 8 9 10 11 12

22 22 17 19 21 14 15 18 14 11 3 9 25

0 1 2 3 4 5 6 7 8 9 10 11 12

9 22 17 19 21 14 15 18 14 11 3 22 250 1 2 3 4 5 6 7 8 9 10 11 12

H S t Al ith


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Heap Sort Algorithm

Repeat the above step until only one node value isleft out . . .

3 9 11 14 14 15 17 18 19 21 22 22 25

0 1 2 3 4 5 6 7 8 9 10 11 12

ds and algorithms oct 25th and 26th classes

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