Slides on Data Structures tree and graph

8/13/2019 Slides on Data Structures tree and graph

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Non inear


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A brief introduction on Trees

Graphs

Review and Examine key properties of Trees and variants

Graph and variants

Discuss usage for solving important problems

(search, sort, selection).

Searching And Sorting Algorithms


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linear data structures

oA linear data structure traverses the data elements

sequentially, in which only one data element can directly

be reached. Ex: Arrays and Linked Lists,

Non-linear data structures

owhereas in non-Linear data structure every data item is

attached to several other data items in a way that is

specific for reflecting relationships. The data items are notarranged in a sequential structure. Ex: Trees, Graphs


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A tree is a collection of nodes and directed

edges, satisfying the following properties:

There is one specially designated node called the

root, which has no edges pointing to it. Every node except the roothas exactly one edge

pointing to it.

There is a uniquepath (of nodes and edges) from

the rootto each node.


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Node user-defined data structure that that containspointers to data and pointers to other nodes: RootNode from which all other nodes descend

Parenthas child nodes arranged in sub-trees.

Childnodes in a tree have 0 or more children.

Leafnode without descendants

Degree# of direct children a tree/sub-tree has.

In-degree:# of branches directed toward the node

Out-degree:# of branches directed away

Degree of node:. Sum (in-degree and out-degree)

Degree of tree: maximum degree of any node

Forest:Collection of trees

Path:sequence of adjacent nodes


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Height# of edges on the

longestpath from the root to a leaf.

LevelRoot is at level 0, its

direct children are at level 1, etc.

Recursive definition for height:1+ max(height(TL), height(TR))

A

B C

D E

GF

H


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If an edge goes from node ato node b, then ais called

the parentof b, and bis called a childof a.

Children of the same parent are called siblings.

If there is a path from a to b, then a is called anancestorof b, and bis called a descendentof a.

A node with all of its descendants is called a subtree.

If a node has no children, then it is called a leafof the

tree. If a node has no parent (there will be exactly one of

these), then it is the rootof the tree.


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A

B C

D E

F

G H

I

J K

A is the root D, E, G, H, J & K are

leaves

B is the parentof D, E &F

D, E & F are siblingsand childrenof B

I, J & K are descendantsof B

A & B are ancestorsof I


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An acyclic connected graph where each node has zero

or more children nodes and there is at most oneparent

node, and at most one node has no parents, which is

root node of the tree.

Figure : General Tree


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A balanced binary tree is commonly defined as a

binary tree in which the height of the two sub trees of

every node never differs by more than 1.

Figure : A Balanced TreeFigure : An Unbalanced Tree


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Heap is an ordered balanced binary tree in which the

value of the node at root of any subtree can be greater

than, less than or equal to the value of either of its

children.

Figure (a): Max-Heap Figure (b): Min-Heap


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A binary tree in which each node is having either

only left sub-tree, either only right sub-tree or no sub-

tree is called as left skewed or right skewed binary

tree respectively.

(a) Left Skewed binary tree

(b) Right Skewed Binary tree

Figure (a) (b)


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A binary tree is a tree in which each node can have

maximum of two children.Thus each node can have no, one or two child.

pointers are used to identify whether it is a left or a right child

Intuitively, a binary treeis a treein which each node

has no more than two children.

Figure : Binary Tree

(These two binarytrees are distinct.)


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A binary search tree is a binary tree in which each node, n,

has a Comparable key (and an associated value)satisfying the

following restriction on the key in any node :

nsvalue is > all values in its left subtree, TL,

nsvalue is < all values in its right sub-tree, TR, and TLand TRare both binary search trees.

John

PeterBrenda

Amy Mary Tom

21

2

3

55

34

135

8

In other words, can weput non-hierarchical

data into a tree. We

will study Binary

Search Trees later.


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A binary tree is ful l if it has no missing nodes.

It is either empty.

Otherwise, the rootssub-trees are fullbinary trees

of height h1.If not empty, each node has 2 children, except the

nodes at level hwhich have no children.

Contains a total of 2h+1-1 nodes (how many leaves?)

This term is

ambiguous, some

indicate that each

node is either full or

empty.


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A binary tree of height his completeif it is fulldown to

level h1, and level his filled from left to right.

All nodes at level h2 and above have 2 children each,

If a node at level h1 has children, all nodes to its left

at the same level have 2 children each, and

If a node at level h1 has 1 child, it is a left child.


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A binary tree is balancedif the difference in height betweenany nodesleft and right sub-tree is 1.

Note that:

A full binary tree is also complete. A complete binary tree is not always full. Full and complete binary trees are also balanced. Balanced binary trees are not always full or complete.


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Complete and Balanced

Not Balanced, Why?


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The number of edges in a tree is n-1. The number of nodes nin a full binary tree is: n = 2h + 1 1

where his the height of the tree. The number of nodes nin a complete binary tree is:

minimum: n = 2h

maximum: n = 2

h + 1

1where his the height of the tree. The number of nodes nin a full or perfect binary tree is:

n = 2L 1whereLis the number of leaf nodes in the tree.

The number of leaf nodes nin a full or perfect binary tree is: n = 2hwhere his the height of the tree.

The number of leaf nodes in a Complete Binary Tree with nnodes is UpperBound(n/ 2). For any non-empty binary tree with n0leaf nodes and n2nodes of

degree 2, n0= n2+ 1.


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Basic Idea:

Instead of using pointersto the left and right child of a

node, use indices into an array of nodes representing

the binary tree.

Also, use variable freeas an index to the first position

in the array that is available for a new entry. Use either

the left or r ight child indices to indicate additional,

available positions. Together, the list of available positions in the array is

called the f ree list.


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Index Item Left

Child

Right

Child0 Jane 1 2

1 Bob 3 4

2 Tom 5 -1

3 Alan -1 -1

4 Ellen -1 -1

5 Nancy -1 -1

6 ? -1 7

7 ? -1 8

8 ? -1 9

9 . . . . . . . . .

root0

free

6

Jane

Alan

TomBob

NancyEllen


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Index Item Left

Child

Right

Child

0 Jane 1 2

1 Bob 3 4

2 Tom 5 -1

3 Alan -1 -1

4 Ellen -1 -1

5 Nancy 6 -1

6 Mary -1 -1

7 ? -1 8

8 ? -1 9

9 . . . . . . . . .

root0

free

7

*Mary Added under Nancy.

Jane

Alan

TomBob

NancyEllen

Mary


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Index Item LeftChild

RightChild

0 Jane 1 2

1 Bob 3 -1

2 Tom 5 -13 Alan -1 -1

4 ? -1 7

5 Nancy 6 -1

6 Mary -1 -17 ? -1 8

8 ? -1 9

9 . . . . . . . . .

root0

free

4

*Ellen deleted.

Jane

Alan

TomBob

Nancy

Mary


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1

2 3

4 75 6

Let i, 1 < i< n, be the number assigned to an element of a complete binary tree.


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1

2 3

4 6

61 2 3

5 7


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Array-based representations allow for efficient traversal.Consider the node at index i.

Left Child is at index 2i+1.

Right Child is at index 2i+2.

Parent is at floor( (i-1)/2 ).


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1

3

7

1 3 7

Drawbackof array-based trees: Example has only 3nodes, but uses 2h+1-1 array cells to store it

However, if the array

is just numbers or

pointers, not a huge

concern.


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We can encode an n-ary tree as a binary tree, byhaving a linked-list of children. Hence still two

pointers, one to the first child and one to the next

sibling. Kinda rotates the tree.
http://en.wikipedia.org/wiki/File:N-ary_to_binary.svg


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Insertion In BST Deleting from a BST

Binary Search Tree Traversing

Depth-first Traversal

Preorder In-order

Post-order

Breadth-First Traversal

Level order

Constructing Tree for Given in-Order & Pre/Post Order Traversal of Tree

Searching and Sorting using Binary Search Tree


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Figure. Insert a new element whose value is 42

set x=42, p=36 and found =0;

While (p!=null and !found)

parent = p;

if (p->Info = x) then

Set found = 1;

Else if (x < p->Info) then

Set p = p->lchild;

Else

Set p = p->rchild;

[end of while loop at Step2]

1stiteration

Set p = 40 //p->rchild;

2nditeration

Set p = 44 //p->rchild;

3rditeration

if (42 < 44) then

Set p = 42 //p->lchild;


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Depth-first Traversal

Preorder

Inorder

Postorder

Breadth-First Traversal

Level order


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Basic Idea:

1) Visitthe root.

2) Recursively invoke preorderon the left sub-tree.3) Recursively invoke preorderon the right sub-tree.


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Basic Idea:

1) Recursively invoke inorderon the left sub-tree.

2)Visit

the root.3) Recursively invoke inorderon the right subtree.


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InorderResult: 10, 20, 30, 40, 50, 60, 70

60

4010

7020

30 50

6

4

3

1

2 7

5


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Basic Idea:

1) Recursively invoke postorderon the left sub-tree.

2) Recursively invokepostorder

on the right subtree.3) Visitthe root.


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PostorderResult: 10, 30, 50, 40, 20, 70, 60

60

4010

7020

30 50

7

4

2

1

5 6

3


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f

c j

a d h k

i

Visit the tree in left-to-right, by level, order: Visit the root node and put its children in a queue (left to right).

Dequeue, visit, and put dequeued nodes children into the queue.

Repeat until the queue is empty. Level order traversal of

the figure given below

fcjadhki


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Basic Ideafor a Nonrecursive,InorderTraversal:1) Push a pointer to the rootof the binary tree onto a stack.

2) Follow leftChild pointers, pushing each one onto the stack,

until aNULLleftChildpointer is found.

3) Process (visit) the item in this node.

4) Get the nodesrightChildpointer:

If it is notNULL, then push it onto the stack, and return to

step 2 with the leftChildpointer of this rightChild.

If it is NULL, then pop a node pointer from the stack, andreturn to step 3. If the stack is empty (so nothing could be

popped), then stopthe traversal is done.


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ExampleA binary tree T has 9 nodes. The inorder and preorder traversals of T yield thefollowing sequence of nodes

Inorder E A C K F H D B G

Preorder F A E K C D H G B

Draw the tree.

The tree T is drawn from its root downward in different stages as follows.

The root of T is obtained by choosing the first node in its preorder. Thus F isthe root of T

The inorder of T to find the nodes in the left subtree T of F. Thus T1 consists ofnodes E, A, C, K then. Then the left child of F is obtained by choosing the first

node in the preorder of T1. Thus A is the left child of F. Similarly, the right subtree T2 of F consists of the nodes H, D, B, and G. D is

the right child of F. Repeating the above process with each new node is givenbelow.


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Figure Construction of binary tree from inorder and preorder sequence

The postorder sequence of tree is visit left subtree, visit right subtree then visit rooted node.

E C K A H B G D F


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Tree finite, non-empty set of elements with a root with allremaining elements in sub-trees Real World applicationsBinary space partitioning trees (BSP)

efficient method for determining visibility relationships betweena static group of 3d polygons from an arbitrary viewpoint. Aviewpoint can be composed of a clusters of polygons. A planecan be found that separates on set of clusters from another thussome are visible and some arent. The tree is rooted at the firstpartitioning plane chose, and the internal nodes describe thepartitioning planes and the leaves describe the regions of space.VIDEO GAMESDOOM XXX, Quake X, etc

Compiler constructiondealing with syntax analysis, symboltable construction,arithmetic expression manipulation. You are a child of two parents, who in turn came from two

parents.Hierarchical!!!


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Discuss reasons forusing treesvs. using lists

Efficient retrieval

traversal of listvs. traversal of a tree.

Sorting

treevs. list


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Definition A graph G consists of two sets V and E.

o The set V is a finite, non-empty set of vertices.o The set E is a set of pairs of vertices, these pairs are called edges.

Graph G can be represent a set of G =(V,E)

A graph is nodesjoined by edge i.e. A set of nodes V anda set of edges E

A node is defined by its name or label. An edge is defined by the two nodes which it connects,

plus optionally:An order of the nodes (direction)A weight or cost (usually a number)


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Graph G= (V, E) V= set of vertices E= set of edges (VV)

Types of graphs Undirected:edge (u, v) = (v, u); for all v, (v, v) E(No

self loops.) Directed:(u, v) is edge from uto v, denoted as u v.

Self loops are allowed. Weighted: each edge hasan associated weight, given

by a weight function w : ER. Dense:|E| |V|2.(# of edges/#of nodes is large) Sparse:|E|


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Path:- (u1,u2,u3

v) Vand(u1,u2),(u2,u3),uk,v) Eare in G(E).Degreei. I n undirected graph

A nodesdegree is the number of its edges

i i . In directed graphI ndegree:-is the # of edges for which v is head. Outdegree:- is the # of edges for which v is tail.

length:the number of edges in the path

cost:the sum of the weights on each edge in the pathcycle:a path that starts and finishes at the same node An acyclicgraph contains no cycles


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Two nodes are adjacent if they are connected by anedge.

If (u, v) E, then vertex vis adjacentto vertex u.

Adjacency relationship is:

Symmetric if G is undirected. Not necessarily so, if Gis directed.

If Gis strongly connected:

There is a path between every pair of vertices.

|E| |V|1.

Furthermore, if |E| = |V| 1, then Gis a tree.


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For example :-(undirected graph) in the first Figure V(G)={1,2,3,4}

E(G)={(1,2),(2,4),(4,3),(3,1)}

For example :-( directed graph) See the second Figure. V(G)={1,2,3,4}

E(G)={(1,2),(2,1)(2,4)(4,2),(4,3)(3,4),(3,1)(1,3)}

Figure Undirected graph (unweighted)Figure Directed graph (unweighted)


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Two standard ways. Adjacency Lists.

Adjacency Matrix.

a

d

b

a

d

b1 2

3 4

1 2 3 41 0 1 1 1

2 1 0 1 03 1 1 0 14 1 0 1 0

abcd

b

a

d

d c

c

a b

a c


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Consists of an array Adjof |V| lists. One list per vertex.

For uV, Adj[u] consists of all vertices adjacent to u.

a

d

b abcd

b

c

d

d c

a

d

b

If weighted, store weightsalso in adjacency lists.

abcd

b

a

d

d c

c

a b

a c


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|V| |V| matrix A. Number vertices from 1 to |V| in some arbitrary manner.

Ais then given by:

otherwise0

),(if1],[

EjiajiA ij

a

d

b1 2

3 4

1 2 3 41 0 1 1 12 0 0 1 03 0 0 0 14 0 0 0 0

a

d

b1 2

3 4

1 2 3 41 0 1 1 12 1 0 1 03 1 1 0 14 1 0 1 0

A= ATfor undirected graphs.


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In the given graph G=(V,E), we want to visit all vertices in Gthat are reachable from vertex v, where v V. Problems with the Graph Traversal:

I. No First Node. So there must be a way to find out the starting nodeof the graph. User can enter the starting point or there can be severalother methods to find out the starting point.

II. When traversing a graph, we must be careful to avoid going round incircles.We do this by marking the vertices which have already beenvisited. List of visited nodes can also be maintained.

III. No natural order among the successor of a particular node.IV. Nodes which are not connected or to which there is no path.

We have two common search methods:1. Breadth first search2. Depth first search


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Breadth First Search It is based on FIFO system using a Queue . Depth First Search It is based on LIFO system using a Stack .

Figure Undirected graph, for finding the breadth first search and depth first search


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The steps of over search follow as start node 11. initially, add 1 to queue as follows

Front=1 Rear =1 Queue:12. Delete front 1 and add adjacent nodes of 1 which

is not visited.

Front= 2 Rear=5 Queue:2, 3, 4, 5

3. Delete front 2 and add adjacent nodes of 2 which

is not visited.

Front=3 Rear=6 Queue:3, 4, 5, 6


is not visited.

Front=4 Rear=7 Queue:4, 5, 6, 7


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is not visited.

Front=5 Rear=8 Queue:5, 6, 7, 86. Delete front 5 from queue and add to queue

adjacent nodes of 5 which is not visited.

Front=6 Rear=8 Queue:6, 7, 8

7. Delete front 6 from queue and add to queue


Front=7 Rear=9 Queue:7, 8, 9



Front=8 Rear=9 Queue:8, 9


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Front=9 Queue:9

Rear=9



Front=0 Queue:0

Rear=0Breadth first search is 1, 2, 3, 4, 5, 6, 7, 8, 9


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The steps of over search follow as start node 11 Initially, push 1 onto stack as follows

Stack: 1

2 Pop top 1 and push all adjacent nodes of 1 which is notvisited .

Stack: 2, 3, 4, 5

3 Pop top 5 and push all adjacent nodes of 5 which is not

visited.

Stack : 2, 3, 4

4 pop top 4 and push all adjacent nodes of 4 which is notvisited.

Stack : 2, 3, 7, 8

5 pop top 8 and push all adjacent nodes of 8 which is not

visited.

Stack : 2,3,7


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6 pop top 7 and push all adjacent nodes of 7 which is not

visited.

Stack : 2,3,6

7 pop top 6 and push all adjacent nodes of 6 which is notvisited.

Stack : 2,3,98 pop top 9 and push all adjacent nodes of 9 which is not

visited.

Stack : 2,39 pop top 3 and push all adjacent nodes of 3 which is not

visited.

Stack : 210 pop top 2 and push all adjacent nodes of 2 which is not

visited.

Stack : empty

Depth first search sequence is: 1, 5, 4, 8, 7, 6, 9, 3, 2


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Electrical network problems Electrical networkanalysis and synthesis are mainly the study of

network topology

Topological sort A topological sort of a graph can

be viewed as an orderings of its vertices along a

horizontal lineso that all directed edges go from left

to right.

Directed acyclic graphs are used in manyapplications to indicate precedence among events

such as code optimization techniques of complier.


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Shortest path A path from source vertex vto w is shortest path if there is shortest path

from v to u and u to w with lower costs. The

shortest paths are not necessarily unique.The most common shortest path problem:-

Traveling salesman.

Airline, The shortest path find which routeprovide minimum air fair total time of flights.


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Minimum spanning tree (MST) A spanning treefor a graph, G = (V, E), is a sub-graph of G that is a

tree and contains all the vertices of G.

In a weighted graph, the weight of a graph is the sum

of the weights of the edges of the graph. A MST for a

weighted/cost graph is a spanning tree with minimum

cost.

There are many applications where minimumspanning tree is needed


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To find cheapest way to connect a set of terminals,cities, electrical, electronic components of a circuit,

computers, or premises by using roads, wires or

wireless, or telephone lines.

The solution is a MST, which has an edge for eachpossible connection weighted by the cost of that

connection.

The routing problems to find path among the system

over Internet, also need MST.

Prims and Kruskals algorithm is the most common

solution for MST.


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Searching AlgorithmsSequential SearchBinary Search

Sorting AlgorithmsBubble SortInsertion SortQuick Sort


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ExampleSuppose we have an array of integers.

Lets search for the number 3. We start at from

beginning and check the first element in the array. Is it3?


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ExampleSuppose the following

array of integers is given.

2 6 7 34 76 123 234 567 677 986

We want to seek the

value 123 from this array.


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Suppose the array to be sorted is : 6 1 4 3 9 First Pass


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Suppose the array to be sorted is : 6 1 4 3 9 First Pass 6 4 3 9 -> 64 3 9


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Suppose the array to be sorted is : 6 1 4 3 9 First Pass 6 4 3 9 -> 64 3 9 1 6 4 3 9 -> 1 4 63 9


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Suppose the array to be sorted is : 6 1 4 3 9 First Pass 6 4 3 9 -> 64 3 9 1 6 4 3 9 -> 1 4 63 9 1 4 6 39 -> 1 4 3 69 1 4 3 6 9-> 1 4 3 6 9


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Suppose the array to be sorted is : 6 1 4 3 9 First Pass 6 4 3 9 -> 64 3 9 1 6 4 3 9 -> 1 4 63 9 1 4 6 39 -> 1 4 3 69 1 4 3 6 9-> 1 4 3 6 9 Second Pass


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Suppose the array to be sorted is : 6 1 4 3 9 First Pass 6 4 3 9 -> 64 3 9 1 6 4 3 9 -> 1 4 63 9 1 4 6 39 -> 1 4 3 69 1 4 3 6 9-> 1 4 3 6 9 Second Pass 43 6 9 -> 43 6 9 *No swap takes place


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Suppose the array to be sorted is : 6 1 4 3 9 First Pass 6 4 3 9 -> 64 3 9 1 6 4 3 9 -> 1 4 63 9 1 4 6 39 -> 1 4 3 69 1 4 3 6 9-> 1 4 3 6 9 Second Pass 43 6 9 -> 43 6 9 *No swap takes place 1 4 36 9 -> 1 3 46 9


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Suppose the array to be sorted is : 6 1 4 3 9 First Pass 6 4 3 9 -> 64 3 9 1 6 4 3 9 -> 1 4 63 9

1 4 6 39 -> 1 4 3 69 1 4 3 6 9-> 1 4 3 6 9 Second Pass 43 6 9 -> 43 6 9 *No swap takes place

1 4 36 9 -> 1 3 46 9 1 3 4 69 -> 1 3 4 69 1 3 4 6 9-> 1 3 4 6 9


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Third Pass 34 6 9 -> 34 6 9 *No swap takes place 1 3 46 9 -> 1 3 46 9 1 3 4 69 -> 1 3 4 69 1 3 4 6 9-> 1 3 4 6 9


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Given the array: 6 8 1 4 5 3 7 2 - and the goal is toput it into ascending order.


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6 8 1 4 5 3 7 2 (Consider index 0) 6 8 1 4 5 3 7 2 (Consider indices 0 - 1) 6 84 5 3 7 2 (Consider indices 0 - 2)


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6 8 1 4 5 3 7 2 (Consider index 0) 6 8 1 4 5 3 7 2 (Consider indices 0 - 1) 6 84 5 3 7 2 (Consider indices 0 - 2) 4 6 85 3 7 2 (Consider indices 0 -3)


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6 8 1 4 5 3 7 2 (Consider index 0) 6 8 1 4 5 3 7 2 (Consider indices 0 - 1) 6 84 5 3 7 2 (Consider indices 0 - 2) 4 6 85 3 7 2 (Consider indices 0 -3) 4 5 6 83 7 2 (Consider indices 0 -4)


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6 8 1 4 5 3 7 2 (Consider index 0) 6 8 1 4 5 3 7 2 (Consider indices 0 - 1) 6 84 5 3 7 2 (Consider indices 0 - 2) 4 6 85 3 7 2 (Consider indices 0 -3) 4 5 6 83 7 2 (Consider indices 0 -4) 3 4 5 6 7 82 (Consider indices 0 -5)


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6 8 1 4 5 3 7 2 (Consider index 0) 6 8 1 4 5 3 7 2 (Consider indices 0 - 1) 6 84 5 3 7 2 (Consider indices 0 - 2) 4 6 85 3 7 2 (Consider indices 0 -3) 4 5 6 83 7 2 (Consider indices 0 -4) 3 4 5 6 7 82 (Consider indices 0 -5) 2 3 4 5 6 7 8( the array is sorted!)


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