February 2010

Master of Computer Application (MCA) – Semester 2

MC0068 – Data Structures using C

Assignment Set – 1

1. Write a program in C using one dimensional arrays to sort a given list of n

numbers using any of the sorting techniques.Ans –

#include<conio.h>

#include<stdio.h>

void main()

{ int i,j,a[10],n,temp;

clrscr();

scanf("NO. of elements: %d ",&n);

for (i=0;i<n;i++) scanf(“%d”,&a[i]);

clrscr();

printf("You array...\n");

for (i=0;i<n;i++) printf(&a[i]," ");

for (i=0;i<n;i++)

{ for (j=0;j<(n-1)-i;j++)

if (a[j]>a[j+1])

{ temp=a[j];

a[j]=a[j+1];

a[j+1]=temp;

}

}

printf("\n\nSorted Array...\n");

for (i=0;i<n;i++) cout<<a[i]<<" ";

getch();

}

2. Demonstrate with your own programming example the usage of

structures within an array.Ans –

#include<conio.h>

#include<stdio.h>

void main()

{ struct student

{ int roll;

Char name[20];

}

} std[5]; //declaring an array of structure

For (int i=0;i<5;i++)

{ scanf(“enter roll no: %d”,&std[i].roll);

Scanf(“enter name: %c”,&std[i].name);

}

}

3. Explain the theory of non linear data structures.Ans –

Trees

we consider one of the most Important non-linear Information structures- trees. A tree Is often used to represent a hierarchy. This is because the relationships between the Items In the hierarchy suggest the branches of a botanical tree.

For example, a tree-like organization charts often used to represent the lines of responsibility in a business as shown in Figure. The president of the company is shown at the top of the tree and the vice-presidents are indicated below her. Under the vice-presidents we find the managers and below the managers the rest of the clerks. Each clerk reports to a manager. Each manager reports to a vice-president, and each vice-president reports to the president.

It just takes a little imagination to see the tree in Figure. Of course. The tree is upside-down. However, this is the usual way the data structure is drawn. The president is called the root of the tree and the clerks are the leaves.

A tree is extremely useful for certain kinds of computations. For example. Suppose we wish to determine the total salaries paid to employees by division or by department. The total of the salaries in division A can be found by computing the sum of the salaries paid in departments Al and A2 plus the salary of the vice-president of division A. Similarly. The total of the salaries paid in department Al is the sum of the salaries of the manager of department Al and of the two clerks below her.

Clearly, in order to compute all the totals. It is necessary to consider the salary of every employee. Therefore, an implementation of this computation must visit all the employees in the tree. An algorithm that systematically visits all the items in a tree is called a tree traversal.

In the same chapter we consider several different kinds of trees as well as several different tree traversal algorithms. In addition. We show how trees can be used to represent arithmetic expressions and how we can evaluate an arithmetic expression by doing a tree traversal. The following is a mathematical definition of a tree:

Definition (Tree) A tree T is a finite. Non-empty set of nodes ,

T = {r} U TI, U T2 U …U Tn with the following properties:

3. A designated node of the set, r, is called the root of the tree: and

4. The remaining nodes are partitioned into n≥ O subsets T, T. …Tn each of which is a tree for convenience, we shall use the notation T= {r. T, T, …T} denote the tree T.

Notice that Definition is recursive-a tree is defined in terms of itself! Fortunately, we do not have a problem with infinite recursion because every tree has a finite number of nodes and because in the base case a tree has n=0 subtrees.

It follows from Definition that the minimal tree is a tree comprised of a single root node. For example Ta = {A}.

Finally. The following Tb = {B, {C}} is also a tree

Ta = {D, {E. {F}}, {G.{H,II}}, {J, {K}. {L}}, {M}}}

How do Ta Tb. & Tc resemble their arboreal namesake? The similarity becomes apparent when we consider the graphical representation of these trees shown in Figure. To draw such a pictorial representation of a tree, T = {r. T1 ,T2, …Tn, beside each other below the root. Finally, lines are drawn from rto the roots of each of the subtrees. T1T2…….Tn

Figure : Examples of trees.

Of course, trees drawn in this fashion are upside down. Nevertheless, this is the conventional way in which tree data structures are drawn. In fact, it is understood that when we speak of “up” and “down,” we do so with respect to this pictorial representation. For example, when we move from a root to a subtree, we will say that we are moving down the tree.

The inverted pictorial representation of trees is probably due to the way that genealogical lineal charts are drawn. A lineal chart is a family tree that shows the descendants of some person. And it is from genealogy that much of the terminology associated with tree data structures is taken.

Figure shows one representation of the tree Tc defined in Equation. In this case, the tree is represented as a set of nested regions in the plane. In fact, what we have is a Venn diagram which corresponds to the view that a tree is a set of sets.

Figure: An alternate graphical representation for trees.

Binary Tree

Used to implement lists whose elements have a natural order (e.g. numbers) and either (a) the application would like the list kept in this order or (b) the order of elements is irrelevant to the application (e.g. this list is implementing a set).

Each element in a binary tree is stored in a "node" class (or struct). Each node contains pointers to a left child node and a right child node. In some implementations, it may also contain a pointer to the parent node. A tree may also have an object of a second "tree" class (or struct) which as a header for the tree. The "tree" object contains a pointer to the root of the tree (the node with no parent) and whatever other information the programmer wants to squirrel away in it (e.g. number of nodes currently in the tree).

In a binary tree, elements are kept sorted in left to right order across the tree. That is if N is a node, then the value stored in N must be larger than the value stored in left-

child(N) and less than the value stored in right-child(N). Variant trees may have the opposite order (smaller values to the right rather than to the left) or may allow two different nodes to contain equal values.

Hash Tables

A very common paradigm in data processing involves storing information in a table and then later retrieving the information stored there. For example, consider a database of driver’s license records. The database contains one record for each driver’s license issued. Given a driver’s license number. we can look up the information associated with that number. Similar operations are done by the C compiler. The compiler uses a symbol table to keep track of the user-defined symbols in a Java program. As it compiles a program, the compiler inserts an entry in the symbol table every time a new symbol is declared. In addition, every time a symbol is used, the compiler looks up the attributes associated with that symbol to see that it is being used correctly.

Typically the database comprises a collection of key-and-value pairs. Information is retrieved from the database by searching for a given key. In the case of the driver’~ license database, the key is the driver’s license number and in the case of the symbol table, the key is the name of the symbol.

In general, an application may perform a large number of insertion and/ or look-up operations. Occasionally it is also necessary to remove items from the database. Because a large number of operations will be done we want to do them as quickly as possible.

Hash tables are a very practical way to maintain a dictionary. As with bucket sort, it assumes we know that the distribution of keys is fairly well-behaved.

Once you have its index. A hash function is a mathematical function which maps keys to integers.

In bucket sort, our hash function mapped the key to a bucket based on the first letters of the key. "Collisions" were the set of keys mapped to the same bucket. If the keys were uniformly distributed. then each bucket contains very few keys!

The resulting short lists were easily sorted, and could just as easily be searched

We examine data structures which are designed specifically with the objective of providing efficient insertion and find operations. In order to meet the design objective certain concessions are made. Specifically, we do not require that there be any specific ordering of the items in the container. In addition, while we still require the ability to remove items from the container, it is not our primary objective to make removal as efficient as the insertion and find operations.

Ideally we would’ build a data structure for which both the insertion and find operations are 0(1) in the worst case. However, this kind of performance can only be achieved with complete a priori knowledge. We need to know beforehand specifically which items are to be inserted into the container. Unfortunately, we do not have this information in the general case. So, if we cannot guarantee 0(1) performance in the worst case, then we make it our design objective to achieve 0(1) performance in the average case.

The constant time performance objective immediately leads us to the following conclusion: Our implementation must be based in some way Kh element of an array in constant time, whereas the same operation in a linked list takes O{k) time.

In the previous section, we consider two searchable containers-the ordered list and the sorted list. In the case of an ordered list, the cost of an insertion is 0(1) and the cost of the find operation is O(n). For a sorted list the cost of insertion is O(n) and the cost of the find operation is O(log n) for the array implementation.

Clearly, neither the ordered list nor the sorted list meets our performance objectives. The essential problem is that a search, either linear or binary, is always necessary. In the ordered list, the find operation uses a linear search to locate the item. In the sorted list, a binary search can be used to locate the item because the data is sorted. However, in order to keep the data sorted, insertion becomes O(n).

In order to meet the performance objective of constant time insert and find operations. we need a way to do them without performing a search. That is, given an item x, we need to be able to determine directly from x the array position where it is to be stored.

4. Write a program in C showing the implementation of stack operations

using arrays.Ans –

#include<stdio.h> #include<conio.h>#define Max 5 int a [Max] , top=-l;- void push(){ int ele;char ch; if(top= =-l)top=0;do{ if(top>=5){ printf("Stack if full");

break;else{ clrscr(); printf ( "Enter element to be insertedn" ) ; scanf(“%d”,&ele); a[top++] =ele; } printf ( "Do you want to add more elements: ?n") ; scanf ( "n%c" , &ch);printf(“%c”,ch)}while((ch= = ‘Y’)||(ch==’Y’));}void pop( ) { if(top= =-l) {printf ( "stack is underflown");}else { for(int i=top-l;i>=0;i–) printf ("%dn", a [i] ) ; }} void main() {clrscr ( ) ;char c;int choice; do{ clrscr( ); printf ("Enter Your Choicen");printf("l -> Pushn");printf ("2 -> Popn");scanf ("%d", &Choice);if(choice= =l) push( ); else if (choice= = 2)pop( );elseprintf(“invalid choice”);printf(“Do You Want to continuen”);scanf(“n%c”,&c);}while((c= = ‘y’)||(c= = ‘Y’));}

5. Write the code blocks in C to perform the following operations on Circular

Linked lists:

A) Insertion of a node at the front

B) Deletion of a node at the front

Ans –

A) Insertion of a node at the front end

NODE insert_ front (int item, NODE last){ NODE temp; temp = getnode( ); /* Create a new node to be inserted */temp->info = item; if (last = = NULL) /* Make temp as the first node */last = temp; else /* Insert at the front end */temp->link = last->link; last->link = temp; /* link last node to first node */ return last; /* Return the last node */}

B) Deletion of a node at the front end

NODE delete_ front(NODE last){ NODE temp, first; if ( last = = NULL ){ printf("List is emptyn");return NULL; } if ( last->link = = last) /* Only one node is present */{ printf("The item deleted is %dn", last->info);freenode (last);return NULL;} /* List contains more than one node */first = last->link; /* obtain node to be deleted */ last->link = last->link; /*Store new first node in link of last */ printf ("The item deleted is %dn", first->info); freenode (first); /* delete the old first node */

return last;} 6. With the help of a suitable numerical example, explain the inorder,

preorder and postorder traversals on a Binary Search Tree.Ans –

With the linear, contiguous ordering of an array’s elements, iterating through an array is a straightforward manner: start at the first array element, and step through each element sequentially. For BSTs there are three different kinds of traversals that are commonly used:

· Preorder traversal

· Inorder traversal

· Postorder traversal

Essentially, all three traversals work in roughly the same manner. They start at the root and visit that node and its children. The difference among these three traversal methods is the order with which they visit the node itself versus visiting its children.

Preorder Traversal

Preorder traversal starts by visiting the current node – call it c, then its left child, and then its right child. Starting with the BST’s root as c, this algorithm can be written out as:

Step 1: Visit c. This might mean printing out the value of the node, adding the node to a List, or something else. It depends on what you want to accomplish by traversing the BST.

Step 2: Repeat step 1 using c’s left child.

Step 3: Repeat step 1 using c’s right child.

Imagine that in step 1 of the algorithm we were printing out the value of c. In this case, what would the output be for a preorder traversal of the BST in Figure 1.8? Well, starting with step 1 we would print the root’s value. Step 2 would have us repeat step 1 with the root’s left child, so we’d print 50. Step 2 would have us repeat step 1 with the root’s left child’s left child, so we’d print 20. This would repeat all the way down the left side of the tree. When 5 were reached, we’d first print out its value (step 1). Since there are no left or right children of 5, we’d return to node 20, and perform its step 3, which is repeating step 1 with 20’s right child, or 25. Since 25 has no children, we’d return to 20, but we’ve done all three steps for 20, so we’d return to 50, and then take on step 3 for node 50, which is repeating step 1 for node 50’s right child. This process would continue on until each node in the BST had been visited.

Inorder Traversal

Inorder traversal starts by visiting the current node’s left child, then the current node, and then its right child. Starting with the BST’s root as c, this algorithm can be written out as:

Step 1: Repeat step 1 using c’s left child.

Step 2: Visit c. This might mean printing out the value of the node, adding the node to an ArrayList, or something else. It depends on what you want to accomplish by traversing the BST.

Step 3: Repeat step 1 using c’s right child.

Postorder Traversal

Finally, postorder traversal starts by visiting the current node’s left child, then its right child, and finally the current node itself. Starting with the BST’s root as c, this algorithm can be written out as:

Step 1: Repeat step 1 using c’s left child.

Step 2: Repeat step 1 using c’s right child.

Step 3: Visit c.

This might mean printing out the value of the node, adding the node to an ArrayList, or something else. It depends on what you want to accomplish by traversing the BST.

Realize that all three traversal times exhibit linear asymptotic running time. This is

because each traversal option visits each and every node in the BST precisely once. So, if

the numbers of nodes in the BST are doubled, the amount of work for a traversal doubles

as well.

7. Explain the theory of Minimum spanning trees.Ans –

Tree is one of the most efficient data structure used in a computer program. There are many types of tree. Binary tree is a tree that always have two branches, Red-Black-Trees, 2-3-4 Trees, AVL Trees, etc. A well balanced tree can be used to design a good searching algorithm.

A Tree is an undirected graph that contains no cycles and is connected. Many trees are what is called rooted, where there is a notion of the "top" node, which is called the

root. Thus, each node has one parent, which is the adjacent node which is closer to the root, and may have any number of children, which are the rest of the nodes adjacent to it. The tree above was drawn as a rooted tree.

Spanning trees

A spanning tree of a graph is just a subgraph that contains all the vertices and is a tree. A graph may have many spanning trees; for instance the complete graph on four vertices

Minimum spanning trees

Now suppose the edges of the graph have weights or lengths. The weight of a tree is just the sum of weights of its edges. Obviously, different trees have different lengths. The problem: how to find the minimum length spanning tree?

The standard application is to a problem like phone network design. You have a business with several offices; you want to lease phone lines to connect them up with each other; and the phone company charges different amounts of money to connect different pairs of cities. You want a set of lines that connects all your offices with a minimum total cost. It should be a spanning tree, since if a network isn’t a tree you can always remove some edges and save money.

A less obvious application is that the minimum spanning tree can be used to approximately solve the traveling salesman problem. A convenient formal way of defining this problem is to find the shortest path that visits each point at least once. Note that if you have a path visiting all points exactly once, it’s a special kind of tree. For instance, in the example above, twelve of sixteen spanning trees are actually paths. If you have a path visiting some vertices more than once, you can always drop some edges to get a tree. So in general the MST weight is less than the TSP weight, because it’s a minimization over a strictly larger set.

On the other hand, if you draw a path tracing around the minimum spanning tree, you trace each edge twice and visit all points, so the TSP weight is less than twice the MST weight. Therefore this tour is within a factor of two of optimal.

8. Explain the following graph problems:

A) Telecommunication problem

B) Knight MovesAns –

Telecommunication problem

Given a set of computers and a set of wires running between pairs of computers, what is the minimum number of machines whose crash causes two given machines to be unable to communicate? (The two given machines will not crash.)

Graph: The vertices of the graph are the computers. The edges are the wires between the computers. Graph problem: minimum dominating sub-graph.

Knight moves

Given: Two squares on an 8×8 chessboard. Determine the shortest sequence of knight moves from one square to the other.

Graph: The graph here is harder to see. Each location on the chessboard represents a vertex. There is an edge between two positions if it is a legal knight move. Graph Problem: Single Source Shortest Path.