EE2209 - DATA STRUCTURES AND ALGORITHMS LABORATORY

LAB MANUAL

Department of Information Technology, Rajalakshmi Engineering College, Thandalam.

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INDEXI II III LAB SYLLABUS LAB PLAN PSEUDO CODES FOR LAB EXPERIMENTS1. (a) Implementation of Linked List

(b) Implementation of Doubly Linked list 2. Represent a polynomial as a linked list and write functions For polynomial addition. 3. Implementation of tree traversal 4. Implementation of stack ( infix to postfix conversion) 5. Implementation of Binary search Tree 6. Implementation of insertion in AVL trees 7. Implementation of hashing techniques 8. Implementation of backtracking algorithm for knapsack problem 9. Implementation of prims and kruskals algorithm 10.Implementation of dijktras algorithm using priority queues

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11.Implementation of array based circular queue 12. Implementation of priority queues using heaps 13. Implementation of branch and bound algorithm 14. Implementation of Randomized algorithm 15. Implementation of Topological sort on a Directed graph to decide is it is cyclic

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EE2209 32

DATA STRUCTURES AND ALGORITHMS LABORATORY

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(Common to EEE, EIE& ICE) Aim: To develop skills in design and implementation of data structures and their applications. 1. Implement singly and doubly linked lists. 2. Represent a polynomial as a linked list and write functions for polynomial addition. 3. Implement stack and use it to convert infix to postfix expression 4. Implement array-based circular queue and use it to simulate a producer-consumer problem. 5. Implement an expression tree. Produce its pre-order, in-order, and post-order traversals. 6. Implement binary search tree. 7. Implement insertion in AVL trees. 8. Implement priority queue using heaps 9. Implement hashing techniques 10. Perform topological sort on a directed graph to decide if it is acyclic. 11. Implement Dijkstra's algorithm using priority queues 12. Implement Prim's and Kruskal's algorithms 13. Implement a backtracking algorithm for Knapsack problem 14. Implement a branch and bound algorithm for traveling salesperson problem 15. Implement any randomized algorithm.TOTAL : 45 PERIODS

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LAB PLANSI.No 1 Name of the Experiment (a) Implementation of Linked List (b) Implementation of Doubly Linked List Represent a polynomial as a linked list and write functions for polynomial addition. Implementation of tree traversal Implementation of stack (converting infix to postfix expression) Implementation of Binary search tree Implementation of hashing techniques Implementation of insertion in AVL Tree Implementation of Backtracking algorithm for Knapsack problem Implementation of Prims and Kruskals algorithm Implementation of Dijktras algorithm using priority queues Implementation of array based circular queue Implementation of priority queues using heaps Implementation of Branch and bound algorithm Implementation of Randomized algorithm Implementation of Topological sort on a Date

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3 4 5 6 7 8 9 10 11 12 13 14 15

6 Directed graph to decide is it is cyclic

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IMPLEMENTATION OF A LINKED LIST Ex. No : 1(a)AIM: To implement a linked list and do all operations on it. ALGORITHM: Step 1: Start the process. Step 2: Initialize and declare variables. Step 3: Enter the choice. INSERT / DELETE. Step 4: If choice is INSERT then a) Enter the element to be inserted. b) Get a new node and set DATA[NEWNODE] = ITEM. c) Find the node after which the new node is to be inserted.

d) Adjust the link fields. e) Print the linked list after insertion.

Step 5: If choice is DELETE then a) Enter the element to be deleted.

b) Find the node containing the element (LOC) and its preceding node (PAR).

c) Set ITEM = DATA [LOC] and delete the node LOC.d) Adjust the link fields so that PAR points to the next element. ie LINK [PAR] = LINK [LOC]. e) Print the linked list after deletion.

Step 6: Stop the process.

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IMPLEMENTATION OF DOUBLY LINKED LIST Ex. No : 1(b)AIM: To implement a doubly linked list and do all operations on it. ALGORITHM: Step 1: Data type declarations record Node { data prev } record List { Node firstNode // points to first node of list; null for empty list Node lastNode // points to last node of list; null for empty list } Step 2: Iterating over the nodes Iterating through a doubly linked list can be done in either direction. In fact, direction can change many times, if desired. node := list.firstNode while node null node := node.next node := list.lastNode while node null node := node.prev Step 3:Inserting a node function insertAfter(List list, Node node, Node newNode) newNode.prev := node newNode.next := node.next if node.next == null list.lastNode := newNode else node.next.prev := newNode node.next := newNode function insertBefore(List list, Node node, Node newNode) newNode.prev := node.prev newNode.next := node if node.prev is null list.firstNode := newNode

9else node.prev.next := newNode node.prev := newNode Function to insert a node at the beginning of a possibly-empty list: function insertBeginning(List list, Node newNode) if list.firstNode == null list.firstNode := newNode list.lastNode := newNode newNode.prev := null newNode.next := null else insertBefore(list, list.firstNode, newNode) A symmetric function inserts at the end: function insertEnd(List list, Node newNode) if list.lastNode == null insertBeginning(list, newNode) else insertAfter(list, list.lastNode, newNode) Step 5:Deleting a node Removing a node is easier, only requiring care with the firstNode and lastNode: function remove(List list, Node node) if node.prev == null list.firstNode := node.next else node.prev.next := node.next if node.next == null list.lastNode := node.prev else node.next.prev := node.prev destroy node

Viva Questions:

A Linked list can grow and shrink in size dynamically at _______. Write an algorithm to detect loop in a linked list. Reverse a linked list. Delete an element from a doubly linked list. Implement an algorithm to reverse a singly linked list. (with and without recursion) Implement an algorithm to reverse a doubly linked list

10 How would you find a cycle in a linked list? Try to do it in O(n) time. Try it using a constant amount of memory. REPRESENT A POLYNOMIAL AS LINKED LIST AND IMPLEMENT POLYNOMIAL ADDITION Ex. No: 2AIM: To represent polynomial as linked list and perform polynomial addition.

ALGORITHM:Step 1: Creating Node. /*Inserting an element in a sorted linked list. Let the data be sorted and put in a singly linked linear list which is being pointed by address head .Let the new data to be entered be d.Use malloc get a node with address pNew. Suppose we want to write a code to enter data d into the node and insert it into its proper place in the list.*/ typedef struct node { int data; struct node *next; }; struct node* pNew = (struct node*) (malloc(sizeof(struct node))); pNew -> data = d; pNew ->next = NULL; pCur = head ; /* check if data is smaller than smallest item on the list*/ if (pNew -> data < pCur -> data ) { pNew ->next = pCur ; head = pNew; } /* now examine the remaining list */ p = pCur -> next ; while(p!=NULL ||p->data < pNew->data ) { pCur = pCur -> next ; p = p -> next ; } pNew -> next = pCur -> next; pCur -> next = pNew ; A polynomial can be represented in an array or in a linked list by simply storing the coefficient and exponent of each term. However, for any polynomial operation, such as addition or multiplication of polynomials, Step 2: Addition of two polynomials

11Consider addition of the following polynomials 5 x12 + 2 x9 + 4x7 + 6x6 + x3 7 x8 + 2 x7 + 8x6 + 6x4 + 2x2 + 3 x + 40 The resulting polynomial is going to be 5 x12 + 2 x9 + 7 x8 + 6 x7 + 14x6 + 6x4 +x3 2x2 + 3 x + 40 Step 3: Result of addition is going to be stored in a third list. Step 4: Started with the highest power in any polynomial. Step 5: If there was no item having same exponent, simply appended the term to the new list, and continued with the process. Step 6: Wherever the exponents were matching, simply added the coefficients and then stored the term in the new list. Step 6: If one list gets exhausted earlier and the other list still contains some lower order terms, then simply append the remaining terms to the new list. //Pseudo code Let phead1 , phead2 and phead3 represent the pointers of the three lists under consideration. Let each node contain two integers exp and coff . Let us assume that the two linked lists already contain relevant data about the two polynomials. Also assume that we have got a function append to insert a new node at the end of the given list. p1 = phead1; p2 = phead2; Let us call malloc to create a new node p3 to build the third list p3 = phead3; /* now traverse the lists till one list gets exhausted */ while ((p1 != NULL) || (p2 != NULL)) { / * if the exponent of p1 is higher than that of p2 then the next term in final list is going to be the node of p1* / while (p1 ->exp > p2 -> exp ) { p3 -> exp = p1 -> exp; p3 -> coff = p1 -> coff ; append (p3, phead3); /* now move to the next term in list 1*/ p1 = p1 -> next; } / * if p2 exponent turns out to be higher then make p3 same as p2 and append to final list * / while (p1 ->exp < p2 -> exp ) { p3 -> exp = p2 -> exp; p3 -> coff = p2 -> coff ; append (p3, phead3); p2 = p2 -> next; } /* now consider the possibility that both exponents are same , then we must add the coefficients to get the term for

12the final list */ while (p1 ->exp = p2 -> exp ) { p3-> exp = p1-> exp; p3->coff = p1->coff + p2-> coff ; append (p3, phead3) ; p1 = p1->next ; p2 = p2->next ; } } /* now consider the possibility that list2 gets exhausted, and there are terms remaining only in list1. So all those terms have to be appended to end of list3. However, you do not have to do it term by term, as p1 is already pointing to remaining terms, so simply append the pointer p1 to phead3 */ if ( p1 != NULL) append (p1, phead3) ; else append (p2, phead3);

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IMPLEMENTATION OF TREE TRAVERSALS Ex. No. :3 DateAIM: To implement tree traversals using linked list. ALGORITHM: Step 1: Start the process. Step 2: Initialize and declare variables. Step 3: Enter the choice. Inorder / Preorder / Postorder. Step 4: If choice is Inorder then a) Traverse the left subtree in inorder. b) Process the root node. c) Traverse the right subtree in inorder. Step 5: If choice is Preorder then a) Process the root node. b) Traverse the left subtree in preorder. c) Traverse the right subtree in preorder. Step 6: If choice is postorder then a) Traverse the left subtree in postorder. b) Traverse the right subtree in postorder. c) Process the root node. Step7: Print the Inorder / Preorder / Postorder traversal. Step 8: Stop the process.

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Viva Questions:

Write a function and the node data structure to visit all of the nodes in a binary tree.

14 What is a balanced tree Do a breadth first traversal of a tree. How would you print out the data in a binary tree, level by level, starting at the top? Write a function to find the depth of a binary tree. How do you represent an n-ary tree? Write a program to print the nodes of such a tree in breadth first order. What is a spanning Tree?

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STACK IMPLEMENTATION FOR CONVERTING INFIX TO POSTFIX EXPRESSION Ex. No. :4 AIM: To implement stack for converting infix to postfix expression ALGORITHM: 1. Start by initializing an empty stack. This will be for holding our operators. 2. Read the string, getting the first operand and add it to the postfix string. 3. Read the next object in the string, normally a operator, and push the operator to the stack. Is the operator of greater, equal, or lesser precedence than the top most operator on the stack? If the operator is greater than the top most operator, push the current operator and continue. If the operator is lesser to the top most operator, pop the stack adding it to the postfix string, and then push the current operator to the stack. Test the next operator to see if it is also lesser and repeat step 2. If not continue. If the operator is equal to the top most operator pop the top of the stack adding it to the postfix string and then push the current operator to the stack. 4. Repeat these steps until all of the operands and operators are taken care of. 5. Afterwards look at the stack one last time to see if any operators remain, pop them all and add them to the end of the postfix string. 6. Stop the process. \

Viva questions:

16 Given an expression tree with no parentheses in it, write the program to give equivalent infix expression with parentheses inserted where necessary. How would you implement a queue from a stack? What is stack? List the difference between queue and stack. How would you implement a queue from a stack?

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IMPLEMENTATION OF BINARY SEARCH TREE Ex. No. :5

AIM: To implement binary search tree and to do BST operations ALGORITHM:find() Operation //Purpose: find Item X in the Tree //Inputs: data object X (object to be found), binary-search-tree node node // Output: bst-node n containing X, if it exists; NULL otherwise. find(X, node)f if(node = NULL) return NULL if(X = node:data) return node else if(X < node:data) return find(X,node:leftChild) else // X > node:data return find(X,node:rightChild) { findMinimum() Operation //Purpose: return least data object X in the Tree //Inputs: binary-search-tree node node // Output: bst-node n containing least data object X, if it exists; NULL otherwise. findMin(node)f if(node = NULL) //empty tree return NULL if(node:leftChild = NULL) return node

18 return findMin(node:leftChild) } insert() Operation //Purpose: insert data object X into the Tree //Inputs: data object X (to be inserted), binary-search-tree node node //Effect: do nothing if tree already contains X; // otherwise, update binary search tree by adding a new node containing data object X insert(X, node)f if(node = NULL)f node = new binaryNode(X,NULL,NULL) return {if(X = node:data) return else if(X < node:data) insert(X, node:leftChild) else // X > node:data insert(X, node:rightChild) } delete() Operation //Purpose: delete data object X from the Tree //Inputs: data object X (to be deleted), binary-search-tree node node //Effect: do nothing if tree does not contain X; // otherwise, update binary search tree by deleting the node containing data object X delete(X, node)f if(node = NULL) //nothing to do return if(X < node:data) delete(X, node:leftChild) else if(X > node:data) delete(X, node:rightChild) else f // found the node to be deleted! Take action based on number of node children if(node:leftChild = NULL and node:rightChild = NULL)f delete node node = NULL return

19 {else if(node:leftChild = NULL)f tempNode = node node = node:rightChild delete tempNode {else if(node:rightChild = NULL)f (similar to the case when node:leftChild = NULL) {else f //replace node:data with minimum data from right subtree tempNode = findMin(node.rightChild) node:data = tempNode:data delete(node:data,node:rightChild) } } }

Viva questions: What is tree? What are the advantages of BST? Write a function and the node data structure to visit all of the nodes in a binary tree.

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IMPLEMENTATION OF INSERTION IN AVL TREE Ex. No. :6 AIM: To implement Insertion in AVL tree.

ALGORITHM: int avl_insert(node *treep, value_t target) { /* insert the target into the tree, returning 1 on success or 0 if it * already existed */ node tree = *treep; node *path_top = treep; while (tree && target != tree->value) { direction next_step = (target > tree->value); if (!Balanced(tree)) path_top = treep; treep = &tree->next[next_step]; tree = *treep; } if (tree) return 0;

21 tree = malloc(sizeof(*tree)); tree->next[0] = tree->next[1] = NULL; tree->longer = NEITHER; tree->value = target; *treep = tree; avl_rebalance(path_top, target); return 1; } void avl_rebalance_path(node path, value_t target) { /* Each node in path is currently balanced. * Until we find target, mark each node as longer * in the direction of target because we know we have * inserted target there */ while (path && target != path->value) { direction next_step = (target > path->value); path->longer node avl_rotate_2(node *path_top, direction dir) { node B, C, D, E; B = *path_top; D = B->next[dir]; C = D->next[1-dir];

22 E = D->next[dir]; *path_top = D; D->next[1-dir] = B; B->next[dir] = C; B->longer = NEITHER; D->longer = NEITHER; return E; } node avl_rotate_3(node *path_top, direction dir, direction third) { node B, F, D, C, E; B = *path_top; F = B->next[dir]; D = F->next[1-dir]; /* node: C and E can be NULL */ C = D->next[1-dir]; E = D->next[dir]; *path_top = D; D->next[1-dir] = B; D->next[dir] = F; B>next[dir] = C; F->next[1-dir] = E; D->longer = NEITHER; /* assume both trees are balanced */ B->longer = F->longer = NEITHER; if (third == NEITHER) return NULL; if (third == dir) { /* E holds the insertion so B is unbalanced */ B->longer = 1-dir; return E; } else { /* C holds the insertion so F is unbalanced */ F->longer = dir; return C; } } and there you have an AVL insertion, in non-recursive style, using only constant space, but sometimes performing more comparisons that absolutely needed.

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IMPLEMENTATION OF HASHING TECHNIQUES Ex. No. :7 AIM: To implement hashing techniques ALGORITHM: For insertion:

HASH-INSERT (T, k) i=0 Repeat j