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4. Dec 2013
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Nächste SlideShare
Unit I- Data structures Introduction, Evaluation of Algorithms, Arrays, Spars...
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1. Advanced C Programming Introduction to Data Structures Data Structures: A data structure is an arrangement of data in a computer's memory or even disk storage. Data structures can be classified into two types • Linear Data Structures • Non Linear Data Structures Linear Data Structures: Linear data structures are those data structures in which data elements are accessed (read and written) in sequential fashion ( one by one) Eg: Stacks , Queues, Lists, Arrays Non Linear Data Structures: Non Linear Data Structures are those in which data elements are not accessed in sequential fashion. Eg: trees, graphs Algorithm: Step by Step process of representing solution to a problem in words is called an Algorithm. Characteristics of an Algorithm: • • Output: An algorithm should have one or more outputs • Finiteness: Every step in an algorithm should end in finite amount of time • Unambiguous: Each step in an algorithm should clearly stated • 1 Input : An algorithm should have zero or more inputs Effectiveness: Each step in an algorithm should be effective
3. Advanced C Programming Stack : Stack is a Linear Data Structure which follows Last in First Out mechanism. It means: the first element inserted is the last one to be removed Stack uses a variable called top which points topmost element in the stack. top is incremented while pushing (inserting) an element in to the stack and decremented while poping (deleting) an element from the stack A top Push(A) B A Push(B) C B A top top Push(C) D C B A Push(D) top top C BA Pop() Valid Operations on Stack: • Inserting an element in to the stack (Push) • Deleting an element in to the stack (Pop) • Displaying the elements in the queue (Display) Note: While pushing an element into the stack, stack is full condition should be checked While deleting an element from the stack, stack is empty condition should be checked Applications of Stack: • • Stacks are used in function calls • 3 Stacks are used in recursion programs Stacks are used in interrupt implementation
4. Advanced C Programming Queue: Queue is a Linear Data Structure which follows First in First out mechanism. It means: the first element inserted is the first one to be removed Queue uses two variables rear and front. Rear is incremented while inserting an element into the queue and front is incremented while deleting element from the queue rear front A Insert(A) B A Insert(B) rear front C B A Insert(C) rear front D C B A Insert(D) rear D C B front Delete() Valid Operations on Queue: • Inserting an element in to the queue • Deleting an element in to the queue • Displaying the elements in the queue Note: While inserting an element into the queue, queue is full condition should be checked While deleting an element from the queue, queue is empty condition should be checked Applications of Queues: • Real life examples • • • Waiting in line Waiting on hold for tech support Applications related to Computer Science • • 4 Threads Job scheduling (e.g. Round-Robin algorithm for CPU allocation) rear front
6. Advanced C Programming Trees : A tree is a Non-Linear Data Structure which consists of set of nodes called vertices and set of edges which links vertices Terminology: • Root Node: The starting node of a tree is called Root node of that tree • Terminal Nodes: The node which has no children is said to be terminal node or leaf . • Non-Terminal Node: The nodes which have children is said to be Non-Terminal Nodes • Degree: The degree of a node is number of sub trees of that node • Depth: The length of largest path from root to terminals is said to be depth or height of the tree • Siblings: The children of same parent are said to be siblings • Ancestors: The ancestors of a node are all the nodes along the path from the root to the node A Property C B D E F G H 6 I Number of nodes Height Root Node Leaves Interior nodes Number of levels Ancestors of H Descendants of B Siblings of E Value : : : : : : : : : 9 4 A ED, H, I, F, C D, E, G 5 I D,E, F D, F
7. Advanced C Programming Binary Trees: Binary trees are special class of trees in which max degree for each node is 2 Recursive definition: A binary tree is a finite set of nodes that is either empty or consists of a root and two disjoint binary trees called the left subtree and the right subtree. Any tree can be transformed into binary tree. By left child-right sibling representation. A B C E K F G D Binary Tree Traversal Techniques: There are three binary tree traversing techniques • Inorder • Preorder • Postorder Inorder: In inorder traversing first left subtree is visited followed by root and right subtree Preorder: In preorder traversing first root is visited followed by left subtree and right subtree. Postorder: In post order traversing first left tree is visited followed by right subtree and root. 7
8. Advanced C Programming Binary Search Tree: A Binary Search Tree (BST) is a binary tree which follows the following conditons • Every element has a unique key. • The keys in a nonempty left subtree are smaller than the key in the root of subtree. • The keys in a nonempty right subtree are grater than the key in the root of subtree. • The left and right subtrees are also binary search trees. 63 89 41 34 56 Valid Operations on Binary Search Tree: • • Deleting an element • Searching for an element • 8 Inserting an element Traversing 72 95
9. Advanced C Programming Avl Tree: If in a binary search tree, the elements are inserted in sorted order then the height will be n, where n is number of elements. To overcome this disadvantage balanced trees were introduced. • Balanced binary search trees • An AVL Tree is a binary search tree such that for every internal node v of T, the heights of the children of v can differ by at most 1. 44 4 2 17 78 1 2 32 88 50 1 48 Operations of Avl tree: • Inserting an element • Deleting an element • Searching for an element • Traversing • 9 3 Height balancing 62 1 1
10. Advanced C Programming Graphs A graph is a Non-Linear Data Structure which consists of set of nodes called vertices V and set of edges E which links vertices Note: A tree is a graph with out loops 0 0 1 1 2 3 Graph 3 2 5 4 6 Tree Graph Traversal: • Problem: Search for a certain node or traverse all nodes in the graph • Depth First Search • • Breadth First Search • 10 Once a possible path is found, continue the search until the end of the path Start several paths at a time, and advance in each one step at a time
11. Advanced C Programming Introduction to Object Oriented Programming Object Oriented Programming: You've heard it a lot in the past several years. Everybody is saying it. What is all the fuss about objects and object-oriented technology? Is it real? Or is it hype? Well, the truth is--it's a little bit of both. Object-oriented technology does, in fact, provide many benefits to software developers and their products. However, historically a lot of hype has surrounded this technology, causing confusion in both managers and programmers alike. Many companies fell victim to this hardship (or took advantage of it) and claimed that their software products were object-oriented when, in fact, they weren't. These false claims confused consumers, causing widespread misinformation and mistrust of object-oriented technology. Object: As the name object-oriented implies, objects are key to understanding object-oriented technology. You can look around you now and see many examples of real-world objects: your dog, your desk, your television set, your bicycle. Definition: An object is a software bundle of variables and related methods Class: 11
12. Advanced C Programming In the real world, you often have many objects of the same kind. For example, your bicycle is just one of many bicycles in the world. Using object-oriented terminology, we say that your bicycle object is an instance of the class of objects known as bicycles. Bicycles have some state (current gear, current cadence, two wheels) and behavior (change gears, brake) in common. However, each bicycle's state is independent of and can be different from other bicycles. Definition: A class is a blueprint or prototype that defines the variables and methods common to all objects of a certain kind. Inheritance: Acquiring the properties of one class in another class is called inheritance The Benefits of Inheritance • Subclasses provide specialized behaviors from the basis of common elements provided by the super class. Through the use of inheritance, programmers can reuse the code in the superclass many times. • Programmers can implement superclasses called abstract classes that define "generic" behaviors. The abstract superclass defines and may partially implement the behavior but much of the class is undefined and unimplemented. Other programmers fill in the details with specialized subclasses. Data Abstraction: The essential element of object oriented programming in abstraction. The complexity of programming in object oriented programming is maintained through abstraction. For example, the program consist of data and code which work over data. While executing a program we don’t thing in which location that data is being stored how the input device is transferring the input to the memory etc. this abstraction allows us to execute the program without thinking deeply about the complexity of execution of program. Encapsulation: 12
13. Advanced C Programming Encapsulation is the mechanism that binds together code and the data and keeps them safe from outside world. In the sense it is a protective wrapper that prevents the code and data from being accessed by other code defied outside the wrapper. Access is controlled through a well defined interface. Polymorphism: Existing in more that one form is called polymorphism. Polymorphism means the ability to take more that one form. For example an operation may exhibit different behavior in different behavior in different instances. For example consider operation of addition. For two numbers the operation will generate a sum. If the operands are string the operation would produces a third string by concatenation. C++ supports polymorphism through method overloading and operator overloading Method overloading: if the same method name used for different procedures that the method is said to be overloaded. Dynamic Binding: Binding refer to the linking of a procedure call to the code to be executed in response to the call. Dynamic binding means that the code associated with a given procedure call is not know until the time of the call at runtime. It is associated with a polymorphism reference depends on the dynamic type of that reference. Message communication: An object oriented program consists of objects that communicate with each other. The process of programming in an object oriented language therefore involves the following basic steps: 1. creating classes that define objects and their behaviors. 2. creating objects from class definitions. 3. establishing communication among objects. 13
14. Advanced C Programming Abstract Data Types: An Abstract Data Type (ADT) is more a way of looking at a data structure: focusing on what it does and ignoring how it does its job. A stack or a queue is an example of an ADT. It is important to understand that both stacks and queues can be implemented using an array. It is also possible to implement stacks and queues using a linked list. This demonstrates the "abstract" nature of stacks and queues: how they can be considered separately from their implementation. To best describe the term Abstract Data Type, it is best to break the term down into "data type" and then "abstract". Data type: When we consider a primitive type we are actually referring to two things: a data item with certain characteristics and the permissible operations on that data. An int in Java, for example, can contain any whole-number value from -2,147,483,648 to +2,147,483,647. It can also be used with the operators +, -, *, and /. The data type's permissible operations are an inseparable part of its identity; understanding the type means understanding what operations can be performed on it. In C++, any class represents a data type, in the sense that a class is made up of data (fields) and permissible operations on that data (methods). By extension, when a data storage structure like a stack or queue is represented by a class, it too can be referred to as a data type. A stack is different in many ways from an int, but they are both defined as a certain arrangement of data and a set of operations on that data. abstract Now lets look at the "abstract" portion of the phrase. The word abstract in our context stands for "considered apart from the detailed specifications or implementation". In C++, an Abstract Data Type is a class considered without regard to its implementation. It can be thought of as a "description" of the data in the class and a list of operations that can be carried out on that data and instructions on how to use these operations. What is excluded though, is the 14
15. Advanced C Programming details of how the methods carry out their tasks. An end user (or class user), you should be told what methods to call, how to call them, and the results that should be expected, but not HOW they work. We can further extend the meaning of the ADT when applying it to data structures such as a stack and queue. In Java, as with any class, it means the data and the operations that can be performed on it. In this context, although, even the fundamentals of how the data is stored should be invisible to the user. Users not only should not know how the methods work, they should also not know what structures are being used to store the data. Consider for example the stack class. The end user knows that push() and pop() (amoung other similar methods) exist and how they work. The user doesn't and shouldn't have to know how push() and pop() work, or whether data is stored in an array, a linked list, or some other data structure like a tree. 15
16. Advanced C Programming Stack ADT Algorithms Push(item) { If (stack is full) print “ stack over flow” else Increment top ; Stack [top]= item; } Pop() { If( stack is empty) print” stack under flow” else Decrement top } Display() { If ( stack is empty) print” no element to display” else for i= top to 0 step -1 Print satck[i]; } 16
17. Advanced C Programming Stack ADT #include<iostream.h> #include<conio.h> #include<stdlib.h> class stack { int stk[5]; int top; public: stack() { top=-1; } void push(int x) { if(top > 4) { cout <<"stack over flow"; return; } stk[++top]=x; cout <<"inserted" <<x; } void pop() { if(top <0) { cout <<"stack under flow"; return; } cout <<"deleted" <<stk[top--]; } void display() { if(top<0) { cout <<" stack empty"; return; } 17
18. Advanced C Programming for(int i=top;i>=0;i--) cout <<stk[i] <<" "; } }; void main() { int ch; stack st; clrscr(); while(1) { cout <<"n1.push 2.pop 3.display 4.exitnEnter ur choice"; cin >> ch; switch(ch) { case 1: cout <<"enter the element"; cin >> ch; st.push(ch); break; case 2: st.pop(); break; case 3: st.display();break; case 4: exit(0); } } } OUTPUTS 1.push 2.pop 3.display Enter ur choice2 stack under flow 1.push 2.pop 3.display Enter ur choice1 enter the element2 inserted2 1.push 2.pop 3.display Enter ur choice1 enter the element3 inserted3 1.push 2.pop 3.display Enter ur choice2 deleted3 1.push 2.pop 3.display Enter ur choice1 enter the element5 18 4.exit 4.exit 4.exit 4.exit 4.exit
19. Advanced C Programming Queue ADT Algorithms Insert ( item) { If rear = max -1 then print “ queue is full” else { Increment rear Queue [rear]=item; } } Delete() { If front = rear print “queue is empty” else Increment front } Display() { If front=rear print “queue is empty “ else For i =front to rear Print queue[i]; } 19
20. Advanced C Programming Queue ADT #include<iostream.h> #include<conio.h> #include<stdlib.h> class queue { int queue[5]; int rear,front; public: queue() { rear=-1; front=-1; } void insert(int x) { if(rear > 4) { cout <<"queue over flow"; front=rear=-1; return; } queue[++rear]=x; cout <<"inserted" <<x; } void delet() { if(front==rear) { cout <<"queue under flow"; return; } cout <<"deleted" <<queue[++front]; } void display() { if(rear==front) { 20
21. Advanced C Programming cout <<" queue empty"; return; } for(int i=front+1;i<=rear;i++) cout <<queue[i]<<" "; } }; void main() { int ch; queue qu; clrscr(); while(1){ cout <<"n1.insert 2.delet 3.display 4.exitnEnter ur choice"; cin >> ch; switch(ch) { case 1: cout <<"enter the element"; cin >> ch; qu.insert(ch); break; case 2: qu.delet(); break; case 3: qu.display();break; case 4: exit(0); } } } OUTPUT 1.insert 2.delet 3.display Enter ur choice1 enter the element21 inserted21 1.insert 2.delet 3.display Enter ur choice1 enter the element22 inserted22 1.insert 2.delet 3.display Enter ur choice1 enter the element16 inserted16 1.insert 2.delet 3.display Enter ur choice3 21 4.exit 4.exit 4.exit 4.exit
22. Advanced C Programming 21 22 16 1.insert 2.delet 3.display 4.exit 22
23. Advanced C Programming Algorithm for Stack Using Linked List Push(item) { If (stack is full) print “ stack over flow” else goto end of list and let it be temp temp->next=item item->next=NULL; } Pop() { If(head is null) print” stack under flow” else goto last but one node and let it be temp temp->next=NULL } Display() { If ( head=NULL) print” no element to display” else { Temp=head; While(temp!=NULL) { Print(“temp->data) Temp=temp->next; } } 23
24. Advanced C Programming Stack Using Linked List #include<iostream.h> #include<conio.h> #include<stdlib.h> class node { public: class node *next; int data; }; class stack : public node { node *head; int tos; public: stack() { os=-1; } void push(int x) { if (tos < 0 ) { head =new node; head->next=NULL; head->data=x; tos ++; } else { node *temp,*temp1; temp=head; if(tos >= 4) { cout <<"stack over flow"; return; } tos++; while(temp->next != NULL) temp=temp->next; temp1=new node; 24
25. Advanced C Programming temp->next=temp1; temp1->next=NULL; temp1->data=x; } } void display() { node *temp; temp=head; if (tos < 0) { cout <<" stack under flow"; return; } while(temp != NULL) { cout <<temp->data<< " "; temp=temp->next; } } void pop() { node *temp; temp=head; if( tos < 0 ) { cout <<"stack under flow"; return; } tos--; while(temp->next->next!=NULL) { temp=temp->next; } temp->next=NULL; } }; void main() { stack s1; int ch; clrscr(); while(1) { 25
26. Advanced C Programming cout <<"n1.PUSHn2.POPn3.DISPLAYn4.EXITn enter ru choice:"; cin >> ch; switch(ch) { case 1: cout <<"n enter a element"; cin >> ch; s1.push(ch); break; case 2: s1.pop();break; case 3: s1.display(); break; case 4: exit(0); } } } OUTPUT 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:1 enter a element23 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:1 enter a element67 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:3 23 67 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:2 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:3 23 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:2 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:2 stack under flow 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:4 26
27. Advanced C Programming Algorithm Queue using Linked List Insert ( item) { If rear = max -1 then print “ queue is full” else { Increment rear Create a new node called item goto last node in the list and let it be temp temp-next=item; item-next=NULL; } } Delete() { If front = rear print “queue is empty” else { Increment front head=head-next; } } Display() { If front=rear print “queue is empty “ else Temp=head; While(temp!=NULL) { Print(“temp-data) Temp=temp-next; } } 27
28. Advanced C Programming Queue using Linked List #include<iostream.h> #include<conio.h> #include<stdlib.h> class node { public: class node *next; int data; }; class queue : public node { node *head; int front,rare; public: queue() { front=-1; rare=-1; } void push(int x) { if (rare < 0 ) { head =new node; head->next=NULL; head->data=x; rare ++; } else { node *temp,*temp1; temp=head; if(rare >= 4) { cout <<"queue over flow"; return; } rare++; while(temp->next != NULL) temp=temp->next; 28
29. Advanced C Programming temp1=new node; temp->next=temp1; temp1->next=NULL; temp1->data=x; } } void display() { node *temp; temp=head; if (rare < 0) { cout <<" queue under flow"; return; } while(temp != NULL) { cout <<temp->data<< " "; temp=temp->next; } } void pop() { node *temp; temp=head; if( rare < 0) { cout <<"queue under flow"; return; } if(front == rare) { front = rare =-1; head=NULL; return; } front++; head=head->next; } }; void main() { queue s1; 29
30. Advanced C Programming int ch; clrscr(); while(1) { cout <<"n1.PUSHn2.POPn3.DISPLAYn4.EXITn enter ru choice:"; cin >> ch; switch(ch) { case 1: cout <<"n enter a element"; cin >> ch; s1.push(ch); break; case 2: s1.pop();break; case 3: s1.display(); break; case 4: exit(0); } } } OUTPUT 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:1 enter a element23 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:1 enter a element54 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:3 23 54 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:2 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:2 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:2 queue under flow 1.PUSH 2.POP 3.DISPLAY 4.EXIT enter ru choice:4 30
31. Advanced C Programming Algorithms fo DeQueue Using Double Linked List Algorithm Insertfirst(item) { if dequeue is empty { Item-next=item-prev=NULL; tail=head=item; } else if(dequeue is full) print” insertion is not possible” else { item-next=head; item-prev=NULL; head=item; } } Algorithm Insertlast (item) { if dequeue is empty { Item-next=item-prev=NULL; tail=head=item; } else if(dequeue is full) print” insertion is not possible” else { tail-next=head; item-prev=tail; tail=item; } } Deletefirst() { If (dequeue is empty) print” no node to delete”; else { Head=head-next; Head-prev=NULL; } } 31
32. Advanced C Programming Deletelast() { if (dequeue is empty) print” no node to delete”; else { tail=tail-prev; tail-next=NULL; } } Displayfirst() { if( dequeue is empty) print “ no node to display” else { temp=head; while(temp-next!=null) then do { print(temp-data); temp=temp-next; } } } Displaylast() { if( dequeue is empty) print “ no node to display” else { temp=tail while(temp-prevt!=null) then do { print(temp-data); temp=temp-prev; } } } 32
33. Advanced C Programming Implementation of DeQueue Using Double Linked List #include<iostream.h> #include<conio.h> #include<stdlib.h> class node { public: int data; class node *next; class node *prev; }; class dqueue: public node { node *head,*tail; int top1,top2; public: dqueue() { top1=0; top2=0; head=NULL; tail=NULL; } void push(int x) { node *temp; int ch; if(top1+top2 >=5) { cout <<"dqueue overflow"; return ; } if( top1+top2 == 0) { head = new node; head->data=x; head->next=NULL; head->prev=NULL; tail=head; top1++; 33
34. Advanced C Programming } else { cout <<" Add element 1.FIRST 2.LASTn enter ur choice:"; cin >> ch; if(ch==1) { top1++; temp=new node; temp->data=x; temp->next=head; temp->prev=NULL; head->prev=temp; head=temp; } else { top2++; temp=new node; temp->data=x; temp->next=NULL; temp->prev=tail; tail->next=temp; tail=temp; } } } void pop() { int ch; cout <<"Delete 1.First Node 2.Last Noden Enter ur choice:"; cin >>ch; if(top1 + top2 <=0) { cout <<"nDqueue under flow"; return; } if(ch==1) { head=head->next; head->prev=NULL; top1--; } 34
35. Advanced C Programming else { top2--; tail=tail->prev; tail->next=NULL; } } void display() { int ch; node *temp; cout <<"display from 1.Staring 2.Endingn Enter ur choice"; cin >>ch; if(top1+top2 <=0) { cout <<"under flow"; return ; } if (ch==1) { temp=head; while(temp!=NULL) { cout << temp->data <<" "; temp=temp->next; } } else { temp=tail; while( temp!=NULL) { cout <<temp->data << " "; temp=temp->prev; } } } }; void main() { dqueue d1; int ch; clrscr(); 35
36. Advanced C Programming while (1) { cout <<"1.INSERT 2.DELETE 3.DISPLAU 4.EXITn Enter ur choice:"; cin >>ch; switch(ch) { case 1: cout <<"enter element"; cin >> ch; d1.push(ch); break; case 2: d1.pop(); break; case 3: d1.display(); break; case 4: exit(1); } } } OUTPUT 1.INSERT 2.DELETE 3.DISPLAU 4.EXIT Enter ur choice:1 enter element4 1.INSERT 2.DELETE 3.DISPLAU 4.EXIT Enter ur choice:1 enter element5 Add element 1.FIRST 2.LAST enter ur choice:1 1.INSERT 2.DELETE 3.DISPLAU 4.EXIT Enter ur choice:1 enter element6 Add element 1.FIRST 2.LAST enter ur choice:2 1.INSERT 2.DELETE 3.DISPLAU 4.EXIT Enter ur choice:3 display from 1.Staring 2.Ending Enter ur choice1 5 4 6 1.INSERT 2.DELETE 3.DISPLAU 4.EXIT Enter ur choice:2 Delete 1.First Node 2.Last Node Enter ur choice:1 1.INSERT 2.DELETE 3.DISPLAU 4.EXIT Enter ur choice:3 display from 1.Staring 2.Ending Enter ur choice1 4 6 1.INSERT 2.DELETE 3.DISPLAU 4.EXIT Enter ur choice:4 36
37. Advanced C Programming Algorithm for Circular Queue Algorithm Insertfirst(item) { if cqueue is empty then head=item; else if(cqueue is full) print” insertion is not possible” else { Rear=(rear +1) mod max } cqueue[rear]=x; } Algorithm Deletet() { If (dequeue is empty) print” no node to delete”; else { Front=(front+1) mod max } } Algorithm display() { If (front >rear) display elements for front to max and 0 to rear Else display elements from front to rear } 37
38. Advanced C Programming Implementation of Circular Queue Using Array #include<iostream.h> #include<conio.h> #include<stdlib.h> class cqueue { int q[5],front,rare; public: cqueue() { front=-1; rare=-1; } void push(int x) { if(front ==-1 && rare == -1) { q[++rare]=x; front=rare; return; } else if(front == (rare+1)%5 ) { cout <<" Circular Queue over flow"; return; } rare= (rare+1)%5; q[rare]=x; } void pop() { if(front==-1 && rare== -1) { cout <<"under flow"; return; } else if( front== rare ) { front=rare=-1; return; 38
39. Advanced C Programming } front= (front+1)%5; } void display() { int i; if( front <= rare) { for(i=front; i<=rare;i++) cout << q[i]<<" "; } else { for(i=front;i<=4;i++) { cout <<q[i] << " "; } for(i=0;i<=rare;i++) { cout << q[i]<< " "; } } } }; void main(){ int ch; cqueue q1; clrscr(); while( 1) { cout<<"n1.INSERT 2.DELETE 3.DISPLAY cin >> ch; switch(ch) { case 1: cout<<"enter element"; cin >> ch; q1.push(ch); break; case 2: q1.pop(); break; case 3: q1.display(); break; case 4: exit(0); } } } 39 4.EXITnEnter ur choice";
40. Advanced C Programming OUTPUT 1.INSERT 2.DELETE 3.DISPLAY Enter ur choice1 enter element4 4.EXIT 1.INSERT 2.DELETE 3.DISPLAY Enter ur choice1 enter element5 4.EXIT 1.INSERT 2.DELETE 3.DISPLAY Enter ur choice1 enter element3 4.EXIT 1.INSERT 2.DELETE 3.DISPLAY Enter ur choice3 453 1.INSERT 2.DELETE 3.DISPLAY Enter ur choice2 4.EXIT 1.INSERT 2.DELETE 3.DISPLAY Enter ur choice3 53 1.INSERT 2.DELETE 3.DISPLAY Enter ur choice4 4.EXIT 40 4.EXIT 4.EXIT
41. Advanced C Programming Program to Algorithm for Dictionary Dictionary Implement Functions of a #include<iostream.h> #include<conio.h> #include<stdlib.h> # define max 10 typedef struct list { int data; struct list *next; }node_type; node_type *ptr[max],*root[max],*temp[max]; class Dictionary { public: int index; Dictionary(); void insert(int); void search(int); void delete_ele(int); }; Dictionary::Dictionary() { index=-1; for(int i=0;i<max;i++) { root[i]=NULL; ptr[i]=NULL; temp[i]=NULL; } } void Dictionary::insert(int key) { index=int(key%max); ptr[index]=(node_type*)malloc(sizeof(node_type)); ptr[index]->data=key; if(root[index]==NULL) { 41
42. Advanced C Programming root[index]=ptr[index]; root[index]->next=NULL; temp[index]=ptr[index]; } else { temp[index]=root[index]; while(temp[index]->next!=NULL) temp[index]=temp[index]->next; temp[index]->next=ptr[index]; } } void Dictionary::search(int key) { int flag=0; index=int(key%max); temp[index]=root[index]; while(temp[index]!=NULL) { if(temp[index]->data==key) { cout<<"nSearch key is found!!"; flag=1; break; } else temp[index]=temp[index]->next; } if (flag==0) cout<<"nsearch key not found......."; } void Dictionary::delete_ele(int key) { index=int(key%max); temp[index]=root[index]; while(temp[index]->data!=key && temp[index]!=NULL) { ptr[index]=temp[index]; temp[index]=temp[index]->next; } ptr[index]->next=temp[index]->next; cout<<"n"<<temp[index]->data<<" has been deleted."; temp[index]->data=-1; 42
43. Advanced C Programming temp[index]=NULL; free(temp[index]); } void main() { int val,ch,n,num; char c; Dictionary d; clrscr(); do { cout<<"nMENU:n1.Create"; cout<<"n2.Search for a valuen3.Delete an value"; cout<<"nEnter your choice:"; cin>>ch; switch(ch) { case 1: cout<<"nEnter the number of elements to be inserted:"; cin>>n; cout<<"nEnter the elements to be inserted:"; for(int i=0;i<n;i++) { cin>>num; d.insert(num); } break; case 2: cout<<"nEnter the element to be searched:"; cin>>n; d.search(n); case 3: cout<<"nEnter the element to be deleted:"; cin>>n; d.delete_ele(n); break; default: cout<<"nInvalid choice...."; } cout<<"nEnter y to continue......"; cin>>c; }while(c=='y'); getch(); } 43
44. Advanced C Programming OUTPUT MENU: 1.Create 2.Search for a value 3.Delete an value Enter your choice:1 Enter the number of elements to be inserted:8 Enter the elements to be inserted:10 4 5 8 7 12 6 1 Enter y to continue......y MENU: 1.Create 2.Search for a value 3.Delete an value Enter your choice:2 Enter the element to be searched:12 Search key is found!! Enter the element to be deleted:1 1 has been deleted. Enter y to continue......y 44
45. Advanced C Programming AVL TREE Algorithm insertion(int x) { If(tree is empty) then root is empty Otherwise { temp=search(item); // temp is the node where search for the item halts if( item > temp) then temp-right=item; otherwise temp-left =item • • • • • Reconstruction procedure: rotating tree left rotation and right rotation Suppose that the rotation occurs at node x Left rotation: certain nodes from the right subtree of x move to its left subtree; the root of the right subtree of x becomes the new root of the reconstructed subtree Right rotation at x: certain nodes from the left subtree of x move to its right subtree; the root of the left subtree of x becomes the new root of the reconstructed subtree } Algorithm Search(int x) Algorithm delete() { • Case 1: the node to be deleted • Case 2: the node to be deleted is, its right subtree is empty • Case 3: the node to be deleted is, its left subtree is empty • Case 4: the node to be deleted right child } is a leaf has no right child, that has no left child, that has a left child and a Algorithm Search(x, root) { if(tree is empty ) then print” tree is empty” otherwise If(x grater than root) search(root-right); Otherwise if(x less than root ) search(root-left) Otherwise return true } } 45
46. Advanced C Programming AVL TREE #include<iostream.h> #include<conio.h> #include<stdlib.h> #include<math.h> void insert(int,int ); void delte(int); void display(int); int search(int); int search1(int,int); int avltree[40],t=1,s,x,i; void main() { int ch,y; for(i=1;i<40;i++) avltree[i]=-1; while(1) { cout <<"1.INSERTn2.DELETEn3.DISPLAYn4.SEARCHn5.EXITnEnter your choice:"; cin >> ch; switch(ch) { case 1: cout <<"enter the element to insert"; cin >> ch; insert(1,ch); break; case 2: cout <<"enter the element to delete"; cin >>x; y=search(1); if(y!=-1) delte(y); else cout<<"no such element in avlavltree"; break; case 3: display(1); cout<<"n"; for(int i=0;i<=32;i++) cout <<i; 46
47. Advanced C Programming cout <<"n"; break; case 4: cout <<"enter the element to search:"; cin >> x; y=search(1); if(y == -1) cout <<"no such element in avltree"; else cout <<x << "is in" <<y <<"position"; break; case 5: exit(0); } } } void insert(int s,int ch ) { int x,y; if(t==1) { avltree[t++]=ch; return; } x=search1(s,ch); if(avltree[x]>ch) { avltree[2*x]=ch; y=log(2*x)/log(2); if(height(1,y)) { if( x%2==0 ) update1(); else update2(); } } else { avltree[2*x+1]=ch; y=log(2*x)/log(2); if(height(1,y)) { if(x%2==1) update1(); else update2(); } 47
48. Advanced C Programming } t++; } void delte(int x) { if( avltree[2*x]==-1 && avltree[2*x+1]==-1) avltree[x]=-1; else if(avltree[2*x]==-1) { avltree[x]=avltree[2*x+1]; avltree[2*x+1]=-1; } else if(avltree[2*x+1]==-1) { avltree[x]=avltree[2*x]; avltree[2*x]=-1; } else { avltree[x]=avltree[2*x]; delte(2*x); } t--; } int search(int s) { if(t==1) { cout <<"no element in avltree"; return -1; } if(avltree[s]==-1) return avltree[s]; if(avltree[s]>x) search(2*s); else if(avltree[s]<x) search(2*s+1); else return s; } void display(int s) { if(t==1) { cout <<"no element in avltree:"; 48
49. Advanced C Programming return; } for(int i=1;i<40;i++) if(avltree[i]==-1) cout <<" "; else cout <<avltree[i]; return ; } int search1(int s,int ch) { if(t==1) { cout <<"no element in avltree"; return -1; } if(avltree[s]==-1) return s/2; if(avltree[s] > ch) search1(2*s,ch); else search1(2*s+1,ch); } int height(int s,int y) { if(avltree[s]==-1) return; } 49
50. Advanced C Programming OUTPUT 1.insert 2.display 3.delete 4.search 5.exit Enter u r choice to perform on AVL tree1 Enter an element to insert into tree4 do u want to continuey 1.insert 2.display 3.delete 4.search 5.exit Enter u r choice to perform on AVL tree1 Enter an element to insert into tree5 do u want to continuey 1.insert 2.display 3.delete 4.search 5.exit Enter u r choice to perform on AVL tree3 Enter an item to deletion5 itemfound do u want to continuey 1.insert 2.display 3.delete 4.search 5.exit Enter u r choice to perform on AVL tree2 4 do u want to continue4 50
51. Advanced C Programming Breath First Search Algorithm Algorithm BFS(s): Input: A vertex s in a graph Output: A labeling of the edges as “discovery” edges and “cross edges” initialize container L0 to contain vertex s i ¬ 0 while Li is not empty do create container Li+1 to initially be empty for each vertex v in Li do if edge e incident on v do let w be the other endpoint of e if vertex w is unexplored then label e as a discovery edge insert w into Li+1 else label e as a cross edge i ¬ i + 1 51
52. Advanced C Programming Breath First Search Implementation #include<iostream.h> #include<conio.h> #include<stdlib.h> int cost[10][10],i,j,k,n,qu[10],front,rare,v,visit[10],visited[10]; void main() { clrscr(); int m; cout <<"enterno of vertices"; cin >> n; cout <<"ente no of edges"; cin >> m; cout <<"nEDGES n"; for(k=1;k<=m;k++) { cin >>i>>j; cost[i][j]=1; } cout <<"enter initial vertex"; cin >>v; cout <<"Visitied verticesn"; cout << v; xvisited[v]=1; k=1; while(k<n) { for(j=1;j<=n;j++) if(cost[v][j]!=0 && visited[j]!=1 && visit[j]!=1) { visit[j]=1; qu[rare++]=j; } v=qu[front++]; cout<<v << " "; k++; visit[v]=0; visited[v]=1; } } 52
53. Advanced C Programming OUTPUT enterno of vertices9 ente no of edges9 EDGES 12 23 15 14 47 78 89 26 57 enter initial vertex1 Visited vertices 12 4 5 3 6 7 8 9 53
54. Advanced C Programming Depth First Search Algorithm Algorithm DFS(v); Input: A vertex v in a graph Output: A labeling of the edges as “discovery” edges and “backedges” for each edge e incident on v do if edge e is unexplored then let w be the other endpoint of e if vertex w is unexplored then label e as a discovery edge recursively call DFS(w) else label e as a backedge 54
55. Advanced C Programming Depth First Search #include<iostream.h> #include<conio.h> #include<stdlib.h> int cost[10][10],i,j,k,n,stk[10],top,v,visit[10],visited[10]; void main() { int m; clrscr(); cout <<"enterno of vertices"; cin >> n; cout <<"ente no of edges"; cin >> m; cout <<"nEDGES n"; for(k=1;k<=m;k++) { cin >>i>>j; cost[i][j]=1; } cout <<"enter initial vertex"; cin >>v; cout <<"ORDER OF VISITED VERTICES"; cout << v <<" "; visited[v]=1; k=1; while(k<n) { for(j=n;j>=1;j--) if(cost[v][j]!=0 && visited[j]!=1 && visit[j]!=1) { visit[j]=1; stk[top]=j; top++; } v=stk[--top]; cout<<v << " "; k++; visit[v]=0; visited[v]=1; } } 55
56. Advanced C Programming OUTPUT enterno of vertices9 ente no of edges9 EDGES 12 23 26 15 14 47 57 78 89 enter initial vertex1 ORDER OF VISITED VERTICES1 2 3 6 4 7 8 9 5 56
57. Advanced C Programming Prim’s Algorithm Algorithm Prim(E,Cost,n,t) { Let (k, l) be an edge of minimum cost in E; Mincost= cost[k,l]; t[1,1]=k; t[1,2]=l; for i=1 to n do { If (cost[i, l]<cost[k,l]) then near[i]=l; Else Near[i]=k; Near[k]=near[j]=0; } For i=2 to n -1 do { Let j be an index such that nearpj]!= 0 and cost[j,near[j]] is minimum T[I,1]=j ; t[I,2]=near[j] mincost=mincost + cost[j,near[j]]; near[j]=0; for k=1 to n do if(near[k] !=0 ) and cost[k,near[k]]) then near[j]=k } } 57 >cost[k,j])
58. Advanced C Programming Prim’s Algorithm #include<iostream.h> #include<conio.h> #include<stdlib.h> int cost[10][10],i,j,k,n,stk[10],top,v,visit[10],visited[10],u; void main() { int m,c; clrscr(); cout <<"enterno of vertices"; cin >> n; cout <<"ente no of edges"; cin >> m; cout <<"nEDGES Costn"; for(k=1;k<=m;k++) { cin >>i>>j>>c; cost[i][j]=c; } for(i=1;i<=n;i++) for(j=1;j<=n;j++) if(cost[i][j]==0) cost[i][j]=31999; cout <<"ORDER OF VISITED VERTICES"; k=1; while(k<n) { m=31999; if(k==1) { for(i=1;i<=n;i++) for(j=1;j<=m;j++) if(cost[i][j]<m) { m=cost[i][j]; u=i; } } else { for(j=n;j>=1;j--) 58
59. Advanced C Programming if(cost[v][j]<m && visited[j]!=1 && visit[j]!=1) { visit[j]=1; stk[top]=j; top++; m=cost[v][j]; u=j; } } cost[v][u]=31999; v=u; cout<<v << " "; k++; visit[v]=0; visited[v]=1; } } OUTPUT enterno of vertices7 ente no of edges9 EDGES Cost 1 6 10 6 5 25 5 4 22 4 3 12 3 2 16 2 7 14 5 7 24 4 7 18 1 2 28 ORDER OF VISITED VERTICES1 6 5 4 3 2 59
60. Advanced C Programming Kruskal’s Algorithm Algorithm Krushkal(E, cost,n,t) { for i=1 to n do parent[i]=-1; i=0; mincost=0; while( I < n-1) { Delete a minimum coast edge (u,v) form the heap and reheapfy using adjust J=find(u); K=find(v); If(j!=k) then { i=i+1; t[I,1]=u; t[I,2]=v; mincost=mincost+ cost[u,v]; union(j,k) } If( i != n-1) the write (“ no spanning tree”); else Return mincost } } 60
61. Advanced C Programming Kruskal’s Algorithm #include<iostream.h> #include<conio.h> #include<stdlib.h> int cost[10][10],i,j,k,n,m,c,visit,visited[10],l,v,count,count1,vst,p; main() { int dup1,dup2; cout<<"enter no of vertices"; cin >> n; cout <<"enter no of edges"; cin >>m; cout <<"EDGE Cost"; for(k=1;k<=m;k++) { cin >>i >>j >>c; cost[i][j]=c; cost[j][i]=c; } for(i=1;i<=n;i++) for(j=1;j<=n;j++) if(cost[i][j]==0) cost[i][j]=31999; visit=1; while(visit<n) { v=31999; for(i=1;i<=n;i++) for(j=1;j<=n;j++) if(cost[i][j]!=31999 && cost[i][j]<v && cost[i][j]!=-1 ) { count =0; for(p=1;p<=n;p++) { if(visited[p]==i || visited[p]==j) count++; } if(count >= 2) { for(p=1;p<=n;p++) if(cost[i][p]!=31999 && p!=j) 61
62. Advanced C Programming dup1=p; for(p=1;p<=n;p++) if(cost[j][p]!=31999 && p!=i) dup2=p; if(cost[dup1][dup2]==-1) continue; } l=i; k=j; v=cost[i][j]; } cout <<"edge from " <<l <<"-->"<<k; cost[l][k]=-1; cost[k][l]=-1; visit++; count=0; count1 =0; for(i=1;i<=n;i++) { if(visited[i]==l) count++; if(visited[i]==k) count1++; } if(count==0) visited[++vst]=l; if(count1==0) visited[++vst]=k; } } 62
63. Advanced C Programming Single Source Shortest Path #include<iostream.h> #include<conio.h> #include<stdio.h> int shortest(int ,int); int cost[10][10],dist[20],i,j,n,k,m,S[20],v,totcost,path[20],p; void main() { int c; clrscr(); cout <<"enter no of vertices"; cin >> n; cout <<"enter no of edges"; cin >>m; cout <<"nenternEDGE Costn"; for(k=1;k<=m;k++) { cin >> i >> j >>c; cost[i][j]=c; } for(i=1;i<=n;i++) for(j=1;j<=n;j++) if(cost[i][j]==0) cost[i][j]=31999; cout <<"enter initial vertex"; cin >>v; cout << v<<"n"; shortest(v,n); } shortest(int v,int n) { int min; for(i=1;i<=n;i++) { S[i]=0; dist[i]=cost[v][i]; } path[++p]=v; 63
64. Advanced C Programming S[v]=1; dist[v]=0; for(i=2;i<=n-1;i++) { k=-1; min=31999; for(j=1;j<=n;j++) { if(dist[j]<min && S[j]!=1) { min=dist[j]; k=j; } } if(cost[v][k]<=dist[k]) p=1; path[++p]=k; for(j=1;j<=p;j++) cout<<path[j]; cout <<"n"; //cout <<k; S[k]=1; for(j=1;j<=n;j++) if(cost[k][j]!=31999 && dist[j]>=dist[k]+cost[k][j] && S[j]!=1) dist[j]=dist[k]+cost[k][j]; } } OUTPUT enter no of vertices6 enter no of edges11 enter EDGE Cost 1 2 50 1 3 45 1 4 10 2 3 10 2 4 15 3 5 30 64
65. Advanced C Programming 4 1 10 4 5 15 5 2 20 5 3 35 653 enter initial vertex1 1 14 145 1452 13 65
66. Advanced C Programming Non recursive Pre order Traversing Algorithm Algorithm preorder( root) { 1. current = root; //start the traversal at the root node 2. while(current is not NULL or stack is nonempty) if(current is not NULL) { visit current; push current onto stack; current = current->llink; } else { pop stack into current; current = current->rlink; //prepare to visit //the right subtree } 66
67. Advanced C Programming Non recursive Pre order Traversing #include<iostream.h> #include<conio.h> #include<stdlib.h> class node { public: class node *left; class node *right; int data; }; class tree: public node { public: int stk[50],top; node *root; tree() { root=NULL; top=0; } void insert(int ch) { node *temp,*temp1; if(root== NULL) { root=new node; root->data=ch; root->left=NULL; root->right=NULL; return; } temp1=new node; temp1->data=ch; temp1->right=temp1->left=NULL; temp=search(root,ch); if(temp->data>ch) temp->left=temp1; else temp->right=temp1; 67
68. Advanced C Programming } node *search(node *temp,int ch) { if(root== NULL) { cout <<"no node present"; return NULL; } if(temp->left==NULL && temp->right== NULL) return temp; if(temp->data>ch) { if(temp->left==NULL) return temp; search(temp->left,ch);} else { if(temp->right==NULL) return temp; search(temp->right,ch); } } void display(node *temp) { if(temp==NULL) return ; display(temp->left); cout<<temp->data <<" "; display(temp->right); } void preorder( node *root) { node *p,*q; p=root; q=NULL; top=0; while(p!=NULL) { cout <<p->data << " "; if(p->right!=NULL) { stk[top]=p->right->data; top++; } p=p->left; if(p==NULL && top>0) 68
69. Advanced C Programming { p=pop(root); } } } node * pop(node *p) { int ch; ch=stk[top-1]; if(p->data==ch) { top--; return p; } if(p->data>ch) pop(p->left); else pop(p->right); } }; void main() { tree t1; int ch,n,i; while(1) { cout <<"n1.INSERTn2.DISPLAY 3.PREORDER TRAVERSEn4.EXITnEnter your choice:"; cin >> ch; switch(ch) { case 1: cout <<"enter no of elements to insert:"; cout<<"n enter the elements"; cin >> n; for(i=1;i<=n;i++) { cin >> ch; t1.insert(ch); } break; case 2: t1.display(t1.root);break; case 3: t1.preorder(t1.root); break; case 4: exit(1); } } } 69
70. Advanced C Programming OUTPUT 1.INSERT 2.DISPLAY 3.PREORDER TRAVERSE 4.EXIT Enter your choice:1 enter no of elements to insert enter the elements7 5 24 36 11 44 2 21 1.INSERT 2.DISPLAY 3.PREORDER TRAVERSE 4.EXIT Enter your choice:2 2 5 11 21 24 36 44 1.INSERT 2.DISPLAY 3.PREORDER TRAVERSE 4.EXIT Enter your choice:3 5 2 24 11 21 36 44 1.INSERT 2.DISPLAY 3.PREORDER TRAVERSE 4.EXIT Enter your choice:4 70
71. Advanced C Programming Non recursive In order Traversing Algorithm inorder( root) { 1. current = root; //start traversing the binary tree at // the root node 2. while(current is not NULL or stack is nonempty) if(current is not NULL) { push current onto stack; current = current->llink; } else { pop stack into current; visit current; //visit the node current = current->rlink; //move to the //right child } } 71
72. Advanced C Programming Non recursive In order Traversing #include<iostream.h> #include<conio.h> #include<stdlib.h> class node { public: class node *left; class node *right; int data; }; class tree: public node { public: int stk[50],top; node *root; tree() { root=NULL; top=0; } void insert(int ch) { node *temp,*temp1; if(root== NULL) { root=new node; root->data=ch; root->left=NULL; root->right=NULL; return; } temp1=new node; temp1->data=ch; temp1->right=temp1->left=NULL; temp=search(root,ch); if(temp->data>ch) temp->left=temp1; else temp->right=temp1; 72
73. Advanced C Programming } node *search(node *temp,int ch) { if(root== NULL) { cout <<"no node present"; return NULL; } if(temp->left==NULL && temp->right== NULL) return temp; if(temp->data>ch) { if(temp->left==NULL) return temp; search(temp->left,ch);} else { if(temp->right==NULL) return temp; search(temp->right,ch); } } void display(node *temp) { if(temp==NULL) return ; display(temp->left); cout<<temp->data; display(temp->right); } void inorder( node *root) { node *p; p=root; top=0; do { while(p!=NULL) { stk[top]=p->data; top++; p=p->left; } if(top>0) { 73
74. Advanced C Programming p=pop(root); cout << p->data; p=p->right; } }while(top!=0 || p!=NULL); } node * pop(node *p) { int ch; ch=stk[top-1]; if(p->data==ch) { top--; return p; } if(p->data>ch) pop(p->left); else pop(p->right); } }; void main() { tree t1; int ch,n,i; while(1) { cout <<"n1.INSERTn2.DISPLAY 3.INORDER TRAVERSEn4.EXITnEnter your choice:"; cin >> ch; switch(ch) { case 1: cout <<"enter no of elements to insert:"; cin >> n; for(i=1;i<=n;i++) { cin >> ch; t1.insert(ch); } break; case 2: t1.display(t1.root);break; case 3: t1.inorder(t1.root); break; case 4: exit(1); } 74
75. Advanced C Programming } } OUTPUT 1.INSERT 2.DISPLAY 3.INORDER TRAVERSE 4.EXIT Enter your choice:1 enter no of elements to inser 5 24 36 11 44 2 21 1.INSERT 2.DISPLAY 3.INORDER TRAVERSE 4.EXIT Enter your choice:3 251121243644 1.INSERT 2.DISPLAY 3.INORDER TRAVERSE 4.EXIT Enter your choice:3 251121243644 1.INSERT 2.DISPLAY 3.INORDER TRAVERSE 4.EXIT Enter your choice:4 75
76. Advanced C Programming Non recursive Post order Traversing Algorithm Algorithm postorder( node root) { 1. current = root; //start traversal at root node 2. v = 0; 3. if(current is NULL) the binary tree is empty 4. if(current is not NULL) a. push current into stack; b. push 1 onto stack; c. current = current->llink; d. while(stack is not empty) if(current is not NULL and v is 0) { push current and 1 onto stack; current = current->llink; } else { pop stack into current and v; if(v == 1) { push current and 2 onto stack; current = current->rlink; v = 0; } else visit current; }} 76
77. Advanced C Programming Non recursive Post order Traversing #include<iostream.h> #include<conio.h> #include<stdlib.h> class node { public: class node *left; class node *right; int data; }; class tree: public node { public: int stk[50],top; node *root; tree() { root=NULL; top=0; } void insert(int ch) { node *temp,*temp1; if(root== NULL) { root=new node; root->data=ch; root->left=NULL; root->right=NULL; return; } temp1=new node; temp1->data=ch; temp1->right=temp1->left=NULL; temp=search(root,ch); if(temp->data>ch) temp->left=temp1; else temp->right=temp1; 77
78. Advanced C Programming } node *search(node *temp,int ch) { if(root== NULL) { cout <<"no node present"; return NULL; } if(temp->left==NULL && temp->right== NULL) return temp; if(temp->data>ch) { if(temp->left==NULL) return temp; search(temp->left,ch);} else { if(temp->right==NULL) return temp; search(temp->right,ch); } } void display(node *temp) { if(temp==NULL) return ; display(temp->left); cout<<temp->data << " "; display(temp->right); } void postorder( node *root) { node *p; p=root; top=0; while(1) { while(p!=NULL) { stk[top]=p->data; top++; if(p->right!=NULL) stk[top++]=-p->right->data; p=p->left; 78
79. Advanced C Programming } } while(stk[top-1] > 0 || top==0) { if(top==0) return; cout << stk[top-1] <<" "; p=pop(root); } if(stk[top-1]<0) { stk[top-1]=-stk[top-1]; p=pop(root); } } node * pop(node *p) { int ch; ch=stk[top-1]; if(p->data==ch) { top--; return p; } if(p->data>ch) pop(p->left); else pop(p->right); } }; void main() { tree t1; int ch,n,i; clrscr(); while(1) { cout <<"n1.INSERTn2.DISPLAY 3.POSTORDER TRAVERSEn4.EXITnEnter your choice:"; cin >> ch; switch(ch) { case 1: cout <<"enter no of elements to insert:"; cout<<"n enter the elements"; cin >> n; for(i=1;i<=n;i++) { cin >> ch; 79
80. Advanced C Programming t1.insert(ch); } break; case 2: t1.display(t1.root);break; case 3: t1.postorder(t1.root); break; case 4: exit(1); } } } OUTPUT 1.INSERT 2.DISPLAY 3.POSTORDER TRAVERSE 4.EXIT Enter your choice:1 enter no of elements to insert: enter the elements7 5 24 36 11 44 2 21 1.INSERT 2.DISPLAY 3.POSTORDER TRAVERSE 4.EXIT Enter your choice:2 2 5 11 21 24 36 44 1.INSERT 2.DISPLAY 3.POSTORDER TRAVERSE 4.EXIT Enter your choice:3 2 21 11 44 36 24 5 1.INSERT 2.DISPLAY 3.POSTORDER TRAVERSE 4.EXIT Enter your choice:4 80
81. Advanced C Programming Quick Sort Algorithm Algorithm quicksort(a[],p,q) { V=a[p]; i=p; j=q; if(i<j) { repeat { repeat I=i+1; Until( a[i]> v); Repeat J=j-1; Until(a[j]<v); If(i<j) then interchange ( a[i],a[j]) }until (i<=j); interchange(a[j], a[p]) } quicksort(a[],p,j); quicksort(a[],j+1,q); } 81
82. Advanced C Programming Quick Sort #include<iostream.h> #include<conio.h> int a[10],l,u,i,j; void quick(int *,int,int); void main() { clrscr(); cout <<"enter 10 elements"; for(i=0;i<10;i++) cin >> a[i]; l=0; u=9; quick(a,l,u); cout <<"sorted elements"; for(i=0;i<10;i++) cout << a[i] << " "; getch(); } void quick(int a[],int l,int u) { int p,temp; if(l<u) { p=a[l]; i=l; j=u; while(i<j) { while(a[i] <= p && i<j ) i++; while(a[j]>p && i<=j ) j--; if(i<=j) { temp=a[i]; a[i]=a[j]; a[j]=temp;} } temp=a[j]; 82
83. Advanced C Programming a[j]=a[l]; a[l]=temp; cout <<"n"; for(i=0;i<10;i++) cout <<a[i]<<" "; quick(a,l,j-1); quick(a,j+1,u); } } OUTPUT enter 10 elements5 2 3 16 25 1 20 7 8 61 14 1 2 3 5 25 16 20 7 8 61 1 2 3 5 25 16 20 7 8 61 1 2 3 5 25 16 20 7 8 61 1 2 3 5 25 16 20 7 8 61 1 2 3 5 25 16 20 7 8 61 1 2 3 5 8 16 20 7 25 61 1 2 3 5 7 8 20 16 25 61 1 2 3 5 7 8 16 20 25 61 1 2 3 5 7 8 16 20 25 61 sorted elements1 2 3 5 7 8 16 20 25 61 83
84. Advanced C Programming Merge Sort Algorithm Algorithm Mergesort(low,high) { If(low<high) { Mid=(low+high)/2; Mergesort(low,mid); Mergesort(mid+1,high) Merge(low,mid,high); } } Algorithm Merge(low,mid,high) { h=low; i=low; j=mid+1; While(h<=mid and j<=high) do { If(a[h]<a[j]) then { b[i]=a[h]; h=h+1; } else { b[i]=a[j]; J=j+1; } if(h>mid) { For k= j to high do b[i]=a[k]; i=i+1; } Else { For k=h to mid do b[i]=a[k]; i=i+1; } For k= low to high do a[k]=b[k]; } } 84
85. Advanced C Programming Merge Sort #include<iostream.h> #include<conio.h> void mergesort(int *,int,int); void merge(int *,int,int,int); int a[20],i,n,b[20]; void main() { clrscr(); cout <<"N enter no of elements"; cin >> n; cout <<"enter the elements"; for(i=0;i<n;i++) cin >> a[i]; mergesort(a,0,n-1); cout <<" numbers after sort"; for(i=0;i<n;i++) cout << a[i] << " "; getch(); } void mergesort(int a[],int i,int j) { int mid; if(i<j) { mid=(i+j)/2; mergesort(a,i,mid); mergesort(a,mid+1,j); merge(a,i,mid,j); } } void merge(int a[],int low,int mid ,int high) { int h,i,j,k; h=low; i=low; j=mid+1; 85
86. Advanced C Programming while(h<=mid && j<=high) { if(a[h]<=a[j]) b[i]=a[h++]; else b[i]=a[j++]; i++; } if( h > mid) for(k=j;k<=high;k++) b[i++]=a[k]; else for(k=h;k<=mid;k++) b[i++]=a[k]; cout <<"n"; for(k=low;k<=high;k++) { a[k]=b[k]; cout << a[k] <<" "; } } OUTPUT N enter no of elements8 12 5 61 60 50 1 70 81 enter the elements 5 12 60 61 5 12 60 61 1 50 70 81 1 50 70 81 1 5 12 50 60 61 70 81 numbers after sort1 5 12 50 60 61 70 81 86
87. Advanced C Programming Heap Sort Algorithm Definition: A heap is a list in which each element contains a key, such that the key in the element at position k in the list is at least as large as the key in the element at position 2k + 1 (if it exists), and 2k + 2 (if it exists) Algorithm Heapify(a[],n) { For i=n/2 to 1 step -1 Adjustify (a,i,n); } Algorithm Adjustify(a[],i,n) { Repeat { J=leftchild(i) Compare left and right child of a[i] and store the index of grater number in j Compare a[j] and a[i] If (a[j]>a[i]) then Copy a[j] to a[i] and move to next level }until(j<n) } 87
88. Advanced C Programming Heap Sort #include<stdio.h> #include<conio.h> int a[20],n; main() { int i,j,temp; clrscr(); printf("ente n"); scanf("%d",&n); printf("enter the elements"); for(i=1;i<=n;i++) scanf("%d",&a[i]); heapify(a,n); for(j=1;j<=n;j++) printf("%d",a[j]); for(i=n;i>=2;i--) { temp=a[i]; a[i]=a[1]; a[1]=temp; adjust(a,1,i-1); printf("n"); for(j=1;j<=n;j++) printf("%d ",a[j]); } printf("nelements after sort"); for(i=1;i<=n;i++) printf("%d ",a[i]); } heapify(int a[],int n) { int i; for( i=n/2;i>=1;i--) adjust(a,i,n); } 88
89. Advanced C Programming adjust(int a[],int i,int n) { int j,iteam; j=2*i; iteam=a[i]; while(j<=n) { if(j<n && a[j]<a[j+1]) j=j+1; if(iteam>=a[j]) break; a[j/2]=a[j]; j=2*j; } a[j/2]=iteam; } 89
90. Advanced C Programming All Paris Shortest Path #include<iostream.h> #include<conio.h> int min(int a,int b); int cost[10][10],a[10][10],i,j,k,c; void main() { int n,m; cout <<"enter no of vertices"; cin >> n; cout <<"enter no od edges"; cin >> m; cout<<"enter thenEDGE Costn"; for(k=1;k<=m;k++) { cin>>i>>j>>c; a[i][j]=cost[i][j]=c; } for(i=1;i<=n;i++) for(j=1;j<=n;j++) { if(a[i][j]== 0 && i !=j) a[i][j]=31999; } for(k=1;k<=n;k++) for(i=1;i<=n;i++) for(j=1;j<=n;j++) a[i][j]=min(a[i][j],a[i][k]+a[k][j]); cout <<"Resultant adj matrixn"; for(i=1;i<=n;i++) { for( j=1;j<=n;j++) { if(a[i][j] !=31999) cout << a[i][j] <<" "; } cout <<"n"; } 90
91. Advanced C Programming getch(); } int min(int a,int b) { if(a<b) return a; else return b; } OUTPUT enter no of vertices3 enter no od edges5 enter the EDGE Cost 124 216 1 3 11 313 232 Resultant adj matrix 046 502 370 91
92. Advanced C Programming Algorithms for Binary Search Tree Algorithm Insert(item) { If(tree is empty) then root is empty Otherwise { temp=search(item); // temp is the node where search for the item halts if( item > temp) then temp-right=item; otherwise temp-left =item } } Algorithm Search(x, root) { if(tree is empty ) then print” tree is empty” otherwise If(x grater than root) search(root-right); Otherwise if(x less than root ) search(root-left) Otherwise return true } } Algorithm Delete(x) { Search for x in the tree If (not found) print” not found” Otherwise{ If ( x has no child) delete x; If(x has left child) move the left child to x position If(x has right child) move the right child to x position If(x has both left and right children) replace ‘x’ with greatest of left subtree of ‘x ‘ or smallest of right subtree of ‘x’ and delete selected node in the subtree } } 92
93. Advanced C Programming Binary Search Tree #include<iostream.h> #include<conio.h> #include<stdlib.h> class node { public: class node *left; class node *right; int data; }; class tree: public node { public: int stk[50],top; node *root; tree() { root=NULL; top=0; } void insert(int ch) { node *temp,*temp1; if(root== NULL) { root=new node; root->data=ch; root->left=NULL; root->right=NULL; return; } temp1=new node; temp1->data=ch; temp1->right=temp1->left=NULL; temp=search(root,ch); if(temp->data>ch) temp->left=temp1; else temp->right=temp1; 93
94. Advanced C Programming } node *search(node *temp,int ch) { if(root== NULL) { cout <<"no node present"; return NULL; } if(temp->left==NULL && temp->right== NULL) return temp; if(temp->data>ch) { if(temp->left==NULL) return temp; search(temp->left,ch);} else { if(temp->right==NULL) return temp; search(temp->right,ch); } } void display(node *temp) { if(temp==NULL) return ; display(temp->left); cout<<temp->data << " "; display(temp->right); } node * pop(node *p) { int ch; ch=stk[top-1]; if(p->data==ch) { top--; return p; } if(p->data>ch) pop(p->left); else pop(p->right); } }; 94
95. Advanced C Programming void main() { tree t1; int ch,n,i; clrscr(); while(1) { cout <<"n1.INSERTn2.POPn3.DISPLAYn4.EXITnEnter your choice:"; cin >> ch; switch(ch) { case 1: cout <<"enter no of elements to insert:"; cout<<"n enter the elements"; cin >> n; for(i=1;i<=n;i++) { cin >> ch; t1.insert(ch); } break; case 2: t1.pop(); break; case 3: t1.display(t1.root);break; case 4: exit(1); } } } 1.INSERT 2.POP 3.DISPLAY 4.EXIT Enter your choice:1 enter no of elements to insert: enter the elements7 5 24 36 11 44 2 21 1.INSERT 2.POP 3.DISPLAY 4.EXIT Enter your choice:3 2 5 11 21 24 36 44 1.INSERT 2.POP 3.DISPLAY 4.EXIT Enter your choice2 2 11 21 24 36 44 1.INSERT 2.POP 3.DISPLAY 4.EXIT 4.EXITEnter your choice:4 95
96. Advanced C Programming Optimal Binary Search Tree #include<iostream.h> #include<conio.h> #include<stdio.h> #define MAX 10 int find(int i,int j); void print(int,int); int p[MAX],q[MAX],w[10][10],c[10][10],r[10][10],i,j,k,n,m; char idnt[7][10]; void main() { clrscr(); cout << "enter the no, of identifiers"; cin >>n; cout <<"enter identifiers"; for(i=1;i<=n;i++) gets(idnt[i]); cout <<"enter success propability for identifiers"; for(i=1;i<=n;i++) cin >>p[i]; cout << "enter failure propability for identifiers"; for(i=0;i<=n;i++) cin >> q[i]; for(i=0;i<=n;i++) { w[i][i]=q[i]; c[i][i]=r[i][i]=0; w[i][i+1]=q[i]+q[i+1]+p[i+1]; r[i][i+1]=i+1; c[i][i+1]=q[i]+q[i+1]+p[i+1]; } w[n][n]=q[n]; r[n][n]=c[n][n]=0; for(m=2;m<=n;m++) { for(i=0;i<=n-m;i++) { j=i+m; w[i][j]=w[i][j-1]+p[j]+q[j]; k=find(i,j); r[i][j]=k; c[i][j]=w[i][j]+c[i][k-1]+c[k][j]; } 96
97. Advanced C Programming } cout <<"n"; print(0,n); } int find(int i,int j) { int min=2000,m,l; for(m=i+1;m<=j;m++) if(c[i][m-1]+c[m][j]<min) { min=c[i][m-1]+c[m][j]; l=m; } return l; } void print(int i,int j) { if(i<j) puts(idnt[r[i][j]]); else return; print(i,r[i][j]-1); print(r[i][j],j); } OUTPUT enter the no, of identifiers4 enter identifiersdo if int while enter success propability for identifiers3 3 1 1 enter failure propability for identifiers2 3 1 1 1 tree in preorder form if do int while 97
98. Advanced C Programming Viva Voice Questions 1. What is the difference between an ARRAY and a LIST? 2. What is faster : access the element in an ARRAY or in a LIST? 3. Define a constructor - what it is and how it might be called (2 methods). 4. Describe PRIVATE, PROTECTED and PUBLIC - the differences and give examples. 5. What is a COPY CONSTRUCTOR and when is it called (this is a frequent question !)? 6. Explain term POLIMORPHISM and give an example using eg. SHAPE object: If I have a base class SHAPE, how would I define DRAW methods for two objects CIRCLE and SQUARE. 7. What is the word you will use when defining a function in base class to allow this function to be a polimorphic function? 8. You have two pairs: new() and delete() and another pair : alloc() and free(). Explain differences between eg. new() and malloc() 9. Difference between “C structure” and “C++ structure”. 10. Diffrence between a “assignment operator” and a “copy constructor” 11. What is the difference between “overloading” and “overridding”? 12. Explain the need for “Virtual Destructor”. 13. Can we have “Virtual Constructors”? 14. What are the different types of polymorphism? 15. What are Virtual Functions? How to implement virtual functions in “C” 16. What are the different types of Storage classes? 17. What is Namespace? 98
99. Advanced C Programming 18. Difference between “vector” and “array”? 19. How to write a program such that it will delete itself after exectution? 20. Can we generate a C++ source code from the binary file? 21. What are inline functions? 22. What is “strstream” ? 23. Explain “passing by value”, “passing by pointer” and “passing by reference” 24. Have you heard of “mutable” keyword? 25. Is there something that I can do in C and not in C++? 26. What is the difference between “calloc” and “malloc”? 27. What will happen if I allocate memory using “new” and free it using “free” or allocate sing “calloc” and free it using “delete”? 28. When shall I use Multiple Inheritance? 29. How to write Multithreaded applications using C++? 30. Write any small program that will compile in “C” but not in “C++” 31. What is Memory Alignment? 32. what is the difference between a tree and a graph? 33. How to insert an element in a binary search tree? 34. How to delete an element from a binary search tree? 35. How to search an element in a binary search tree? 36. what is the disadvantage in binary search tree? 37. what is ment by height balanced tree? 38. Give examples for height blanced tree? 39. What is a 2-3 tree? 99
100. Advanced C Programming 40. what is a dictonary? 41.What is a binary search tree? 42. what is an AVL tree? 43. how height balancing is performed in AVL tree? 44. what is a Red Black tree? 45. what is difference between linked list and an array? 46. how dynamic memory allocation is performed in c++? 47. what are tree traversing techniques? 48. what are graph traversing techniques? 49. what is the technique in quick sort.? 50 what is the technique in merge sort? 51. what is data structure. 52. how to implement two stacks in an array? 53. what is ment by generic programming? 54. write the syntax for function templet? 55. write the syntax for class templet? 56. what is ment by stream? 57. what is the base class for all the streams? 58. how to create a file in c++? 59. how do you read a file in c++? 60. how do you write a file in c++? Finite- State Machines Derived from the on-line notes by Dr. Matt Stallmann & Suzanne Balik. 100
101. Advanced C Programming A finite-state machine (FSM) is an abstract model of a system (physical, biological, mechanical, electronic, or software). An FSM consists of • • • • • States: a finite number of states that represent the internal "memory" of the system. Inputs to the system (e.g., a user doing something, a character read from the input) Transitions, which represent the "response" of the system to its environment. Transitions depend upon the current state of the machine as well as the current input and often result in a change of state. An initial state, which is the state the system is in before it accepts any input. Final states, states that represent a legal end to a transaction. Simple example: A soda- pop machine Consider an FSM that models a soda-pop machine that dispenses soda-pop for 30 cents. The possible inputs to the machine are n - nickel, d - dime, q - quarter, s - select soda. The states of the machine are designated by circles. Each state is labeled with the amount of money that has been deposited so far. State 00 is the initial or start state. This is indicated by the incoming arrow. States which represent a total input of 30 or more cents are considered final states and are designated by double circles. The transitions from state to state are shown as arcs (lines with arrows). Each transition is labeled with the input that caused it. Let’s take an example. Suppose the user inserts a sequence of coins … q takes the machine to state 25; another q takes it to state 50. It dispenses soda pop & change and transitions back to state 00. At this point, the user could select a drink. The machine would also have to give change. 101
102. Advanced C Programming An FSM for an ATM A finite-state machine can also be used to model a simple automated teller machine (ATM). In this case, each transition is labeled with an input— • from the user (such as insert card), or • from a central database (such as PIN OK versus bad PIN). Each of the transitions may, in actuality, involve a number of complex steps which are not shown on the diagram. 102
103. Advanced C Programming Recognizing input streams FSMs can also be used to precisely define programming-language syntax. Another application of finite-state machines is as a notation for the precise description of a programming language syntax. Consider this description of real constants in Pascal: A real number in PASCAL is written as a string of digits containing a decimal point. There must be at least one digit before and after the decimal point. 103
104. Advanced C Programming Real data may also be represented using PASCAL scientific notation. A real data item written in scientific notation consists of a sign followed by a real number, followed by the letter E, another sign and an integer (+ signs may be omitted). The English description is imprecise. Are the strings .9 and 9. valid, or do you have to say 0.9 and 9.0, respectively? This FSM answers the question: A valid real constant in Pascal is any string that leads the FSM from the start state to a final (double-circled) state. What happens when we feed the FSM .9? Starts out in State 0. There’s no transition for a dot, so that’s an illegal input. What happens when we feed it 9.? Starts out in State 0, and then goes to State 2 and then to State 3. But that’s an error, because State 3 is not an accepting state. What about the string 9.0? Starts out in State 0, and then goes to State 2 and then to State 3, and then Step 4. This is an accepting state, so 9.0 is a legal real constant. (Note that .9 and 9. are valid floating-point constants in Java.) Text Processing with FSMs Finite-state machines are often used in text processing. The grep utility takes a string or regular expression and converts it to an FSM before doing a search. Simplifed example: Suppose a file consists of as and bs only and the search is for the string "abba". Here’s an FSM for doing this search: If, for example, this FSM were used to locate the string "abba" in a file containing "aababbbabba...", it would move through these states: Input: 104 a a b a b b b a b b a
105. Advanced C Programming State: 0 1 1 2 1 2 3 0 1 2 3 4 The states listed above are the states of the machine after the corresponding input. When the final state 4 is reached, the search is successful. We say that the machine has recognized the string. Exercise: Design an FSM that recognizes a string consisting of alternating as and bs, beginning with an a. If the input consists of anything else (e.g., abababaab, the FSM should continue to read it but not recognize it. Here is how an FSM can be developed and translated directly into a program. Consider the following specifications: • A word is a maximal sequence of zero or more upper- and zero or more lower-case letters. Under this definition, what about the sequence "Go State!"? Is "State!" a word? No; it contains a punctuation mark. Is "tat" a word? No, because it is not maximal. How about "irregardless" Yes, because it’s a sequence of (0 or more) upper & lower-case letters. • wc is the word count, initially 0 • lc is the line count, initially 0 • cc is the character count, initially 0 Here’s an FSM to recognize "words." Note that each transition is labeled with an action as well as an input. In state 0, the FSM remembers that we're not currently in the middle of a word, while state 1 remembers that we are. Question: Why are there no final states in this FSM? It’s not trying to recognize anything; it’s just counting characters. An FSM’s behavior can also be specified by a table. Each row corresponds to a state Each column corresponds to an input symbol (or category of input symbols). The table version for the word counting FSM is given below. 105
106. Advanced C Programming State 0 1 A–Z a–z 1: ++wc ++cc 1: ++cc n 0: ++lc, ++cc 0: ++lc, ++cc other 0: ++cc 0: ++cc A standard idiom for translating an FSM into a program is the while-switch idiom. • A while loop gets one character at a time from the input stream. • Inside the loop is a switch statement that causes different code to be executed based on the current state. Here is the code. import java.io.*; public class WordCounter{ public static void main(String[] args) { if (args.length == 1) { try { BufferedReader br = new BufferedReader( new FileReader(args[0])); int wc = 0, lc = 0, cc = 0; char ch; int state = 0; int next; while ((next = br.read()) != −1) { ch = (char) next; switch (state) { case 0: if ( ch == ’n’ ) { ++lc; ++cc; } else if ( Character.isLetter(ch) ) { state = 1; ++wc; ++cc; } else ++cc; break; case 1: if ( ch == ’n’ ) { state = 0; ++lc; ++cc; } else if ( Character.isLetter(ch) ) ++cc; else { state = 0; ++cc; 106
107. Advanced C Programming } break; default: System.out.println("Invalid state: " + state); } } System.out.println( lc + "t" + wc + "t"+ cc); } catch(IOException e) { System.out.println("File error: " + e); System.out.println( "usage: java WordCounter filename"); } } else System.out.println("usage: java WordCounter filename"); } } Within each case, the program makes a decision based on the current input character. Simplifying the program The program can be simplifed by noting that— • ++cc is done on every transition, • whenever the current input is n we do ++lc and go to (or stay in) state 0, and • otherwise, the state and counters change only when a letter is encountered in state 0 or a character other than a letter (or a newline) is encountered in state 1. The simplified program looks like: import java.io.*; public class WordCounter{ public static void main(String[] args) { if (args.length == 1) { try { BufferedReader br = new BufferedReader( new FileReader(args[0])); int wc = 0, lc = 0, cc = 0; char ch; int state = 0; int next; while ((next = br.read()) != −1) { ch = (char) next; ++cc; if (ch == ‘n’) { ++lc; 107
108. Advanced C Programming state = 0; } else if (state == 0 && Character.isLetter(ch)) { ++wc; state = 1; } else if (state == 1 && !Character.isLetter(ch)) { state = 0; } } System.out.println( lc + "t" + wc + "t"+ cc); } catch(IOException e) { System.out.println("File error: " + e); System.out.println( "usage: java WordCounter filename"); } } else System.out.println("usage: java WordCounter filename"); } } So FSMs can sometimes be "optimized" considerably. But, to avoid bugs, it is best to start with the straightforward implementation! Summary 1. Finite-state machines can be used to model the interaction between a system and its environment. 2. The state of an FSM remembers what has occurred so far. In addition to the FSM state, there may be variables that remember other details. The designer has to use judgment to decide what to model with an FSM state and what to leave as a variable. 3. A transition occurs when an event in the environment causes the system to change state (either the FSM state or variables). 4. A FSM can be depicted either by a bubble diagram or a transition table. 5. In text stream applications each transition corresponds to a single input character. 6. The while-switch idiom gives a method for mechanically translating a FSM to a program. Simplifications can be made by taking advantage of special features of the particular FSM. 108
109. Advanced C Programming Real Time Kernel This is a project to implement a real-time kernel for the ARM processor. The main features of this kernel will be: 1) 2) 3) 4) 5) Priority based preemptive scheduling Dynamic creation and deletion of tasks. Provision for all types of synchronization mechanisms. Mechanism for large number of timers. A simple best user API. Prerequisites: The following skills are essential for implementing this project.: Operating System Concepts A students must have a thorough understanding of Operating Systems concepts. In case the students don’t have this knowledge, a one week course on OS fundamentals must be conducted by the guide which must cover the following topics: 1) 2) 3) 4) 5) 6) Scheduling Synchronization Mutual Exclusion I/O Management Process Management Memory Management The discussion must be slanted towards Real-time systems. And implementation points must be discussed. The following text books must used : Operating System Concepts : Silbersatcz and Galvin Operating System : Design and Implementation : Tannenbaumm and Wood hull C Programming in the Linux Environment The students must have a very good knowledge of C programming language as it is the language for the project implementation. Further , the students must have moderate familiarity with the Linux application development process. The usage of GCC,MAKE and VI editor must be known by the students. Compilation , Linking , Loading and Object Files Formats 109
110. Advanced C Programming A familiarity with object file formats and compilation process is essential. If the students don’t have any knowledge in this field , then a 1 day session ca be taken by the guide. DESIGN OF THE KERNEL The following are the steps for running a kernel(The detailed steps are given in the project schedule) 1) BOOTING 2) SYSTEM INITIALISATION The steps are considered one by one : BOOTING Operating Systems are unlike other programs. It has to be put into memory by a separate program called boot-loader. It is because the kernel is the first program to run in the system. There should be an understanding between the boot loader and operating system. The is given in a standard called multi boot. In Linux GRUB is the boot loader program which implements the multi boot standard. We have to understand this standard if we have to boot out own kernel. The complete documentation of multi boot and GRUB is 1) Available on the Internet. 2) Available as online documentation on the Linux system[info grub]. SYSTEM INITIALIZATION After booting the system has to be initialized. First the memory has to be initialized. Next the interrupts have to be setup. Finally if there is hardware to be initialized it is done. In particular the Intel 80386 must be initialized as follows : (for the flat memory mode). 1) 2) 3) 4) 5) Disable all interrupts Setup GDT – kernel ‘s code, data, stack and other segments. Setup IDT – manage all exceptions and interrupts. Load GDT register and IDT register. Initialize Hardware – setup timer hardware to generate periodic interrupts. 6) Enable Interrupts. DATA STRUCTURES AND ALGORITHMS 110