C – Practical System Programming for Graphics in C.

Practical System Programming for Graphics in C

By Shritam Bhowmick
Web Application Penetration Tester
LinkedIn: https://www.linkedin.com/profile/view?id=281014248&trk=nav_responsive_tab_profile
Academia: https://independent.academia.edu/ShritamBhowmick
Facebook: https://www.facebook.com/coded32

This is an introduction post for “Graphics and Internet” for WBUT 2014, which includes C as system programming and the underlying source code concepts for drawing basic graphics using C. Earlier in this post, I discussed data-structures in C for WBUT University in details which covered the practical aspects of year 2014. This post in contrary will mention the graphical aspect of the C programming keeping in mind the practical labs. I am glad I could post as far as the syllabus is concerned (which is really outdated as discussed in the last post), but this will help the current students who could have a look here after 2014 passes.

===========================================================================

1.) Draw a Basic Circle:

#include
#include

main()
{
int gd = DETECT, gm;

initgraph(&gd, &gm, "C:TCBGI");

circle(100, 100, 50);

getch();
closegraph();
return 0;
}

===========================================================================

2.) Draw a Rainbow

#include
#include
#include
#include

void main()
{
int gdriver = DETECT,gmode;
int x,y,i;
initgraph(&gdriver,&gmode,"C:TurboC4BGI");
x=getmaxx()/2;
y=getmaxy()/2;
for(i=30;i<200;i++)
{
delay(100);
setcolor(i/10);
arc(x,y,0,180,i-10);
}
getch();
}

===========================================================================

3.) Draw a Smiley

#include
#include
#include
#include

void main()

{
int gd=DETECT,gm;
initgraph(&gd,&gm,"C:BGI");
outtextxy(150,10,"This is a program to draw a Smiley Asshole!");
circle(200,200,80);
circle(160,180,8);
circle(240,180,8);
arc(200,230,180,360,20);
line(190,245,210,260);
line(210,245,210,260)
arc(200,260,180,360,9);
line(200,180,195,210);
arc(202,210,190,10,8);
getch();
closegraph();

}

===========================================================================

4.) Draw a Diagonal Line

#include < stdio.h >
#include < conio.h >
#include < graphics.h >

void main()
{
int gdriver=DETECT,gmode=0;
initgraph(&gdriver, &gmode, "c:tcbgi");
line(0,0,getmaxx(),getmaxy());
getch();
closegraph();

}

===========================================================================

5.) Draw an Eclipse

#include
#include

main()
{
int gd = DETECT, gm;

initgraph(&gd, &gm, "C:TCBGI");

ellipse(100, 100, 0, 360, 50, 25);

getch();
closegraph();
return 0;
}

===========================================================================

5.) Draw a Triangle

#include
#include
#include

void main() {
int gd = DETECT, gm;
initgraph(&gd, &gm, "c:tcbgi");

line(300, 100, 200, 200);
line(300, 100, 400, 200);
line(200, 200, 400, 200);

getch();
closegraph();
}

There are others as well which are related to Web Development. The posts belonging to Web-development could already be accessed from here. The section is for everything related to ‘code’ and working with code in different languages. HTML could be found there. I would appreciate if the readers leave behind a feedback. Roger out./-

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Data Structures in C Practicals and Why Colleges with their IDE’s are Outdated. A Wiser Proof of Concept.

Data Structures in C and Why Colleges with their IDE’s are Outdated!

By Shritam Bhowmick
Web Application Penetration Tester
LinkedIn: https://www.linkedin.com/profile/view?id=281014248&trk=nav_responsive_tab_profile
Academia: https://independent.academia.edu/ShritamBhowmick
Facebook: https://www.facebook.com/coded32

The Introduction Preface

Hi, this is about all the practical lab C  for WBUT Winters (Semester 3rd of the University prescribed Syllabus) section data-structures questions solved at once place for a ready reference to the students who might seek help and also on an information as to why colleges now should be moving from the stone age. All the programs are complied under Tubro C++ compiler and would not be/should not be expected to be executed/processed to create an object under GNU C compiler or  any Windows variant compilers (such as Visual C compiler). Since the syllabus itself is too outdated and people had been talking about them in various forums; I came up with a compilation of the code as well as the questions which could help them understand the core with the ready service code to be executed only within from Turbo C++ compiler. This is C code and not C++ code. The source code for each of them had been practically tested on Windows 8.1 platform using DOSBOX as an emulator to run Turbo C.

The Recommendations

It’s recommended to move on with the syllabus and pay less attention there, if anyone really is serious on C and C++ programming. Python, Ruby, Delphi, Lua and Perl are the modern spectacular languages to choose from and Java is really not that friendly as you might think if you have to do a certain task. Java isn’t that flexible in terms that Java would really need long source code for a simple problem. See “Python to other languages – A comparison“. Get to know more at Stackoverflow. To refernce from the original source, here’s is the justification why someone would prefer to code in Python rather than any other older languages in a nutshell:

  1. Speed of development. I can crank out working code before the Java folks have stuff that will compile.
  2. Flexibility. I can refactor and rework a module in an afternoon. The Java folks have to redesign things on paper to be sure they can get some of the bigger refactorings to recompile. In Java, the changes aren’t as small and focused.I should probably blog about some of the nastier refactorings that would be a huge pain — even using the refactoring tools in Eclipse or NetBeans. The most recent example was a major change to the top-most superclass of a class hierarchy that would have invalidated a mountain of Java code. But the method was used in exactly two places, allowing a trivial change to the superclass and those two places. 100’s of unit tests passed, so I’m confident in the change without the compiler overhead.
  3. Simplicity. Python doesn’t need a sophisticated IDE and super-complex WS libraries. Python code can be written in Komodo Edit. A WS interface between your Java front-end and PostgresSQL can be knocked together in Werkzeug in a day. You can plug your transactions into this without much overhead.

The post primarily covered Python against Java, but there are instances why some one could just prefer Python over C which could be referenced from here in this post. This does not at all mean to compare C and Python on basis of what they have to deliver but rather compares them on an average on the end-users opinion perspective. There are other considerations as well into modern computing which is resolved below in the list.

Why Would Python be Slow but equivalently faster ever than C Deployments?

Python is a higher level language than C, which means it abstracts the details of the computer from you – memory management, pointers, etc, and allows you to write programs in a way which is closer to how humans think. It is true that C code usually runs 10 to 100 times faster than Python code if you measure only the execution time. However if you also include the development time Python often beats C. For many projects the development time is far more critical than the run time performance. Longer development time converts directly into extra costs, fewer features and slower time to market.

Why use C contrary to the switching to any other programming languages, a hope out of no-where!?

“If there ever were a quote that described programming with C, it would be this. To many programmers, this makes C scary and evil. It is the Devil, Satan, the trickster Loki come to destroy your productivity with his seductive talk of pointers and direct access to the machine. Then, once this computational Lucifer has you hooked, he destroys your world with the evil “segfault” and laughs as he reveals the trickery in your bargain with him.

But, C is not to blame for this state of affairs. No my friends, your computer and the Operating System controlling it are the real tricksters. They conspire to hide their true inner workings from you so that you can never really know what is going on. The C programming language’s only failing is giving you access to what is really there, and telling you the cold hard raw truth. C gives you the red pill. C pulls the curtain back to show you the wizard. C is truth. Why use C then if it’s so dangerous? Because C gives you power over the false reality of abstraction and liberates you from stupidity.”

The Coverage

Now since we are already hopped into C with Turbo C as a compiler. I’d go writing the source code of the practicals here and try to keep it updated as much as I could to my disposal. Instead this is the responsibility at your side to post back with comments on what else its to be added so this could help others in the process.

Notes

The source code has replaced all the instances of \n to n. Take care of that plus the include headers have not been defined to be kept away from copy/paste practices. This in my opinion will lead to a better coding practices and would lead to a great start with self-research.

The Question Set

The questions I have dug so far follows these:

  1. Doubly Linked List with insertion and deletion.
  2. Priority Queue with addition and deletion of elements.
  3. Dequeue or Double Ended Queue using Linked List.
  4. Dequeue or Double Ended Queue using Circular Array.
  5. Priority Queue Implementation using Arrays.
  6. Circular Queue using Linked List with Circular Queue Concept Notes.
  7. Circular Queue Implementation in C Data-Structures using Arrays.
  8. Stack for PUSH and POP operations in C both via recursive and regular methods.
  9. Linear Search Implementation in C using Arrays.
  10. Binary Search Implementation in C using Arrays.
  11. Insertion Sort Implementation in C using Arrays.
  12. Merge Sort Implementation in C using Arrays.
  13. Quick Sort Implementation in C using Arrays.
  14. Heap Sort Implementation in C using Arrays.

Source Code Solutions

1.) Doubly Linked List with Insertion, Deletion and Reverse Display

Here’s the source code (to avoid copy/paste gimmicks, I have stripped off all the includes. Feel free to add them at your own expense) for Doubly Linked List:

#include
#include
#include
struct node
{
struct node *previous;
int data;
struct node *next;
}*head, *last;

void insert_beginning(int value)
{
struct node *var,*temp;
var=(struct node *)malloc(sizeof(struct node));
var->data=value;
if(head==NULL)
{
head=var;
head->previous=NULL;
head->next=NULL;
last=head;
}
else
{
temp=var;
temp->previous=NULL;
temp->next=head;
head->previous=temp;
head=temp;
}
}

void insert_end(int value)
{
struct node *var,*temp;
var=(struct node *)malloc(sizeof(struct node));
var->data=value;
if(head==NULL)
{
head=var;
head->previous=NULL;
head->next=NULL;
last=head;
}
else
{
last=head;
while(last!=NULL)
{
temp=last;
last=last->next;
}
last=var;
temp->next=last;
last->previous=temp;
last->next=NULL;
}
}

int insert_after(int value, int loc)
{
struct node *temp,*var,*temp1;
var=(struct node *)malloc(sizeof(struct node));
var->data=value;
if(head==NULL)
{
head=var;
head->previous=NULL;
head->next=NULL;
}
else
{
temp=head;
while(temp!=NULL && temp->data!=loc)
{
temp=temp->next;
}
if(temp==NULL)
{
printf("n%d is not present in list ",loc);
}
else
{
temp1=temp->next;
temp->next=var;
var->previous=temp;
var->next=temp1;
temp1->previous=var;
}
}
last=head;
while(last->next!=NULL)
{
last=last->next;
}
}
int delete_from_end()
{
struct node *temp;
temp=last;
if(temp->previous==NULL)
{
free(temp);
head=NULL;
last=NULL;
return 0;
}
printf("nData deleted from list is %d n",last->data);
last=temp->previous;
last->next=NULL;
free(temp);
return 0;
}

int delete_from_middle(int value)
{
struct node *temp,*var,*t, *temp1;
temp=head;
while(temp!=NULL)
{
if(temp->data == value)
{
if(temp->previous==NULL)
{
free(temp);
head=NULL;
last=NULL;
return 0;
}
else
{
var->next=temp1;
temp1->previous=var;
free(temp);
return 0;
}
}
else
{
var=temp;
temp=temp->next;
temp1=temp->next;
}
}
printf("data deleted from list is %d",value);
}

void display()
{
struct node *temp;
temp=head;
if(temp==NULL)
{
printf("List is Empty");
}
while(temp!=NULL)
{
printf("-> %d ",temp->data);
temp=temp->next;
}
}

int main()
{
int value, i, loc;
head=NULL;
printf("Select the choice of operation on link list");
printf("n1.) insert at begningn2.) insert at atn3.) insert at middle");
printf("n4.) delete from endn5.) reverse the link listn6.) display listn7.)exit");
while(1)
{
printf("nnenter the choice of operation you want to do ");
scanf("%d",&i);
switch(i)
{
case 1:
{
printf("enter the value you want to insert in node ");
scanf("%d",&value);
insert_beginning(value);
display();
break;
}
case 2:
{
printf("enter the value you want to insert in node at last ");
scanf("%d",&value);
insert_end(value);
display();
break;
}
case 3:
{
printf("after which data you want to insert data ");
scanf("%d",&loc);
printf("enter the data you want to insert in list ");
scanf("%d",&value);
insert_after(value,loc);
display();
break;
}
case 4:
{
delete_from_end();
display();
break;
}
case 5:
{
printf("enter the value you want to delete");
scanf("%d",value);
delete_from_middle(value);
display();
break;
}
case 6 :
{
display();
break;
}
case 7 :
{
exit(0);
break;
}
}
}
printf("nn%d",last->data);
display();
getch();
}

2.) Priority Queue with Addition and Deletion of Elements

#include
#include
#define MAX 5

void insert_by_priority(int);
void delete_by_priority(int);
void create();
void check(int);
void display_pqueue();

int pri_que[MAX];
int front, rear;

void main()

{

int n, ch;
printf("n1 - Insert an element into queue");
printf("n2 - Delete an element from queue");
printf("n3 - Display queue elements");
printf("n4 - Exit");
create();

while (1)

{

printf("nEnter your choice : ");

scanf("%d", &ch);

switch (ch)

{

case 1:

printf("nEnter value to be inserted : ");

scanf("%d",&n);

insert_by_priority(n);

break;

case 2:

printf("nEnter value to delete : ");

scanf("%d",&n);

delete_by_priority(n);

break;

case 3:

display_pqueue();

break;

case 4:

exit(0);

default:

printf("nChoice is incorrect, Enter a correct choice");

}

}

}

/* Function to create an empty priority queue */

void create()

{

front = rear = -1;

}

/* Function to insert value into priority queue */

void insert_by_priority(int data)

{

if (rear >= MAX - 1)

{

printf("nQueue overflow no more elements can be inserted");

return;

}

if ((front == -1) && (rear == -1))

{

front++;

rear++;

pri_que[rear] = data;

return;

}

else

check(data);

rear++;

}

/* Function to check priority and place element */

void check(int data)

{

int i,j;

for (i = 0; i <= rear; i++) { if (data >= pri_que[i])

{

for (j = rear + 1; j > i; j--)

{

pri_que[j] = pri_que[j - 1];

}

pri_que[i] = data;

return;

}

}

pri_que[i] = data;

}

/* Function to delete an element from queue */

void delete_by_priority(int data)

{

int i;

if ((front==-1) && (rear==-1))

{

printf("nQueue is empty no elements to delete");

return;

}

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

{

if (data == pri_que[i])

{

for (; i < rear; i++)

{

pri_que[i] = pri_que[i + 1];

}

pri_que[i] = -99;

rear--;

if (rear == -1)

front = -1;

return;

}

}

printf("n%d not found in queue to delete", data);

}

/* Function to display queue elements */

void display_pqueue()

{

if ((front == -1) && (rear == -1))

{

printf("nQueue is empty");

return;

}

for (; front <= rear; front++)

{

printf(" %d ", pri_que[front]);

}

front = 0;

}

3.) Dequeue or Double Ended Queue using Linked List

What is Dequeue?

The word dequeue is short form of double ended queue. In a dequeue , insertion as well as deletion can be carried out either at the rear end or the front end. The following diagram illustrates a version of the same concept:

Dequeue Represented Using a Circular Array

The source code is available below:

#define MAX 100
#include
#include

void insert(int);
int delStart();
int delEnd();
int queue[MAX];
int rear =0, front =0;
void display();
void insertEnd(int);
void insertStart(int);

int main()
{
char ch , a='y';
int choice, c, token;
printf("Enter your choice for which deque operation you want to perform operation n");
do
{
printf("n1.Input-restricted deque n");
printf("2.output-restricted deque n");
printf("nEnter your choice for the operation : ");
scanf("%d",&c);
switch(c)
{
case 1:
printf("nDo operation in Input-Restricted c dequen");
printf("1.Insert");
printf("n2.Delete from end");
printf("n3.Delete from begning");
printf("n4.show or display");
do
{
printf("nEnter your choice for the operation in c deque: ");
scanf("%d",&choice);
switch(choice)
{
case 1: insertEnd(token);
display();
break;
case 2: token=delEnd();
printf("nThe token deleted is %d",token);
display();
break;
case 3: token=delStart();
printf("nThe token deleted is %d",token);
display();
break;
case 4: display();
break;
default:printf("Wrong choice");
break;
}
printf("nDo you want to continue(y/n) to do operation in input-restricted c deque: ");
ch=getch();
}
while(ch=='y'||ch=='Y');
getch();
break;

case 2 :
printf("nDo operation in Output-Restricted c dequen");
printf("1.Insert at the End");
printf("n2.Insert at the begning");
printf("n3.Delete the element");
printf("n4.show or display");
do
{
printf("nEnter your choice for the operation: ");
scanf("%d",&choice);
switch(choice)
{
case 1: insertEnd(token);
display();
break;
case 2: insertStart(token);
display();
break;
case 3: token=delStart();
printf("nThe token deleted is %d",token);
display();
break;
case 4: display();
break;
default:printf("Wrong choice");
break;
}
printf("nDo you want to continue(y/n):");
ch=getch();
}
while(ch=='y'||ch=='Y');
getch();
break ;
}

printf("nDo you want to continue(y/n):");
ch=getch();
}
while(ch=='y'||ch=='Y');
getch();
}

void display()
{
int i;
printf("nThe queue elements are:");
for(i=rear;i<front;i++)
{
printf("%d  ",queue[i]);
}
}

void insertEnd(int token)
{
char a;
if(front==MAX/2)
{
printf("nQueue fullnyou cant enter more elements at the end of c queue");
return;
}
do
{
printf("nEnter the token to be inserted:");
scanf("%d",&token);
queue[front]=token;
front=front+1;
printf("do you want to continue insertion Y/N");
a=getch();
}
while(a=='y');
}

void insertStart(int token)
{
char a;
if(front==MAX/2)
{
printf("nQueue fullnyou cant enter more elements at the start of queue");
return;
}
do
{
printf("nEnter the token to be inserted:");
scanf("%d",&token);
rear=rear-1;
queue[rear]=token;
printf("do you want to continue insertion Y/N");
a=getch();
}
while(a=='y');
}
int delEnd()
{
int t;
if(front==rear)
{
printf("nQueue empty");
return 0;
}
front=front-1;
t=queue[front+1];
return t;
}
int delStart()
{
int t;
if(front==rear)
{
printf("nQueue empty");
return 0;
}
rear=rear+1;
t=queue[rear-1];
return t;
}

4.) Dequeue or Double Ended Queue using Circular Array

The available source code is different from the previous one which used linked list. This time the code uses the concept for circular array and implements a Dequeue or Double Ended Queue implementation around circular array. The source code is below as per the references.

#include
#include

#define MAX 30

typedef struct dequeue
{
int data[MAX];
int rear,front;
}dequeue;

void initialize(dequeue *p);
int empty(dequeue *p);
int full(dequeue *p);
void enqueueR(dequeue *p,int x);
void enqueueF(dequeue *p,int x);
int dequeueF(dequeue *p);
int dequeueR(dequeue *p);
void print(dequeue *p);

void main()
{
int i,x,op,n;
dequeue q;
initialize(&q);

do
{
printf("n1.Createn2.Insert(rear)n3.Insert(front)n4.Delete(rear)n5.Delete(front)");
printf("n6.Printn7.ExitnnEnter your choice:");
scanf("%d",&op);

switch(op)
{
case 1: printf("nEnter number of elements:");
scanf("%d",&n);
initialize(&q);
printf("nEnter the data:");

for(i=0;i<n;i++) { scanf("%d",&x); if(full(&q)) { printf("nQueue is full!!"); exit(0); } enqueueR(&q,x); } break; case 2: printf("nEnter element to be inserted:"); scanf("%d",&x); if(full(&q)) { printf("nQueue is full!!"); exit(0); } enqueueR(&q,x); break; case 3: printf("nEnter the element to be inserted:"); scanf("%d",&x); if(full(&q)) { printf("nQueue is full!!"); exit(0); } enqueueF(&q,x); break; case 4: if(empty(&q)) { printf("nQueue is empty!!"); exit(0); } x=dequeueR(&q); printf("nElement deleted is %dn",x); break; case 5: if(empty(&q)) { printf("nQueue is empty!!"); exit(0); } x=dequeueF(&q); printf("nElement deleted is %dn",x); break; case 6: print(&q); break; default: break; } }while(op!=7); } void initialize(dequeue *P) { P->rear=-1;
P->front=-1;
}

int empty(dequeue *P)
{
if(P->rear==-1)
return(1);
return(0);
}

int full(dequeue *P)
{
if((P->rear+1)%MAX==P->front)
return(1);
return(0);
}

void enqueueR(dequeue *P,int x)
{
if(empty(P))
{
P->rear=0;
P->front=0;
P->data[0]=x;
}
else
{
P->rear=(P->rear+1)%MAX;
P->data[P->rear]=x;
}
}

void enqueueF(dequeue *P,int x)
{
if(empty(P))
{
P->rear=0;
P->front=0;
P->data[0]=x;
}
else
{
P->front=(P->front-1+MAX)%MAX;
P->data[P->front]=x;
}
}

int dequeueF(dequeue *P)
{
int x;
x=P->data[P->front];
if(P->rear==P->front)   //delete the last element
initialize(P);
else
P->front=(P->front+1)%MAX;
return(x);
}

int dequeueR(dequeue *P)
{
int x;
x=P->data[P->rear];
if(P->rear==P->front)
initialize(P);
else
P->rear=(P->rear-1+MAX)%MAX;
return(x);
}

void print(dequeue *P)
{
if(empty(P))
{
printf("nQueue is empty!!");
exit(0);
}

int i;
i=P->front;
while(i!=P->rear)
{
printf("n%d",P->data[i]);
i=(i+1)%MAX;
}
printf("n%dn",P->data[P->rear]);
}

5.) Priority Queue using Array

I noticed I didn’t updated a priority queue implementation using arrays. This is the reason the source code is here for a ready reference to the readers of the blog. Here is the original version of the source code which is tested against the terms and conditions posted before which should be able to run via a Turbo C compiler. Any other compiler just won’t work because the syllabus itself is outdated, and people (and universities) must realize this fast.

#include
#include
#define size 5

int queue[5][2] = {0};
int top = -1;
int bottom;

void push(int value, int pr)

{

int i,j,k;

if(top < size-1) { if(queue[top][1] > pr)

{

for(i=0;i<top;i++) { if(queue[i][1] > pr)

{

break;

}

}

for(j=top;j>=i;j--)

{

queue[j+1][0] = queue[j][0];

queue[j+1][1] = queue[j][1];

}

top++;

queue[i][0] = value;

queue[i][1] = pr;

}

else

{

top++;

queue[top][0] = value;

queue[top][1] = pr;

}

}

else

{

printf("queue overflow n");

}

}

void pop()

{

int i;

if(queue[0][0] == 0)

{

printf("n The queue is empty  n");

}

else

{

printf("After , dequeue the following value is erased n  %d n", queue[0][0]);

for(i=0;i<top;i++) { queue[i][0] = queue[i+1][0]; queue[i][1] = queue[i+1][1]; } queue[top][0] = 0; queue[top][1] = 0; top--; } } void display() { int i,j; printf("ElementtPriority n"); for(i=size - 1;i>=0;i--)

{

for(j=0;j<2;j++)

{

printf(" %dt",queue[i][j]);

}

printf("n");

}

}

int main()

{

int i,j, ch=0 ,value = 0,pr=0;

while(1)

{

printf("n Please Enter the choice. n");

printf("1 for Enqueue n 2 for Dequeue n 3 for displayn  5 for exit: t n");

scanf("%d",&ch);

switch(ch)

{

case 1:

printf("n Please Enter the number to be inserted: t ");

scanf("%d", &value);

printf("n Please Enter the priority: t ");

scanf("%d", &pr);

push(value,pr);

break;

case 2:

pop();

break;

case 3:

display();

break;

case 5:

exit(0);

default:

printf("You entered wrong choice!");

}
}
}

6.) Circular Queue using Linked List in C Data-structures with Notes

In a standard queue data structure re-buffering problem occurs for each dequeue operation. To solve this problem by joining the front and rear ends of a queue to make the queue as a circular queue
Circular queue is a linear data structure. It follows FIFO principle.

  • In circular queue the last node is connected back to the first node to make a circle.
  • Circular linked list fallow the First In First Out principle
  • Elements are added at the rear end and the elements are deleted at front end of the queue
  • Both the front and the rear pointers points to the beginning of the array.
  • It is also called as “Ring buffer”.
  • Items can inserted and deleted from a queue in O(1) time.

Circular Queue can be created in three ways. They are:

  • · Using single linked list
  • · Using double linked list
  • · Using arrays

Follow is the source code for implementing a Circular Queue in C using Linked List:

#include
#include
#define max 5
int q[max];
int rear = -1, front = -1;
void insert();
void delet();
void display();

void main()
{
char ch;
int choice;
do
{
clrscr();
printf("n1 -> Insert");
printf("n2 -> Delete");
printf("n3 -> Display");
printf("n0 -> Exit");

printf("nnEnter Your Choice -> ");
scanf("%d", &choice);
switch (choice)
{
case 1:
insert();
break;
case 2:
delet();
break;
case 3:
display();
break;
case 0:
exit();
default:
printf("nnInvalid Choice");
}
printf("nnDo u Want to Continue -> ");
ch = getche();
} while (ch == 'y' || ch == 'Y');
}

void insert()
{
int data;
if ((rear == max - 1 && front == 0) || rear == front - 1)
{
printf("nSorry! Q is Full");
}
else
{
printf("nEnter data You want to insert ->");
scanf("%d", &data);
if (front == -1)
{
front++;
rear++;
}
else if (rear == max - 1)
{
rear = 0;
}
else
{
rear++;
}
q[rear] = data;
printf("nnData inserted successfully");
}
}
void delet()
{
if (front == -1)
{
printf("nnSorry! Q is Empty");
}
else
{
if (front == rear)
{
front = rear = -1;
}
else if (front == max - 1)
{
front = 0;
}
else
{
front++;
}
printf("nElement deleted Successfully");
}
}

void display()
{
int i;
if (front == -1)
{
printf("nnSorry! Q is Empty");
}
else
{
printf("nn:: Queue Elements are ::n");
if (front <= rear)
{
for (i = front; i <= rear; i++)
{
printf("n%d", q[i]);
}
}
else
{
for (i = front; i < max; i++)
{
printf("n%d", q[i]);
}
for (i = 0; i <= rear; i++)
{
printf("n%d", q[i]);
}
}
}
}

7.) Circular Queue Implementation in C using Arrays

Earlier in the post, we talked about implementation circular queue using Linked List. This section talks about the same implementation but using arrays. The source code below could be used for the purposes to demonstrate the concepts:

#include
#include
#include
#define MAXSIZE 5
int cq[MAXSIZE];
int front,rear;
void main()
{
void add(int,int);
void del(int);
int will=1,i,num;
front = -1;
rear = -1;
clrscr();
printf("nProgram for Circular Queue demonstration through array");
while(1)
{
printf("nnMAIN MENUn1.INSERTIONn2.DELETIONn3.EXIT");
printf("nnENTER YOUR CHOICE : ");
scanf("%d",&will);
switch(will)
{
case 1:
printf("nnENTER THE QUEUE ELEMENT : ");
scanf("%d",&num);
add(num,MAXSIZE);
break;
case 2:
del(MAXSIZE);
break;
case 3:
exit(0);
default: printf("nnInvalid Choice . ");
}

} //end of  outer while
}               //end of main
void add(int item,int MAX)
{
//rear++;
//rear= (rear%MAX);
if(front ==(rear+1)%MAX)
{
printf("nnCIRCULAR QUEUE IS OVERFLOW");
}
else
{
if(front==-1)
front=rear=0;
else
rear=(rear+1)%MAX;
cq[rear]=item;
printf("nnRear = %d    Front = %d ",rear,front);
}
}
void del(int MAX)
{
int a;
if(front == -1)
{
printf("nnCIRCULAR QUEUE IS UNDERFLOW");
}
else
{
a=cq[front];
if(front==rear)
front=rear=-1;
else
front = (front+1)%MAX;
printf("nnDELETED ELEMENT FROM QUEUE IS : %d ",a);
printf("nnRear = %d    Front = %d ",rear,front);

}
}

Here is a similar implementation of a queue using array. The source code could be referenced from the below:

#include
#include
#define N 6

int queue[N]={0};
int rear=0,front=0;

void insert(void);
void del(void);
void disp(void);
void cre(void);

void main()
{
int user=0;

clrscr();
while(user!=4)
{
clrscr();
printf("nnnttt THE SIZE OF QUEUE IS %d",N);
printf("nt 1.INSERT");
printf("nt 2.DELETE");
printf("nt 3.DISPLAY");
printf("nt 4.EXIT");
printf("nt 5.CREATE");
scanf("%d",&user);
switch(user)
{
case 1:
insert();
break;
case 2:
del();
break;
case 3:
disp();
break;
case 4:
printf("nt THANK U");
break;
case 5:
cre();
break;
}
getch();

}
getch();
}

void insert(void)
{
int t;
if(rear<N)
{
printf("nt ENTER A VALUE IN QUEUE");
scanf("%d",&t);
queue[rear]=t;
rear++;
}
else
{
printf("nt Q OVERFLOW!!!!!!!!!!!!!!!");
}
}
void del(void)
{
int i;
printf("nt %d gets deleted.........",queue[front]);
queue[front]=0;
front++;
}
void disp(void)
{
int i;
for(i=front;i<rear;i++)
{
printf("nt %d",queue[i]);
}
}
void cre(void)
{

int t;
printf("nt ENTER A VALUE IN QUEUE");
scanf("%d",&t);
front=0;
queue[front]=t;
rear=front+1;
}

8.) Stack Implementation to perform PUSH and POP Operations

After having gone through the Queue utilities in C and discussing queue with circular based and using Linked List, here is the source code to implement Stack for PUSH and POP operations in C.

#include
#define MAXSIZE 5

struct stack
{
int stk[MAXSIZE];
int top;
};
typedef struct stack STACK;
STACK s;

void push(void);
int  pop(void);
void display(void);

void main ()
{
int choice;
int option = 1;
s.top = -1;

printf ("STACK OPERATIONn");
while (option)
{
printf ("------------------------------------------n");
printf ("      1    -->    PUSH               n");
printf ("      2    -->    POP               n");
printf ("      3    -->    DISPLAY               n");
printf ("      4    -->    EXIT           n");
printf ("------------------------------------------n");

printf ("Enter your choicen");
scanf    ("%d", &choice);
switch (choice)
{
case 1:
push();
break;
case 2:
pop();
break;
case 3:
display();
break;
case 4:
return;
}
fflush (stdin);
printf ("Do you want to continue(Type 0 or 1)?n");
scanf    ("%d", &option);
}
}
/*  Function to add an element to the stack */
void push ()
{
int num;
if (s.top == (MAXSIZE - 1))
{
printf ("Stack is Fulln");
return;
}
else
{
printf ("Enter the element to be pushedn");
scanf ("%d", &num);
s.top = s.top + 1;
s.stk[s.top] = num;
}
return;
}
/*  Function to delete an element from the stack */
int pop ()
{
int num;
if (s.top == - 1)
{
printf ("Stack is Emptyn");
return (s.top);
}
else
{
num = s.stk[s.top];
printf ("poped element is = %dn", s.stk[s.top]);
s.top = s.top - 1;
}
return(num);
}
/*  Function to display the status of the stack */
void display ()
{
int i;
if (s.top == -1)
{
printf ("Stack is emptyn");
return;
}
else
{
printf ("n The status of the stack is n");
for (i = s.top; i >= 0; i--)
{
printf ("%dn", s.stk[i]);
}
}
printf ("n");
}

Here is yet another version of Stack implementation but using recursive method:

#include

#include

struct node

{

int a;

struct node *next;

};

void generate(struct node **);

void display(struct node *);

void stack_reverse(struct node **, struct node **);

void delete(struct node **);

int main()

{

struct node *head = NULL;

generate(&head);

printf("nThe sequence of contents in stackn");

display(head);

printf("nInversing the contents of the stackn");

if (head != NULL)

{

stack_reverse(&head, &(head->next));

}

printf("nThe contents in stack after reversaln");

display(head);

delete(&head);

return 0;

}

void stack_reverse(struct node **head, struct node **head_next)

{

struct node *temp;

if (*head_next != NULL)

{

temp = (*head_next)->next;

(*head_next)->next = (*head);

*head = *head_next;

*head_next = temp;

stack_reverse(head, head_next);

}

}

void display(struct node *head)

{

if (head != NULL)

{

printf("%d  ", head->a);

display(head->next);

}

}

void generate(struct node **head)

{

int num, i;

struct node *temp;

printf("Enter length of list: ");

scanf("%d", &num);

for (i = num; i > 0; i--)

{

temp = (struct node *)malloc(sizeof(struct node));

temp->a = i;

if (*head == NULL)

{

*head = temp;

(*head)->next = NULL;

}

else

{

temp->next = *head;

*head = temp;

}

}

}

void delete(struct node **head)

{

struct node *temp;

while (*head != NULL)

{

temp = *head;

*head = (*head)->next;

free(temp);

}

}

9.) Linear Search Implementation in C

This is an abstract source code of Linear search utility in C. This could be used along in the Turbo C compiler. Others were specifically not tested and not recommended.

#include

int main()
{
int array[100], search, c, n;

printf("Enter the number of elements in arrayn");
scanf("%d",&n);

printf("Enter %d integer(s)n", n);

for (c = 0; c < n; c++)
scanf("%d", &array[c]);

printf("Enter the number to searchn");
scanf("%d", &search);

for (c = 0; c < n; c++)
{
if (array[c] == search)     /* if required element found */
{
printf("%d is present at location %d.n", search, c+1);
break;
}
}
if (c == n)
printf("%d is not present in array.n", search);

return 0;
}

10.) Binary Search using Arrays

The source code below is the implementation of binary search algorithm in C using arrays.

#include
int main(){

int a[10],i,n,m,c=0,l,u,mid;

printf("Enter the size of an array: ");
scanf("%d",&n);

printf("Enter the elements in ascending order: ");
for(i=0;i<n;i++){
scanf("%d",&a[i]);
}

printf("Enter the number to be search: ");
scanf("%d",&m);

l=0,u=n-1;
while(l<=u){
mid=(l+u)/2;
if(m==a[mid]){
c=1;
break;
}
else if(m<a[mid]){
u=mid-1;
}
else
l=mid+1;
}
if(c==0)
printf("The number is not found.");
else
printf("The number is found.");

return 0;
}

11.) Insertion Sort Implementation in C using Arrays.

This is insertion sort in C using Arrays. The source code is available below:

#include

int main()
{
int n, array[1000], c, d, t;

printf("Enter number of elementsn");
scanf("%d", &n);

printf("Enter %d integersn", n);

for (c = 0; c < n; c++) {
scanf("%d", &array[c]);
}

for (c = 1 ; c <= n - 1; c++) { d = c; while ( d > 0 && array[d] < array[d-1]) {
t          = array[d];
array[d]   = array[d-1];
array[d-1] = t;

d--;
}
}

printf("Sorted list in ascending order:n");

for (c = 0; c <= n - 1; c++) {
printf("%dn", array[c]);
}

return 0;
}

12.) Merge Sort in C using Arrays

Here is the source code for Merge sort in C using Arrays:

#include
#define MAX 50

void mergeSort(int arr[],int low,int mid,int high);
void partition(int arr[],int low,int high);

int main(){

int merge[MAX],i,n;

printf("Enter the total number of elements: ");
scanf("%d",&n);

printf("Enter the elements which to be sort: ");
for(i=0;i<n;i++){
scanf("%d",&merge[i]);
}

partition(merge,0,n-1);

printf("After merge sorting elements are: ");
for(i=0;i<n;i++){
printf("%d ",merge[i]);
}

return 0;
}

void partition(int arr[],int low,int high){

int mid;

if(low<high){
mid=(low+high)/2;
partition(arr,low,mid);
partition(arr,mid+1,high);
mergeSort(arr,low,mid,high);
}
}

void mergeSort(int arr[],int low,int mid,int high){

int i,m,k,l,temp[MAX];

l=low;
i=low;
m=mid+1;

while((l<=mid)&&(m<=high)){

if(arr[l]<=arr[m]){ temp[i]=arr[l]; l++; } else{ temp[i]=arr[m]; m++; } i++; } if(l>mid){
for(k=m;k<=high;k++){
temp[i]=arr[k];
i++;
}
}
else{
for(k=l;k<=mid;k++){
temp[i]=arr[k];
i++;
}
}

for(k=low;k<=high;k++){
arr[k]=temp[k];
}
}

13.) Quick Sort Implementation in C using Arrays

Below is the raw source code for Quick Sort implementation in C using Arrays:

#include

void quicksort(int [10],int,int);

int main(){
int x[20],size,i;

printf("Enter size of the array: ");
scanf("%d",&size);

printf("Enter %d elements: ",size);
for(i=0;i<size;i++)
scanf("%d",&x[i]);

quicksort(x,0,size-1);

printf("Sorted elements: ");
for(i=0;i<size;i++)
printf(" %d",x[i]);

return 0;
}

void quicksort(int x[10],int first,int last){
int pivot,j,temp,i;

if(first<last){
pivot=first;
i=first;
j=last;

while(i<j){
while(x[i]<=x[pivot]&&i<last) i++; while(x[j]>x[pivot])
j--;
if(i<j){
temp=x[i];
x[i]=x[j];
x[j]=temp;
}
}

temp=x[pivot];
x[pivot]=x[j];
x[j]=temp;
quicksort(x,first,j-1);
quicksort(x,j+1,last);

}
}

14.) Heap Sort Implementation in C using Arrays

Below is the source code for Heap Sort in C using Arrays:

#include
void main()

{

int heap[10], no, i, j, c, root, temp;

printf("n Enter no of elements :");

scanf("%d", &no);

printf("n Enter the nos : ");

for (i = 0; i < no; i++)

scanf("%d", &heap[i]);

for (i = 1; i < no; i++)

{

c = i;

do

{

root = (c - 1) / 2;

if (heap[root] < heap[c])   /* to create MAX heap array */

{

temp = heap[root];

heap[root] = heap[c];

heap[c] = temp;

}

c = root;

} while (c != 0);

}

printf("Heap array : ");

for (i = 0; i < no; i++) printf("%dt ", heap[i]); for (j = no - 1; j >= 0; j--)

{

temp = heap[0];

heap[0] = heap[j    /* swap max element with rightmost leaf element */

heap[j] = temp;

root = 0;

do

{

c = 2 * root + 1;    /* left node of root element */

if ((heap[c] < heap[c + 1]) && c < j-1)

c++;

if (heap[root]<heap[c] && c<j)    /* again rearrange to max heap array */

{

temp = heap[root];

heap[root] = heap[c];

heap[c] = temp;

}

root = c;

} while (c < j);

}

printf("n The sorted array is : ");

for (i = 0; i < no; i++)

printf("t %d", heap[i]);

printf("n Complexity : n Best case = Avg case = Worst case = O(n logn) n");

}

15.) Bubble Sort Implementation in C using Arrays

Below is the source code for bubble sort in C using Arrays:

#include

int main()
{
int array[100], n, c, d, swap;

printf("Enter number of elementsn");
scanf("%d", &n);

printf("Enter %d integersn", n);

for (c = 0; c < n; c++)
scanf("%d", &array[c]);

for (c = 0 ; c < ( n - 1 ); c++)
{
for (d = 0 ; d < n - c - 1; d++) { if (array[d] > array[d+1]) /* For decreasing order use < */
{
swap       = array[d];
array[d]   = array[d+1];
array[d+1] = swap;
}
}
}

printf("Sorted list in ascending order:n");

for ( c = 0 ; c < n ; c++ )
printf("%dn", array[c]);

return 0;
}

5. Operating System – Round Robin CPU Scheduling and Multilevel Queue

Operating System – Round Robin Scheduling

By Shritam Bhowmick
Web Application Penetration Tester
LinkedIn: https://www.linkedin.com/profile/view?id=281014248&trk=nav_responsive_tab_profile
Academia: https://independent.academia.edu/ShritamBhowmick
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Round Robin Scheduling

Previous post discussed the two types of CPU Scheduling, Preemptive and Non-Preemptive CPU Scheduling and was focused with FCFS, SJF and Priority Scheduling techniques. We had discussed in the last post how SJF and Priority Scheduling could be modified to attain preemptive scheduling and why FCFS scheduling cannot be used as preemptive scheduling. This post will go further with another preemptive type of CPU Scheduling which is known as Round Robin CPU Scheduling. Now, as we already know Preemptive Scheduling means forcibly stopping a job from the execution and keep it in the waiting state/blocked state for a period of time until it finishes executing another job execution processing, Round Robin is another instance of Preemptive Scheduling wherein it is possible to stop the job and pick up another job from the READY QUEUE  and start executing it. But there is a difference in the way Round Robin Scheduling in implemented.

In SJF (modified to SRT {Shortest Remaining Time First Scheduling}) and Priority (modified with LONG TERM and SHORT TERM Scheduler), it was possible to attain preemptive CPU Scheduling, and used time (CPU BURST TIME) and priority respectively; this however isn’t the case with Round Robin Scheduling (does not depend on CPU BURST Time of the job, or the priority of the job. All the jobs are treated equally). Round Robin Scheduling, the CPU time is divided into number of quanta. Assume the CPU quanta to be duration = 2 ms (Milli-Second). Now, whenever a job is allocated to the CPU for execution, the job will be executed for a maximum of 2 ms. If the CPU BURST time required for the job execution to be completed is more than 2 ms, after 2 ms the job will be forcibly terminated and pushed back to the READY QUEUE and the new job from the READY QUEUE will be allocated to the CPU for execution and again that this new job will be only until 2 ms since that is the quanta defined in terms of Round Robin CPU Scheduling. In case the job execution requirement for any job is less than 2 ms, the job terminates normally and not forcefully. So after termination of the the job in case the job finishes before 2 ms, the Job might go to the I/O WAITING STATE/BLOCK STATE for the CPU to handle I/O operations or might go to the HALTED STATE/TERMINATED STATE. The CPU can’t stay idle for the remaining time if the jobs has finished before the determined quanta, hence the best efficient applied as per Round Robin implementation is to take a new job and start executing the job from that point of time and hence save time with efficiency. This means, assume if the quanta is 2 ms and the job was finished in 1 ms, a new job from the READY QUEUE is taken by the CPU Scheduler and executed, this new job quanta time starts from the earlier saved time which was 1 ms (2-1 = 1 ms). So for an example, there are 4 jobs in the READY QUEUE to be executed by the CPU :

J1 = 4 ms
J2 = 2 ms
J3 = 3 ms
J4 = 6 ms

Assumption is CPU quanta to be 2 ms. That is 1 quanta = 2 ms. Now for Round Robin CPU Scheduling, the job is taken as FCFS basis, but a limited quanta is available for all the jobs per execution cycle, that is in this case 2 ms (1 quanta). The timeline, these jobs will be served is as follows:

J1    J2    J3    J4

|—–|—–|—–|—–|
2ms 2ms 2ms 2ms

Hence, on the first execution cycle:

J1 = 2 ms completed, remaining: 2 ms of CPU BURST time.
J2 = 2 ms completed, remaining: 0 ms of CPU BURST time.
J3 = 2 ms completed, remaining: 1 ms of CPU BURST time.
J4 = 2 ms completed, remaining: 4 ms of CPU BURST time.

Since, Round Robin CPU Scheduler is taking the jobs in FCFS (First Come First Served) basis, there is no priority or CPU time burst minimum time required job first assigned with the scheduling process. The allocation is done purely on first come, first served basis this way and each execution cycle for each job spends 1 quanta of time that is 2 ms which is in this case. After 1 quanta of time spent on a particular job, the job if remaining CPU BURST remains, get’s to the READY QUEUE and next job is taken. If this next job is processed and only required the exact processing time equal to that of the 1 quanta time, the job can either be pushed to the I/O WAITING STATE to cover I/O operation or might reach the HALTED STATE/TERMINATED STATE which is normal Job execution termination. If the Job has finished before the 1 quanta time, the CPU Scheduler without wasting any time pushes the next job from the READY QUEUE to the ACTIVE STATE and starts executing it until the 1 quanta time period which is being assigned similarly to other jobs. When jobs are pushed to the READY QUEUE, they always are pushed to the back of the READY QUEUE since other jobs which are finished first should be at the first to be taken by the CPU. This means the CPU takes the job from the READY QUEUE head, which again means it is following all the rules of the FCFS basis. It has also to be noted that jobs could be pushed to the READY QUEUE from the I/O WAIT STATE, NEW STATE, or ACTIVE STATE (in-case of forcible termination). Here, we need to look at the process diagram to verify if the jobs could be taken to the READY QUEUE in three ways:

ifwelook

The process state diagram suggests, that indeed the READY QUEUE (READY STATE as per process state diagram) get’s the jobs from all the three directions, that is the active state when forcible termination of job is done by the CPU, the new state when new arrivals of jobs are scheduled to be transferred from the new state to the ready state and the wait or blocked state when the jobs are awaiting I/O operations. Hence, after the first execution cycle, the timeline for the above example would look like this:

J1    J2    J3    J4   J1   J3    J4    J4

|—–|—–|—–|—–|—–|—|——|——|
2ms 2ms 2ms 2ms 2ms 1ms 2ms 2ms (assuming no new jobs have arrived at the READY QUEUE)

Now, after this timeline, all the JOBS will be completed assuming no new job arrivals were in the READY QUEUE from the NEW STATE. Hence all the jobs are treated equally in the Round Robin CPU Scheduling. There must be a timer in the system to remind the CPU what 1 quanta is over for a job, this brings the topic to a new curve. The timer we are talking about here could be programmable as well since in some installations, an administrator might want a small quanta, in other installations, the same administrator might want a higher quanta. SO this timer should be programmable, to interrupt the processor reminding the CPU that 1 quanta is past and a job must be stopped.

Timer (Programmable)

Whenever the CPU Scheduler decides that a new job has to be allocated to the CPU or the CPU allocates a new job for execution, immediately the timer has to be reset. At the end of the time pointer, the timer will generate an interrupt when again the CPU will go to the system program and the responsibility of that system program will be to terminate the current executed job, get a new job from the READY QUEUE, give it to the CPU for execution and re-start the timer. That is what happens when the required CPU BURST time in more than the quanta set in the timer. If the CPU BURST time is less then that of the quanta described in the timer, the process is not executed completely till the timer gives an interrupt. The execution is completed before the timer gives an interrupt but the process for the whole cyclic execution process is not yet completed wholly. So the last statement of every CPU BURST must be a system call. The system call for performing an I/O operation or a system call indicating that the process is complete. Again following the system call, the CPU Scheduler has to fetch a new process (job) from the READY QUEUE, give it to the CPU for execution and at the same time, the timer has to be reset so that the next time quanta starts from that point.

Variations of Round Robin Scheduling

There could be different variations of the Round Robin Scheduling, earlier there were no considerations on the Round Robin Scheduling but there could be Jobs which require more CPU BURST time duration depending if the CPU BURST time is more for I/O Operations (I/O Bound Jobs) or CPU Time (CPU Bound Jobs). The variation will depend on these CPU BURST time duration requirements! The main algorithm used will be Round Robin Scheduling.

To implement these concepts, there is a need to know MULTI-LEVEL QUEUE

MULTI-LEVEL QUEUE

There could be multiple READY QUEUE’s as such the following demonstrating how it would work:

multi-levelqueueThe entire diagram suggests that there could be multiple READY QUEUE which could be taken into consideration depending upon the jobs since the variation of jobs have either I/O operation taking the CPU BURST or CPU time taking the time duration of the CPU Processing. This way, if efficiency has to be maintained such that I/O Operations are given the highest priority, the multiple READY QUEUE would have (for an example) 1 READY QUEUE divided into 3 READY QUEUE, they could be:

  1. Q1
  2. Q2
  3. Q3

Here are the Jobs which would be as per the priority of execution (Round Robin Scheduling Implementation does not use Priority, here we are only giving the job prioritizing the jobs in the READY QUEUE not in the Scheduling Algorithm).

  1. Q1 – Handles the I/O Bound Jobs which require more I/O CPU BURST time duration.
  2. Q2 – Handles the moderate CPU Requirement time duration.
  3. Q3 – Handles the CPU Bound Jobs which require more CPU time duration.

Now the execution of such jobs are in the format as below:

  1. Q1- any jobs which are here must be executed at first preference.
  2. Q2 – only after Q1 is empty, the jobs in Q2 are taken. That is all I/O Bound jobs have to be completed first.
  3. Q3 – only after both Q1 and Q2 are empty, the jobs pending on Low priority that is CPU Bound jobs are taken.

There are disadvantages here. Consider that the jobs could be dynamic in nature which means, a job which is an I/O Bound Job could be changed into requiring CPU time (converts itself into being CPU Bound Job). That way, with the current MULTI-LEVEL QUEUE implementation, the jobs in the Q1 (requiring to be executed first in order) in-spite the jobs being transformed to CPU Bound Job by being of Dynamic Nature has to be in the Q1 and executed in first preference. This is unwanted and there has to be some techniques to resolve this. This is where MULTI-LEVEL FEEDBACK QUEUE is used.

MULTILEVEL FEEDBACK QUEUE

In Multilevel feedback queue, there would be again multiple number of queue’s. Assume there are 3 queues as shown below in the diagram (but with a provision that the jobs could be taken from one queue to another queue).

pushq1toq3Whenever the jobs are pushed into the READY QUEUE, it’s always taken to the QUEUE number 1 which is Q1. The time quanta for all the queue’s are:

  1. Q1 = 2 ms
  2. Q2 = 5 ms
  3. Q3 = 10 ms

Whenever a job is entered the first time, let’s assume the nature of the jobs isn’t known that is whether the Job is I/O Bound Job or a CPU Bound Job (what will be the CPU BURST Duration). In queue number 1, the job is to be put. In queue number 1 (Q1) the jobs will be allocated to the CPU using Round Robin Scheduling, for execution of the job. Once it is allocated to the CPU, it will be executed for a minimum of 2 ms, if one fins that the job (The CPU BURST) is over is finished before 2 ms, the job will be kept in Q1 and would not be moved to Q2. This is because the first it has executed, it has shown that the CPU BURST requirement is less than 2 ms, so it’s predicted that the next CPU BURST time will also be less than 2 ms. If the CPU BURST time of the Job is less than 2 ms, the Job remains in Q1. If the timer trigger has occurred and the job is greater than 2 ms, the job will be pushed to next queue which is Q2 which has the time quanta of more than 2 ms. This will again follow the Round Robin CPU Scheduling technique for the job to be allocated to the CPU for execution. This time the time quanta is 5 ms in Q2. In Q2, if one fins that the CPU time requirement for the job is less than 5 ms, it will remain in Q2 itself. Otherwise if it is greater than 5 ms, it will be pushed to the next queue which is Q3, with time quanta 10 ms. All of these queue will follow the Round Robing CU Scheduling algorithm for CPU jobs allocation for job execution.

So conclusively, the first time the job is pushed to the Q1, if the job CPU BURST duration is grater than the current quanta threshold of that particular queue, it would be pushed to the second queue or else be retained there itself. This depends on the nature of the jobs depending upon the requirement of the jobs for the time duration needed for a complete execution of the job with levels of time quanta divided into the READY QUEUE since the QUEUE itself now has been divided according to the time duration using quanta’s. In a similar fashion, the jobs which could be pushed downwards, should be the job require more I/O Bound Operations, the same should be the procedure, to push back the job upwards that is from Q3 to Q2 and then Q2 to Q1 as per the requirement and the nature of the job (if dynamic!). This means if the job at some point of time remain to be I/O Bound, the job remains in the Q1, if at another point of time duration. the job changes it’s nature from I/O Bound to CPU Bound, it could be pushed downward (lower preference). Similarly, low priority jobs (CPU Bound Jobs) could be pushed upwards by increasing it’s priority as per the dynamic nature of the job (changes from CPU Bound job to I/O Bound Job). Hence Jobs are kept at the queue level according to the nature of the job. These are the variations of the Round Robin Technique.

downtoupq3toq2toq1

The diagram above shows clearly how the jobs are now flexibly aligned in accordance to the time duration required and the dynamic nature of the job. If the jobs are changed from CPU Bound job to the I/O bound, it’d go upwards the direction and if the I/O bound jobs are changed to CPU Bound jobs, it’d move downwards the direction as per the requirement. I/O Bound Jobs have the highest priority since it requires less CPU BURST time duration than the CPU bound Jobs which requires more CPU BURST time duration and hence has a lower priority.

A feedforward is also used, since there is a concept called switch time which is the extra time required by the CPU to change from a queue to another queue. The queue could be switched directly to the appropriate queue without having them to move through the intermediate queue. This means if a job has to be transfer from the Q3 to Q1, it is directly possible but needs switch time and some extra CPU processing. This is demonstrated in the diagram below:

feedforwardWe have concluded over preemptive and non-preemptive CPU Scheduling from the last post and concluded this with Round Robin CPU Scheduling. All the posts which covered process management are:

  1. FCFS (First Come First Served) – Nonpreemptive
  2. SJF (Shortest Job First) – Nonpreemptive
  3. Priority Scheduling – Nonpreemptive
  4. SRT (Shortest Remaining Time First) – Preemptive
  5. Round Robin Scheduling – Preemptive
  6. Multilevel Queue – Preemptive

But all these CPU Scheduling which we have discussed are based on a single CPU. The modern operating systems should not remain satisfied with single processor CPU. There are distributing computing nature which is used in modern operating system be it networked processors among many systems or a single set-up having multiple core processors. We discussed single resource with multiple processes among which the resource is to be shared. What we will see next is a setup wherein there are multiple resource which is to be shared with multiple processes.

DISTRIBUTED COMPUTING

There are models which could go under distributed computing:

  1. WORKSTATION MODEL
  2. PROCESSOR POOL MODEL

Workstation Model could be considered as such every user which has one full fledged computer having it’s own memory, hard-disk, etc but shared on for example a LAN network without which the the computer remains functional. Every user has the processing power.

Processor Pool Model could be considered wherein one can have a high-end server having multiple processors and to the users, there could be high-end terminals (example a graphics terminal), so that the user terminal doesn’t have any processing capabilities. Users does not have processing capabilities, the processing is centralized.

Naturally for these two types of models, the approach taken for CPU sharing must be different. There hence must be two different kind of allocation techniques:

  1. NON-MIGRATORY
  2. MIGRATORY

NON-MIGRATORY to some extend is static in nature. In NON-MIGRATORY CPU allocation, once the process is created, it is decided on which of the processors the process has to be executed; once decided it is fixed and that process is executed on that processor until termination.

MIGRATORY is dynamic wherein the process could be migrated from one processor to another depending upon the requirement. The process itself hence could be terminated in another processor and could be originated or allocated to another processor during initiation of the process.

The next post would cover these aspects of CPU process sharing since this post was dedicated to cover Round Robin Scheduling and Multilevel Queue. Single CPU Management and Process Management hence has been covered in these five posts which could be used for a ready reference:

  1. Operating System Introductory
  2. Operating System  – Importance of an Operating System
  3. Operating System – CPU Management
  4. Operating System – Process State Diagram and Basic CPU Scheduling
  5. Operating System – Round Robin CPU Scheduling and Multilevel Queue

Use the links to keep updated on the Process Management of Operating System, next concurrent processing would be covered. Thank you and I bid good-bye to the followers of the blog. Roger out.

3. Operating System – CPU Management

CPU Management – A step towards Process Management

 

By Shritam Bhowmick
Web Application Penetration Tester
LinkedIn: https://www.linkedin.com/profile/view?id=281014248&trk=nav_responsive_tab_profile
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Last post, I talked about the coverage of the series on operating system and the introductory post with the importance of operating system, as well as defining the operating system. In this post, I’d discuss CPU management and hence go closer into process management. The CPU is known by Central Processing Unit and it is hence obvious CPU is a process manager of the computer system. It processes the given data into desired output which is useful for the user. The process which takes place in the CPU should be in the main memory, that is the data which is to be executed via processing should be in the main memory (Random Access Memory or RAM).  A multi-user system uses processes, and a process does not mean a multiple CPU handling these processes, it means that the processes are handled by the same CPU but with optimization in time and processing efficiency. The data provided by the user is fed to the CPU which is a ‘Job’, this Job needs processing to be done and hence result in execution in the CPU, the Jobs require the dependency data to be present in the main memory. Jobs are the tasks which the CPU has to undertake in order to process them, and then execute and write the results to the secondary devices (special I/O devices as discussed in y previous posts) or directly as a output to a video terminal, or printer, etc.

The terms which would be required for CPU management are:

  1. Waiting Time
  2. Turnaround Time

What is Waiting Time?

The gap in time between the given job for execution to the CPU and the actual time when the CPU begins to execute the given job is known as the waiting time. Or in other words the time the processing for the given job to be executed/processed remains in the ready queue for the CPU to begin processing or executing is termed as the waiting time.

Let t(0) be the time the user has given a certain job for processing or execution to the CPU. And let the actual time when the CPU begins processing/executing the given job be t(i). The time difference between t(i) and t(0) i.e: t(i) – t(0) = t(w) {Waiting time} is termed as the waiting time. So the equations are:

T(i) - T(0) = T(w)

Where,

  1. T(i) = Time at which the user had supplied the job for processing or execution to the CPU.
  2. T(0) = Time at which the CPU actually starts processing or execution of the supplied job.
  3. T(w) = Waiting Time.

Now for the CPU to give an output, there must be some time taken by the CPU in order to process the job. This time on a timeline of the whole processing from job given, to job initialization by the CPU and the output could be shown as below:

 

Waiting Time Demonstration

 

As shown, the processed job by the CPU is at the last entry which is marked as T (p), this would be the output time stamp when the job is done by the CPU.  Now, the concept of turnaround time hops in.

What is Turnaround Time?

The time taken for a job to complete it’s processing or execution from the point of the job submission to the CPU until it’s processing has been completed is termed as turnaround time. Or in other words, turnaround time is the total time taken between the submission of a program/process/thread/task  for execution and the return of the complete output to the user.

Now, as we had assumed the time stamp when the job is completed to be T (p), the turnaround time would be as following described by the equation:


T (t) = T (p) - T (0)

Where,

  1. T (t) = Turnaround Time
  2. T (p) = Processed or executed job by the CPU returned to the user or written in the special I/O device or outputted.
  3. T (0) = The time stamp when the user submitted the job/task to the CPU.

Diagrammatically, it would be:

Turnover Time

Having said that, the users would always want the waiting time and the turnaround time to be minimum in order to accomplish processing and outputting of the given jobs. But the CPU handles multi-jobs simultaneously and hence this isn’t possible or realistic. Jobs should be executed in first come first served (FCFS) basis and hence certain waiting time has to be allotted to the incoming jobs. Let’s assign these incoming jobs to be:

J1, J2, J3, J4 which would be incoming from ascending order, i.e: 1st J1 is fed to the CPU, then J2, J3 and finally J4. Hence the processing by the CPU should also be in accordance. That is if FCFS (First Come First Served) method is applied, J1 should be undertaken first and then the rest and hence the processing of J1 will be first then the rest. Applying that:

J1 = 15 time units of execution.
J2 = 8 time units of execution.
J3 = 10 time units of execution.
J4 = 3 time units of execution.

The jobs have been assigned with there respective time units of execution, that is each job is actually undertaken by the CPU at a certain time lapse, which until then it’s a waiting time. Assuming the jobs have arrived to the CPU progressively as J1, J2, J3 and then J4, I draw a general timeline:

 

Calculate Average Waiting Time

 

The above diagram shows the calculation part of the average job wait time. The calculation is simple, ignore the numbers and because it’s FCFS, start from ‘0’ for the Job 1 since Job 1 never waited, the others were in queue. The numbers given are for execution time that is the time interval the job took to get over with execution and hence the waiting time for each would be calculated as:

W (j1) = 0
W (j2) = 0 + 15 = 15
W (j3) = 15 + 8 = 23
W (j4) = 23 + 10 = 33


Total Wait time = 71
______________

The last 3 units of time wasn’t a wait time, it was the time taken for the 4th job to get over with execution, the 4th job already waited for 33 units of time for it’s turn to get execution scheduled. This process undertaken by the CPU is called Job Scheduling or simply Scheduling. That comes deep later, but to calculate the average waiting time, I need to consider all the jobs which are 4 items so:

71/4 = 17.75 is the average waiting time i.e: T (wa). Hence, T (Wa) = 17.75 time units for the above example on wait time average. Now, changing the sequential order of the incoming jobs. The CPU has no control over the incoming of the jobs, but what the CPU has control of is the order in which the jobs will be given a priority for execution. This could be decided by the CPU. So, consider the order below for jobs: (considering the same queue of jobs, the situation here is re-scheduling the jobs

J4 = 3 units of time
J2 = 8 units of time
J3 = 10 units of time
J1 = 15 units of time

In this case, the execution time remains the same, however the jobs are re-shuffled according to the CPU’s preference and the priority in taking up the job. Previously, the jobs were taken first come first served basis, but here even though the jobs arrived sequentially, the CPU has the authority and control of the execution land the processing of the jobs which were given by the user. The CPU decides to execute the job in the order such that J4 was executed first, next was J2, and then J3 and finally J4 was executed.

Considering the above situation, I now wanted to determine the average waiting time for such a schedule prescribed by the CPU because the CPU has the control on the scheduling part but does not have the control over the incoming jobs. To draw a rough sketch of just occurrences with a job execution and waiting timeline:

 

Scheduling without FCFS. Rescheduling with CPU’s control over Scheduling

 

Considering the image below I had drawn (MS paint, mind not!), the time taken for execution of the jobs in previous scenario were the same, but the scheduling was done by the CPU without FCFS method but taking in priority to which jobs has to be executed first to perform optimization. The magic happens here. Now if we calculate the waiting time:

W (J4) = 0 (since it arrived first)
W (J2) = 0 + 3 = 3 (since J2 had to wait for 3 units of time because 3 units of time were wasted {rather not!} in execution of job J4)
W (J3) = 3 + 8 = 11 (since J3 had to wait until J2 was finished executing with time lapse of 8 units)
W (J1) = 11 + 10 = 21 (since J1 had to wait for J3 to finish up with execution with time lapse of 10 units)

Hence;

W (J4) = 0
W (J2) = 3
W (J3) = 11
W (J1) = 21


Total Wait Time = 35 time units
________________________________

Now, if I calculate the average waiting time T (Wa) for this, that is 35/4, it comes to 8.75 time units. So:

T (Wa) = 8.75

which is far lesser in waiting time than the previous scenario. Calculating the optimization which had occurred here and the difference, our previous FCFS achieved 17.75 of waiting time, and this scheduling by the CPU achieved 8.75. The difference in wait time optimization being:

17.75 - 8.75 = 9.00 time units

A budget of total 9 time units.  So we find that just by changing the order, the same jobs were optimized. The average waiting time were reduced drastically so naturally the CPU would now prefer the second type of scheduling rather than the first scheduling. The user wold always prefer the first scenario but the CPU won’t allow that because it has to follow optimization rules. In the second scenario, there is a pattern why the average waiting time were drastically reduced. This is achieved by executing the jobs first which take lesser time for execution because the next queue of jobs will have to ‘wait’ lesser and hence optimization could occur.There are various objectives which could be achieved, the objective achieved here was to minimize the average waiting time for the jobs together to be finished in an optimized way.

So that’d be the very touch towards process management in this post. Next post, I’d bring some more fruits to the blog. The concepts of process management is vast and does not end with this basic post.  To keep updated if you like these posts, subscribe via mail or keep visiting the blog. Cheers out!