Computer Programming

/*Title : Creating Student Database Using Singly linked list */

#include<stdio.h>

#include<conio.h>

struct Student

{

int roll;

char name[30];

float marks;

struct Student *next; //Self referential pointer...

};

typedef struct Student Node;

void Linkfloat()

{

float a=0,*b; //To create link of float to some compiler ...

b=&a;

a=*b;

}

void Display(Node *head)

{

Node *p;

int i;

if(head==NULL)

{

printf("

There is no records in database.

");

}

else

{

p=head;

for(i=0;i<80;i++)

{

printf("-");

}

printf("

Updated Student Database

");

for(i=0;i<80;i++)

{

printf("-");

}

printf("

");

printf("Roll No. Name Marks

");

for(i=0;i<80;i++)

{

printf("-");

}

printf("

");

while(p!=NULL)

{

printf("%d %s %0.2f",p->roll,p->name,p->marks);

printf("

");

for(i=0;i<80;i++)

{

printf("-");

}

printf("

");

flushall();

p=p->next; //Go to next node...

}

}

}

void DReverse(Node * head)

{

Node *p;

int i;

if(head==NULL)

{

printf("

There is no records in database.

");

}

else

{

p=head;

if(p->next!=NULL)

{

DReverse(p->next); //Recursive call...

}

printf("%d %s %0.2f",p->roll,p->name,p->marks);

printf("

");

for(i=0;i<80;i++)

{

printf("-");

}

printf("

");

flushall();

printf("

");

}

}

Node* Create(Node *head)

{

int n,i;

Node *nn,*p;

printf("

How many Entries to Create Database???

");

scanf("%d",&n);

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

{

if(head==NULL)

{

nn=(Node*)malloc(sizeof(Node)); //Creating first node...

printf("

Enter Roll No. , Name & Marks

");

scanf("%d",&(nn->roll));

flushall();

gets(nn->name);

scanf("%f",&(nn->marks));

nn->next=NULL; //Set next to NULL...

head=nn;

}

else

{

p=nn=head;

while(nn->next!=NULL)

{

nn=nn->next;

}

nn->next=(Node*)malloc(sizeof(Node)); //Creating further nodes...

nn=nn->next;

printf("

Enter Roll NO.,Name & Marks

");

scanf("%d",&(nn->roll));

flushall();

gets(nn->name);

scanf("%f",&(nn->marks));

nn->next=NULL;

nn=p;

}

}

return head;

}

Node* Insert(Node *head)

{

int ch,r;

char ans;

Node *p,*nn,*q;

do

{

printf("

Whwre do you want to enter new entry???

");

printf("

1.At the Begining

2.At the middle

3.At the end

");

printf("

Enter your choice:

");

scanf("%d",&ch);

switch(ch)

{

case 1:

/* Insert at Begining */

p=head;

nn=(Node*)malloc(sizeof(Node));

printf("

Enter Roll NO., Name & Marks

");

scanf("%d",&(nn->roll));

flushall();

gets(nn->name);

scanf("%f",&(nn->marks));

nn->next=NULL;

nn->next=p;

head=nn; //set first node as head...

printf("

Entry is Created successfully.

");

Display(head);

break;

case 2:

/* Insert at Middle */

if(head==NULL)

{

printf("

Yet database is not created.");

printf("

Database is empty.

");

printf("

First Create Database.

");

}

else

{

printf("

After which Roll NO. You want to insert new Data???

");

scanf("%d",&r);

p=head;

while(p->roll!=r && p->next!=NULL)

{

p=p->next; //Go upto that roll no....

}

if(p->roll!=r)

{

printf("

There is no such entry.

");

}

else

{

nn=(Node*)malloc(sizeof(Node));

printf("

Enter Roll NO.,Name & Marks

");

scanf("%d",&(nn->roll));

flushall();

gets(nn->name);

scanf("%f",&(nn->marks));

nn->next=NULL;

q=p->next;

p->next=nn;

nn->next=q;

printf("

Entry is Created successfully.

");

Display(head);

}

}

break;

case 3:

/* Insert at end */

if(head==NULL)

{

printf("

Yet database is not created.");

printf("

Database is empty.

");

printf("

First Create Database.

");

}

else

{

p=head;

nn=(Node*)malloc(sizeof(Node));

printf("

Enter Roll NO.,Name & Marks

");

scanf("%d",&(nn->roll));

flushall();

gets(nn->name);

scanf("%f",&(nn->marks));

nn->next=NULL;

while(p->next!=NULL)

{

p=p->next; //Go upto last node...

}

p->next=nn;

printf("

Entry is Created successfully.

");

Display(head);

}

break;

}

printf("

Do you want to Insert more data(Y/N)???");

flushall();

scanf("%c",&ans);

}while(ans=='y' ans=='Y');

return head;

}

Node* Delete(Node* head)

{

Node *p,*q,*r;

char ans;

int ch,n;

do{

printf("

Which Entry you want to Delete???

");

printf("

1.First

2.Middle

3.End

");

scanf("%d",&ch);

if(head==NULL)

{

printf("

Yet database is not created.");

printf("

Database is empty.

");

printf("

First Create Database.

");

}

else

{

switch(ch)

{

case 1:

/*Delete first node */

p=head;

head=head->next; //Set second node as head...

free(p);

printf("

First entry is deleted.

");

Display(head);

break;

case 2:

/*Delete middle Node*/

p=head;

printf("

Enter roll no. which you want to delete:

");

scanf("%d",&n);

while((p->next)->roll!=n && p->next->next!=NULL)

{

p=p->next; //Go upto -1 node which you want to delete...

}

if(p->next->next==NULL)

{

printf("

There is no such entry.

");

}

else

{

q=p->next;

r=q->next;

p->next=r;

free(q); //Delete that node...

printf("

Entry is deleted.

");

Display(head);

}

break;

case 3:

/* Delete last node */

p=head;

while(p->next->next!=NULL)

{

p=p->next; //Go upto -1 node which you want to delete...

}

q=p->next;

free(q); //Delete last node...

p->next=NULL;

printf("

Last entry is deleted.

");

Display(head);

break;

}

}

printf("

Do you want to delete more data(Y/N)???

");

flushall();

scanf("%c",&ans);

}while(ans=='y' ans=='Y');

return head;

}

Search(Node *head)

{

Node *p;

int r,cnt=0;

if(head==NULL)

{

printf("

Yet database is not created.");

printf("

Database is empty.

");

printf("

First Create Database.

");

}

else

{

p=head;

printf("

Enter roll no. which you want to Search:

");

scanf("%d",&r);

while(p->roll!=r && p->next!=NULL) //Search for roll no...

{

p=p->next;

cnt++;

}

if(p->roll!=r)

printf("

There is no such entry.

");

else

{

printf("

Roll NO. %d is at %d th Position.",r,(cnt+1));

printf("

Roll No. Name Marks

");

printf("%d %s %0.2f",p->roll,p->name,p->marks);

}

}

}

Modify(Node * head)

{

Node *p;

int r;

if(head==NULL)

{

printf("

Yet database is not created.");

printf("

Database is empty.

");

printf("

First Create Database.

");

}

else

{

p=head;

printf("

Enter the Roll no. whose data you want to modify:

");

scanf("%d",&r);

while(p->roll!=r && p->next!=NULL)

{

p=p->next;

}

if(p->roll!=r)

{

printf("

Thre is no such record in Database.

");

}

else

{

printf("

Entered roll no's Data is:

");

printf("Roll No. Name Marks

"); //Displaying Data who is going to modify....

printf("%d %s %f",p->roll,p->name,p->marks);

printf("

Enter New roll no ,New name & Marks for this entry:

");

scanf("%d",&p->roll);

flushall();

gets(p->name); //Enter new data...

scanf("%f",&(p->marks));

printf("

Entered New Data is:

");

printf("Roll No. Name Marks

");

printf("%d %s %f",p->roll,p->name,p->marks);

Display(head);

}

}

}

Count(Node *head)

{

Node *p;

int cnt=0;

if(head==NULL)

{

printf("

Yet database is not created.");

printf("

Database is empty.

");

printf("

First Create Database.

");

printf("

There are 0 records in Database.

");

}

else

{

p=head;

while(p->next!=NULL)

{

p=p->next;

cnt++; //Counting records...

}

printf("

There are %d records in Database.

",(cnt+1));

}

}

void main()

{

int ch,i;

char op;

Node *head;

head=NULL;

printf("

*----------Studednt Database-----------*

");

do

{

printf("

Menu

1.Create Database

2.Insert

3.Delete

4.Search

5.Modify

6.Display

7.Display Reverse

8.Count Records

9.Exit

");

printf("

Enter your choice

");

scanf("%d",&ch);

switch(ch)

{

case 1:

head=Create(head); //Call to Create...

break;

case 2:

head=Insert(head); //Call to Insert...

break;

case 3:

head=Delete(head); //Call to Delete...

break;

case 4:

Search(head); //Call to Search...

break;

case 5:

Modify(head); //Call to Modify...

break;

case 6:

Display(head); //Call to Display...

break;

case 7:

for(i=0;i<80;i++)

{

printf("-");

}

printf("

Updated Student Database

");

for(i=0;i<80;i++)

{

printf("-");

}

printf("

");

printf("Roll No. Name Marks

");

for(i=0;i<80;i++)

{

printf("-");

}

printf("

");

DReverse(head); //Call to displaying reversre...

break;

case 8:

Count(head); //Call to counting records...

break;

case 9:

exit(); //Exit...

default :

printf("

You entered wrong choice.

");

}

printf("

Do you want to Exit(Y/N)???

");

flushall();

scanf("%c",&op);

}while(op=='n' op=='N');

}

Add another pointer to the nodes for the previous node:

struct node {

struct node *next;

struct node *previous;

void *data;

};

typedef struct node node;

Then change the logic for insertion and removal to make sure you set the previous pointer as well as the next one.

1. DDA algorithm involves floating-point operations, while Bresenham algorithm uses only integer operations.
2. DDA algorithm calculates the exact position of each pixel, while Bresenham algorithm determines the closest pixel to the ideal line path.
3. DDA algorithm can suffer from precision issues due to floating-point calculations, while Bresenham algorithm is more accurate and efficient.
4. DDA algorithm is simpler to implement but slower than Bresenham algorithm.
5. DDA algorithm is susceptible to rounding errors, while Bresenham algorithm is not.
6. DDA algorithm can produce jagged lines due to rounding errors, while Bresenham algorithm generates smoother lines.
7. DDA algorithm is suitable for both lines and circles, while Bresenham algorithm is primarily used for drawing lines.
8. DDA algorithm can handle lines with any slope, while Bresenham algorithm is more efficient for lines with slopes close to 0 or 1.
9. DDA algorithm involves multiplication and division operations, while Bresenham algorithm uses addition and subtraction operations.
10. DDA algorithm is a general line drawing algorithm, while Bresenham algorithm is specialized for line drawing and rasterization.

Examples of pattern recognition include detecting faces in images, identifying fraudulent behavior in financial transactions, and recognizing speech in audio recordings. These tasks involve recognizing consistent and repeating patterns within data to make accurate predictions or classifications.

1. Eye to eye contact is lost while writing.

2. The written material cannot be stored and reused.

3. Advance preparation of material is not possible.

To determine if a linked list is circular, you can use the Floyd's cycle detection algorithm. This algorithm involves using two pointers moving at different speeds through the list, and if there is a cycle, the two pointers will eventually meet at the same node. If they don't meet and one of the pointers reaches the end of the list, then the list is not circular.

The main difference between c and c++ is the concept of 'Object Oriented Programming' (OOPS).

Thus c does not have the benefits of oops like:

1. abstraction

2. encapsulation

3. inheritance

4. polymorphism etc.

Common operations on a singly linked list include insertion (at the beginning, end, or specific position), deletion (from the beginning, end, or specific position), traversal (visiting each node in the list), searching (finding a specific value), and updating (modifying the value of a node).

A ruler is a line segment because it ends at a one point when a line goes on forever.

Conditional statements are used in programming to make decisions based on certain conditions. They allow the program to execute different code blocks depending on whether a condition is true or false. Common conditional statements include if, else, and else if.

Some disadvantages of custom written software include higher development costs, longer implementation times, potential compatibility issues with other systems, and difficulties in scaling and maintaining the software as business needs evolve.

An artificial neural network is a structure which will attempt to find a relationship i.e. a function between the inputs, and the provided output(s), in order that when the net be provided with unseen inputs, and according with the recorded internal data (named "weights"), will try to find a correct answer for the new inputs. Hidden Markov models, are used for find the states for which a given stochastic process went through. The main difference could be this: In order to use a markov chain, the process must depend only in it´s last state. For use a neural network, you need a lot of past data. After training process, neural networks are capable of predicting next states of the system based only on the last state. In addition, given the ability to measure the prediction error (for example, after actual event, signal or state has happend and was compared to prediction), the neural network is capable of adapting itself and capture online changes in the undergoing process to improve the model of prediction and decrease the estimation error for the next states. Theoretically such approach can eliminate the need in initial training, as the network started from some random model will eventually adapt itself to the actual process it tries to estimate given this feedback error loop and will start to make correct estimations / predictions after a certain amount of steps. In such setup one can assume that neural network can be used when no past data is available at all. In this case neural network build the model of the ongoing process "from scratch" based on the observations in the "online" mode.

Now it is! This is how every word is "invented". Someone like you comes up with a new word to meet a new need, and eventually it finds its way to a dictionary. A dictionary is a follower, but you can be a leader. Good answer. Yes it is a word

To print "thank you" in inverted commas in C programming, you can use the following code:

#include <stdio.h>
int main() {
    printf("\"thank you\"\n");
    return 0;
}

This code will display the output as: "thank you"

HTML, CSS, & JavaScript.

Go to the Help menu and look at the About option which will tell you. The version of Excel you have is linked to the version of Office you have. If you know what version of Office you have, then you know what version of Excel you have.

.exe and .msi

To calculate the position of the 12 squares aligned around the edges of a perfect dodecagon, you can use the following steps:

1. Calculate the radius of the dodecagon:

   • The edges of the dodecagon are 576 units long, and the dodecagon is a regular polygon, so the radius can be calculated as:

   • Radius = (Edge Length) / (2 × tan(π/12)) = 576 / (2 × tan(π/12)) ≈ 500.00 units

2. Calculate the position of the centers of the squares:

   • The squares are 576 × 576 units, so their centers will be 288 units away from the edge of the dodecagon.

   • The centers of the squares will be evenly spaced around the dodecagon, with an angular separation of 30 degrees (360 degrees / 12 squares).

   • The coordinates of the centers of the squares can be calculated using the following formulas:

     • x = (Radius + 288) × cos(θ)

     • y = (Radius + 288) × sin(θ)

   • Where θ is the angle of the square, starting from the positive x-axis and increasing counterclockwise.

Applying these formulas, the coordinates of the centers of the 12 squares are:

1. (681.40, 1801.21)

2. (1365.45, 1117.77)

3. (1365.45, -1117.77)

4. (681.40, -1801.21)

5. (-681.40, -1801.21)

6. (-1365.45, -1117.77)

7. (-1365.45, 1117.77)

8. (-681.40, 1801.21)

9. (0.00, 2000.00)

10. (1000.00, 1732.05)

11. (1000.00, -1732.05)

12. (0.00, -2000.00)

Okay, let's solve this step-by-step:

1. We have a regular dodecagon with edge length of 576 units.

2. The dodecagon is centered at (0, 0).

3. You want to place 12 squares of size 576 x 576 units around the edges of the dodecagon.

4. The squares should be rotated 30 degrees more than the previous one.

To calculate the position of the centers of the 12 squares:

1. The angle between each square is 360/12 = 30 degrees.

2. The radius of the dodecagon is 576 / (2 * tan(π/12)) = 500 units.

3. The center of the first square is at:

   • x = 500 * cos(0) = 500

   • y = 500 * sin(0) = 0

4. The center of the second square is at:

   • x = 500 * cos(30 * π/180) = 433.01

   • y = 500 * sin(30 * π/180) = 250.00

5. Continuing this pattern, the coordinates of the 12 square centers are:

| Square | X | Y |

| --- | --- | --- |

| 1 | 500.00 | 0.00 |

| 2 | 433.01 | 250.00 |

| 3 | 250.00 | 433.01 |

| 4 | 0.00 | 500.00 |

| 5 | -250.00 | 433.01 |

| 6 | -433.01 | 250.00 |

| 7 | -500.00 | 0.00 |

| 8 | -433.01 | -250.00 |

| 9 | -250.00 | -433.01 |

| 10 | 0.00 | -500.00 |

| 11 | 250.00 | -433.01 |

| 12 | 433.01 | -355.o1

The coordinates are rounded to two decimal places, as requested.

Easier to learn: Second-generation programming languages are easier to learn than first-generation languages. They are closer to human language and are more intuitive

i don’t know

The first thing you must do is find somewhere to post it. The next thing you need to do is copy and paste it. After that you must save it to documents and then when you go to the website you want to post it on just double click and then it will say paste and then if you click it, it will pop up on the website and ask you if you want to post this image and if the answer yesthen you will click sure if not please click no thanks.

An example of a DFA that resembles an ATM can be constructed with states representing different transaction phases (e.g., idle, card insertion, PIN entry, transaction processing) and transitions triggered by user inputs (e.g., card swipe, key press). This DFA would validate the sequence of inputs based on the current state to determine if a valid transaction has occurred.

Artificial Intelligence is self awareness, sentience in a programmed computer. Such is impossible with digital technology and the limits placed on digital platforms by the relatively miniscule memory capacities available. So, no roles for business.