you tube download

This post is for downloading the videos from you tube without using any downloader for the same.

Following is the page of a youtube website.

Now for downloading the video just write ss before the url.so it would become

www.ssyoutube.com/watch?v=LvMpWn2CNzs

following page will be open

Now you can select either MP4 360p, MP4 720p and more as available.This can be used to download youtube video direct without any downloader.

CUT THE STICKS

If there are N sticks, A cut operation is performed such that length of all of them are reduced by the length of the smallest stick.

Suppose we have 6 sticks of length

5 4 4 2 2 8

then in one cut operation we make a cut of length 2 from each of the 6 sticks. For next cut operation 4 sticks are left (of non-zero length), whose length are

3 2 2 6

Above step is repeated till no sticks are left.

Input Format 
The first line contains a single integer N. 
The next line contains N integers: a₀, a₁,...a_N-1 separated by space, where a_i represents the length of i^th stick.

Output Format 
 output will be the number of sticks that are cut in separate lines

#include <stdio.h>

#include <string.h>

#include <math.h>

#include <stdlib.h>

#include <stdbool.h>

int findmin(int * b,int n);

int check(int *a,int n);

int main() {

int n,i,min;

 scanf("%d",&n);

    int arr[n],count=0;

    for(i=0;i<n;i++)

        {

        scanf ("%d",&arr[i]);

        }

     while(check(arr,n))

    {

         min = findmin(arr,n);

         for(i=0;i<n;i++)

             { if(arr[i]>0){

              arr[i]=arr[i]-min;

                 count++;}

              }

         printf("%d\n",count);

         count=0;

        

    }

   

    return 0;

}

  int  check(int *a,int n)

    {

    int i;

    for(i=0;i<n;i++)

        {

        if(a[i] > 0)

            {return 1;

            }

        }

       return 0;

   }

   

  int findmin(int * b,int n) 

      { int i, min=1000;

      for(i=0;i<n;i++)

          {

          if((b[i]>0) && b[i] < min)

              min=b[i];

          }

       return min;

      

      }

LONELY INTEGER

Program:to find number occurring only once

If there are integers in an array,you have to find out the numer that occurs only once

Input Format

The first line of the input indicates the number of integers. 
The next line is integers that form the array.

#include <stdio.h>

#include <string.h>

#include <math.h>

#include <stdlib.h>

int lonelyinteger(int a_size, int* a) {

    int a_count[100],i,count=0,c;

   for(i=0;i< 100;i++)

       {

       a_count[i]=0;

       }

    for(i=0;i< a_size;i++)

        {c=*(a+i);

        a_count[c]=(a_count[c])+1;

        }

    for(i=0;i< 100;i++)

        {

           if(a_count[i]==1)

            {return (i); }

        }

   

   

return count;

}

int main() {

    int res;

   

    int _a_size, _a_i;

    scanf("%d", &_a_size);

    int _a[_a_size];

    for(_a_i = 0; _a_i < _a_size; _a_i++) {

        int _a_item;

        scanf("%d", &_a_item);

       

        _a[_a_i] = _a_item;

    }

   

    res = lonelyinteger(_a_size, _a);

    printf("%d", res);

   

    return 0;

}

hiding folder completely

Following post is for hiding a folder completely.Folder is here at D:\test\ with name abc as shown in the image

Now for hiding the folder open command prompt and go to folder test and then type folloeing command at cmd prompt to hide it

Attrib +r +h +s abc

Folder will get hidden

And if you try to view it by even using folder option show hidden files and folder,it will not be unhide.

Now if you want to view the folder,you can do by this unchecking hide protected operating system files as shown

Either you can do it by using command prompt by using following

Attrib –r –s –h abc

IRCTC TATKAAL RESERVATION TRICK

Magic auto fill is a great tool for reservation during peak hours of tatkaal.It allows you to fill all the details of passenger in advance using magic auto fill,which you can use to enter in irctc form during reservation.
First visit the following link to find the http://ctrlq.org/irctc/  and click the reservation form

following page will be displayed

now fill the details of all the passenger you want to enter,and click on the i m felling lucky button.

then following screen will be displayed.

now drag this magic auto fill to your bookmark bar.if bookmark bar is not visible,use ctrl+shift+b (in chrome).Drag it as shown below

Bookmark of Magic auto fill will be attached to your bookmark bar.Now open the reservation page fill the journey to ,from and date and after clicking the book now fill all the details by just clicking on the magic auto fill bookmark

following will be very helpful during tatkaal hours,you just have to fill the captcha and make payment.

Fibonacci program in java

/*CSEMATTER.BLOGSPOT.IN
Program-fibonacci series in java*/

import java.util.Scanner;
class fib
{
public static void main(String a[])
{
Scanner sc=new Scanner(System.in);
int prev=0,next=1,sum,num,i;
System.out.println("ENTER THE NO.");
num=sc.nextInt();
System.out.println(""+prev);
System.out.println(""+next);
for(i=2;i<num;i++)
{
sum=prev+next;
prev=next;
next=sum;
System.out.println(""+sum);
}
}
}

FACTORIAL JAVA PROGRAM

/*CSEMATTER.BLOGSPOT.IN
Program-java program to calculate factorial*/

import java.util.Scanner;
class factorial
{
public static void main(String a[])
{
Scanner sc=new Scanner(System.in);
int i,num,f=1;
System.out.println("ENTER THE NO. TO CALCULATE FACTORIAL");
num=sc.nextInt();
for(i=1;i<num;i++)
{
f=f*i;
}
System.out.println("factorial is"+f);

}
}

Concurrent checkpoint

/*CSEMATTER.BLOGSPOT.IN
Program-to implement concurrent checkpoint*/


#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<time.h>

void delay(unsigned int mseconds)
{
    clock_t goal = mseconds + clock();
    while (goal > clock());
}
int main()
{
    int pr[10],p[10],n,i,j,count,temp;

    printf("Enter the no. of processes: ");
    scanf("%d",&n);
    for(i=0; i<n; i++)
        p[i]=i+1;
    printf("Enter the priority of processes \n");
    for(i=0; i<n; i++)
    {
        printf("Priority of process P%d : \n",i+1);
        scanf("%d",&pr[i]);
    }

    for(i=0; i<n-1; i++)
    {
        for(j=i+1; j<n; j++)
        {
            if(pr[i]>pr[j])
            {
                temp=pr[i];
                pr[i]=pr[j];
                pr[j]=temp;

                temp=p[i];
                p[i]=p[j];
                p[j]=temp;
            }
        }
    }
    for(i=0; i<n; i++)
    {
        count=0;
        for(j=0; j<n; j++)
        {
            if(i!=j)
                printf("P%d is sending request to P%d \n",p[i],p[j]);
             else
                 continue;
        }
        for(j=0; j<n; j++)
        {
            if(i!=j)
            {
                printf("P%d is sending ack to P%d \n",p[j],p[i]);
                count++;
            }
            else
                continue;
        }
        if(count==n-1)
        {
            printf("Ack from all the processes has been received \n");
            printf("P%d is forming a Consistent Checkpoint point \n",p[i]);
             delay(10000);
            for(j=0; j<n; j++)
            {
                if(i!=j)
                    printf("P%d sending release msg to P%d \n",p[i],p[j]);
                else
                    continue;
            }
            printf("\n \n");
        }
        else
            printf("Ack from all the processes not received hence Process P%d cann't form a consistent checkpoint \n \n",p[i]);
    }
    getch();

}

/* OUTPUT:

Enter the no. of processes: 3
Enter the no. of processes: 3
Enter the priority of processes
Priority of process P1 :
3
Priority of process P2 :
2
Priority of process P3 :
1
P3 is sending request to P2
P3 is sending request to P1
P2 is sending ack to P3
P1 is sending ack to P3
Ack from all the processes has been received
P3 is forming a Consistent Checkpoint point
P3 sending release msg to P2
P3 sending release msg to P1


P2 is sending request to P3
P2 is sending request to P1
P3 is sending ack to P2
P1 is sending ack to P2
Ack from all the processes has been received
P2 is forming a Consistent Checkpoint point
P2 sending release msg to P3
P2 sending release msg to P1


P1 is sending request to P3
P1 is sending request to P2
P3 is sending ack to P1
P2 is sending ack to P1
Ack from all the processes has been received
P1 is forming a Consistent Checkpoint point
P1 sending release msg to P3
P1 sending release msg to P2    */

bayzentine program

/*CSEMATTER.BLOGSPOT.IN
Program-implement bayzentine program in c*/

#include<stdio.h>
#include<conio.h>
#include<time.h>
#include<stdlib.h>


int main()
{
int i,j,p[20][20],n,nf,cnt0=0,cnt1=1;
printf("Enter the no. of processes:");
scanf("%d",&n);
printf("Enter the process no. of faulty process:");
scanf("%d",&nf);
for(i=0;i<n;i++)
{
if(i==nf)
{
for(j=0;j<n;j++)
{
if(j!=i)
p[i][j]=random(2);
else
p[i][j]=NULL;}
}
else
{
for(j=0;j<n;j++)
{
if(j!=i)
p[i][j]=1;
else
p[i][j]=NULL;
}}
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(p[i][j]==1)
++cnt1;
else if(p[i][j]==0)
++cnt0;
else
continue;
}
if(cnt1>cnt0)
continue;
else
{
printf("Processes are not synchronised \n \n");

}
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
printf("%d \t",p[i][j]);
}
printf("\n");
}
getch();
}

/*OUTPUT:
Enter the no. of processes:3
Enter the process no. of faulty process:2
Processes are not synchronised

    1   1
1       1
1   0          */

Encryption using Hashing

/* CSEMATTER.BLOGSPOT.IN
    Encryption using Hashing         */


#include <iostream>
#include<cstring>
#include<cmath>
using namespace std;
int evaluate(string str);

int myatoi(string num)
{
int temp=1,res=0,i=0;
while(num[i]!='\0')
{
res=res+temp*num[i++]-'0';
temp=temp*10;
}
return res;
}
int main()
{
string num;
getline(cin,num);
int num=myatoi(num);
cout<<"\n";
while(num>0)
{
int var;
string str1;
getline(cin,str1);
cout << str1<< "\n";
var=evaluate(str1);
--num;
}
return 0;
}

int evaluate(string str)
{
   // cout << "Hello World!" << endl;
   int n,i,k,l,int1,int2;
 
   //string str;
 
    // getline(cin,str);
    int arr[256]={0},arr_cpyi[256]={0},arr_cpy[256]={0};
    for(int j=0;j<str.size();j++)
    {
    ++arr[str[j]];
    ++arr_cpyi[str[j]];
    ++arr_cpy[str[j]];
    }
    // int *ini=&arr_cpy[0];
   
   
       for(k=0;k<256;k++)
    {
   
    for(l=0;l<255;l++)
    {
    if(arr_cpyi[l]>arr_cpyi[l+1])
    {
    // cout<<"*";
    int temp1=arr_cpyi[l];
    arr_cpyi[l]=arr_cpyi[l+1];
    arr_cpyi[l+1]=temp1;
    }
    }
    }
    int z=0;
    for(i=0;i<256;i++)
    {
    if(arr_cpyi[i]!=0)
    ++z;
    else
    continue;
    }
   
    int j,q,v=0;
    int y=ceil(+(z/2))+1;
    // cout<<y<<"\n";
    // int y=str.size()-2;
   
   
    //for(i=255;i>=(256-str.size());i--)
    while(y>0)
    {
    j=0;
    q=255;
    char ch1,ch2;
    // while(arr_cpyi[i]!=arr_cpy[j])
    // j++;
   
    while(arr_cpyi[256-z]!=arr_cpy[j])
    j++;
    while(arr_cpyi[255-v]!=arr_cpy[q])
    q--;
    // cout<<"j="<<j;
    // cout<<"q="<<q<<"\n";
    arr_cpy[j]=0;
    arr_cpy[q]=0;
   
   
    // cout<<"j="<<j<<"\n";
    // cout<<"q="<<q<<"\n";
   
    // ch=*temp-*ini;
    int1=&arr_cpy[j]-&arr_cpy[0];
    int2=&arr_cpy[q]-&arr_cpy[0];
    ch1=(char)int1;
    ch2=(char)int2;
    for(int x=0;x<str.size();)
    {
    if(ch1==str[x])
    {
    str[x]=ch2;
   
    ++x;
    }
    else if(ch2==str[x])
    {
    str[x]=ch1;
   
    ++x;
    }
    else
    ++x;
    }
    --z;
    ++v;
    --y;
   
   
    }
   
   
    cout<< str << endl;
   
    return 0;
}
   

MAEKAWA ALGORITHM DISTRIBUTED SYSTEM

/*CSEMATTER.BLOGSPOT..IN
Program name-Maekawa algorithm in distributed system*/

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<time.h>

void process1();
void process2();
void delay(unsigned int mseconds)
{
    clock_t goal = mseconds + clock();
    while (goal > clock());
}
int p[10],pr[10],n,i,j,count=0,ran_pr,temp,max_lmt,min_lmt;

int main()
{
    printf("Enter the no. of processses \n");
    scanf("%d",&n);
    for(i=0; i<n; i++)
        p[i]=i+1;
      do
       {
           ran_pr=random(n);
       }
       while(ran_pr!=0);
    ran_pr=3;
    printf("Control process is P%d \n",ran_pr);
    printf("Enter the priority of processes \n");
    for(i=0; i<n; i++)
    {
        printf("Priority of process P%d : \n",p[i]);
        scanf("%d",&pr[i]);
    }
    for(i=0; i<n-1; i++)
    {
        for(j=i+1; j<n; j++)
        {
            if(pr[i]>pr[j])
            {
                temp=pr[i];
                pr[i]=pr[j];
                pr[j]=temp;

                temp=p[i];
                p[i]=p[j];
                p[j]=temp;
            }
        }
    }
    for(i=0; i<n-1; i++)
    {
        if(p[i]<ran_pr)
        {
            max_lmt=ran_pr;
            process1();
        }
        else
        {
            min_lmt=ran_pr+1;
            process2();
        }
    }

    getch();
}

void process1()
{
    for(j=0; j<ran_pr; j++)
    {
        if(i!=j)
            printf("Process P%d is sending request to P%d \n",p[i],p[j]);
        else
            continue;
    }
    printf("\n");
    for(j=0; j<ran_pr; j++)
    {
        if(i!=j)
        {
            printf("Process P%d is acknowledging P%d \n",p[j],p[i]);
            count++;
        }
        else
            continue;
    }
    printf("\n");
    if(count==(ran_pr-1))
    {
        printf("Process P%d is entering CS \n",p[i]);
        delay(10000);
    }
    printf("\n");
    for(j=0; j<ran_pr; j++)
    {
        if(i!=j)
            printf("Process P%d is sending reply to P%d \n",p[i],p[j]);
        else
            continue;
    }
}

void process2()
{
    for(j=ran_pr+1; j<n; j++)
    {
        if(i!=j)
            printf("Process P%d is sending request to P%d \n",p[i],p[j]);
        else
            continue;
    }
    printf("\n");
    for(j=ran_pr+1; j<n; j++)
    {
        if(i!=j)
        {
            printf("Process P%d is acknowledging P%d \n",p[j],p[i]);
            count++;
        }
        else
            continue;
    }
    printf("\n");

    if(count==(n-ran_pr-1))
    {
        printf("Process P%d is entering CS \n",p[i]);
        delay(10000);
    }
    printf("\n");
    for(j=ran_pr+1; j<n; j++)
    {
        if(i!=j)
            printf("Process P%d is sending reply to P%d \n",p[i],p[j]);
        else
            continue;
    }
}

/*OUTPUT:
Enter the no. of processses 5
Control process is P3
Enter the priority of processes
Priority of process P1 :2
Priority of process P2 :1
Priority of process P3 :5
Priority of process P4 :3
Priority of process P5 :4

Process P2 is sending request to P1
Process P2 is sending request to P3
Process P1 is acknowledging P2
Process P3 is acknowledging P2
Process P2 is entering CS
Process P2 is sending reply to P1
Process P2 is sending reply to P3

Process P1 is sending request to P2
Process P1 is sending request to P3
Process P2 is acknowledging P1
Process P3 is acknowledging P1
Process P1 is entering CS
Process P1 is sending reply to P2
Process P1 is sending reply to P3

Process P4 is sending request to P5
Process P5 is acknowledging P4
Process P4 is entering CS
Process P4 is sending reply to P5

Process P5 is sending request to P4
Process P4 is acknowledging P5
Process P5 is entering CS
Process P5 is sending reply to P4

Process P3 is sending request to P1
Process P3 is sending request to P2
Process P1 is acknowledging P3
Process P2 is acknowledging P3
Process P5 is entering CS
Process P3 is sending reply to P1
Process P3 is sending reply to P2
*/

MIN FILTER IN MATLAB

/*CSEMATTER.BLOGSPOT.IN
PROGRAM FOR MIN FILTER IN MATLAB*/

A = imread('atul1.jpg');
A=imresize(A,[300,300]);
A = rgb2gray(A)
figure,imshow(A),title('ORIGINAL IMAGE');
B=zeros(size(A));
modifyA=padarray(A,[1 1]);

        x=[1:3];
        y=[1:3];
     
for i= 1:size(modifyA,1)-2
    for j=1:size(modifyA,2)-2
     
     
       window=reshape(modifyA(i+x-1,j+y-1),[],1);
               B(i,j)=min(window);

    end
end
B=uint8(B);
figure,imshow(B),title('IMAGE AFTER MIN FILTERING');

MEDIAN FILTER IN MATLAB

/*CSEMATTER.BLOGSPOT.IN
PROGRAM FOR MEDIAN FILTER IN MATLAB*/

A = imread('atul1.jpg');

A=imresize(A,[300,300]);

 A = rgb2gray(A)

figure,imshow(A),title('ORIGINAL IMAGE');

modifyA=zeros(size(A)+2);

B=zeros(size(A));

        for x=1:size(A,1)

            for y=1:size(A,2)

                modifyA(x+1,y+1)=A(x,y);

            end

        end

    

for i= 1:size(modifyA,1)-2

    for j=1:size(modifyA,2)-2

        window=zeros(9,1);

        inc=1;

        for x=1:3

            for y=1:3

                window(inc)=modifyA(i+x-1,j+y-1);

                inc=inc+1;

            end

        end

      

        med=sort(window);

        B(i,j)=med(5);

      

    end

end

B=uint8(B);

figure,imshow(B),title('IMAGE AFTER MEDIAN FILTERING');

MEAN FILTER IN MATLAB PROGRAM

/*CSEMATTER.BLOGSPOT.IN
PROGRAM FOR MEAN FILTER IN MATLAB*/


A = imread('atul1.jpg');
A=imresize(A,[300,300]);
A = rgb2gray(A)
figure,imshow(A),title('ORIGINAL IMAGE');

modifyA=zeros(size(A)+2);
B=zeros(size(A));
        for x=1:size(A,1)
            for y=1:size(A,2)
                modifyA(x+1,y+1)=A(x,y);
            end
        end
   
for i= 1:size(modifyA,1)-2
    for j=1:size(modifyA,2)-2
        window=zeros(9,1);
        inc=1;
        sum=0;
        for x=1:3
            for y=1:3
                window(inc)=modifyA(i+x-1,j+y-1);
     
                sum=sum+window(inc);
                inc=inc+1;
            end
        end
       sum=sum/9;
        B(i,j)=sum;
     
    end
end
B=uint8(B);
figure,imshow(B),title('IMAGE AFTER MEAN FILTERING');

MAX FILTER MATLAB PROGRAM

/*CSEMATTER.BLOGSPOT.IN
PROGRAM FOR MAX FILTER IN MATLAB*/


A = imread('atul1.jpg');
A=imresize(A,[300,300]);
 A = rgb2gray(A)
figure,imshow(A),title('ORIGINAL IMAGE');
B=zeros(size(A));
modifyA=padarray(A,[1 1]);

        x=[1:3];
        y=[1:3];
     
for i= 1:size(modifyA,1)-2
    for j=1:size(modifyA,2)-2
       window=reshape(modifyA(i+x-1,j+y-1),[],1);
               B(i,j)=max(window);
 
    end
end
B=uint8(B);
figure,imshow(B),title('IMAGE AFTER MAX FILTERING');

LINEAR FILTER IN MATLAB PROGRAM

/*CSEMATTER.BLOGSPOT.IN
PROGRAM FOR LINEAR FILTER IN MATLAB*/


A = imread('atul1.jpg');
A=imresize(A,[300,300]);
A = rgb2gray(A)
figure,imshow(A),title('ORIGINAL IMAGE');
modifyA=zeros(size(A)+2);
B=[1,2,4;1,2,4;1,2,4];
        for x=1:size(A,1)
            for y=1:size(A,2)
                modifyA(x+1,y+1)=A(x,y);
            end
        end
   
for i= 1:size(modifyA,1)-2
    for j=1:size(modifyA,2)-2
        sum=0;
            for x=1:3
               for y=1:3
                sum=sum +((modifyA(i+x-1,j+y-1))*B(x,y));
            end
        end
       sum=sum/9;
        A(i,j)=sum;
     
    end
end

figure,imshow(A),title('IMAGE AFTER LINEAR FILTERING');

HISTOGRAM EQUALIZATION IN MATLAB

/*CSEMATTER.BLOGSPOT.IN
PROGRAM FOR HISTOGRAM EQUALIZATION IN MATLAB*/


im1=imread('atul1.jpg');
im=rgb2gray(im1);
im=imresize(im,[256,256]);
numofpixels=size(im,1)*size(im,2);
figure(1);
subplot(1,2,1);
imshow(im);
subplot(1,2,2);
imhist(im);
HIm=uint8(zeros(size(im,1),size(im,2)));
freq=zeros(256,1);
probf=zeros(256,1);
probc=zeros(256,1);
cum=zeros(256,1);
output=zeros(256,1);
for i=1:size(im,1)
    for j=1:size(im,2)
        value=im(i,j);
        freq(value+1)=freq(value+1)+1;
        probf(value+1)=freq(value+1)/numofpixels;
   end
end
sum=0;
no_=255;
for i=1:size(probf)
   sum=sum+freq(i);
   cum(i)=sum;
   probc(i)=cum(i)/numofpixels;
   output(i)=round(probc(i)*no_);
end

for i=1:size(im,1)
    for j=1:size(im,2)
     HIm(i,j)=output(im(i,j)+1);
    end
end
figure(2);
subplot(1,2,1);
imshow(HIm);
subplot(1,2,2);
imhist(HIm);

OPERATIONS ON MATRIX IN MATLAB

/*CSEMATTER.BLOGSPOT.IN*/

DIGITAL IMAGE PROCESSING LAB
/*PROGRAM TO PERFORM ADDITION SUBTRACTION ETC. IN MATLAB*/

a=[1,2,3;4,5,6;7,8,9]
b=[2,4,6;1,3,5;3,4,1]
d=[1,2,3]
c=a+b;
disp('addition is ');
disp(c);
c=a-b;
disp('difference is ');
disp(c);
c=a*b;
disp('multiplication is');
disp(c);
disp('dot product');
c=a.*b;
disp(c);
disp('divison is');
c=a/d;
disp(c);
c=inv(a);
disp('inverse is');
disp(c);
disp('powerof matrix');
c=a^2;
disp(c);
disp('scalar divison');
c=a/2;
disp(c);

SHIFT REDUCE PARSER

/*CSEMATTER.BLOGSPOT.IN
PROGRAM-SHIFT REDUCE PARSER*/


#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
#include<string.h>

char ip_sym[15],stack[15];
int ip_ptr=0,st_ptr=0,len,i;
char temp[2],temp2[2];
char act[15];
void check();
int main()
{
printf("\n\t\t SHIFT REDUCE PARSER\n");
printf("\n GRAMMER\n");
printf("\n E->E+E\n ");
printf("\n E->E*E\n E->a");
printf("\n enter the input symbol:\t");
gets(ip_sym);
printf("\n\t stack implementation table");
printf("\n stack\t\t input symbol\t\t action");
printf("\n______\t\t ____________\t\t ______\n");
printf("\n $\t\t%s$\t\t\t--",ip_sym);
strcpy(act,"shift  ");
temp[0]=ip_sym[ip_ptr];
temp[1]='\0';
strcat(act,temp);
len=strlen(ip_sym);
for(i=0;i<=len-1;i++)
{
stack[st_ptr]=ip_sym[ip_ptr];
stack[st_ptr+1]='\0';
ip_sym[ip_ptr]=' ';
ip_ptr++;
printf("\n $%s\t\t%s$\t\t\t%s",stack,ip_sym,act);
strcpy(act,"shift ");
temp[0]=ip_sym[ip_ptr];
temp[1]='\0';
strcat(act,temp);
check();
st_ptr++;
}
st_ptr++;
check();
}
void check()
{
int flag=0;
temp2[0]=stack[st_ptr];
temp2[1]='\0';
if((!strcmpi(temp2,"a")))
{
stack[st_ptr]='E';
if(!strcmpi(temp2,"a"))
printf("\n $%s\t\t%s$\t\t\tE->a",stack, ip_sym);
flag=1;
}
if((!strcmpi(temp2,"+"))||(strcmpi(temp2,"*")))
{
flag=1;
}
if((!strcmpi(stack,"E*E"))||(!strcmpi(stack,"E+E")))
{
strcpy(stack,"E");
st_ptr=0;
if(!strcmpi(stack,"E+E"))
printf("\n $%s\t\t%s$\t\t\tE->E+E",stack,ip_sym);
if(!strcmpi(stack,"E*E"))
printf("\n $%s\t\t%s$\t\t\tE->E*E",stack,ip_sym);
flag=1;
}

if(!strcmpi(stack,"E")&& ip_ptr ==len)
{
 printf("\n $%s\t\t%s$\t\t\tACCEPT",stack,ip_sym);
getch();
 exit(0);
}
if(flag==0)
{
 printf("\n%s\t\t\t%s\t\t reject",stack,ip_sym);
 exit(0);
}
return;
}

THREE ADDRESS CODE

/*CSEMATTER.BLOGSPOT.IN
PROGRAM-TO GENERATE THREE ADDRESS CODE*/

#include<stdio.h>
#include<string.h>
void pm();
void plus();
void div();
int i,ch,j,l,addr=100;
char ex[10],exp[10],exp1[10],exp2[10],id1[5],op[5],id2[5];
void main()
{
clrscr();
while(1)
{
printf("\n1.assignment\n2.arithmetic\n3.relational\n4.Exit\nEnter the choice:");
scanf("%d",&ch);
switch(ch)
{
case 1:
printf("\nEnter the expression with assignment operator:");
scanf("%s",exp);
l=strlen(exp);
exp2[0]='\0';
i=0;
while(exp[i]!='=')
{
i++;
}
strncat(exp2,exp,i);
strrev(exp);
exp1[0]='\0';
strncat(exp1,exp,l-(i+1));
strrev(exp1);
printf("Three address code:\ntemp=%s\n%s=temp\n",exp1,exp2);
break;

case 2:
printf("\nEnter the expression with arithmetic operator:");
scanf("%s",ex);
strcpy(exp,ex);
l=strlen(exp);
exp1[0]='\0';

for(i=0;i<l;i++)
{
if(exp[i]=='+'||exp[i]=='-')
{
if(exp[i+2]=='/'||exp[i+2]=='*')
{
pm();
break;
}
else
{
plus();
break;
}
}
else if(exp[i]=='/'||exp[i]=='*')
{
div();
break;
}
}
break;

case 3:
printf("Enter the expression with relational operator");
scanf("%s%s%s",&id1,&op,&id2);
if(((strcmp(op,"<")==0)||(strcmp(op,">")==0)||(strcmp(op,"<=")==0)||(strcmp(op,">=")==0)||(strcmp(op,"==")==0)||(strcmp(op,"!=")==0))==0)
printf("Expression is error");
else
{
printf("\n%d\tif %s%s%s goto %d",addr,id1,op,id2,addr+3);
addr++;
printf("\n%d\t T:=0",addr);
addr++;
printf("\n%d\t goto %d",addr,addr+2);
addr++;
printf("\n%d\t T:=1",addr);
}
break;
case 4:
exit(0);
}
}
}
void pm()
{
strrev(exp);
j=l-i-1;
strncat(exp1,exp,j);
strrev(exp1);
printf("Three address code:\ntemp=%s\ntemp1=%c%ctemp\n",exp1,exp[j+1],exp[j]);
}
void div()
{
strncat(exp1,exp,i+2);
printf("Three address code:\ntemp=%s\ntemp1=temp%c%c\n",exp1,exp[i+2],exp[i+3]);
}
void plus()
{
strncat(exp1,exp,i+2);
printf("Three address code:\ntemp=%s\ntemp1=temp%c%c\n",exp1,exp[i+2],exp[i+3]);
}


/*OUTPUT
Example Generation of Three Address Project Output Result

1. assignment
2. arithmetic
3. relational
4. Exit
Enter the choice:1
Enter the expression with assignment operator:
a=b
Three address code:
temp=b
a=temp

1.assignment
2.arithmetic
3.relational
4.Exit
Enter the choice:2
Enter the expression with arithmetic operator:
a+b-c
Three address code:
temp=a+b
temp1=temp-c

1.assignment
2.arithmetic
3.relational
4.Exit
Enter the choice:2
Enter the expression with arithmetic operator:
a-b/c
Three address code:
temp=b/c
temp1=a-temp

1.assignment
2.arithmetic
3.relational
4.Exit
Enter the choice:2
Enter the expression with arithmetic operator:
a*b-c
Three address code:
temp=a*b
temp1=temp-c

1.assignment
2.arithmetic
3.relational
4.Exit
Enter the choice:2
Enter the expression with arithmetic operator:a/b*c
Three address code:
temp=a/b
temp1=temp*c
1.assignment
2.arithmetic
3.relational
4.Exit
Enter the choice:3
Enter the expression with relational operator
a
<=
b

100 if a<=b goto 103
101 T:=0
102 goto 104
103 T:=1

1.assignment
2.arithmetic
3.relational
4.Exit
Enter the choice:4

*/

PREDICTIVE PARSER PROGRAM

/*CSEMATTER.BLOGSPOT.IN
PROGRAM NAME-STACK IMPLEMENTATION OF PREDICTIVE PARSER*/

#include<ctype.h>
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<string.h>
#define non_term_count 5
#define term_count 6
#define rightHandMarker term_count-1
#define no_of_production 8

char nt[]={'E','A','T','B','F'},ter[]={'i','+','*','(',')','$'};
char arr[20][20][20]={
   {"TA","","","TA","",""},
    {"","+TA","","","@","@"},
    {"FB","","","FB","",""},
    {"","@","*FB","","@","@"},
    {"i","","","(E)","",""}
};
char ipstr[20];
char stack[40],prod[10];
int i=0,top=1,ia,ix;

struct predictive_table
{
 char prod[10];
}table[non_term_count][term_count]={"\0"};

struct
{
 int isNull;
 char term[10];
}first[non_term_count]={{0,"(i"},{1,"+"},{0,"(i"},{1,"*"},{0,"(i"}};
struct
{
 int isMarker;
 char term[10];
}follow[non_term_count]={{1,")"},{1,")"},{1,"+)"},{1,"+)"},{1,"+*)"}};
char terminals[term_count-1][2]={"i","+","*","(",")"};
char nonTerminals[non_term_count][3]={"E","A","T","B","F"};
char grammar[no_of_production][10]={"E T A","A + T A","A @","T F B","B * F B","B @","F ( E )","F i"};
char pgrammar[no_of_production][10]={"E->TA","A->+TA","A->@","T->FB","B->*FB","B->@","F->(E)","F->i"};

int matchTerminals(char term[])
{
 int i=-1;
 for(i=0; i<term_count-1; i++)
if(!strcmp(term,terminals[i]))
break;
 return i;
}

void copy_follow(int index,int loc)
{
 char term[2];
 int j,ind;
 if(follow[index].isMarker==1)
 {
  if(strlen(table[index][rightHandMarker].prod)==0)
   strcpy(table[index][rightHandMarker].prod,pgrammar[loc]);
  else
   printf("\nGiven Grammar is not LL(1)");
 }
 j=0;

 while(follow[index].term[j]!='\0')
 {
  term[0]=follow[index].term[j];
  term[1]='\0';
  ind = matchTerminals(term);
  if(strlen(table[index][ind].prod)==0)
   strcpy(table[index][ind].prod,pgrammar[loc]);
  else
   printf("\nGiven Grammar is not LL(1)");
  j++;
 }
}

int matchNonTerminals(char non_term[])
{
 int i=-1;
 for(i=0; i<non_term_count; i++)
if(!strcmp(non_term,nonTerminals[i]))
break;
 return i;
}



void show_table()
{
 int k,i,j;
 printf("\n\t\t\tPredictive Parser Table \n\n");
 printf("   |");
 for(i=0; i<term_count-1; i++)
  printf("    %s    |",terminals[i]);
 printf("    $    |");
 for(i=0; i<non_term_count; i++)
 {
  printf("\n----------------------------------------------------------------");
  printf("\n%s",nonTerminals[i]);
  for(k=strlen(nonTerminals[i]); k<3; k++)
   printf(" ");
  printf("|");
  for(j=0; j<term_count; j++)
  {
    printf(" %s",table[i][j].prod);
    for(k=strlen(table[i][j].prod); k<8; k++)
     printf(" ");
    printf("|");
  }
 }
}
void fun();
int main()
{
 int i,k,j,t,flag=0,indexTerm,indexNonTerm,index;
 char nextChar[10]="\0",nonterm[10]="\0",term[10]="\0";
 for(i=0; i<no_of_production; i++)
 {
  t=0;
  j=0;
  flag=0;
  while(grammar[i][j]!=' '&&grammar[i][j]!='\0')
   nonterm[t++]=grammar[i][j++];
  nonterm[t]='\0';
  indexNonTerm=matchNonTerminals(nonterm);
X:  j++;
  t=0;
  while(grammar[i][j]!=' '&&grammar[i][j]!='\0')
   nextChar[t++]=grammar[i][j++];
  nextChar[t]='\0';
  if(!strcmp(nextChar,"@"))
   copy_follow(indexNonTerm,i);
  else
  {
   index=matchTerminals(nextChar);
   if(index!=term_count-1)
   {
    if(strlen(table[indexNonTerm][index].prod)==0)
     strcpy(table[indexNonTerm][index].prod,pgrammar[i]);
    else
    {
      printf("\nGiven Grammar is not LL(1)");
      break;
    }
   }
   else
   {
    index=matchNonTerminals(nextChar);
    if(index!=non_term_count)
    {
     k=0;
     while(first[index].term[k]!='\0')
     {
       term[0]=first[index].term[k];
       term[1]='\0';
       indexTerm=matchTerminals(term);
       if(strlen(table[indexNonTerm][indexTerm].prod)==0)
strcpy(table[indexNonTerm][indexTerm].prod,pgrammar[i]);
       else
       {
printf("\n Given Grammar is not LL(1) ");
break;
       }
       k++;
     } //end while
     if(first[index].isNull==1)
     {
       flag=1;
       goto X;
     }
     else
      flag=0;
    }
   }
  }
  if(flag==1)
   copy_follow(indexNonTerm,i);
 }
 show_table();
 fun();
 getch();
}


void fun()
{

       void pop();
void push(char);
int resolve_nt(char);
int resolve_t(char);
void advance();
char a,x;
int len,k;
stack[0]='$';
stack[1]='E';
printf("\nenter the input string:\n");
scanf("%s",ipstr);
printf("I/p string\t\tStack\t\tProduction Used\n");
while(1)
{
a=ipstr[i];
x=stack[top];
for(k=i;ipstr[k]!='$';k++)
printf("%c",ipstr[k]);
printf("$\t\t");
if(x==a)
{
if(x=='$')
{
printf("input is accepted");
break;
}
else
{
pop();
advance();
}
}
else if(isupper(x))
{
ix=resolve_nt(x);
ia=resolve_t(a);
strcpy(prod,arr[ix][ia]);
len=strlen(prod);
pop();
for(k=1;k<=len;k++)
push(prod[len-k]);
if(stack[top]=='@')
pop();
}
else
{
printf("error");
break;
}
for(k=0;k<=top;k++)
printf("%c",stack[k]);
printf("\t\t\t%s\n",prod);
}
}

void push(char t)
{
top+=1;
stack[top]=t;
}
void pop()
{
top--;
}
void advance()
{
i++;
}
int resolve_nt(char t)
{
int k,index;
for(k=0;k<5;k++)
{
if(t==nt[k])
{
index=k;
break;
}
}
return index;
}

int resolve_t(char t)
{
int k,index;
for(k=0;k<6;k++)
{
if(t==ter[k])
{
index=k;
break;
}
}
return index;
}

/*OUTPUT


                        Predictive Parser Table

   |    i    |    +    |    *    |    (    |    )    |    $    |
----------------------------------------------------------------
E  | E->TA   |         |         | E->TA   |         |         |
----------------------------------------------------------------
A  |         | A->+TA  |         |         | A->@    | A->@    |
----------------------------------------------------------------
T  | T->FB   |         |         | T->FB   |         |         |
----------------------------------------------------------------
B  |         | B->@    | B->*FB  |         | B->@    | B->@    |
----------------------------------------------------------------
F  | F->i    |         |         | F->(E)  |         |         |
enter the input string:
i+i*i$
I/p string              Stack           Production Used
i+i*i$          $AT                     TA
i+i*i$          $ABF                    FB
i+i*i$          $ABi                    i
i+i*i$          $AB                     i
+i*i$           $A                      @
+i*i$           $AT+                    +TA
+i*i$           $AT                     +TA
i*i$            $ABF                    FB
i*i$            $ABi                    i
i*i$            $AB                     i
*i$             $ABF*                   *FB
*i$             $ABF                    *FB
i$              $ABi                    i
i$              $AB                     i
$               $A                      @
$               $                       @
$               input is accepted

*/