Write a program to remove left recursion?

Answer 1

#include"stdio.h"
#include"conio.h"
int fabo(int);
void main()
{
int result=0,a=1,b=1,c;
printf("enter upto which you want to generate the series");
scanf("%d",&c);
result=fabo(c);
printf("%d\n%d\n",a,b);
printf("the fabonnaci series is %d\n",result);
getch();
}
int fabo(int n)
{
if (n==1);
return 1;
else if(n==2);
return 1;
else
return fabo(n-1)+fabo(n-2);
}

Answer 2

#include<stdio.h>

#include<conio.h>

#include<string.h>

#define SIZE 10

main () {

char non_terminal;

char beta,alpha;

char production[SIZE];

int index=3; /* starting of the string following "->" */

printf("Enter the grammar:\n");

scanf("%s",production);

non_terminal=production[0];

if(non_terminal==production[index]) {

alpha=production[index+1];

printf("Grammar is left recursive.\n");

while(production[index]!=0 && production[index]!='|')

index++;

if(production[index]!=0) {

beta=production[index+1];

printf("Grammar without left recursion:\n");

printf("%c->%c%c\'",non_terminal,beta,non_terminal);

printf("\n%c\'->%c%c\'|E\n",non_terminal,alpha,non_terminal);

}

else

printf("Grammar can't be reduced\n");

}

else

printf("Grammar is not left recursive.\n");

}

getch();

}

Answer 3

#include

int factorial(int n){
if(n 1)
return 1;
return n*factorial(n-1);
}

int main(int argc, char *argv[]){

if(argc != 2){
printf("%s number", argv[0]);
exit(1);
}

printf("%s! = %d", argv[0], factorial(atoi(argv[1])));
return 0;
}

Answer 4

Following function calculates factorial of a number using recursion. unsigned int factorial(unsigned int a)

{

if (a == 1)

return 1;

else

{

a *= factorial(a-1);

return a;

}

} recursion n. see recursion.

Answer 5

import java.io.*;

class bin

{

InputStreamReader in=new InputStreamReader(System.in);

BufferedReader br=new BufferedReader(System.in);

int a[]=new int[5];

int i,j,r;

System.out.println("Enter range");

r=Integer.parse.Int(br.readLine());

System.out.println("Enter number");

for(i=0;i<5;i++)

{

a[i]=Integer.parse.Int(br.readLine());

}

for(i=0;i<5;i++)

for(j=i+1;j<i;j++)

{

if(a[j]>a[i])

System.out.println(a[i]);

}

for(i=0;i<5;i++)

{

System.out.println(a[i]);

}

}

Answer 6

/*Algorithm - First, we will create a Max heap. Then, we pick the top

element (max element) from the heap and swap it with the last element

of the array. Now, the max element is in its correct place. Next, we

consider the array to be of size 'n-1' and call heapify on the

0th element so as to create the heap again. We do the same process

until all the elements are in their correct place */

void heapsort(int a[], int n)

{

/* To find the first node from which we will start heapifying.

Usually, this is the last node in the 2nd last layer of the binary

heap*/

int templog = (int)pow(2,(ceil(log((float)n))-1)) - 1;

//Create a min heap

for(int i = templog ;i >= 0 ; i--)

{

heapify(a, int i);

}

//Extract top element from heap and swap with last element

int temp_n = n;

while(temp_n>0)

{

swap(a,0,temp_n);

temp_n--;

heapify(a,0,temp_n);

}

}

void heapify(int a[], int v, int n)

{

//For a node v, its children are 2*v+1 and 2*v+2

//The case where node has no left or right child

if(2*v+1 >= n)

return;

//The case where node has a left child but not a right child

else if(2*v+2 >= n)

{

if(a[2*v+1] > a[v]) //if left child greater than parent

{

swap(a,v,2*v+1);

heapify(a,2*v+1);

}

}

//The case where node has both left and right child

else

{

//If left child greater than parent and right child

if((a[2*v+1] > a[v]) && (a[2*v+1] > a[2*v+2]))

{

swap(a,v,2*v+1);

heapify(a,2*v+1);

}

//If right child greater than parent and left child

else if((a[2*v+2] > a[v]) && (a[2*v+2] > a[2*v+1]))

{

swap(a,v,2*v+2);

heapify(a,2*v+1);

}

}

}

void swap(int a[],int left,int right)

{

int temp = a[left];

a[left] = a[right];

a[right] = temp;

}