#### Aim:

Write a C Program to Implement Non-Restoring Division Algorithm.

__Theory:__

**Non-Restoring Division:**

- In each cycle content of the register, A is first shifted and then the divisor is added or subtracted with the content of register a depending upon the sign of A. In this, there is no need of restoring, but if the remainder is negative then there is a need of restoring the remainder. This is the faster algorithm of division.

#### Algorithm:

- At each step, left shift the dividend by 1 position.
- Subtract the divisor from A (A-M).
- If the result is positive then the step is said to be “successful”.
- In this case, the quotient bit will be “1” and the restoration is NOT Required. So, the next step will also be subtraction.
- If the result is negative then the step is said to be “unsuccessful”.
- In this case, the quotient bit will be “0”.
- Here, the restoration is NOT performed like the restoration division algorithm.
- Instead, the next step will be ADDITION in place of subtraction.
- Repeat steps 1 to 4 for all bits of the Dividend.

#### Program:

#include<stdio.h> #include<stdlib.h> int acum[100]={0}; void add(int acum[],int b[],int n); int q[100],b[100],l; int main() { int x,y; printf("Enter the Number : "); scanf("%d%d",&x,&y); int i=0; while(x>0||y>0) { if(x>0) { q[i]=x%2; x=x/2; } else { q[i]=0; } if(y>0) { b[i]=y%2; y=y/2; } else { b[i]=0; } i++; } int n=i; int bc[50]; printf("\n"); for(i=0;i<n;i++) { if(b[i]==0) { bc[i]=1; } else { bc[i]=0; } } bc[n]=1; for(i=0;i<=n;i++) { if(bc[i]==0) { bc[i]=1; i=n+2; } else { bc[i]=0; } } b[n]=0; int j; for(i=n;i!=0;i--) { if(acum[n]==0) { for(j=n;j>0;j--) { acum[j]=acum[j-1]; } acum[0]=q[n-1]; for(j=n-1;j>0;j--) { q[j]=q[j-1]; } add(acum,bc,n+1); } else { for(j=n;j>0;j--) { acum[j]=acum[j-1]; } acum[0]=q[n-1]; for(j=n-1;j>0;j--) { q[j]=q[j-1]; } add(acum,b,n+1); } if(acum[n]==1) { q[0]=0; } else { q[0]=1; } } if(acum[n]==1) { add(acum,b,n+1); } printf("\nQuoient : "); for( l=n-1;l>=0;l--) { printf("%d",q[l]); } printf("\nRemainder : "); for( l=n;l>=0;l--) { printf("%d",acum[l]); } return 0; } void add(int acum[],int bo[],int n) { int i=0,temp=0,sum=0; for(i=0;i<n;i++) { sum=0; sum=acum[i]+bo[i]+temp; if(sum==0) { acum[i]=0; temp=0; } else if(sum==2) { acum[i]=0; temp=1; } else if(sum==1) { acum[i]=1; temp=0; } else if(sum==3) { acum[i]=1; temp=1; } } }

#### Execution:

Input: 15 7 Output: Enter the Number : Quoient: 0010 Remainder: 00001

#### Result:

Thus the Program for Non-Restoring Division was executed Successfully.