Implementation of Non-Restoring Division Algorithm for Unsigned Integer

In the previous article, we have already discussed the Non-Restoring Division Algorithm. In this article, we will discuss the implementation of this algorithm. Non-restoring division algorithm is used to divide two unsigned integers. The other form of this algorithm is Restoring Division. This algorithm is different from the other algorithm because here, there is no concept of restoration and this algorithm is less complex than the restoring division algorithm. Let the dividend Q = 0110 and the divisor M = 0100. The following table demonstrates the step by step solution for the given values:

	Accumulator-A(0)	Dividend-Q(6)	Status
Initial Values	0000	0111	0101(M)
Step1:Left-Shift	0000	111_
Operation:A – M	1011	1110	Unsuccessful(-ve) A+M in Next Step
Step2:Left-Shift	0111	110_
Operation:A + M	1100	1100	Unsuccessful(-ve) A+M in Next Step
Step3:Left-Shift	1001	100_
Operation:A + M	1110	1000	Unsuccessful(-ve) A+M in Next Step
Step4:Left-Shift	1101	000_
Operation:A + M	0010	0001	Successful(+ve)
	Remainder(2)	Quotient(1)

Approach: From the above solution, the idea is to observe that the number of steps required to compute the required quotient and remainder is equal to the number of bits in the dividend. Initially, let the dividend be Q and the divisor be M and the accumulator A = 0. Therefore:

1. At each step, left shift the dividend by 1 position.
2. Subtract the divisor from A (A – M).
3. 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.
4. 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.
5. Repeat steps 1 to 4 for all bits of the Dividend.

Below is the implementation of the above approach: 

C++

// C++ program to divide two
// unsigned integers using
// Non-Restoring Division Algorithm

#include <iostream>
#include <string>
using namespace std;

// Function to add two binary numbers
string add(string A, string M)
{
    int carry = 0;
    string Sum = ""; // Iterating through the number
    // A. Here, it is assumed that
    // the length of both the numbers
    // is same
    for (int i = A.length() - 1; i >= 0; i--) {
        // Adding the values at both
        // the indices along with the
        // carry
        int temp = (A[i] - '0') + (M[i] - '0') + carry;

        // If the binary number exceeds 1
        if (temp > 1) {
            Sum += to_string(temp % 2);
            carry = 1;
        }
        else {
            Sum += to_string(temp);
            carry = 0;
        }
    }

    // Returning the sum from
    // MSB to LSB
    return string(Sum.rbegin(), Sum.rend());
}

// Function to find the compliment
// of the given binary number
string compliment(string m)
{
    string M = ""; // Iterating through the number
    for (int i = 0; i < m.length(); i++) {
        // Computing the compliment
        M += to_string((m[i] - '0' + 1) % 2);
    }

    // Adding 1 to the computed
    // value
    M = add(M, "0001");
    return M;
}

// Function to find the quotient
// and remainder using the
// Non-Restoring Division Algorithm
void nonRestoringDivision(string Q, string M, string A)
{
    // Computing the length of the
    // number
    int count = M.length();
    string comp_M = compliment(M);

    // Variable to determine whether
    // addition or subtraction has
    // to be computed for the next step
    string flag = "successful";

    // Printing the initial values
    // of the accumulator, dividend
    // and divisor
    cout << "Initial Values: A: " << A << " Q: " << Q
         << " M: " << M << endl;

    // The number of steps is equal to the
    // length of the binary number
    while (count) {
        // Printing the values at every step
        cout << "\nstep: " << M.length() - count + 1;

        // Step1: Left Shift, assigning LSB of Q
        // to MSB of A.
        cout << " Left Shift and ";

        A = A.substr(1) + Q[0];

        // Choosing the addition
        // or subtraction based on the
        // result of the previous step
        if (flag == "successful") {
            A = add(A, comp_M);
            cout << "subtract: ";
        }
        else {
            A = add(A, M);
            cout << "Addition: ";
        }

        cout << "A: " << A << " Q: " << Q.substr(1) << "_";

        if (A[0] == '1') {
            // Step is unsuccessful and the
            // quotient bit will be '0'
            Q = Q.substr(1) + "0";
            cout << " -Unsuccessful";

            flag = "unsuccessful";
            cout << " A: " << A << " Q: " << Q
                 << " -Addition in next Step" << endl;
        }
        else {
            // Step is successful and the quotient
            // bit will be '1'
            Q = Q.substr(1) + "1";
            cout << " Successful";
            flag = "successful";
            cout << " A: " << A << " Q: " << Q
                 << " -Subtraction in next step" << endl;
        }
        count--;
    }
    cout << "\nQuotient(Q): " << Q << " Remainder(A): " << A
         << endl;
}

// Driver code
int main()
{
    string dividend = "0111";
    string divisor = "0101";
    string accumulator = string(dividend.size(), '0');
    nonRestoringDivision(dividend, divisor, accumulator);

    return 0;
}

Java

// This class implements non-restoring division algorithm
public class NonRestoringDivision {
// This method performs binary addition of two binary numbers
public static String add(String A, String M) {
    int carry = 0;
    StringBuilder sum = new StringBuilder();

    // Start from the least significant bit and go up to the most significant bit
    for (int i = A.length()-1; i >= 0; i--) {
        // Calculate the sum of two bits and the carry
        int temp = Character.getNumericValue(A.charAt(i)) + Character.getNumericValue(M.charAt(i)) + carry;

        // If the sum is greater than 1, append the least significant bit to the sum and set the carry to 1
        if (temp > 1) {
            sum.append(temp % 2);
            carry = 1;
        }
        // Otherwise, append the sum to the result and set the carry to 0
        else {
            sum.append(temp);
            carry = 0;
        }
    }

    // Reverse the result and return as a string
    return sum.reverse().toString();
}

// This method calculates the 2's complement of a binary number
public static String compliment(String m) {
    StringBuilder M = new StringBuilder();

    // Invert each bit of the input string
    for (int i = 0; i < m.length(); i++) {
        M.append((Character.getNumericValue(m.charAt(i)) + 1) % 2);
    }

    // Add 1 to the inverted value and return as a string
    M = new StringBuilder(add(M.toString(), "0001"));
    return M.toString();
}

// This method performs non-restoring division algorithm
public static void nonRestoringDivision(String Q, String M, String A) {
    int count = M.length();
    String comp_M = compliment(M);
    String flag = "successful";

    // Print the initial values of A, Q, and M
    System.out.println("Initial Values: A: " + A + " Q: " + Q + " M: " + M);

    // Repeat the division process for each bit of M
    while (count > 0) {
        // Shift the contents of A and Q to the left by one bit
        System.out.print("\nstep: " + (M.length() - count + 1) + " Left Shift and ");
        A = A.substring(1) + Q.charAt(0);

        // If the previous step was successful, subtract M from A; otherwise, add M to A
        if (flag.equals("successful")) {
            A = add(A, comp_M);
            System.out.print("subtract: ");
        }
        else {
            A = add(A, M);
            System.out.print("Addition: ");
        }

        // Update the value of Q based on the sign of A and print the current values of A and Q
        System.out.print("A: " + A + " Q: " + Q.substring(1) + "_");
        if (A.charAt(0) == '1') {
            Q = Q.substring(1) + '0';
            System.out.println(" -Unsuccessful");
            flag = "unsuccessful";
            System.out.println("A: " + A + " Q: " + Q + " -Addition in next Step");
        }
        else {
            Q = Q.substring(1) + '1';
            System.out.println(" Successful");
            flag = "successful";
            System.out.println("A: " + A + " Q: " + Q + " -Subtraction in next step");
        }

        // Decrement count
            count--;
        }

        System.out.println("\nQuotient(Q): " + Q + " Remainder(A): " + A);
    }
//Driver code
    public static void main(String[] args) {
        String dividend = "0111";
        String divisor = "0101";
        String accumulator = "0".repeat(dividend.length());

        nonRestoringDivision(dividend, divisor, accumulator);
    }
}

Python3

# Python program to divide two
# unsigned integers using
# Non-Restoring Division Algorithm

# Function to add two binary numbers
def add(A, M):
    carry = 0
    Sum = ''

    # Iterating through the number
    # A. Here, it is assumed that
    # the length of both the numbers
    # is same
    for i in range (len(A)-1, -1, -1):

        # Adding the values at both
        # the indices along with the
        # carry
        temp = int(A[i]) + int(M[i]) + carry

        # If the binary number exceeds 1
        if (temp>1):
            Sum += str(temp % 2)
            carry = 1
        else:
            Sum += str(temp)
            carry = 0

    # Returning the sum from
    # MSB to LSB
    return Sum[::-1]

# Function to find the compliment
# of the given binary number
def compliment(m):
    M = ''

    # Iterating through the number
    for i in range (0, len(m)):

        # Computing the compliment
        M += str((int(m[i]) + 1) % 2)

    # Adding 1 to the computed
    # value
    M = add(M, '0001')
    return M

# Function to find the quotient
# and remainder using the
# Non-Restoring Division Algorithm
def nonRestoringDivision(Q, M, A):

    # Computing the length of the
    # number
    count = len(M)


    comp_M = compliment(M)

    # Variable to determine whether
    # addition or subtraction has
    # to be computed for the next step
    flag = 'successful'

    # Printing the initial values
    # of the accumulator, dividend
    # and divisor
    print ('Initial Values: A:', A,
           ' Q:', Q, ' M:', M)

    # The number of steps is equal to the
    # length of the binary number
    while (count):

        # Printing the values at every step
        print ("\nstep:", len(M)-count + 1,
               end = '')

        # Step1: Left Shift, assigning LSB of Q
        # to MSB of A.
        print (' Left Shift and ', end = '')
        A = A[1:] + Q[0]

        # Choosing the addition
        # or subtraction based on the
        # result of the previous step
        if (flag == 'successful'):
            A = add(A, comp_M)
            print ('subtract: ')
        else:
            A = add(A, M)
            print ('Addition: ')

        print('A:', A, ' Q:',
              Q[1:]+'_', end ='')

        if (A[0] == '1'):


            # Step is unsuccessful and the
            # quotient bit will be '0'
            Q = Q[1:] + '0'
            print ('  -Unsuccessful')

            flag = 'unsuccessful'
            print ('A:', A, ' Q:', Q,
                   ' -Addition in next Step')

        else:

            # Step is successful and the quotient
            # bit will be '1'
            Q = Q[1:] + '1'
            print ('  Successful')

            flag = 'successful'
            print ('A:', A, ' Q:', Q,
                   ' -Subtraction in next step')
        count -= 1
    print ('\nQuotient(Q):', Q,
           ' Remainder(A):', A)

# Driver code
if __name__ == "__main__":

    dividend = '0111'
    divisor = '0101'

    accumulator = '0' * len(dividend)

    nonRestoringDivision(dividend,
                         divisor,
                         accumulator)

C#

// C# program to divide two
// unsigned integers using
// Non-Restoring Division Algorithm

using System;

class Program
{
    // Function to add two binary numbers
    static string Add(string A, string M)
    {
        int carry = 0;
        string Sum = "";

        // Iterating through the number
        // A. Here, it is assumed that
        // the length of both the numbers
        // is same
        for (int i = A.Length - 1; i >= 0; i--)
        {

            // Adding the values at both
            // the indices along with the
            // carry
            int temp = int.Parse(A[i].ToString()) + int.Parse(M[i].ToString()) + carry;

            // If the binary number exceeds 1
            if (temp > 1)
            {
                Sum += (temp % 2).ToString();
                carry = 1;
            }
            else
            {
                Sum += temp.ToString();
                carry = 0;
            }
        }

        char[] charArray = Sum.ToCharArray();
        Array.Reverse(charArray);

        // Returning the sum from
        // MSB to LSB
        return new string(charArray);
    }


    // Function to find the compliment
    // of the given binary number
    static string Compliment(string m)
    {
        string M = "";

        // Iterating through the number
        for (int i = 0; i < m.Length; i++)
        {
            // Computing the compliment
            M += ((int.Parse(m[i].ToString()) + 1) % 2).ToString();
        }

        // Adding 1 to the computed
        // value
        M = Add(M, "0001");

        return M;
    }

    // Function to find the quotient
    // and remainder using the
    // Non-Restoring Division Algorithm
    static void NonRestoringDivision(string Q, string M, string A)
    {

        // Computing the length of the
        // number
        int count = M.Length;

        string comp_M = Compliment(M);


        // Variable to determine whether
        // addition or subtraction has
        // to be computed for the next step
        string flag = "successful";

        // Printing the initial values
        // of the accumulator, dividend
        // and divisor
        Console.WriteLine("Initial Values: A: {0}, Q: {1}, M: {2}", A, Q, M);

        // The number of steps is equal to the
        // length of the binary number
        while (count > 0)
        {
            // Printing the values at every step
            // Step1: Left Shift, assigning LSB of Q
            // to MSB of A.
            Console.Write("\nstep: {0} Left Shift and ", M.Length - count + 1);

            A = A.Substring(1) + Q[0];

            // Choosing the addition
            // or subtraction based on the
            // result of the previous step
            if (flag == "successful")
            {
                A = Add(A, comp_M);
                Console.Write("subtract: ");
            }
            else
            {
                A = Add(A, M);
                Console.Write("Addition: ");
            }

            Console.Write("A: {0}, Q: {1}_{2}", A, Q.Substring(1), "");

            if (A[0] == '1')
            {

                // Step is unsuccessful and the
                // quotient bit will be '0'
                Q = Q.Substring(1) + '0';
                Console.WriteLine(" -Unsuccessful");
                flag = "unsuccessful";
                Console.WriteLine("A: {0}, Q: {1} -Addition in next Step", A, Q);
            }
            else
            {

                // Step is successful and the quotient
                // bit will be '1'
                Q = Q.Substring(1) + '1';
                Console.WriteLine(" Successful");
                flag = "successful";
                Console.WriteLine("A: {0}, Q: {1} -Subtraction in next step", A, Q);
            }

            count--;
        }

        Console.WriteLine("\nQuotient(Q): {0}, Remainder(A): {1}", Q, A);
    }

    // Driver code
    static void Main(string[] args)
    {
        string dividend = "0111";
        string divisor = "0101";

        string accumulator = new string('0', dividend.Length);

        NonRestoringDivision(dividend, divisor, accumulator);
    }
}


//This code contributed by shivhack999

Javascript

// JavaScript program to divide two
// unsigned integers using
// Non-Restoring Division Algorithm

// Function to add two binary numbers
function add(A, M)
{
    let carry = 0
    let Sum = ''

    // Iterating through the number
    // A. Here, it is assumed that
    // the length of both the numbers
    // is same
    for (var i = A.length - 1; i > -1; i--)
    {

        // Adding the values at both
        // the indices along with the
        // carry
        let temp = parseInt(A[i]) + parseInt(M[i]) + carry

        // If the binary number exceeds 1
        if (temp > 1)
        {
            Sum += (temp % 2).toString()
            carry = 1
        }
        else
        {
            Sum += (temp).toString()
            carry = 0
        }
    }

    // Returning the sum from
    // MSB to LSB

    return Sum.split("").reverse().join("")
}

// Function to find the compliment
// of the given binary number
function compliment(m)
{
    let M = ''

    // Iterating through the number
    for (var i = 0; i < m.length; i++)

        // Computing the compliment
        M += ((parseInt(m.charAt(i)) + 1) % 2).toString()

    // Adding 1 to the computed
    // value
    M = add(M, '0001')
    return M
}

// Function to find the quotient
// and remainder using the
// Non-Restoring Division Algorithm
function nonRestoringDivision(Q, M, A)
{

    // Computing the length of the
    // number
    let count = (M).length


    let comp_M = compliment(M)

    // Variable to determine whether
    // addition or subtraction has
    // to be computed for the next step
    let flag = 'successful'

    // Printing the initial values
    // of the accumulator, dividend
    // and divisor
    console.log ('Initial Values: A:', A,
           ' Q:', Q, ' M:', M)

    // The number of steps is equal to the
    // length of the binary number
    while (count > 0)
    {
        Q = Q.split()
        // Printing the values at every step
        process.stdout.write ("\n step: " + (M.length-count + 1))

        // Step1: Left Shift, assigning LSB of Q
        // to MSB of A.
        process.stdout.write (' Left Shift and ')
        A = Array.from(A)
        A.shift()
        A.push(Q[0])
        A = A.join("")

        // Choosing the addition
        // or subtraction based on the
        // result of the previous step
        if (flag == 'successful')
        {
            A = add(A, comp_M)
            console.log ('subtract: ')
        }
        else
        {
            A = add(A, M)
            console.log ('Addition: ')
        }

        Q.shift()

        Q = Q.join("")
        process.stdout.write('A: '+ A + ' Q: ' + Q + " ")

        if (A[0] == '1')
        {

            // Step is unsuccessful and the
            // quotient bit will be '0'
            Q = Q + '0'
            console.log ('  -Unsuccessful')

            flag = 'unsuccessful'
            console.log ('A:', A, ' Q:', Q,
                   ' -Addition in next Step')
        }

        else
        {
            // Step is successful and the quotient
            // bit will be '1'
            Q = Q + '1'
            console.log ('  Successful')

            flag = 'successful'
            console.log ('A:', A, ' Q:', Q,
                   ' -Subtraction in next step')
        }
        count -= 1
    }
    console.log ('\nQuotient(Q):', Q,  ' Remainder(A):', A)
}

// Driver code
let dividend = '0111'
let divisor = '0101'

let accumulator = '0' * (dividend).length

nonRestoringDivision(dividend,
                         divisor,
                         accumulator)

// This code is contributed by phasing17.

Output:

Initial Values: A: 0000  Q: 0111  M: 0101

step: 1 Left Shift and subtract: 
A: 1011  Q: 111_  -Unsuccessful
A: 1011  Q: 1110  -Addition in next Step

step: 2 Left Shift and Addition: 
A: 1100  Q: 110_  -Unsuccessful
A: 1100  Q: 1100  -Addition in next Step

step: 3 Left Shift and Addition: 
A: 1110  Q: 100_  -Unsuccessful
A: 1110  Q: 1000  -Addition in next Step

step: 4 Left Shift and Addition: 
A: 0010  Q: 000_  Successful
A: 0010  Q: 0001  -Subtraction in next step

Quotient(Q): 0001  Remainder(A): 0010