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:

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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)

chevron_right


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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.