Non-Restoring Division For Unsigned Integer

Last Updated : 23 Sep, 2025

The non-restoring division algorithm is a faster method to divide binary numbers. Unlike traditional division, it avoids repeatedly adding back the divisor, making it more efficient for computers to perform.

  • It uses repeated subtraction and addition to find the quotient bits.
  • If subtraction gives a negative result, the algorithm just adds in the next step instead of fixing it.

Flow Chart of Non-Restoring Division For Unsigned Integer

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Steps Involved in the Non-Restoring Division Algorithm

  • Step-1: First the registers are initialized with corresponding values (Q = Dividend, M = Divisor, A = 0, n = number of bits in dividend)
  • Step-2: Check the sign bit of register A
  • Step-3: If it is 1 shift left content of AQ and perform A = A+M, otherwise shift left AQ and perform A = A-M (means add 2's complement of M to A and store it to A)
  • Step-4: Again the sign bit of register A
  • Step-5: If sign bit is 1 Q[0] become 0 otherwise Q[0] become 1 (Q[0] means least significant bit of register Q)
  • Step-6: Decrements value of N by 1
  • Step-7: If N is not equal to zero go to Step 2 otherwise go to next step
  • Step-8: If sign bit of A is 1 then perform A = A+M
  • Step-9: Register Q contains quotient and A contains remainder.

Example

Let’s divide the binary number 1011 (which is 11 in decimal) by 0011 (which is 3 in decimal) using the Non-Restoring Division Algorithm.

Initialization:

  • Dividend (Q) = 1011
  • Divisor (M) = 0011
  • Accumulator (A) = 0000
  • Number of bits (n) = 4

Step-by-Step Solution

1. Initial Setup:

  • Q = 1011
  • M = 0011
  • A = 0000
  • n = 4

2. First Iteration:

  • Shift Left AQ: A = 0000, Q = 1011 becomes A = 0000, Q = 0110
  • Perform Operation: A = A - M = 0000 - 0011 = 1101 (2's complement of 0011)
  • Sign Bit of A: 1
  • Update Q[0]: 0
  • Decrement N: N = 3

3. Second Iteration:

  • Shift Left AQ: A = 1101, Q = 0110 becomes A = 1010, Q = 1100
  • Perform Operation: A = A + M = 1010 + 0011 = 1101
  • Sign Bit of A: 1
  • Update Q[0]: 0
  • Decrement N: N = 2

4. Third Iteration:

  • Shift Left AQ: A = 1101, Q = 1100 becomes A = 1011, Q = 1000
  • Perform Operation: A = A - M = 1011 - 0011 = 1000 (2's complement of 0011)
  • Sign Bit of A: 1
  • Update Q[0]: 0
  • Decrement N: N = 1

5. Fourth Iteration:

  • Shift Left AQ: A = 1000, Q = 1000 becomes A = 0001, Q = 0000
  • Perform Operation: A = A + M = 0001 + 0011 = 0010
  • Sign Bit of A: 0
  • Update Q[0]: 1
  • Decrement N: N = 0

6. Final Adjustment:

  • Sign Bit of A: 0 (no additional adjustment needed)
  • Final Result: Quotient (Q) = 0011 (3 in decimal)
  • Remainder (A) = 0010 (2 in decimal)
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