--- title: "Restoring vs non-restoring division" subject: "Computer Organization and Architecture" chapter: 5 key: "semester_5/computer_organization_and_architecture:restoring-vs-nonrestoring" source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/restoring-vs-nonrestoring site: BCT Engineering --- # Restoring vs non-restoring division Computer Organization and Architecture · Chapter 5: Computer Arithmetic - Asked in: 2076 Chaitra, 6 marks; 2076 Asoj, 6 marks; 2075 Asoj, 6 marks; 2071 Shrawan, 4 marks; 2068 Baisakh (old course), 8 marks (asked 5 times) - Video: [3. Binary Division method (Restoring and Non-restoring Division Algorithm)](https://www.youtube.com/watch?v=6ToR6vuRb3M) — Always Learn More - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/restoring-vs-nonrestoring - Whole chapter: https://www.bctengineer.com/md/semester_5/computer_organization_and_architecture/chapter-5 ## Question Differentiate between restoring and non-restoring division. ## Short answer Restoring division adds back the divisor if subtraction yields negative remainder. Non-restoring division does not restore; it adds or subtracts in the next step based on the remainder sign. Non-restoring is faster with one operation per bit, and corrects remainder if final A is negative. ## Answer Both methods divide a dividend Q (in the A:Q registers) by a divisor M, giving a quotient in Q and a remainder in A. In restoring division, a failed subtraction is undone by adding M back; in non-restoring division the correction is delayed to the next step. | Basis | Restoring division | Non-restoring division | |---|---|---| | After A - M gives a negative result | Restore: A = A + M, then Q0 = 0 | Do not restore; Q0 = 0 and next step adds M | | Operation per step | Shift, subtract, (restore if needed) | Shift, then subtract or add depending on sign of A | | Steps per bit | 1 or 2 (subtract, and add if negative) | Exactly 1 add/subtract | | Speed | Slower (extra restoring addition) | Faster (fewer operations) | | Hardware / control | Simple, but needs restore path | Slightly more complex sign test, no restore step | | Final remainder | Always correct in A | If A is negative at the end, add M once to correct | | Quotient bit | 1 if A - M ≥ 0 | 1 if A (after op) ≥ 0, else 0 | | Number of additions | More | Fewer | #### Example: 11 ÷ 3 (n = 4, M = 0011) **Restoring:** | Step | A (after shift) | Q | A - M | Action | A | Q (after) | |---|---|---|---|---|---|---| | init | 00000 | 1011 | | | | | | 1 | 00001 | 0110 | -00010 (neg) | negative: Q0=0, restore A | 00001 | 0110 | | 2 | 00010 | 1100 | -00001 (neg) | negative: Q0=0, restore A | 00010 | 1100 | | 3 | 00101 | 1000 | 00010 | positive: Q0=1 | 00010 | 1001 | | 4 | 00101 | 0010 | 00010 | positive: Q0=1 | 00010 | 0011 | Quotient = 0011 = 3, remainder = 2. **Non-restoring** (negative values shown with a minus sign for clarity): | Step | Shift AQ left: A | Q | Operation | A (new) | Q0 set | Q | |---|---|---|---|---|---|---| | init | 00000 | 1011 | | | | | | 1 | 00001 | 0110 | A = A - M | -00010 | 0 | 0110 | | 2 | -00100 | 1100 | A = A + M | -00001 | 0 | 1100 | | 3 | -00001 | 1000 | A = A + M | 00010 | 1 | 1001 | | 4 | 00101 | 0010 | A = A - M | 00010 | 1 | 0011 | Quotient = 0011 = 3, remainder = 2. The final A is positive, so no correction is needed. Both give 11 = 3 × 3 + 2. The non-restoring method used no restore steps. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. After A - M gives a negative result, restoring division: A. Adds M back (restores A) and sets Q0 = 0 B. Leaves A and adds M next step C. Sets Q0 = 1 D. Stops Answer: A. Non-restoring delays the correction by adding M in the next step. (Key point: Handling negative result) 2. Why is non-restoring division faster? A. It has exactly one add/subtract per step and no restore addition B. It has a faster clock C. It uses fewer bits D. It skips the shift Answer: A. Restoring may need two operations per bit; non-restoring needs one. (Key point: Speed) 3. In non-restoring division, if A is negative at the end: A. Add M once to correct the remainder B. Subtract M C. Quotient is wrong and must be redone D. Nothing needed Answer: A. Restoring always gives a correct remainder in A, non-restoring may need a final correction. (Key point: Final remainder) 4. 11 divided by 3 gives: A. Quotient 3, remainder 2 B. Quotient 2, remainder 3 C. Quotient 4, remainder 1 D. Quotient 3, remainder 1 Answer: A. 3 x 3 + 2 = 11 in both methods. 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