--- title: "Booth algorithm and 8 × 4" subject: "Computer Organization and Architecture" chapter: 5 key: "semester_5/computer_organization_and_architecture:booth-8x4" source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/booth-8x4 site: BCT Engineering --- # Booth algorithm and 8 × 4 Computer Organization and Architecture · Chapter 5: Computer Arithmetic - Asked in: 2076 Asoj, 10 marks - Video: [Booth's Algorithm With Example | booths | booths algo](https://www.youtube.com/watch?v=cWfaw7b3jKY) — 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/booth-8x4 - Whole chapter: https://www.bctengineer.com/md/semester_5/computer_organization_and_architecture/chapter-5 ## Question Write an algorithm of Booth multiplication. Perform 8×4 using Booth multiplication algorithm. ## Short answer Booth multiplication uses multiplier bit pairs Q0,Q-1: subtract for 10, add for 01, shift right otherwise. For 8 times 4 in 5 bits, A:Q becomes 0000100000, which equals 32. ## Answer #### Algorithm of Booth multiplication Booth's algorithm multiplies two signed numbers in 2's complement form. It looks at the multiplier bit pair (Q0, Q-1) and treats a run of 1s as (next higher power − lowest power), so a block of 1s needs only one subtraction and one addition. 1. Load M = multiplicand, Q = multiplier, A = 0, Q-1 = 0, SC = n (number of bits). 2. Look at (Q0, Q-1): - 00 or 11: no arithmetic. - 10 (start of a run of 1s): A = A − M. - 01 (end of a run of 1s): A = A + M. 3. Arithmetic shift right the combined A, Q, Q-1 by one bit (the sign bit of A is kept). 4. SC = SC − 1. If SC ≠ 0 go to step 2; otherwise stop. 5. The 2n-bit product is in A:Q. #### Flowchart ```text +-----------------------+ | A=0, Q-1=0, SC=n | | M=multiplicand Q=mult | +-----------+-----------+ v +-----------------------+<-----------+ | Q0 Q-1 = ? | | +--+---------+--------+-+ | 10| 01| 00/11| | v v | | A = A - M A = A + M | | | | | | +----+----+---------+ | v | +-----------------------+ | | ASR (A, Q, Q-1) | | | SC = SC - 1 | | +-----------+-----------+ | v | SC = 0 ? --- No --------------+ | Yes v Product in A,Q ``` #### 8 × 4 Five bits are used so that 8 is a valid positive signed number. Multiplicand M = 8 = 01000, multiplier Q = 4 = 00100, -M = 11000, n = 5 bits. ```text Step Action A Q Q-1 - Initial 00000 00100 0 1 Q0Q-1=00: no change 00000 00100 0 ASR 00000 00010 0 2 Q0Q-1=00: no change 00000 00010 0 ASR 00000 00001 0 3 Q0Q-1=10: A = A - M 11000 00001 0 ASR 11100 00000 1 4 Q0Q-1=01: A = A + M 00100 00000 1 ASR 00010 00000 0 5 Q0Q-1=00: no change 00010 00000 0 ASR 00001 00000 0 ``` Result in A,Q = 00001 00000 = 32. **Answer: 8 × 4 = 32** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In Booth's algorithm, what is done when (Q0, Q-1) = 10? A. A = A - M B. A = A + M C. No arithmetic, only shift D. A = A - Q Answer: A. A pair 10 marks the start of a run of 1s in the multiplier, so M is subtracted from A before the shift. (Key point: Booth pair 10) 2. Why does Booth's algorithm need fewer operations for a multiplier with a block of 1s? A. Every 1 bit is added once, but zeros are skipped B. The block of 1s is replaced by 0s C. The multiplicand is shifted left instead of the product shifting right D. A run of 1s is treated as (next higher power - lowest power): one subtraction and one addition Answer: D. The block of 1s is the difference of two powers of 2, so only one subtract (at 10) and one add (at 01) are needed. (Key point: Run of 1s idea) 3. For Q = 4 = 00100, which steps do arithmetic? A. Add at step 1 and subtract at step 5 B. Subtract at step 3 (pair 10) and add at step 4 (pair 01) C. Only a subtract at step 3 D. No steps, only shifts Answer: B. The single 1 bit gives one 10 pair then one 01 pair. (Key point: Single 1 multiplier) 4. What is 8 x 4? A. 16 = 00000 10000 B. 32 = 00010 00000 C. 32 = 00001 00000 D. 64 = 00010 00000 Answer: C. A:Q = 00001 00000 = 32. (Key point: Result) ## Close variants set in other exams - [Booth multiplication 7×3](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-7x3) — Computer Organization and Architecture, chapter 5, asked 2× ## Related in Computer Organization and Architecture - [Booth multiplication algorithm and hardware](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-algorithm) — Computer Organization and Architecture, chapter 5, asked 8× - [Booth multiplication of -5 × -9](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-5x-9) — Computer Organization and Architecture, chapter 5, asked 5× - [Booth multiplication of 5 × -6](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-5x-6) — Computer Organization and Architecture, chapter 5, asked 5× - [Booth multiplication of -7 × -10](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-7x-10) — Computer Organization and Architecture, chapter 5, asked 5× - [Booth multiplication of 10 × (-5)](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-10x-5) — Computer Organization and Architecture, chapter 5, asked 5× - [Booth multiplication of -6 × -11](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-6x-11) — Computer Organization and Architecture, chapter 5, asked 5×