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Booth multiplication of 23 × -21 (6 bits)

Short answer

In Booth multiplication, inspect multiplier bit pair Q0,Q-1; subtract M on 10, add M on 01, else do nothing, then arithmetic-shift right. For 23 times -21 in 6 bits, the product A:Q is 111000011101, which equals -483.

Explain Booth algorithm. Use the Booth algorithm to multiply 23 (multiplicand) by -21 (multiplier), where each number is represented using 6 bits.

Booth algorithm

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.

23 × (−21), 6 bits

Multiplicand M = 23 = 010111, multiplier Q = −21 = 101011 (2's complement of 21 = 010101), −M = 101001, n = 6 bits.

Step  Action               A       Q       Q-1
-     Initial              000000  101011  0
1     Q0Q-1=10: A = A - M  101001  101011  0
      ASR                  110100  110101  1
2     Q0Q-1=11: no change  110100  110101  1
      ASR                  111010  011010  1
3     Q0Q-1=01: A = A + M  010001  011010  1
      ASR                  001000  101101  0
4     Q0Q-1=10: A = A - M  110001  101101  0
      ASR                  111000  110110  1
5     Q0Q-1=01: A = A + M  001111  110110  1
      ASR                  000111  111011  0
6     Q0Q-1=10: A = A - M  110000  111011  0
      ASR                  111000  011101  1

Result in A,Q = 111000 011101 = -483.

Answer: 23 × -21 = -483

✓ Answer checked by Claude Opus 5.5 (high effort)

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