Booth's Algorithm: $20 \times (-19) = -380$

Take $n=6$ bits (sign inclusive). Result $2n=12$ bits.

• Multiplicand $M = 20_{10}=010100_2$, $-M = 101100_2$ (2's complement)
• Multiplier $Q = -19_{10}=101101_2$ (since $19=010011$, invert+1)
• Registers: $A=000000$, $Q_{-1}=0$, count $=6$

Rule per step: inspect $Q_0 Q_{-1}$:

• $00$ or $11$: no add, ASHR $A,Q,Q_{-1}$
• $01$: $A \leftarrow A+M$
• $10$: $A \leftarrow A-M$, then ASHR

Step	$Q_0Q_{-1}$	Action	$A$	$Q$	$Q_{-1}$
Init	-	$A=0$	000000	101101	0
1	10	$A=A-M$	101100	101101	0
		ASHR	110110	010110	1
2	01	$A=A+M$	001010	010110	1
		ASHR	000101	001011	0
3	10	$A=A-M$	110001	001011	0
		ASHR	111000	100101	1
4	11	none	111000	100101	1
		ASHR	111100	010010	1
5	01	$A=A+M$	010000	010010	1
		ASHR	001000	001001	0
6	10	$A=A-M$	110100	001001	0
		ASHR	111010	000100	1

After 6 shifts, count=0.

$$A\,Q = 111010\,000100_2 = 12\text{-bit 2's complement}$$

Check: $111010000100_2$ → invert $000101111011+1=000101111100_2=380_{10}$ → value $-380_{10}$.

$$\mathbf{20 \times (-19) = -380}$$

If using $6$-bit unsigned representation: $111010000100$ ($0xE84$). In decimal $380 = 000101111100$, $-380 =111010000100$.

For $5$-bit demo ($M=010100$ still 6-bit needed because $20>15$), same result. Any $n \ge 6$ yields same 12-bit product with sign extension.