Using Booth$'$s Algorithm show the multiplication of $7 \times 5$

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written 7.9 years ago by

teamques10 ★ 66k

Using the flowchart, we can solve the given question as follows:

$(5)_{10}= 0101$ (in 2’s complement)

$(7)_{10} =0111$ (in 2’s complement)

Multiplicand (B) = 0101

Multiplier (Q) = 0111

And initially $Q_{-1}$= 0

Count =4

Steps	A	Q	$Q_1$	Operation
Initial	0000	0111	0	-
1	1011 1101	0111 1011	0 1	A=A-B Right shift
2	1110	1101	1	Right shift
3	1111	0110	1	Right Shift
4	0100 0010	0110 0011	1 0	A=A+B <right shift="">
Result	0010	0011	-	-

The result to the problem is $(0010 0011)_2$ or $(35)_{10}$

Here a property of Arithmetic right shift can be seen. When after addition or subtraction(A-B or A+B) a right shift occurs,the left most bit of $A(A_{n-1})$ is shifted to the second left most bit i.e. $A_{n-2}$ the $(n-1)^{th}$ bit is retained as shown in the above example. This is done to preserve the sign of A.

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