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When multiplicand $Y$ is multiplied by multiplier $X = x_{n - 1}x_{n-2} \dots x_0$ using bit-pair recoding in Booth's algorithm, partial products are generated according to the following table.

$${\begin{array}{c|ccc|c}\hline
\textbf{Row}&  \bf{  x_{i+1}}&\bf { x_i}&\bf{ x_{i-1}}&  \textbf{Partial Product} \\\hline
1&0&0&0&0 \\ 2&0&0&1& \text{Y} \\     3&0&1&0& \text{Y} \\   4&0&1&1& \text{2Y} \\    5&1&0&0& \text{?} \\    6&1&0&1& \text{-Y}  \\    7&1&1&0& \text{-Y} \\    8&1&1&1& \text{?} \\ \hline   
 \end{array}}$$

The partial products for rows $5$ and $8$ are

  1. $2Y$ and $Y$
  2. $-2Y$ and $2Y$
  3. $-2Y$ and $0$
  4. $0$ and $Y$
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3 Answers

Best answer
10 votes
10 votes
We can have $1$ bit or $2$ bit Booth codes. This question is about $2$ bit Booth codes. In Booth's algorithm, we get partial products by multiplying specific codes with the multiplicand. These partial products are used to get the actual product. We initially calculate $2$ bit booth code of the multiplier in this question. Then each bit of the code is multiplied with the multiplicand to get the partial product as shown in the last column of the given table.
Here, the multiplicand is $Y.$ So, notice that each row of partial product column is multiplied with $Y.$

Now, the question is how to get these codes i.e., how to represent a multiplier with a $2$ bit booth code. For that we need to look at the pair of $3$ bits as shown in the table below. To get code $C_{i},$ look for $3$ bits as shown.
$${\begin{array}{ccc|c}
\bf{x_{i+1}}&    \bf{x_i}&  \bf{x_{i+1}}&\bf{ Booth code (C_i)}\\\hline
0&0&0&0 \\ 0&0&1&1 \\    0&1&0&1 \\   0&1&1&2\\    1&0&0&-2\\    1&0&1&-1  \\    1&1&0&- 1 \\   1&1&1&0\\   
\end{array}}$$
Now, multiply $i^{\text{th}}$ code to get partial product.
Therefore, Option C is correct.
edited by
14 votes
14 votes

Partial product is calculated by using bit pair recording in booths algorithm, which is improvement technique used in booths algorithm. Here we consider 3 bits at a time for getting the partial product. This eliminates the worst case behaviour of normal Booth's algorithm. Partial Product calculation can be seen in the below link:

http://www.geoffknagge.com/fyp/booth.shtml
"100" corresponds to -2M and "111" corresponds to 0
-2 X(i+1) + x (i) + X(i-1)

ANS : C

1 votes
1 votes

Please refer to this, it will help you.

https://www.techtud.com/short-notes/basics-booths-multiplication-booth-re-coding

In the given question they just want to know how you are decoding a three-bit number. In each row from the right-hand side, you take three bits and decode it in booth multiplier form(2's complement number) and simply find the weighted decimal equivalent of booth multiplier.

Answer:

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