Computer Organization and Architecture

Indian Institute of Technology, Guwahati

Assignment:

Booth’s Algorithm for Signed Multiplication

Booth’s Algorithm is an elegant way to multiply signed numbers using the same

hardware as before and save cycles and can handle multiple bits at a time.

Group Details: Bhavya Madan 11010113 Dept. CSE

Chaitanya Agarwal 11010115 Dept. CSE

Sanket Garg 11012335 Dept. MNC

Assignment Guide: Dr. Santosh Biswas Dept CSE IIT Guwahati

Booth’s Algorithm for Signed Multiplication

Since, the multiplication of numbers gives the same absolute value whether

they are positive or negative. So we just have to take care whether the numbers

finally give a positive output or negative output. This is taken care by following

preprocessing and post processing. Other than that we follow the same

algorithm for normal unsigned multiplication.

Preprocessing:

Find the Signed bit of the Product AQ. This can be done using an Ex-OR gate

with signed bits of both the numbers as the input for the Ex-OR gate.

Let the signed bits of numbers be F1 and F2 respectively, and F as the signed bit

for product.

F = F1 ⊕ F2

Post processing:

If F = 0, the product is what is calculated.

If F = 1, the product we get is the 2’s complement of the product (as a negative

number), it can be converted back to the normal representation by putting ‘-’ as

its prefix.

We have an alternate representation for the numbers to avoid preprocessing

and post processing.

Booth’s Algorithm for Signed Multiplication

Use the 2’s complement signed representation for the negative numbers.

Example: -7 is represented as 1001 (1 in the MSB signifies the negative number).

How to have this representation working for Booth’s Algorithm ?

Algorithm:

Let the multiplicand be M, Multiplier be Q and a four-bit register be A.

Add a zero to the end of the number Q (an Extra bit)

Example: 0111 is written as 0111| 0

Read the last two bits of Q including that extra bit, and corresponding symbol

on right is used:

00 0

01 1

10 Ī

11 0

When we get 1, we take A = A + M (ADD)

When we get Ī, we take A = A – M (Subtract)

Right Shift the cumulative register made up of A and Q.

Note: after shifting, the MSB of the number A gets the same serial input as the

original signed bit.

Repeat the process until the MSB of Q reaches the extra bit.

Flowchart of the Booths Algorithm

The following Illustrations will help you understand the Algorithm better.

Illustration 1

1. Multiplicand = 0100

Multiplier = 0111

Here,

We are multiplying 4 and 7

So,

M = 0100

Q = 0111

We start by initializing A with 0000

M A Q

0100 0000 0111 0 Putting

zero

(2’s comp.

Of -4)

Last 2 digits of Q are 10 so we do A=A-M

Therefore A = (0000)-(0100) = 1100

We shift one position.

We put 1 in the MSB as the no. before shifting was negative

1110 0011 1

Last 2 digits of Q are 11.We just shift and not

add. We put 1 in the MSB as the no. before shifting was negative

1111 0001 1

Last 2 digits of Q are 11.We just shift and not

add. We put 1 in the MSB as the no. before shifting was negative

1111 1000 1

Last 2 digits of Q are 01 so we do A=A+M

A=(1111)+(0100)=0011

We shift one position

0001 1100 0

00011100 is 28 in binary our desired product.

Illustration 2

2. Multiplicand = 1100

Multiplier = 0111

Here,

We are multiplying -4 and 7

(1100 is 2’s complement of -4)

So,

M = 1100

Q = 0111

We start by initializing A with 0000

M A Q

1100 0000 0111 0 Putting zero

(2’s comp.

Of -4)

Last 2 digits of Q are 10 so we do A=A-M

Therefore A = (0000)-(1100) = 0100

We shift one position

0010 0011 1

Last 2 digits of Q are 11.We just shift and not

add.

0001 0001 1

Last 2 digits of Q are 11.We just shift and not

add.

0000 1000 1

Last 2 digits of Q are 01 so we do A=A+M

A=(0000)+(1100)=1100

We shift one postion

We put 1 in the MSB as the no. before shifting was negative

1110 0100 0

11100100 is -28 in 2’s complement representation and our

desired product

Illustration 3

3. Multiplicand = 0111

Multiplier = 1100

Here,

We are multiplying -4 and 7

(1100 is 2’s complement of -4)

So,

M = 0111

Q = 1100

We start by initializing A with 0000

M A Q

0111 0000 1100 0 Putting zero

(2’s complement Of -4)

Last 2 digits of Q are 00 so we just shift and not

add.

0000 0110 0

Last 2 digits of Q are 00.We just shift and not

add.

0000 0011 0

Last 2 digits of Q are 10 so we do A=A-M

A=(0000)-(0111)=1001

We shift one position

We put 1 in the MSB as the no. before shifting was negative

1100 1001 1

Last 2 digits of Q are 11. We just shift and not

add.

We put 1 in the MSB as the no. before shifting was

negative

1110 0100 1

11100100 is -28 in 2’s complement representation and our

desired product

Illustration 4

4. Multiplicand = 1100

Multiplier = 1101

Here,

We are multiplying -4 and -3

(1100 is 2’s complement of -4 and 1101 is 2’s complement of -3)

So,

M = 1100

Q = 1101

We start by initializing A with 0000

M A Q

1100 0000 1101 0 Putting zero

(2’s complement Of -4)

Last 2 digits of Q are 10 so we do A=A-M

Therefore A = (0000)-(1100) = 0100

We shift one position

0010 0110 1

Last 2 digits of Q are 01 so we do A=A+M.

A = (0010)+(1100) = 1110

We shift one position

We put 1 in the MSB as the no. before shifting was negative

1111 0011 0

Last 2 digits of Q are 10 so we do A=A-M

Therefore A = (1111)-(1100) = 0011

We shift one position

0001 1001 1

Last 2 digits of Q are 11. We just shift and not add

0000 1100 1

00001100 which is 12 in binary is our desired product.

Hope you would have developed a good understanding of how the Booth

algorithm works. Now, Let us have a look at its Control Path design.

Basics of Control Path Design for Booth Multiplier

We have already seen the Data path which tells us all about how the data flows

in the Booth’s Algorithm and what Arithmetic or Shifting operations are

performed. Now how all this stuff in managed in the Hardware?

For this we’ll see the Control Path Design for the Booth’s Algorithm.

Types of Hardware Blocks required:

A. Adder – Subtractor

B. Counter

C. Multi-Function Register

D. Comparator

Let’s start with an example:

M A Q L

1100 0000 01110 100

-4 7

1100 0100 01110 011

0010 00111

1100 0001 00011 010

1100 0000 10001 001

1100 1100 10001 000

1100 1110 01000

L is counter as we need to make sure we do only four iterations.

We need a control signal to freeze the value in A and Q when L is 000. In the

beginning, we need to reset A to 0000, load M and Q. So we need a control path

design.

Value of M is available is in external memory or incoming data bus.

The algorithm:

A<-0; M<-Inbus; L<-4;

Rest Load Load

Q [4:1] Inbus; Q[0] 0;

Load Load

Loop: if Q[1:0]=01 go to Add

If Q[1:0]=10 go to Sub

Go to Rshift (for 00 or 11)

Add: A A+M; go to Rshift

Sub: A A-M go to Rshift

Rshift: Shift (AQ); LL-1;

shift puts MSB as old MSB Decrement

If L<>0 (Comparator)

go to loop;

Output: AQ; HALT;

So we require a register which can load, reset, intelligent shift, decrememnt,

although not all.

C L R D

1 0 0 0 Clear

0 1 0 0 Load

0 0 1 0 Right shift

0 0 0 1 Decrement by

1

0 0 0 0 Freeze

Multifunction register

Although Q required only load and shift, L required load, decrement, etc. then

also we use a register with all five operations in all registers for simplicity.

Although we have only five operations, we could have done using 3 bits also but

then we would require a decoder. So we use a flat design. Also it’s easier to

debug.

Tri state buffer

The terminals of AQ are always connected to output trio but required stable

output is obtained only after freeze (i.e. Halt). As we don’t want output to

change as AQ changes, we cut the signal using tri-state buffer, like a switch.

Bidirectional bus

When C=0, 1 gets control signal 0 and 2 gets the inverted signal i.e. 1. So 2 allow

flow. Hence data flows from Y1 to X1. When C=1, data flows from X1 to Y1.