BOOTH MULTIPLICATION, Assignments of Computer science

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Booth's Algorithm

Example

Points to remember

When using Booth's Algorithm:



You will need twice as many bits in your

product as you have in your original two

operands.



The leftmost bit of your operands (both

your multiplicand and multiplier) is a SIGN

bit, and cannot be used as part of the

value.

Beginning Product

The multiplier is:

Add 5 leading zeros to the multiplier to

get the beginning product :

A Q

Step 1 for each pass



Use the LSB (least significant bit) and the

previous LSB to determine the arithmetic

action.



If it is the FIRST pass, use 0 as the previous LSB.



Possible arithmetic actions:

00  no arithmetic operation

01  add multiplicand to left half of product

10  subtract multiplicand from left half of product

ASR-Arithmetic Shift Right

Negative integers where the high order bit is the "sign bit." An arithmetic right shift replicates the sign bit as needed to fill bit positions.

Booth’s Algorithm

Example



Example of multiplying 2 x (-5)



For our example, let's reverse the operation,

and multiply ( -5) x 2

 The numerically larger operand (5) would require 3 bits to represent in binary (101). So we must use AT LEAST 4 bits to represent the operands, to

allow for the sign bit.



Let's use 5-bit 2's complement:

 -5 is 11011 (multiplier)  2 is 00010 (multiplicand)

Example: -5x2;

A Q QN-1 M Operations Count 0000 0010 0 1011 Initialization 4 1011 3 1011 1011 2 1011 1 1011 1011 0

No Op; ASR (^001) ASR 0

1 AA +M 110 1

AA - M ASR

Example



Let's continue with our example of multiplying

( -5) x 2



Remember:



- 5 is 11011 (multiplier)



2 is 00010 (multiplicand)



And we added 5 leading zeros to the

multiplier to get the beginning product :

Example continued

Initial Product and previous LSB

(Note: Since this is the first pass, we use 0 for the previous LSB)

Pass 1, Step 1: Examine the last 2

bits

The last two bits are 10 , so we need to: subtract the multiplicand from left half of product

Example: Pass 1 continued

Pass 1, Step 2: ASR (arithmetic shift right)

Before ASR

After ASR

(left-most bit was 1, so a 1 was shifted in on the left)

Pass 1 is complete.

Example: Pass 2

Current Product and previous LSB

Pass 2, Step 1: Examine the last 2

bits

The last two bits are 11 , so we do NOT need to perform an arithmetic action -- just proceed to step 2.

Example: Pass 3

Current Product and previous LSB

Pass 3, Step 1: Examine the last 2

bits

The last two bits are 01 , so we need to: add the multiplicand to the left half of the product

Example: Pass 3

continued

Pass 3, Step 1: Arithmetic action

(1) 11111 (left half of product)

+00010 (mulitplicand)

00001 (drop the leftmost carry)

Place result into left half of product