Computer Arithmetic Algorithms, Israel Koren; Pub A K Peters, 2002. Computer Arithmetic: Algorithms and Hardware Designs, B. Parhami; Oxford University Press, 2001. Other text, lecture slide handouts and notes will be on line. Lecture Outline. Jan 7 lecture 1.ps.pdf Introduction, the integers.

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Negative Number Representation

This is a book about algorithms for performing arithmetic, and their imple-mentation on modern computers. We are concerned with software more than hardware — we do not cover computer architecture or the design of computer hardware since good books are. Cations to computer arithmetic. We consider different number representations, fast algorithms for multiplication, division and exponentiation, and the use of the Chinese remainder theorem (CRT). Chapter 3 covers floating-point arithmetic. Our concern is with high-precision floating-point arithmetic, implemented in software if the precision.

• Sign Magnitude

Sign magnitude is a very simple representation of negative numbers. In sign magnitude the first bit is dedicated to represent the sign and hence it is called sign bit.

Sign bit ‘1’ represents negative sign.

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Sign bit ‘0’ represents positive sign.

In sign magnitude representation of a n – bit number, the first bit will represent sign and rest n-1 bits represent magnitude of number.

For example,

• +25 = 011001

Where 11001 = 25

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And 0 for ‘+’

• -25 = 111001

Where 11001 = 25

And 1 for ‘-‘.

Range of number represented by sign magnitude method = -(2^n-1-1) to +(2^n-1-1) (for n bit number)

But there is one problem in sign magnitude and that is we have two representations of 0

+0 = 000000

– 0 = 100000

• 2’s complement method

To represent a negative number in this form, first we need to take the 1’s complement of the number represented in simple positive binary form and then add 1 to it.

For example:

(8)10 = (1000)2

1’s complement of 1000 = 0111

Adding 1 to it, 0111 + 1 = 1000

So, (-8)10 = (1000)2

Please don’t get confused with (8)10 =1000 and (-8)10=1000 as with 4 bits, we can’t represent a positive number more than 7. So, 1000 is representing -8 only.

Range of number represented by 2’s complement = (-2^n-1 to 2^n-1 – 1)

Floating point representation of numbers

• 32-bit representation floating point numbers IEEE standard

Normalization

• Floating point numbers are usually normalized
• Exponent is adjusted so that leading bit (MSB) of mantissa is 1
• Since it is always 1 there is no need to store it
• Scientific notation where numbers are normalized to give a single digit before the decimal point like in decimal system e.g. 3.123 x 10³

For example, we represent 3.625 in 32 bit format.

Changing 3 in binary=11

Changing .625 in binary

Writing in binary exponent form

3.625=11.101 X 2⁰

On normalizing

11.101 X 2⁰=1.1101 X 2¹

On biasing exponent = 127 + 1 = 128

(128)10=(10000000) 2

For getting significand

Digits after decimal = 1101

Expanding to 23 bit = 11010000000000000000000

Setting sign bit

As it is a positive number, sign bit = 0

Finally we arrange according to representation

• 64-bit representation floating point numbers IEEE standard

Again we follow the same procedure upto normalization. After that, we add 1023 to bias the exponent.

For example, we represent -3.625 in 64 bit format.

Changing 3 in binary = 11

Changing .625 in binary

Writing in binary exponent form

3.625 = 11.101 X 2⁰

On normalizing

11.101 X 2⁰ = 1.1101 X 2¹

On biasing exponent 1023 + 1 = 1024

(1024)10 = (10000000000)2

So 11 bit exponent = 10000000000

52 bit significand = 110100000000 …………. making total 52 bits

Setting sign bit = 1 (number is negative)

So, final representation

1 10000000000 110100000000 …………. making total 52 bits by adding further 0’s

Converting floating point into decimal

Let’s convert a FP number into decimal

1 01111100 11000000000000000000000

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The decimal value of an IEEE number is given by the formula:

(1 -2s) * (1 + f) * 2^{( e – bias )}

where

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• s, f and e fields are taken as decimal here.
• (1 -2s) is 1 or -1, depending upon sign bit 0 and 1
• add an implicit 1 to the significand (fraction field f), as in formula

Again, the bias is either 127 or 1023, for single or double precision respectively.

First convert each individual field to decimal.

• The sign bit s is 1
• The e field contains 01111100 = (124)10
• The mantissa is 0.11000 … = (0.75)10

Putting these values in formula

(1 – 2) * (1 + 0.75) * 2124 – 127 = ( – 1.75 * 2^-3 ) = – 0.21875

This article has been contributed by Anuj Batham.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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