General Fixed Radix Number Systems

• Nonredundant• Positive radix, ß• n digits in digit set• Vector:

1 2 0( , , , )n n { 1,1}i

i ii allows or

1

0

ni

i ii

X x

i

i iw

General Fixed Radix Number Systems• For Given Radix and n, how many number systems?

• ANSWER: Number equal to all possible permutationsof n choose –1 (or +1),

0 1 2

n n n n

n

0

2n

n

i

n

i

Choose -1

Positive Radix

Negative Radix

• Of these, 1 is Pos. Radix 1 is Neg. Radix

• The following is the Radix-complement:( 1,1,1,1, ,1,1)

General FR Number Systems - Properties

P Largest Representable Integer

1 2 0( )n nP p p p

N Smallest Representable Integer

pi are Digits of P

1, 1

0 ,i

i

ifp

otherwise

1( 1)( 1)

2i ip

General FR Number Systems - Properties

1 1 1

0 0 0

1 1( 1)( 1) ( 1) ( 1)

2 2

n n ni i i

i ii i i

P

Using the pi Expression and Forming the Radix Polynomial for P

1 11

0 0

1 1( 1) ( )

2 2

n ni i i

i i

Q Q

digit

weightDefine as Q

1n

1( 1)

2nP Q

General FR Number Systems - Properties

( 1, 1, 1, , 1)

Q is the value represented by the following n-tuple if all i=1

For N, the Smallest Representable Value:

1 2 1 0( )n nN y y y y

1, 1

0 ,i

i

ify

otherwise

1( 1)( 1)

2i iy

General FR Number Systems - Properties

1 1 1

0 0 0

1 1( 1)( 1) ( 1) ( 1)

2 2

n n ni i i

i ii i i

N

Using Similar Analysis as With the Case of P:

1 11

0 0

1 1( 1) ( )

2 2

n ni i i

i i

Q Q

digit

weightDefine as Q

1n

1( 1)

2nN Q

General FR Number Systems –Symmetry

Summarizing:

N X P Where 1 nP N

• In General These Bounds are Asymmetric

• Measure of Asymmetry is:

| |P N P N Q

• Therefore, Q is a Measure of Asymmetry for GeneralizedFixed Radix Number Systems

GFRNS – Asymmetry Examples

Consider the Negative Radix System:

Asymmetric Range:n even times as many negative as positive values n odd times as many positive as negative values

| |P N P N Q ulp

2’s Complement:

( , 1, 1, 1, 1, )

2 ( 1, 1, 1, 1, , 1)

(1 more negative number)

| |P N P N Q ulp

System:

2 ( 1, 1, 1, 1, , 1)

(1 more positive number)

GFRNS – Complement

Recall that a complement of a digit, xi, is:

The Complement of a Value, X, is Calculated as:

( 1)i ix x

1 1 1

0 0 0

( 1)n n n

i i ii i i i i

i i i

X x x

Q X

X Q X Thus,

( )X X Q X Q

Signed-Digit Number Systems

• Fixed radix (positional)• Allows each digit to carry a sign

{ 1, 2,....,1,0,1,...., 1}ix

i i

example

This signed digit (SD) is a new definition of the digit complement

Signed-Digit Example

{9,8,....,1,0,1,....,8,9}ix

min max99, 99

[ 99,99]

1

X X

X

ulp

for a total of 19 possible digits

If n = 2

199 values, however there are 192 = 361 representations possible which implies this is a redundant number system

10

Signed-Digit Example - Redundancy

[ 99,99]X

19 possible digits

199 values and 192 = 361 representations implies redundancy

{9,8,....,1,0,1,....,8,9}ix

361 199% 81.4%

199redundancy

10(12) (08) ( 8)ix Example redundant representation:

For n = 2, range is

Redundancy Index, = + + 1 – r for digit set is [- , ]Here, = 9 + 9 + 1 – 10 = 9, but if = 0 & =9, then = 0.

Restricting Redundancy{ , 1, ,1,0,1 , }

11, 5 9 ( 10)

2

ix a a a

where

ra r a r

2

6

66 66 133

(13) 169

(12) 8

169 133% 27%, 6 6 1 10 3

133

Let a

Range of x includes numbers

representations

unique

redundancy

Signed-Digit Characteristics

• Positive radix, ß > 0

• X = 0 is unique

• Easy to convert

• Constant Delay for Add/Sub Regardless of Word Size

Breaking the Carry Chain Using SDCan make sum only a function of two digit positions

),,,( 11 iiiii yxyxfs

1

1 ( )

1 ( )

0 )

i i

i i i

i i i

if x y a

t if x y a

if x y a

1i i i iw x y rt

 

 

Carry-Free Addition AlgorithmStep 1: Find interim sum wi and transfer digit ti+1 where

and

Step 2: Find final sum si

0: 0i i is w t note t

positional sum pi

Signed Digit Addition Hardware

ix

1ix

1ix

iy

1iy

1iy

is

1is

1is

it

1it

2it

2it

iw

1iw

1iw

SD Addition Example

Let a = 6 for r = 10

1

1

{6, 5, , 1,0,1, ,5,6}

( ) 10

1 ( ) 6

1 ( ) 6

0

i

i i i i

i i

i i i

i i i

x

w x y t

if x y

t if x y

otherwise

s w t

SD Addition Example (Continued)

Let X = 1634, Y = 3366Using normal addition produces a carry chain

1 1 1

1 6 3 4

3 3 6 6

5 0 0 0

X

Y

But by the carry-free algorithm

1 6 3 4

3 3 6 60 1 1 1 0

4 1 1 0

5 0 0 0

i

i

i

X

Yt

w

s

Converting Decimal to SD

Let r = 10, a = 6Consider the value as xi + yi and use algorithm

Converting from SD to decimal – just sum plus and minus weights

2030 – 204 = 1826

1 8 2 6

0 1 0 1 0

1 2 2 4

2 2 3 4

i i

i

i

i

x y

t

w

s

Selecting a to Eliminate Carry Chain in SD

For no carry, require

{ , 1, ,1,0,1, , 1, }

1

1 (1)

, 1

i

i i i

i

i

i i i

s a a a a

s w t a

worst case t restricting

w a equation

must be true for all x y where t

Selecting a to Eliminate Carry Chain in SD1: 2

(2 ) (1) 2

(1)

2 1 1 ( )

2 :

( ) (1) 0

(1)

1

12 1 ( )

2

11

2

i i

i

i i

i

i

Case x y a

w a r a r

By

a r a or a r upper bound

Case x y a

w a r a r

w r a

By

r a a

ra r or a lower bound

ra r

Binary SD Addition

2 , 1, ( , 1,0,1 )

11 2 !

2

For r a only one digit set

rr a impossible for digit set

Implies no guarantee that si = wi + ti will not produce a carry

Looking at algorithm:Step 1:

1

1

( ) 2

1 1

1 1

0 0

i i i i

i i

i i i

i i

w x y t

if x y

t if x y

if x y

Unmodified Binary SD Addition Table

xi,yi

ti+1

wi

11 1111010100

10

0 0 0 0

01

1

11

1

Step 2: Based on calculation of wi and ti+1

10

1 1

0 1 0 1

i i is w t

Note: redundancy allows choices for wi and ti+1

How Useful is Unmodified Table?

Works if operands do not contain If operands contain only 0’s and 1’s, no carry generated.

1

0 1 0 1or or

i i is w t

1 1 .... 1 1 1

0 0 .... 0 0 11 1 1 .... 1 1

1 1 .... 1 1 0

1 0 0 .... 0 0 0

i

i

i

t

w

s

Example

Why not use this approach to break carry chain for unsigned binary number?

Limitations of Table

1 1

1 01

1 01, 11

.

i i i

i i i

i

If w x y

and t x y

s has carry

0 1 1 1 1 1

1 0 0 1 0 1

1 1 1 1 1 1

1 1 1 0 1 0

* * * * 1 0 0

i

i

i

t

w

s

carries

Example (-9)10 + (29)10

Does not work if operands contain 1

SD Addition Table Choices

11

01

01

1 0

0

1

1

1

1000

1011

11

11

01

01

00

00

ix iy

i it w

i it w

1 1 0,1i ix or y

1 1 1i ix or y

Takagi, 1985

Modified Binary SD Addition Table

xi,yi

xi-1,yi-1 -neither is at least

one is neither is at least

one is - -

ti+1

wi

1101 11010100,11

1

0

0

0

0 0

1

1 1

1

1

0 11

01

1

11

1

Repeating Example with Modified Table

Example (-9)10 + (29)10

0 1 1 1 1 1

1 0 0 1 0 1

0 0 0 1 1 1

1 1 1 0 1 0

1 1 0 1 0 0

i

i

i

t

w

s

no carries

Two SD Encodings

x

Encoding 1

xh xl

Encoding 2

xh xl

0 00 00

1 01 01

1 10 11

4!=24 possible encodingsOnly nine are distinct under permutation and logical negation two’s complement

Encoding 1

Satisfies simple relation

x = xl - xh

and 11 has a valid numerical value of 0.

SD to two’s complement conversion performed by two’s complement subtraction

1 2 0

1 2 0

, , ,

, , ,

l l ln n

h h hn n

x x x

x x x

Encoding 2

Satisfies relation

xi = -2xih + xi

l

This means that xil and xi-1

h have the same weight

Also simplified addition table possible by regrouping bits

1 ˆ, ,l hi i ix and x can be regrouped to form x

Two’s Complement/BSD Conversion

Two’s Complement to SDBits can be encoded directly with MSB negative one

BSD to Two’s Complement

1

0 0 0 0

0 1 1 1

1 0 1 0

1 1 0 0

1 0 1 1

1 1 0 1

i i i iy c z c

12i i i iy c z c

One algorithm simpler than complete binary adder zi is two’s complement result c0 = 0

0 1 0 1 01 1 1 1 0 0

1 0 1 1 0

i

i

i

yc

z

Example -1010

Binary SD RepresentationsRepresentation of a value with the minimum number of non-zero digits – Important in multiplication and division since each zero eliminates an operation

 8 4 2 1

0 1 0 1

1 0 1 1 .

0 1 1 1

1 1 0 1

1 1 1 1

Minimal SD representation of X = 5

X = 5, n = 4, r = 2

Alternate Class of BSD Addition Tables*

*see M. Thornton, A Signed Binary Addition Circuit Based on an Alternative Class of Addition Tables

xi,yi

ti+1

wi

11 1111010100

10

0 0 0 0

01

1

11

1

Motivation: Previous tables based on calculation ofwhere wi and ti require 2-bit encoding

i i is w t

Alternative restricts these values to {0,1} or {1,0}

Note: In the discussion ci will be used in place of ti and ui for wi

Basic Idea for Alternate SBD Representation

• Restrict

1 1

0,1 1,0

1,0 0,1

2( )

2, 1,0,1,2

i i

i i

i i i i i i

i i i

i i

c u

or

c u

where

x y u c b b

s u c

x y

• Add 2bi+1 since it is “borrowed” from i+1 column• Subtract bi since it is “borrowed” from i-1 column

Basic SD Addition Tables

1 0 1

1 2 1 0

0 1 0 1

1 0 1 2

1 0 1

1 10 01 00

0 01 00 01

1 00 01 10

i i ix y s iy

ix

12i i i ix y u c iy

ix 0 1

1 1

1 0

0 1

0 * 0

Both have inherent propagation limitations

Alternative 1 BSD Addition Table*

iy

ix 1 0 1

11

0 11

0 01

1 1

0 0 1 1 1 0

01

0 00

0 10

0 0

1 1 0 0 1 1

11

1 10

0 00

1 1

1 0 1 1 1 0

bi+1

ci+1 ui

ci+1 ui

bi=1

bi=0

1 1

0,1 1,0 1,0

1 1 2

2( )

1 2 1 2

i i i

i i

i i i i

c u b

x y

u c b b

* Table 2 of Thornton paper

Alternative 2 BSD Addition Table*

iy

ix 1 0 1

10

1 00

1 11

1 0

1 1 0 0 1 1

00

1 10

0 01

1 1

0 0 0 1 0 0

11

1 01

1 11

0 0

1 1 0 0 0 1

bi+1

ci+1 ui

ci+1 ui

bi=1

bi=0

1,0 0,1 0,1

1 0 1 1 1 0 0

0 1 1 0 1 0 11 0 0 0 1 0 0

1 1 0 0

0

0

0 0 1

1 1 1 0 0 0 1

0 0 0 0 0 1

1

1

0

0 0

i i ic u b

X

YBU

C

S

* Table 4 of Thornton paper

Signed Binary Addition Hardware Using Intermediate Borrow and Carry

ix

1ix

1ix

iy

1iy

1iyis

1is

ic

1iciu

1iu

2ix2iy

2ib

2ix2iy

1iu

2ic1

1

1

1

1

1

12

22

2

2

22

2

2

2

2

2

2

2

2

2

2

2

1

1

1

Encoding Scheme For Even Parity*

• Choose 1 00

0 01 11

1 10

or

• Then each successive pair of signed binary digits canbe grouped with even parity

• Single bit error coverage over digit pairs

3 11 0000

2 10 0011

1 01 1100

0 00 0101 1111

1 01 0110

2 10 1001

3 11 1010

value pair encoding

or

*see M. Thornton, Signed Binary Addition Circuitry with Inherent Even Parity Outputs