Coping With the Carry Problem 1. Limit Carry to Small Number of Bits Hybrid Redundant Residue Number...

Coping With the Carry Problem

1. Limit Carry to Small Number of Bits• Hybrid Redundant• Residue Number Systems

2. Detect the End of Propagation Rather Than Wait for Worst-case Time• Asynchronous (Self-Timed) Design

3. Speed-up Propagation Using Carry Lookahead and Other Methods• Lookahead • Carry-skip• Ling Adder • Carry-select• Prefix Adders • Conditional Sum

4. Eliminate Carry Propagation Altogether• Redundant Number Systems• Signed-Digit Representations

Residue Number Systems (RNS)

• Convert Arithmetic on Large Numbers to Arithmetic on Small Numbers

• Significant Speedup in Some Signal Processing Algorithms

• Valuable Tool for Theoretical Studies of the Limits of Fast Arithmetic


• Integer System

• Addition, Subtraction, Multiplication Carry Free !!!

• Division, Comparison, Sign Detection Complex and Slow

• Inconvenient For Fractional Representations• Generally Used For Special Purpose Applications such as DSP Filters


• Radix is n-tuple of Integers (mn,mn-1,...,m1)

Not a Single Base Value

• Integer X Represented by n-tuple (xn,xn-1,...,x1)

• qi is Largest Integer Such That:

• xi is the Residue of X mod mi

i i iX m q x

0 ( 1)i ix m

RNS Example Problem

Chinese Scholar, Sun Tzu wrote (1500 years ago):What number has the remainders of 2, 3 and 2 when divided by the values 7, 5 and 3 respectively?

NOTATION:modi i i ix X m x m

Sun Tzu’s Problem:

(7|5|3)

2 | 3 | 2RNS

X

Residue (Modulo) of a Number

11mod 3 2

1mod3 ( 1 3) mod3

2mod3

(2 3) mod 3

5mod3

2

Many Examples in Chapter 4 of Text Use:

RNS 8 | 7 | 5 | 3

Moduli Selection

• Dynamic Range – Product of k Relatively Prime Moduli

• Product, M, is Number of Different Representable Values

in the RNS

DEFINITON

mi and mj are Relatively Prime if gcd(mi,mj) = 1

EXAMPLE

mi = 4 and mj = 9, gcd(4,9) = 1

Although Neither 4 Nor 9 is Prime, They are Relatively Prime

RNS Representation

• Consider RNS(8|7|5|3) (our default RNS in this class)

• 840 Distinct Representable Values

• Since

• Can Represent

• Any Interval of 840 Consecutive Values

8 7 5 3 840M

i iX m M X m

[0,839],[ 420,419],X

Example RNS ValuesRNS=(8|7|5|3)

RNS

RNS

RNS

RNS

RNS

R

RNS

NS

RNS

(5 | 0 |1|

(0 | 0 | 0 | 0) 0,840,1680,

(1|1|1|1) 1,841,1681,

(2 | 2 | 2 | 2) 2,842,1682,

(0 |1| 3 | 2) 8,848,1696,

21,861, ,

(0 |1| 4 |1) 64,904,1744,

(2 | 0 | 0 | 2) 70,770,1610,

(7 | 6 | 4 | 2) 1,839

0) 7

,

01

1

1

679,

RNS Example170110 RNS=(8|7|5|3)

10 10 10 10

10 10 10 10

10 10 10 10

10 10 10 10

1701 / 8 212 5

1701 / 7 243 0

1701 / 5 340 1

1701 / 3 567 0

q remainder

q remainder

q remainder

q remainder

RNS Complementation

• Given RNS Representation of X, -X is Obtained by

Complementing Each Digit. Zero Digits are unchanged.

10 RNS

10 RNS

RNS

21 (5 | 0 |1| 0)

21 (8 5 | 0 | 5 1| 0)

(3 | 0 | 4 | 0)

EXAMPLE

10 10

10 10

10 10

10 10

21 / 8 2 5, 5mod8 3

21 / 7 3 0

21 / 5 4 1, 1mod5 4

21 / 3 7 0

q remainder

q remainder

q remainder

q remainder

CHECK

Chinese Remainder Theorem

• RNS can be viewed as a weighted system.

10 10 10 10

RNS (8 | 7 | 5 | 3)

(105 ,120 ,336 ,280 )iw

EXAMPLE

RNS

840

840

(1| 2 | 4 | 0)

(1 105) (2 120) (4 336) (0 280)

1689 9

RNS Encoding Efficiency

• Example Requires 11 Bits

84041%

2048Efficiency

mod 8 mod 7 mod 5 mod 3

• 840 Different Values Represented

• 211=2048

lg2(840)=9.71411-9.714=1.3 Bits Wasted

RNS Arithmetic • Addition, Subtraction, Multiplication Can be Performed with

Independent Operations on Each Digit • Following Examples Show This Process

10 RNS

10 RNS

8 7 5 3

RNS

8 7 5 3

RNS

8 7 5 3

RNS

5 (5 | 5 | 0 | 2)

1 (7 | 6 | 4 | 2)

( 5 7 | 5 6 | 0 4 | 2 2 )

(4 | 4 | 4 |1)

( 5 7 | 5 6 | 0 4 | 2 2 )

(6 | 6 |1| 0)

( 5 7 | 5 6 | 0 4 | 2 2 )

(3 | 2 | 0 |1)

X

Y

X Y

X Y

X Y

• For Subtraction, Can Complement the Number and Add Also

RNS Circuit Structure

mod 8 mod 7 mod 5 mod 3

mod-8unit

mod-7unit

mod-5unit

mod-3unit

Choosing RNS Moduli

• Assume we wish to represent 100,00010 Values

• Standard Binary lg2(100,000)10 = 16.609610 =17 bits

• RNS(13|11|7|5|3|2), Dynamic RangeM=30,03010

–Insufficient Dynamic Range

–Maximum Digit Width = 4 bits, Total = 17 bits

• RNS(17|13|11|7|5|3|2), Dynamic RangeM=510,51010

– Dynamic Range 5.1 Times Too Large

– Maximum Digit Width = 5 bits, Total = 22 bits

• Adding More Prime Moduli is Inefficient

Choosing RNS Moduli• Remove mi=5 From RNS(17|13|11|7|5|3|2)

• RNS(17|13|11|7|3|2), Dynamic RangeM=102,10210

• Still Have Relatively Prime Moduli


– 1 5-bit, 2 4-bit, 1 3-bit, 1 2-bit and 1 1-bit Modulo Units Required

• Maximum Delay 5-bit Carry-Propagate

• Can Combine (3,7) and (2,13) Moduli With no

Speed Penalty

• RNS(26|21|17|11), Dynamic RangeM=102,10210


– 3 5-bit and 1 4-bit Modulo Units Required

Relatively Prime Values• Powers of Smaller Primes are Relatively Prime

Example

• gcd(32, 22) = 1 But gcd(32,3) = 3

– Can REPLACE a Modulus With its Power

– Try Use Sequence of SMALLEST Valued Moduli

RNS(22 |3), Dynamic RangeM=1210

RNS(32 |23 |7|5), Dynamic RangeM=2,52010

RNS(11|32 |23 |7|5), Dynamic RangeM=27,72010

RNS(13|11|32 |23 |7|5), Dynamic RangeM=360,36010


– Dynamic Range 3.6 times that Needed

Relatively Prime Values RNS(13|11|32 |23 |7|5), Dynamic RangeM=360,36010



• Reduce the Above by Factor of 3

• Replace 32 with 3 and Combine 3 and 5 to Get 15

RNS(15|13|11 |23 |7), Dynamic RangeM=120,12010



• Using This Strategy Can Generally Find the “Best”

Moduli in Terms of Speed and Representation

Efficiency

Moduli Choice for Simple Arithmetic Unit Design

• Simple Units Also Lead to Speed and Cost Benefits• Modulo-ADD,SUBTRACT, MULTIPLY Units

Simple to Design if mi=2ai or 2ai-1

• Power of 2 Moduli Lead to Simple Design– Standard a-bit Binary Adder– Example: Use 16 Instead of 13– Exception in Case of Lookup Table Implementation

•Power of 2a-1 Moduli Lead to Simple Design– Standard a-bit Binary Adder with End-around Carry– Referred to as “Low-cost” Moduli

RNS Low-Cost Moduli

Theorem:A sufficient condition for 2a-1 and 2b-1 to be a relatively prime pair is that a and b are relatively prime.

• Any List of Relatively Prime Numbers:

ak-2> ...>a1>a0

• Can be Used as a BASIS of k-modulus RNS:

RNS(2ak-2|2ak-2 -1|...|2a1-1|2a0-1)

• Widest Residues (Longest Carry-chain) is ak-2-bit Values

Low-Cost Moduli Example

• Consider the Example From Earlier

X=[0,100,000]

• Choosing the Moduli From Smallest to Largest:

RNS(23 | 23 -1| 22 -1) Basis:3, 2 M=16810

RNS(24 | 24 -1| 23 -1) Basis:4, 3 M=168010

RNS(25 | 25 -1 | 23 -1| 22 -1) Basis:5, 3, 2 M=20,83210

RNS(25 | 25 -1 | 24 -1| 23 -1) Basis:5, 4, 3 M=104,16010

• Can’t Include 2 and 4 in Same Basis Set, gcd(2,4)=2

Low-Cost Moduli Example

RNS(25 | 25 -1 | 24 -1| 23 -1) Basis:5, 4, 3 M=104,16010

= RNS(32 | 31 | 15| 7)

• Requires 5+5+4+3=17 bits

• Requires 2 5-bit, 1 4-bit and 1 3-bit Module

• 4 RNS Digits

• Efficiency = (100,001/104,160)=0.96004100%

• Comparing With Unrestricted Moduli: RNS(25 | 25 -1 | 24 -1| 23 -1) 17 bits M=104,16010

5-bit Carry-ripple but Simpler Circuit, Fewer Digits

RNS(15|13|11 |23 |7) 18 bits M=120,12010

4-bit Carry-ripple , 1 Extra Digit

Encoding and Decoding

• Advantages of Alternative Number Systems

Must Not be Outweighed By Conversions

to/from the System

• Encoding From Fixed Positional System to

RNS Easily Accomplished Using a Table-

Lookup and Modulo Addition Circuits

Encoding with Lookup Table

• Conversion of Signed-Magnitude or 2’s

Complement Accomplished by Converting

Magnitude and Taking RNS Complement• Consider the Following Identity:

1 1 01 2 1 0 1 1 02

2 2 2i i i i

i

kk k km m m m m

y y y y y y y

• Idea is to Compute a Table of All Terms and

Store in a Table for all i, j Then Add2

i

j

m

Example Lookup Table

• Use Default RNS=(8|7|5|3)

j 2 j 7

2 j 5

2 j 3

2 j

0 1 1 1 1 1 2 2 2 2 2 4 4 4 1 3 8 1 3 2 4 16 2 1 1 5 32 4 2 2 6 64 1 4 1 7 128 2 3 2 8 256 4 1 1 9 512 1 2 2

• For mi=8 We Can Use 3 LSbs of Value

Example Encoding

2 10

3 2 1 0 2 2 10

7 5 2

2 107

1 105

0 103

3 2 1 0 RNS RNS

10100100 164 RNS (8 | 7 | 5 | 3)

mod 8 ( ) 100 4

2 2 2 7,5,2

mod7 2 4 4 3

mod5 3 2 4 4

mod3 2 2 1 2

( | | | ) (4 | 3 | 4 | 2)

Y

x Y y y y

Y j

x Y

x Y

x Y

Y x x x x

1 1 01 2 1 0 1 1 02

2 2 2i i i i

i

kk k km m m m m

y y y y y y y

RNS to Mixed-Radix Form

• CRT States That a Mixed-Radix Number System

(MRS) is Associated with any RNS

• Solves comparison, sign detection, and overflow problems

• MRS is k-digit Weighted Positional Number System

(mk-1|mk-2|...|m2|m1|m0)

• MRS Weights are Products:

(mk-2...m2m1m0, ...,m2m1m0, m1m0, m0,1)

• MRS Digit Sets in Each of k Positions:

[0, mk-1-1],...,[0, m2-1],[0, m1-1],[0, m0-1]

• MRS Digits in Same Range as RNS Digits

RNS to MRS Example

• Example Position Weights MRS (8|7|5|3)

(7)(5)(3)=105, (5)(3)=15, 3, 1

• (0|3|1|0)MRS(8|7|5|3)

=(0)(105)+(3)(15)+(1)(3)+(0)(1)=4810

• RNS to MRS Conversion Requires Finding the zi

that Correspond to the yi in:

1 2 1 0 RNS 1 2 1 0 MRS( | | | | ) ( | | | | )k kY y y y y z z z z

RNS to MRS Conversion

• From MRS Definition we Have:

1 2 2 1 0 2 1 0 1 1 0( ) ( ) ( ) (1)k kY z m m m m z m m z m z

• Easy to See that z0 = y0, Subtracting This Value From

RNS and MRS Values Results in:

0 1 2 1 RNS 1 2 1 MRS( ' | | ' | ' | 0) ( | | | | 0)k kY y y y y z z z

0'j

j j my y y

RNS to MRS Conversion (cont)

• Thus, if We Can Divide by m0, We Have an

Iterative Approach for Conversion

• Dividing y' (a Multiple of m0) by m0 is SCALING

Easier Than Normal RNS Division

• Accomplished by Multiplying by Muliplicative

Inverse of m0

1 2 1 1 2 1( '' | | '' | '' ) ( | | | )k RNS k MRSy y y z z z • Next, Divide Both Representations by m0:

Multiplicative Inverses

• Multiplicative Inverse is a Value When Multiplied

by Given Quantity Yields a Product of 1

• Example Multiplicative Inverses of 3 Relative to

mi=8, 7, 5:8

7

5

3 3 1

3 5 1

3 2 1

• Thus, Multiplicative Inverses are 3, 5 and 2

• Can Build a Lookup Table Circuit to Store Inverses

CRT LUTi im ix

ii i i m

MM x

3 8 0 0 1 1 0 5 2 2 1 0 3 3 1 5 4 4 2 0 5 5 2 5 6 6 3 0 7 7 3 5

2 7 0 0 1 1 2 0 2 2 4 0 3 3 6 0 4 4 8 0 5 6 0 0 6 7 2 0

1 5 0 0 1 3 3 6 2 6 7 2 3 1 6 8 4 5 0 4

0 3 0 0 1 2 8 0 2 5 6 0

Multiplicative Inverses Example

• Divide the Number Y' = (0|6|3|0)RNS by 3

• Accomplish Through Multiplication by (3|5|2|-)RNS

RNSRNS RNS

8 7 5

RNS

(0 | 6 | 3 | 0)(0 | 6 | 3 | 0) (3 | 5 | 2 | )

3( 0 3 | 6 5 | 3 2 | )

(0 | 2 |1| )

RNS/MRS Conversion Example

• Convert Y=(0|6|3|0)RNS to MRS z0 = y0 = 0

• Divide by 3RNS

RNS

(0 | 6 | 3 | 0)(0 | 2 |1| )

3

• Now, We Have z1=1, Subtract by 1 and Divide by 5

RNSRNS RNS

RNS

(7 |1| 0 | )(7 |1| 0 | ) (5 | 3 | | )

5(3 | 3 | | )

• This Gives z2 = 3, Subtract by 3 and Divide by 7

RNS/MRS Conversion Example

RNS

RNS

(0 | 0 | | )(0 | 0 | | )

7

• Thus Y=(0|6|3|0)RNS is (0|3|1|0)MRS

• Position Weights MRS (8|7|5|3)

(7)(5)(3)=105, (5)(3)=15, 3, 1

So, Y=(0|6|3|0)RNS = (0|3|1|0)MRS = (48)10

RNS/MRS Conversion

•Consider Conversion of (3|2|4|2)RNS from RNS(8|7|5|3)

to DecimalRNS RNS RNS

RNS RNS

RNS RNS

RNS RNS

(3 | 2 | 4 | 2) (3 | 0 | 0 | 0) + (0 | 2 | 0 | 0)

+ (0 | 0 | 4 | 0) + (0 | 0 | 0 | 2)

= 3×(1| 0 | 0 | 0) + 2×(0 |1| 0 | 0)

+ 4×(0 | 0 |1| 0) + 2×(0 | 0 | 0 |1)

•Need to Determine Values of (1|0|0|0)RNS, (0|1|0|0)RNS,

(0|0|1|0)RNS and (0|0|0|1)RNS

RNS/MRS Conversion

•From Definition of RNS, Positions with 0 are Multiples

of RNS(8|7|5|3) and Position with 1 are <Y>mi=1

RNS

RNS

RNS

RNS

(1| 0 | 0 | 0) = 105

(0 |1| 0 | 0) = 120

(0 | 0 |1| 0) = 336

(0 | 0 | 0 |1) = 280

RNS

840

(3 | 2 | 4 | 2)

3 105 2 120 4 336 2 280

779


• How Did We Find w3 = (1|0|0|0)RNS = 105?

• Since Digits in 7, 5, 3 Places are 0, w3 Must be a

Multiple of (7)(5)(3)=105

• Must Pick the Multiple of 105 Such That its Residue

With Respect to 8 is 1

• Accomplished by Multiplying 105 by its’ Multiplicative

Inverse with Respect to 8

• This Process is Formalized in Chinese Remainder

Theorem


THEOREM: Chinese Remainder Theorem (CRT)

The magnitude of an RNS number can be obtained

from the CRT formula:1

1 2 1 0 RNS0

( | | | | )i

k

k i i ii m M

Y y y y y M y

where, by definition, Mi = M/mi and i = < Mi-1>mi is

the multiplicative inverse of Mi with respect to mi.


1

1 2 1 0 RNS0

( | | | | )i

k

k i i ii m M

Y y y y y M y

• Can Avoid Multiplications in Conversion Process

by Storing <Mi<iyi>mi>M in a Table

• Example Table Given on page 64 of Textbook (and

also in slide 33)

Difficult RNS Operations

• Sign Test

• Magnitude Comparison

• Overflow Detection

• Generalized Division

Suffices to discuss first three in context of being able to

do magnitude comparison since they are essentially same

if M is such that M=N+P+1 where the values represented

are in interval [-N,P].

Difficult RNS Operations

• Sign Test same as Comparison with P

• Overflow Detection accomplished using Signs

of Operands and Results

Focus On:

• Magnitude Comparison

• Generalized Division

Magnitude Comparison•Could Convert to Weighted Representation Using CRT

Too Complicated – too much Overhead

Use Approximate CRT Instead

Divide CRT Equality by M1

1 2 1 0 RNS0

( | | | | )i

k

k i i ii m M

Y y y y y M y

111 2 1 0 RNS

01

( | | | | )

i

kk

i i ii m

y y y yYm y

M M

/i iM M m by Definition

Approximate CRT

• Addition of Terms is Modulo-1

• All mi-1<iyi>mi Are in [0,1)

• Whole Part of Result Discarded and Fractional Part Kept

• Much Easier than CRT Modulo-M Addition

• mi-1<iyi>mi Can be Precomputed for all y and i

• Use Table Lookup Circuit and Fractional Adder

(ignore carry-outs)

111 2 1 0 RNS

01

( | | | | )

i

kk

i i ii m

y y y yYm y

M M

Approximate CRT LUTi im iy 1

ii i i m

m y

3 8 0 . 0 0 0 0 1 . 1 2 5 0 2 . 2 5 0 0 3 . 3 7 5 0 4 . 5 0 0 0 5 . 6 2 5 0 6 . 7 5 0 0 7 . 8 7 5 0

2 7 0 . 0 0 0 0 1 . 1 4 2 9 2 . 2 8 5 7 3 . 4 2 8 6 4 . 5 7 1 4 5 . 7 1 4 3 6 . 8 5 7 1

1 5 0 . 0 0 0 0 1 . 4 0 0 0 2 . 8 0 0 0 3 . 2 0 0 0 4 . 6 0 0 0

0 3 0 . 0 0 0 0 1 . 3 3 3 3 2 . 6 6 6 7

Magnitude Comparison ExampleUse approximate CRT decoding to determine the larger

of the two numbers.RNS RNS(0 | 6 | 3 | 0) (5 | 3 | 0 | 0)X Y

1.6250 .4286 .0000 .0000 .0536

Y

M

X Y

Reading the Values from the Tables:

1.0000 .8571 .2000 .0000 .0571

X

M

Thus, we conclude that:

Approximate CRT ErrorIf Maximum Error in Approximate CRT Table is , then

Approximate CRT Decoding Yields Scaled Value of

RNS Number with Error No Greater than k

0.0571 - 0.0536 = 0.0035 > 4 = 0.0002, so X > Y is Safe

Previous Example Table Entries Rounded to 4 Digits

Maximum Error in Each Entry is = 0.00005

k = 4 Digits Error is 4 = 0.0002

Redundant RNS Representations

• Do Not Have Restrict Digits in RNS to Set [0, mi -1]

• If [0, i] Where i mi Then RNS is Redundant

• Redundant RNS Simplifies Modular Reduction Step for Each Arithmetic Operation

Redundant RNS Example• Consider mod-13 with [0,15]• Redundant since: 0 mod13 0 0 mod13 13

1 mod13 1 1 mod13 14

2 mod13 2 2 mod13 15

• Addition Using Pseudo-redundancies Can be Done with Two 4-bit Adders

00

Ignore

Cout

X Y

SUM

1 1 0 1 (13)

1 1 0 1 (13)1 1 0 1 0 (26)

0 0 1 1

1 0 1 01 1 0 1 (13)

Coping With the Carry Problem 1. Limit Carry to Small Number of Bits Hybrid Redundant Residue Number...

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