SIMULATION BOUNDS FOR EQUIVALENCE VERIFICATION OF ARITHMETIC DATAPATHS WITH FINITE WORD-LENGTH

OPERANDS

Namrata ShekharNamrata Shekhar1, Priyank Kalla, Priyank Kalla1, M. Brandon Meredith, M. Brandon Meredith22, Florian Enescu, Florian Enescu2

1Department of Electrical and Computer Engineering,

University of Utah, Salt Lake City, UT-84112.

2Department of Mathematics and Statistics,

Georgia State University, Atlanta, GA-30303

OutlineOutline

Problem Overview

Application: Arithmetic datapaths in DSP designs

Problem modeling

Polynomial functions over finite integer rings

Limitations of previous work

Theory and Applications

Results and Conclusions

Future work

The Equivalence Verification ProblemThe Equivalence Verification Problem

Fixed Bit-width OperandsFixed Bit-width Operands

Control the datapath size: Fixed size bit-vectors (m)

* *

8-bit

8-bit

16-bit32-bit

* *

8-bit

8-bit

8-bit8-bit

Bit-vector of size m: Integer values in 0,…, 2m-1

Fixed-size (m) bit-vector

arithmetic

Polynomials reduced %2m

Algebra over the ring Z2m

General Datapath ModelGeneral Datapath Model

Bit-vector operands with different word-lengths

Input variables : {x1,…, xd} Output variables : f

Input bit-widths: {n1,…, nd} Output width : m

.

.

Model f as polynomial function

* *

8-bit

12-bit

20-bit32-bit

n n n m1 2 d 22 2 2Z × Z × × Z Z

1 21 2 22 2n n mx Z ,x Z f Z

1n m

1 {0, ..., 2 -1} {0, ..., 2 -1}x , f

Arithmetic Data-path: ImplementationArithmetic Data-path: Implementation

Signal Truncation

Keep lower-order bits, ignore higher bits

.

Fractional Arithmetic with rounding

Keep higher-order m-bits, round lower order bits

.

Saturation Arithmetic

Saturate at overflow

Used in image-processing applications

m m

1 d 1 d( ,.., ) % 2 ( ,.., ) % 2f x x g x x

m m

m m

- % 2 - % 2

2 2

f f g g

Motivation: Motivation: Convolution of A and B

Datapath size is fixed

How many simulation vectors required to prove C = C’ ?

FFT(A)

invFFT(FAB)

FFT(B)

a0a1a2

a3

b0b1b2

b3

FAB0

FAB1

FAB2

FAB3

c’0c’1c’2

c’3

j i j0

3

ij

c a b

0 3 iC = (c0, c1, c2, c3) where

C′ = (c′0, c′1, c′2, c′3) = DFT-1 (DFT(A)·DFT(B))

Example: Anti-Aliasing FunctionExample: Anti-Aliasing Function

F = 1 = 1 = 2√a2 + b2 2√x [Peymandoust et al, TCAD‘03]

Expand into Taylor series

F ≈ 1 x6 – 9 x5 + 115 x4

64 32 64

– 75 x3 + 279 x2 – 81 x 16 64 32

+ 85 64

Scale coefficients;

Implement as bit-vectors

MAC

x = a2 + b2

coefficients

a b

x

F

DFF

coefficients

Example: Anti-Aliasing FunctionExample: Anti-Aliasing Function

Implemented as a fixed size datapath in x F1[15:0], F2[15:0], x[15:0]

F1 = 156x6 + 62724x5 + 17968x4 + 18661x3 + 43593 x2

+ 40244x + 13281

F2 = 156x6 + 5380x5 + 1584x4 + 10469x3 + 27209 x2

+ 7456x + 13281

F1 ≠ F2 ; F1[15:0] = F2[15:0]

How many vectors required to prove: F1 % 216 ≡ F2 % 216

Contributions Contributions

Abstract the design as a polynomial function

Exhaustive simulation is not necessary to prove

equivalence

Upper bound on the number of vectors

To prove equivalence

Sufficient to catch errors

Bound corresponds to a function in number theory

Previous WorkPrevious Work

Bit/Word level canonical diagrams

BDDs, ZBDDs, BMDs, TEDs

SAT and MILP-based techniques

Bit-vector decision procedures, Word-level ATPG, SMT

Theorem-Proving, term-rewriting

Problem model is algebraic

Previous Work: SimulationPrevious Work: Simulation

Reduction in simulation complexity

Three-valued logic simulation [Bryant, ACM ’91]

Utilizing structural information [Brand, ICCAD ’92]

Automated approach using BDDs [Yuan , ICCAD ‘99]

Polynomial methods using the fundamental theorem of

algebra

Generate simulation vectors [Sanchez, HLDVT ‘99]

Reduce the complexity of model checking

[ Raudvere, ICCAD ‘05]

Fundamental Theorem of AlgebraFundamental Theorem of Algebra

A degree-k polynomial F(x) has exactly k roots

F(x) = x2 + 6x = x ( x + 6 )

If F(x) = 0 for k + 1 values, then F(x) is a zero polynomial

Can also be extended for multi-variate polynomials

Limitations:

Results applicable only over unique factorization domains

(UFD): Z, Zp, C

Z2m is a non-UFD

Why is the Problem Difficult?Why is the Problem Difficult?

Consider F(x) = x2 + 6x in Z8

Degree-2 polynomial has 4 unique roots

F(x) = 0 for 4 vectors, but F(x) ≠ 0 in Z8

Not applicable to bit-vector arithmetic

F

x+6

F

x+4 x+2x

Previous Work: Finite Ring AlgebraPrevious Work: Finite Ring Algebra

f (x1, …, xd) % n ≡ g(x1, …, xd) % n

Proving equivalence is NP-hard [Ibarra et al, ACM ‘83]

Previous approaches

f (x1, x2, …, xd) – g (x1, x2, …, xd) ≡ 0 % 2m : Zero Equivalence

[ICCD ’05]

Reduction to canonical forms [ICCAD ’05]

Limitations:

Intermediate expression swell

No error trace is provided

Simulation vector generation: Based on zero equivalence

Zero EquivalenceZero Equivalence

module fixed_bit_width (x, f, g);

input [2:0] x;

output [2:0] f, g;

assign f[2:0] = x2 + 6x – 3;

assign g[2:0] = 5x2 + 2x + 5;

h(x) = f (x) – g(x) = 4x2 + 4x

h(x) ≡ 0 for all values of x in {0,…,7}: Vanishing polynomial

Required: To find if any given expression vanishes

Use concepts from ideal membership testing

Ideal Membership TestingIdeal Membership Testing

h:Z2m[x] → Z2m defined by % 2m

Ideal members map to 0

Test for membership in

Representative expression for members of this ideal [Chen,

Disc. Math ‘96]

Use concepts from Number theory and polynomial algebra

Idealxi

xi % 2m

h: % 2m

0f

gf – g ?Z2m[x]

Ideal of Vanishing Polynomials

Z2m

Results From Number TheoryResults From Number Theory

Find least n such that 2m|n!

Smarandache Function: λ = SF (2m)

λ = SF(23) = 4, since 23|4!

n! divides a product of n consecutive numbers

4! divides 99 X 100 X 101 X 102

2m divides the product of n consecutive numbers

23 divides the product of 4 consecutive numbers

Results From Number TheoryResults From Number Theory

F ≡ G in Z23 or (F - G) ≡ 0 % 23

23|(F - G) in Z23

23 divides the product of 4 consecutive numbers

If (F-G) is a product of 4 consecutive numbers then 23|(F - G)

A polynomial as a product of 4 consecutive numbers?

. (x-1) (x-2) (x-3)(x)

Basis for FactorizationBasis for Factorization

Y0(x) = 1

Y1(x) = (x)

Y2(x) = (x)(x - 1) = Product of 2 consecutive numbers

Y3(x) = (x)(x - 1)(x - 2) = Product of 3 consecutive numbers

…

…

Yk(x) = (x – k + 1) Yk-1(x) = Product of k consecutive numbersRule 1: Factorize into at least Yλ (x) to vanish, where

λ = SF(2m)

Example 1: Vanishing PolynomialExample 1: Vanishing Polynomial

4th degree polynomial F over Z23

λ = 4

.

Degree (x) = k = 4 = λ

F can be written as a product of 4 consecutive numbers

in x

.

F is a vanishing polynomial

4 3 2F x 2x 3x 2x

3

4F = (x) (x-1) (x-2) (x-3) = Y ( ) % 2x

Constraints on the CoefficientConstraints on the Coefficient

F(x) = 4x2 + 4x = (x)(x-1) % 23

Y4(x) = (x)(x-1)

Rule 2: Coefficient has to be a multiple of bk = 2m/gcd(k!, 2m)

Here, Coefficient of F(x) = 4, Degree of F(x) = 2

b2 = 23/gcd(2!, 23) = 4 is a multiple of the coefficient

Use Rule 1 and Rule 2 to determine if any F(x) = 0 % 2m

compensated by constant

missing factor (x-2)(x-3)

4

Deciding Vanishing PolynomialsDeciding Vanishing Polynomials

Fλ is an arbitrary polynomial over

Yλ = Yk is as defined earlier: λ = SF(2m)

ak is an arbitrary integer

.

Polynomial F in vanishes if

Rule 1 Rule 2

F = F Y + a b k k kk Y

m2Z

m2Z

m

k m

2b =

gcd(k!,2 )

Polynomial RepresentationsPolynomial Representations

: Forward difference operator

p is the degree of F(x)

Newton’s interpolation formula: Any polynomial F(x) can be written as

0

F( ) ( F)(0)

pk

k

xx

k

( F)( ) F( 1) F( )x x x

Polynomial RepresentationsPolynomial Representations

: Forward difference operator

p is the degree of F(x)

Newton’s interpolation formula: Any polynomial F(x) can be written as

0

F( ) ( F)(0)

p

k

k

x

kx

( F)( ) F( 1) F( )x x x

Reinterpret Vanishing idealReinterpret Vanishing ideal

: Forward difference operator

Yk(x) is as defined earlier

p is the degree of F(x)

Newton’s interpolation formula: Any polynomial F(x) can be written as

F( ) x

( F)( ) F( 1) F( )x x x

0

( F)(0)Y ( )

!

kp

kk

xk

Reinterpret Vanishing idealReinterpret Vanishing ideal

: Forward difference operator

Yk(x) is as defined earlier

p is the degree of F(x)

Newton’s interpolation formula: Any polynomial F(x) can be written as

F( ) x

( F)( ) F( 1) F( )x x x

0

Y (( F)(0)

!)

p

kk

k

kx

Polynomial RepresentationsPolynomial Representations

is any arbitrary integer

Yk(x) is as defined earlier

p is the degree of F(x)

Newton’s interpolation formula: Any polynomial F(x) can be written as

kk 0

F( ) cp

kx Y (x)

k

k

( F)(0)c

k!

Polynomial RepresentationsPolynomial Representations

Fλ is an arbitrary polynomial over

Yλ = Yk is as defined earlier: λ = SF(2m)

is an arbitrary integer

Any polynomial F(x) in can be written as

m2Z

m2Z

F(x) =

Rule 1

c+ k kk YF Y 0 0 k

k

k

( F)(0)c

k!

Polynomial RepresentationsPolynomial Representations

Fλ is an arbitrary polynomial over

Yλ = Yk is as defined earlier: λ = SF(2m)

is an arbitrary integer

Any polynomial F(x) in can now be reduced to

m2Z

m2Z

F = c k kk Y 0 k

k

k

( F)(0)c

k!

Polynomial RepresentationsPolynomial Representations

Apply Rule 2

F(x) vanishes iff ck is a multiple of

Check for all ck, where

ck evaluated no more than λ times

F(x) evaluated no more than λ times

Any polynomial F(x) in can now be reduced to

m2Z

F = c k kk Y 0 k

m

k m

2b =

gcd(k!,2 )0 k

ResultsResults

By extension,

If F(x) ≡ 0 for any λ consecutive values of x in

.

Further,

F(x) – G(x) = 0 → F(x) = G(x)

Any λ consecutive values of x are sufficient to prove

equivalence

m2Z

2F( ) 0 Z mx x

ExampleExample

f = x4 + x2

Simulating,

x=0, f=0

x=1, f=2

x=2, f=4

x=3, f=2

x=4, f=0

x=5, f=2

x=6, f=4

x=7, f=2

Consider f, g over Z23

λ = SF(23) = 4

g = 2x2

Simulating,

x=0, g=0

x=1, g=2

x=2, g=0

x=3, g=2

x=4, g=0

x=5, g=2

x=6, g=0

x=7, g=2

Extension to Multiple VariablesExtension to Multiple Variables

Given d variables x = <x1, …, xd> with degrees

k = < k1, …, kd> over Z2m

Basis for d variables:

.

d

k i i 0 0 1 1 d di=1

Y = Y ( ) Y ( ) Y ( ) Y ( )x x x x

Rule 1 (for d variables): Factorize into Yλ (x), such that

ki ≥ λ for any xi ; λ = SF(2m)

Example: Vanishing PolynomialExample: Vanishing Polynomial

4th degree polynomial F(x, y) over Z23

SF(23) = 4

.

Degree (x) = k1 = 4 = SF(23)

Degree (y) = k2 = 1

F can be written as a product of 4 consecutive numbers

in x

.

F is a vanishing polynomial

4 3 2F x y 2x y 3x y 2xy

3

4 1F = (x) (x-1) (x-2) (x-3) y = Y ( ) Y ( ) 2 %x y

Effect of Bit-vectorsEffect of Bit-vectors

4th degree polynomial F(x, y) in Z21 × Z22 → Z23

λ = 4

.

Define

.

F = Y2(x) · Y1(y) ≡ 0 % 23

2F = x y - x y = x (x-1) (y)

1 3 2 3

1 2min{2 ,SF(2 )} 2; min{2 ,SF(2 )} 4

in

i min{2 , } 12

0 1x Z ,x , 220 1 2 3y Z , y , , ,

Rule 1(extended): Factorize into Yk(x), such that ki ≥ μi for any xi

ResultsResults

By extension, for F(x) over

If F(x) ≡ 0 for any μi consecutive values of xi in

.

Total number of vectors:

Further,

F(x) – G(x) = 0 → F(x) = G(x) over

To prove equivalence, we need vectors

i 2F( ) 0 x Z

inx in2

Z

n n m1 d 22 2Z × × Z Z

n n m1 d 22 2Z × × Z Z

i1

d

i

i1

d

i

ExampleExample

f = x4y+ x2y

Simulating,

x=0, y=0, f=0

x=1, y=0, f=0

x=2, y=0, f=0

x=3, y=0, f=0

x=0, y=1, f=0

x=1, y=1, f=2

x=2, y=1, f=4

x=3, y=1, f=2

Consider f, g over Z22 × Z2 → Z23

λ = 4, μ1 = 4 ; μ2 = 2

Required: μ1.μ2 = 8 vectors

g = 2x2y

Simulating,

x=0, y=0, g=0

x=1, y=0, g=0

x=2, y=0, g=0

x=3, y=0, g=0

x=0, y=1, g=0

x=1, y=1, g=2

x=2, y=1, g=0

x=3, y=1, g=2

Experimental SetupExperimental Setup

Distinct RTL designs are input to GAUT [U. de LESTER, 2004]

Extract data-flow graphs for RTL designs

Construct the corresponding polynomial representations (F, G)

Extract bit-vector sizes for inputs and outputs

Determine the maximum number of simulation vectors required

Check for equivalence or determine bugs

Simulation Results for Equivalent DesignsSimulation Results for Equivalent Designs

1.00E+001.00E+051.00E+101.00E+151.00E+201.00E+251.00E+301.00E+351.00E+401.00E+451.00E+501.00E+551.00E+601.00E+651.00E+70

IRR PSK Degree-4Filter 1

Degree-4Filter 2

Savitzky-GolayFilter

4th orderIIR

MIBENCH

Total Test Vectors Required Test Vectors

Simulation Results for Buggy DesignsSimulation Results for Buggy Designs

1

10

100

1000

Anti-alias Function PSK 4th Order IIR VanishingPolynomial

Our approach Random Simulation

LimitationsLimitations

a = 127 b = 1 f1 = 383

c = 255

≠

a = 127 b = 1 f2 = 127c = 255

a[7:0] b[7:0]

t1[7:0]

+

c[7:0]

+

f1[8:0]

a[7:0] b[7:0]

t2[7:0]

+

c[7:0]

+

f2[8:0]

ConclusionsConclusions

Technique to verify equivalence of polynomial RTL

computations

Bit-vector arithmetic is polynomial algebra over the system

of finite integer rings

Exhaustive simulation is not necessary to prove

Results based on concepts from number theory and

polynomial algebra

n n n m1 2 d 22 2 2Z × Z × × Z Z

( ) % 2 ( ) % 2 m m

1 2 d 1 2 df x , x , , x = g x , x , , x

Questions ?Questions ?

Polynomial AbstractionPolynomial Abstraction

If (x > 2b’10)

then y = x * x * x

Else y = x*x

Traditional modeling

.

Proposed modeling: y as a polyfunction from

: Unique representation

Issues with the proposed abstraction:

Scalability

4 32Z Z3 2y 3x 8x 22x

3 2[0,2] y = x and 3 y = x x x =