Exploiting Vanishing Polynomials for Equivalence Verification of Fixed-Size Arithmetic Datapaths Namrata Shekhar, Priyank Kalla, Florian Enescu, Sivaram

Exploiting Vanishing Polynomials for Exploiting Vanishing Polynomials for Equivalence Verification of Fixed-Size Equivalence Verification of Fixed-Size

Arithmetic Datapaths Arithmetic Datapaths

Namrata ShekharNamrata Shekhar1, Priyank Kalla, Priyank Kalla1, Florian Enescu, Florian Enescu2, Sivaram Gopalakrishnan, Sivaram Gopalakrishnan1

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

Overall Verification ProblemOverall Verification Problem

• Our Focus: Equivalence Verification of Fixed-size Arithmetic DatapathsOur Focus: Equivalence Verification of Fixed-size Arithmetic Datapaths

• Applications: Polynomial Signal Processing, etc. Applications: Polynomial Signal Processing, etc. Problem Modeling and ApproachProblem Modeling and Approach

• Polynomials over Finite Integer RingsPolynomials over Finite Integer Rings Limitations of Previous WorkLimitations of Previous Work Our ContributionsOur Contributions

• Exploiting Vanishing Polynomials for Equivalence TestExploiting Vanishing Polynomials for Equivalence Test

• Related to Ideal Membership TestingRelated to Ideal Membership Testing Algorithm Design and Experimental Verification RunsAlgorithm Design and Experimental Verification Runs Results Results Conclusions & Future WorkConclusions & Future Work

The Equivalence Verification ProblemThe Equivalence Verification Problem

Fixed-Size (m) Data-path: ModelingFixed-Size (m) Data-path: Modeling

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

Fixed-Size Data-path: ImplementationFixed-Size Data-path: Implementation

Signal Truncation

• Keep lower order m-bits, ignore higher bits• f % 2m ≡ g % 2m

Fractional Arithmetic with rounding

• Keep higher order m-bits, round lower order bits• f - f %2m ≡ g - g%2m

2m 2m

Saturation Arithmetic

• Saturate at overflow

• Used in image-processing applications

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 coefficients

a b

x

F

DFF

Example: Anti-Aliasing FunctionExample: Anti-Aliasing Function

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]

Transform the problem • F1 - F2 = 57344x5 + 16384x4 + 8192x3 + 16384x2 + 32768x

≡ 0 % 216

F1 - F2 : Vanishing polynomial

Previous Work: Function RepresentationsPrevious Work: Function Representations

Boolean Representations (f: B → B)

• BDDs, MTBDDs, ADDs etc.

Moment Diagrams (f: B → Z)

• BMDs, K*BMDs, HDDs etc.

Canonical DAGs for Polynomials (f: Z → Z)

• Taylor Expansion Diagrams (TEDs)

Required: Representation for f: Z2m → Z2m

Previous Work: OthersPrevious Work: Others

SAT and MILP-based techniques

• Suitable for linear/multi-linear forms

Word-level ATPG, congruence closure based techniques,

co-operative decision procedures

• Solve linear congruences under modulo arithmetic

Theorem-Proving (HOL), term-rewriting

• Works when datapath size can be abstracted using

data dependence, symmetry, and other abstractions

• Here, datapath size (m) defines ring cardinality (Z2m)

Previous Work: Symbolic AlgebraPrevious Work: Symbolic Algebra

MODDs: Based on finite fields [Pradhan et al, DATE ‘04]

• Literal based decomposition

Symbolic Algebra Tools: Singular, Macaulay, Maple,

Mathematica, Zen, etc.

• Polynomial equivalence over R, Q, C, Zp

• Unique Factorization Domains (UFDs) : Uniquely

factorize into irreducibles

• Match corresponding coefficients to prove

equivalence

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

Z2m is a non-UFD

• f = x2 + x in Z6 can be factorized as

Atypical approach required to prove equivalence

f

x x+1

f

x+3 x+4

Proposed SolutionProposed Solution

Equivalence can be proved by

• f (x) - g(x) ≡ 0 % 2m: Zero Equivalence

Exploit Vanishing polynomials

An instance of Ideal Membership Testing

Proposed solution based on:

• Niven & Warren’s work [Proc. Amer. Math.Soc, 1957]

• Singmaster’s work [J. Num. Th, 1974]

Ideal Membership TestingIdeal Membership Testing

h:P →Q defined by % 2m

• x in P maps to x %2m in Q

• Ideal members map to 0

Derive Ideal of Vanishing Polynomials in Z2m

Grobner's basis? Buchberger's algorithm?

Known for UFDs, but for Z2m ?

P Q

Ideal xx % 2m

h: % 2m

0f

gf – g ?

Ideal Membership Testing in ZIdeal Membership Testing in Zpp

Fermat’s Little Theorem:

• x p ≡ x (mod p) or

• x p – x ≡ 0 (mod p)

• x p –x generates the vanishing ideal in Zp[x]

f(x) = 0 % p iff f(x) = (xp-x)g(x)

Zp: Principal Ideal Domain

This does not follow in Z2m

Ideal Membership TestingIdeal Membership Testing

Generate the ideal of vanishing polynomials % 2m ?

• Vanishing Ideal is finitely generated

[Niven et al, Am. Math. Soc., ‘57]

Need an algorithm for membership testing

• Representative expression for members of this ideal

[Singmaster, J. Num. Th ‘74]

P Q

Ideal xx % 2m

h: % 2m

0f

gf – g ?

Results From Number TheoryResults From Number Theory

n! divides a product of n consecutive numbers.

• 4! divides 99 X 100 X 101 X 102

Find least n such that 2m|n!

• Smarandache Function (SF).

• SF(23) = 4, since 23|4!

2m divides the product of n = SF(2m) consecutive numbers


f ≡ g in Z23 or (f - g) ≡ 0 % 23

• 23|(f - g) in Z23

• 23|4!

• 4! divides the product of 4 consecutive numbers

A polynomial as a product of 4 consecutive numbers? • (x+1)

Write (f-g) as a product of SF(23) = 4 consecutive numbers


f ≡ g in Z23 or (f - g) ≡ 0 % 23

• 23|(f - g) in Z23

• 23|4!


A polynomial as a product of 4 consecutive numbers? • (x+1)(x+2)



f ≡ g in Z23 or (f - g) ≡ 0 % 23

• 23|(f - g) in Z23

• 23|4!


A polynomial as a product of 4 consecutive numbers? • (x+1)(x+2)(x+3)



f ≡ g in Z23 or (f - g) ≡ 0 % 23

• 23|(f - g) in Z23

• 23|4!


A polynomial as a product of 4 consecutive numbers? • (x+1)(x+2)(x+3)(x+4)


Basis for factorizationBasis for factorization

S0(x) = 1

S1(x) = (x + 1)

S2(x) = (x + 1)(x + 2) = Product of 2 consecutive numbers

S3(x) = (x + 1)(x + 2)(x + 3) = Product of 3 consecutive numbers

… …

Sn(x) = (x + n) Sn-1(x) = Product of n consecutive numbers

Rule 1: Factorize into atleast Sn(x) to vanish, where n = SF(2m).

Example 1: Vanishing polynomialExample 1: Vanishing polynomial

4th degree polynomial p in Z23 ; SF(23) = 4

p = x4 +2x3 + 3x2 + 2x

p can be written as a product of 4 consecutive numbers.

• or p = (x+1)(x+2)(x+3)(x+4) = S4(x) in Z23.

p is a vanishing polynomial.

Example 2: Vanishing polynomialExample 2: Vanishing polynomial

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}

4x2+4x not equal to (x+1)(x+2)(x+3)(x+4) Required: To show that h(x) is a vanishing polynomial

in Z23

Constraints on the CoefficientConstraints on the Coefficient

h(x) = 4x2 + 4x = 4(x+1)(x+2)

In Z23 , SF(23) = 4. Product of 4 consecutive numbers:

• S4(x) = (x+1) (x+2) (x+3) (x+4)

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

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

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

Constraints on the CoefficientConstraints on the Coefficient

h(x) = 4x2 + 4x = 4(x+1)(x+2)

compensated by constant

In Z23 , SF(23) = 4. Product of 4 consecutive numbers:

• S4(x) = (x+1) (x+2) (x+3) (x+4)

missing factors

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

Here, Coefficient of h(x) = 4, Degree of h(x) = k = 2

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

Deciding Vanishing PolynomialsDeciding Vanishing Polynomials

n = SF(2m), i.e. the least n such that 2m|n!

Fn is an arbitrary polynomial in Z2m[x]

ak is an arbitrary integer

bk = 2m/gcd(k!,2m)

Polynomial F in Z2m vanishes if

F = FnSn + Σ n-1ak bk Sk

k=0

Rule 1 Rule 2

AlgorithmAlgorithm

Input: poly, 2m

1. Calculate n = SF(2m)

2. k = n: Reduce according to Rule 1• Divide by Sn

• If remainder is zero, F = FnSn, else Continue

poly = 4x2 + 4x in Z23

1. n = SF(23) = 4

2. k = 4: Divide by S4

• Degree (poly) = 2

< degree(S4) = 4• quo = 0, rem = 4x2 + 4x

• F4 = 0; Continue

Example 1Example 1F = FnSn + Σ n-1ak bk Sk k=0

AlgorithmAlgorithm

Input: poly, 2m

1. Calculate n = SF(2m)

2. k = n: Reduce according to Rule 1• Divide by Sn

• If remainder is zero, F = FnSn, else Continue

poly = 4x2 + 4x in Z23

1. n = SF(23) = 4


• Degree (poly) = 2

< degree(S4) = 4• quo = 0, rem = 4x2 + 4x

• F4 = 0; Continue

Example 1Example 1F = FnSn + Σ n-1ak bk Sk k=0

AlgorithmAlgorithm

3. Reduce according to Rule 2. Divide by Sn-1 to S0

• Check if quotient is a multiple of

bk = 2m/gcd(k!,2m)• If remainder is zero,

stop. Else, continue


• degree (poly) = 2 < degree(S3) = 3

• quo= 0, rem = 4x2 + 4x continue


• quo = 4; rem = 0• b2 = 23/gcd(2!,23) = 4• a2 = quo/ b2 = 1 Є Z

Example 1Example 1

poly = a2.b2.S2 = 1.4.(x+1)(x+2) ≡ 0 in Z23

F = FnSn + Σ n-1ak bk Sk k=0

AlgorithmAlgorithm

3. Reduce according to Rule 2. Divide by Sn-1 to S0

• Check if quotient is a multiple of

bk = 2m/gcd(k!,2m)• If remainder is zero,

stop. Else, continue


• degree (poly) = 2 < degree(S3) = 3

• quo= 0, rem = 4x2 + 4x continue


• quo = 4; rem = 0• b2 = 23/gcd(2!,23) = 4• a2 = quo/ b2 = 1 Є Z

Example 1Example 1

poly = a2.b2.S2 = 1.4.(x+1)(x+2) ≡ 0 in Z23

F = FnSn + Σ n-1ak bk Sk k=0

Example 2Example 2

poly = 5x2 + 3x + 7 in Z23 n = SF(23) = 4 degree (poly) = 2 < n. Skip Rule 1 and goto Rule 2 Divide by S2

• quo = 5; rem = 5 + 4x

• b2 = 23/gcd(2!,23) = 4

• a2 = quo/ b2 is not in Z poly does not satisfy Rule 2

poly is not a vanishing polynomial in Z23

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 size

Find the difference (f-g) and invoke the algorithm

Algorithm implemented in MAPLE

Compare with BMD, SAT and MILP

Complexity: O(n+1), n = SF(Z2m)

ResultsResults

1

10

100

1000

Anti-Alia

s Funct

ion

Horner

Poly

1

Horner

Poly

2

Horner

Poly

3

Poly U

nopt.

cos

x

cot-1

xer

f x

ln(1

+x/1-

x)

Vanis

hing P

oly.

Our Approach (< 10 s)

BMD

SAT-Zchaff

Applications to SynthesisApplications to Synthesis

Datapath size (m): 8 bits

SF(28) = 10

Polynomial can be

factorized into S10(x)

ConclusionsConclusions

Technique to verify equivalence of univariate polynomial

RTL computations

Fixed-size bit-vector arithmetic is polynomial algebra over

finite integer rings

f(x) % 2m ≡ g(x) % 2m is transformed into f(x) - g(x) ≡ 0 % 2m

Efficient algorithm to determine vanishing polynomials

Future WorkFuture Work

Future Work involves extensions for -

• Multivariate datapaths with fixed bit-widths

[To Appear, ICCAD ‘05]

• Multiple Word-length Implementations

[In Review, DATE ‘06]

Verification of Rounding and Saturation Arithmetic

Formal Error Analysis

Applications to Synthesis –

• Need cost models for low power, area, delay

Questions?Questions?

Documents

Exploiting Vanishing Polynomials for Equivalence Verification of Fixed-Size Arithmetic Datapaths Namrata Shekhar, Priyank Kalla, Florian Enescu, Sivaram