Verification & Synthesis of Arithmetic Datapaths using Finite Ring Algebra Priyank Kalla Priyank Kalla Electrical and Computer Engineering University of

Verification & Synthesis of Verification & Synthesis of Arithmetic Datapaths using Arithmetic Datapaths using

Finite Ring AlgebraFinite Ring Algebra

Priyank KallaPriyank Kalla

Electrical and Computer Engineering

University of Utah

Salt Lake City, UT-84112

Research Group MembersResearch Group Members

Graduate StudentsGraduate Students::

• Namrata ShekharNamrata Shekhar – PhD – PhD RTL Verification of Arithmetic DatapathsRTL Verification of Arithmetic Datapaths

• Sivaram GopalakrishnanSivaram Gopalakrishnan – PhD – PhD RTL Synthesis of Arithmetic DatapathsRTL Synthesis of Arithmetic Datapaths

• Vijay DurairajVijay Durairaj – PhD + – PhD + Chris CondratChris Condrat – BS/MS – BS/MS SAT & SAT-based Decision ProceduresSAT & SAT-based Decision Procedures

Collaborators:Collaborators:

• Prof.Prof. Florian Enescu + BrandonFlorian Enescu + Brandon Mathematics & Statistics, Georgia State Univ.Mathematics & Statistics, Georgia State Univ.

OutlineOutline Introducing the ProblemsIntroducing the Problems

• Equivalence Verification & High-Level SynthesisEquivalence Verification & High-Level Synthesis

• Applications: Fixed Point Arithmetic, Polynomial Signal Applications: Fixed Point Arithmetic, Polynomial Signal Processing, Audio/Video/Multimedia DSPProcessing, Audio/Video/Multimedia DSP

ModelingModeling: Poly-Functions over Finite Integer Rings: Poly-Functions over Finite Integer Rings

Previous WorkPrevious Work: CAD & Symbolic Computer Algebra: CAD & Symbolic Computer Algebra

ContributionsContributions• Vanishing Polynomials, Canonical Forms for VerificationVanishing Polynomials, Canonical Forms for Verification

• Polynomial Reduction + Decomposition for SynthesisPolynomial Reduction + Decomposition for Synthesis

Algorithm Design & Experimental ResultsAlgorithm Design & Experimental Results Open Problems & ChallengesOpen Problems & Challenges: CAD + Algebra: CAD + Algebra

The Verification & Synthesis ProblemsThe Verification & Synthesis Problems

Polynomials over Bit-Vectors?Polynomials over Bit-Vectors?

Quadratic filter design for polynomial signal processing

y = a0 . x12 + a1 . x1 + b0 . x0

2 + b1 . x0 + c . x0 . x1

Bit-Vector Arithmetic = %2Bit-Vector Arithmetic = %2mm Algebra Algebra

Represent integers as a vector of bits Bit x0 represents values 0 or 1

Vector X[1:0] = {x1, x0} represents integers

00, 01, 10, 11 (4 values from 0 to 3) Bit-vector of size m: integer values in 0,…, 2m-1 Vector X[m-1 : 0] represents integers reduced % 2m

+2-bit

2-bit

3-bit*

2-bit

2-bit

4-bit

ADDER MULTIPLIER

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

* *

16-bit

16-bit

16-bit16-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

• If( x[7:0] > 255 ) then x[7:0] = 255;

• 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]; F1 % 216 = F2 % 216

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

≡ 0 % 216

F1 - F2 : Vanishing polynomial

Multiple Word-Length OperandsMultiple Word-Length Operands Bit-vector operands with different word-lengthsBit-vector operands with different word-lengths

Input variables : Input variables : {{xx11,…, ,…, xxdd} } Output variablesOutput variables: : ff, , gg

Input bit-widths: Input bit-widths: {{nn11,…, ,…, nndd} } Output width : Output width : mm

Model as polynomial function Model as polynomial function

**

8-bit

12-bit

20-bit

32-bit

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

n n n m1 2 d 22 2 2Z × Z × × Z Z

Example: Digital Image Rejection UnitExample: Digital Image Rejection Unit

YY11 ≠ Y ≠ Y22

• YY11[15:0] = Y[15:0] = Y22[15:0][15:0]

• YY11 % 2 % 21616 ≡ Y ≡ Y22 % 2 % 21616

4 4 2 2

1Y 16384(A B ) 64767(A B ) A B

57344AB(A B)

input A[11:0], B[7:0]; output Y1[15:0], Y2[15:0];

2 2 2

2

2

Y 24576A B 15615A 8192AB 32768AB

A 17153B 65535B

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, co-

operative decision procedures

• Solve linear congruences under modulo arithmetic

Theorem-Proving (HOL), term-rewriting

• Abstract away the data-path size using data

independence, symmetry, other abstractions

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

Previous Work: Symbolic AlgebraPrevious Work: Symbolic Algebra

Galois Field Decomposition of Boolean

Functions: GF(2m) [Pradhan, 1978 + recent work]

Symbolic Algebra Tools: Singular, Macaulay,

Maple, Mathematica, ZEN, NTL, CoCoA

• Polynomial equivalence over R, Q, C, Zp

• Unique Factorization Domains (UFDs) : Uniquely

factorize into irreducibles

• Match corresponding coefficients to prove

equivalence

Symbolic Algebra in CADSymbolic Algebra in CAD

MODD: DAG representation of polynomials over GF(2m)

[Pradhan, IWLS 05, DATE 04]

Guiding Synthesis engines using Groebner’s basis

[De Micheli, TCAD 02]

• Given polynomial F and Library elements <I1, …, In>

• F = h1 I1 + …… + hn In

• Again, works over UFDs

• Approximate RTL as polynomials over Reals

Theorem Proving [Clarke et al.]

• HOL + Mathematica = Analytica

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

Z2m is a non-UFD

• f = x2 + 6x in Z8 can be factorized as

Atypical approach required to prove equivalence

f

x x+6

f

x+2 x+4

Problem Formulations + SolutionsProblem Formulations + Solutions

f (x1, …, xd) % n ≡ g(x1, …, xd) % n

• Proving equivalence is NP-hard [Ibarra, J. ACM ’83]

Vanishing polynomials [ICCD ’05, DATE’06]

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

• An instance of Ideal Membership Testing

• Efficient solutions over fields (Groebner’s bases): Z2m[x1,…, xd] ?

Canonical forms [ICCAD ’05]

• Unique representations for polyfunctions over Z2m

• Equivalence by coefficient matching

• Concepts from Hungerbuhler et al. [J. Sm. Not., ‘06 ]

Ideal Membership TestingIdeal Membership Testing

( f – g ) % 2m = 0 or ( f – g ) vanishes % 2m

Membership in the Ideal of all Vanishing Polynomials in Z2m

Grobner's basis? Buchberger's algorithm?

Generate the Ideal!

Z2m[x1, …, xd]

Z2m

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

Generalize the result from:

%p to %pm to % (any integer) n

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

(f-g) % 2m = 0 means that 2m | (f-g)

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: Vanishing polynomialExample: 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: Vanishing polynomialExample: 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, try Rule 2 Divide by S2

• quo = 5; rem = 5 + 4x

• b2 = 23/gcd(2!,23) = 4

• a2 = quo/ b2 = 5/4 is not in Z poly does not satisfy Rule 2 poly is not a vanishing polynomial in Z23

Status of our Work - ExtensionsStatus of our Work - Extensions

Multiple Variables: Multiple Variables: Z2m[x1, …, xd]

• Given polynomials (Given polynomials (f, gf, g) ) dd variables variables

xx = < = <xx11, …, , …, xxdd> > over over ZZ22mm

Prove that Prove that (f-g) = 0 %(f-g) = 0 % 2 2mm What if word-lengths are different too?What if word-lengths are different too?

xx1 1 Z Z22nn

1 1 , …… , , …… , xxd d Z Z22nn

dd

No problems!No problems! Straight-forward extensions of previousStraight-forward extensions of previous conceptsconcepts

Other approach: Other approach: Canonical formsCanonical forms of poly- of poly-functionsfunctions

Polyfunctions over Polyfunctions over ZZ22mm

Polynomials over Z2m[x1, …, xd]

• Represented by polyfunctions from Z2m[x1, …, xd] to Z2m

F1 % 2m ≡ F2 % 2m => they have the same underlying polyfunction

(f )

Use equivalence classes of polynomials

• Derive representative for each class: Canonical form

f

g

Equivalence classes

Z2m[x1, …, xd] Z2m

F2

F1

G2

G1

Motivating our ApproachMotivating our Approachmodule fixed_bit_width (x, f, g);

input [2:0] x;

output [2:0] f, g;

assign f[2:0] = 5x2 + 6x - 3;

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

f (x) = 5x2 + 6x - 3 = (x2 + 2x + 5) + (4x2 + 4x) • f (x) = g(x) + V (x) in Z23

V (x) = 4x2 + 4x ≡ 0 % 23 ; for x in {0,…,7}• f (x) = g (x) + 0 in Z23

Required: To identify and eliminate such redundant sub-expressions(vanishing)

Vanishing Polynomials for ReducibilityVanishing Polynomials for Reducibility

In Z23, say f (x) = 4x2

• f (x) = f (x) - V(x)

• Generate V(x) of degree 2

• V(x) = 4x2 + 4x ≡ 0 % 23

Reduce by subtraction: • 4x2 f (x)

– 4x2 + 4x V(x)= - 4x = - 4x % 8 = 4x

• 4x2 can be reduced to 4x

• Degree reduction

Degree Reduction: RequirementDegree Reduction: Requirement Generate appropriate vanishing polynomial , V(x) f (x) = axk + a1xk-1 + …

V(x) = axk + a2xk-1 + …

f (x) – V(x) = bxk-1 + …

V(x): axk is the leading term Identify constraints on

• Degree : k; Coefficient : a f (x) = axk + …

If 2m|ak!, then V(x) = ak! x + k ≡ 0 % 2m k

= axk + a1xk-1…..

Coefficient Reduction: ExampleCoefficient Reduction: Example

Degree is not always reducible

In Z23, f (x) = 6x2

• a = 6, k = 2

• 8 does not divide 6 × 2!

Divide and subtract

• 6x2 = 2x2 + 4x2 % 23

• 4x2 can be reduced to 4x f (x) = 2x2 + 4x : Lower Coefficient

Our ApproachOur Approach

Say f (x) = akxk + ak-1xk-1 + …+ a0

• In decreasing lexicographic order

Required: f (x) in reduced, minimal, unique form

• Check if degree can be reduced

• Check if coefficient can be reduced

• Perform corresponding reductions

• Repeat for all monomials …

Experimental SetupExperimental Setup

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

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 zero-testing algorithm

Reduce f, g to canonical forms….

Algorithm implemented in MAPLE

Compare with BMD, SAT and MILP

Complexity: O(kd)

ResultsResults

1

10

100

1000

PSK

Anti-Alia

s Funct

ion

Cubic F

ilter

Degre

e-4

Filter

Savitz

y-Gola

y Filt

er

MIB

ENCH

Horner

Poly

1

Horner

Poly

2

Horner

Poly

3

Poly U

nopt.

Vanis

hing P

oly.

Our Approach (< 30 s)

BDD-VIS

BMD

SAT-Zchaff

Limitations: Mathematics versus CADLimitations: Mathematics versus CAD

Finite Ring Algebra = Cannot DIVIDE!Right shift breaks the model

Overflow arithmetic versus Saturation Comparators = non-polynomial?

Internal bit-vectors of arbitrary word-lengths

Only Yes/No Equivalence? Find bugs too!

Couple Verification with Simulation!

Arithmetic interfaced with Boolean……

Intermediate Signal TruncationIntermediate Signal Truncation

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]

Proposed SolutionProposed Solution

≠

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]

Canonize

Canonize

Canonize

Canonize

Required: Algorithm for reduction to canonical form over

n n n m1 2 d 22 2 2Z × Z × × Z Z

Polynomial Abstraction from RTLPolynomial Abstraction from RTL

If If (x > 2b’10)(x > 2b’10)

then then y = x * x * xy = x * x * x

Else Else y = x*xy = 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 =

Verification via Simulation?Verification via Simulation?

Simulation Vector GenerationSimulation Vector Generation

f = f = xx44 + x + x2 2 (specification)(specification)

• x=0, f=0x=0, f=0

• x=1, f=2x=1, f=2

• x=2, f=4x=2, f=4

• x=3, f=2x=3, f=2

• x=4, f=0x=4, f=0

• x=5, f=2x=5, f=2

• x=6, f=4x=6, f=4

• x=7, f=2x=7, f=2

Prove Prove f f = = gg over over ZZ88 via Simulationvia Simulation

• How many vectors to Simulate? Exhaustive Simulation?How many vectors to Simulate? Exhaustive Simulation?

• NO!NO! Simulate onlySimulate only n = n = SF(2SF(233) = ) = 4 CONSECUTIVE vectors4 CONSECUTIVE vectors

g = g = 2x 2x2 2 (implementation)(implementation)

• x=0, g=0x=0, g=0

• x=1, g=2x=1, g=2

• x=2, g=0x=2, g=0

• x=3, g=2x=3, g=2

• x=4, g=0x=4, g=0

• x=5, g=2x=5, g=2

• x=6, g=0x=6, g=0

• x=7, g=2x=7, g=2

Applications to SynthesisApplications to Synthesis

Datapath size (m): 8 bits

SF(28) = 10

Polynomial can be

factorized into S10(x)

F[7:0] = (x+1)(x+2)...(x+10)

Polynomial DecompositionPolynomial Decomposition

F = 8xyF = 8xy22 + 11y + 11y33 + 9y + 9y2 2 + 1 + 1 X, Y [3:0] are 4-bit vectors (%16)X, Y [3:0] are 4-bit vectors (%16) Synthesize: Area = Synthesize: Area = 320320

U = 8x + 3yU = 8x + 3y F = UF = U33 + U + U2 2 + 1 + 1 Synthesize: Area = 181+20 = Synthesize: Area = 181+20 = 201201

ConclusionsConclusions

Finite Word-Length Bit-Vector Arithmetic is Finite Ring

Algebra

Computations reduced % integer power of 2

NON-UFDs! Number Theory, Commutative Algebra

Mathematicians should help us here…….

Techniques to verify equivalence of polynomial RTL

computations

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

Ideal Membership Testing, Canonical Forms….

Tremendous scope in high-level synthesis….

Questions?Questions?

Documents

Verification & Synthesis of Arithmetic Datapaths using Finite Ring Algebra Priyank Kalla Priyank Kalla Electrical and Computer Engineering University of