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

Exploiting Vanishing Polynomials for Equivalence Verification of Fixed-Size Arithmetic Datapaths Namrata Shekhar1, Priyank Kalla1, Florian Enescu2, 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

Outline • Overall Verification Problem • Our Focus: Equivalence Verification of Fixed-size Arithmetic Datapaths • Applications: Polynomial Signal Processing, etc. • Problem Modeling and Approach • Polynomials over Finite Integer Rings • Limitations of Previous Work • Our Contributions • Exploiting Vanishing Polynomials for Equivalence Test • Related to Ideal Membership Testing • Algorithm Design and Experimental Verification Runs • Results • Conclusions & Future Work

The Equivalence Verification Problem

8-bit * 16-bit 8-bit 8-bit * * 32-bit 8-bit * 8-bit 8-bit Fixed-size (m) bit-vector arithmetic Polynomials reduced %2m Algebra over the ring Z2m Fixed-Size (m) Data-path: Modeling • Control the datapath size: Fixed size bit-vectors (m) • Bit-vector of size m: integer values in 0,…, 2m-1

Fixed-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

DFF Example: 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 coefficients coefficients a b x = a2 + b2 x MAC F

Example: 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 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: 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 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? • Z2mis a non-UFD • f = x2 + xinZ6can be factorized as f f x x+1 x+3 x+4 • Atypical approach required to prove equivalence

Proposed 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]

h: % 2m Ideal x x % 2m 0 f g f – g ? Ideal Membership Testing Q P • 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 ?

Ideal Membership Testing in Zp • Fermat’s Little Theorem: • x p≡ x (mod p) or • x p – x ≡ 0 (mod p) • x p –xgenerates the vanishing ideal in Zp[x] • f(x) = 0 % p iff f(x) = (xp-x)g(x) • Zp: Principal Ideal Domain • This does not followin Z2m

h: % 2m Ideal x x % 2m f g f – g ? Ideal Membership Testing 0 Q P • Generate the idealof 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]

Results From Number Theory • n! divides a product of n consecutive numbers. • 4! divides 99X 100 X 101 X 102 • Find least n such that 2m|n! • Smarandache Function (SF). • SF(23) = 4, since 23|4! • 2mdivides the product of n = SF(2m) consecutive numbers

Results From Number Theory • f ≡ gin Z23or(f - g) ≡ 0 % 23 • 23|(f - g)inZ23 • 23|4! • 4! divides the product of 4 consecutive numbers Write (f-g) as a product of SF(23) = 4 consecutive numbers • A polynomial as a product of 4 consecutive numbers? • (x+1)

Results From Number Theory • f ≡ gin Z23or(f - g) ≡ 0 % 23 • 23|(f - g)inZ23 • 23|4! • 4! divides the product of 4 consecutive numbers Write (f-g) as a product of SF(23) = 4 consecutive numbers • A polynomial as a product of 4 consecutive numbers? • (x+1)(x+2)

Results From Number Theory • f ≡ gin Z23or(f - g) ≡ 0 % 23 • 23|(f - g)inZ23 • 23|4! • 4! divides the product of 4 consecutive numbers Write (f-g) as a product of SF(23) = 4 consecutive numbers • A polynomial as a product of 4 consecutive numbers? • (x+1)(x+2)(x+3)

Results From Number Theory • f ≡ gin Z23or(f - g) ≡ 0 % 23 • 23|(f - g)inZ23 • 23|4! • 4! divides the product of 4 consecutive numbers Write (f-g) as a product of SF(23) = 4 consecutive numbers • A polynomial as a product of 4 consecutive numbers? • (x+1)(x+2)(x+3)(x+4)

Basis 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 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) inZ23. • p is a vanishing polynomial.

Example 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 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 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 Polynomials • Polynomial F in Z2m vanishes if • n = SF(2m),i.e. the least n such that 2m|n! • Fn is an arbitrary polynomial in Z2m[x] • akis an arbitrary integer • bk = 2m/gcd(k!,2m) F = FnSn + Σn-1ak bk Sk k=0 Rule 1 Rule 2

Input: poly, 2m Calculate n = SF(2m) k = n: Reduce according to Rule 1 Divide by Sn If remainder is zero, F = FnSn, else Continue poly = 4x2 + 4xin Z23 n = SF(23) = 4 k = 4: Divide by S4 Degree (poly) = 2 < degree(S4) = 4 quo = 0, rem = 4x2 + 4x F4 = 0; Continue Algorithm Example 1 F = FnSn + Σn-1ak bk Sk k=0

Reduce according to Rule 2. Divide by Sn-1to S0 Check if quotient is a multiple of bk = 2m/gcd(k!,2m) If remainder is zero, stop. Else, continue k = 3: Divide by S3 degree (poly) = 2 < degree(S3) = 3 quo= 0, rem = 4x2 + 4xcontinue k = 2: Divide by S2 quo = 4; rem = 0 b2 = 23/gcd(2!,23) = 4 a2 = quo/ b2 = 1 ЄZ Algorithm Example 1 F = FnSn + Σn-1ak bk Sk k=0 poly = a2.b2.S2 = 1.4.(x+1)(x+2) ≡ 0 in Z23

Example 2 • poly = 5x2 + 3x+ 7 inZ23 • 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 inZ • poly does not satisfy Rule 2 • poly is not a vanishing polynomial inZ23

Experimental 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)

Results

Applications to Synthesis • Datapath size (m): 8 bits • SF(28) = 10 • Polynomial can be factorized into S10(x)

Conclusions • 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) % 2mis transformed into f(x) - g(x) ≡ 0 % 2m • Efficient algorithm to determine vanishing polynomials

Future 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?

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