Namrata Shekhar 1 , Priyank Kalla 1 , Florian Enescu 2 , Sivaram Gopalakrishnan 1

1. Equivalence Verification of Polynomial Datapaths with Fixed-Size Bit-Vectors using Finite Ring Algebra 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

2. Outline • Overall Verification Problem • Our Focus: Equivalence Verification of Fixed-size Arithmetic Datapaths • Problem Modeling • Polynomial Functions over Finite Integer Rings • Limitations of Previous Work • Approach and Contributions • Canonical form for Polynomials over Finite Rings • Algorithm Design and Experimental Verification Runs • Results, Conclusions & Future Work

3. The Equivalence Verification Problem

4. Motivation • Quadratic filter design for polynomial signal processing • y = a0 . x12 + a1 . x1 + b0 . x02 + b1 . x0 + c . x0 . x1

5. 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

6. 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

7. coefficients coefficients a b x = a2 + b2 x MAC DFF F 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

8. Example 1: 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] • To Prove: F1 % 216 ≡ F2 % 216 • F1≡ F2 in Z2m[x1, …, xd]

9. Previous Work: Function Representations • Boolean Representations (f: B → B): BDDs, ZBDDs 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 • SAT, MILP, Word-level ATPG, … • Theorem-Proving (HOL), term-rewriting • Works when datapath size can be abstracted away

10. Previous Work: Symbolic Algebra • Symbolic Algebra Tools: Singular, Macaulay, Maple, Mathematica, Zen, Dagwood etc. • Polynomial representations: Sparse, Dense, Recursive, Straight-line programs, DAGs, etc. • Polynomial equivalence over R, Q, C, Zp • Unique Factorization Domains (UFDs) : Uniquely factorize into irreducibles • Match corresponding irreducibles to prove equivalence

11. Why is the Problem Difficult? • Z2mis a non-UFD • f = x2 + 6xinZ8can be factorized as f f x x+6 x+4 x+2 • Atypical approach required to prove equivalence

12. Proposed Solution • f (x1, …, xd) % n≡g(x1, …, xd) % n • Proving equivalence is NP-hard • Vanishing polynomials [ICCD ‘05] • 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: Current focus • Unique representations for polyfunctionsover Z2m • Equivalence by coefficient matching • Concepts from Hungerbuhler etal. [To appear J. Sm. Not., ‘06 ]

13. Equivalence classes F1 G1 Polyfunctions over Z2m f F2 g G2 Z2m[x1, …, xd] Z2m • Polynomials over Z2m[x1, …, xd] • Represented by polyfunctions from Z2m[x1, …, xd]to Z2m • F1 % 2m≡ F2 % 2m=> they have thesame underlying polyfunction (f ) • Use equivalence classes of polynomials • Derive representative for each class: Canonical form

14. Motivating our Approach module 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)

15. h: % 2m Vanishing polynomials: Requirement Set of all Vanishing polynomials 0 f g Z2m Z2m[x1, …, xd] • Generate vanishing expressions V(x) • Test if f (x) = g (x) + V(x) • f (x) = f (x)– V(x) • Challenge: Infinite number of vanishing polynomials • Required: To generate V(x) specific to given f (x)

16. Vanishing Polynomials for Reducibility • In Z23, sayf (x) = 4x2 • f (x) = f (x) - V(x) • Generate V(x) of degree 2 • V(x) = 4x2 + 4x ≡ 0 % 23 • Reduce by subtraction: • 4x2f (x) – 4x2 + 4x V(x) = - 4x = - 4x % 8 = 4x • 4x2can be reduced to 4x • Degree reduction

17. Coefficient Reduction: Example • Degree is not always reducible • In Z23, f (x) = 6x2 • a = 6 • k = 2 • Divide and subtract • 6x2 = 2x2 + 4x2 % 23 • 4x2 can bereduced to 4x • f (x) = 2x2 + 4x : Lower Coefficient

18. Our 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 …

19. Degree 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):axkis the leading term • Identify constraints on • Degree : k • Coefficient : a • Use concepts from number theory

20. 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) of 2m = n • SF(23) = 4, since 23|4! • 2mdivides the product of n = SF(2m) consecutive numbers • Use SF(2m) to generate vanishing polynomial V(x)

21. Results From Number Theory • V (x) ≡ 0 % 23 • 23| V (x)inZ23 • 23| 4! , sinceSF(23) = 4 • 4! divides the product of 4 consecutive numbers Write V(x) 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) = 4! x + 4 ≡ 0 % 23 4

22. Constraints on the Coefficient • In Z23, SF(23) = 4. Product of 4 consecutive numbers: • (x+1) (x+2) missing factors • V (x) = 4x2 + 4x = (x+1)(x+2) ≡ 0 % 23 compensated by constant • 4(x+1)(x+2) = 4·2!· x + 2 2 (x+3) (x+4) 4 • f (x) = axk + … Rule 1: If 2m|ak!, then V(x) = ak! x + k ≡ 0 % 2m k = axk + a1xk-1…..

23. Example: Vanishing Polynomial • Considerf (x) = 4x2in Z23 • a = 4 • k = 2 • V (x) = ? • Rule 1: 2m|a·k! => 23 | 4·2! • V (x) = a·k! x + k = 4. 2! x + 2 = 4. 2! (x+2) (x+1) k 2 2! = 4x2 + 4x ≡ 0 % 23 • f (x) = 4x2 V(x) = - 4x2 + 4x = - 4x = - 4x % 8 = 4x

24. Coefficient Reduction: Requirement • Define v2(k!) = {max x Є N: 2x| k!} • number-of-factors-2ink! • v2(4!) = v2(4 * 3 * 2* 1) = 3 • Rule1: 2m| a·k! • Number-of-factors-2 in a·k! ≥ m or • v2(a·k!) ≥ m • If 2mdoes not dividea·k! • v2(a·k!) < m • v2(k!) < m and a < 2m- v2(k!) Constraint on degree Constraint on coefficient Rule 2: Coefficient (a) has to be in the range {0, …, 2m- v2(k!)-1}

25. Notation: Multivariate Polynomials • Given d variables x = <x1, …, xd> with degrees k =<k1, …, kd> • Replace in Rule 1 and Rule 2 • xk= ∏di=1 xiki • k! = ∏di=1ki! • x = ∏di=1 xi kki • v2(k!) = ∏dv2(ki!) • Use lexicographic term ordering for variables

26. Uniqueness • Theorem: Any polynomial F in Z2mcan be uniquely written as F = ΣkЄNd αk xk • d is the number of variables • αkЄ {0, …, 2m- v2(k!)-1}is the coefficient • v2(k!)< m

27. Reduction Procedure • Given a monomial f (x) = a·xk • Degree reduction: Determine if 2m| a·k! • If yes, generate V(x) of with a·xk as the leading term • f (x) = f (x) – V(x) • If 2m| a·k!, check if coefficient (a) is in {0, …, 2m- v2(k!)-1} • If not, perform division according to a·xk = q. 2m- v2(k!)·xk + r·xk Degree reducible Reduced form r <2m- v2(k!)

28. Reducing 6x2y Degree Reduction: a = 6; k! = kx!.ky! = 2!1! = 2 23 does not divide (a ·k! ) = 12; Degree reduction not possible Perform coefficient reduction Example: Reduction Given poly = 6x2y + 4xyin Z23

29. Coefficient Reduction: v2(k!) = v2(2!·1!) = 1 Range = {0, …, 2m- v2(k!) – 1} = {0,…, 3} a = 6 > 3; Can reduce coefficient 6x2y = 4x2y + 2x2y Degree reduction for 4x2y: Rule 1: 23 | (4· 2! ·1!) 4x2y – 4 . 2!·1! x + 2 y + 1 = 4xy 2 1 poly = (4xy + 2x2y) + 4xy≡ 2x2yin Z23 Example: Reduction Given poly = 6x2y + 4xyin Z23

30. Experimental Setup • Distinct RTL designs are input to GAUT [U. de LESTER, ‘04] • Extract data-flow graphs for RTL designs • Construct the corresponding polynomial representations (f, g) • Extract bit-vector size • Reduce f and g to canonical form • Equivalence check by coefficient matching • Algorithm implemented in MAPLE • Complexity: O(kd); wherekis the total degree anddis the number of variables • Compare with BDD, BMD and SAT

31. Results

32. Conclusions & Future Work • Technique to verify equivalence of multivariate polynomial RTL computations • Fixed-size bit-vector arithmetic is polynomial algebra over the finite integer ring, Z2m • f (x1,…, xd) % 2m≡ g (x1, …, xd) % 2mis proved by reduction to canonical form • Efficient algorithm to determine unique representations • Future Work involves extensions for - • Multiple Word-length Implementations [DATE ‘06] • Verification of Rounding and Saturation Arithmetic

33. Questions?

34. Comparison