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Namrata Shekhar 1 , Priyank Kalla 1 , M. Brandon Meredith 2 , Florian Enescu 2

SIMULATION BOUNDS FOR EQUIVALENCE VERIFICATION OF ARITHMETIC DATAPATHS WITH FINITE WORD-LENGTH OPERANDS. Namrata Shekhar 1 , Priyank Kalla 1 , M. Brandon Meredith 2 , Florian Enescu 2 1 Department of Electrical and Computer Engineering, University of Utah, Salt Lake City, UT-84112.

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Namrata Shekhar 1 , Priyank Kalla 1 , M. Brandon Meredith 2 , Florian Enescu 2

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  1. SIMULATION BOUNDS FOR EQUIVALENCE VERIFICATION OF ARITHMETIC DATAPATHS WITH FINITE WORD-LENGTH OPERANDS Namrata Shekhar1, Priyank Kalla1, M. Brandon Meredith2, 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

  2. Outline • 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

  3. The Equivalence Verification Problem

  4. 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 Bit-width Operands • Control the datapath size: Fixed size bit-vectors (m) • Bit-vector of size m: Integer values in 0,…, 2m-1

  5. 20-bit 8-bit * 32-bit * 12-bit General 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

  6. Arithmetic 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

  7. a0 FFT(A) a1 FAB0 invFFT(FAB) a2 c’0 a3 FAB1 c’1 FAB2 c’2 FAB3 b0 FFT(B) c’3 b1 b2 b3 Motivation: Convolution of A and B C = (c0, c1, c2, c3) where C′ = (c′0, c′1, c′2, c′3) = DFT-1 (DFT(A)·DFT(B)) Datapath size is fixed How many simulation vectors required to prove C = C’ ?

  8. 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

  9. Example: 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

  10. 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

  11. Previous 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

  12. Previous 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]

  13. Fundamental Theorem of Algebra • A degree-k polynomial F(x) has exactly k roots • F(x) = x2 + 6x = x ( x + 6 ) • IfF(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 • Z2mis a non-UFD

  14. F F x x+6 x+4 x+2 Why is the Problem Difficult? • Consider F(x) = x2 + 6xinZ8 • Degree-2 polynomial has 4 unique roots • F(x) = 0 for 4 vectors, but F(x) ≠ 0 inZ8 • Not applicable to bit-vector arithmetic

  15. Previous 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

  16. Zero 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

  17. h: % 2m Ideal xi xi % 2m f 0 g f – g ? Ideal Membership Testing • h:Z2m[x]→ Z2m defined by % 2m • Ideal members map to0 • Test for membership in • Representative expression for members of this ideal [Chen, Disc. Math ‘96] • Use concepts from Number theory and polynomial algebra Z2m[x] Z2m Ideal of Vanishing Polynomials

  18. Results 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 99X 100 X 101 X 102 • 2mdivides the product ofnconsecutive numbers • 23 divides the product of 4 consecutive numbers

  19. Results From Number Theory • F ≡ G in Z23or(F - G) ≡ 0 % 23 • 23|(F - G)inZ23 • 23divides 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) (x-1) (x-2) (x-3)

  20. Basis 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 numbers Rule 1: Factorize into at least Yλ(x) to vanish, where λ = SF(2m)

  21. Example 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

  22. Constraints on the Coefficient • F(x) = 4x2 + 4x = (x)(x-1) % 23 • Y4(x) = (x)(x-1) 4 compensated by constant (x-2)(x-3) missing factor 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

  23. Deciding Vanishing Polynomials • Polynomial F in vanishes if • Fλ is an arbitrary polynomial over • Yλ= Ykis as defined earlier: λ = SF(2m) • akis an arbitrary integer • . Rule 1 Rule 2

  24. Polynomial Representations • Newton’s interpolation formula: Any polynomial F(x) can be written as • : Forward difference operator • pis the degree of F(x)

  25. Polynomial Representations • Newton’s interpolation formula: Any polynomial F(x) can be written as • : Forward difference operator • pis the degree of F(x)

  26. Reinterpret Vanishing ideal • Newton’s interpolation formula: Any polynomial F(x) can be written as • : Forward difference operator • Yk(x)is as defined earlier • pis the degree of F(x)

  27. Reinterpret Vanishing ideal • Newton’s interpolation formula: Any polynomial F(x) can be written as • : Forward difference operator • Yk(x)is as defined earlier • pis the degree of F(x)

  28. Polynomial Representations • Newton’s interpolation formula: Any polynomial F(x) can be written as • is any arbitrary integer • Yk(x)is as defined earlier • pis the degree of F(x)

  29. Rule 1 Polynomial Representations • Any polynomial F(x) in can be written as • Fλ is an arbitrary polynomial over • Yλ= Ykis as defined earlier: λ = SF(2m) • is an arbitrary integer 0

  30. Polynomial Representations • Any polynomial F(x) in can now be reduced to • Fλ is an arbitrary polynomial over • Yλ= Ykis as defined earlier: λ = SF(2m) • is an arbitrary integer

  31. Polynomial Representations • Any polynomial F(x) in can now be reduced to • Apply Rule 2 • F(x) vanishes iff ck is a multiple of • Check for all ck, where • ckevaluated no more thanλtimes • F(x) evaluated no more thanλtimes

  32. Results • 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

  33. 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 Example • 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

  34. Extension to Multiple Variables • Given d variables x = <x1, …, xd> with degrees k =< k1, …, kd> over Z2m • Basis for d variables: • . Rule 1 (for d variables): Factorize into Yλ(x), such that ki≥λfor anyxi ; λ= SF(2m)

  35. Example: 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

  36. Effect of Bit-vectors • 4th degree polynomial F(x, y) in Z21× Z22→ Z23 • λ= 4 • . • Define • . • F = Y2(x) · Y1(y) ≡ 0 % 23 Rule 1(extended): Factorize into Yk(x), such that ki≥μifor anyxi

  37. Results • 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

  38. 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 Example • 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

  39. 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 sizes for inputs and outputs • Determine the maximum number of simulation vectors required • Check for equivalence or determine bugs

  40. Simulation Results for Equivalent Designs

  41. Simulation Results for Buggy Designs

  42. Limitations a[7:0] b[7:0] c[7:0] a[7:0] b[7:0] c[7:0] + + ≠ t1[7:0] t2[7:0] + + f1[8:0] f2[8:0] a = 127 b = 1 f1 = 383 c = 255 a = 127 b = 1 f2 = 127 c = 255

  43. Conclusions • 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

  44. Questions ?

  45. Polynomial 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

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