1. Equivalence Verification of Polynomial Datapaths with Fixed-Size Bit-Vectors using Finite Ring Algebra Namrata Shekhar, Priyank Kalla, Florian Enescu, Sivaram Gopalakrishnan Namrata Shekhar 1, Priyank Kalla 1, Florian Enescu 2, Sivaram Gopalakrishnan 1 1 Department of Electrical and Computer Engineering, University of Utah, Salt Lake City, UT-84112. 2 Department of Mathematics and Statistics, Georgia State University, Atlanta, GA-30303

2. Outline  Overall Verification Problem Our Focus: Equivalence Verification of Fixed-size Arithmetic DatapathsOur Focus: Equivalence Verification of Fixed-size Arithmetic Datapaths  Problem Modeling Polynomial Functions over Finite Integer RingsPolynomial Functions over Finite Integer Rings  Limitations of Previous Work  Approach and Contributions Canonical form for Polynomials over Finite RingsCanonical 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 = a 0. x 1 2 + a 1. x 1 + b 0. x 0 2 + b 1. x 0 + c. x 0. x 1

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

6. Fixed-Size Data-path: Implementation   Signal Truncation Keep lower order m-bits, ignore higher bits f % 2 m ≡ g % 2 m   Fractional Arithmetic with rounding Keep higher order m-bits, round lower order bits f - f %2 m ≡ g - g%2 m 2 m 2 m   Saturation Arithmetic Saturate at overflow Used in image-processing applications

7. Example: Anti-Aliasing Function   F = 1 = 1 = 2√a 2 + b 2 2√x [Peymandoust et al, TCAD‘03]   Expand into Taylor series F ≈ 1 x 6 – 9 x 5 + 115 x 4 64 32 64 – 75 x 3 + 279 x 2 – 81 x 16 64 32 + 85 64   Scale coefficients; Implement as bit-vectors MAC x = a 2 + b 2 coefficients ab x F DFF

8. Example 1: Anti-Aliasing Function   F 1 [15:0], F 2 [15:0], x[15:0]   F 1 = 156x 6 + 62724x 5 + 17968x 4 + 18661x 3 + 43593 x 2 + 40244x +13281   F 2 = 156x 6 + 5380x 5 + 1584x 4 + 10469x 3 + 27209 x 2 + 7456x + 13281   F 1 ≠ F 2 ; F 1 [15:0] = F 2 [15:0]   To Prove: F 1 % 2 16 ≡ F 2 % 2 16 F 1 ≡ F 2 in Z 2 m [x 1, …, x d ]

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: Z 2 m → Z 2 m   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, Z p Unique Factorization Domains (UFDs) : Uniquely factorize into irreducibles Match corresponding irreducibles to prove equivalence

11. Why is the Problem Difficult?   Z 2 m is a non-UFD f = x 2 + 6x in Z 8 can be factorized as  Atypical approach required to prove equivalence f x+6 f x+4x+2x

12. Proposed Solution   f (x 1, …, x d ) % n ≡ g(x 1, …, x d ) % n Proving equivalence is NP-hard   Vanishing polynomials [ICCD ‘05] f(x) – g(x) ≡ 0 % 2 m : Zero Equivalence An instance of Ideal Membership Testing Efficient solutions over fields (Groebner’s bases): Z 2 m [x 1,…, x d ] ?   Canonical forms: Current focus Unique representations for polyfunctions over Z 2 m Equivalence by coefficient matching Concepts from Hungerbuhler et al. [To appear J. Sm. Not., ‘06 ]

13. Polyfunctions over Z 2 m   Polynomials over Z 2 m [x 1, …, x d ] Represented by polyfunctions from Z 2 m [x 1, …, x d ] to Z 2 m   F 1 % 2 m ≡ F 2 % 2 m => they have the same underlying polyfunction ( f )   Use equivalence classes of polynomials Derive representative for each class: Canonical form f g Equivalence classes Z 2 m [x 1, …, x d ] Z2mZ2m F2F2 F1F1 G2G2 G1G1

14. Motivating our Approach module fixed_bit_width (x, f, g); input [2:0] x; output [2:0] f, g; assign f[2:0] = 5x 2 + 6x - 3; assign g[2:0] = x 2 + 2x + 5;  f (x) = 5x 2 + 6x - 3 = (x 2 + 2x + 5) + (4x 2 + 4x) f (x) = g(x) + V (x) in Z 2 3  V (x) = 4x 2 + 4x ≡ 0 % 2 3 ; for x in {0,…,7} f (x) = g (x) + 0 in Z 2 3  Required: To identify and eliminate such redundant sub-expressions (vanishing)

15. Vanishing polynomials: Requirement  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) Z2mZ2m Set of all Vanishing polynomials f g Z 2 m [x 1, …, x d ] 0 h: % 2 m

16. Vanishing Polynomials for Reducibility  In Z 2 3, say f (x) = 4x 2 f (x) = f (x) - V(x) Generate V(x) of degree 2 V(x) = 4x 2 + 4x ≡ 0 % 2 3  Reduce by subtraction: 4x 2 f (x) – 4x 2 + 4x V(x) = - 4x = - 4x % 8 = 4x 4x 2 can be reduced to 4x Degree reduction

17. Coefficient Reduction: Example  Degree is not always reducible  In Z 2 3, f (x) = 6x 2 a = 6 k = 2  Divide and subtract 6x 2 = 2x 2 + 4x 2 % 2 3 4x 2 can be reduced to 4x  f (x) = 2x 2 + 4x : Lower Coefficient

18. Our Approach  Say f (x) = a k x k + a k-1 x k-1 + …+ a 0 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) = ax k + a 1 x k-1 + … V(x) = ax k + a 2 x k-1 + … f (x) – V(x) = bx k-1 + …  V(x): ax k is 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 99 X 100 X 101 X 102   Find least n such that 2 m |n! Smarandache Function (SF) of 2 m = n SF(2 3 ) = 4, since 2 3 |4!   2 m divides the product of n = SF(2 m ) consecutive numbers   Use SF(2 m ) to generate vanishing polynomial V(x)

21. Results From Number Theory   V (x) ≡ 0 % 2 3 2 3 | V (x) in Z 2 3 2 3 | 4!, since SF(2 3 ) = 4 4! divides the product of 4 consecutive numbers  A polynomial as a product of 4 consecutive numbers? (x+1)(x+2)(x+3)(x+4) = 4! x + 4 ≡ 0 % 2 3 4 Write V(x) as a product of SF(2 3 ) = 4 consecutive numbers

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

23. Example: Vanishing Polynomial  Consider f (x) = 4x 2 in Z 2 3 a = 4 k = 2 V (x) = ?  Rule 1: 2 m |a·k! => 2 3 | 4 · 2! V (x) = a · k! x + k = 4. 2! x + 2 = 4. 2! (x+2) (x+1) k 2 2! = 4x 2 + 4x ≡ 0 % 2 3 f (x) = 4x 2 V(x) = - 4x 2 + 4x = - 4x = - 4x % 8 = 4x

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

25. Notation: Multivariate Polynomials  Given d variables x = with degrees k =  Replace in Rule 1 and Rule 2 x k = ∏ d i=1 x i k i k! = ∏ d i=1 k i ! x = ∏ d i=1 x i k k i v 2 (k!) = ∏ d v 2 (k i !)  Use lexicographic term ordering for variables

26. Uniqueness  Theorem: Any polynomial F in Z 2 m can be uniquely written as F = Σ kЄN d α k x k  d is the number of variables  α k Є {0, …, 2 m- v2(k!) -1} is the coefficient  v 2 (k!) < m

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

28. 1. 1.Reducing 6x 2 y 2. 2.Degree Reduction: a = 6; k! = k x !.k y ! = 2!1! = 2 2 3 does not divide (a ·k! ) = 12; Degree reduction not possible Perform coefficient reduction Example: Reduction Given poly = 6x 2 y + 4xy in Z 2 3

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

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(k d ); where k is the total degree and d is 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, Z 2 m   f (x 1,…, x d ) % 2 m ≡ g (x 1, …, x d ) % 2 m is 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