Exploiting Vanishing Polynomials for Equivalence Verification of Fixed-Size Arithmetic Datapaths 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 Department of Mathematics and Statistics, Georgia State University, Atlanta, GA-30303

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

The Equivalence Verification Problem

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

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

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 x – 75 x x 2 – 81 x   Scale coefficients; Implement as bit-vectors MAC x = a 2 + b 2 coefficients ab x F DFF

Example: Anti-Aliasing Function   F 1 [15:0], F 2 [15:0], x[15:0]   F 1 = 156x x x x x x   F 2 = 156x x x x x x   F 1 ≠ F 2 ; F 1 [15:0] = F 2 [15:0]   Transform the problem F 1 - F 2 = 57344x x x x x ≡ 0 % 2 16   F 1 - F 2 : 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: Z 2 m → Z 2 m

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 ( Z 2 m )

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, Z p Unique Factorization Domains (UFDs) : Uniquely factorize into irreducibles Match corresponding coefficients to prove equivalence

Why is the Problem Difficult?   Z 2 m is a non-UFD f = x 2 + x in Z 6 can be factorized as  Atypical approach required to prove equivalence f xx+1 f x+3x+4

Proposed Solution   Equivalence can be proved by f (x) - g(x) ≡ 0 % 2 m : 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]

Ideal Membership Testing   h:P →Q defined by % 2 m x in P maps to x %2 m in Q Ideal members map to 0   Derive Ideal of Vanishing Polynomials in Z 2 m   Grobner's basis? Buchberger's algorithm?   Known for UFDs, but for Z 2 m ? P Q Ideal x x % 2 m h: % 2 m 0 f g f – g ?

Ideal Membership Testing in Z p   Fermat’s Little Theorem: x p ≡ x (mod p) or x p – x ≡ 0 (mod p) x p –x generates the vanishing ideal in Z p [x]   f(x) = 0 % p iff f(x) = (x p -x)g(x)   Z p : Principal Ideal Domain   This does not follow in Z 2 m

Ideal Membership Testing   Generate the ideal of vanishing polynomials % 2 m ? 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 x x % 2 m h: % 2 m 0 f g f – g ?

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). SF(2 3 ) = 4, since 2 3 |4!   2 m divides the product of n = SF(2 m ) consecutive numbers

Results From Number Theory   f ≡ g in Z 2 3 or (f - g) ≡ 0 % |(f - g) in Z |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(2 3 ) = 4 consecutive numbers

Results From Number Theory   f ≡ g in Z 2 3 or (f - g) ≡ 0 % |(f - g) in Z |4! 4! divides the product of 4 consecutive numbers  A polynomial as a product of 4 consecutive numbers? (x+1)(x+2) Write (f-g) as a product of SF(2 3 ) = 4 consecutive numbers

Results From Number Theory   f ≡ g in Z 2 3 or (f - g) ≡ 0 % |(f - g) in Z |4! 4! divides the product of 4 consecutive numbers  A polynomial as a product of 4 consecutive numbers? (x+1)(x+2)(x+3) Write (f-g) as a product of SF(2 3 ) = 4 consecutive numbers

Results From Number Theory   f ≡ g in Z 2 3 or (f - g) ≡ 0 % |(f - g) in Z |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) Write (f-g) as a product of SF(2 3 ) = 4 consecutive numbers

Basis for factorization   S 0 (x) = 1   S 1 (x) = (x + 1)   S 2 (x) = (x + 1)(x + 2) = Product of 2 consecutive numbers   S 3 (x) = (x + 1)(x + 2)(x + 3) = Product of 3 consecutive numbers   …   S n (x) = (x + n) S n-1 (x) = Product of n consecutive numbers Rule 1: Factorize into atleast S n (x) to vanish, where n = SF(2 m ).

Example 1: Vanishing polynomial   4 th degree polynomial p in Z 2 3 ; SF(2 3 ) = 4   p = x 4 +2x 3 + 3x 2 + 2x   p can be written as a product of 4 consecutive numbers. or p = (x+1)(x+2)(x+3)(x+4) = S 4 (x) in Z 2 3.   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] = x 2 + 6x – 3; assign g[2:0] = 5x 2 + 2x + 5;  h(x) = f(x) – g(x) = 4x 2 + 4x  h(x) ≡ 0 for all values of x in {0,…,7}  4x 2 +4x not equal to (x+1)(x+2)(x+3)(x+4)  Required: To show that h(x) is a vanishing polynomial in Z 2 3

Constraints on the Coefficient   h(x) = 4x 2 + 4x = 4(x+1)(x+2)   In Z 2 3, SF(2 3 ) = 4. Product of 4 consecutive numbers: S 4 (x) = (x+1) (x+2) (x+3) (x+4) Rule 2: Coefficient has to be a multiple of b k = 2 m /gcd(k!, 2 m )  Here, Coefficient of h(x) = 4, Degree of h(x) = 2  b 2 = 2 3 /gcd(2!, 2 3 ) = 4 is a multiple of the coefficient

Constraints on the Coefficient   h(x) = 4x 2 + 4x = 4(x+1)(x+2) compensated by constant   In Z 2 3, SF(2 3 ) = 4. Product of 4 consecutive numbers: S 4 (x) = (x+1) (x+2) (x+3) (x+4) missing factors Rule 2: Coefficient has to be a multiple of b k = 2 m /gcd(k!, 2 m )  Here, Coefficient of h(x) = 4, Degree of h(x) = k = 2  b 2 = 2 3 /gcd(2!, 2 3 ) = 4 is a multiple of the coefficient

Deciding Vanishing Polynomials   n = SF(2 m ), i.e. the least n such that 2 m |n!   F n is an arbitrary polynomial in Z 2 m [x]   a k is an arbitrary integer   b k = 2 m /gcd(k!,2 m )  Polynomial F in Z 2 m vanishes if F = F n S n + Σ n-1 a k b k S k k=0 Rule 1 Rule 2

Algorithm Input: poly, 2 m Calculate n = SF(2 m ) k = n: Reduce according to Rule 1 Divide by S n If remainder is zero, F = F n S n, else Continue poly = 4x 2 + 4x in Z n = SF(2 3 ) = k = 4: Divide by S 4 Degree (poly) = 2 < degree(S 4 ) = 4 quo = 0, rem = 4x 2 + 4x F 4 = 0; Continue Example 1 F = F n S n + Σ n-1 a k b k S k k=0

Algorithm Input: poly, 2 m Calculate n = SF(2 m ) k = n: Reduce according to Rule 1 Divide by S n If remainder is zero, F = F n S n, else Continue poly = 4x 2 + 4x in Z n = SF(2 3 ) = k = 4: Divide by S 4 Degree (poly) = 2 < degree(S 4 ) = 4 quo = 0, rem = 4x 2 + 4x F 4 = 0; Continue Example 1 F = F n S n + Σ n-1 a k b k S k k=0

Algorithm Reduce according to Rule 2. Divide by S n-1 to S 0 Check if quotient is a multiple of b k = 2 m /gcd(k!,2 m ) If remainder is zero, stop. Else, continue k = 3: Divide by S 3 degree (poly) = 2 < degree(S 3 ) = 3 quo= 0, rem = 4x 2 + 4x continue k = 2: Divide by S 2 quo = 4; rem = 0 b 2 = 2 3 /gcd(2!,2 3 ) = 4 a 2 = quo/ b 2 = 1 Є Z Example 1 poly = a 2.b 2.S 2 = 1.4.(x+1)(x+2) ≡ 0 in Z 2 3 F = F n S n + Σ n-1 a k b k S k k=0

Algorithm Reduce according to Rule 2. Divide by S n-1 to S 0 Check if quotient is a multiple of b k = 2 m /gcd(k!,2 m ) If remainder is zero, stop. Else, continue k = 3: Divide by S 3 degree (poly) = 2 < degree(S 3 ) = 3 quo= 0, rem = 4x 2 + 4x continue k = 2: Divide by S 2 quo = 4; rem = 0 b 2 = 2 3 /gcd(2!,2 3 ) = 4 a 2 = quo/ b 2 = 1 Є Z Example 1 poly = a 2.b 2.S 2 = 1.4.(x+1)(x+2) ≡ 0 in Z 2 3 F = F n S n + Σ n-1 a k b k S k k=0

Example 2   poly = 5x 2 + 3x + 7 in Z 2 3   n = SF(2 3 ) = 4   degree (poly) = 2 < n. Skip Rule 1 and goto Rule 2   Divide by S 2 quo = 5; rem = 5 + 4x b 2 = 2 3 /gcd(2!,2 3 ) = 4 a 2 = quo/ b 2 is not in Z   poly does not satisfy Rule 2   poly is not a vanishing polynomial in Z 2 3

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(Z 2 m )

Results

Applications to Synthesis  Datapath size ( m ): 8 bits  SF(2 8 ) = 10  Polynomial can be factorized into S 10 (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) % 2 m ≡ g(x) % 2 m is transformed into f(x) - g(x) ≡ 0 % 2 m   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?