Exploiting Vanishing Polynomials for Equivalence Verification of Fixed-Size Arithmetic Datapaths GIEE, NTU ALCom Lab Presenter: 陳炳元

Outline Introduction Modeling Univariate Vanishing Polynomials Algorithm of Univariate Multi-varite Vanishing Polynomials Algorithm of Multi-varite Conclusions and Future Work

Outline Introduction Modeling Univariate Vanishing Polynomials Algorithm of Univariate Multi-varite Vanishing Polynomials Algorithm of Multi-varite Conclusions and Future Work

Bit-Vector Arithmetic = %2 m Algebra  Represent integers as a vector of bits  Bit x 0 represents values 0 or 1  Vector X[1:0] = {x 1, x 0 } represents integers  00, 01, 10, 11  Bit-vector of size m : integer values in 0,…, 2 m -1  Vector X[m-1 : 0] represents integers reduced % 2 m

Fixed-Size (m) Data-path: Modeling Control the datapath size: Fixed size bit-vectors ( m ) 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

Anti-Aliasing Function  F 1 [15:0] = 156x x x x x x  F 2 [15:0] = 156x x x x x x Now that polynomially F 1  F 2 because they have different coefficients; But because the datapath size is fixed to 16 bits F 1 [15:0]= F 2 [15:0],or in other words F 1 % 2 16 = F 2 % 2 16

Outline Introduction Modeling Univariate Vanishing Polynomials Algorithm of Univariate Multi-varite Vanishing Polynomials Algorithm of Multi-varite Conclusions and Future Work

Binary Relation Let A be a set,R is a subset of A  A,we say that R is a binary relation on A Let R be a binary relation. We sometimes write aRb for (a,b)  R A binary relation R on A is called reflexive if(x,x)  R for all x  A A binary relation R on A is called symmetric if(x,y)  R  (y,x)  R for all x,y  A A binary relation R on A is called transitive if(x,y)  R and (y,z)  R  (x,z)  R for all x,y,z  A

Equivalence Relation A binary relation R on A is called equivalence relation if it is reflexive,symmetric,transitive A equivalence relation R on A.We define the equivalence class of a,[a]={b  A| (a,b)  R } Example: Let A= ,if (a,b)  R  a  b mod n. Then [0],[1],…,[n-1] are all equivalence class

Partition Let A be a set. A partition of A is a collection of disjoint nonempty subsets of A. The equivalence classes of R on A form a partition of A. Lemma1: Let R be an equivalence relation on A. The following statements are equivalence: (1)(a,b)  R (2)[a]=[b] (3)[a]  [b]  {}

Partition Theorem1: Let R be an equivalence relation on A.P={[a]| a  A}  P is partition of A pf: 顯然 Claim:[a]  [b],then [a]  [b]={} If [a]  [b]  {} By the lemma1,we have [a]=[b]  故 P 為 A 之一 partition

Partition Example: Let A= ,if (a,b)  R  a  b mod n. Then corresponding to equivalence class set  n ={[0],[1],…,[n-1]} Note: We sometimes write  n ={0,1,…,n-1}

Binary Operation Let S be a set.  :S  S  S function,then we say that  is a binary operation on S if  is a binary operation on S then we will write a  b rather than  (a,b). (S,  1,  2,…,  K ) is a set S together with k binary operation  1,  2,…,  K on S

Introduction to Rings A ring (R,+, ‧ ) is a set,R together with two binary operation +, ‧ on R called addition and multiplication,satisfying the following properties: (1)(R,+) is an abelin group. We write the identity element 0 (2)Multiplication is associative,means that a  (b  c)=(a  b)  c  a,b,c  R (3)The left and right distributive laws hold,means that a  (b+c)=a  b+a  c and (a+b)  c=a  c+b  c  a,b,c  R

Introduction to Rings Example: 我們在  上定義一個 equivalence relation, 若 (a,b)  R  a  b mod n 對應之 equivalence class set  n ={[0],[1],…,[n-1]}. 在  n 上定義兩個 binary operation +,  by [a]+[b]=[a+b mod n] 與 [a]  [b]=[ab mod n]  (  n +, ‧ ):ring NOTE: (1)  [a],[b]  n,[a]+[b]=[a+b mod n]=[b+a mod n]=[b]+[a] and [a]  [b]=[ab mod n]=[ba mod n]=[b]  [a] (2)  [a]  n,[a]+[0]=]=[a+0 mod n]=[a mod n]=[a] and [a]  [1]=[a  1 mod n]=[a mod n]=[a]

Introduction to Rings A ring R is called a commutative ring if ab=ba  a,b  R A ring R is called a ring with unity 1 if there is an element 1  0 in R such that 1  a=a=a  1  a  R Example: (1)  Q,R,C:commutative ring with unity 1 where the ring operations are the usual addition and multiplication. (2)Let n be a positive integer. Then the set  n ={0,1,…,n-1},under addition and multiplication modulo n is commutative ring with unity 1.

Introduction to Rings Let (R,+, ‧ ) be a ring,define by R[X]={a n x n +a n-1 x n-1 +· · ·+a 1 x+a 0 |  n ,a i  R,  i=1, 2,...,n} is called the ring of polynomials over R In fact,R[X] is ring Example: (  3,+, ‧ ),f(x)=2x 3 +1, g(x)=x+2  f(x)+g(x)= 2x 3 +x and f(x)  g(x)=2x 4 +x 3 +x+2=2x 4 +2x+2

Outline Introduction Modeling Univariate Vanishing Polynomials Algorithm of Univariate Multi-varite Vanishing Polynomials Algorithm of Multi-varite Conclusions and Future Work

Vanishing Polynomials Vanishing Polynomial f(x)  2 m [X],  x  2 m  f(x)=0 f(x),g(x)  2 m [X] if(f-g): vanishing means that 2 m | (f-g) n! divides a product of n consecutive numbers 4! divides 99 X 100 X 101 X 102 Find least n  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

Basic Number Theory Let f(x),g(x)  2 3 [X], if (f-g): vanishing (1)2 3 |(f - g) in  2 3 (2) 2 3 |4! (3) 4! divides the product of 4 consecutive numbers (4)  x  2 3, 4!|(x+1)(x+2)(x+3)(x+4) Write (f-g) as a product of SF(2 m ) = n consecutive numbers

Basis for factorization  Y 0 (x) = 1  Y 1 (x) = (x + 1)  Y 2 (x) = (x + 1)(x + 2) : Product of 2 consecutive numbers  Y 3 (x) = (x + 1)(x + 2)(x + 3) : Product of 3 consecutive numbers  …  Y n (x) = Y n-1 (x) (x + n) : Product of n consecutive numbers

Basis for factorization Theorem2: S n (x) is vanishing in  2 m [X], where n = SF(2 m ). Example: f = x 4 +2x 3 + 3x 2 + 2x  2 3 [X]; SF(2 3 ) = 4 f can be written as a product of 4 consecutive numbers. f= (x+1)(x+2)(x+3)(x+4) = Y 4 (x) f is a vanishing polynomial.

Constraints on the Coefficient Example: h(x) = 4x 2 + 4x = 4(x+1)(x+2) = 4  S 2 (x)  2 3 [X]; SF(2 3 ) = 4 h(x) =0  x  2 3 h is a vanishing polynomial. But h(x) not equal to S 4 (x) Theorem3: 2 m /gcd(k!, 2 m )  Y k (x)=0 in  2 m [X],0  k  SF(2 m ),2 m /gcd(k!, 2 m ) 為使 2 m /gcd(k!, 2 m )  Y k (x)=0 之 min NOTE: b  Y k (x)=0 in  2 m [X]  2 m /gcd(k!, 2 m ) |b

Constraints on the Coefficient Example: h(x) = 4x 2 + 4x = 4(x+1)(x+2) = 4  Y 2 (x)  2 3 [X] 2 3 /gcd(2!, 2 3 ) | 4  h is a vanishing polynomial.

Deciding Vanishing Polynomials Theorem4: Let F be a polynomial in  2 m [X].Then F is vanishes  F = F n Y n + Σ n-1 a k b k Y k  n = SF(2 m ), i.e. the least n such that 2 m |n!  F n is an arbitrary polynomial in  2 m [x]  a k is an arbitrary integer  b k = 2 m /gcd(k!,2 m ) k=0 Theorem3Theorem2

Outline Introduction Modeling Univariate Vanishing Polynomials Algorithm of Univariate Multi-varite Vanishing Polynomials Algorithm of Multi-varite Conclusions and Future Work

Algorithm Procedure zero_Identifi( polynomial, 2 m ) 1.Calculate n = SF(2 m ) 2.k = n: Reduce according to Throrem2 Divide by S n If remainder is zero,then F = F n Y n, else Continue

Algorithm 3.Reduce according to Theorem3. Divide by Y n-1 to Y 0 Check if quotient is a multiple of b k = 2 m /gcd(k!,2 m ) If remainder is zero,then stop. else continue

Example 1 p= 4x 2 + 4x in Z n = SF(2 3 ) = 4 2.k = 4: Divide by Y 4 deg(p) = 2< deg(Y 4 )= 4 quo = 0, rem = 4x 2 + 4x F 4 = 0; Continue 3.k = 3: Divide by S 3 deg(p) = 2<deg(S 3 ) = 3 quo= 0, rem = 4x 2 + 4x continue 4.k = 2: Divide by Y 2 quo = 4; rem = 0 b 2 = 2 3 /gcd(2!,2 3 ) = 4 a 2 = quo/ b 2 =1  Z p is a vanishing polynomial in Z 2 3

Example 2 p= 5x 2 + 3x + 7 in Z n = SF(2 3 ) = 4 2.k = 4: Divide by Y 4 deg(p) = 2< deg(Y 4 ) = 4 quo = 0,rem=5x 2 +3x + 7 F 4 = 0; Continue 3.k = 3: Divide by Y 3 deg(p) = 2<deg(Y 3 ) = 3 quo= 0, rem = 5x 2 +3x + 7 continue 4.k = 2: Divide by Y 2 quo = 5; rem = 4x+5 b 2 = 2 3 /gcd(2!,2 3 ) = 4 a 2 = quo/ b 2 =5/4  Z p is not a vanishing polynomial in Z 2 3

Outline Introduction Modeling Univariate Vanishing Polynomials Algorithm of Univariate Multi-varite Vanishing Polynomials Algorithm of Multi-varite Conclusions and Future Work

Multiple Polynomial Definition: (1) Let R be a ring, define the ring R[x 1,…,x n ] of polynomial in the indeterminates x 1,…,x n recursively as follows: R[x 1,…,x n ] = (R[x 1,…,x n-1 ])[x n ] (2) x 1 i1 …x n in is called monomial, ax 1 i1 …x n in is called monomial term, where a 為此 term 之 coefficient (3) The degree of x 1 i1 …x n in is (i1,…,in)  n denote deg(x 1 i1 …x n in )

Multiple Polynomial If R is a commutative and f(x 1,…,x n )  R[x 1,…,x n ], then the mapping  : R  R given by (a 1,…,a n )  (a 1,…,a n ) The mapping is called polyfunction. In this problem,  :  2 n1  …  2 nd  2 m Definition: Let k=(k 1,k 2,…,k d )  n, we define Y k =y k1 (x 1 ) … y kd (x d ), where y ki (x i ) is the falling factor of degree k i in x i

Example F(x 1, x 2 )=x 1 4 x 2 + 2x 1 3 x 2 + 3x 1 2 x 2 + 2x 1 x 2 over  2 2 [x 1,…,x n ] 1.sf(2 2 )=4 2.F=Y (4,1) (x 1, x 2 )=Y 4 (x 1 )Y 1 (x 2 )=0 in  2 2 We wish to generalize these results to analyze polyfunctions over  2 n1  …  2 nd  2 m

Lemma8 (1)   2 ni then Y 2 ni (x i )=0,  i=1, 2, …, d (2)Let n=sf(2 m ) then Y n (x)=0 Take  i =min{2 ni,sf(2 m )},  i=1, 2, …, d then Y  i   2 ni is vanshing polynomial Lemma: Let k=(k 1,k 2,…,k d )  n, then Y k =0  k i  i, for some i

Example F(x 1, x 2 )=x 1 2 x 2 - x 1 x 2 be a polynomial corresponding to the polyfunction  :  2 1  2 2  2 3. We show that F is a vanshing polynomial. 1.sf(2 3 )=4,  1=2,  2=4 2.x 1 2 x 2 - x 1 x 2 =x 1 (x 1 -1) x 2 =Y (2,1) (x 1, x 2 ) 3.k1   1=2 The condition in lemma is satisfied, Hence F is vanshing polynomial

Constraints on the Coefficient Theorem8: The expression c k  Y k =0 in  2 m [X]  2 m /gcd(  i=1 k i !, 2 m ) | c k  c k  2 m  k=(k 1,k 2,…,k d )  n such that k i  i,  i=1,2,…,d d

Example F(x 1, x 2 )=4x 1 x x 1 x 2 be a polynomial corresponding to the polyfunction  :  2 1  2 2  2 3. We show that F is a vanshing polynomial. 1.2 n1 =2, 2 n2 =4, sf(2 3 )=4,  1=2,  2=4, 2.4x 1 x x 1 x 2 =4x 1 x 2 (x 1 -1) = c (1,2) Y (1,2) (x 1, x 2 ) 3.8/gcd( 1!x2!, 8) | c (1,2) 且 k1  1=2, k2  2=4 The condition in lemma is satisfied, Hence F is vanshing polynomial

Deciding Vanishing Polynomials Theorem: Let F be a polynomial representation for the function  from  2 n1  …  2 nd to  2 m. Then F is vanishes  F =  i=1 Q i Y  (i) + Σa k b k Y k  n = SF(2 m ), i.e. the least n such that 2 m |n!  k=(k 1,k 2,…,k d )  n, for each k i =1,2,…,  i  Y  (i) is falling factor of degree  i  Q i is an arbitrary polynomial  a k is an arbitrary integer  b k = 2 m /gcd(k!,2 m ) Theorem8 Lemma8 d

Outline Introduction Modeling Univariate Vanishing Polynomials Algorithm of Univariate Multi-varite Vanishing Polynomials Algorithm of Multi-varite Conclusions and Future Work

Example

Outline Introduction Modeling Univariate Vanishing Polynomials Algorithm of Univariate Multi-varite Vanishing Polynomials Algorithm of Multi-varite Conclusions and Future Work

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