Alebraic division: { Let f dividend = ac + ad + bc + bd + e and f divisor = a + b { Then f quotient = c + d f remainder = e { Because (a + b) (c + d)

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1 ALGEBRAIC METHODS Outline cgiovanni De Micheli Stanford University Alebraic model. Division and substitution. Kernel theory. { Kernel and cube extraction. Decomposition. Boolean alebra: { Complement. Alebraic model Alebraic division Given two alebraic expressions: { Symmetric distribution laws. { Don't care sets. Alebraic methods: { Boolean functions!polynomials. { Expressions (sum of product forms). f quotient = f dividend =f divisor when: { f dividend = f divisor f quotient + f remainder { f divisor f quotient 6= 0 { and the support of f divisor and f quotient is disjoint.

2 Alebraic division: { Let f dividend = ac + ad + bc + bd + e and f divisor = a + b { Then f quotient = c + d f remainder = e { Because (a + b) (c + d) + e = f dividend and fa b \ fc d =. Non-alebraic division: { Let f i = a + bc and f j = a + b. An alorithm for division A = fcj A j = 1 2 ::: l set of cubes (monomials) of the dividend. B = fci B i = 1 2 ::: n set of cubes (monomials) of the divisor. Quotient Q and remainder R are sum of cubes (monomials). { Then (a + b) (a + c) = f i but fa b \ fa c 6=. An alorithm for division ALGEBRAIC DIV ISION(A B) f for (i = 1 to n) f D = fc A j such that C A j CB i if ( D == ) return( A) D i = D with var. in sup(c B i ) dropped if i = 1 Q = D i else Q = Q \ D i R = A ; Q B return(q R) f dividend = ac + ad + bc + bd + e f divisor = a + b A = fac ad bc bd eand B = fa b. i = 1: { C B 1 = a, D = fac ad and D 1 = fc d. { Then Q = fc d. i = 2 = n: { C B 2 = b, D = fbc bd and D 2 = fc d. { Then Q = fc d \ fc d = fc d. Result: { Q = fc d and R = fe. f quotient = c + d and f remainder = e.

3 Theorem Substitution Consider expression pairs. Given f i and f j, then f i =f j is empty when: { f j contains a variable not in f i. { f j contains a cube whose support is not contained in that of any cube of f i. { f j contains more terms than f i. { The count of any variable in f j than in f i. Apply division (in any order). If quotient is not void: { Evaluate area/delay ain { Substitute f dividend by jf quotient +f remainder where j = f divisor. Use lters to reduce divisions. Substitution alorithm Extraction SUBSTITUTE( G n (V E) )f for (i = 1 2 ::: jv j) f for (j = 1 2 ::: jv j j 6= i) f A = set of cubes of f i B = set of cubes of f j if (A B pass the lter test ) f (Q R) = ALGEBRAIC DIV ISION(A B) if (Q 6= ) f f quotient = sum of cubes of Q f remainder = sum of cubes of R if ( substitution is favorable) f i = j f quotient + f remainder Search for common sub-expressions: { Sinle-cube extraction: monomial. { Multiple-cube (kernel) extraction. Search for appropriate divisors.

4 Denitions Cube-free expression: { Cannot be factored by a cube. Kernel of an expression: f x = ace + bce + de + Divide f x by a. Get ce. Not cube free. Divide f x by b. Get ce. Not cube free. Divide f x by c. Get ae + be. Not cube free. Divide f x by ce. Get a + b. Cube free. Kernel! { Cube-free quotient of the expression divided by a cube, called co-kernel. Divide f x by d. Get e. Not cube free. Divide f x by e. Get ac + bc + d. Cube free. Kernel! Kernel set K(f) of an expression: Divide f x by. Get 1. Not cube free. Expression f x is a kernel of itself because cube free. { Set of kernels. K(f x ) = f(a + b) (ac + bc + d) (ace + bce + de + ). Theorem (Brayton and McMullen) Two expressions f a and f b have a common multiple-cube divisor f d if and only if: f x = ace + bce + de + f y = ad + bd + cde + e f z = abc { there exist kernels k a 2 K(f a ) and k b 2 K(f b ) s.t. f d is the sum of 2 (or more) cubes in k a \ k b. Consequence: { If kernel intersection is void, then the search for common sub-expression can be dropped. K(f x ) = f(a + b) (ac + bc + d) (ace + bce + de + ). K(f y ) = f(a + b + ce) (cd + ) (ad + bd + cde + e). The kernel set of f z is empty. Select intersection (a + b) f w = a + b f x = wce + de + f y = wd + cde + e f z = abc

5 Naive method: Kernel set computation { Divide function by elements in power set of its support set. { Weed out non cube-free quotients. Smart way: { Use recursion: Kernels of kernels are kernels. { Exploit commutativity of multiplication. Recursive kernel computation simple alorithm R KERNELS(f)f K = foreach variable x 2 sup(f) f if(jcubes(f x)j 2) f f C = larest cube containin x, s.t. CUBES(f C) = CUBES(f x) K = K [ R KERNELS(f=f C ) K = K [ f return(k) CUBES(f C)f return the cubes of f whose support C Analysis Recursive kernel computation Some computation may be redundant: { : Divide by a and then by b. Divide by b and then by a. { Obtain duplicate kernels. Improvement: { Keep a pointer to literals used so far. KERNELS(f j)f K = for i = j to n f if(jcubes(f x i )j 2) f f C = larest cube containin x, s.t. CUBES(f C) = CUBES(f x i ) if (x k 2n C 8k < i) K = K [ KERNELS(f=f C i+ 1) K = K [ f return(k)

6 f = ace + bce + de + Literals a or b. No action required. Literal c. Select cube ce: { Recursive call with aruments: (ace + bce)=ce = a + b pointer j = { Call considers variables fd e. No kernel. { Adds a + b to the kernel set at the last step. Literal d. No action required. Literal e. Select cube e: { Recursive call with aruments: ac + bc + d and pointer j = { Call considers variable f. No kernel. { Adds ac + bc + d to the kernel set at the last step. Literal. No action required. Adds ace + bce + de + to the kernel set. K = f(ace + bce + de + ) (ac + bc + d) (a + b). Matrix representation of kernels Boolean matrix: { Rows: cubes. Columns: variables. Rectanle (R C): { Subset of rows and columns with all entries equal to 1. Prime rectanle: { Rectanle not inside any other rectanle. Co-rectanle (R C 0 ) of a rectanle (R C): { C 0 are the columns not in C. A co-kernel corresponds to a prime rectanle with at least two rows. f x = ace + bce + de + var a b c d e cube RnC ace bce de x = ace + bce + de + s = cde + b Sinle-cube extraction t = ce x = t (a+b) +de + s = td + b Rectanle (prime): (f1 2 f3 5) { Co-kernel ce. Co-rectanle: (f1 2 f ). { Kernel a + b.

7 Sinle-cube extraction Form auxiliary function: { Sum of all local functions. Form matrix representation: { A rectanle with two rows represents a common cube. { Best choice is a prime rectanle. Use function ID for cubes: { Cube intersection from dierent functions. Expressions: { f x = ace + bce + de + { f s = cde + b Auxiliary function: { f aux = ace + bce + de + + cde + b Matrix: var a b c d e cube ID RnC ace x bce x de x x cde s b s Prime rectanle: (f1 2 5 f3 5) Extract cube ce. Cube extraction alorithm Multiple-cube extraction CUBE EXTRACT( G n (V E) )f while (some favorable common cube exist) f C = select common cube to extract Generate new label l Add to network v l and f l = f C Replace all functions f, where f l is a divisor, by l f quotient + f remainder x = ace+bce+de+ y = ad+bd+cde+e z = abc w = a+ b x = wce+de+ y = wd+cde+e z = abc

8 Multiple-cube extraction We need a kernel/cube matrix. Relabelin: { Cubes by new variables. { Kernels by cubes. Form auxiliary function: { Sum of all kernels. Extend cube intersection alorithm. f p = ace + bce. { K(f p ) = f(a + b). f q = ae + be + d. { K(f q ) = f(a + b) (ae + be + d). Relabelin: { x a = a x b = b x ae = ae x be = be x d = d K(f p ) = ffx a x b K(f q ) = ffx a x b fx ae x be x d. (2) f aux = x a x b + x a x b + x ae x be x d. Co-kernel: x a x b. { x a x b corresponds to kernel intersection a+ b. { Extract a + b from f p and f q. Kernel extraction alorithm KERNEL EXT RACT ( G n (V E), n k)f while (some favorable common kernel intersection exist) f Compute kernel set of level k for (i = 1 to n) f Compute kernel intersections f = select kernel intersection to extract Generate new label l Add v l to the network with expression f l = f Replace all functions f where f l is a divisor by l f quotient + f remainder

9 Decomposition Decomposition Dierent ways: x = ace + bce + de + t = ac+bc+d x = te + s = a+b t = sc+d x = te + { Method of Ashenhurst and Curtis. { NAND/NOR decomposition. Kernel-based decomposition: { Divide expression recursively. f x = ace + bce + de + Select kernel ac + bc + d. Decompose: f x = te + Recur on the quotient f t : { Select kernel a + b: f t = ac + bc + d { Decompose: f t = sc + d f s = a + b Decomposition alorithm DECOMPOSE( G n (V E), k)f repeat f v x = selected vertex with expression whose size is above k if (v x = ) return decompose expression f x

10 Summary Alebraic transformations View Boolean functions as alebraic expression. Fast manipulation alorithms. Some optimality lost, because Boolean properties are nelected. Useful to reduce lare networks.