Multilevel Logic Synthesis -- Algebraic Methods

1. Multilevel Logic Synthesis-- Algebraic Methods

2. Why Algebraic Methods? • What is our goal? minimization, maximally factored form. • Algebraic methods provide fast algorithms. • Algebraic methods treat logic function like a polynomial • Fast algorithms for polynomial manipulation exist. • Algebraic methods may lose the optimality, but the results are quite good in general. • Algebraic methods can be used to iterate and interleave with Boolean operations. ENEE 644

3. Division: Outline • What is Division? • 7 = 2 x 3 + 1 • Divisor: what to divide with? • Quotient and remainder: how to divide? • F = G · H + R • Why Division? Division is central in many operations: • factoring and decomposition • resubstitution • extraction ENEE 644

4. Product • The product of two algebraic expressions F = X1+X2+…+Xm and G = Y1+Y2+…+Yn is the SOP expression obtained by simplifying the result of (X1+X2+…+Xm)(Y1+Y2+…+Yn) using xx’=0 and xx=x. • Example: (a+b)(a+c) = a + ab + ac + bc • The product FG is algebraic if F and G are orthogonal, otherwise, it is called a Boolean product. • Example: (a+b)(c+d) = ac + ad + bc + bd (algebraic) (a+b)(a+c) = aa+ab+ac+bc (Boolean) ENEE 644

5. Recall: Algebraic and Boolean Expressions • f is an algebraic expression if f is a set of cubes (SOP), such that no single cube contains another (minimal with respect to single cube containment). Otherwise, f is called a Boolean expression. • Example: a+bc is algebraic, a+ab is Boolean. • The support of an expression f, supp(f), is the set of variables that f explicitly depends on. Two expressions f and g are said to be orthogonal if supp(f)supp(g)=, denoted byfg. • Example: a+b  c+d ENEE 644

6. Division • Division is an operation that takes two SOP expressions F and P, and generates SOP expressions Q and R, s.t. F = PQ +R. • The division is algebraic is PQ is an algebraic product; otherwise it is a Boolean division. • Q is called the quotient, R is called the remainder. • If R=0, P is a factor, otherwise it is called a divisor. • Example: F = ad + ae + bcd + j • For P = a, Q = d+e, R = bcd+j • For P = a+bc, Q=d, R=ae+j • Both a and a+bc are divisors, not factors. ENEE 644

7. Weak Division • For division F = PQ+R, it is weak if • PQ is an algebraic product; • R has as few cubes as possible; • PQ+R and F are the same expression (same cubes). {Q,R} is unique given F and P, denote Q=F/P. • Example: • For F=ad+abc+bcd, P=a+bc: Q=d, R=abc • For F=ac+ad+bc+bd+e, P=a+b: Q=c+d,R=e • For F=ad+aef+ab+b’cd+b’cef, P=a+b’c • If P=a: Q=d+ef+b • If P=b’c, Q=d+ef Q=d+ef,R=ab ENEE 644

8. Kernel and Co-Kernel: Motivation • We know how to divide a given expression F by another expression G. (division) • How to find G? • Too many Boolean divisors: a+bc’ • Restrict to algebraic divisors, e.g. cubes • Problem: Given a set of functions, find common weak (algebraic) divisors. ENEE 644

9. Kernels: Definitions • An expression is cube-free if no cube divides the expression evenly (i.e. no remainder, or there is no literal that is common to all the cubes). • Example: ab+c is cube-free ab+ac is not (a is a common factor) abc is not (must have two or more cubes) • The primary divisorsof an expression F are the set of expressions D(F) = {F/c | c is a cube}. • Example: F = (a+b+c)(d+e)f + bfg + h df+ef (=F/a) is a primary divisor d+e+g (=F/bf) is a primary divisor ENEE 644

10. Kernels: Definitions (cont’d) • The kernelsof an expression F are K(F) = {G | G  D(F) and G is cube-free}. • Example: F = (a+b+c)(d+e)f + bfg + h df+ef (=F/a)  D(F), but not a kernel. d+e+g (=F/bf)  D(F) and is cube-free. • A cube c used to obtain the kernel K=F/c is called a co-kernel of K. C(F) = {c | F/c  K(F)}. • A kernel is of level 0 if it has no kernels except itself. A kernel of level n has at least one kernel of level n-1, but no kernel (except itself) of level n or greater. ENEE 644

11. Kernels: Example F = adf + aef + bdf + bef + cdf + cef + bfg + h = (a+b+c)(d+e)f + bfg + h ENEE 644

12. Kernel Computation: Cube Intersection Given F = X1+X2+…+Xn, compute K(F) and C(F) • C*(F) = {c | c= XiXjfor some i and j} (This contains all the level-0 co-kernels.) • weak_div(F,c) = F/c is a kernel with c as the corresponding co-kernel. Example: F = abcd + abce + efg • C*(F) = {abc, e} • F/abc = d+e, F/e = abc+fg • Problem: gives only level-0 co-kernels, to get for example level-1 co-kernels, ENEE 644

13. cube primary divisor level ab c+d’ 0 bc a+d’ 0 abc abd’ bcd’ ade cde a’ce a bc+bd’+de 1 abc - c ab+bd’+de+a’e 1 abd’ ab - bcd’ bc bd’ - bd’ a+c 0 ade a a - - de a+c 0 cde c - c de - e ad+cd+a’c 1 a’ce c - c e ce - ce a’+d 0 Kernel Computation: Example Find all the level-0 co-kernels for abc+abd’+bcd’+ade+cde+a’ce How about level-1 kernels? Repeat Above Recursively ENEE 644

14. Fundamental Theorem • Theorem: If two expressions F and G have the property that kFK(F), kGK(G)  | kG  kF |1 (kG and kF have at most one term in common), then F and G have no common nontrivial algebraic divisors (i.e. with at least two terms) else it does. • Importance: If we “kernel” all functions and there are no nontrivial intersections, then the only common algebraic divisors left are single cube divisors. This is used for quickly finding common subfunctions. ENEE 644

15. factoring F = ab+ac+bc F = a(b+c)+bc goal: minimize the number of literals factored form SOP form Factoring and Factoring Algorithm • Factoring is the process of deriving a factored form for a given logic function in SOP form. • A factoring algorithm is algebraic if it guarantees an algebraic factored form from an algebraic SOP expression; otherwise it is Boolean. ENEE 644

16. Generic Factoring Algorithm Factor(F) { if (F has no factor) return F; D = Divisor(F); (Q,R) = Divide(F,D); return Factor(Q)Factor(D) + Factor(R); } Example: F = abc+abd+ae+af+g D = c+d F = DQ+R Q = ab R = ae+af+g = a(e+f) + g (Factor(R)) O = ab(c+d) + a(e+f)+g ENEE 644

17. Recall: + · · a a c b Maximal Factorizations • A factored form is maximally factored if • For every sum of products, there are no two syntactically equivalent factors in the products; • For every product of sums, there are no two syntactically equivalent factors in the sums. • Example: • ab+ac is not maximally factored, a(b+c) is. • (a+b)(a+c) is not maximally factored, a+bc is. ENEE 644

18. Problem with Algorithm Factor • O = ab(c+d) + a(e+f)+g is not optimal since it is not maximally factored. a(b(c+d)+e+f)+g • The problem occurs when • Quotient Q is a single cube, and • Some of the literals of Q appear in the remainder R • Solution to the problem: • If the quotient Q is not a single cube, done; else • Pick a literal l in Q that occurs most frequently in the cubes of F; • Divide F by l to obtain a new divisor D1; F = l D1 + R1, where literal l does not appear in R1. Lemma: partial factored form l D1 + R1 is maximally factored if Q is a single cube ENEE 644

19. Another Example F =ace+ade+bce+bde+cf+df D =a+b Q =ce+de R =cf+df P =QD+R =(ce+de)(a+b)+(cf+df) O =e(c+d)(a+b)+(c+d)f Problem: O=e(c+d)(a+b)+(c+d)f is not maximally factored because it can be further factored to (c+d)(e(a+b)+f). The problem occurs when Q and R have a common factor. (c+d in this case). ENEE 644

20. Solution to the Problem • Make Q cube-free to obtain Q1; • Divide F by Q1 to obtain a new divisor D1; • If D1 is cube-free, we get a partial factored form F=Q1D1+R1 and recursively factor Q1,D1, and R1; • If D1 is not cube-free, let D1=cD2 and D3=Q1D2, we have the partial factoring F=cD3+R1 and can recursivelyfactor D3 and R1. Lemma: if Q is not a single cube, F is maximally factored at this level. ENEE 644

21. Application: Decomposition • Given an expression F and one of its divisors D, the decomposition associates a new variable, G, with the divisor and reduce F to QG+R, where Q=F/G and R is the remainder in the division. • Decomposition vs. factorization • Divisor is added as a new variable. • The new variable may fan out elsewhere in both regular or complement forms. • Why decomposition: reduce the size of the expression (e.g. to typical library cells). ENEE 644

22. decomposition F = Ge+g G=ac+bc+d F=ace+bce+de+g goal: minimize the size of the expression factoring F = ab+ac+bc F = a(b+c)+bc SOP form factored form goal: minimize the number of literals factored form SOP form Example: Decomposition ENEE 644

23. fj fj fi Application: Algebraic Resubstitution • Goal: reduce number of literals • Idea: (Reuse)For multi-output function, one output (or its complement) may be a useful divisor in another output. • Procedure: substitution of Fj into Fi • Divide Fi by Fj: Fi = QFj + R • Divide R by F’j: R = QCF’j+RC • Result: Fi = QFj + QCF’j+RC • In practice: this is tried for each output pair, therefore requires O(n2) divisions when there are n outputs. • Filters:criteria that reduce the number of divisions. For example, don’t try Fi/Fj if 1) Fj contains a literal not in Fi, or 2) Fj has more terms than Fi, or 3) there is a literal occurs more often in Fj than in Fi. ENEE 644

24. 21 literals saving of 12 literals 9 literals Example: Algebraic Resubstitution • F/G: • F/ab’ = d+fg • F/c = d+fg F = (d+fg)G + (a’c’e+bc’e+aef) = QG + R • R/G’: • G’ = (ab’+c)’ = (a’+b)c’ = a’c’ + bc’ • R/a’c’ = e • R/bc’ = e R = eG’ + aef F = (d+fg)G + eG’ + aef (#23 on Page 448) Resubstitution of G=ab’+c into F=ab’d+cd+ab’fg+cfg+a’c’e+bc’e+aef ENEE 644

25. Application: Node Elimination • Idea: replace a literal (and its complement) in an expression by its algebraic SOP form. • Goal: prepare for a different factorization or decomposition. • Elimination value: Fi = QiG + QiCG’ + Ri • Let ni be the number of times that G or G’ appears in the factored form Fi, LG be the number of literals in the factored form for G, then approximately, (niLG-ni)-literals will be added after the elimination. This value is defined as the elimination value of G. • In practice, don’t eliminate node with large e_value. ENEE 644

26. Example: Node Elimination • F = e(af+c’G) + (d+fg)(c+G’) • G = a’+b • e_valueG =(niLG-ni)= 2x2-2 = 2 elimination of G (and G’) ENEE 644

27. Application: Extraction • Idea: identifies common sub-expressions and manipulates the Boolean network. • We can combine decomposition, substitution, and node elimination to provide an effective extraction algorithm: Quick_Extraction • For each node in the network, apply decomposition; • For all profitable pairs, apply algebraic substitution; • Eliminate all single literal functions; • Eliminate all functions with small e_value; ENEE 644

28. Example: Extraction F1 = abc+abd+ae+af+g F2= ace+ade+bce+bde+cf+df (11+16 = 27 literals) Decomposition: O1 = aY1 + g Y1 = bX1 + e + f X1 = c + d O2 = X2Z2 Z2 = eY2 + f Y2 = a + b X2 = c + d (18 literals) Resubstitution: (X1 = X2) O1 = aY1 + g Y1 = bX1 + e + f X1 = c + d O2 = X1Z2 Z2 = eY2 + f Y2 = a + b (16 literals) Node Elimination: (e_valueY2=-1) O1 = aY1 + g Y1 = bX1 + e + f X1 = c + d O2 = X1Z2 Z2 = e(a+b) + f (15 literals) ENEE 644