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ECE 697B (667) Fall 2004 Synthesis and Verification of Digital Systems

ECE 697B (667) Fall 2004 Synthesis and Verification of Digital Systems. Two-level Boolean Functions SOP Representation. Slides adopted (with permission) from A. Kuehlmann, UC Berkeley 2003. Representation of Boolean Functions. Sum of Products:

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ECE 697B (667) Fall 2004 Synthesis and Verification of Digital Systems

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  1. ECE 697B (667)Fall 2004Synthesis and Verificationof Digital Systems Two-level Boolean Functions SOP Representation Slides adopted (with permission) from A. Kuehlmann, UC Berkeley 2003

  2. Representation of Boolean Functions • Sum of Products: • A function can be represented by a sum of cubes (products): f = ab + ac + bc Since each cube is a product of literals, this is a “sum of products” (SOP) representation • A SOP can be thought of as a set of cubes F F = {ab, ac, bc} • A set of cubes that represents f is called a cover of f. F1={ab, ac, bc} and F2={abc,abc’,ab’c,ab’c’,bc} are covers of f = ab + ac + bc. ECE 667 - Synthesis & Verification - Lecture 3/4

  3. SOP • Covers (SOP’s) can efficiently represent many practical logic functions (i.e. for many, there exist small covers). • Two-level minimization seeks the minimum size cover (least number of cubes) ac bc = onset minterm Note that each onset minterm is “covered” by at least one of the cubes! None of the offset minterms is covered ab c b a ECE 667 - Synthesis & Verification - Lecture 3/4

  4. Irredundant Cubes • Let F = {c1, c2, …, ck} be a cover for f, i.e. f = ik=1ci A cube ci F is irredundant if F\{ci}  f Example: f = ab + ac + bc ac bc bc Not covered ab ac c b F\{ab}  f a ECE 667 - Synthesis & Verification - Lecture 3/4

  5. Prime Cubes • A literal j of cube ci F ( =f ) is prime if (F \ {ci })  {c”i }  f where c”i is ci with literal j of ci deleted. • A cube of F is prime if all its literals are prime. Example f = ab + ac + bc ci = ab; c”i = a (literal b deleted) F \ {ci }  {c”i } = a + ac + bc F=ac + bc + a = F \{ci }  {c”i } bc a ac c b Not equal to f since offset vertex is covered a ECE 667 - Synthesis & Verification - Lecture 3/4

  6. Definitions - Prime, Irredundant, Essential • Definition 1A cube is primeif it is not contained in any other cube. A cover is primeif all its cubes are prime. • Definition 2A cover is irredundant if all its cubes are irredundant. • Definition 3A prime of f is essential (essential prime) if there is a minterm (essential vertex) in that prime that is in no other prime. • Definition 4 Two cubes are orthogonal if they do not have any minterm in common • E.g. c1= ab c2 = b’c are orthogonal c1= ab’ c2 = bc are not orthogonal ECE 667 - Synthesis & Verification - Lecture 3/4

  7. abc bd c b d cd a Prime and Irredundant Covers Example: f = abc + b’d + c’d is prime and irredundant. abc is essential since abcd’abc, but not in b’d or c’d or ad Why is abcd not an essential vertex of abc? What is an essential vertex of abc? What other cube is essential? What prime is not essential? abcd abcd’ ECE 667 - Synthesis & Verification - Lecture 3/4

  8. n=3, m=3 a a b b c c f1 f2 f3 PLA’s - Multiple Output Functions • A PLA is a function f : Bn Bm represented in SOP form: Personality Matrix abc f1f2f3 10- 1 - - -11 1 - - 0-0 - 1 - 111 - 1 1 00- - - 1 ECE 667 - Synthesis & Verification - Lecture 3/4

  9. PLA’s (cont.) • Each distinct cube appears just once in the AND-plane, and can be shared by (multiple) outputs in the OR-plane, e.g., cube (abc). • Extensions from single output to multiple output minimization theory are straightforward. • Multi-level logic can be viewed mathematically as a collection of single output functions. ECE 667 - Synthesis & Verification - Lecture 3/4

  10. f = abc + abc fa = bc c c b b a a Shannon (Boole) Cofactors Let f : Bn B be a Boolean function, and x= (x1, x2, …, xn) the variables in the support of f. The cofactor fa of f w.r.t literal a=xi or a=x’i is: fxi (x1, x2, …, xn) = f (x1, …, xi-1, 1, xi+1,…, xn) fx’i (x1, x2, …, xn) = f (x1, …, xi-1, 0, xi+1,…, xn) The computation of the cofactor is a fundamental operation in Boolean reasoning ! Example: ECE 667 - Synthesis & Verification - Lecture 3/4

  11. Generalized Cofactor • The generalized cofactor fC of f by a cube C is f with the fixed values indicated by the literals of C, e.g. if C=xi x’j, then xi =1, and xj =0. • if C= x1 x’4 x6 fC is just the function f restricted to the subspace where x1 =x6 =1 and x4 =0. • As a function, fC does not depend on x1,x4 or x6 anymore (However, we still consider fC as a function of all n variables, it just happens to be independent of x1,x4 and x6). • x1f  fx1 Example: f= ac + a’c , af = ac, fa=c ECE 667 - Synthesis & Verification - Lecture 3/4

  12. Fundamental Theorem Theorem:Let c be a cube and f a function. Then c  f  fc  1. Proof.We use the fact that xfx = xf, and fx is independent of x. If : Suppose fc  1. Then cf=fcc=c. Thus, c  f. f c ECE 667 - Synthesis & Verification - Lecture 3/4

  13. C=xz m= 000 m’= 101 m’ Proof (cont.) Only if. Assume c  f Then c  cf = cfc. But fc is independent of literals i  c. If fc 1, then  m  Bn, fc(m)=0. We will construct a m’ from m and c in the following manner: mi’=mi, if xic and xic, mi’=1, if xi  c, mi’=0, if xi  c. i.e. we made the literals of m’ agree with c, i.e. m’  c Þ c(m’)=1 Also, fc is independent of literals xi,xi  c Þ fc(m’) =fc(m) = 0 Þ fc(m’) c(m’)= 0 contradicting c  cfc. m ECE 667 - Synthesis & Verification - Lecture 3/4

  14. Cofactor of Covers Definition: The cofactor of a cover F is the sum of the cofactors of each of the cubes of F. Note: If F={c1, c2,…, ck} is a cover of f, then Fc= {(c1)c, (c2)c,…, (ck)c} is a cover of fc. Suppose F(x) is a cover of f(x), i.e. Then for 1jn, is a cover of fxj(x) ECE 667 - Synthesis & Verification - Lecture 3/4

  15. Cofactor of Cubes Definition: The cofactor Cxj of a cube C with respect to a literal xj is • C if xj and x’j do not appear in C • C\{xj} if xj appears positively in C, i.e. xjC •  if xj appears negatively in C, i.e. xjC Example 1: C = x1 x’4 x6, Cx2 = C (x2 and x2 do not appear in C ) Cx1 = x’4 x6 (x1 appears positively in C) Cx4 =  (x4 appears negatively in C) Example 2: F = abc + b’d + c’d Fb = ac + c’d (Just drop b everywhere and throw away cubes containing literalb) ECE 667 - Synthesis & Verification - Lecture 3/4

  16. Shannon Expansion f : Bn B Shannon Expansion: Theorem: F is a cover of f. Then We say that f (F) is expanded about xi. xi is called the splitting variable. ECE 667 - Synthesis & Verification - Lecture 3/4

  17. ac bc ab c c b b a a Shannon Expansion (cont.) Example Cube bc got split into two cubes ECE 667 - Synthesis & Verification - Lecture 3/4

  18. List of Cubes (Cover Matrix) We often use a matrix notation to represent a cover: Example:F = ac + c’d = a b c d a b c d ac 1 2 1 2 or 1 - 1 - c’d 2 2 0 1 - - 0 1 • Each row represents a cube • 1 means that the positive literal appears in the cube • 0 means that the negative literal appears in the cube • The 2 (or -) here represents that the variable doesnot appearin the cube. It implicitly represents both 0 and 1 values. ECE 667 - Synthesis & Verification - Lecture 3/4

  19. Operations on Lists of Cubes • AND operation: • take two lists of cubes • computes pair-wise AND between individual cubes and put result on new list • represent cubes as pairs of computer words • set operations are implemented as bit-vector operations Algorithm AND(List_of_Cubes C1,List_of_Cubes C2) { C = Æ foreach c1Î C1 { foreach c2Î C2 { c = c1 Ç c2 C = C È c } } return C } ECE 667 - Synthesis & Verification - Lecture 3/4

  20. Operations on Lists of Cubes • OR operation: • take two lists of cubes • computes union of both lists • Naive implementation: • On-the-fly optimizations: • remove cubes that are completely covered by other cubes • complexity is O(m2); m is length of list • conjoin adjacent cubes • remove redundant cubes? • complexity is O(2n); n is number of variables • too expensive for non-orthogonal lists of cubes Algorithm OR(List_of_Cubes C1, List_of_Cubes C2) { return C1È C2 } ECE 667 - Synthesis & Verification - Lecture 3/4

  21. Operations on Lists of Cubes • Simple trick: • keep cubes in lists orthogonal • check for redundancy becomes O(m2) • but lists become significantly larger (worst case: exponential) Example: 01-0 01-0 0-1- 1-01 1-01 1-11 001- 0111 1-11 OR = ECE 667 - Synthesis & Verification - Lecture 3/4

  22. Operations on Lists of Cubes • Adding cubes to orthogonal list: • How can the above procedure be further improved? • What about the AND operation, does it gain from orthogonal cube lists? Algorithm ADD_CUBE(List_of_Cubes C, Cube c) { if(C = Æ) return {c} c’ = TOP(C) Cres =c-c’ /* chopping off minterms */ foreach cresÎCres { C = ADD_CUBE(C\{c’},cres) È {c’} } return C } ECE 667 - Synthesis & Verification - Lecture 3/4

  23. Operation on Lists of Cubes • Naive implementation of COMPLEMENT operation • apply De’Morgan’s law to SOP • complement each cube and use AND operation • Example: • Naive implementation of TAUTOLOGY check • complement function using the COMPLEMENT operator and check for emptiness • We will show that we can do better than that ! Input non-orth. orthogonal 01-10 => 1---- => 1---- -0--- 00--- ---0- 01-0- ----1 01-11 ECE 667 - Synthesis & Verification - Lecture 3/4

  24. A Possible Solution? • Let A be an orthogonal cover matrix. Let all cubes of A be pair-wise distinguished by at least two literals (this can be achieved by an on-the-fly resolution of cube pairs that are distinguished by only literal). Does the following conjecture hold? A  1 Û A has a row of all “-”s ? This would dramatically simplify the tautology check!!! ECE 667 - Synthesis & Verification - Lecture 3/4

  25. Generic Tautology Check Algorithm CHECK_TAUTOLOGY(List_of_Cubes C) { if(C == Æ) return FALSE; if(C == {-...-})return TRUE; // cube with all ‘-’ xi = SELECT_VARIABLE(C) C0 = COFACTOR(C,^Xi) if(CHECK_TAUTOLOGY(C0) == FALSE) { print xi = 0 return FALSE; } C1 = COFACTOR(C,Xi) if(CHECK_TAUTOLOGY(C1) == FALSE) { print xi = 1 return FALSE; } return TRUE; } ECE 667 - Synthesis & Verification - Lecture 3/4

  26. Improvements • Variable ordering: • pick variable that minimizes the two sub-cases (“-”s get replicated into both cases) • Quick decision at leaf: • return TRUE if C contains at least one complete “-” cube among others (case 1) • return FALSE if number of minterms in onset is < 2n (case 2) • return FALSE if C contains same literal in every cube (case 3) ECE 667 - Synthesis & Verification - Lecture 3/4

  27. -1-0 --10 ---- --10 --11 ---0 ---- ---- Example -1-0 --10 1-11 0--- x1’ Tautology (case 1) x1 -1-0 --10 --11 x2’ No tautology (case 3) x2 No tautology (case 3) x3’ ---0 --10 --11 tautology(case 1) x4 x3 ---0 ---1 x4’ tautology(case 1) ECE 667 - Synthesis & Verification - Lecture 3/4

  28. Some Special Functions Definition: A function f : Bn B issymmetricwith respect tovariables xi and xjiff f(x1,…,xi,…,xj,…,xn) = f(x1,…,xj,…,xi,…,xn) Definition: A function f : Bn B istotally symmetric iff any permutation of the variables in f does not change the function • Symmetry can be exploited in searching the binary decision tree • because: • - That means we can skip one of four sub-cases • - used in automatic variable ordering for BDDs ECE 667 - Synthesis & Verification - Lecture 3/4

  29. Unate Functions Definition:A function f : Bn B ispositive unateinvariable xiiff This is equivalent tomonotone increasingin xi: for all min-term pairs (m-, m+) where For example, m-3=1001, m+3=1011(where i=3) ECE 667 - Synthesis & Verification - Lecture 3/4

  30. Unate Functions Similarly fornegative unate monotone decreasing: A function isunatein xi if it is either positive unate or negative unate in xi. Definition:A function isunateif it is unate in each variable. Definition:A cover F ispositive unatein xi iff xi  cj for all cubes cjF ECE 667 - Synthesis & Verification - Lecture 3/4

  31. Example positive unate in a,b negative unate in c m+ c f(m-)=1  f(m+)=0 b m- a ECE 667 - Synthesis & Verification - Lecture 3/4

  32. The Unate Recursive Paradigm • Key pruning technique is based on exploiting the properties ofunatefunctions • based on the fact that unate leaf cases can be solved efficiently • New case splitting heuristic • splitting variable is chosen so that the functions at lower nodes of the recursion tree become unate ECE 667 - Synthesis & Verification - Lecture 3/4

  33. The Unate Recursive Paradigm Unate covers F have many extraordinary properties: • If a cover F is minimal with respect to single-cube containment, all of its cubes are essential primes. • In this case F is the unique minimum cube representation of its logic function. • A unate cover represents the tautology iff it contains a cube with no literals, i.e. a single tautologous cube. This type of implicit enumeration applies to many sub-problems (prime generation, reduction, complementation, etc.). Hence, we refer to it as the Unate Recursive Paradigm. ECE 667 - Synthesis & Verification - Lecture 3/4

  34. a merge b c Unate Recursive Paradigm • Create cofactoring tree stopping atunate covers • choose, at each node, the “most binate” variable for splitting • recurse till no binate variable left (unate leaf) • “Operate” on the unate cover at each leaf to obtain the result for that leaf. Return the result • At each non-leaf node,merge(appropriately) the results of the two children. • Main idea: “Operation” on unate leaf is computationally less complex • Operations: complement, simplify,tautology,generate-primes,... ECE 667 - Synthesis & Verification - Lecture 3/4

  35. The Binate Select Heuristic Tautology and other programs based on the unate recursive paradigm use a heuristic calledBINATE_SELECTto choose the splitting variable in recursive Shannon expansion. The idea is for a given cover F, choose the variable which occurs, both positively and negatively, most often in the cubes of F. ECE 667 - Synthesis & Verification - Lecture 3/4

  36. The Binate Select Heuristic Example Unate and non-unate covers: a b c d G = ac+cd’ 1 - 1 - - - 1 0 a b c d F = ac+c’d+bcd’ 1 - 1 - - - 0 1 - 1 1 0 => Choose c for splitting! The binate variables of a cover are those with both 1’s and 0’s in the corresponding column. • In the unate recursive paradigm, the BINATE_SELECT heuristic chooses a (most) binate variable for splitting, which is thus eliminated from the sub-covers. is unate is not unate ECE 667 - Synthesis & Verification - Lecture 3/4

  37. Example c 1 0 1 - 1 - F= - - 0 1 - 1 1 0 1--- -1-0 unate FC FC ---1 unate c 1 0 1 - 1 - 0 F= - - 0 1 - - 1 1 0 1 1---0 -1-01 ---1- unate e 0 1 -1-0- unate 1---- unate ECE 667 - Synthesis & Verification - Lecture 3/4

  38. Two Useful Theorems Theorem: Theorem:Let A be a unate cover matrix. Then A1 if and only if A has a row of all “-”s. Proof: If. A row of all “-”s is the tautology cube. Only if. Assume no row of all “-”s. Without loss of generality, suppose function is positive unate. Then each row has at least one “1” in it. Consider the point (0,0,…,0). This is not contained in any row of A. Hence A1. ECE 667 - Synthesis & Verification - Lecture 3/4

  39. Unate Reduction Let F(x) be a cover. Let (a,x’) be a partition of the variables x, and let where • the columns of A correspond to variables a of x • T is a matrix of all “-”s. Theorem:Assume A 1. Then F1  F’1 ECE 667 - Synthesis & Verification - Lecture 3/4

  40. Example We pick for the partitioning unate variables because it is easy to decide that A¹1 ECE 667 - Synthesis & Verification - Lecture 3/4

  41. Unate Reduction Result: Only have to look at D1 to test if this is a tautology. Note: A1, A2 has no row of all “-”s. Hence is a unate cover. Hence (A1, A2)1 A1 B1 C2 C1 D1 B2 A2 ECE 667 - Synthesis & Verification - Lecture 3/4

  42. Proof Theorem: Assume A 1. Then F1F’1 Proof: if: Assume F’1. Then we can replace F’ by all -’s. Then last row of F becomes a row of all “-”s, so tautology. A1 T=“-”s ECE 667 - Synthesis & Verification - Lecture 3/4

  43. Proof (contd) Only if: Assume F’ 1. Then there is a minterm m2 such that F’(m2)=0, i.e. m2cube of F’. Similarly, m1 exists where A(m1)=0, i.e. m1cube of A. Now the minterm (m1,m2) in the full space satisfies F(m1,m2)=0 since m1m2 AX and m1m2TF’. (a, x’) is any row of first part a(m1) ^ x’(m2)=0 ^ x’(m2)=0 (t,f’) is any row of the last part t(m1) ^ f’(m2)=t(m1) ^ 0 = 0 So m1m2 is not in any cube of F. ECE 667 - Synthesis & Verification - Lecture 3/4

  44. Improved Tautology Check Algorithm CHECK_TAUTOLOGY(List_of_Cubes C) { if(C == Æ) return FALSE; if(C == {-...-})return TRUE; // cube with all ‘-’ C = UNATE_REDUCTION(C) xi = BINATE_SELECT(C) C0 = COFACTOR(C,^xi) if(CHECK_TAUTOLOGY(C0) == FALSE) { return FALSE; } C1 = COFACTOR(C,xi) if(CHECK_TAUTOLOGY(C1) == FALSE) { return FALSE; } return TRUE; } ECE 667 - Synthesis & Verification - Lecture 3/4

  45. -1-0 --10 1-11 0--- x1 -1-0 --10 ---- -1-0 --10 --11 x2 --10 --11 ---0 --10 --11 ---0 x3 ---- ---0 ---1 ---- Previous Example No tautology(case 2) Unate reduction 0--- x1’ Tautology (case 1) x2’ No tautology (case 3) x3’ No tautology (case 3) x4 Tautology (case 1) x4’ Tautology (case 1) ECE 667 - Synthesis & Verification - Lecture 3/4 ^x4

  46. Recursive Complement Operation • We have shown how tautology check (SAT check) can be implemented recursively using the Binary Decision Tree • Similarly, we can implement Boolean operations recursively • Example COMPLEMENT operation: Theorem: Proof: ECE 667 - Synthesis & Verification - Lecture 3/4

  47. COMPLEMENT Operation Algorithm COMPLEMENT(List_of_Cubes C) { if(C contains single cube c) { Cres = complement_cube(c) // generate one cube per return Cres // literal l in c with ^l } else { xi = SELECT_VARIABLE(C) C0 = COMPLEMENT(COFACTOR(C,^xi)) Ù ^xi C1 = COMPLEMENT(COFACTOR(C,xi)) Ù xi return OR(C0,C1) } } ECE 667 - Synthesis & Verification - Lecture 3/4

  48. Complement of a Unate Cover • Complement of a unate cover can be computed more efficiently • Idea: • variables appear only in one polarity on the original cover (ab + bc + ac) = (a+b)(b+c)(a+c) • when multiplied out, a number of products are redundant aba + abc + aca + acc + bba + bbc + bca + bcc = ab + ac + bc • we just need to look at the combinations for which the variables cover all original minterms • this works independent of the polarity of the variables because of symmetry to the (1,1,1,…,1) case ECE 667 - Synthesis & Verification - Lecture 3/4

  49. Complement on a Unate Cover • Map cube matrix F into Boolean matrix B convert: “0”,”1” to “1” (literal is present) “-” to “0” (literal is not present) ECE 667 - Synthesis & Verification - Lecture 3/4

  50. Complement of a Unate Cover Find all minimal column covers of B. • A column cover is a set of columns J such that for each row i,  jJ such that Bij = 1 Example: {1,4} is aminimal column coverfor All rows “covered” by at least one 1. ECE 667 - Synthesis & Verification - Lecture 3/4

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