Synthesis of Two-Level Circuits

Synthesis of Two-Level Circuits Define Two-Level Circuit Prime Implicants (PIs) Methods of Finding PIs Selecting PIs Multiple Output Functions Read HS 127-165

Introduction • We now begin applying the ideas from general Boolean algebras to formula minimization • We are interested in circuit implementations that are optimal in some sense • We consider the Boolean function algebra over the switching algebra ({0,1},·) • We will need some sense of optimality • We will initially focus on “two-level” circuits • For such circuits there are at most two gates between inputs and outputs (not counting possible inverters used form complements of the inputs • All Boolean functions over the switching algebra can be implemented by two-level circuits • Such circuits are generally very fast because of the small propagation delay from input to output • Such circuits are very simple to analyze and minimum cost implementations can be found directly • The methods we develop will form a starting point for multi-level circuits

Two-Level Formulas • Definition: A letter is a constant or a variable. A literal is a letter or its complement. • For example, let x1 and x2 be variables defined over the switching algebra. Then 0, 1, x1, and x2 are letters. Also, 0, 1, x1, x10, x2, and x2$ are literals. • Definition: A product term is a formula with one of the following forms • The constant 1 • A nonconstant literal • A conjunction (meet) of nonconstant literals where no letter appears more than once • Definition: A sum term is a formula with one of the following forms • The constant 0 • A nonconstant literal • A disjunction (join) of nonconstant literals where no letter appears more than once • For example, xy0z is a product term, x+y0+z is a sum term, x is both, and x(y+z) is neither

Two-Level Formulas (cont.) • Definition: A sum of products or disjunctive normal formula (abbreviated SOP or ) is a formula with one of the following forms • The constant 1 • A product term • A disjunction (join) of product terms • Definition: A product of sums or conjunctive normal formula (abbreviated POS or  is a formula with one of the following forms • The constant 0 • A sum term • A conjunction (meet) of sum terms • For example, xy+xy0+z is a sum of products, (x+y)(x+y0)z is a product of sums, x is both, and xy+(x+z)(y+z) is neither. • We define the cost of a formula to be determined by the number of terms and literals it contains • For two SOP formulas g and h, the cost of g is less than the cost of h if g has fewer product terms. If they have the same number of product terms, then the cost of g is lower if it has fewer literals • For two POS formulas g and h, the cost of g is less than that of h if g has fewer sum terms. If they have the same number of sum terms, then the cost of g is lower if it has fewer literals • We compare a POS formula with a SOP formula by comparing the number of terms and literals in a similar fashion • Definition: A two-level formula is either a SOP or POS formula

Implicants • Definition: An implicant of a function f is a product term p that is covered by f. That is, p·f. As an example, xy0, xyz, and yz are implicants of xy0+yz • A key to thinking about what an implicant is is to look at the related word “implies.” That is, if p·f then whenever p=1 it must be that f=1 so that “p implies f.” • Definition: A prime implicant of f is an implicant of f that is not covered by any other implicant of f. That is, if p is a prime implicant of f, then for any other implicant q of f, then p£q.As an example, xy0, yz, and xz are prime implicants of xy0 + yz, but xyz is not. • Note that if p is an implicant that is not prime, then there exists a literal in p that can be removed to yield another implicant • Definition: An essential prime implicant is a prime implicant that covers a minterm which is not covered by any other implicant For example, xy0 and yz are essential prime implicants of xy0+yz, but xz is not:xy0 contains xy0z and xy0z0yz contains xyz and x0yzXz contains xyz and xy0z each of which is covered by one of the above implicants

Quine’s Theorem • Quine's Prime Implicant Theorem: A minimal SOP formula must always consist of a sum of prime implicants if any definition of cost is used in which the addition of a single literal to any formula increases the cost of the formula. Proof: follows by assuming that a minimum cost SOP formula exists that contains a non-prime implicant. This implicant can be replaced by an implicant with one fewer literal without changing the function. This contradicts the assertion that the original formula was minimum cost. • The consequence of Quine's theorem for us is that when we want to find a minimum cost SOP formula we need only look at sums of prime implicants and not all possible sums of minterms

Consensus Theorem • Consider the relationship between implication and the consensus theorem • A syllogism is a logical form consisting of: • a major premise (such as “all humans are mortal”) • a minor premise (such as “I am human”) • conclusion (“I am mortal”) • A premise can be represented in terms of letters representing predicates. For example, “all humans are mortal” can be written h)m where h is predicate which may be true or false depending upon the individual in question and m is a predicate which also may be true or false • The proposition x)y can be considered a function that maps {TRUE, FALSE}2 to {TRUE, FALSE} in such a way that the function is true if x is false or if y is true whenever x is true. • We can represent this Boolean function via the formula x)y=x0+y • The syllogism can be written via the formula (h)m)¢(I)h))(I)m) which is equivalent to (h0+m)(I+h))(I0+m) • Note that the implied term is the consensus of the other two terms from the consensus theorem for Boolean algebras that states (x+y)(x0+z)(y+z)=(x+y)(x0+z) • A special case of the consensus theorem xy+x0z+yz=xy+x0z follows when z=y:xY+x0Y+Y=xY+x0Y. Since Y covers both xY and x0Y, this is Y=xY+x0Y. • The capital Y is meant to indicate that Y may represent a product of literals not including x or x0 • Though a convoluted way of proving that Y=xY+x0Y, it justifies calling Y the consensus term of xY and x0Y

yz 00 01 11 10 x 0 1 0 0 1 0 0 0 1 0 Adjacency and Minimization • Definition: Two product terms are said to be adjacent if the just defined version of the consensus theorem can be applied to them • That is, two product terms are said to be adjacent if they are of the form xY and x’Y. • If two product terms are adjacent, and both terms are implicants for some Boolean function f, then Y is also an implicant of f and any function using only Y will be of lower cost than one using both xY and x’Y. • This is the basis of Boolean function minimization and is what we all used to minimize using Karnaugh maps: Y=yz: xyz and x’yz are both implicants of f.Hence, we eliminate x and say the minimalfunction is simply f=yz.

Minimization Problem • We will break the minimization problem up into two sub-problems: • Determining all prime implicants of a function • Determining a sum of those implicants which is a minimum cost representation of the function • There are many ways to find prime implicants: • Mathematically, • Via Karnaugh Maps, • Iteratively. • Once we have all the prime implicants, we can find a sum of them that will provide a minimum cost function.

PIs by Iterative Methods • Definition: A complete sum is a SOP formula composed of all the prime implicants of the function it represents • A complete sum for a function f can be determined in the following steps: • Write the function f in minterm canonical form • Consider all pairs of adjacent terms, mark them as not prime, and form their consensus term • Repeat the above step until no new terms can be found • The terms that are not marked are the prime implicants of f, the marked terms are not prime and may be removed from the formula • This can be more easily tracked in tabular form:The first column contains the minterms of f grouped by the number of uncomplemented literals • For example if then the table would be constructed as shown on the right: • Recall that if the switching function f is incompletely specified, it corresponds to the interval [f1,f1+f-] where f1 is the sum of minterms for which the function must be 1 and f- is the sum of minterms for which the function is don't care

PIs and Don’t Cares • Definition: A prime implicant of an incompletely specified function given by (f1,f-) is a product term p such that • p·f1+f- • p¢f10 • p is not contained in any other implicant of (f1,f-) • In other words, p is a prime implicant of (f1,f-) if it is a prime implicant of f1+f- that covers at least one minterm of f1 • We can find the prime implicants of an incompletely specified function (f1,f-) by finding the prime implicants of the completely specified function f1+f- and removing those that don't cover any minterms of f1 • As an example, let f1=yz0+xy0z and f-=x0z • The minterms of f1 are xyz0, x0yz0, xy0z} • The minterms of f- are {x0y0z, x0yz} • In the table, we mark those terms that do not cover any minterms of f1 with the letter d, and when we're finished, the prime implicants of f1+f- that are marked with d are ignored because they are not prime implicants of (f1,f-)

Complete Sums • The above method uses the simplified form of consensus Y=xY+x0Y. We don't need to go all the way down to minterm canonical form if we are willing to apply the more general form consensus xY+x0Z=xY+x0Z+YZ • Theorem: A SOP formula is a complete sum if and only if • No term includes any other term • If the consensus of two terms in the formula exist, it is covered by some term in the formula • This suggests that given a SOP formula, we can build a complete sum by adding the consensus terms from all possible combinations of terms and removing any terms covered by another term As an example, consider the SOP formula f=wx+x0y+xyz • Combining the first two terms yields f=wx+x0y+xyz+wy • Combining the second and third terms yields f=wx+x0y+xyz+wy+yyz=wx+x0y+xyz+wy+yz • The last term contains the third term and no other consensus terms can be formed, so the complete sum for f is f=wx+x0y+wy+yz

PIs by Recursive Methods • If F and F are formulas which are complete sums, then a complete sum for F¢ G can be formed by: • Multiplying out F and G, applying the idempotent and distributive properties and using x¢x0=0 • Eliminate all terms contained in some other term • Since a POS formula consists of products of complete sums, we can repeatedly apply the above theorem to generate a complete sum directly from a POS formula • We can always derive a POS formula from Boole's expansion theorem which gives a recursive algorithm • If we apply the theorem above to each level in the recursion, we can generate a complete sum by recursively applying a very simple algorithm CS(F): • Determine a “splitting variable” x that appears in the most terms of F • Let Fx0 be the return value from CS when passed the formula F with x=0 • Let Fx be the return value from CS when passed the formula F with x=1 • Multiply out and simplify G=(x+Fx0)(x0+Fx) • Return G

Find Sum of PIs • Thus far, we have found the PIs but that’s only step 1. • The question that remains is how to select a subset of the prime implicants such that the sum covers the function and is of minimum cost • In general, this problem is conceptually simple to solve but requires checking each of a large number of possible solutions and picking the best one • We base the approach on a key assumption: We just require one optimal solution, even though more may exist that are also optimal • We break the problem into two steps: • Do as much simplification as possible • Enumerate all possible solutions of the simplified problem and pick the best one • The basic problem we are trying to solve is to find the subset of prime implicants whose sum covers all the minterms and is the lowest cost • The problem is commonly formulated as the selection of columns out of a constraint matrix or assigning values in a constraint equation

Constraint Matrix • The constraint matrix for the covering problem is a matrix with component values of 0 and 1 with • One row for every minterm • One column for every prime implicant • The value 1 at row i and column j if prime implicant pj covers minterm mi and the value 0 otherwise • The problem is to choose a subset of the columns such that each row has a value 1 in at least one of the chosen columns and the total cost of the columns is minimal • Formally, let C be a matrix of m rows and n columns such that Cij2{0,1} and such that 8i9j3Cij=1. Let the cost associated with a set S of prime implicants be denoted c(S). The covering problem is to find a subset Sµ {1,2,...,n} that satisfies

Constraint Matrix Example • As an example, let f=yz+x0y+y0z0+xyz+x0z0 for which the complete sum is f=x0y+x0z0+y0z0+yz. • There are four prime implicants and we name them p1 through p4 • The function covers 5 minterms (x0y0z0, x0yz0, x0yz,xyz,xy0z0} • The constraint matrix C has the following form • Notice that p3 and p4 are essential prime implicants (they each cover a minterm that is not covered by any other prime implicant • If all prime implicants are the same cost, selecting {p1,p3,p4} is a minimum cost covering

Constraint Equation • The constraint equation represents each prime implicant by a variable pj in a switching algebra with: • The interpretation that pj=1 indicates that the prime implicant is included in the final solution whereas pj=0 indicates that it is not • Each minterm is represented by a sum term i = j=1n Cij pj where Cij=1 if the jth prime implicant covers the ith minterm • The goal is to find an assignment of 0 and 1 values to the variables pj that the product of all  terms equals one and the total cost is minimized • Formally, let i be a sum term consisting of a subset of the (uncomplemented) letters p1 through pj. Let the cost associated with the assignment vector p=(p1, p2, ..., pn) be c(p). The covering problem is to find a vector p=(p1, p2, ..., pn) as • As an example, the function f from the previous example would yield a constraint equation (p2+p3)(p1+p2)p3p4=1 • Again note that p3 and p4 are essential primes (must equal one) • If all prime implicants are of equal cost, selecting p1=1, p2=0, p3=1, and p4=1 is a minimal cost solution

Reduction Techniques • The previous example was simple enough to solve almost by inspection, in general we must consider (explicitly or implicitly) all possible solutions • We first consider methods of implicitly eliminating as many solutions as possible before we resort to a full enumeration • All essential primes must be included in the solution because they cover minterms that are otherwise uncovered • If a row of a constraint matrix has a single 1 value in it, the corresponding column must be retained • The column and all rows it covers can be removed from the constraint matrix yielding a simpler problem • If a sum term in a constraint equation consists of a single letter, then it must be assigned the value 1 and the equation can be simplified noting that (1+pi)=1 • From the previous example, p3 and p4 are essential

Row Dominance • Row or constraint dominance occurs if one minterm is covered whenever another minterm is covered • Row k is said to dominate row l in a constraint matrix if row k has the value 1 in every column for which row l has a 1 (row k may have other 1's as well) • Since row l must be covered and doing so also covers row k, we can eliminate row k to yield a simpler problem • For the following constraint matrix, row 1 dominates row 2 so row 1 can be eliminated from the problem • One constraint term dominates another if the absorption property can be applied. That is, if we can simplify the equation by applying x(x+y)=x • The constraint matrix above can be represented by the constraint equation (p1+p2+p4)(p1+p4)(p2+p3)=1. We can apply consensus to the first two terms yielding (p1+p4)(p2+p3)=1.

Column Dominance • Column or variable dominance occurs if a prime implicant pk costs no more than prime implicant pl but covers all the minterms that pl covers (pk may cover more minterms as well) • In this case, pl can be eliminated without fear of loosing the ability to find an optimal solution • Column pk dominates column pl in a constraint matrix if it has 1's in all rows where pl has 1s and prime pk is of no greater cost • For example, in matrix (4), column p2 dominates p3 if the cost of p2 is no greater than the cost of p3 • In a constraint equation, variable pk dominates pl if the cost of pk is no greater than pl and pk appears in every term that pl appears in • For example, in the previous constraint equation ((p1+p2+p4)(p1+p4)(p2+p3)=1), p2 occurs in every term that p3 occurs in

Exact Solutions • Once all possible simplifications have been applied, we are forced to begin enumerating possibilities to pick the cheapest • This can be done, for instance, by multiplying out the constraint equation to SOP form, simplifying via the idempotent property, and choosing the minterms of the lowest cost product term • The above approach can be quite tedious for constraint equations with more than a few sum terms • Instead, a solution can be found in a systematic way • Simplify the problem accounting for essential primes and dominance • Pick a splitting variable pj • Assign pj=0 and solve the reduced problem by first simplifying (accounting for essential primes and dominance) and then enumerating the solutions to the reduced problem • Assign pj=1 and solve the reduced problem by first simplifying (accounting for essential primes and dominance) and then enumerating the solutions to the reduced problem • With a clever bounding technique, we can stop exploring entire branches of possibilities whenever we find that they cannot yield a cheaper solution than the best one we've already found • To apply the branch-and-bound idea, we need a way to lower-bound the cost of a solution

Multiple-Output Functions • A multiple-output function is a vector-valued function with components that are scalar-valued switching functions over a common set of inputs, f=(f1,f2,..., fn) • For example, f=(xz+yz,x0y+x0z0) is a multiple-output function • It is often of interest to find an implementation of a multiple-output function that minimizes the total cost over all outputs and not the cost of any single output. Here we address finding two-level formulas that minimize the total number of gates • Finding a minimum cost formula for each of the components is not necessarily a minimum cost formula for the multiple-output function because it does not take into account implicants which may be shared between the functions • For example, in the function f above, the formula given is a minimum cost sum of prime implicants for each of the components. However, the formula f=(xz+x0yz, x0z0+x0yz) involves fewer terms • Note that the term x0yz need be formulated only once and it is not a prime implicant of either output • The solution is to consider the covering problem using not only the prime implicants of the individual outputs, but of all possible products of the outputs • That is, consider the prime implicants of {f1, f2, ..., fn, f1f2, f1f3, ..., f1f2f3, ..., f1f2f3...fn}

Shorthand Notation • We will use a shorthand notation for representing the implicants of multiple-output functions • First choose some ordering of the inputs as in x1, x2, x3, ..., xm and some ordering of the outputs as in f1,f2,f3,..., fn • An implicant of the multiple-output function is a string of m symbols representing values of the inputs taken from the set {0,1,-}, a vertical line, and a string of n symbols representing values of the outputs taken from the set {0,1,-} • A value of 1 in the ith position from the left indicates that the input xi must be 1 for the implicant to apply. Similarly, a value of 0 indicates that xi must be 0 for the implicant to apply, and - indicates that implicant applies regardless of the value of xi • A value of 1 in the ith position right of the vertical bar indicates that the output fjmust be 1 when the implicant applies and a value of - indicates that the output is don't care when the implicant applies. On the other hand, a value of 0 implies that the output is not necessarily 1 when the implicant applies • Because of the don't care possibility for inputs, more than one implicant may apply for a given set of input values and in this case, the outputs are given as the join of the corresponding outputs for all matching implicants • For example, the previous function f=(xz+yz,x0y+x0z0) can be written as • The input x1=x=0, x2=y=1, x3=z=1 matches both middle implicants, and for this combination of inputs the output is f1=1+0=1 and f2=0+1=1

Tabular Method Revisited • We can find the prime implicants through the tabular method appropriately modified • Group the implicants by the number of 1s in the input sequence • When applying consensus to two minterms, the output sequence for the new implicant is formed by the meet of the output sequences for the two minterms • An implicant used to form a consensus term is checked only if its output sequence equals that of the consensus term • As an example, the function f (1-1|10+-11|10+01-|01+0-0|01) has prime implicants found by the table

Covering Problem Revisited • The covering problem is handled through a slightly modified method • Minterms for which multiple outputs are 1 are expanded into a sum of minterms with a single output function given the value 1 (because some prime implicants may cover only some of the output functions) • For example, the minterm 011|11 is expanded as 011|10 + 011|01 and each are listed individually in the constraint matrix • With p1=011|11, p2=0-0|01, p3=01- |01, p4=-11|10, and p5=111|10, the constraint matrix becomes

Synthesis of Two-Level Circuits