The University of Texas at Arlington. Algebraic Methods

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1 The University of Texas at Arlington Algebraic Methods CSE 244 Introduction to Digital Logic Dr. Gergely Záruba Bridge from Discrete Structures To describe we use the description tools of Axiomatic Theory : Unambiguous definitions Axioms (Postulates) the starting truths Theorems the provably correct statements We began acquiring tools of description in Discrete Structurest Here we acquire more such tools, and put some of the theory in practical perspective Vocabulary is crucial (makes the material simple and makes sure we all talk the same language) 2

2 BOOLEAN ALGEBRA 3 A Model to Represent Logic We need a model that can be used to represent tlogical lthought htand reason. George Bool (849) has come up with an abstract algebraic structure that has been used since. This includes: A set of postulates A set of theorems (or rules) 4 2

3 Boolean Algebra Postulate- Defintion: Boolean algebra is a closed algebraic system, containing a set of K (two or more) elements and two operators * and + It is closed as any element a in K and b in K, the sum or OR (a+b) and the product or AND ( a*b) both will belong to K Precedence is given to the * operator 5 Boolean Algebra Postulate-2 Existence of and elements There exist unique elements and such that : a + = a a * = a We call the identity element for the OR operation and the identity element for the AND operation 6 3

4 Boolean Algebra Postulate -3 Commutativity of the operators: For a,b K: a + b = b + a a * b = b * a 7 Boolean Algebra Postulate -4 Associativity of the operators: For a,b,c K: a + (b +c) = (a + b) + c a * (b * c) = (a * b) * c 8 4

5 Boolean Algebra Postulate -5 Distributivity of the + over * and * over +: For a,b,c K: a + (b * c) = (a + b) * (a + c) a * (b + c) = (a * b) + (a * c) 9 Boolean Algebra Postulate -6 Existence of the complement: For a K, there exists an ā K such that: a + ā = a * ā = 5

6 E.g., a+b Venn Diagrams for Visualization K a b Could be used to visualize more than 2 variables Could be used to visualize the postulates The Principle of Duality If an expression is valid then the dual of that t expression is also valid The dual can be found by swapping all operators against their counterparts (i.e., + with * and vice versa) and by swapping the identity elements with each other (i.e., by and vice versa) E.g.,: the dual of (a+b)(a+c) is (ab+ac) The following theories are numbered arbitrarily so theory- is not going to be called theory- by other textbooks 2 6

7 Boolean Algebra Theorem - Idempotency: a + a = a a * a = a (notice that they are duals so it is enough to prove one of them) 3 Boolean Algebra Theorem -2 Null elements: a + = a * = (notice that they are duals so it is enough to prove one of them) 4 7

8 Boolean Algebra Theorem -3 Involution: a = a 5 Boolean Algebra Theorem -4 Absorption: a + ab = a a * (a+b) = a (notice that they are duals so it is enough to prove one of them) 6 8

9 Boolean Algebra Theorem -5 Absorption-2: a + āb = a+b a * (ā+b) = ab (notice that they are duals so it is enough to prove one of them) 7 Boolean Algebra Theorem -6 Absorption-3: ab + āb = b (a+b) * (ā+b) = b (notice that they are duals so it is enough to prove one of them) 8 9

10 Boolean Algebra Theorem -7 Absorption-4: ab + ābc = ab + bc (a + b) * (ā + b + c) = (a + b) (b + c) (notice that they are duals so it is enough to prove one of them) 9 Boolean Algebra Theorem -8 DeMorgan Theorems: a + b = a * b a * b = a + b (notice that they are duals so it is enough to prove one of them) They can be generalized for an arbitrary (but fininte) number of variables 2

11 Boolean Algebra Theorem -9 Consensus: ab + āc +bc = ab + āc (a + b) * (ā + c) * (b + c) = (a + b) (ā + c) (notice that they are duals so it is enough to prove one of them) 2 Boolean Algebra Theorem - Shanon s Expansion: f(x,x 2,,x n )=x f(,x 2,,x n )+x f(,x 2,,x n ) f(x,x 2,,x n )=[x +f(,x 2,,x n )]+[x +f(,x 2,,x n )] (notice that they are duals) 22

12 Boolean Algebra Cheat Sheet Textbook Table 2.2 (pg. 9) 23 SWITCHING ALGEBRA/FUNCTIONS 24 2

13 From Boolean Algebra to Switching Algebra For switching algebra, we specify K to contain exactly two elements K={,} We can refer to these values as {false,true}, or {low,high} as well. (In general capital letters represent variables and small letters represent values that they take, ie i.e., X vs. x ) 25 Number of Switching Functions A function f(x, X 2,.., X n ) has n variables. Since each of these variables can take one of exactly two values, there are 2 n different ways variables can be assigned to functions. Since each(!) one of those assignments can result in two possible outcomes, there are 2 2n different switching functions For n=, there are two functions: f = and f = For n=, there are four functions f(a): f (A)=, f (A)=A, f 2 (A)=A, and f (A)= 26 3

14 Switching Functions for n=2 There are thus 6 two-variable switching functions, f (A,B) to f 5 (A,B) A B f f f 2 f 3 f 4 f 5 f 6 f 7 f 8 f 9 f f f 2 f 3 f 4 f 5 Any two variable function can be reduced to one of these Note: f 8 is AND, f 4 is OR, f 6 is XOR, f 7 is NAND, and f is NOR 27 Truth Tables Previous slide contains 6 truth tables in one. This is still doable for n=2. For n=3, there are a total of 256 different switching functions, for n=4, ~4.29B for n=5, and ~.6* 77 for n=8 (the number of atoms in the observable universe is about 8 ) A truth table has four sections: ABC f(a,b,c)=ab+ac 28 4

15 Algebraic Forms The same switching function can be given in many ways. There are some forms that are more desirable SOP form (Sum of Products), the OR-ing of several AND-ed literals (uncomplemented or complemented variables) E.g., f(a,b,c,d)=abc+dc+acd (true heavy) POS form (Product of Sums) E.g., f(a,b,c,d,e) =(A+E)(C+D+E)(B+D) (false heavy) 29 Algebraic Forms: Canonical SOP a.k.a. disjunctive normal form If a product (AND) contains all variables of the function in either complemented or uncomplemented form then this product is called a minterm A canonical SOP is a form where all products are minterms E.g., f(a,b,c) = ABC + ABC + ABC + ABC If we looked at the literals as a n-bit binary code (n is the number of variables in the function) and assigned to a literal if it is complemented and otherwise, then each of the minterms could be uniquely numbered. E.g., ABC would be and thus minterm- or m E.g., ABC would be and thus minterm-5 or m 5 Thus the above f(a,b,c) = m +m 2 +m 5 +m 7 = m(,2,5,7) (Note, that this is a notation only!) The order of the variables in this notation becomes important; the minterms for f(a,b,c) and f(b,a,c) are not the same 3 5

16 Algebraic Forms: Canonical POS a.k.a. conjunctive normal form If a product (OR) contains all variables of the function in either complemented or uncomplemented form then this product is called a maxterm A canonical POS is a form where all products are maxterms E.g., f(a,b,c) = (A+B+C) (A+B+C) (A+B+C) (A+B+C) If we looked at the literals as a n-bit binary code (n is the number of variables in the function) and assigned to a literal if it is complemented and otherwise(!!!), then each of the minterms could be uniquely numbered. E.g., (A+B+C) would be and thus maxterm-6 or M 6 E.g., (A+B+C) would be and thus maxterm-52or M 2 Thus the above f(a,b,c) = M * M 2 * M 5 * M 6 = M(,2,5,6) (Note, that this is a notation only!) Notice, that order of variables is important again 3 Duality of Canonical POS and SOP Minterms and maxterms are complements of each other i.e., M i =m i =M i and m i =M i =m i e.g., if f(a,b,c)= m(,2,5,7) then f(a,b,c) = M(,2,5,7) And since f(a,b,c)= m(,3,4,6) it becomes apparent that in order to convert from canonical SOP to POS (or vice versa) the complementing min/maxterms need to be used. 32 6

17 So, How Do We Obtain Canonical SOP Forms? Recall Shanon s expansion theorem (T) f(x,x 2,,x n )=x f(,x 2,,x n )+x f(,x 2,,x n ) This can be used one after the other for all variables to obtain canonical SOP Alternatively Theory-6 could be used ab+ab=a a use this for each product not containing a variable How about canonical POS? Same tricks could be used (b versions of theories) or just use the duality rule by finding canonical SOP first. 33 Incompletely Specified Functions In real life, many times, functions do need to be true for some minterms, do need to be false for some others, but there may be some minterms for which the designer does not care (don t care) if the function is false or not. For example if we know that some input combinations will never happen (e.g., BCD input) or if for some input combination it really does not matter what the output is. In canonical SOP we can represent don t care minterms as d i instead of m i. In canonical POS we can use Di. E.g., a function working on BCD input could be: f(a,b,c,d) = m(,2,4,6,8)+d(,,2,3,4,5) Don t cares can actually be helpful when minimizing functions (to make them as simple as possible). 34 7

18 Simplifying Switching Functions Normal forms (POS,SOP, and cannonicals) are not necessarily the simplest forms of a function. E.g., f(a,b,c,d)=abc+d= m(,3,5,7,9,,3,4,5) Former requires one 3-input AND and one 2-input OR gates, latter requires nine 4-input AND and one 9-input OR gates. If we want our function to be realized with the least amount of logic gates, then we need to reduce the number of operators and the number of inputs to those operators. Having said that, in order to simplify (or minimize) switching functions we usually (but not always) first put them in canonical SOP We can then use switching algebra (or Boolean algebra) to try to reduce them. This is cumbersome and requires a lot of experience. It is also not easily automated. Would be nice to have some simple methods 35 KARNAUGH MAPS 36 8

19 The Need for Karnaugh Maps Is a semi-systematic way to reduce switching functions Can be used to up to five (becomes somewhat tricky) maybe six (becomes really tricky) variable switching functions Recall Theory-6: ab+ab=a This is the best way to reduce functions Thus, if we could organize minterms, so that the ones differing in just a single literal (complemented vs. uncomplemnted) would be placed next to each other then these two could be simplified E.g., f(a,b,c)=abc+abc=bc = m 7 +m 3 How could we graphically put adjacent minterms next to 37 each other? Two/Three Variable K-Maps Organizing minterms into such a map is somewhat straight forward for small n-s B A m m 2 m m 3 C AB m m 2 m 6 m 4 Note, that m 4 and m are adjacent And that m 5 and m are adjacent as well m m 3 m 7 m

20 Four Variable K-Map Graph theory indicates we can still do this for four variables AB CD m m 4 m 2 m 8 Adjacencies wrap around borders! m m 5 m 3 m 9 m 3 m 7 m 5 m m 2 m 6 m 4 m Where are the white arrows? Order has to be learned, not really straight forward 39 Five Variable K-Map Graph theory indicates we can t really do this so it becomes tricky we need two tables f(a,b,c,d,e) DE BC DE BC m m 4 m 2 m 8 m 6 m 2 m 28 m 24 m m 5 m 3 m 9 m 7 m 2 m 29 m 25 m 3 m 7 m 5 m m 9 m 23 m 3 m 27 m 2 m 6 m 4 m m 8 m 22 m 3 m 26 A= A= What are those adjacencies again? 4 2

21 Mahoney Maps Can we rearrange the variables so there is more method to the madness? BD AC m m m 5 m 4 m 2 m 3 m 7 m 6 m m m 5 m 4 m 8 m 9 m 3 m 2 Adjacencies wrap around borders! Order is somewhat simpler 4 So How Do We Plot Switching Functions? E.g., f(a,b,c,d)= m(,2,4,6,8,2) AB CD m m 4 m 2 m 8 m m 5 m 3 m 9 m 3 m 7 m 5 m m 2 m 6 m 4 m 42 2

22 Minterm and Maxterm K-Maps E.g., f(a,b,c,d)= m(,2,4,6,8,2) AB MINTERM CD m m 4 m 2 m 8 m m 5 m 3 m 9 AB MAXTERM CD m m 4 m 2 m 8 m m 5 m 3 m 9 m 3 m 7 m 5 m m 3 m 7 m 5 m m 2 m 6 m 4 m m 2 m 6 m 4 m E.g., f(a,b,c,d)= M(,3,5,7,9,,3,4,5) 43 Why Did We Do This Again? SIMPLIFICATION! So, how do we simplify? We look for representatives of T6 by circling -s next to each other. Pairs work, 4-neighbors work (if they are either in line or in box), 8-neighbors work, 6-neighbors work, etc

23 K-Map Based Simplification We need to find groups (implicants) that cover the entire map. AB CD m m 4 m 2 m m 5 m 3 m 9 m 8 m 3 m 7 m 5 m F(A,B,C,D)=CD+ACD Some terms: Implicant Prime implicant Essential prime implicant Cover m 2 m 6 m 4 m 45 K-Map Based Simplification We need to find groups (implicants) that cover the entire map. AB CD m m 4 m 2 m 8 m m 5 m 3 m 9 m 3 m 7 m 5 m F(A,B,C,D)=AD+CAD Some terms: Implicant Prime implicant Essential prime implicant Cover m 2 m 6 m 4 m 46 23

24 K-Map Based Simplification We need to find groups (implicants) that cover the entire map. AB CD m m 4 m 2 m m 5 m 3 m 9 m 8 m 3 m 7 m 5 m F(A,B,C,D)=AD+CD Some terms: Implicant Prime implicant Essential prime implicant Cover m 2 m 6 m 4 m 47 Algorithm- for Finding the Best Cover. Count the adjacencies for each minterm 2. Select an uncovered minterm with lowest adjacency number (if several are the same, pick randomly) 3. Check which prime implicant covers most other uncovered minterms and select that. 4. Go to 2 Is this optimal? We are making random choices, how can we be sure? 48 24

25 Algorithm-2 for Finding the Best Cover. Circle all prime implicants 2. Select all essential prime implicants 3. Select the minimum cover from the remaining prime implicants to finish the cover Is this optimal? Step three is still random How could we make it optimal? An exhaustive search on step-3 would make this algorithm optimal. This is expensive. 49 How about POS Forms? Terminology is different: instead of implicant we have implicate We can do the same operations over zeros. When writing the POS form keep in mind that variables need to be complemented It is just easier to to do the complemented function using SOP form and apply DeMorgan 5 25

26 Incomplete Functions If the function is incomplete then we can decide to include or ignore don t cares. We include them only if they make a prime implicant larger. For example: f(abcd)= m(24568)+d(7 f(a,b,c,d)= m(2,4,5,6,8,)+d(7,9,2,3,4) ) = M(,,3,,6)+D(7,9,2,3,4) Note that the two minimizations may not result in equivalent functions! 5 QUINE-MCCLUSKY TABULAR METHOD 52 26

27 Quine-McCluskey Method Systematic method (can be easily computerized) Minterms are written up in a table (n-variable minterms). Based on this (n-)-variable implicants are identified. Based on this (n-2) variable implicants are identified and so on. Then we identify essential prime implicants, and prime implicants. The cover is based on these. 53 Q-M Method Step- Step. Write all minterms in a two column table; organize minterms in groups based on how many -s they have (not neccesarily the minterm order) E.g, f(a,b,c,d)= m(,,2,4,5,7,,2,4) Minterm # ABCD Group- 2 Group- 4 5 Group Group

28 Q-M Method Step-2 Do an exhaustive search between adjacent groups to see if minterms can be combined; repeat this for the new columns as well. List- List-2 List-3 Minterm ABCD Minterms ABCD Minterms ABCD, -,4,,5 --,2-2,4 - - s have to be in the same place 4,5 - and they can 5 2, - differ in only one 4,5 - digit 2 4,2-7 5,7-4,4-55 2,4 - Q-M Method Step-2 Do an exhaustive search between adjacent groups to see if minterms can be combined; repeat this for the new columns as well. List- List-2 List-3 Minterm ABCD Minterms ABCD Minterms ABCD, - PI 2,4,,5 -- PI,2 - PI 32 2,4-4,5-5 2, - PI 43 5,7 4,5 - - PI 5 2,4 4,2 - - PI ,4 5,7 - - PI 75 4,4 - PI 6 2,4 - PI

29 Q-M Method Step-3 Now need to find a minimum cover. For that we need to create another cross reference table PI X X X X PI 2 X X PI 3 X X PI 4 X X PI 5 X X PI 6 X X PI 7 X X Separating Pis with different number of implicants 57 Q-M Method Step-4 Identify essential prime implicants (minterm is only covered by them). They need to be chosen **PI X X X PI 2 X X PI 3 X X PI 4 X X **PI 5 X PI 6 X X PI 7 X X 58 29

30 Q-M Method Step-4.2 Select PIs to finish cover; (remove essentials and their minterms for more visibility) **PI X X X PI 2 X X PI 3 X X PI 4 X X **PI 5 X 5 PI 6 X X PI 7 X X PI 2 X PI 3 X X PI 4 X PI 6 X X PI 7 X X But even then, how to choose? 59 Q-M Method Step **PI X X X PI 2 X X PI 3 X X PI 4 X X **PI 5 X PI 6 X X PI 7 X X List- List-2 List-3 Minterm ABCD Minterms ABCD Minterms ABCD, -,4,,5 -- PI,2 - PI 2 2,4-4,5-5 2, - PI 3 4,5-2 4,2 - PI 4 7 5,7 - PI 5 4,4 - PI PI 2 X *PI 3 X X PI 4 X PI 6 X X *PI 7 X X Cyclic chart Thus, PI,PI 3,PI 5,PI 7 f(a,b,c,d)= AC+BCD+ABD+ABD 6 2,4 - PI 7 3

31 How About Q-M and Incomplete Functions? Same first and second steps, handling d s as if they were m s In step three do not create columns for d s 6 What if We have Multiple Outputs? If we have multiple outputs (more than one function), we could benefit from shared implicants. We can extend the Q-M tables to include more functions. We just need one more column for the lists that indicates in which function they are used. For step 2: terms can only be combined if they have a common flag (and the combination can carry only those) For step 3: each function should be represented by columns of those minterms that they have PIs for. 62 3