K-Maps, Logical Adjacency – Study Guide | CPSC 5155G, Study notes of Computer Architecture and Organization

Download K-Maps, Logical Adjacency – Study Guide | CPSC 5155G and more Study notes Computer Architecture and Organization in PDF only on Docsity! Karnaugh Maps Logical Adjacency K–Maps are a standard method for simplification of Boolean expressions. The difference between K–Maps and Quine–McCluskey (Q–M) is that K–M is well suited to manual solutions of small problems, and Q–M lends itself to automated solutions. Logical adjacency is the basis for all Boolean simplification methods. The K–Map approach is a manual procedure that transforms logical adjacency into physical adjacency on the paper. Simplification is done by inspection. The key idea behind logical adjacency is expressed in the following simplifications of Boolean expressions. XY + XY = X(Y + Y ) = X1 = X (X + Y)(X + Y ) = XX + XY + YX +YY = XX + XY + XY + 0 = XX + XY + XY = X + X(Y + Y) = X1 = X + X = X Karnaugh Maps Logical Adjacency (Part 2) Two Boolean terms are said to be logically adjacent when they contain the same variables and differ in the form of exactly one variable. Put another way, only one literal is different; a variable appears negated in one term and is not negated in the other term. All other variables appear in the same way. Consider the following lists of terms. List 1 X X List 2 XY XY XY XY List 3 (X + Y) (X + Y ) (X + Y ) (X + Y) In each of the lists, each term is logically adjacent not only to the term before it but also to the term following. In particular XY is adjacent to both XY and XY (X + Y) is adjacent to both (X + Y ) and (X + Y). In other words, the adjacency list is “circular”; the first item follows the last one. Karnaugh Maps Creating a K–Map The standard K–map is used to simplify Boolean expressions in canonical form; each term has exactly one literal for every Boolean variable. The method can be expanded to Boolean expressions not in canonical form. SOP Expressions Place a “1” in the K–Map for every term in the expression. The positions not filled with a “1” are assumed to have a “0”. POS Expressions Place a “0” in the K–Map for every term in the expression. The positions not filled with a “0” are assumed to have a “1”. POS K–Map SOP K–Map Karnaugh Maps Locating Terms for Sum of Product Expressions In SOP, each term is to be represented by a “1” in the K–map. Where should that term be put? Apply the “SOP Copy Rule” to each of the terms. 1. Write the variables in a standard uniform order. 2. Replace each complemented variable with a 0 Replace each plain variable with a 1. Consider F(X, Y, Z) the term XYZ corresponds to 0 0 0 the term XYZ corresponds to 0 1 1 the term XYZ corresponds to 1 0 1 the term XYZ corresponds to 1 1 0 the term XYZ corresponds to 1 1 1 Karnaugh Maps Locating the SOP Terms (Part 2) Let us examine the SOP expression F(X, Y, Z) = XYZ + XYZ + XYZ + XYZ We place four “1”, one for each of the terms. For XYZ, the “1” is placed at X = 0, Y = 1, Z = 1. For XYZ, the “1” is placed at X = 1, Y = 0, Z = 1. For XYZ , the “1” is placed at X = 1, Y = 1, Z = 0. For XYZ, the “1” is placed at X = 1, Y = 1, Z = 1. Karnaugh Maps Locating Terms for Product of Sums Expressions In POS, each term is to be represented by a “0” in the K–map. Where should that term be put? Apply the “POS Copy Rule” to each of the terms. 1. Write the variables in a standard uniform order. 2. Replace each complemented variable with a 1 Replace each plain variable with a 0. Consider F(X, Y, Z) the term X+Y+Z corresponds to 0 0 0 the term X + Y + Z corresponds to 0 0 1 the term X + Y + Z corresponds to 0 1 0 the term X + Y + Z corresponds to 1 0 0 the term X+Y+Z corresponds to 1 1 1 Karnaugh Maps Locating the POS Terms (Part 2) Let us examine the POS expression F(X, Y, Z) = (X + Y + Z)(X + Y + Z )(X + Y + Z)(X + Y + Z) We place four “1”, one for each of the terms. For (X + Y + Z), the “0” is placed at X = 0, Y = 0, Z = 0. For (X + Y + Z ), the “0” is placed at X = 0, Y = 0, Z = 1. For (X + Y + Z), the “0” is placed at X = 0, Y = 1, Z = 0. For (X + Y + Z), the “0” is placed at X = 1, Y = 0, Z = 0. Karnaugh Maps Example K–Map on 3 Variables Consider the K–map shown above. It can be shown to represent the Boolean expression F(X, Y, Z) = (X + Y + Z)(X + Y + Z)(X + Y + Z)(X + Y + Z ). This is a K–map representation of a POS (Product of Sums) expression, also called an expression in Conjunctive Normal Form. The K–map procedure works by grouping adjacent squares. Here are the adjacencies: 000 and 010 group to form 0–0 representing the term (X + Z) 100 and 000 group to form –00 representing the term (Y + Z) 000 and 001 group to form 00– representing the term (X + Y) The expression is F(X, Y, Z) = (X + Y)(X + Z)(Y + Z) Karnaugh Maps Another Look at the K–Map We continue to examine the above expression, with the encoding 0000, 0001, 1000, and 1001. In this slide, we show another view that explains the “wrap around”. The absolute requirement for a K–Map is that the rows and columns be encoded with a Gray code. The standard sequence is 00, 01, 11, 00. Others are possible. Here is the identical K–Map with a new Gray code sequence for X, Y. The sequence is 01 00 10 11. Note that each term differs by 1 from each of its predecessor and successor. We now see the required square grouping to produce the term X’Y’. Karnaugh Maps Another Example Four–Variable K–Map Consider the Boolean function F(W, X, Y, Z) = W’X’Y’Z’ + W’X’YZ’ + WX’Y’Z’ + WX’YZ’. Using the SOP encoding method, these are terms 0000, 0010, 1000, and 1010. The four corners are logically adjacent. The first row reduces to -000 The last row reduces to -010. The two reduce to -0-0, or X’Z’.