1. CS151 Introduction to Digital Design Chapter 2-4-1 Map Simplification

2. CS 151 Circuit Optimization Goal: To obtain the simplest implementation for a given function Optimization is a more formal approach to simplification that is performed using a specific procedure or algorithm Optimization requires a cost criterion to measure the simplicity of a circuit To distinct cost criteria we will use:  Literal cost (L)  Gate input cost (G)  Gate input cost with NOTs (GN)

3. CS 151 Literal – a variable or its complement Literal cost – the number of literal appearances in a Boolean expression corresponding to the logic circuit diagram Examples:  F = BD + A C + A L = 8  F = BD + A C + A + AB L =  F = (A + B)(A + D)(B + C + )( + + D) L =  Which solution is best? Literal Cost D DBC B BD C BC

4. CS 151 Gate Input Cost Gate input costs - the number of inputs to the gates in the implementation corresponding exactly to the given equation or equations. (G - inverters not counted, GN - inverters counted) For SOP and POS equations, it can be found from the equation(s) by finding the sum of:  all literal appearances  the number of terms excluding terms consisting only of a single literal,(G) and  optionally, the number of distinct complemented single literals (GN). Example:  F = BD + A C + A G = 11, GN = 14  F = BD + A C + A + AB G =, GN =  F = (A + )(A + D)(B + C + )( + + D) G =, GN =  Which solution is best? D B C B B D C B D B C

5. CS 151 Example 1: F = A + B C + Cost Criteria (continued) A B C F B C L = 5  L (literal count) counts the AND inputs and the single literal OR input. G = L + 2 = 7  G (gate input count) adds the remaining OR gate inputs GN = G + 2 = 9  GN(gate input count with NOTs) adds the inverter inputs

6. CS 151 Example 2: F = A B C + L = 6 G = 8 GN = 11 F = (A + )( + C)( + B) L = 6 G = 9 GN = 12 Same function and same literal cost But first circuit has better gate input count and better gate input count with NOTs Select it! Cost Criteria (continued) B C A A B C F CB F A B C A

7. CS 151 Map Simplification Simplifying Boolean Expressions  Algebraic manipulation Lacks rules for predicting steps in the process Difficult to determine if the simplest expression has been reached  Map simplification (Karnough map or K-map) Straightforward procedure for function of up to 4 variables Provides a pictorial form of the truth table Maps for five and six variables can be drawn, but are more cumbersome to use

8. CS 151 Karnaugh Maps (K-map) A K-map is a collection of squares  Each square represents a minterm  Adjacent squares differ in the value of one variable  Visual diagram of all possible ways a function may be expressed in standard forms Sum-of-Products Product-of-Sums  Alternative algebraic expressions for the same function are derived by recognizing patterns of squares  select simplest.

9. CS 151 Two-Variable Map 2 variables  4 minterms  4 squares. Y’Y XX’

10. CS 151 K-Map Function Representation Example: F(X,Y) = XY’ + XY From the map, we see that F (X,Y) = X. This can be justified using algebraic manipulations: F(X,Y) = XY’ + XY = X(Y’ +Y) = X.1 = X 1

11. CS 151 K-Map Function Representation Example: G(x,y) = m1 + m2 + m3 1 11 G(x,y) = m1 + m2 + m3 = X’Y + XY’ + XY From the map, we can see that G = X + Y

12. CS 151 How can we locate a minterm square on the map?  Use figure (a) OR  use column # and row # from figure (b) E.g. m 5 is in row 1 column 01 (5 10 = 101 2 ) Three-Variable Maps 3 variables  8 minterms (m0 – m7). Q. Show the area representing X’? Y’? Z’?

13. CS 151 Alternative Map Labeling Map use largely involves:  Entering values into the map, and  Reading off product terms from the map. Alternate labelings are useful: y z x 1 0 2 4 3 5 6 7 x y z z y y z z 1 0 2 4 3 5 6 7 x 0 1 00011110 x

14. CS 151 Example Functions By convention, we represent the minterms of F by a "1" in the map and leave the minterms of F’ blank Example: Learn the locations of the 8 indices based on the variable order shown (x, most significant and z, least significant) on the map boundaries 00101101 1 0 X Z Y X YZ 11 11 00101101 1 0 X Z Y X YZ 1 111