1. PHY 201 (Blum)1 Karnaugh Maps References: Chapters 4 and 5 in Digital Principles (Tokheim) Chapter 3 in Introduction to Digital Systems (Palmer and Perlman)

2. PHY 201 (Blum)2 Review: Expressing truth tables Every truth table can be expressed in terms of the basic Boolean operators AND, OR and NOT operators. E.g. using sum of products or product of sums. The circuits corresponding to those truth tables can be build using AND, OR and NOT gates, which can be made out of transistors. In the sum of products approach, the input in each line of a truth table can be expressed in terms of AND’s and NOT’s (though we will only need the rows that have 1 as an output).

3. PHY 201 (Blum)3 Algebra  Gate A’ means NOT A high  low  high input low output Red probe indicator

4. PHY 201 (Blum)4 Algebra  Gates AB means A AND B

5. PHY 201 (Blum)5 Algebra  Gates A+B means A OR B

6. PHY 201 (Blum)6 Review: Line by Line Inputs Expression AB 00 (Not A) AND (NOT B) A´B´ A´B´ is true for the first line and false for the rest 01 (Not A) AND B A´B A´B is true for the second line and false for the rest 10 A AND (NOT B) AB´ AB´ is true for the third line and false for the rest 11 A AND B ABAB A´B´ is true for the fourth line and false for the rest This is not yet a truth table. It has no outputs.

7. PHY 201 (Blum)7 Writing the expression To express a truth table as a sum of products (minterm expression), take the input lines that correspond to true (high, 1) outputs. Write the expressions for each of those input lines (as shown on the previous slide). This step will involve NOTs and ANDs Then feed all of those expressions into an OR gate.

8. PHY 201 (Blum)8 Example 1 ABCOut 0001 0010 0101 0110 1000 1011 1100 1111

9. PHY 201 (Blum)9 Example 1 (Cont.) A’B’C’ + A’BC’ + AB’C + ABC The expression one arrives at in this way is known as the sum of products. You take the product (the AND operation) first to represent a given line. Then you sum (the OR operation) together those expressions. It’s also called the minterm expression.

10. PHY 201 (Blum)10 Simplifying Boolean algebra expressions Recall that (A’B’C + A’BC’ + A’BC + AB’C’ + AB’C + ABC’ + ABC) and (A+B+C) correspond to the same truth table. Before building a circuit that realizes a Boolean expression, we would like to simplify that expression as much as possible. Fewer gates means Fewer transistors Less space required Less power required (less heat generated) More money made

11. PHY 201 (Blum)11 A few fundamental theorems A + 1 = 1 A + 0 = A A·1 = A A·0= 0 A + A = A A·A = A A + A’ = 1 A·A’ = 0

12. PHY 201 (Blum)12 A Trivial Simplification Example ABCOutExpressions 0000 0010 0101A’ B C’ 0111A’ B C 1000 1010 1101A B C’ 1111A B C

13. PHY 201 (Blum)13 Simplifying a trivial example A´BC´ + A´BC + ABC´ + ABC A´B (C´ + C) + AB (C´ + C) A´B + AB (A´ + A) B B C+C’ means C OR (NOT C) In other words, we don’t care about C

14. PHY 201 (Blum)14 How simplification occurs Note that simplification occurs when two terms differ by only one factor. For example, the terms A´BC´ and A´BC have A’B in common and differ only in the C factor. A’BC’ + A’BC  A’B(C’+C)  A’B If the two terms differ by more than one factor, there is no simplification For example, the terms A’BC’ and A’B’C have A’ in common and differ in the B and C factors A’BC’ + A’B’C  A’(BC’ + B’C)  no simplification

15. PHY 201 (Blum)15 Majority Rules Example ABCMajority 0000 0010 0100 0111 1000 1011 1101 1111

16. PHY 201 (Blum)16 Row Expressions ABCRow expressions 000A’B’C’ 001A’B’C 010A’BC’ 011A’BC 100AB’C’ 101AB’C 110ABC’ 111ABC

17. PHY 201 (Blum)17 Majority rules (sum of products) without simplification A´BC + AB´C + ABC´ + ABC NOTs ANDs OR

18. PHY 201 (Blum)18 Majority Rules: Boolean Algebra Simplification A´BC + AB´C + ABC´ + ABC The term A’BC can be combined with ABC since they differ by one and only one term Same for AB’C and ABC Same for ABC’ and ABC In logic, ABC = ABC + ABC + ABC A´BC+AB´C+ABC´+ABC+ABC+ABC A´BC+ABC + AB´C+ABC + ABC´+ABC (A´+A)BC + A(B´+B)C + AB(C´+C) BC + AC + AB

19. PHY 201 (Blum)19 Majority rules after simplification

20. PHY 201 (Blum)20 Majority Rules Comparison Gates: 3 NOTs, 4 3-input ANDs, 1 4-input OR Gates: 0 NOTs, 3 2-input ANDs, 1 3-input OR

21. PHY 201 (Blum)21 Simplifying made easy Simplifying Boolean expressions is not always easy. So we introduce next a method (a Karnaugh or K map) that is supposed to make simplification more visual. The first step is to rearrange the inputs into what is called “Gray code” order. Here, Gray is a guy not a color.

22. PHY 201 (Blum)22 Frank Gray in Wikipedia

23. PHY 201 (Blum)23

24. PHY 201 (Blum)24 Gray code In addition to binary numbers, there is another way of representing numbers using 1’s and 0’s. Put another way, there is another useful ordering of the combinations of 1’s and 0’s. It is not useful for doing arithmetic, but has other purposes. In gray code the numbers are ordered such that consecutive numbers differ by one bit only.

25. PHY 201 (Blum)25 Gray code (Cont.) 000 001 011 010 110 111 101 100 Each row different by one bit only

26. PHY 201 (Blum)26 Constructing Gray code (a.k.a. reflected binary code) 0 1

27. PHY 201 (Blum)27 Reflect lower bits and add 0’s in front of the original rows and 1’s in front of the new rows 00 01 11 10 Lower bits Reflect lower bits through red line Add 0’s Add 1’s

28. PHY 201 (Blum)28 Reflect lower bits and 0’s then 1’s in front (again) 000 001 011 010 110 111 101 100 Reflect lower bits through red line Add 0’s Add 1’s

29. PHY 201 (Blum)29 An important property In gray-code order, two consecutive rows of a truth table differ by one bit only. Thus if a truth table is put in gray code order and if two consecutive rows contain a 1, then a simplification of the Boolean expression is possible. A term like X + X’ can be factored out.

30. PHY 201 (Blum)30 Trivial Example in Gray code ABCOut 0000 0010 0111 0101 1101 1111 1010 1000 Note: Gray code ordered inputs

31. PHY 201 (Blum)31 Improving Some combinations that differ only by a single bit are not in consecutive rows. Thus there may be a simplification associated with such a combination and we might miss it. So we put some of the inputs in as columns.

32. PHY 201 (Blum)32 Two rows that differ by one bit but are not consecutive ABCOut 0000 0010 0111 0101 1101 1111 1010 1000

33. PHY 201 (Blum)33 A row-column version AB\C01 0000 0111 1111 1000 Place the C inputs across the top. All inputs are filled in with light blue. In this version, more inputs differing by one bit only are in adjacent positions. This output corresponds to the input A=0, B=1 and C=0

34. PHY 201 (Blum)34 Karnaugh-map This way of arranging truth tables combined with the rules for simplifying Boolean expressions goes by the name Karnaugh map or K map. Named for Maurice Karnaugh.

35. PHY 201 (Blum)35 Maurice Karnaugh

36. PHY 201 (Blum)36 The rules Put the truth table into a form with inputs in Gray code order. Then one identifies output “blocks” (as large as possible). A block must be a rectangle containing 1’s and only 1’s. The simplification rules require that the number of 1’s in a block should be a power of 2 (1, 2, 4, 8, …). However, a given output 1 can belong to more than one block.

37. PHY 201 (Blum)37 Wrapping There are still cases in which inputs differing by only one bit are not adjacent (e.g. the first and last row). Imagine that the rows wrap around, so for instance, a block can include the top and bottom rows (without intermediate rows). Similarly for columns.

38. PHY 201 (Blum)38 WXYZOutput 00001 00010 00100 00110 01001 01011 01101 01110 10001 10010 10100 10110 11001 11011 11101 11110 Karnaugh Example

39. PHY 201 (Blum)39 Karnaugh Example (Unsimplified Boolean algebra expression) WXY’Z + W’XY’Z + WX’Y’Z’ + W’X’Y’Z’ + WXYZ’ + WXY’Z’ + W’XY’Z’ + W’XYZ’

40. PHY 201 (Blum)40 Example in Karnaugh (identifying block in gray code truth table) Z0110 WX\Y0011 001 W’X’Y’Z’ 000 011 W’XY’Z’ 1 W’XY’Z 01 W’XYZ’ 111 WXY’Z’ 1 WXY’Z 01 WXYZ’ 101 WX’Y’Z’ 000

41. PHY 201 (Blum)41 For Yellow Group: W and X inputs change; Y and Z inputs don’t change from zeros. Group represented by Y’Z’ Z0110 WX\Y0011 001 W’X’Y’Z’ 000 011 W’XY’Z’ 1 W’XY’Z 01 W’XYZ’ 111 WXY’Z’ 1 WXY’Z 01 WXYZ’ 101 WX’Y’Z’ 000

42. PHY 201 (Blum)42 For Red Group: W and Z inputs change; X input does not change from 1; Y input does not change from 0. Group represented by XY’ Z0110 WX\Y0011 001 W’X’Y’Z’ 000 011 W’XY’Z’ 1 W’XY’Z 01 W’XYZ’ 111 WXY’Z’ 1 WXY’Z 01 WXYZ’ 101 WX’Y’Z’ 000

43. PHY 201 (Blum)43 For Green group: W and Y inputs change; X input does not change from 1; Z input does not change from 0. Group represented by XZ’ Z0110 WX\Y0011 001 W’X’Y’Z’ 000 011 W’XY’Z’ 1 W’XY’Z 01 W’XYZ’ 111 WXY’Z’ 1 WXY’Z 01 WXYZ’ 101 WX’Y’Z’ 000

44. PHY 201 (Blum)44 Result Y’Z’ + XY’ + X Z’ A block of size two eliminates one Boolean variable; a block of four eliminates two Boolean variables; and so on. To find the expression for a block, identify the inputs for that block that don’t change, AND them together, that’s your expression for the block. Obtain an expression for each block and OR them together. Every 1 must belong to at least one block (even if it is a block onto itself).

45. PHY 201 (Blum)45 From Binary order to Gray code order

46. PHY 201 (Blum)46 From Binary order to Gray code order

47. PHY 201 (Blum)47 Online References http://www.facstaff.bucknell.edu/masta scu/eLessonsHTML/Logic/Logic3.html http://www.cs.usm.maine.edu/~welty/k arnaugh.htm http://en.wikipedia.org/wiki/Frank_Gray _(researcher) http://en.wikipedia.org/wiki/Maurice_Ka rnaugh