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Transcript
Boolean Algebra & Digital Logic Boolean algebra was developed by the Englishman George Boole, who published the basic principles in the 1854 treatise An Investigation of the Laws of Thought on Which to Found the Mathematical Theories of Logic and Probabilities. The applicability to computing machines was discovered by three Americans Claude Shannon Symbolic Analysis of Relay and Switching Circuits, 1938. George Stibitz An employee of Bell Labs, he developed a binary adder using mechanical relays in 1937, the model “K 1” adder because he built it at home on his kitchen table. John Atanasoff He was probably the first to use purely electronic relays (vacuum tubes) to build a binary adder. Boolean algebra is a two–valued algebra based on the constant values denoted as either FALSE, TRUE 0, 1 The use of this algebra for computation is based on the fact that binary arithmetic is based on two values, always called “0” and “1”. Basic Boolean Operators Boolean algebra is defined in terms of two constants (defined above), which we call “0” and “1”. Other courses will call these values “F” and “T”. Boolean algebra is defined in terms of three basic operators, to which we shall add a useful fourth operator. The three operators are NOT, AND, & OR. Each of these three basic operators is implemented by a basic electronic device called a “logic gate”. We present the gates along with the definition. NOT This function takes one input and produces one output. The gate is shown below. The circle at the right end of the triangle is important. Algebraically, this function is denoted f(X) = X’ or f(X) = X . The notation X’ is done for typesetting convenience only; the notation The evaluation of the function is simple: 0 = 1 and 1 = 0. X is better. Basic Boolean Operators (Part 2) Logic OR This is a function of two Boolean variables. We denote the logical OR of two Boolean variables X and Y by “X + Y”. Some logic books will use “X Y”. The evaluation of the logical OR function is shown by a truth table X 0 0 1 1 Y 0 1 0 1 X+Y 0 1 1 1 Basic Boolean Operators (Part 3) Logic AND This is a function of two Boolean variables. We denote the logical AND of two Boolean variables X and Y by “X Y”. Some logic books will use “X Y”. The evaluation of the logical AND function is shown by a truth table X 0 0 1 1 Y 0 1 0 1 XY 0 0 0 1 Another Boolean Operator While not a basic Boolean operator, the exclusive OR is very handy. Logic XOR This is a function of two Boolean variables. We denote the logical XOR of two Boolean variables X and Y by “X Y”. Most logic books seem to ignore this function. The evaluation of the logical XOR function is shown by a truth table X Y 0 0 0 1 1 0 1 1 From this last table, we see immediately that X 0 = X and X 1 = X XY 0 1 1 0 Bitwise Operations on Binary Integers These operations are defined for Boolean values, but can be extended to integer values viewed as binary sequences X Y XY X Y X+Y X Y XY 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 1 0 1 0 1 1 0 1 1 0 0 1 1 0 1 1 1 1 1 1 As binary integer examples, I propose to use the ASCII codes for alphabetic characters. ‘A’ Upper case ASCII 0100 0001 Pattern for OR 0010 0000 This gives lower case 0110 0001 ‘N’ 0100 1110 0010 0000 0110 1110 Lower case ASCII 0110 0001 0110 1110 Pattern for AND 1101 1111 1101 1111 Upper case 0100 0001 0100 1110 More Boolean Operations on Integers Consider the 8–bit binary number 0110 1001 The logical NOT of this number is 1001 0110. The second value is the one’s–complement of the first value. We could show more examples. Here are a few with 8–bit binary numbers. 0101 1000 + 0011 0100 0111 1100 0101 1000 0011 0100 0001 0000 In programming languages, these set or clear individual bits in a binary value. One might write the following code in some variant of Jave lower_case = upper_case | 0x20 The lower case representation of a character is the logical OR of the upper case representation and binary 0010 0000 (0x20). Truth Tables The fact that any Boolean variable can assume only one of two possibly values can be shown, by induction, to imply the following. For N > 0, N Boolean variables can take only 2N different combinations of values. For small values of N, we can use this to specify a function using a truth table with 2N rows, plus a header row to label the variables and the function. N 2N 1 2 2 4 3 8 4 16 5 32 6 64 Four–variable truth tables have 17 rows total. This is just manageable. Five–variable truth tables have 33 rows total. This is excessive. N–variable truth tables, for N > 5, are almost useless. Sample Truth Table A B C F1(A, B, C) 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1 This truth table for 3 variables has 23 = 8 rows, plus a label row. This truth table forms a complete definition of the function. We shall later give it another name, but can base all our discussions on this table. Another Sample Truth Table A B C F2(A, B, C) 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1 Two Truth Tables in One A B C F1(A, B, C) F2(A, B, C) 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 Truth tables are often used to show pairs of functions, such as these two, which will later be shown to be related. This is easier than two complete tables. Truth tables rarely show more than two functions, just because large truth tables are “messy” and hard to read. Labeling Rows in a Truth Table The row numbers are just labels. They are not really a part of the truth table, but aid in our discussions and conversions to Boolean expressions. The row numbers are the decimal equivalents of the variable values viewed as binary numbers. The first row is always “row 0”. Row Number X Y Z F(X, Y, Z) 0 0 0 0 1 0 = 04 + 02 + 01 1 0 0 1 1 1 = 04 + 02 + 11 2 0 1 0 0 2 = 04 + 12 + 01 3 0 1 1 1 3 = 04 + 12 + 11 4 1 0 0 1 4 = 14 + 02 + 01 5 1 0 1 0 5 = 14 + 02 + 11 6 1 1 0 1 6 = 14 + 12 + 01 7 1 1 1 1 7 = 14 + 12 + 11 The and Notations These can be viewed as shorthand notation for expressing truth tables. notation: Give the row numbers in which the function has value 1. notation: Give the row numbers in which the function has value 0. Example: The Exclusive OR (XOR) function Row Number 0 1 2 3 X Y = ( 1, 2 ) X 0 0 1 1 Y 0 1 0 1 XY 0 1 1 0 X Y = ( 0, 3 ) Exercise: Convert F(X, Y, Z) = ( 0, 2, 4, 6 ) to the notation. Answer: The function has three variables, so the truth table has 23 = 8 rows, numbered 0 through 7. If rows 0, 2, 4, and 6 have ones, then the rows containing zeroes must be 1, 3, 5, and 7. F(X, Y, Z) = ( 1, 3, 5, 7 ). Examples: F1 and F2 Consider the two functions (F1 and F2), which we shall explain later. Row A B C F1(A, B, C) F2(A, B, C) 0 0 0 0 0 0 1 0 0 1 1 0 2 0 1 0 1 0 3 0 1 1 0 1 4 1 0 0 1 0 5 1 0 1 0 1 6 1 1 0 0 1 7 1 1 1 1 1 F1(X, Y, Z) = ( 1, 2, 4, 7 ) = ( 0, 3, 5, 6 ) F2(X, Y, Z) = ( 3, 5, 6, 7 ) = ( 0, 1, 2, 4 ) Evaluation of Boolean Expressions The relative precedence of the operators is: 1) NOT do this first 2) AND 3) OR do this last As in the usual algebra, parentheses take precedence. AB + CD, often written as AB + CD, is read as (AB) + (CD) A B C D is read as A B C D . The latter is really messy. AB + CD = 10 + 11 = 0 + 1 = 1 A(B + C)D = 1(0 + 1)1 = 1 1 1 = 1 A B C D = 1 0 11 = 0 0 + 1 0 = 0 + 0 = 0 A B = 1 0 = 0 = 1 A B = 1 0 = 0 1 = 0 Question: Where to Put the 0’s and 1’s We now ask how to fill a truth table by evaluating a Boolean expression. Example: F(X, Y, Z) = X Y Y Z X Y Z Step 1: There are three Boolean variables, so the truth table will have 8 rows. Let’s evaluate this function for all eight possible values of X, Y, Z. Row 0 X = 0 Y=0 Z=0 F(X, Y, Z) = 00 + 00 + 010 = 0 + 0 + 0 = 0 Row 1 X = 0 Y=0 Z=1 F(X, Y, Z) = 00 + 01 + 011 = 0 + 0 + 0 = 0 Row 2 X = 0 Y=1 Z=0 F(X, Y, Z) = 01 + 10 + 000 = 0 + 0 + 0 = 0 Row 3 X = 0 Y=1 Z=1 F(X, Y, Z) = 01 + 11 + 001 = 0 + 1 + 0 = 1 Row 4 X = 1 Y=0 Z=0 F(X, Y, Z) = 10 + 00 + 110 = 0 + 0 + 0 = 0 Row 5 X = 1 Y=0 Z=1 F(X, Y, Z) = 10 + 01 + 111 = 0 + 0 + 1 = 1 Row 6 X = 1 Y=1 Z=0 F(X, Y, Z) = 11 + 10 + 100 = 1 + 0 + 0 = 1 Row 7 X = 1 Y=1 Z=1 F(X, Y, Z) = 11 + 11 + 101 = 1 + 1 + 0 = 1 This function can be written as F(X, Y, Z) = (3, 5, 6, 7) where the 1’s are = (0, 1, 2, 4) where the 0’s are Some Basic Identities of Boolean Algebra Here are a few identities basic to Boolean algebra, presented in two forms: the “AND form” and the “OR form”. Some resemble standard algebra and some do not. Identity Name Identity Law Null (Dominance) Law Idempotent Law Inverse Law Commutative Law Associative Law Distributive Law Absorption Law DeMorgan’s Law AND form 1X = X 0X = 0 XX = X XX’ = 0 XY = YX X(YZ) = (XY)Z X + (YZ) = (X+Y)(X+Z) X(X + Y) = X (XY)’ = X’ + Y’ OR form 0+X = X 1+X = 1 X+X = X X+X’ = 1 X+Y = Y+X X+(Y + Z) = (X + Y)+Z X(Y + Z) = (XY) + (XZ) X+(XY) = X (X + Y)’ = X’Y’ It is due to the associative law, one can unambiguously write expressions such as XYZ and X + Y + Z. The Basic Unusual Boolean Theorem Here are two sets of theorems in Boolean algebra. For all X For all X For all X For all X 0X 1X 0+X 1+X = = = = 0 X X 1 OK OK OK What? Consider the following truth tables W 0 0 1 1 X 0 1 0 1 W+X 0 1 1 1 From this, we derive the truth table proving the last two theorems. X 0 1 0+X 0 1 1+X 1 1 The Principle of Duality If a statement in Boolean algebra is true, so is its dual. To take the dual of an expression do the following: change all logical AND to logical OR and all logical OR to logical AND change all 0 to 1 and 1 to 0. Postulate 0X = 1X = 0+X = 1+X = 0 X X 1 An Unexpected Pair: Dual 1+X 0+X 1X 0X = = = = 1 X X 0 Two distributive laws, each the dual of the other. For all Boolean values of Boolean variables A, B, C: A(B + C) = AB + AC For all Boolean values of Boolean variables A, B, C: A + BC = (A + B)(A + C) If A = 1, the statement becomes 1 + BC = (1 + B)(1 + C), or 1 = 11. If A = 0, the statement becomes 0 + BC = (0 + B)(0 + C), or BC = BC. Another Look at the Second Distributive Law Let’s use the truth table approach to proving the equality A + BC = (A + B)(A + C). A 0 0 0 0 1 1 1 1 B 0 0 1 1 0 0 1 1 C 0 1 0 1 0 1 0 1 (A + B) 0 0 1 1 1 1 1 1 (A + C) 0 1 0 1 1 1 1 1 BC 0 0 0 1 0 0 0 1 A + BC 0 0 0 1 1 1 1 1 (A + B)(A + C) 0 0 0 1 1 1 1 1 Note that the last two columns show values that are identical for all possible combinations or X, Y, and Z. For that reason, the two functions are identical. We now show a few tricks for generating truth tables. Example (Page 1) Consider the truth table for the Boolean expression A + BC. We start with the basic truth table, which has eight rows. A B C 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 BC A + BC To fill the column BC, we enter a 0 when B = 0 and the value for C when B = 1. Example (Page 2) A B C BC 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 A + BC To fill the column A + BC, we place a 1 when A = 1, and the value for BC when A = 0. Example (Page 3) A B C BC A + BC 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 1 1 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 Other Logic Gates The top gate shows the NOR gate and its logical equivalent. The bottom line shows the NAND gate and its logical equivalent. In my notes, I call these “derived gates” as they are composites of Boolean gates that are more basic from the purely theoretical approach. X Y OR NOR X Y AND NAND 0 0 0 1 0 0 0 1 0 1 1 0 0 1 0 1 1 0 1 0 1 0 0 1 1 1 1 0 1 1 1 0 AND Gates and OR Gates: The Real Way In actual fact, the NAND and NOR gates are more primitive than the AND, OR, and NOT gates in that they are easier to build from transistors. AND is NOT (NAND) X 0 0 1 1 Y NAND AND 0 1 0 1 1 0 0 1 0 1 0 1 OR is NOT (NOR) X 0 0 1 1 Y NOR OR 0 1 0 1 0 1 0 0 1 1 0 1 Multiple–Input Gates The standard definitions of the AND and OR gates call for two inputs. 3–input and 4–input varieties of these gates are quite common. Here we give informal, but precise, definitions. Gate Number of Inputs Output NOT Exactly 1 0 if input is 1, 1 if input is 0 AND 2 or more 0 if any input is 0 1 if and only if all inputs are 1. OR 2 or more 1 if any input is 1 0 if and only if all inputs are 0. NAND 2 or more 1 if any input is 0 0 if and only if all inputs are 1 NOR 0 if any input is 1 1 if and only if all inputs are 0. 2 or more Example: “Changing the Number of Inputs” Some lab experiments call for gates with input counts other than what we have. We begin with two ways to fabricate a 4–input AND gate from 2–input ANDs. Another Example We now consider how to take a 4–input AND gate and make it act as if it were a 2–input AND gate. There are always multiple solutions. Here are two solutions. There are many others.