Fundamentals of Computer Systems
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1 Fundamentals of Computer Systems Boolean Logic Stephen A. Edwards Columbia University Fall 2012
2 Boolean Logic George Boole
3 Boole s Intuition Behind Boolean Logic Variables X, Y,... represent classes of things No imprecision: A thing either is or is not in a class If X is sheep and Y is white things, XY are all white sheep, and XY = YX XX = X. If X is men and Y is women, X + Y is both men and women, and X + Y = Y + X X + X = X. If X is men, Y is women, and Z is European, Z(X + Y) is European men and women and Z(X+Y) = ZX+ZY.
4 The Axioms of (Any) Boolean Algebra A Boolean Algebra consists of A set of values A An and operator An or operator + A not operator X A false value 0 A A true value 1 A
5 The Axioms of (Any) Boolean Algebra A Boolean Algebra consists of A set of values A An and operator An or operator + Axioms A not operator X A false value 0 A A true value 1 A X + Y = Y + X X Y = Y X X + (Y + Z) = (X + Y) + Z X (Y Z) = (X Y) Z X + (X Y) = X X (X + Y) = X X (Y + Z) = (X Y) + (X Z) X + (Y Z) = (X + Y) (X + Z) X + X = 1 X X = 0
6 The Axioms of (Any) Boolean Algebra A Boolean Algebra consists of A set of values A An and operator An or operator + Axioms A not operator X A false value 0 A A true value 1 A X + Y = Y + X X Y = Y X X + (Y + Z) = (X + Y) + Z X (Y Z) = (X Y) Z X + (X Y) = X X (X + Y) = X X (Y + Z) = (X Y) + (X Z) X + (Y Z) = (X + Y) (X + Z) X + X = 1 X X = 0 We will use the first non-trivial Boolean Algebra: A = {0, 1}. This adds the law of excluded middle: if X = 0 then X = 1 and if X = 1 then X = 0.
7 Simplifying a Boolean Expression You are a New Yorker if you were born in New York or were not born in New York and lived here ten years. Axioms X + (X Y) X + Y = Y + X X Y = Y X X + (Y + Z) = (X + Y) + Z X (Y Z) = (X Y) Z X + (X Y) = X X (X + Y) = X X (Y + Z) = (X Y) + (X Z) X + (Y Z) = (X + Y) (X + Z) X + X = 1 X X = 0 Lemma: X 1 = X (X + X) = X (X + Y) if Y = X = X
8 Simplifying a Boolean Expression You are a New Yorker if you were born in New York or were not born in New York and lived here ten years. Axioms X + (X Y) = (X + X) (X + Y) X + Y = Y + X X Y = Y X X + (Y + Z) = (X + Y) + Z X (Y Z) = (X Y) Z X + (X Y) = X X (X + Y) = X X (Y + Z) = (X Y) + (X Z) X + (Y Z) = (X + Y) (X + Z) X + X = 1 X X = 0 Lemma: X 1 = X (X + X) = X (X + Y) if Y = X = X
9 Simplifying a Boolean Expression You are a New Yorker if you were born in New York or were not born in New York and lived here ten years. Axioms X + (X Y) = (X + X) (X + Y) = 1 (X + Y) X + Y = Y + X X Y = Y X X + (Y + Z) = (X + Y) + Z X (Y Z) = (X Y) Z X + (X Y) = X X (X + Y) = X X (Y + Z) = (X Y) + (X Z) X + (Y Z) = (X + Y) (X + Z) X + X = 1 X X = 0 Lemma: X 1 = X (X + X) = X (X + Y) if Y = X = X
10 Simplifying a Boolean Expression You are a New Yorker if you were born in New York or were not born in New York and lived here ten years. Axioms X + (X Y) = (X + X) (X + Y) = 1 (X + Y) = X + Y Lemma: X + Y = Y + X X Y = Y X X + (Y + Z) = (X + Y) + Z X (Y Z) = (X Y) Z X + (X Y) = X X (X + Y) = X X (Y + Z) = (X Y) + (X Z) X + (Y Z) = (X + Y) (X + Z) X + X = 1 X X = 0 X 1 = X (X + X) = X (X + Y) if Y = X = X
11 More properties = = = = = 1 X + 0 = X X + 1 = 1 X + X = X X + XY = X X + XY = X + Y 0 0 = = = = = 1 X 0 = 0 X 1 = X X X = X X (X + Y) = X X (X + Y) = XY
12 More Examples XY + YZ(Y + Z) = XY + YZY + YZZ = XY + YZ = Y(X + Z) X + Y(X + Z) + XZ = X + YX + YZ + XZ = X + YZ + XZ = X + YZ
13 More Examples XYZ + X(Y + Z) = XYZ + XY + XZ Expand = X(YZ + Y + Z) Factor w.r.t. X = X(YZ + Y + Z + YZ) Z YZ = X(YZ + YZ + Y + Z) Reorder = X Y(Z + Z) + Y + Z Factor w.r.t. Y = X(Y + Y + Z) Y + Y = 1 = X(1 + Z) 1 + Z = 1 = X X1 = X (X + Y + Z)(X + YZ) = XX + XYZ + YX + Y YZ + ZX + Z YZ = X + XYZ + XY + YZ + XZ = X + YZ
14 Sum-of-products form Can always reduce a complex Boolean expression to a sum of product terms: XY + X X + Y(Z + XY) + Z = XY + X(X + YZ + YXY + Z) = XY + XX + XYZ + XYXY + X Z = XY + XYZ + X Z (can do better) = Y(X + XZ) + X Z = Y(X + Z) + X Z = YX Z + X Z = Y + X Z
15 What Does This Have To Do With Logic Circuits? Claude Shannon
16 Shannon s MS Thesis We shall limit our treatment to circuits containing only relay contacts and switches, and therefore at any given time the circuit between any two terminals must be either open (infinite impedance) or closed (zero impedance).
17 Shannon s MS Thesis It is evident that with the above definitions the following postulates hold. 0 0 = 0 A closed circuit in parallel with a closed circuit is a closed circuit = 1 An open circuit in series with an open circuit is an open circuit = = 1 An open circuit in series with a closed circuit in either order is an open circuit. 0 1 = 1 0 = 0 A closed circuit in parallel with an open circuit in either order is an closed circuit = 0 A closed circuit in series with a closed circuit is a closed circuit. 1 1 = 1 An open circuit in parallel with an open circuit is an open circuit. At any give time either X = 0 or X = 1
18 Alternate Notations for Boolean Logic Operator Math Engineer Schematic Copy x X X or X X Complement x X X X AND x y XY or X Y X Y XY OR x y X + Y X Y X + Y
19 Definitions Literal: a Boolean variable or its complement E.g., X X Y Y Implicant: A product of literals E.g., X XY XYZ Minterm: An implicant with each variable once E.g., XYZ XYZ X YZ Maxterm: A sum of literals with each variable once E.g., X + Y + Z X + Y + Z X + Y + Z
20 Be Careful with Bars X Y = XY
21 Be Careful with Bars X Y = XY Let s check all the combinations of X and Y: X Y X Y X Y XY XY
22 Truth Tables A truth table is a canonical representation of a Boolean function X Y Minterm Maxterm X XY XY X + Y X + Y 0 0 X Y X + Y X Y X + Y X Y X + Y X Y X + Y Each row has a unique minterm and maxterm The minterm is 1 maxterm is 0 for only its row
23 Sum-of-minterms and Product-of-maxterms Two mechanical ways to translate a function s truth table into an expression: X Y Minterm Maxterm F 0 0 X Y X + Y X Y X + Y X Y X + Y X Y X + Y 0 The sum of the minterms where the function is 1: F = X Y + X Y The product of the maxterms where the function is 0: F = (X + Y)(X + Y)
24 Expressions to Schematics F = X Y + X Y X Y
25 Expressions to Schematics F = X Y + X Y X X Y Y
26 Expressions to Schematics F = X Y + X Y X X X Y Y Y
27 Expressions to Schematics F = X Y + X Y X X X Y Y Y X Y
28 Expressions to Schematics F = X Y + X Y X X X Y X Y + X Y = F Y Y X Y
29 Expressions to Schematics F = X Y + X Y = (X + Y)(X + Y) X X X Y X Y + X Y = F Y Y X Y (X + Y)(X + Y) = F
30 Minterms and Maxterms: Another Example The minterm and maxterm representation of functions may look very different: X Y Minterm Maxterm F 0 0 X Y X + Y X Y X + Y X Y X + Y X Y X + Y 1 The sum of the minterms where the function is 1: F = X Y + X Y + X Y The product of the maxterms where the function is 0: F = X + Y
31 Expressions to Schematics 2 F = X Y + X Y + X Y = X + Y X X Y + X Y + X Y = F Y X + Y = F
32 The Menagerie of Gates
33 The Menagerie of Gates Buffer Inverter AND NAND OR NOR XOR XNOR
34 De Morgan s Theorem X + Y = X Y X Y = X + Y Proof by Truth Table: X Y X + Y X Y X Y X + Y
35 De Morgan s Theorem in Gates AB = A + B = A + B = A B =
36 Bubble Pushing A B C D F Apply De Morgan s Theorem: Transform NAND into OR with inverted inputs
37 Bubble Pushing A B C D F Apply De Morgan s Theorem: Transform NAND into OR with inverted inputs Two bubbles on a wire cancel
38 Bubble Pushing A B C D F Apply De Morgan s Theorem: Transform NAND into OR with inverted inputs Two bubbles on a wire cancel
39 PONG PONG, Atari 1973 Built from TTL logic gates; no computer, no software Launched the video arcade game revolution
40 Horizontal Ball Control in PONG M L R A B X X X X X X X X The ball moves either left or right. Part of the control circuit has three inputs: M ( move ), L ( left ), and R ( right ). It produces two outputs A and B. Here, X means I don t care what the output is; I never expect this input combination to occur.
41 Horizontal Ball Control in PONG M L R A B E.g., assume all the X s are 0 s and use Minterms: A = MLR + MLR B = M LR + M LR + MLR 3 inv + 4 AND3 + 1 OR2 + 1 OR3
42 Horizontal Ball Control in PONG M L R A B Assume all the X s are 1 s and use Maxterms: A = (M + L + R)(M + L + R) B = M + L + R 3 inv + 3 OR3 + 1 AND2
43 Horizontal Ball Control in PONG M L R A B Choosing better values for the X s and being much more clever: A = M B = MR 1 NAND2 (!)
44 Karnaugh Maps Basic trick: put similar variable values near each other so simple functions are obvious M L R A B X X X X X X X X The M s are already arranged nicely
45 Karnaugh Maps Basic trick: put similar variable values near each other so simple functions are obvious M L R A B X X Let s rearrange the L s by permuting two pairs of rows X X X X X X
46 Karnaugh Maps Basic trick: put similar variable values near each other so simple functions are obvious M L R A B X X X X X X Let s rearrange the L s by permuting two pairs of rows X X
47 Karnaugh Maps Basic trick: put similar variable values near each other so simple functions are obvious M L R A B X X X X Let s rearrange the L s by permuting two pairs of rows X X X X
48 Karnaugh Maps Basic trick: put similar variable values near each other so simple functions are obvious M L R A B X X X X Let s rearrange the L s by permuting two pairs of rows X X X X
49 Karnaugh Maps Basic trick: put similar variable values near each other so simple functions are obvious M L R A B X X X X X X Let s rearrange the L s by permuting two pairs of rows X X
50 Karnaugh Maps Basic trick: put similar variable values near each other so simple functions are obvious M L R A B X X X X X X X X Let s rearrange the L s by permuting two pairs of rows
51 Karnaugh Maps Basic trick: put similar variable values near each other so simple functions are obvious M L R A B X X X X X X X X Let s rearrange the L s by permuting two pairs of rows
52 Karnaugh Maps Basic trick: put similar variable values near each other so simple functions are obvious M L R A B X X X X X X X X Let s rearrange the L s by permuting two pairs of rows
53 Karnaugh Maps Basic trick: put similar variable values near each other so simple functions are obvious M L R A B X X X X X X X X The R s are really crazy; let s use the second dimension
54 Karnaugh Maps Basic trick: put similar variable values near each other so simple functions are obvious M L R A B X X X X X X X X The R s are really crazy; let s use the second dimension
55 Karnaugh Maps Basic trick: put similar variable values near each other so simple functions are obvious M L R A B X0 X X 1 X The R s are really crazy; let s use the second dimension X 1 X X1 X0
56 Karnaugh Maps Basic trick: put similar variable values near each other so simple functions are obvious M L R A B X0 X X 1 X MR X 1 X X1 X0 M
57 Maurice Karnaugh s Maps Transactions of the AIEE, 1953
58 The Seven-Segment Decoder Example a W X Y Z a b c d e f g X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X e f d g c b
59 Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X 0 X 1 1 X X Y W The Karnaugh Map Sum-of-Products Challenge Cover all the 1 s and none of the 0 s using as few literals (gate inputs) as possible. Few, large rectangles are good. Covering X s is optional.
60 Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X 0 X 1 1 X X Y W The minterm solution: cover each 1 with a single implicant. a = W X Y Z + W X Y Z + W X Y Z + W X Y Z + W X Y Z + W X Y Z + W X Y Z + W X Y Z 8 4 = 32 literals 4 inv + 8 AND4 + 1 OR8
61 Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X 0 X 1 1 X X Merging implicants helps Y W Recall the distributive law: AB + AC = A(B + C) a = W X Y Z + W Y + W X Z + W X Y = 12 literals 4 inv + 1 AND4 + 2 AND3 + 1 AND2 + 1 OR4
62 Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X 0 X 1 1 X X Y W Missed one: Remember this is actually a torus. a = X Y Z + W Y + W X Z + W X Y = 11 literals 4 inv + 3 AND3 + 1 AND2 + 1 OR4
63 Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X 0 X 1 1 X X Y W Taking don t-cares into account, we can enlarge two implicants: a = X Z + W Y + W X Z + W X = 9 literals 3 inv + 1 AND3 + 3 AND2 + 1 OR4
64 Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X 0 X 1 1 X X Y W Can also compute the complement of the function and invert the result. Covering the 0 s instead of the 1 s: a = W X Y Z + X Y Z + W Y = 9 literals 5 inv + 1 AND4 + 1 AND3 + 1 AND2 + 1 OR3
65 Karnaugh Map for Seg. a W X Y Z a X X X X X X Z X X 0 X 1 1 X X Y W To display the score, PONG used a TTL chip with this solution in it: (13) (12) OUTPUT a OUTPUT b
66 Boolean Laws and Karnaugh Maps W Y Z W X Y Z + W X Y Z+ W X Y Z + W X Y Z+ W X Y Z + W X Y Z+ W X Y Z + W X Y Z X Factor out the W s
67 Boolean Laws and Karnaugh Maps W Y Z (W + W) X Y Z+ (W + W) X Y Z+ (W + W) X Y Z+ (W + W) X Y Z X Use the identities W + W = 1 and 1 X = X.
68 Boolean Laws and Karnaugh Maps W Y X Z X Y Z+ X Y Z+ X Y Z+ X Y Z Factor out the Y s
69 Boolean Laws and Karnaugh Maps W Y Z (Y + Y) X Z+ (Y + Y) X Z Apply the identities again X
70 Boolean Laws and Karnaugh Maps W Y Z X Z+ X Z Factor out Z X
71 Boolean Laws and Karnaugh Maps W Y Z X (Z + Z) Simplify X
72 Boolean Laws and Karnaugh Maps W Y Z X Done X
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