Boolean Logic Theory.pptx
ECSE 323 Digital System Design Winter 2011
Boolean Logic Theory
Prof. Warren Gross
Material used in this set of slides was based on Fundamentals of Digital Logic with VHDL Design by S. Brown and Z. Vranesic
1
Textbook Reading • This topic will cover Chapters 2 (IntroducEon to Logic Circuits) and parts of Chapter 4 (OpEmized ImplementaEon of Logic FuncEons) in the textbook. • First, we will look at the basics of logic circuits. We will cover all of Chapter 2. – We will not cover SecEons 2.9 and 2.10 NOW, but we will come back and cover them in a couple of weeks when we study design using CAD tools and the VHDL hardware descripEon language. 2
x = 0
x = 1
S x
S Battery
x
Light
3
S Power supply
x
Light
4
Power supply
S
S
x1
x2
Light
(a) The logical AND function (series connection) S x1
Power supply
S
Light
x2 (b) The logical OR function (parallel connection) 5
S
Power supply
X 1
S
S
X 3
Light
X2
L(x1, x2, x3) = ?
6
R
Power supply
x
S
Light
L(x) = ?
7
x 1 x 2
x 1 x 2 x 1 ⋅ x 2 ⋅ … ⋅ x n
x 1 ⋅ x 2 x n (a) AND gates
x 1 x 2
x 1 x 2
x 1 + x 2
x 1 + x 2 + …+ x n
x n (b) OR gates x
x
(c) NOT gate 8
Truth Tables
• Completely defines a logic funcEon • Size grows exponenEally in number of input variables • N input variables à 2N rows in the truth table
Truth Table for Three-‐input AND and OR operaEons
9
→
→
→
x1
0
0
1
1
x2
0→
1
→
0
→
1
1→
1
→
0
→
0
A
1→
1
→
0
→
1
f
0→
0
→
0
→
1
B
Network that implements
f(x1, x2)
=
x
1
0
x
1
2
0
A
1
0
B
1
0
f
1
0
Time
Timing diagram
• functional behavior
• no delay between change of input values and corresponding change in output values
10
x1
f
x2
f
=
x1
+ x
1
⋅ x
2
x
1
g
x
2
g
=
x
+ x
1
2
• FuncEonally-‐equivalent networks – Many ways to implement the same logic funcEon, each one has a cost – Example cost: # of logic gates + # inputs to all of the logic gates 11
Boolean FuncEons • Need a way to manipulate Boolean funcEons – E. g. to show
x¯1 + x1 · x2 = x¯1 + x2 • Truth tables too cumbersome (large and Eme-‐ consuming to produce for complex funcEons) • Goal: Manipulate and transform logic funcEons to find the least costly implementaEon (logic minimiza?on) 12
Boolean Algebra • George Boole (1849) – applied to circuits built out of switches by Claude Shannon in the 1930’s
• All Boolean operaEons are performed on the set {0,1} – AND, OR, and NOT
• HunEngton’s basic postulates are sufficient to define the algebra 13
HunEngton’s Basic Postulates 5a.
x * 0 = 0
5b.
x + 1 = 1
8a.
x * x’ =0
8b.
x + x’ = 1
10a.
x * y = y * x
10b.
x + y = y + x
12a.
x * (y + z) = x * y + x * z
12b.
x + y * z = (x + y) * (x + z)
DominaEon
CommutaEve
DistribuEve
14
Principle of Duality • Why are the basic postulates listed in pairs (e.g. 5a, 5b?) • The Dual of any statement in Boolean algebra is a statement obtained from the original one by interchanging operands + and * , and their idenEty elements 0 and 1
• The polarity of the variables are unchanged • Example: – If S = AB + AC + B C then Sd = (A+B)(A+C)(B +C)
• Theorem: Dual of any theorem in Boolean algebra is also a theorem – If any statement is a consequence of axioms of Boolean algebra, then dual is also a consequence of those axioms. Duals can be proven by using duals of each step of proof of original statement • Note: In general S �= S d . Also, in general Sd is not a negaEon of S. 15
Examples of Principle of Duality • Write the dual of each of the Boolean equaEons: – a) (A * 1) * (0 + A ) = 0 • SoluEon: To obtain the dual equaEon, interchange + and * , and interchange 0 and 1 . (A + 0) + (1 * A ) = 1
– b) A + A B = A + B • SoluEon: First write equaEon using * to obtain A + (A * B) = A + B. Then dual is A * (A + B) = A * B, which can be wriien as A(A + B) = AB
• Use of brackets – When no brackets used, then the precedence of operaEons is: • NOT ( ) – highest priority • AND (*) • OR (+) – lowest priority
– When brackets are used, then operaEons in brackets have the highest priority • The most inner brackets is executed first 16
• The textbook lists 17 properFes in SecFon 2.5 • learn them well, they are very useful !
Boolean Axioms 1a.
0 * 0 = 0
1b. 1 + 1 = 1 2a. 1 * 1 = 1
Single Variable Theorems 5a. x * 0 = 0 5b. x + 1 = 1 6a. x * 1 = x 6b. x + 0 = x
2b. 0 + 0 = 0
7a. x * x = x
3a. 0 * 1 = 1 * 0 = 0
7b. x + x = x
3b. 1 + 0 = 0 + 1 = 1
8a. x * x’ = 0
4a. If x = 0, then x’ = 1
8b. x + x’ = 1
4b. If x = 1 then x’ = 0
9.
x’’ = x 17
Two and Three Variable ProperEes 10a. 10b. 11a. 11b. 12a. 12b. 13a. 13b.
x * y = y * x x + y = y + x x * (y * z) = (x * y) *z x + (y + z) = (x + y) + z x * (y + z) = x * y + x * z x + y * z = (x + y) * (x + z) x + x * y = x x * (x + y) = x
CommutaEve AssociaEve DistribuEve AbsorpEon
18
Two and Three Variable ProperEes 14a. x * y + x * y’ = x 14b. (x + y) * (x + y’) = x 15a. (x * y )’ = x’ + y’ 15b. 16a. 16b. 17a. 17b.
(x + y)’ = x’ * y’ x + x’ * y = x + y x * (x’ + y) = x * y x * y + y * z + x’ * z = x * y + x’ * z (x + y) * (y + z) * (x’+z) = (x + y) * (x’ +z)
Combining DeMorgan’s Theorem
Consensus
19
ComplementaEon De Morgan’s Theorem (x * y)’ = x’ + y’
(x + y)’ = x’ * y’
The complement !F of a funcEon F is obtained by changing – – – – –
True variables by complemented variables Complemented variables by true variables Boolean constants 0, 1 by 1, 0 respecEvely Operator “•“ by “+“ Operator “+“ by “•“
EXAMPLE:
F = A B + !C !F = (!A + !B) • C
20
Examples •
Prove: (x 1
+ x3 ) · (x¯1 + x¯3 ) = x1 · x¯3 + x¯1 · x3
21
f (x1 , x2 , x3 ) = x1 · x¯3 + x¯2 · x¯3 + x1 · x3 + x¯2 · x3 = x¯1 · x¯2 + x1 · x2 + x1 · x¯2
(1) (2)
22
Venn Diagrams • Visual aid • Venn diagrams provide a good intuiEve understanding of how logic expressions may be equivalent
(a) Constant 1
(b) Constant 0
x x
x x
(c) Variable x x
y
(e) x ⋅ y x
y
(g) x ⋅ y
(d) x x
y
(f) x + y x
y z
(h) x⋅ y + z 23
Synthesis using AND/OR/NOT • Create a product term that has a value of 1 for each valuaEon for which the output funcEon f has to be 1. – Take the logical sum of these product terms to realize f. x 0 0 1 1
y 0 1 0 1
f 0 1 1 0
x y
f(x,y) =
x y
f 24
Synthesis of Logic FuncEons • We will now develop a formal synthesis method • Minterms – A product term in which each of the n variables appears once
• Maxterm – The complement of minterms • Use De Morgans theorem 25
Three-‐variable minterms and maxterms. 26
Sum-‐of-‐Products Form • A funcEon f can be represented by an expression that is a sum of minterms (products) – Each minterm is ANDed with the value of f for the corresponding valuaEon of the input variable
• If each product term is a minterm à canonical sum-‐of-‐products form • SOP or canonical SOP are not necessarily minimal implementaEons
f = m0 * 1 + m1 * 1 + m2 * 0 + m3 * 1 = m0 + m1 + m3 = x1’x2’ + x1’x2 + x1x2 27
28
Product-‐of-‐Sums Form • •
Dual of sum-‐of-‐products f ’ can be represented by sum of minterms for which f ’ = 1 à rows where f = 0 – Then take complement to find a representaEon of f
29
x2
f
x 1 x3
(a) A minimal sum-of-products realization
x 3 x1
f
x2
(b) A minimal product-of-sums realization
In general, the cost of the SOP and POS forms are NOT equal !!
30
Synthesis using NAND / NOR Gates A B Z
A B Z
0 0 1
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 0
NOR
NAND
NAND / NOR gates are simpler to realize in CMOS technology than AND/OR gates (we will see this in the next topic) 31
x 1 x 2
x 1
x 1 x 2
x 2 (a) x 1 x 2 =
x 1 x 2
x 1 + x 2
x 1
x 1 x 2
x 2 (b) x 1 +
x 2 = x 1 x 2
DeMorgan’s theorem in terms of logic gates. 32
x3 x4 x5 x1 x2
x3 x4 x5 x1 x2
x3 x4 x5 x1 x2
NAND-‐NAND network Using NAND gates to implement a sum-‐of-‐products.
33
x 1
x 1
x 3
x 3
x 5
x 5
x 2 x 4
x 2 x 4
x 1 x 2 x 3 x 4 x 5
NOR-‐NOR network Using NOR gates to implement a product-‐of sums.
34
x1 x2
f
x3 (a) POS implementation
x1 x2
f
x3 (b) NOR implementation 35
x1 f
x2 x3 (a) SOP implementation
x1 f
x2 x3 (b) NAND implementation
36
Examples • The textbook has several other examples in Chapter 2, it is worthwhile to go over them • SecEons 2.8 and 2.12 have some more useful examples
37
Logic MinimizaEon • This sub-‐topic covers parts of Chapter 4 of the textbook (4.1 – 4.5, 4.8, 4.9, 4.11) • We will come back and cover 4.6 and 4.7 when we look at combinaEonal circuits
38
Logic MinimizaEon • It is easy to find the canonical SOP or POS form of a logic funcEon • Finding minimal realizaEons not so straighporward – Boolean algebra transformaEon can be tedious and are impracEcal for funcEons with a large number of input varables
• We will study opEmizaEon techniques – Karnaugh map – Tabular method (more suitable for CAD tools) 39
Karnaugh Map • Key idea in minimizaEon is to reduce the number of product (or sum) terms, and the number of variables in each term. • Karnaugh map is a graphical technique that makes it easy to use the combining property: 14a. x * y + x * y’ = x 14b. (x + y) * (x + y’) = x 40
41
Two-‐Variable Map x1 x2 0 0
m0
0 1
m1
1 0
m2
1 1
m3
(a) Truth table
x x2 1
0
1
0 m0 m2 1 m1 m3
(b) Karnaugh map
LocaEon of two-‐variable minterms. 42
x2
f = x1’x2’ + x1’x2 + x1x2
x1 1
0
1
0
1
0
1
1
1
f = x2 + x1
• If a variable is included in all its possible values in a grouping, then it can be removed • Find the smallest number of product terms that cover the funcEon • Draw the largest possible rectangular groupings • cost of each of the product terms should be as low as possible • Two variable map à can have groups of 1, 2 or 4 minterms • group of 1: 2 literals in the product term • group of 2: 1 literal • group of 4: 0 literals (the trivial funcEon f = 1)
43
Three-‐Variable Map x1 x2 x3 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0
m0 m1 m2
1 1 1
m7
m3 m4 m5 m6
(a) Truth table
x3
x1 x 2 00
01
11
10
0
m0
m2
m6
m4
1
m1
m3
m7
m5
(b) Karnaugh map • Label with Gray Code • Columns 1 and 4 differ only in x1 à adjacent • Groups of 1, 2, 4, (8) 44
x1 x2
x3
00
01
11
10
0
0
0
1
1
1
1
0
0
1
00
01
11
10
0
1
1
1
1
1
0
0
0
1
f = x1 x3 + x2 x3
x1 x2
x3
f = x3 + x1 x2
45
Four-‐Variable Map
x3 x 4
x3
x1 x2 00
01
11
10
00 m0
m4 m12 m8
01 m1
m5 m13 m9
11 m3
m7 m15 m11
m2
m6 m14 m10
10 • Groups of 1, 2, 4, 8, (16) • 4, 3, 2, 1, (0) literals
x1
x2
x4
46
x1 x 2 x3 x 4 00 01 11
10
x1 x 2 x3 x 4 00 01 11
10
00 0
0
0
0
00 0
0
0
0
01 0
0
1
1
01 0
0
1
1
11 1
0
0
1
11 1
1
1
1
10
0
0
1
10
1
1
1
1
f 1 = x2 x 3 + x 1 x 3 x 4 x1 x 2 x3 x 4 00 01 11
10
1
f 2 = x3 + x 1 x 4 x1 x 2 x3 x 4 00 01 11
10
00 1
0
0
1
00 1
1
1
0
01 0
0
0
0
01 1
1
1
0
11 1
1
1
0
11 0
0
1
1
10
1
0
1
10
0
1
1
1
f 3 = x2 x 4 + x 1 x 3 + x 2 x 3 x 4
0
x1 x 2 f 4 = x1 x 3 + x 1 x 3 + or x2 x 3 47
Five-‐Variable Map x 3 x4
x1 x2 00
01
11
10
00
x3 x4
x1 x 2 00
01
11
00
01
1
1
1
01
1
11
1
1
11
1
1
10
1
1
10
1
1
x5 = 0
10
1
x5 = 1
f 1 = x1 x3 + x1 x3 x4 + x 1 x2 x3 x 5 48
MinimizaEon Strategy Terminology • Literal: Each appearance of a variable either in true or complemented form – e.g. x1x2’x3 has 3 literals – x1’x3x4’x6 has 4 literals
• Implicant: A product term that indicates the input valuaEon(s) for which a given funcEon f is equal to 1 is called an implicant of f. – Example : the minterms are the most basic implicants (that consist of n literals)
x3
x1 x 2 00
01
11
10
0
1
1
0
0
1
1
1
1
0
x1
x2 x3
11 possible implicants
49
Terminology • Prime Implicant: An implicant is called a prime implicant if it cannot be combined into another implicant that has fewer literals. – i.e. if you delete a literal in a prime implicant, it is no longer a valid implicant – Karnaugh map is one way to find the prime implicants (the largest groupings)
x3
x1 x 2 00
01
11
10
0
1
1
0
0
1
1
1
1
0
x1
x2 x3
x1’ and x2x3 are prime implicants x1’x3 is not a prime implicant (e.g. you can delete x3)
50
Terminology • A cover is a collecEon of implicants that account for all valuaEons for which a given funcEon is equal to 1. • Most funcEons have a number of different covers – E.g. Set of all minterms for which f = 1 is a cover – E.g. Set of all prime implicants is a cover
• Want to choose a cover that has the lowest cost – # gates + total # of inputs to all gates – Assume primary inputs are available in both true and complemented form at zero cost – If an inversion is needed inside the circuit, the corresponding NOT gate and its input is included in the cost 51
MinimizaEon Procedure • Find the minimum-‐cost subset of the prime implicants that will cover the funcEon – Some may have to be included (essen3al prime implicants) while for others there may be a choice – To choose which of the non-‐essenEal prime implicants to include in the cover we usually use a heurisEc • A heurisEc does not consider ALL the possibiliEes, but results in a good result most of the Eme • CAD tools make use of many heurisEcs 52
x3 x4
f ( x1,…, x4) = Σ m(2, 3, 5, 6, 7, 10, 11, 13, 14)
x 1 x2 00
01
11
10
00
x 1 x2 x4
01
1
11
1
1
10
1
1
x1 x3
x2 x3 x4
1 1 1
1
x3 x4
x2 x 3 • 5 prime implicants • EssenEal prime implicants (include a minterm only covered by one PI) • m7 not covered by the essenEal prime implicants • m7 can be covered by either x1’x3 or x1’x2x4. Choose x1’x3 because of lower cost. 53
MinimizaEon Procedure 1. Generate the set of prime implicants for f 2. Find the set of essenEal prime implicants 3. If the set of essenEal prime implicants covers f then choose this set. Otherwise determine the nonessenEal prime implicants that should be added to for a complete minimum-‐cost cover • A heurisEc: select one non-‐essenEal prime implicant in the cover and find the rest of the cover. • Next, assume it is not in the cover and find another cover. Choose the cover with the lowest cost. 54
f ( x1,…, x4) = Σ m(0, 4, 8, 10, 11, 12, 13, 15) Only x3’x4’ is essenEal – include in cover
x3 x4
x x2 1
00 00
1
01 1
11 1
01
1
11
1
10
10 1
x3 x4 x1 x2 x3
1
x1 x2 x4 x1 x3 x4
1
x 1 x2 x 3
x1 x 2 x4 One choice (include x1x2x3’): f = x3’x4’ + x1x2x3’ + x1x3x4 + x1x2’x3
Another choice (do not include x1x2x3’): f = x3’x4’ + x1x2x4 + x1x2’x3 (lower cost)
55
x 1 x 2
x 3 x 4
00
00
01
1
1
11
x 1 x 3 x 4
1
01
x 2 x 3 x 4
1 1
11 10
10
1
1
x 1 x 3 x 4
1 x 2 x 3 x 4
x 1 x 2 x 4
x1 x 2 x 4 x 1 x 2 x 3
x 1 x 2 x 3
• f ( x1,…, x4) = Σ m(0, 2, 4, 5, 10, 11, 13, 15) • All prime implicants are non-‐essenEal • Choose any prime implicant and then exclude it – both alternaEves (blue and black sets of prime implicants) are of equal cost in this example
56
MinimizaEon of POS Forms • CAD tools usually synthesize both the SOP and POS forms and select the one with the lowest cost
f (x1, x2, x3) = Π M(4, 5, 6) = (x1’ + x2)(x1’ + x3)
x3
x1 x 2 00
01
11
10
0
1
1
0
0
1
1
1
1
0
( x1 + x 3 )
( x1 + x2 )
57
Incompletely Specified FuncEons x x x 3 x 4 1 2 00
01
11
10
00
0
1
d
0
01
0
1
d
0
11
0
0
d
0
10
1
1
d
1
x 2 x 3 x 3 x 4
10
00 0
1
d
0
01 0
1
d
0
11
0
0
d
0
10
1
1
d
1
– “Don’t-‐care” condiEon
• Can assign the funcEon value to give the lowest cost circuit
– (can make different choices for the SOP and POS forms)
(a) SOP implementation
x x x3 x 4 1 2 00 01 11
• Certain input combinaEons many be known to never happen
( x 2 + x 3 ) ( x 3 + x 4 )
(b) POS implementation
f ( x1,…, x4) = Σ m(2, 4, 5, 6, 10) + D(12, 13, 14, 15)
58
MulEple-‐Output Circuits x2 x3 x4
• SomeEmes, circuits share some of the same logic gates and can be combined
x1 x3 x1 x3
• One strategy: find minimum-‐ cost realizaEon of f1 and f2 and the share the common product terms x x x 3 x 4 1 2 00
00
01
01
11 1
10 1
1
1
1
11
1
1
10
1
1
(a) Function
f 1
f 1
(c) Combined circuit for f 1 and f 2
x x x 3 x 4 1 2 00
f 2
x2 x3 x4
00
01
01 11
1
1
10
1
1
(b) Function
11 1
10 1
1
1
1 f 2
59
x3 x4
00
01
00 01
1
1
11
1
1
10
1
11
x1 x2
00
01
00 01
1
1
11
1
1
10
1
x1 x4 x1 x2 x4 x1 x2 x3 x4 x2 x4
00
01
00
1
11
x1 x2
x3 x4
10
1
(a) Optimal realization of f x3 x4
x1 x2
01
1
11
1
10
1
11
10
1
1
1
1
(b) Optimal realization of
3
x1 x2
x3 x4
10
00
01
00
1 1
01
1
11
1
10
(c) Optimal realization of f
3
and
f
4
together
1
f
f
Cost = 29 (no sharing)
f4
11
10
1
1
1
1
Cost = 23 (sharing)
3
4
(d) Combined circuit for f
3
and
f
4
60
Another strategy: finding opEmal realizaEons of f3 and f4 independently is not a good idea in this case
Cubical RepresentaEon • Karnaugh map is only useful for minimizing funcEons of a small number of variables • We will study more a systemaEc algebraic opEmizaEon technique • There are many techniques based on cubical representa3on – Truth table – Algebraic expression – Venn Diagram – Karnaugh map – Cubical representaEon
• Cubical representaEon is used in modern CAD tools 61
2-‐Cube • • •
Map a funcEon of n variables onto an n-‐ dimensional cube (n-‐cube) An edge joins two verEces for which the labels differ in the value of only one variable Finding edges for which f = 1 is equivalent to applying the combining property (14a) – Same as finding pairs of adjacent cells for which f = 1 in a Karnaugh map
•
f = {1x, x1} à f = x1 + x2
01 x2
x1
11
1x x1
00
10
x1 x2
f
0 0 1 1
0 1 1 1
0 1 0 1
f (x1, x2) = Σ m(1, 2, 3). 62
3-‐Cube f = {xx0, 10x} = x3’ + x1x2’ is least-‐expensive
f (x1, x2, x3) = Σ m(0, 2, 4, 5, 6)
Some possible representaEons: f = {000,010,100,101,110} = {0x0, 1x0, 101} = {x00,x10,101} = {x00,x10,10x} = {xx0,10x}
63
4-‐Cube
64
n-‐cube • It is impracEcal to draw a cube of more than 4-‐ dimensions, but it is easy to extend the concept to n-‐ dimensions. • n-‐cubes contain lower dimensional cubes (which contain lower dimensional cubes…) • 2k adjacent verEces – k-‐cube – Vertex: 0-‐cube – Edge: 1-‐cube – Side: 2-‐cube – etc…
• The largest possible k-‐cubes that exist for a given funcEon are its prime implicants ! 65
Quine-‐McCluskey Method • A tabular method based on cubical representaEon • The tabular form is more suited to implementaEon in a computer algorithm • Two-‐steps 1. GeneraEon of Prime Implicants 2. DeterminaEon of a Minimum Cover
66
GeneraEon of Prime Implicants • Prime Implicants are the largest possible k-‐cubes for which either f = 1 or f is unspecified (don’t-‐care) • Assume f and it’s don’t-‐cares are specified as minterms – Write down the verEces of f
• Key idea: compare verEces in pairwise fashion to see if they can be combined into larger cubes
– Then see if these cubes can be combined into larger cubes, etc…keep going unEl you find the prime implicants
• Two cubes that are idenEcal in all variables (coordinates) except one, for which one cube has the value 0 and the other has 1, then these cubes can be combined into a larger cube.
67
Example 1 • f(x1,x2,x3,x4) = {1000,1001,1010,1011} • Cubes 1000 and 1001 differ only in one variable (x4) – Combine into new cube 100x
• Cubes 1010 and 1011 can be combined into 101x • Combine 100x and 101x into 10xx • f = x1x2’ 68
Example 2 • Group the minterms such that the cubes in each group have the same number of 1s and sort the groups by the number of 1s • Only need to compare each cube in a group with all the cubes in the immediately preceding group • Place a check if a (k-‐1)-‐cube is included in a k cube
List 1(0-‐cubes) 0
0 0 0 0
4 8
0 1 0 0 1 0 0 0
10 12
1 0 1 0 1 1 0 0
11 13
1 0 1 1 1 1 0 1
15
1 1 1 1
List 2 (1-‐cubes) 0,4 0,8
0 x 0 0 x 0 0 0
8,10 4,12 8,12
1 0 x 0 x 1 0 0 1 x 0 0
10,11 12,13
1 0 1 x 1 1 0 x
11,15 13,15
1 x 1 1 1 1 x 1
f ( x1,…, x4) = Σ m(0,4,8,10,11,12,13,15)
List 3(2-‐cubes) 0,4,8,12
x x 0 0
The cubes without a check are the prime implicants of f P = {10x0, 101x, 110x, 1x11, 11x1, xx00} = {p1,p2,p3,p4,p5,p6}
69
DeterminaEon of a Minimum Cover • Find a minimum-‐cost cover for f • Assume that the cost is directly proporEonal to the number of inputs to all gates – i.e. the number of literals in the prime implicants
• Construct a prime-‐implicant cover table
– A row for each prime implicant – A column for each minterm to be covered – Put checks to indicate the minterms covered by each prime implicant
• If there is a single check in some column, then the prime implicant that covers the minterm of this column is essen?al and it must be included in the final cover
– Remove the rows corresponding to the essenEal prime implicants and the columns of the minterms associated with them
70
Prime implicant p 1
Minterm 0 4 8 10 11 12 13 15
1 0 x 0
p 2
1 0 1 x
p 4
1 x 1 1
p 3 p 5 p 6
1 1 0 x 1 1 x 1 x x 0 0 Initial prime implicant cover table
Prime implicant p 1
Minterm 10 11 13 15
p 2 p 3 p 4 p 5 After the removal of essential prime implicants 71
Row-‐Dominance •
p1 only covers minterm 10 while p2 covers both minterm 10 and minterm 11 – p2 dominates p1 – Cost of p2 is the same as p1 therefore choose p2 rather than p1 and remove p1 from the table
• •
• • •
p5 dominates p3 – Remove p3
Must choose p2 to cover minterm 10 and p5 to cover minterm 13 which takes care of covering minterm 11 and minterm 15 Final cover is C = {p2,p5,p6} = {101x, 11x1,xx00} f = x1x2’x3 + x1x2x4 + x3’x4 Do not remove a dominated row if the cost of its prime implicant is less than that of the dominaEng row’s prime implicant
Prime implicant p 1
Minterm 10 11 13 15
p 2 p 3 p 4 p 5
Prime implicant
Minterm 10 11 13 15
p 2 p 4 p 5 After the removal of dominated rows 72
Example 3 – Column Dominance • f ( x1,…, x4) = Σ m(0,2,5,6,7,8,9,13) + D(1,12,15) • P = {00x0,0x10,011x,x00x,xx01,1x0x,x1x1} = {p1,p2,p3,p4,p5,p6,p7} • Do not include don’t care minterms in the prime implicant cover table since they don’t need to be covered List 1
List 2
0
0 0 0 0
1 2 8
0 0 0 1 0 0 1 0 1 0 0 0
5 6 9 12
0 0 1 1
7 13
0 1 1 1 1 1 0 1
15
1 1 1 1
1 1 0 1
0 1 0 0
1 0 1 0
List 3
0,1 0,2 0,8
0 0 0 x 0 0 x 0 x 0 0 0
1,5 2,6 1,9 8,9 8,12
0 0 x 1 1
x x 0 0 x
0 1 0 0 0
1 0 1 x 0
5,7 6,7 5,13 9,13 12,13
0 0 x 1 1
1 1 1 x 1
x 1 0 0 0
1 x 1 1 x
7,15 13,15
x 1 1 1 1 1 x 1
0,1,8,9
x 0 0 x
1,5,9,13 8,9,12,13
x x 0 1 1 x 0 x
5,7,13,15
x 1 x 1
73
0 0 x 0 0 x 1 0 0 1 1 x x 0 0 x x x 0 1 1 x 0 x x 1 x 1
Prime implicant
p1 p2 p3 p4 p5 p6 p7
0
2
5
Minterm 6 7
0 0 x 0 0 x 1 0 0 1 1 x x 0 0 x x x 0 1 1 x 0 x x 1 x 1
Prime implicant
Row p4 dominates p6 and p7 dominates p5
0
2
0
2
(c) After the removal of rows
9
13
Minterm 5 6
7
p 5 and
8
p 6
• Column 13 dominates column 5
Minterm 5 6
(b) After the removal of columns 9 and 13 Prime implicant p1 p2 p3 p4 p7
• Remove the dominaFng column (9)
(a) Initial prime implicant cover table
p1 p2 p3 p4 p5 p6 p7
8
• Column 9 dominates column 8
7
8
• Remove column 13
Prime implicant p1 p2 p3
Minterm 2 6
(d) After including in the cover
p
p 4 and
C = {p2,p4,p7} = {0x10,x00x,x1x1}
7
f = x1’x3x4’ + x2x3’ + x2x4
74
Example 4 • No essenEal prime implicants or dominaEng rows or columns • All prime implicants have equal cost • Use branching – Choose any prime implicant, say p3 and include it in cover, compute rest of cover and its cost – C1 = {p1,p3,p4} – Now, find a new cover excluding p3 and compare costs – C2 = {p1, p5} = Cmin
75
Quine-‐McCluskey Summary 1. 2. 3. 4.
5. 6.
Start with list of cubes that represent minterms where f = 1 or don’t care. Generate the prime implicants by successive pairwise comparisons of the cubes. Derive a cover table which indicates minterms where f = 1 that are covered by each prime implicant. Include the essenEal prime implicants (if any) in the final cover and reduce the table by removing both these prime implicants and the covered minterms. Use the concept of row and column dominance to reduce the table further. A dominated row is only removed if the cost of its prime implicant is greater than or equal to the cost of the dominaEng row’s prime implicant. Repeat steps 3 and 4 unEl the cover table is empty or no further reducEon of the table is possible. If the reduced cover table is not empty, then use the branching approach to determine the remaining prime implicants that should be included in a minimum cost cover. 76
PracEcal LimitaEons of Quine-‐ McCluskey • FuncEons are seldom defined in the form of minterms, the are usually given as algebraic expressions or sets of cubes – List of minterms may be very large
• Many comparisons – Slow
• Cover table is computaEonally expensive
77
Boolean Logic Theory
Prof. Warren Gross
Material used in this set of slides was based on Fundamentals of Digital Logic with VHDL Design by S. Brown and Z. Vranesic
1
Textbook Reading • This topic will cover Chapters 2 (IntroducEon to Logic Circuits) and parts of Chapter 4 (OpEmized ImplementaEon of Logic FuncEons) in the textbook. • First, we will look at the basics of logic circuits. We will cover all of Chapter 2. – We will not cover SecEons 2.9 and 2.10 NOW, but we will come back and cover them in a couple of weeks when we study design using CAD tools and the VHDL hardware descripEon language. 2
x = 0
x = 1
S x
S Battery
x
Light
3
S Power supply
x
Light
4
Power supply
S
S
x1
x2
Light
(a) The logical AND function (series connection) S x1
Power supply
S
Light
x2 (b) The logical OR function (parallel connection) 5
S
Power supply
X 1
S
S
X 3
Light
X2
L(x1, x2, x3) = ?
6
R
Power supply
x
S
Light
L(x) = ?
7
x 1 x 2
x 1 x 2 x 1 ⋅ x 2 ⋅ … ⋅ x n
x 1 ⋅ x 2 x n (a) AND gates
x 1 x 2
x 1 x 2
x 1 + x 2
x 1 + x 2 + …+ x n
x n (b) OR gates x
x
(c) NOT gate 8
Truth Tables
• Completely defines a logic funcEon • Size grows exponenEally in number of input variables • N input variables à 2N rows in the truth table
Truth Table for Three-‐input AND and OR operaEons
9
→
→
→
x1
0
0
1
1
x2
0→
1
→
0
→
1
1→
1
→
0
→
0
A
1→
1
→
0
→
1
f
0→
0
→
0
→
1
B
Network that implements
f(x1, x2)
=
x
1
0
x
1
2
0
A
1
0
B
1
0
f
1
0
Time
Timing diagram
• functional behavior
• no delay between change of input values and corresponding change in output values
10
x1
f
x2
f
=
x1
+ x
1
⋅ x
2
x
1
g
x
2
g
=
x
+ x
1
2
• FuncEonally-‐equivalent networks – Many ways to implement the same logic funcEon, each one has a cost – Example cost: # of logic gates + # inputs to all of the logic gates 11
Boolean FuncEons • Need a way to manipulate Boolean funcEons – E. g. to show
x¯1 + x1 · x2 = x¯1 + x2 • Truth tables too cumbersome (large and Eme-‐ consuming to produce for complex funcEons) • Goal: Manipulate and transform logic funcEons to find the least costly implementaEon (logic minimiza?on) 12
Boolean Algebra • George Boole (1849) – applied to circuits built out of switches by Claude Shannon in the 1930’s
• All Boolean operaEons are performed on the set {0,1} – AND, OR, and NOT
• HunEngton’s basic postulates are sufficient to define the algebra 13
HunEngton’s Basic Postulates 5a.
x * 0 = 0
5b.
x + 1 = 1
8a.
x * x’ =0
8b.
x + x’ = 1
10a.
x * y = y * x
10b.
x + y = y + x
12a.
x * (y + z) = x * y + x * z
12b.
x + y * z = (x + y) * (x + z)
DominaEon
CommutaEve
DistribuEve
14
Principle of Duality • Why are the basic postulates listed in pairs (e.g. 5a, 5b?) • The Dual of any statement in Boolean algebra is a statement obtained from the original one by interchanging operands + and * , and their idenEty elements 0 and 1
• The polarity of the variables are unchanged • Example: – If S = AB + AC + B C then Sd = (A+B)(A+C)(B +C)
• Theorem: Dual of any theorem in Boolean algebra is also a theorem – If any statement is a consequence of axioms of Boolean algebra, then dual is also a consequence of those axioms. Duals can be proven by using duals of each step of proof of original statement • Note: In general S �= S d . Also, in general Sd is not a negaEon of S. 15
Examples of Principle of Duality • Write the dual of each of the Boolean equaEons: – a) (A * 1) * (0 + A ) = 0 • SoluEon: To obtain the dual equaEon, interchange + and * , and interchange 0 and 1 . (A + 0) + (1 * A ) = 1
– b) A + A B = A + B • SoluEon: First write equaEon using * to obtain A + (A * B) = A + B. Then dual is A * (A + B) = A * B, which can be wriien as A(A + B) = AB
• Use of brackets – When no brackets used, then the precedence of operaEons is: • NOT ( ) – highest priority • AND (*) • OR (+) – lowest priority
– When brackets are used, then operaEons in brackets have the highest priority • The most inner brackets is executed first 16
• The textbook lists 17 properFes in SecFon 2.5 • learn them well, they are very useful !
Boolean Axioms 1a.
0 * 0 = 0
1b. 1 + 1 = 1 2a. 1 * 1 = 1
Single Variable Theorems 5a. x * 0 = 0 5b. x + 1 = 1 6a. x * 1 = x 6b. x + 0 = x
2b. 0 + 0 = 0
7a. x * x = x
3a. 0 * 1 = 1 * 0 = 0
7b. x + x = x
3b. 1 + 0 = 0 + 1 = 1
8a. x * x’ = 0
4a. If x = 0, then x’ = 1
8b. x + x’ = 1
4b. If x = 1 then x’ = 0
9.
x’’ = x 17
Two and Three Variable ProperEes 10a. 10b. 11a. 11b. 12a. 12b. 13a. 13b.
x * y = y * x x + y = y + x x * (y * z) = (x * y) *z x + (y + z) = (x + y) + z x * (y + z) = x * y + x * z x + y * z = (x + y) * (x + z) x + x * y = x x * (x + y) = x
CommutaEve AssociaEve DistribuEve AbsorpEon
18
Two and Three Variable ProperEes 14a. x * y + x * y’ = x 14b. (x + y) * (x + y’) = x 15a. (x * y )’ = x’ + y’ 15b. 16a. 16b. 17a. 17b.
(x + y)’ = x’ * y’ x + x’ * y = x + y x * (x’ + y) = x * y x * y + y * z + x’ * z = x * y + x’ * z (x + y) * (y + z) * (x’+z) = (x + y) * (x’ +z)
Combining DeMorgan’s Theorem
Consensus
19
ComplementaEon De Morgan’s Theorem (x * y)’ = x’ + y’
(x + y)’ = x’ * y’
The complement !F of a funcEon F is obtained by changing – – – – –
True variables by complemented variables Complemented variables by true variables Boolean constants 0, 1 by 1, 0 respecEvely Operator “•“ by “+“ Operator “+“ by “•“
EXAMPLE:
F = A B + !C !F = (!A + !B) • C
20
Examples •
Prove: (x 1
+ x3 ) · (x¯1 + x¯3 ) = x1 · x¯3 + x¯1 · x3
21
f (x1 , x2 , x3 ) = x1 · x¯3 + x¯2 · x¯3 + x1 · x3 + x¯2 · x3 = x¯1 · x¯2 + x1 · x2 + x1 · x¯2
(1) (2)
22
Venn Diagrams • Visual aid • Venn diagrams provide a good intuiEve understanding of how logic expressions may be equivalent
(a) Constant 1
(b) Constant 0
x x
x x
(c) Variable x x
y
(e) x ⋅ y x
y
(g) x ⋅ y
(d) x x
y
(f) x + y x
y z
(h) x⋅ y + z 23
Synthesis using AND/OR/NOT • Create a product term that has a value of 1 for each valuaEon for which the output funcEon f has to be 1. – Take the logical sum of these product terms to realize f. x 0 0 1 1
y 0 1 0 1
f 0 1 1 0
x y
f(x,y) =
x y
f 24
Synthesis of Logic FuncEons • We will now develop a formal synthesis method • Minterms – A product term in which each of the n variables appears once
• Maxterm – The complement of minterms • Use De Morgans theorem 25
Three-‐variable minterms and maxterms. 26
Sum-‐of-‐Products Form • A funcEon f can be represented by an expression that is a sum of minterms (products) – Each minterm is ANDed with the value of f for the corresponding valuaEon of the input variable
• If each product term is a minterm à canonical sum-‐of-‐products form • SOP or canonical SOP are not necessarily minimal implementaEons
f = m0 * 1 + m1 * 1 + m2 * 0 + m3 * 1 = m0 + m1 + m3 = x1’x2’ + x1’x2 + x1x2 27
28
Product-‐of-‐Sums Form • •
Dual of sum-‐of-‐products f ’ can be represented by sum of minterms for which f ’ = 1 à rows where f = 0 – Then take complement to find a representaEon of f
29
x2
f
x 1 x3
(a) A minimal sum-of-products realization
x 3 x1
f
x2
(b) A minimal product-of-sums realization
In general, the cost of the SOP and POS forms are NOT equal !!
30
Synthesis using NAND / NOR Gates A B Z
A B Z
0 0 1
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 0
NOR
NAND
NAND / NOR gates are simpler to realize in CMOS technology than AND/OR gates (we will see this in the next topic) 31
x 1 x 2
x 1
x 1 x 2
x 2 (a) x 1 x 2 =
x 1 x 2
x 1 + x 2
x 1
x 1 x 2
x 2 (b) x 1 +
x 2 = x 1 x 2
DeMorgan’s theorem in terms of logic gates. 32
x3 x4 x5 x1 x2
x3 x4 x5 x1 x2
x3 x4 x5 x1 x2
NAND-‐NAND network Using NAND gates to implement a sum-‐of-‐products.
33
x 1
x 1
x 3
x 3
x 5
x 5
x 2 x 4
x 2 x 4
x 1 x 2 x 3 x 4 x 5
NOR-‐NOR network Using NOR gates to implement a product-‐of sums.
34
x1 x2
f
x3 (a) POS implementation
x1 x2
f
x3 (b) NOR implementation 35
x1 f
x2 x3 (a) SOP implementation
x1 f
x2 x3 (b) NAND implementation
36
Examples • The textbook has several other examples in Chapter 2, it is worthwhile to go over them • SecEons 2.8 and 2.12 have some more useful examples
37
Logic MinimizaEon • This sub-‐topic covers parts of Chapter 4 of the textbook (4.1 – 4.5, 4.8, 4.9, 4.11) • We will come back and cover 4.6 and 4.7 when we look at combinaEonal circuits
38
Logic MinimizaEon • It is easy to find the canonical SOP or POS form of a logic funcEon • Finding minimal realizaEons not so straighporward – Boolean algebra transformaEon can be tedious and are impracEcal for funcEons with a large number of input varables
• We will study opEmizaEon techniques – Karnaugh map – Tabular method (more suitable for CAD tools) 39
Karnaugh Map • Key idea in minimizaEon is to reduce the number of product (or sum) terms, and the number of variables in each term. • Karnaugh map is a graphical technique that makes it easy to use the combining property: 14a. x * y + x * y’ = x 14b. (x + y) * (x + y’) = x 40
41
Two-‐Variable Map x1 x2 0 0
m0
0 1
m1
1 0
m2
1 1
m3
(a) Truth table
x x2 1
0
1
0 m0 m2 1 m1 m3
(b) Karnaugh map
LocaEon of two-‐variable minterms. 42
x2
f = x1’x2’ + x1’x2 + x1x2
x1 1
0
1
0
1
0
1
1
1
f = x2 + x1
• If a variable is included in all its possible values in a grouping, then it can be removed • Find the smallest number of product terms that cover the funcEon • Draw the largest possible rectangular groupings • cost of each of the product terms should be as low as possible • Two variable map à can have groups of 1, 2 or 4 minterms • group of 1: 2 literals in the product term • group of 2: 1 literal • group of 4: 0 literals (the trivial funcEon f = 1)
43
Three-‐Variable Map x1 x2 x3 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0
m0 m1 m2
1 1 1
m7
m3 m4 m5 m6
(a) Truth table
x3
x1 x 2 00
01
11
10
0
m0
m2
m6
m4
1
m1
m3
m7
m5
(b) Karnaugh map • Label with Gray Code • Columns 1 and 4 differ only in x1 à adjacent • Groups of 1, 2, 4, (8) 44
x1 x2
x3
00
01
11
10
0
0
0
1
1
1
1
0
0
1
00
01
11
10
0
1
1
1
1
1
0
0
0
1
f = x1 x3 + x2 x3
x1 x2
x3
f = x3 + x1 x2
45
Four-‐Variable Map
x3 x 4
x3
x1 x2 00
01
11
10
00 m0
m4 m12 m8
01 m1
m5 m13 m9
11 m3
m7 m15 m11
m2
m6 m14 m10
10 • Groups of 1, 2, 4, 8, (16) • 4, 3, 2, 1, (0) literals
x1
x2
x4
46
x1 x 2 x3 x 4 00 01 11
10
x1 x 2 x3 x 4 00 01 11
10
00 0
0
0
0
00 0
0
0
0
01 0
0
1
1
01 0
0
1
1
11 1
0
0
1
11 1
1
1
1
10
0
0
1
10
1
1
1
1
f 1 = x2 x 3 + x 1 x 3 x 4 x1 x 2 x3 x 4 00 01 11
10
1
f 2 = x3 + x 1 x 4 x1 x 2 x3 x 4 00 01 11
10
00 1
0
0
1
00 1
1
1
0
01 0
0
0
0
01 1
1
1
0
11 1
1
1
0
11 0
0
1
1
10
1
0
1
10
0
1
1
1
f 3 = x2 x 4 + x 1 x 3 + x 2 x 3 x 4
0
x1 x 2 f 4 = x1 x 3 + x 1 x 3 + or x2 x 3 47
Five-‐Variable Map x 3 x4
x1 x2 00
01
11
10
00
x3 x4
x1 x 2 00
01
11
00
01
1
1
1
01
1
11
1
1
11
1
1
10
1
1
10
1
1
x5 = 0
10
1
x5 = 1
f 1 = x1 x3 + x1 x3 x4 + x 1 x2 x3 x 5 48
MinimizaEon Strategy Terminology • Literal: Each appearance of a variable either in true or complemented form – e.g. x1x2’x3 has 3 literals – x1’x3x4’x6 has 4 literals
• Implicant: A product term that indicates the input valuaEon(s) for which a given funcEon f is equal to 1 is called an implicant of f. – Example : the minterms are the most basic implicants (that consist of n literals)
x3
x1 x 2 00
01
11
10
0
1
1
0
0
1
1
1
1
0
x1
x2 x3
11 possible implicants
49
Terminology • Prime Implicant: An implicant is called a prime implicant if it cannot be combined into another implicant that has fewer literals. – i.e. if you delete a literal in a prime implicant, it is no longer a valid implicant – Karnaugh map is one way to find the prime implicants (the largest groupings)
x3
x1 x 2 00
01
11
10
0
1
1
0
0
1
1
1
1
0
x1
x2 x3
x1’ and x2x3 are prime implicants x1’x3 is not a prime implicant (e.g. you can delete x3)
50
Terminology • A cover is a collecEon of implicants that account for all valuaEons for which a given funcEon is equal to 1. • Most funcEons have a number of different covers – E.g. Set of all minterms for which f = 1 is a cover – E.g. Set of all prime implicants is a cover
• Want to choose a cover that has the lowest cost – # gates + total # of inputs to all gates – Assume primary inputs are available in both true and complemented form at zero cost – If an inversion is needed inside the circuit, the corresponding NOT gate and its input is included in the cost 51
MinimizaEon Procedure • Find the minimum-‐cost subset of the prime implicants that will cover the funcEon – Some may have to be included (essen3al prime implicants) while for others there may be a choice – To choose which of the non-‐essenEal prime implicants to include in the cover we usually use a heurisEc • A heurisEc does not consider ALL the possibiliEes, but results in a good result most of the Eme • CAD tools make use of many heurisEcs 52
x3 x4
f ( x1,…, x4) = Σ m(2, 3, 5, 6, 7, 10, 11, 13, 14)
x 1 x2 00
01
11
10
00
x 1 x2 x4
01
1
11
1
1
10
1
1
x1 x3
x2 x3 x4
1 1 1
1
x3 x4
x2 x 3 • 5 prime implicants • EssenEal prime implicants (include a minterm only covered by one PI) • m7 not covered by the essenEal prime implicants • m7 can be covered by either x1’x3 or x1’x2x4. Choose x1’x3 because of lower cost. 53
MinimizaEon Procedure 1. Generate the set of prime implicants for f 2. Find the set of essenEal prime implicants 3. If the set of essenEal prime implicants covers f then choose this set. Otherwise determine the nonessenEal prime implicants that should be added to for a complete minimum-‐cost cover • A heurisEc: select one non-‐essenEal prime implicant in the cover and find the rest of the cover. • Next, assume it is not in the cover and find another cover. Choose the cover with the lowest cost. 54
f ( x1,…, x4) = Σ m(0, 4, 8, 10, 11, 12, 13, 15) Only x3’x4’ is essenEal – include in cover
x3 x4
x x2 1
00 00
1
01 1
11 1
01
1
11
1
10
10 1
x3 x4 x1 x2 x3
1
x1 x2 x4 x1 x3 x4
1
x 1 x2 x 3
x1 x 2 x4 One choice (include x1x2x3’): f = x3’x4’ + x1x2x3’ + x1x3x4 + x1x2’x3
Another choice (do not include x1x2x3’): f = x3’x4’ + x1x2x4 + x1x2’x3 (lower cost)
55
x 1 x 2
x 3 x 4
00
00
01
1
1
11
x 1 x 3 x 4
1
01
x 2 x 3 x 4
1 1
11 10
10
1
1
x 1 x 3 x 4
1 x 2 x 3 x 4
x 1 x 2 x 4
x1 x 2 x 4 x 1 x 2 x 3
x 1 x 2 x 3
• f ( x1,…, x4) = Σ m(0, 2, 4, 5, 10, 11, 13, 15) • All prime implicants are non-‐essenEal • Choose any prime implicant and then exclude it – both alternaEves (blue and black sets of prime implicants) are of equal cost in this example
56
MinimizaEon of POS Forms • CAD tools usually synthesize both the SOP and POS forms and select the one with the lowest cost
f (x1, x2, x3) = Π M(4, 5, 6) = (x1’ + x2)(x1’ + x3)
x3
x1 x 2 00
01
11
10
0
1
1
0
0
1
1
1
1
0
( x1 + x 3 )
( x1 + x2 )
57
Incompletely Specified FuncEons x x x 3 x 4 1 2 00
01
11
10
00
0
1
d
0
01
0
1
d
0
11
0
0
d
0
10
1
1
d
1
x 2 x 3 x 3 x 4
10
00 0
1
d
0
01 0
1
d
0
11
0
0
d
0
10
1
1
d
1
– “Don’t-‐care” condiEon
• Can assign the funcEon value to give the lowest cost circuit
– (can make different choices for the SOP and POS forms)
(a) SOP implementation
x x x3 x 4 1 2 00 01 11
• Certain input combinaEons many be known to never happen
( x 2 + x 3 ) ( x 3 + x 4 )
(b) POS implementation
f ( x1,…, x4) = Σ m(2, 4, 5, 6, 10) + D(12, 13, 14, 15)
58
MulEple-‐Output Circuits x2 x3 x4
• SomeEmes, circuits share some of the same logic gates and can be combined
x1 x3 x1 x3
• One strategy: find minimum-‐ cost realizaEon of f1 and f2 and the share the common product terms x x x 3 x 4 1 2 00
00
01
01
11 1
10 1
1
1
1
11
1
1
10
1
1
(a) Function
f 1
f 1
(c) Combined circuit for f 1 and f 2
x x x 3 x 4 1 2 00
f 2
x2 x3 x4
00
01
01 11
1
1
10
1
1
(b) Function
11 1
10 1
1
1
1 f 2
59
x3 x4
00
01
00 01
1
1
11
1
1
10
1
11
x1 x2
00
01
00 01
1
1
11
1
1
10
1
x1 x4 x1 x2 x4 x1 x2 x3 x4 x2 x4
00
01
00
1
11
x1 x2
x3 x4
10
1
(a) Optimal realization of f x3 x4
x1 x2
01
1
11
1
10
1
11
10
1
1
1
1
(b) Optimal realization of
3
x1 x2
x3 x4
10
00
01
00
1 1
01
1
11
1
10
(c) Optimal realization of f
3
and
f
4
together
1
f
f
Cost = 29 (no sharing)
f4
11
10
1
1
1
1
Cost = 23 (sharing)
3
4
(d) Combined circuit for f
3
and
f
4
60
Another strategy: finding opEmal realizaEons of f3 and f4 independently is not a good idea in this case
Cubical RepresentaEon • Karnaugh map is only useful for minimizing funcEons of a small number of variables • We will study more a systemaEc algebraic opEmizaEon technique • There are many techniques based on cubical representa3on – Truth table – Algebraic expression – Venn Diagram – Karnaugh map – Cubical representaEon
• Cubical representaEon is used in modern CAD tools 61
2-‐Cube • • •
Map a funcEon of n variables onto an n-‐ dimensional cube (n-‐cube) An edge joins two verEces for which the labels differ in the value of only one variable Finding edges for which f = 1 is equivalent to applying the combining property (14a) – Same as finding pairs of adjacent cells for which f = 1 in a Karnaugh map
•
f = {1x, x1} à f = x1 + x2
01 x2
x1
11
1x x1
00
10
x1 x2
f
0 0 1 1
0 1 1 1
0 1 0 1
f (x1, x2) = Σ m(1, 2, 3). 62
3-‐Cube f = {xx0, 10x} = x3’ + x1x2’ is least-‐expensive
f (x1, x2, x3) = Σ m(0, 2, 4, 5, 6)
Some possible representaEons: f = {000,010,100,101,110} = {0x0, 1x0, 101} = {x00,x10,101} = {x00,x10,10x} = {xx0,10x}
63
4-‐Cube
64
n-‐cube • It is impracEcal to draw a cube of more than 4-‐ dimensions, but it is easy to extend the concept to n-‐ dimensions. • n-‐cubes contain lower dimensional cubes (which contain lower dimensional cubes…) • 2k adjacent verEces – k-‐cube – Vertex: 0-‐cube – Edge: 1-‐cube – Side: 2-‐cube – etc…
• The largest possible k-‐cubes that exist for a given funcEon are its prime implicants ! 65
Quine-‐McCluskey Method • A tabular method based on cubical representaEon • The tabular form is more suited to implementaEon in a computer algorithm • Two-‐steps 1. GeneraEon of Prime Implicants 2. DeterminaEon of a Minimum Cover
66
GeneraEon of Prime Implicants • Prime Implicants are the largest possible k-‐cubes for which either f = 1 or f is unspecified (don’t-‐care) • Assume f and it’s don’t-‐cares are specified as minterms – Write down the verEces of f
• Key idea: compare verEces in pairwise fashion to see if they can be combined into larger cubes
– Then see if these cubes can be combined into larger cubes, etc…keep going unEl you find the prime implicants
• Two cubes that are idenEcal in all variables (coordinates) except one, for which one cube has the value 0 and the other has 1, then these cubes can be combined into a larger cube.
67
Example 1 • f(x1,x2,x3,x4) = {1000,1001,1010,1011} • Cubes 1000 and 1001 differ only in one variable (x4) – Combine into new cube 100x
• Cubes 1010 and 1011 can be combined into 101x • Combine 100x and 101x into 10xx • f = x1x2’ 68
Example 2 • Group the minterms such that the cubes in each group have the same number of 1s and sort the groups by the number of 1s • Only need to compare each cube in a group with all the cubes in the immediately preceding group • Place a check if a (k-‐1)-‐cube is included in a k cube
List 1(0-‐cubes) 0
0 0 0 0
4 8
0 1 0 0 1 0 0 0
10 12
1 0 1 0 1 1 0 0
11 13
1 0 1 1 1 1 0 1
15
1 1 1 1
List 2 (1-‐cubes) 0,4 0,8
0 x 0 0 x 0 0 0
8,10 4,12 8,12
1 0 x 0 x 1 0 0 1 x 0 0
10,11 12,13
1 0 1 x 1 1 0 x
11,15 13,15
1 x 1 1 1 1 x 1
f ( x1,…, x4) = Σ m(0,4,8,10,11,12,13,15)
List 3(2-‐cubes) 0,4,8,12
x x 0 0
The cubes without a check are the prime implicants of f P = {10x0, 101x, 110x, 1x11, 11x1, xx00} = {p1,p2,p3,p4,p5,p6}
69
DeterminaEon of a Minimum Cover • Find a minimum-‐cost cover for f • Assume that the cost is directly proporEonal to the number of inputs to all gates – i.e. the number of literals in the prime implicants
• Construct a prime-‐implicant cover table
– A row for each prime implicant – A column for each minterm to be covered – Put checks to indicate the minterms covered by each prime implicant
• If there is a single check in some column, then the prime implicant that covers the minterm of this column is essen?al and it must be included in the final cover
– Remove the rows corresponding to the essenEal prime implicants and the columns of the minterms associated with them
70
Prime implicant p 1
Minterm 0 4 8 10 11 12 13 15
1 0 x 0
p 2
1 0 1 x
p 4
1 x 1 1
p 3 p 5 p 6
1 1 0 x 1 1 x 1 x x 0 0 Initial prime implicant cover table
Prime implicant p 1
Minterm 10 11 13 15
p 2 p 3 p 4 p 5 After the removal of essential prime implicants 71
Row-‐Dominance •
p1 only covers minterm 10 while p2 covers both minterm 10 and minterm 11 – p2 dominates p1 – Cost of p2 is the same as p1 therefore choose p2 rather than p1 and remove p1 from the table
• •
• • •
p5 dominates p3 – Remove p3
Must choose p2 to cover minterm 10 and p5 to cover minterm 13 which takes care of covering minterm 11 and minterm 15 Final cover is C = {p2,p5,p6} = {101x, 11x1,xx00} f = x1x2’x3 + x1x2x4 + x3’x4 Do not remove a dominated row if the cost of its prime implicant is less than that of the dominaEng row’s prime implicant
Prime implicant p 1
Minterm 10 11 13 15
p 2 p 3 p 4 p 5
Prime implicant
Minterm 10 11 13 15
p 2 p 4 p 5 After the removal of dominated rows 72
Example 3 – Column Dominance • f ( x1,…, x4) = Σ m(0,2,5,6,7,8,9,13) + D(1,12,15) • P = {00x0,0x10,011x,x00x,xx01,1x0x,x1x1} = {p1,p2,p3,p4,p5,p6,p7} • Do not include don’t care minterms in the prime implicant cover table since they don’t need to be covered List 1
List 2
0
0 0 0 0
1 2 8
0 0 0 1 0 0 1 0 1 0 0 0
5 6 9 12
0 0 1 1
7 13
0 1 1 1 1 1 0 1
15
1 1 1 1
1 1 0 1
0 1 0 0
1 0 1 0
List 3
0,1 0,2 0,8
0 0 0 x 0 0 x 0 x 0 0 0
1,5 2,6 1,9 8,9 8,12
0 0 x 1 1
x x 0 0 x
0 1 0 0 0
1 0 1 x 0
5,7 6,7 5,13 9,13 12,13
0 0 x 1 1
1 1 1 x 1
x 1 0 0 0
1 x 1 1 x
7,15 13,15
x 1 1 1 1 1 x 1
0,1,8,9
x 0 0 x
1,5,9,13 8,9,12,13
x x 0 1 1 x 0 x
5,7,13,15
x 1 x 1
73
0 0 x 0 0 x 1 0 0 1 1 x x 0 0 x x x 0 1 1 x 0 x x 1 x 1
Prime implicant
p1 p2 p3 p4 p5 p6 p7
0
2
5
Minterm 6 7
0 0 x 0 0 x 1 0 0 1 1 x x 0 0 x x x 0 1 1 x 0 x x 1 x 1
Prime implicant
Row p4 dominates p6 and p7 dominates p5
0
2
0
2
(c) After the removal of rows
9
13
Minterm 5 6
7
p 5 and
8
p 6
• Column 13 dominates column 5
Minterm 5 6
(b) After the removal of columns 9 and 13 Prime implicant p1 p2 p3 p4 p7
• Remove the dominaFng column (9)
(a) Initial prime implicant cover table
p1 p2 p3 p4 p5 p6 p7
8
• Column 9 dominates column 8
7
8
• Remove column 13
Prime implicant p1 p2 p3
Minterm 2 6
(d) After including in the cover
p
p 4 and
C = {p2,p4,p7} = {0x10,x00x,x1x1}
7
f = x1’x3x4’ + x2x3’ + x2x4
74
Example 4 • No essenEal prime implicants or dominaEng rows or columns • All prime implicants have equal cost • Use branching – Choose any prime implicant, say p3 and include it in cover, compute rest of cover and its cost – C1 = {p1,p3,p4} – Now, find a new cover excluding p3 and compare costs – C2 = {p1, p5} = Cmin
75
Quine-‐McCluskey Summary 1. 2. 3. 4.
5. 6.
Start with list of cubes that represent minterms where f = 1 or don’t care. Generate the prime implicants by successive pairwise comparisons of the cubes. Derive a cover table which indicates minterms where f = 1 that are covered by each prime implicant. Include the essenEal prime implicants (if any) in the final cover and reduce the table by removing both these prime implicants and the covered minterms. Use the concept of row and column dominance to reduce the table further. A dominated row is only removed if the cost of its prime implicant is greater than or equal to the cost of the dominaEng row’s prime implicant. Repeat steps 3 and 4 unEl the cover table is empty or no further reducEon of the table is possible. If the reduced cover table is not empty, then use the branching approach to determine the remaining prime implicants that should be included in a minimum cost cover. 76
PracEcal LimitaEons of Quine-‐ McCluskey • FuncEons are seldom defined in the form of minterms, the are usually given as algebraic expressions or sets of cubes – List of minterms may be very large
• Many comparisons – Slow
• Cover table is computaEonally expensive
77
