CS 20 Lecture 14 Karnaugh Maps

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CS 20 Lecture 14Karnaugh Maps

• Professor CK Cheng
• CSE Dept.
• UC San Diego

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Karnaugh Maps

• Introduction
• The Maps
• Boolean Optimization

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Introduction
Definitions Literal xi or xi Product
Term x2x1x0 Sum Term x2 x1 x0 Minterm of
n variables A product of n literals in
which every variable appears exactly once.
f(a,b,c,d) abcd, abcd Maxterm of n
variables A sum of n literals in which every
variable appears exactly once. f(a,b,c,d)
(abcd), (abcd)
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Introduction
Input Boolean expression of n binary
variables Goal Simplification of the
expression. E.g. we want to minimize terms and
literals. Applications Logic rule
reduction Hardware Design cost and performance
optimization. Cost (wires, gates) literals,
product terms, sum terms
Performance speed, reliability
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Introduction
An example of 2-variable function f(A,B)AB
ID A B f(A,B) minterm
0 0 0 0
1 0 1 1 AB
2 1 0 1 AB
3 1 1 1 AB
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Function can be represented by sum of
minterms f(A,B) ABABAB This is not
minimal however! We want to minimize the number
of literals and terms. We factor out common terms
ABABAB ABABABAB (AA)BA(BB)BA
Hence, we have f(A,B) AB
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K-Map Truth Table in 2 Dimensions
A 0 A 1
AB
0 2
0 1 1 1
B 0 B 1
1 3
AB
AB
f(A,B) A B
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Another Example f(A,B)B
ID A B f(A,B) minterm
0 0 0 0
1 0 1 1 AB
2 1 0 0
3 1 1 1 AB
f(A,B)ABAB(AA)BB
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On the K-map
A 0 A 1
0 2
0 0 1 1
B 0 B 1
1 3
AB
AB
f(A,B)B
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Using Maxterms
ID A B f(A,B) Maxterm
0 0 0 0 AB
1 0 1 1
2 1 0 0 AB
3 1 1 1
f(A,B)(AB)(AB)(AA)B0BB
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The Maps Representation of k-Variable Functions

• Boolean Expression
• Truth Table
• Cube
• K Map
• Binary Decision Diagram

(1,1,1)
(1,1,0)
A
(0,1,0)
(0,1,1)
B
(1,0,1)
C
(0,0,0)
(0,0,1)
A cube of 3 variables (A,B,C)
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Representation of k-Variable Func.

• Boolean Expression
• Truth Table
• Cube
• K Map
• Binary Decision Diagram

(0,1,1,1)
(0,1,1,0)
(1,1,1,1)
(1,1,1,0)
B
(0,0,1,1)
(0,0,1,0)
(1,0,1,0)
(1,0,1,1)
C
(0,1,0,1)
(1,1,0,1)
D
(0,0,0,0)
(0,0,0,1)
(1,0,0,1)
(1,0,0,0)
A
A cube of 4 variables (A,B,C,D)
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Three-Variable K-Map
Id a b c f (a,b,c) 0 0
0 0 f(0,0,0) 1 0 0 1
f(0,0,1) 2 0 1 0
f(0,1,0) 3 0 1 1 f(0,1,1) 4
1 0 0 f(1,0,0) 5 1 0
1 f(1,0,1) 6 1 1 0
f(1,1,0) 7 1 1 1 f(1,1,1)
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Three-Variable K-Map
Id a b c f (a,b,c) 0 0
0 0 1 1 0 0 1
0 2 0 1 0 1 3 0
1 1 0 4 1 0 0
1 5 1 0 1 0 6 1
1 0 1 7 1 1 1
0
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Corresponding K-map
b 1
Gray code
(0,0) (0,1) (1,1) (1,0)
0 2 6
4
c 0
1 1 1 1
1 3 7
5
c 1
0 0 0 0
a 1
f(a,b,c) c
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Karnaugh Maps (K-Maps)

• Boolean expressions can be minimized by combining
  terms
• K-maps minimize equations graphically

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K-map

• Find rectangles to cover 1s in adjacent entries.
• Rectangles can overlap but should not include
  0s.
• Use the rectangle that corresponds to a product
  term.

y(A,B)ABCABC AB(CC)AB
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Another 3-Input example
Id a b c f (a,b,c) 0 0
0 0 0 1 0 0 1
0 2 0 1 0 1 3 0
1 1 0 4 1 0 0
1 5 1 0 1 1 6 1
1 0 - 7 1 1 1
1
Dont Care Entry - means the entry is not
relevant either at input or output. In other
words, we are free to assign either 0 or 1 to
reduce the Boolean expression.
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Corresponding K-map
b 1
(0,0) (0,1) (1,1) (1,0)
0 2 6
4
c 0
0 1 - 1
1 3 7
5
c 1
0 0 1 1
a 1
f(a,b,c) a bc
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Yet another example
Id a b c f (a,b,c,d) 0 0
0 0 1 1 0 0 1
1 2 0 1 0 - 3 0
1 1 0 4 1 0 0
1 5 1 0 1 1 6 1
1 0 0 7 1 1 1
0
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Corresponding K-map
b 1
(0,0) (0,1) (1,1) (1,0)
0 2 6
4
c 0
1 - 0 1
1 3 7
5
c 1
1 0 0 1
a 1
f(a,b,c) b
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Quiz

• Given Boolean function,
• f(a,b,c) abacbcabbc
• Write the truth table
• Use Karnaugh map to derive the function in a
  minimal expression of sum of product form.

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4-input K-map
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4-input K-map
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4-input K-map
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K-maps with Dont Cares
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K-maps with Dont Cares
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K-maps with Dont Cares