Karnaugh Map
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Karnaugh Map
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Logic Circuits
Transcript of Karnaugh Map
Karnaugh Map
Karnaugh Map
• Proposed by Veitch and modified by
Karnaugh
• A diagram made up of squares, each
squares represents one minterm
• The map presents a visual diagram of all
possible ways a function may be
expressed in a standard form.
Two-Variable Map
A B F
0 0 0
0 1 0
1 0 1
1 1 1
0 1
2 3
A’
A
B’ B
0 0
1 1
A’
A
B’ B
The output 1 correspond to
these minterms: AB’ and AB
Three-Variable Map
A B C F
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 1
0 1
2 3
6 7
4 5
C’ C
A’B’
A’B
AB
AB’
0 0
1 0
1 1
0 0
C’ C
A’B’
A’B
AB
AB’
A’BC
The output 1 correspond to these minterms: A’BC’ , ABC’ and ABC
Four-Variable Map
A B C D F
0 0 0 0 0
0 0 0 1 1
0 0 1 0 0
0 0 1 1 0
0 1 0 0 0
0 1 0 1 0
0 1 1 0 1
0 1 1 1 1
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 0
1 1 0 0 0
1 1 0 1 0
1 1 1 0 1
1 1 1 1 0
0 1 3 2
4 5 7 6
12 13 15 14
8 9 11 10
A’B’
A’B
AB
AB’
0 1 0 0
0 0 1 1
0 0 0 1
0 0 0 0
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
C’D’ C’D CD CD’
Five-Variable Map
A B C D E F
0 0 0 0 0 1
0 0 0 0 1 0
0 0 0 1 0 1
0 0 0 1 1 0
0 0 1 0 0 1
0 0 1 0 1 0
0 0 1 1 0 1
0 0 1 1 1 0
0 1 0 0 0 0
0 1 0 0 1 1
0 1 0 1 0 0
0 1 0 1 1 0
0 1 1 0 0 0
0 1 1 0 1 1
0 1 1 1 0 0
0 1 1 1 1 0
A B C D E F
1 0 0 0 0 0
1 0 0 0 1 0
1 0 0 1 0 0
1 0 0 1 1 0
1 0 1 0 0 0
1 0 1 0 1 1
1 0 1 1 0 0
1 0 1 1 1 1
1 1 0 0 0 0
1 1 0 0 1 1
1 1 0 1 0 0
1 1 0 1 1 1
1 1 1 0 0 0
1 1 1 0 1 1
1 1 1 1 0 0
1 1 1 1 1 0
Five-Variable Map
0 1 3 2
4 5 7 6
12 13 15 14
8 9 11 10
B’C’
B’C
BC
BC’
D’E’ D’E DE DE’
A’
16 17 19 18
20 21 23 22
28 29 31 30
24 25 27 26
B’C’
B’C
BC
BC’
D’E’ D’E DE DE’
A
1 0 0 1
1 0 0 1
0 1 0 0
0 1 0 0
B’C’
B’C
BC
BC’
D’E’ D’E DE DE’
A’
0 0 0 0
0 1 1 0
0 1 0 0
0 1 1 0
B’C’
B’C
BC
BC’
D’E’ D’E DE DE’
A
6-Variable Map
A B C D E F Y
0 0 0 0 0 0 0
0 0 0 0 0 1 0
0 0 0 0 1 0 1
0 0 0 0 1 1 1
0 0 0 1 0 0 0
0 0 0 1 0 1 0
0 0 0 1 1 0 1
0 0 0 1 1 1 1
0 0 1 0 0 0 0
0 0 1 0 0 1 0
0 0 1 0 1 0 0
0 0 1 0 1 1 0
0 0 1 1 0 0 1
0 0 1 1 0 1 1
0 0 1 1 1 0 0
0 0 1 1 1 1 0
A B C D E F Y
0 1 0 0 0 0 0
0 1 0 0 0 1 0
0 1 0 0 1 0 1
0 1 0 0 1 1 1
0 1 0 1 0 0 0
0 1 0 1 0 1 0
0 1 0 1 1 0 1
0 1 0 1 1 1 1
0 1 1 0 0 0 1
0 1 1 0 0 1 0
0 1 1 0 1 0 0
0 1 1 0 1 1 0
0 1 1 1 0 0 1
0 1 1 1 0 1 0
0 1 1 1 1 0 0
0 1 1 1 1 1 0
A B C D E F Y
1 0 0 0 0 0 0
1 0 0 0 0 1 0
1 0 0 0 1 0 1
1 0 0 0 1 1 1
1 0 0 1 0 0 0
1 0 0 1 0 1 0
1 0 0 1 1 0 1
1 0 0 1 1 1 1
1 0 1 0 0 0 0
1 0 1 0 0 1 0
1 0 1 0 1 0 0
1 0 1 0 1 1 0
1 0 1 1 0 0 1
1 0 1 1 0 1 1
1 0 1 1 1 0 0
1 0 1 1 1 1 0
A B C D E F Y
1 1 0 0 0 0 0
1 1 0 0 0 1 0
1 1 0 0 1 0 1
1 1 0 0 1 1 1
1 1 0 1 0 0 0
1 1 0 1 0 1 0
1 1 0 1 1 0 1
1 1 0 1 1 1 1
1 1 1 0 0 0 0
1 1 1 0 0 1 0
1 1 1 0 1 0 0
1 1 1 0 1 1 0
1 1 1 1 0 0 0
1 1 1 1 0 1 0
1 1 1 1 1 0 0
1 1 1 1 1 1 1
0 1 3 2
4 5 7 6
12 13 15 14
8 9 11 10
C’D’
C’D
CD
CD’
E’F’ E’F EF EF’
16 17 19 18
20 21 23 22
28 29 31 30
24 25 27 26
C’D’
C’D
CD
CD’
E’F’ E’F EF EF’
32 33 35 34
36 37 39 38
44 45 47 46
40 41 43 42
C’D’
C’D
CD
CD’
E’F’ E’F EF EF’
48 49 51 50
52 53 55 54
60 61 63 62
56 57 59 58
C’D’
C’D
CD
CD’
E’F’ E’F EF EF’
B B’
A’
A
A’B’ A’B
AB’ AB
Pairs
• Group of two 1s that are vertically or horizontally adjacent.
• It is customary to encircle a pair of adjacent 1s for easy identification.
• If one variable changes its value, that variable can be eliminated. (The variable and its complement will be eliminated.)
• If more than one group exists on a map, you can OR the simplified product to get the simplified Boolean equation.
Examples
0 0 0 0
0 0 1 0
0 0 1 0
0 0 0 0
A’B’
A’B
AB
AB’
0 0 0 0
0 0 1 1
0 1 0 0
0 1 0 0
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
C’D’ C’D CD CD’
0 0 0 0
0 0 1 1
0 0 0 0
0 0 0 0
A’B’
A’B
AB
AB’
0 1 1 0
0 0 0 1
0 0 0 1
0 0 0 0
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
C’D’ C’D CD CD’
a b
c d
Quads
• A group of four 1s that are end to end or in
the form of square.
• Two variables and their complements can
be dropped.
Examples
0 0 1 0
0 0 1 0
0 0 1 0
0 0 1 0
A’B’
A’B
AB
AB’
0 0 0 0
1 1 1 1
0 0 0 0
0 0 0 0
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
C’D’ C’D CD CD’
0 0 0 0
0 0 1 1
0 0 1 1
0 0 0 0
A’B’
A’B
AB
AB’
0 1 1 0
0 1 1 0
0 0 0 0
0 0 0 0
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
C’D’ C’D CD CD’
a b
c d
Octets
• A group of eight adjacent 1s.
• It eliminates three variables and their
complements.
0 0 1 1
0 0 1 1
0 0 1 1
0 0 1 1
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
0 0 0 0
1 1 1 1
1 1 1 1
0 0 0 0
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
a b
Examples:
Overlapping Groups
• When you encircle groups, you are
allowed to use the same 1 more than
once.
• Always overlap groups when possible
to get the largest groups you can.
Overlapping Groups
0 0 1 1
1 1 1 1
0 0 1 1
0 0 1 1
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
0 0 0 0
1 1 1 1
1 1 1 1
0 0 1 0
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
a c
Examples:
0 0 1 1
1 1 1 1
0 0 1 1
0 0 1 1
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
0 0 0 0
1 1 1 1
1 1 1 1
0 0 1 0
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
b d
Rolling the Map
• Group the 1s just like by imagining rolling
the map so that the left side touches the
right side or the upper touches the lower
side.
• Draw half-circles around each group.
Rolling the Map
1 0 0 1
1 0 0 1
1 0 0 1
1 0 0 1
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
0 1 1 0
0 0 0 0
0 0 0 0
0 1 1 0
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
a c
Examples:
b
1 0 0 1
1 0 0 1
1 0 0 1
1 0 0 1
C’D’ C’D CD CD’
A’B’
A’B
AB
AB’
0 1 1 0
0 0 0 0
0 0 0 0
0 1 1 0
A’B’
A’B
AB
AB’
d
C’D’ C’D CD CD’
Redundant Groups
• A group whose 1s are completely
overlapping by other groups.
• The inner pair in (a) are completely
overlapped by the outside pairs, and can
be eliminated to get the simpler map as
shown in (b).
Examples:
0 1 0 0
0 1 1 0
0 0 1 0
0 0 0 0
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’ Redundant 0 1 0 0
0 1 1 0
0 0 1 0
0 0 0 0
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
0 0 0 01
0 0 1 0
1 1 1 0
1 1 0 0
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
a
Ex 1.
Ex 2.
0 0 0 01
0 0 1 0
1 1 1 0
1 1 0 0
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
a
Example 1.
• Draw the Karnaugh map
for the output F of the
given truth table. Simplify
as much as possible then
draw the logic circuit
using the three basic
gates.
x y z F
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
Example 2
• Simulate the figure below, then make a
truth table for it. Derive a simplified
equation using Karnaugh Map
Example 3
• Obtain the simplified expression in sum of
products for the Boolean functions:
(a)F(A,B,C,D,E)=∑(0,1,4, 5,16,17, 21, 25, 29)
(b)F(A,B,C,D,E,F)= ∑(6, 9, 13, 18, 19, 25, 27, 29,
41,45, 57, 61)
Example 4
• Obtain the simplified function in (a) sum of
products and in (b) product of sums for
the Boolean functions below.
A’B’CE’ + A’B’C’D’ + B’D’E’ + B’CD’ + CDE’
+ BDE’
Don’t-Care Conditions
• Represented by x in the truth table, its
because they can be treated as 0s or 1s,
whichever leads to a simpler circuit.
• In K-map, treat x’s as 1s and try to form
the largest groups to include with the real
1s.
• The remaining x’s will be treated as 0’s
Ex. Simplify the function F using
Karnaugh Map.
A B C D F
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1
0 1 0 0 1
0 1 0 1 1
0 1 1 0 0
0 1 1 1 1
1 0 0 0 X
1 0 0 1 X
1 0 1 0 X
1 0 1 1 X
1 1 0 0 X
1 1 0 1 X
1 1 1 0 X
1 1 1 1 X
0 1 3 2
4 5 7 6
12 13 15 14
8 9 11 10
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
1 0 1 0
1 1 1 0
x x x x
x x x x
A’B’
A’B
AB
AB’
C’D’ C’D CD CD’
CD
BD
C’D’
F= C’D’ + CD + BD
Example 2
• Simplify the Boolean function F in sum of
products using the don’t care condition d;
a. F=y’ + x’z’
d=yz + xy
b. F= B’C’D’ + BCD’ + ABC’D
d=B’CD’ + A’BC’D’
Exercises
1. Simplify the following Boolean functions
using K-map:
a. F(w,x,y,z)=∑(2,3,10,11,12,13,14,15)
b. F(A,B,C,D)=Π(0,1,2,3,4,10,11)
c. F=A’B’CE’ + A’B’C’D’ + B’D’E’ + B’CD’ +
CDE’ + BDE’
2. Simplify the Boolean function F together
with the don’t-care condition d in i) sum of
products ii) product of sum
a. F(w,x,y,z)=∑(0,1,2,3,7,8,10)
d(w,x,y,z)= ∑(5,6,11,15)
b. F(A,B,C,D)= ∑(3,4,13,15)
d(A,B,C,D)= ∑(1,2,5,6,8,10,12,14)
