Section 4.1 - Karnaugh Map
4.1
Page 1 of 7
Karnaugh Map
The Karnaugh map method is a simple, straightforward procedure for minimizing the number of operations in
standard-form expressions.
The Boolean n-cubes provide the basis for these maps.
Each vertex in each n-cube represents a minterm of an n-variable Boolean function.
Each Boolean function could be represented visually in the form of an n-cube by marking those vertices in which the
value of that function is a 1, i.e. the 1-minterms.
Ex., for the carry and sum functions:
110
ci
0
0
0
0
1
1
1
1
xi
0
0
1
1
0
0
1
1
yi
0
1
0
1
0
1
0
1
ci+1
0
0
0
1
0
1
1
1
si
0
1
1
0
1
0
0
1
cx
111
110
111
xy
011
010
cy
100
000
101
001
ci+1 = c'xy + cx'y + cxy' + cxy
= xy + cy + cx
011
010
100
000
101
001
si = c'x'y + c'xy' + cx'y' + cxy
To find the standard form of a function that contains the minimal number of operators, we use the concept of a
Boolean subcube.
In general, an m-subcube of an n-cube can be defined as that set of 2m vertices in which n - m of the variables will
have the same value at every vertex, while the remaining m variables will take
110
111
on the 2m possible combinations of the values 0 and 1.
For example, in the 3-cube:
011
The 2-subcube is defined as that set of 2m = 22 = 4 vertices in which n - m = 3 - 2
010
= 1 of the variable will have the same value at every vertex (0xx, or 1xx)
while the remaining variables will take on the 2m = 22 = 4 possible
combinations (x00, x01, x10, and x11).
100
101
The 1-subcube is defined as that set of 2m = 21 = 2 vertices in which n - m = 3 - 1
= 2 of the variables will have the same value at every vertex (00x, 01x, 10x,
000
001
or 11x) while the remaining variable will take on the 2m = 21 = 2 possible
combinations (xx0, and xx1).
Each subcube can be characterized by the n - m variable values that are the same for every vertex in the subcube.
If a Boolean function has a value of 1 in each vertex of the m-subcube, the 2m 1-minterms in that subcube could be
expressed by a single term of n - m literals.
For example, in the carry function ci+1, the two 1-minterms (011, and 111) form a 1-subcube which could be
expressed by the single term of n - m = 3-1 = 2 literals xy.
•
•
To minimize the number of OR operators, we have to choose as few subcubes as possible while still covering
all its 1-minterms.
To minimize the number of AND operators, we have to choose the largest possible subcubes.
We can determine and select the subcubes visually from a cube representation, or we could derive them from the
two-dimensional form of this cube which is called a Karnaugh map, or map for short.
There is a one-to-one correspondence between Boolean cubes and Karnaugh maps. Karnaugh maps are just twodimensional representation of Boolean cubes.
The map is an array of squares. Each square corresponds to one vertex of the cube, i.e. one minterm of the Boolean
function.
Section 4.1 - Karnaugh Map
Page 2 of 7
A two-variable map consists of 4 squares which
correspond to the 4 minterms of a two-variable
Boolean function. The largest subcube in a twovariable map is a 1-cube, which represents a single
variable or its complement.
x
y
0
0
x'y'
1
xy'
minterm
numbers
1
0
x
y
0
1
0
1
2
3
subcube
subcube
subcube
subcube
0
x'y
2
1
3
1
xy
x'
y'
y
x
A three-variable map consists of 8 squares. The largest subcubes are of size 2. These 2-subcubes can be expressed
by a single literal. The 1-subcubes can be expressed by two literals.
x
yz
01
00
11
0
0
10
1
x'y'z'
x'y'z
x
3
x'yz
yz
00
01
0
2
11
1
10
3
x
z
2
0
x'yz'
00
yz
01
11
10
0
1
3
2
4
5
7
6
x'z
yz
y'z'
xy'
0
z'
4
1
5
xy'z'
xy'z
7
xyz
6
4
5
7
6
x
1
xyz'
three-variable map
1
Examples of 2-subcubes
Examples of 1-subcubes
A four-variable map consists of 16 squares.
yz
wx
00
00
w'x'
y'z'
0
w'x
y'z'
11
12
wx
y'z'
wx'
y'z'
w'x'
y'z
4
01
10
11
01
w'x
y'z
1
5
13
wx
y'z
8
wx'
y'z
9
w'x'
yz
w'x
yz
yz
10
3
7
w'x'
yz'
w'x
yz'
wx
2
00
01
0
11
1
yz
10
3
wx
00
6
15
wx
yz
14
wx
yz'
11
wx'
yz
10
wx'
yz'
5
7
6
z'
01
12
13
15
11
3
2
4
5
7
6
w'z
14
12
13
xy
00
8
9
11
8
9
0
1
0
4
1
3
c
2
1
5
1
7
1
ci+1 = xy + cy + cx
6
1
11
10
10
Examples of 3-subcubes
10
14
xy
x'y'
Examples of 2 and 1-subcubes
00
01
0
0
10
3
1
4
1
11
1
1
2
1
5
w'yz'
xy'z'
w
10
10
11
01
15
11
If the minterm is a 1-minterm, a 1 will be placed inside its square, otherwise, it is left empty.
Example, for the carry and sum functions:
c
10
1
01
11
four-variable map
01
0
00
z
4
00
2
7
6
1
si = c'x'y + c'xy' + cx'y' + cxy
Section 4.1 - Karnaugh Map
4.2
Page 3 of 7
Map Method of Simplification
Four steps:
1) Map generation from truth table or functional expression. Place a 1 in the squares that correspond to 1minterms.
2) Prime implicant generation.
A prime implicant (PI) is defined as a subcube that is not contained within any other subcube.
e.g. In figure 4.13 below, w'z' is a PI but w'y'z' is not because w'y'z' is inside w'z'.
A list of prime implicants is generated by inspecting each 1-minterm, finding the largest possible subcube of 1minterms that includes the minterm in question, and then adding that subcube to the list of prime
implicants.
If two or more different subcubes are discovered, they are all added to the list.
If we rediscover a subcube that is already on the list, it will not be added the second time.
3) Essential prime implicant selection.
An essential prime implicant (EPI) is the subcube that includes a 1-minterm that is not included in any
other subcube.
e.g. In figure 4.13 below, w'z' is an EPI because m0 is not in any other PI. However, w'y is not an EPI
because all four of its 1-minterms are in another PI.
Look for any 1-minterm that is included in only one prime implicant. This prime implicant is an EPI and is
added to the cover list.
4) Create minimal cover.
Generate a cover list consisting of the smallest possible number of prime implicants such that every 1-minterm
is contained in at least one prime implicant.
The cover list must include all the EPIs.
Example: Using the map method, simplify the Boolean function F = w'y'z' + wz + xyz + w'y.
yz
wx
00
01
0
00
1
12
13
1
8
10
9
1
14
1
11
5
1
12
8
10
10
step 1)
Figure 4.13.
7
13
1
w'y
6
1
15
14
11
10
1
9
1
2
1
1
11
1
10
3
1
4
01
11
1
6
1
01
0
00
1
15
00
2
7
1
11
wx
1
5
1
yz
10
3
1
1
4
01
11
1
step 2)
w'z'
yz
wz
Step 2) Do the 1-minterms in
order.
m0 is in w'z'.
m2 is in w'z' and w'y.
m3 is in w'y and yz.
m4, m6, m7, same as m0,
m2, m3
m9 and m13 are in wz.
m11 and m15 are in wz and
yz.
Therefore, PI list = w'z', wz,
yz, w'y.
Step 3) EPI list:
m0 and m4 are covered only by the PI w'z'.
m9 and m13 are covered only by the PI wz.
Therefore, EPI list = w'z', wz.
Step 4) Cover list:
The two EPIs w'z' and wz must be in the cover list. w'z' also covers m2 and m6. wz also covers m11 and m15. There
are only two 1-minterms m3 and m7 that are not covered. These two 1-minterms can be covered by either yz or
w'y since both of them are of the same size. (Note that if one was larger than the other, we would only use the
larger one.)
Therefore, there are two cover lists: (1) w'z', wz, yz and (2) w'z', wz, w'y.
Thus, there are two simplified equivalent functional expressions for F = w'y'z' + wz + xyz + w'y:
(1) F = w'z' + wz + yz
(2) F = w'z' + wz + w'y.
Section 4.1 - Karnaugh Map
Page 4 of 7
Example: Using the map method, simplify the Boolean function F = w'x'yz' + w'xy + wxz + wx'y' + w'x'y'z'.
yz
wx
00
01
11
PI list: w'x'z', w'xy, wxz, wx'y', and x'y'z', wy'z, xyz, w'yz'.
10
w'x'z'
0
00
1
3
2
1
4
5
7
01
6
1
12
13
11
1
9
1
w'xy
1
15
14
wxz
11
10
wx'y'
1
8
10
EPI list: empty
1
1
x'y'z'
wy'z
xyz
Cover lists:
(1) w'x'z', w'xy, wxz, wx'y'.
(2) x'y'z', wy'z, xyz, w'yz'.
The two simplified equivalent functional expressions are:
(1) F = w'x'z' + w'xy + wxz + wx'y'.
(2) F = x'y'z' + wy'z + xyz + w'yz'.
w'yz'
However, bad if we had selected the two PIs w'x'z' and xyz, because now we would need 5 PIs to cover all the 1minterms. e.g. w'x'z', xyz, w'xy, wxz, and wx'y'.
4.3
Don't-Care Conditions
Previously, we have considered only those Boolean functions that are completely specified.
There are situations where the outputs are not specified for some input combinations, i.e. we do not specify whether
the output is a 0 or a 1 for an input combination.
Such a Boolean function is called an incompletely specified function, and the minterms for which the function is
not specified are called don't-care minterms (d-minterms) or don't-care conditions.
We can use d-minterms to further simplify the Boolean expression because we can assign either a 0 or a 1 to them in
a way that will make our subcubes larger.
Example: Convert BCD to nine's complements:
Input
Decimal
x3 x 2 x 1 x 0
0
0000
1
0001
2
0010
3
0011
4
0100
5
0101
6
0110
7
0111
8
1000
9
1001
10 - 15
don't care
x1x0
x3x2
00
01
0
11
1
00
10
3
1
4
01
1
11
X
5
x1x0
x3x2
7
8
x1x0
x3x2
15
X
12
11
9
11
X
10
X
10
y2 = x2x1' + x2'x1 = x2 ⊕ x1
X
3
2
5
7
6
13
X
8
15
14
X
11
10
X
X
9
X
y3 = x3' x2' x1'
01
11
1
3
5
12
8
13
X
1
01
1
15
14
X
11
X
11
10
X
10
1
6
01
0
1
X
X
00
00
7
9
y1 = x1
2
1
1
X
x1x0
x3x2
10
1
4
11
1
10
0
14
X
10
1
4
01
13
X
1
11
01
00
6
01
00
0
00
00
2
1
1
12
10
Output
Decimal
y3 y 2 y 1 y 0
9
1001
8
1000
7
0111
6
0110
5
0101
4
0100
3
0011
2
0010
1
0001
0
0000
10 - 15
don't care
11
1
10
3
2
1
4
5
7
6
1
12
8
13
X
9
X
y0 = x0'
15
14
X
11
10
X
X
Section 4.1 - Karnaugh Map
4.4
Page 5 of 7
Tabulation Method
Previously, finding subcubes in a map is essentially a trial-and-error procedure because it requires a person to
recognize subcube patterns in a map. Gets more difficult with more variables.
The tabulation method is an algorithmic exhaustive search method to find the subcubes in a map.
Quite tedious for human but good for computers.
Example: Find the prime implicants for the Boolean function F = w'y'z' + wz + xyz + w'y.
From figure 4.13, we see that there are ten 1-minterms.
Step (1): Generate a list of 0-subcubes.
Group the 1-minterms such that each group Gi
contains those minterms whose 1's count (number
of variables whose values are equal to 1) is equal
to i. Thus, in G0 the minterms have no 1's. In G1,
the minterms have one 1's. In G2, the minterms
have two 1's. etc.
The "yes" and "no" flag indicates whether or not
that subcube is covered by a larger subcube. The
flag will be set to "yes" or "no" after we have
generated the list of next-larger subcubes.
Step (2): Generate a list of 1-subcubes.
In the list of 0-subcubes, compare each of the
minterms in group Gi with each of the minterms in
group Gi+1 and generate a 1-subcube for group Gi
in the list of 1-subcubes if these minterms differ in
one variable.
Note that "-" is not an indication of the value of a
variable but is rather a placeholder for a variable
that has already been eliminated from the
expression.
Group
ID
G0
G1
G2
G3
G4
Group
ID
G0
G1
G2
G3
Step (3): Generate a list of 2-subcubes.
In the list of 1-subcubes, compare each of the
minterms in group Gi with each of the minterms in
group Gi+1 and generate a 2-subcube for group Gi
in the list of 2-subcubes if these minterms differ in
one variable.
Group
ID
G0
G1
G2
Subcube
Subcube Value
Minterms w
x
y
z
m0
0
0
0
0
m2
0
0
1
0
m4
0
1
0
0
m3
0
0
1
1
m6
0
1
1
0
m9
1
0
0
1
m7
0
1
1
1
m11
1
0
1
1
m13
1
1
0
1
m15
1
1
1
1
List of 0-subcubes
Subcube
Covered
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
Subcube Value
Subcube
Minterms w
x
y
z
m0, m2
0
0
0
m0, m4
0
0
0
m2, m3
0
0
1
m2, m6
0
1
0
m4, m6
0
1
0
m3, m7
0
1
1
m3, m11
0
1
1
m6, m7
0
1
1
m9, m11
1
0
1
m9, m13
1
0
1
m7, m15
1
1
1
m11, m15
1
1
1
m13, m15
1
1
1
List of 1-subcubes
Subcube
Covered
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
yes
Subcube Value
Subcube
Minterms
w
x
y
z
m0, m2 m4, m6
0
0
m2, m3 m6, m7
0
1
m3, m7 m11, m15
1
1
m9, m11 m13, m15
1
1
List of 2-subcubes
Subcube
Covered
no
no
no
no
Step (4):
Repeat until all the subcube covered are "no". From the subcube values of all the "no" rows, the prime implicants
are: w'z', w'y, yz, wz.
Section 4.1 - Karnaugh Map
Page 6 of 7
Example: Find the essential prime implicants and the minimal covers for the Boolean function
F = w'y'z' + wz + xyz + w'y.
Using the four PIs (w'z', w'y, yz, wz) obtained in the last example, we construct the EPI table.
Prime
Implicant
Name
Prime
Implicant
Expression
P1
w'z'
(0,2,4,6)
P2
w'y
(2,3,6,7)
P3
yz
(3,7,11,15)
P4
wz
(9,11,13,15)
Implicant
Minterms
EPI covered minterms:
Not covered minterms:
Function Minterms
0
2
⊗
×
×
3
4
6
⊗
×
×
×
×
0
2
9
11
6
×
15
×
⊗
×
⊗
×
9
11
13
15
7
Step (1): Enter an × to indicate where a given minterm is covered by a given PI.
Step (2): Indicate those columns that have only one × by circling that ×.
Step (3): A PI row having a ⊗ is an EPI. Thus, EPI = w'z', wz.
Step (4): The EPIs will cover the other ×'s in the same row.
Step (5): Find the best way to cover the remaining minterms m3 and m7. Can use either w'y or yz.
PI list: w'z', w'y, yz, wz
EPI list: w'z', wz
Cover list:
(1) w'z', wz, w'y
(2) w'z', wz, yz
Minimal-cover expressions:
(1) F = w'z' + wz + w'y
(2) F = w'z' + wz + yz
13
×
×
4
3
7
Section 4.1 - Karnaugh Map
Page 7 of 7
Example: Find the minimal cover for the function with the 1-minterms 2, 6, 7, 8, 9, 13, 15.
Group
ID
G1
G2
G3
G4
Subcube
Subcube Value
Minterms w
x
y
z
m2
0
0
1
0
m8
1
0
0
0
m6
0
1
1
0
m9
1
0
0
1
m7
0
1
1
1
m13
1
1
0
1
m15
1
1
1
1
List of 0-subcubes
Prime
Implicant
Name
Prime
Implicant
Expression
P1
w'yz'
(2, 6)
P2
wx'y'
(8, 9)
P3
w'xy
(6, 7)
P4
wy'z
(9, 13)
P5
xyz
(7, 15)
P6
wxz
(13,15)
Subcube
Covered
yes
yes
yes
yes
yes
yes
yes
Group
ID
G1
G2
G3
Implicant
Minterms
EPI covered minterms:
Not covered minterms:
Subcube
Subcube Value
Minterms w
x
y
z
m2, m6
0
1
0
m8, m9
1
0
0
m6, m7
0
1
1
m9, m13
1
0
1
m7, m15
1
1
1
m13, m15
1
1
1
List of 1-subcubes
Subcube
Covered
no
no
no
no
no
no
Function Minterms
2
6
⊗
×
×
7
8
9
⊗
×
13
×
×
×
×
2
15
6
×
8
×
×
13
15
9
7
Three minterms left uncovered: m7, m13, m15.
m7 can be covered by P3 or P5.
m13 can be covered by P4 or P6.
m15 can be covered by P5 or P6.
This can be expressed in terms of the following product of sums of these PIs:
(P3 + P5)(P4 + P6)(P5 + P6)
Multiplying the second and third sums produces the following expression:
(P3 + P5)(P4P5 + P5P6 + P4P6 + P6)
Since P6 covers the same minterms as do P5P6 (because the set of minterms covered by P5P6 is a subset of the set of
minterms covered by P6) and P5P6 has fewer PIs, we can remove P5P6 from the expression.
Similarly, we can remove P 4P6. This reduces the cover expression to
(P3 + P5)(P4P5 + P6)
Multiplying out this expressions yields
P3P4P5 + P4P5 + P3P6 + P5P6
Again we can remove P3P4P5 because it is a subset of P4P5. Thus the final expression is
P4P5 + P3P6 + P5P6
Finally, by combining these three options with the EPIs, we get the three different minimal covers:
F = P1 + P2 + P4 + P5 = w'yz' + wx'y' + wy'z + xyz.
F = P1 + P2 + P3 + P6 = w'yz' + wx'y' + w'xy + wxz.
F = P1 + P2 + P5 + P6 = w'yz' + wx'y' + xyz + wxz.