124 Unit 5
Study Guide
In this unit we will study the Karnaugh (pronounced “car-no”) map. Just about any
type of algebraic manipulation we have done so far can be facilitated by using the
map, provided the number of variables is small.
1.
Study Section 5.1, Minimum Forms of Switching Functions.
(a) Define a minimum sum of products.
(b) Define a minimum product of sums.
2.
Study Section 5.2, Two- and Three-Variable Karnaugh Maps.
(a) Plot the given truth table on the map. Then, loop two pairs of 1’s on the
map and write the simplified form of F.
PQ
00
01
10
11
F
1
1
0
1
P
Q
0
1
0
1
F
F=
Now simplify F algebraically and verify that your answer is correct.
(b) F(a, b, c) is plotted below. Find the truth table for F.
a
0
1
00
0
1
01
1
1
11
0
1
10
1
0
bc
abc
000
001
010
011
100
101
110
111
F
F
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Karnaugh Maps 125
(c) Plot the following functions on the given Karnaugh maps:
F1(R, S, T) = Σ m(0, 1, 5, 6) F2(R, S, T) = Π M(2, 3, 4, 7)
0
1
0
00
00
01
01
11
11
10
10
1
Why are the two maps the same?
(d) Plot the following function on the given map:
f(x, y, z) = z′ + x′z + yz
Do not make a minterm expansion or a truth table before plotting.
x
yz
0
1
00
01
11
10
(e) For a three-variable map, which squares are “adjacent” to square 2?
_____________
(f ) What theorem is used when two terms in adjacent squares are combined?
(g) What law of Boolean algebra justifies using a given 1 on a map in two or
more loops?
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126 Unit 5
(h) Each of the following solutions is not minimum.
a
0
bc
a
1
00
1
01
1
11
1
0
bc
f = ab′ + abc
10
1
00
1
01
1
11
1
1
10
1
1
g = a′ + ab
In each case, change the looping on the map so that the minimum solution
is obtained.
(i) Work Problem 5.3.
(j) Find two different minimum sum-of-products expressions for the function G, which is plotted below.
a
bc
00
a
0
1
1
1
00
1
01
1
11
1
10
1
01
11
1
10
1
bc
0
1
1
1
1
G
3.
G=
1
G=
G
Study Section 5.3, Four-Variable Karnaugh Maps.
(a) Note the locations of the minterms on three- and four-variable maps
(Figures 5-3(b) and 5-10). Memorize this ordering. This will save you a lot
of time when you are plotting Karnaugh maps.
This ordering is valid only for the order of the variables given. If we label
the maps as shown below, fill in the locations of the minterms:
BC
00
A
01
11
10
CD
00
AB
0
00
1
01
01
11
10
11
10
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Karnaugh Maps 127
(b) Given the following map, write the minterm and maxterm expansions for F
in decimal form:
ab
cd
00
01
00
01
1
11
10
1
1
1
1
1
11
F=
1
1
10
F=
(c) Plot the following functions on the given maps:
(1) f (w, x, y, z) = Σ m(0, 1, 2, 5, 7, 8, 9, 10, 13, 14)
(2) f (w, x, y, z) = x′z′ + y′z + w′xz + wyz′
wx
00
yz
01
11
10
wx
00
yz
00
00
01
01
11
11
10
10
01
11
10
Your answers to (1) and (2) should be the same.
(d) For a four-variable map, which squares are adjacent to square 14? ________
To square 8? _________
(e) When we combine two adjacent 1’s on a map, this corresponds to applying
the theorem xy′ + xy = x to eliminate the variable in which the two terms
differ. Thus, looping the two 1’s as indicated on the following map is equivalent to combining the corresponding minterms algebraically:
ab
cd
00
01
10
1
00
01
1
11
1
10
11
1
a′b′c′d + ab′c′d = b′c′d
[The term b′c′d can be read directly from the
map because it spans the first and last columns
(b′) and because it is in the second row (c′d).]
1
1
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128 Unit 5
Loop two other pairs of adjacent 1’s on this map and state the algebraic
equivalent of looping these terms. Now read the loops directly off the map
and check your algebra.
(f ) When we combine four adjacent 1’s on a map (either four in a line or four
in a square) this is equivalent to applying xy + xy′ = x three times:
ab
cd
00
01
00
1
01
1
11
10
11
1
1
1
10
1
1
1
a′b′cd + a′b′cd′ + ab′cd + ab′cd′ = a′b′c + ab′c = b′c
Loop the other four 1’s on the map and state the algebraic equivalent.
(g) For each of the following maps, loop a minimum number of terms which
will cover all of the 1’s.
ab
01
11
00
1
1
01
1
1
cd
11
10
00
1
10
ab
cd
00
11
1
10
1
00
1
01
1
11
1
1
10
1
f1
01
1
1
1
1
f2
(For each part you should have looped two groups of four 1’s and two
groups of two 1’s).
Write down the minimum sum-of-products expression for f1 and f2 from
these maps.
f1 = __________________________________________________
f2 = __________________________________________________
(h) Why is it not possible to combine three or six minterms together rather
than just two, four, eight, etc.?
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Karnaugh Maps 129
(i) Note the procedure for deriving the minimum product of sums from the
map. You will probably make fewer mistakes if you write down f ′ as a sum
of products first and then complement it, as illustrated by the example in
Figure 5-14.
( j) Work Problems 5.4 and 5.5.
4.
Study Section 5.4, Determination of Minimum Expressions Using Essential
Prime Implicants.
(a) For the map of Figure 5-15, list three implicants of F other than those which
are labeled.
For the same map, is ac′d′ a prime implicant of F?
Why or why not?
(b) For the given map, are any of the circled
terms prime implicants?
Why or why not?
5.
AB
CD
00
01
11
10
1
00
01
1
1
11
1
1
10
1
1
1
Study Figure 5-18 carefully and then answer the following questions for the
given map:
(a) How many 1’s are adjacent to m0?
(b) Are all these 1’s covered by a single prime
implicant?
AB
00
CD
00
1
(c) From your answer to (b), can you determine
whether B′C′ is essential?
01
1
(d) How many 1’s are adjacent to m9?
11
1
(e) Are all of these 1’s covered by a single prime
implicant?
10
1
01
0
1
11
4
2
1
1
1
3
10
1
1
8
9
7
6
1
10
(f ) From your answer to (e), is B′C′ essential?
(g) How many 1’s are adjacent to m7?
(h) Why is A′C essential?
(i) Find two other essential prime implicants and tell which minterm makes
them essential.
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130 Unit 5
6.
(a)How do you determine if a prime implicant is essential using a Karnaugh map?
(b) For the following map, why is A′B′not essential?
Why is BD′ essential?
Is A′D′ essential? Why?
Is BC′ essential? Why?
Is B′CD essential? Why?
Find the minimum sum of products.
AB
00
CD
01
11
00
1
1
1
01
1
1
1
11
1
10
1
10
1
1
1
(c) Work Programmed Exercise 5.1.
(d) List all 1’s and X’s that are adjacent to 10.
AB
00
CD
01
11
10
00
10
14
112
8
01
X1
15
X13
9
11
3
X7
115
111
10
2
6
X14
10
Why is A′C′ an essential prime implicant?
List all 1’s and X’s adjacent to 115.
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Karnaugh Maps 131
Based on this list, why can you not find an essential prime implicant that
covers 115?
Does this mean that there is no essential prime implicant that covers 115?
What essential prime implicant covers 111?
Can you find an essential prime implicant that covers 112? Explain.
Find two prime implicants that cover 112.
Give two minimum expressions for F.
(e) Work Problem 5.6.
(f ) If you have a copy of the LogicAid program available, use the Karnaugh
map tutorial mode to help you learn to find minimum solutions from
­Karnaugh maps. This program will check your work at each step to make
sure that you loop the terms in the correct order. It also will check your
final answer. Work Problem 5.7 using the Karnaugh map tutor.
7.
(a) In Example 4, page 107, we derived the following function:
Z = Σ m(0, 3, 6, 9) + Σ d(10, 11, 12, 13, 14, 15)
Plot Z on the given map using X’s to represent don’t-care terms.
AB
00
CD
01
11
10
00
01
11
10
Z
(b) Show that the minimum sum of products is
Z = A′B′C′D′ + B′CD + AD + BCD′
Which four don’t-care minterms were assigned the value 1 when forming
your solution?
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132 Unit 5
(c) Show that the minimum product of sums for Z is
Z = (B′ + C )(B′ + D′)(A′ + D)(A + C + D′)(B + C′ + D)
Which one don’t-care term of Z was assigned the value 1 when forming
your solution?
(d) Work Problem 5.8.
8.
Study Section 5.5, Five-Variable Karnaugh Maps.
(a) The figure below shows a three-dimensional five-variable map. Plot the 1’s
and loops on the corresponding two-dimensional map, and give the minimum sum-of-products expression for the function.
BC
00
DE
00
A= 1
01
1
1
01
11
BC
DE
10
1
1
00
01
11
10
00
11
10
1
A
1
0
1
01
11
A= 0
1
1
10
F=
(b) On a five-variable map (Figure 5-21), what are the five minterms adjacent
to minterm 24?
(c) Work through all of the examples in this section carefully and make sure
that you understand all of the steps.
(d) Two minimum solutions are given for Figure 5-24. There is a third minimum sum-of-products solution. What is it?
(e) Work Programmed Exercise 5.2.
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Karnaugh Maps 133
(f )
BC
DE
16
00
00
01
20
X
11
X
0
17
A
1
0
01
1
23
1
18
10
3
22
1
2
1
13
31
1
1
30
6
9
27
1
7
1
8
25
5
X
1
12
29
X
1
11
24
4
21
19
10
28
X
15
26
X
14
11
X
10
Find the three 1’s and X’s adjacent to 118. Can these all be looped with a
single loop?
Find the 1’s and X’s adjacent to 124. Loop the essential prime implicant that
covers 124.
Find the 1’s and X’s adjacent to 13. Loop the essential prime implicant that
covers 13.
Can you find an essential prime implicant that covers 122? Explain.
Find and loop two more essential prime implicants.
Find three ways to cover the remaining 1 on the map and give the corresponding minimum solutions.
(g) If you have the LogicAid program available, work Problem 5.9, using the
Karnaugh map tutor.
9.
Study Section 5.6, Other Uses of Karnaugh Maps. Refer to Figure 5-8 and note
that a consensus term exists if there are two adjacent, but nonoverlapping prime
implicants. Observe how this principle is applied in Figure 5-26.
10.
Work Problems 5.10, 5.11, 5.12, and 5.13. When deriving the minimum solution
from the map, always write down the essential prime implicants first. If you do
not, it is quite likely that you will not get the minimum solution. In addition, make
sure you can find all of the prime implicants from the map (see Problem 5.10(b)).
11.
Review the objectives.
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Karnaugh Maps
Switching functions can generally be simplified by using the algebraic techniques
described in Unit 3. However, two problems arise when algebraic procedures are
used:
1.
2.
The procedures are difficult to apply in a systematic way.
It is difficult to tell when you have arrived at a minimum solution.
The Karnaugh map method studied in this unit and the Quine-McCluskey procedure studied in Unit 6 overcome these difficulties by providing systematic methods
for simplifying switching functions. The Karnaugh map is an especially useful tool
for simplifying and manipulating switching functions of three or four variables, but
it can be extended to functions of five or more variables. Generally, you will find
the Karnaugh map method is faster and easier to apply than other simplification
methods.
5.1 Minimum Forms of Switching Functions
When a function is realized using AND and OR gates, the cost of realizing the function is directly related to the number of gates and gate inputs used. The K
­ arnaugh
map techniques developed in this unit lead directly to minimum cost two-level
­circuits composed of AND and OR gates. An expression consisting of a sum of
product terms corresponds directly to a two-level circuit composed of a group of
AND gates feeding a single OR gate (see Figure 2-5). Similarly, a product-of-sums
expression corresponds to a two-level circuit composed of OR gates feeding a single AND gate (see Figure 2-6). Therefore, to find minimum cost two-level AND-OR
gate circuits, we must find minimum expressions in sum-of-products or product-­ofsums form.
A minimum sum-of-products expression for a function is defined as a sum
of ­product terms which (a) has a minimum number of terms and (b) of all those
­expressions which have the same minimum number of terms, has a minimum number
of literals. The minimum sum of products corresponds directly to a minimum t­ wo-level
gate circuit which has (a) a minimum number of gates and (b) a minimum number
134
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