Three-Variable K-Map
A three-variable K-map is shown in Fig. 3.3. There are eight minterms for three binary variables; therefore, the map consists of eight squares. Note that the minterms are arranged, not in a binary sequence, but in a sequence similar to the Gray code ( Table 1.6). The characteristic of this sequence is that only one bit changes in value from one adjacent column to the next.The map drawn in part (b) is marked with numbers in each row and each column to show the relationship between the squares and the three variables. For example, the square assigned to m5 corresponds to row 1 and column 01. When these two numbers are concatenated, they give the binary number 101, whose decimal equivalent is 5. Each cell of the map corresponds to a unique minterm, so another way of looking at square m5 = xyz is to consider it to be in the row marked x and the column belonging to yz (column 01). Note that there are four squares in which each variable is equal to 1 and four in which each is equal to 0. The variable appears unprimed in the former four.
squares and primed in the latter. For
convenience, we write the variable with its letter symbol under the four
squares in which it is unprimed.
To understand the
usefulness of the map in simplifying Boolean functions, we must recognize the
basic property possessed by adjacent squares:
Any two adjacent squares in the
map differ by only one variable, which is primed in one square and unprimed
in the other. For example, m5 and
m7 lie in two adjacent
squares. Variable y is primed in m5 and
unprimed in m7, whereas
the other two variables are the same in both squares. From the postulates of Boolean
algebra, it follows that the sum of two minterms in adjacent squares can be
simplified to a single product term consisting of only two literals. To clarify
this concept, consider the sum of two adjacent squares such as m5 and m7:
m5 + m7 = xyz + xyz = xz(y+ y) = xz
Here, the two squares differ by the
variable y , which can be removed when the sum of the two minterms is
formed. Thus, any two minterms in adjacent squares (vertically or horizontally,
but not diagonally, adjacent) that are ORed together will cause a removal of
the dissimilar variable. The next four examples explain the procedure for
minimizing a Boolean function with a K-map.
Simplify the Boolean function
F (x, y, z) = (2, 3, 4, 5)
F = xy + xy
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In certain cases, two squares in the map are
considered to be adjacent even though they do not touch each other. In Fig. 3.3(b), m0 is adjacent to m2
and m4 is adjacent
to m6 because their
minterms differ by one variable. This difference can be readily verified
algebraically:
m0 + m2 = xyz + xyz= xz(y + y) = xz m4 + m6 = xyz + xyz = xz
+ (y + y) = xz
Consequently, we must modify the definition of
adjacent squares to include this and other similar cases. We do so by
considering the map as being drawn on a surface in which the right and left
edges touch each other to form adjacent squares.
Simplify the Boolean function
F (x, y, z) = (0, 2, 4, 5, 6)
The
map for F is shown in Fig. 3.6.
First, we combine the four adjacent squares in the first and last columns to
give the single literal term z. The
remaining single square, representing minterm 5, is combined with an adjacent
square that has already been used once. This is not only permissible, but
rather desirable, because the two adjacent squares give the two-literal term xy and the single square represents the three-literal minterm xyz. The simplified function is
F = z + xy
If a function is not expressed in sum-of-minterms
form, it is possible to use the map to obtain the minterms of the function and
then simplify the function to an expression with a minimum number of terms. It
is necessary, however, to make sure that the algebraic expression is in
sum-of-products form. Each product term can be plotted in the map in one, two,
or more squares. The minterms of the function are then read directly from the
map.
For the Boolean function
F = AC + AB + ABC + BC
Express
this function as a sum of minterms.
Find
the minimal sum-of-products expression.
Note that F
is a sum of products. Three product terms in the expression have two literals
and are represented in a three-variable map by two squares each. The two
squares corresponding to the first term, AC,
are found in Fig. 3.7 from the
coincidence of A (first row) and C (two middle columns) to give squares
001 and 011. Note that, in marking 1’s in the squares, it is possible to find a
1 already placed there from a preceding term.
This happens with the second term, AB, which has 1’s in squares 011 and
010. Square 011 is common with the first term, AC, though, so only one 1 is marked in it. Continuing in this
fashion, we determine that the term ABC belongs
in square 101, corresponding to minterm 5, and the term BC has
two 1’s in squares 011 and 111. The function has a total of five minterms, as
indicated by the five 1’s in the map of
Fig. 3.7. The minterms are read
directly from the map to be 1, 2, 3, 5, and 7. The function can be expressed in
sum-of-minterms form as
F (A, B, C) = (1, 2, 3, 5, 7)
The sum-of-products expression, as originally
given, has too many terms. It can be simplified, as shown in the map, to an
expression with only two terms:
F = C + AB
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