code 11,10,00,01. The rows represent
products wx, wx, w x, and wx,
respectively, and the columns
correspond to the products yz, yz, y z,
and yz, respectively. Using Gray codes
and considering cells adjacent in the
first and last rows and in the first and
eighth columns, the first and fourth
columns, the second and seventh
columns, the third and sixth columns,
and the fifth and eighth columns are
considered adjacent. 19. Which rows
and which columns of a 4 16 map for
Boolean functions in six variables
usin
cells all containing 1s representing a
single literal. Once all prime
implicants have been identified, the
goal is to find the smallest possible
subset of these prime implicants with
the property that the cells
representing these prime implicants
cover al
visual method for simplifying sum-ofproducts expansions; they are not
suited for mechanizing this process.
We will first illustrate how K-maps are
used to simplify expansions of Boolean
functions in two variables. We will
continue by showing how K-maps ca
variables. The K-maps we used to
minimize Boolean functions in two,
three, and four variables are built
using 2 2, 2 4, and 4 4 rectangles,
respectively. Furthermore,
corresponding cells in the top row and
bottom row and in the leftmost column
and rightmo
the second and seventh columns, the
third and sixth columns, and the fifth
and eighth columns (as the reader
should verify). To use a K-map to
minimize a Boolean function in n
variables, we first draw a K-map of the
appropriate size. We place 1s in all
ce
maps to simplify these sum-ofproducts expansions. (a) wxyz + wxyz
+ wxy z + wxyz + wx yz + wx y z + wxyz
+ w xyz + w xyz (b) wxy z + wxyz +
wxyz + wx y z + wxy z + w xyz + w x y z
(c) wxyz + wxy z + wxyz + wxyz + wx y z
+ wxyz + wxyz + wxy z + wxyz + w xy
these sum-of-products expansions in
three variables. a) xy z b) xyz + x y z c)
xyz + xyz + xyz + x yz 8. Construct a Kmap for F (x, y, z) = xz + yz + xyz. Use
this K-map to find the implicants,
prime implicants, and essential prime
implicants of F (x, y,
one of the minterms. Once we have
found essential prime implicants, we
can simplify the table by eliminating
the rows for minterms covered by
these prime implicants. Furthermore,
we can eliminate any prime implicants
that cover a subset of minterms
covere
generates a copy of its own source code
as its complete output. Producing the
shortest possible quine in a given
programming language is a popular
puzzle for hackers. P1: 1 CH12-7T
Rosen-2311T MHIA017-Rosen-v5.cls
May 13, 2011 10:27 840 12 / Boolean
Algeb
to a block of all 1s in the K-map is
called an implicant of the function
being minimized. It is called a prime
implicant if this block of 1s is not
contained in a larger block of 1s
representing the product of fewer
literals than in this product. The goal
columns. (These K-maps contain 2n
cells because
n/2+ n/2 = n.) The rows and columns
need to be positioned so that the cells
representing minterms differing in just
one literal are adjacent or are
considered adjacent by specifying
additional adjacencies of
(6,7) w xz 001 TABLE 7 wxyz wxyz
wxyz wxyz wxyz wxyz wx yz wz X X X X
wyz X X wxy X X xyz X X Exercises 1. a)
Draw a K-map for a function in two
variables and put a 1 in the cell
representing xy. b) What are the
minterms represented by cells
adjacent to t
minterms by bit strings and then group
these terms together according to the
number of 1s in the bit strings. This is
shown in Table 5. All the Boolean
products that can be formed by taking
Boolean sums of these products are
shown in Table 6. The only pro
of 2, 4, 8, or 16 cells that represent
minterms that can be combined. Each
cell representing a minterm must
either be used to form a product using
fewer literals, or be included in the
expansion. In Figure 9 some examples
of blocks that represent products
and [Ka93]), so the existence of a
polynomial-time algorithm for
minimizing Boolean circuits is unlikely.
The QuineMcCluskey method has
exponential complexity. In practice, it
can be used only when the number of
literals does not exceed ten. Since the
197
follow that x = y if there exists a value z
in cfw_0, 1 such that a) xz = yz? b) x + z = y
+ z? c) x z = y z? d) x z = y z? e) x |
z = y | z? A Boolean function F is called
self-dual if and only if F (x1,.,xn) = F
(x1,., xn). 3. Which of these functions
a
Figure 5(a). This K-map can be thought
of as lying on a cylinder, as shown in
Figure 5(b). On the cylinder, two cells
have a common border if and only if
they are adjacent. (a) (b) xyz xyz xyz
xyz xyz xyz xyz xyz yz yz yz yz x x xyz
xyz xyz xyz xyz xyz xy
final answer is z + x y. As was
illustrated in Example 9, the Quine
McCluskey method uses this sequence
of steps to simplify a sum-of-products
expression. 1. Express each minterm in
n variables by a bit string of length n
with a 1 in the ith position if x
combined differ by exactly one in the
number of 1s in the bit strings that
represent them. When two minterms
are combined into a product, this
product contains two literals. A
product in two literals is represented
using a dash to denote the variable that
identify terms to group. For these
reasons there is a need for a procedure
for simplifying sum-of-products
expansions that can be mechanized.
The QuineMcCluskey method is such a
procedure. It can be used for Boolean
functions in any number of variables. I
present in the sum-of-products
expansion. The three K-maps are
shown in Figure 3. We can identify
minterms that can be combined from
the K-map. Whenever there are 1s in
two adjacent cells in the K-map, the
minterms represented by these cells
can be combin
example, excluding all squares with ds
and forming blocks, as shown in Figure
11(b), produces the expression wx y +
wxy + wxz. Including some of the ds
and excluding others and forming
blocks, as TABLE 1 Digit w x y z F 0 0 0
0001000102001003001104
010005
arbitrarily assigned. In the
minimization process we can assign 1s
as values to those combinations of the
input values that lead to the largest
blocks in the K-map. This is illustrated
in Example 8. EXAMPLE 8 One way to
code decimal expansions using bits
8n1 for n 1, a0 = 1 b) The general
solution of the associated linear
homogeneous recurrence relation is
a(h) n = 6n. A particular solution is
a(p) n = 1 2 8n. Hence, the general
solution is an = 6n + 1 2 8n. Using the
initial condition, it follows that =
Circuits 831 x x y y 1 1 x x y y 1 1 x x y y
1 1 1 (a) (b) (c) FIGURE 3 K-maps for
the Sum-of-Products Expansions in
Example 1. are in all four cells, the four
minterms can be combined into one
term, namely, the Boolean expression 1
that involves none of
xyz + x y z 13. a) Draw a K-map for a
function in four variables. Put a 1 in
the cell that represents wxyz. b) Which
minterms are represented by cells
adjacent to this cell? 14. Use a K-map to
find a minimal expansion as a Boolean
sum of Boolean products
McCluskey method to simplify the
sumof-products expansions in Exercise
12. 24. Use the QuineMcCluskey
method to simplify the sumof-products
expansions in Example 4. 25. Use the
QuineMcCluskey method to simplify
the sumof-products expansions in
Exercise 14
x2) = (1/ 5)[1/(1 x) 1/(1 x)],
where = (1 + 5)/2 and = (1
5)/2. Using the fact that 1/(1 x) =
k=0 kxk, it follows that G(x) = (1/ 5)
k=0(k k)xk. Hence, fk = (1/
5)(k k). 41. a) Let G(x) = n=0
Cnxn be the generating function for
cfw_Cn. Then G(x)2 = n=