Homework 1
Nick Holzman
1. A) individuals right to bear arms
B) Totinos pizza rolls are one of the best foods to have
c) Trying to show how less stressful it can be to own an android phone over an app
p. 84:
1. A is true e is true and I is false
2. e is true I is false and O is false
3. A is true I is True and O is False
4. E is true and A
5. A is false e is false I is true
p. 91:
1. A) Some priest
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 ad
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
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
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 t
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 ro
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
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 + w
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
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
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
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 p
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
cons
(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) W
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 su
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 F
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 th
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
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
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 b
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
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 u
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
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 TABL
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. E
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 +
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,
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