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Assignment 421solutions

Course

Discrete Mathematics  (Mathematics 1271)

23 Documents
Students shared 23 documents in this course
Academic year: 2021/2022
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Lakehead University

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Math 1271

Assignment 4

Solutions

  1. Give the dual of each of the

following Boolean expressions: r'*y, rgtz, rylyz,

and ry +

z'.

Solution: The dual of rl *

y is rty. The dual of ry' z, is r I A'

  • z. The dual of

ry I

yz is

(r *

a)fu

  • z).

The dual of ry *

zt is (r

-l y)zt.

  1. Given the

Boolean functions ct(r,y): rAt+ r'and 0@,A):

r'y* r, evaluate

a(0, 1), B(0,

1), a(1,0),

B(1,0),

a(0,0), and

p(1,

1).

Solution:

We have o(0, 1)

####### :

0'0f 1

####### :7,

####### 0(0,1)

####### :

####### 1'1+

####### :

####### 1,

o(1,0)

####### :

####### 1'1+

####### :

####### 1,

####### B(1,0)

####### :

####### 0*

####### 1

####### :

1, cv(0,0)

####### :0'

1f 1

####### :

1, and B(1,1)

####### :

####### 0'

####### I

####### *

####### 1

####### :

####### 1.

####### 3.

Put the

Boolean expressions

(rz *

y)' I ryz,

ny' +

ny I

zt, and yzt *

nz into

disjunctive

normal form.

Solution:

We have:

and

(rz
*
y)'
*
rYz
:
(rz)'Y' + rYz
:
(r'
*
z')y'
+
ryz
:
,'a'
+
z'y'
+
ryz
:
r'a' (z
+
zt)
+
z' y' (r * r') + nyz
:
tr'
,y'
*
r'z'y'
+
rz'u'
+
r'z'y'
+
ryz
:
trtytz +
r'a'z'
I
ry'z'
+
ryz
ry'
+
ny
*
z'
:
ry'(zl
z')
+
ny(z+ z')
+
(r+r')(y+a')r'
:
ra' z I ny' z' *
nyz
*
ryz'
+
ryz'
+
r'yz'
+
ny' z' + n'y' z'
:
r,at z
*
ry' z' + raz +
rYz'
+
r'Yz'
+
r'Y' z'

and

yz'
+ rz
-
az'(r
+u') +
rz(a
*
y')
:
nazt +
r'uz'
*
r'yz
*
ny'z
  1. Determine if

y'z' is a

prime implicant of r'yz' +

n'A'z' t

ry'z'.

Solution: Let

B:

n'yz'+r'y'z'+ry'z''

Wefirst haveto checkwhethety'z'+p:

p'

We have y'z'

: (r + r')y'z'

####### :

trytzt *

r'y'zl. Since both ry'z'

and n'y'z' occtt in

the disjunctive normal form of B,

we have

y'z' + 0

####### :

F

factors of y'z' are

yl and, z'.We have to checkwhether

y'+

P:

####### P

and whether

z'*p:

B'We

can

do this by considering the

functions they define. We

have

p(0,0, 1)

####### :

0 and

(y'+

13)(0,0,1)

####### :

####### (1 +0)

####### :

1, so

we see

A'+ B I

p.

Similarly, 13(1,1,0)

####### :

6tt

####### Q'

####### +

####### 0):

####### (

####### +0)

####### :

t, so z' + t3 + P

Thts y'z' is a

prime implicant

of

B.

  1. Use Karnough

maps to find minimal forms

for the

expressions ryz *

ry'z' +

rtyz I

r'gz' and ryz' +

ry'z' *

r'y'z' +

r'yz'

####### .

uz ?.'

!1'z

U'

ix I I

trl

####### I I

Solution:

For a

####### :

ryz *

rytzt *

r'yz * r'yz' we

get:

and we have a

####### -

yz +

r'y I

ryz'

For B

####### :

ryz' + ry'z' +

r'y'z' +

rtyzt we

get

and we

have

B

####### :

7t.

  1. Draw circuits to implement

the given Boolean expressions: n' I rA,

(r * A)',

a'

####### *

r'a, and n'yt * r'a.

Solution:

We have:

r' * rU,

%

X

xt+x

(r*a)',

;H

a'
+
fr'a,

and finally,

for r'y' *

rty

####### :

####### ,'

(a +

y')

####### :

r'

,

we have

just

&*b)

2'*/I

yz
az
u'z
r
1 1

tr' I

t

####### I

3

x

F''

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Assignment 421solutions

Course: Discrete Mathematics  (Mathematics 1271)

23 Documents
Students shared 23 documents in this course

University: Lakehead University

Was this document helpful?
Math 1271 Assignment 4
Solutions
1. Give the dual of each of the following Boolean expressions: r'*y, rgtz, rylyz,
and ry + z'.
Solution: The dual of rl * y is rty. The dual of ry' z, is r I A' * z. The dual of
ry I yz is (r * a)fu + z). The dual of ry * zt is (r -l y)zt.
2. Given the Boolean functions ct(r,y): rAt+ r'and 0@,A): r'y* r, evaluate
a(0, 1), B(0, 1), a(1,0), B(1,0), a(0,0), and p(1, 1).
Solution: We have o(0, 1) : 0'0f 1 :7, 0(0,1) : 1'1+0 : 1, o(1,0) : 1'1+0 : 1,
B(1,0) : 0.0* 1 : 1, cv(0,0) :0' 1f 1 : 1, and B(1,1) : 0' I * 1 : 1.
3. Put the Boolean expressions (rz * y)' I ryz, ny' + ny I zt, and yzt * nz into
disjunctive normal form.
Solution: We have:
and
(rz * y)' * rYz : (rz)'Y' + rYz
: (r' * z')y' + ryz
: ,'a' + z'y' + ryz
: r'a' (z + zt)
+ z'
y' (r * r') + nyz
: tr'
,y' * r'z'y' + rz'u' + r'z'y' + ryz
: trtytz + r'a'z' I ry'z' + ryz
ry' + ny * z' : ry'(zl z') + ny(z+ z') + (r+r')(y+a')r'
: ra' z I ny' z' * nyz * ryz' + ryz' + r'yz' + ny' z' + n'y' z'
: r,at
z * ry' z' + raz + rYz' + r'Yz' + r'Y' z'
and
yz' + rz - az'(r +u') + rz(a * y')
: nazt
+ r'uz' * r'yz
* ny'z
4. Determine if y'z' is a prime implicant of r'yz' + n'A'z' t ry'z'.
Solution: Let B: n'yz'+r'y'z'+ry'z'' Wefirst haveto checkwhethety'z'+p: p'
We have y'z' : (r + r')y'z' : trytzt * r'y'zl. Since both ry'z' and n'y'z' occtt in
the disjunctive normal form of B, we have y'z' + 0 : F.The factors of y'z' are
yl and, z'.We have to checkwhether y'+ P: P and whether z'*p: B'We
can do this by considering the functions they define. We have p(0,0, 1) : 0 and
(y'+ 13)(0,0,1) : (1 +0) : 1, so we see A'+ B I p. Similarly, 13(1,1,0) :6 6tt4
Q' + 0): (1 +0) : t, so z' + t3 + P Thts y'z' is a prime implicant of B.
5. Use Karnough maps to find minimal forms for the expressions ryz * ry'z' +
rtyz I r'gz' and ryz' + ry'z' * r'y'z' + r'yz' .

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