Chapter 3 Problem Solutions -y 3.1. (a) Xy + Xy

8/14/2019 Chapter 3 Problem Solutions -y 3.1. (a) Xy + Xy

1/42

- 3.1 -

Chapter 3

Problem Solutions

3.1. (a) xy + xy-+ x-y-= x(y+y

-) + x

-y-

P4b

= x.1 + x

-

y

-

P5a= x + x

-y-

P2b

= x + y-

T7a

(b) (x-z-+ x-y + x

-z + xy)

__________________

= (x-z + x

-z-+ yx + yx

-)

__________________

P3a,P3b

= [x-(z+z

-) + y(x+x

-)]

________________

P4b

= (x-.1 + y.1)

__________

P5a

= (x-+ y)

______

P2b

= (x-)

__

y-

T9a

= xy- T5

(c) (x + y)(x-z-+ z)(y

-+ xz)

_______

= (x + y)(z + z-x-)(y-+ xz)

_______

P3a,P3b

= (x + y)(z + x-)(y-+ xz)

_______

T7a

= (x + y)(z + x-)y(xz)

___

T9a,T5

= (x + y)(z + x-)y(x

-+ z-) T9b

= (x + y)(x-

+ z)(x-+ z-)y P3a,P3b

= (x + y)(x-

+ zz-)y P4a

= (x + y)(x-

+ 0)y P5b

= (x + y)x-y P2a

= x-y(x + y) P3b

= x-[y(x + y)] T8b

= x-[y(y + x)] P3a

= x-y T6b


2/42

- 3.2 -

3.1. (continued)

(d) (xy + yz + xz)

_____________

= (xy)

___

(yz)

___

(xz)

___

T9a

= (x-+ y-)(y-+ z-)(x-

+ z-) T9b

= (x-+ y-)(x-+ z-)(y-

+ z-) P3b

= (x- + y-z-)(y- + z-) P4a

= (x-+ y-z-)y-+ (x

-+ y

-z-)z-

P4b

= y-(x-+ y-z-) + z

-(x-+ y

-z-) P3b

= y-x-+ y-y-z-+ z-x-+ z-y-z-

P4b

= x-y-+ y-y-z-+ x-z-+ y-z-z-

P3b

= x-y-+ y-z-+ x-z-+ y-z-

T4b

= x-y-

+ y-z-+ y-z-+ x-z-

P3a

= x-y-

+ y-z-+ x-z-

T4a

(e) xy + yz + x-z = xy + yz.1 + x

-z P2b

= xy + yz(x + x-) + x

-z P5a

= xy + yzx + yzx-+ x-z P4b

= xyz + xy + x-zy + x

-z P3a,P3b

= xy(z + 1) + x-z(y + 1) P4b

= xy.1 + x-z.1 T2a

= xy + x-z P2b

(f) (x + y)(x-+ z) = (x + y)x

-+ (x + y)z P4b

= x-(x + y) + z(x + y) P3b

= x-x + x

-y + zx + zy P4b

= xx-+ x

-y + zx + zy P3b

= 0 + x-y + zx + zy P5b

= x-y + zx + zy P2a

= x-y + zx + zy.1 P2b

= x-y + zx + zy(x + x

-) P5a

= x-y + zx + zyx + zyx

-P4b

= x-yz + x

-y + xzy + xz P3a,P3b

= x-y(z + 1) + xz(y + 1) P4b

= x-y.1 + xz.1 T2a

= x-y + xz P2b


3/42

- 3.3 -

3.1. (continued)

(g) (x + y)(y + z)(x + z) = [(x + y)y + (x + y)z](x + z) P4b

= [y(x + y) + z(x + y)](x + z) P3b

= (yx + yy + zx + zy)(x + z) P4b

= (yx + y + zx + zy)(x + z) T4b

= (y + yx + yz + zx)(x + z) P3a

= (y + zx)(x + z) T6a

= (y + zx)x + (y + zx)z P4b

= x(y + zx) + z(y + zx) P3b

= xy + xzx + zy + zzx P4b

= xy + xxz + yz + xzz P3b

= xy + xz + yz + xz T4b

= xy + (xz + xz) + yz P3a

= xy + xz + yz T4a

(h) xy-+ yz

-+ x-z = xy

-.1 + yz-.1 + x

-z.1 P2b

= xy-(z + z

-) + yz

-(x + x

-) + x

-z(y + y

-) P5a

= xy-z + xy

-z-+ yz

-x + yz

-x-+ x-zy + x

-zy-

P4b

= x-yz + x

-yz-+ y-zx + y

-zx-+ xz

-y + xz

-y-P3a,P3b

= x-y(z + z

-) + y

-z(x + x

-) + xz

-(y + y

-) P4b

= x-y.1 + y

-z.1 + xz

-.1 P5a

= x-y + y

-z + xz

-P2b

3.2. To prove the cancellation law does not hold, use the method

of contradiction. According to T7b, x(x-+y)=xy. Assuming

the cancellation law holds, it follows that x-+y=y for all x

and y in the Boolean algebra. Since x and y denote any

elements in the Boolean algebra, let x=y. It then follows

that y-+y=y. However, from P5a it is known that y must be 1

in this case and not any arbitrary element in the algebra.

Thus, by contradiction, the cancellation law does not hold.

By applying a dual argument starting with T7a, it can be

shown that x+y=x+z does not imply y=z.


4/42

- 3.4 -

3.3.

(a) (b) (c)

x y z x-z+xy (x

-+y)(x+z) (x+y+z)

______

x-y-z-

xy+yz+x-z xy+x

-z

0 0 0 o 0 1 1 0 0

0 0 1 1 1 0 0 1 1

0 1 0 0 0 0 0 0 0

0 1 1 1 1 0 0 1 1

1 0 0 0 0 0 0 0 0

1 0 1 0 0 0 0 0 0

1 1 0 1 1 0 0 1 1

1 1 1 1 1 0 0 1 1

3.4. Let B={0,1,a} where a0,1. By Postulate P5 of a Boolean

algebra, the complement of the element a, i.e., a-, must

exist and satisfy the relationships a+a-=1 and a.a

-=0. Since

there are just three elements in B, a-

must be 0, 1, or a.

Suppose a-=a, then a+a

-=a+a=a. However, since a1, Postulate

P5a is not satisfied. Thus, a

-

a. Now suppose a-

=1. Then

a.a-=a.1=a. However, since a0, Postulate P5b is not

satisfied. Finally, suppose a-=0. Then, a+a

-=a+0=a. Again

Postulate P5a is not satisfied since a1. Therefore, a-does

not exist and B cannot be a Boolean algebra.


5/42

- 3.5 -

3.5. (a)

x y z x-

z-

x-+y x

-+z-

(x-+y)(x

-+z-) yz f

0 0 0 1 1 1 1 1 0 1

0 0 1 1 0 1 1 1 0 1

0 1 0 1 1 1 1 1 0 1

0 1 1 1 0 1 1 1 1 1

1 0 0 0 1 0 1 0 0 0

1 0 1 0 0 0 0 0 0 0

1 1 0 0 1 1 1 1 0 1

1 1 1 0 0 1 0 0 1 1

(b)

x y z x-

xy x-z xy+x

-z (xy+x

-z)

______

yz f

0 0 0 1 0 0 0 1 0 1

0 0 1 1 0 1 1 0 0 0

0 1 0 1 0 0 0 1 0 1

0 1 1 1 0 1 1 0 1 1

1 0 0 0 0 0 0 1 0 1

1 0 1 0 0 0 0 1 0 1

1 1 0 0 1 0 1 0 0 0

1 1 1 0 1 0 1 0 1 1


6/42

- 3.6 -

3.5. (continued)

(c)

x y z x-

y-

x+y-

y+z x-+z (x+y

-)(y+z) f

0 0 0 1 1 1 0 1 0 0

0 0 1 1 1 1 1 1 1 1

0 1 0 1 0 0 1 1 0 0

0 1 1 1 0 0 1 1 0 0

1 0 0 0 1 1 0 0 0 0

1 0 1 0 1 1 1 1 1 1

1 1 0 0 0 1 1 0 1 0

1 1 1 0 0 1 1 1 1 1


7/42

- 3.7 -

3.5. (continued)

(d)

w x y z w-

y-

wx wxy y-+z w

-(y-+z) f

0 0 0 0 1 1 0 0 1 1 1

0 0 0 1 1 1 0 0 1 1 1

0 0 1 0 1 0 0 0 0 0 0

0 0 1 1 1 0 0 0 1 1 1

0 1 0 0 1 1 0 0 1 1 1

0 1 0 1 1 1 0 0 1 1 1

0 1 1 0 1 0 0 0 0 0 0

0 1 1 1 1 0 0 0 1 1 1

1 0 0 0 0 1 0 0 1 0 0

1 0 0 1 0 1 0 0 1 0 0

1 0 1 0 0 0 0 0 0 0 0

1 0 1 1 0 0 0 0 1 0 0

1 1 0 0 0 1 1 0 1 0 0

1 1 0 1 0 1 1 0 1 0 0

1 1 1 0 0 0 1 1 0 0 1

1 1 1 1 0 0 1 1 1 0 1

3.6. (a) f(x,y,z) = x-y-z-+ x

-y-z + x

-yz + xy

-z + xyz

-

= m0+ m

1+ m

3+ m

5+ m

6

= 3m(0,1,3,5,6)

(b) f(w,x,y,z) = w-x-y-z + w

-x-yz-+ w-xy-z-+ w

-xyz-+ w-xyz + wx

-yz

+ wxy-z- + wxy-z + wxyz

= m1

+ m2+ m

4+ m

6+ m

7+ m

11+ m

12

+ m13

+ m15

= 3m(1,2,4,6,7,11,12,13,15)


8/42

- 3.8 -

3.7. (a) f(x,y,z) = 3m(0,2,4,5,7)

= x-y-z-+ x

-yz-+ xy

-z-+ xy

-z + xyz

x y z f

0 0 0 1

0 0 1 0

0 1 0 1

0 1 1 0

1 0 0 1

1 0 1 1

1 1 0 0

1 1 1 1


9/42

- 3.9 -

3.7. (continued)

(b) f(w,x,y,z) = 3m(1,3,7,8,9,14,15)

= w-x-y-z + w

-x-yz + w

-xyz + wx

-y-z-+ wx

-y-z

+ wxyz-+ wxyz

w x y z f

0 0 0 0 0

0 0 0 1 1

0 0 1 0 0

0 0 1 1 1

0 1 0 0 0

0 1 0 1 0

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 1

1 0 1 0 0

1 0 1 1 0

1 1 0 0 0

1 1 0 1 0

1 1 1 0 1

1 1 1 1 1

3.8. (a) f(x,y,z) = (x+y-+z)(x

-+y+z)(x

-+y-+z-)

= M2M4M7

= JM(2,4,7)

(b) f(w,x,y,z) = (w+x+y+z)(w+x+y-+z-)(w+x

-+y+z

-)(w-+x+y+z)

.(w-+x+y+z

-)(w-+x+y

-+z)(w

-+x-+y-+z)

= M0M3M5M8M9M10M14

= JM(0,3,5,8,9,10,14)


10/42

- 3.10 -

3.9. (a) f(x,y,z) = JM(0,1,2,5,7)

= (x+y+z)(x+y+z-)(x+y

-+z)(x

-+y+z

-)(x-+y-+z-)

x y z f

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 0


11/42

- 3.11 -

3.9. (continued)

(b) f(w,x,y,z) = JM(0,3,6,7,9,10,12,13,15)

= (w+x+y+z)(w+x+y-+z-)(w+x

-+y-+z)(w+x

-+y-+z-)

.(w-+x+y+z

-)(w-+x+y

-+z)(w

-+x-+y+z)

.(w-+x-+y+z-)(w-+x-+y-+z-)

w x y z f

0 0 0 0 0

0 0 0 1 1

0 0 1 0 1

0 0 1 1 0

0 1 0 0 1

0 1 0 1 1

0 1 1 0 0

0 1 1 1 0

1 0 0 0 1

1 0 0 1 0

1 0 1 0 0

1 0 1 1 1

1 1 0 0 0

1 1 0 1 0

1 1 1 0 1

1 1 1 1 0

3.10. (a) [(w+x+y)(w-+xz)+y

-z-]

_________________

= [(w+x+y)(w-+xz)]

______________

(y-z-)

___

= [(w+x+y)

______

+(w-+xz)

_____

](y+z)

= [w

-

x

-

y

-

+w(xz)

___

](y+z)= [w

-x-y-+w(x

-+z-)](y+z)

(b) [x-(w-y-+xyz

-)]

__________

= x + (w-y-+xyz

-)

_______

= x + (w-y-)

___

(xyz-)

____

= x + (w+y)(x-+y-+z)


12/42

- 3.12 -

3.10. (continued)

(c) {wy-[(wy

-)

___

+xz-]}

____________

= w-+ y + [(wy

-)

___

+xz-]

________

= w-+ y + (wy

-)(xz

-)

___

= w

-

+ y + wy

-

(x

-

+z)

(d) {wx+z[(w+x-+y-)

______

+(x+y)

____

]}

____________________

= (wx)

___

{z[(w+x-+y-)

______

+(x+y)

____

]}

_________________

= (w-+x-){z-+[(w+x

-+y-)

______

+(x+y)

____

]

______________

}

= (w-+x-)[z-+(w+x

-+y-)(x+y)]

3.11. (a) f = x[w-.1.yz

-+z(1.y

-+w.1

-)] + x

-[w-.0.yz

-+z(0.y

-+w.0

-)]

= x(w-yz-+y-z) + x

-(wz)

(b) f = [z+w-xy.0-+0.(xy-+wx-)][z-+w-xy.1-+1.(xy-+wx-)]

= (z+w-xy)(z

-+xy-+wx-)

3.12. (a) yz + x-y-z + xyz

-= 1.yz + x

-y-z + xyz

-

= (x+x-)yz + x

-y-z + xyz

-

= xyz + x-yz + x

-y-z + xyz

-

= xy(z+z-) + x

-z(y+y

-)

= xy + x-z

(b) x-y-z-

+ x-y-z + x

-yz-+ xy

-z-+ xy

-z + xyz

-

= y-(x-z-+x-z+xz

-+xz) + yz

-(x-+x)

= y-[x-(z-+z)+x(z

-+z)] + yz

-

= y-(x-+x) + yz

-

= y-+ yz

-

= y-+ z-

(c) xy-+ xz + x

-z-+ xyz

-+ x-y-z = x(y

-+z+yz

-) + x

-(z-+y-z)

= x(y-+z+y) + x

-(z-+y-)

= x + x-(y-+z-)

= x + y-+ z-


13/42

- 3.13 -

3.12. (continued)

(d) (x+yz)

_____

+ y-z = x

-(y-+z-) + y

-z

= x-y-+ x-z-+ y-z

= x-y-(z+z

-) + x

-z-+ y-z

= x-y-z + x-y-z- + x-z- + y-z

= x-z-(y-+1) + y

-z(x-+1)

= x-z-

+ y-z

(e) w-x-y-z + wx

-y-z + xz + xyz

-= x-y-z(w-+w) + x(z+yz

-)

= x-y-z + x(z+y)

= x-y-z + xz + xy

= (x-y-+x)z + xy

= (y-+x)z + xy

= y-z + xz + xy

= y-z + xz(y+y

-) + xy

= y-z + xyz + xy

-z + xy

= y-z(1+x) + xy(z+1)

= y-z + xy

(f) w-x-yz + wxy + w

-y-+ xy

-+ x-y-= w-x-yz + wxy + y

-(w-+x+x

-)

= w-x-yz + wxy + y

-

= (w-x-z+wx)y + y

-

= w-x-z + wx + y

-

(g) (w+x+y-+z)(w+x+y

-+z-)(w+x

-+y-+z)(w+x

-+y-+z-)(w

-+x-+y-+z)(w

-+x-+y-+z-)

= (w+x+y-+zz

-)(w+x

-+y-+zz-)(w-+x-+y-+zz-)

= (w+x+y-)(w+x

-+y-)(w-+x-+y-)

= (w+x+y-)(ww

-+x-+y-)

= (w+x+y-)(x

-+y-)

= y-+ x-(w+x)

= y- + wx- + xx-

= y-+ wx

-

(h) (x+z)(w+x)(y-+z)(w+y

-) = (x+wz)(y

-+wz)

= wz + xy-


14/42

- 3.14 -

3.13. xi.f(x

1,x2,...,x

i,...,x

n)

= xi[xi.f(x

1,x2,...,1,...,x

n)+x-i.f(x

1,x2,...,0,...,x

n)]

= xi.f(x

1,x2,...,1,...,x

n)+x

ix-i.f(x

1,x2,...,0,...,x

n)

= xi.f(x

1,x2,...,1,...,x

n)

Equation (b) follows from the duality principle.

3.14. (a) f(x,y,z) = x-(y-+z) + z

-

= x-y-+ x

-z + z

-

= x-y-(z+z

-) + x

-z(y+y

-) + z

-(x+x

-)(y+y

-)

= x-y-z + x

-y-z-+ x-yz + xyz

-+ xy

-z-+ x-yz-

(b) f(x,y,z) = (x+y-)(x+z)

= x + y

-

z

= x(y+y-)(z+z

-) + (x+x

-)y-z

= xyz + xyz-+ xy

-z + xy

-z-+ x

-y-z

3.15. (a) f(x,y,z) = (y+z-)(xy

-+z)

= (y+z-)(x+z)(y

-+z)

= (xx-+y+z

-)(yy

-+x+z)(xx

-+y-+z)

= (x+y+z-)(x

-+y+z

-)(x+y+z)(x+y

-+z)(x

-+y-+z)

(b) f(x,y,z) = x + x-z-(y+z)

= (x+x-)(x+z

-)(x+y+z)

= (x+z-)(x+y+z)

= (x+z-+yy-)(x+y+z)

= (x+y+z-)(x+y

-+z-)(x+y+z)

3.16. (a) f-(x,y,z) = 3m(1,3,4,6,7) = JM(0,2,5)

(b) f-(x,y,z) = 3m(1,2,5,7) = JM(0,3,4,6)

(c) f-(w,x,y,z) = 3m(0,2,3,5,9,10,11,13,15)

= JM(1,4,6,7,8,12,14)

(d) f-(w,x,y,z) = 3m(3,7,8,10,12,13)

= JM(0,1,2,4,5,6,9,11,14,15)


15/42

- 3.15 -

3.17. (a) f(x,y,z) = 3m(1,3,5)

f-(x,y,z) = JM(1,3,5)

f=(x,y,z) = f(x,y,z) =JM(0,2,4,6,7)

(b) f(x,y,z) = JM(3,4)

f-(x,y,z) = 3m(3,4)

f=(x,y,z) = f(x,y,z) = 3m(0,1,2,5,6,7)

(c) f(w,x,y,z) = 3m(0,1,2,3,7,9,11,12,15)

f-(w,x,y,z) = 3m(4,5,6,8,10,13,14)

f=(w,x,y,z) = f(w,x,y,z) = JM(4,5,6,8,10,13,14)

(d) f(w,x,y,z) = JM(0,2,5,6,7,8,9,11,12)

f-(w,x,y,z) = JM(1,3,4,10,13,14,15)

f=(w,x,y,z) = f(w,x,y,z) = 3m(1,3,4,10,13,14,15)

3.18. (a) f = v(wx-z+y)(w+z) + w

-

(b) f = [(w+x)(v-+y+z)+w

-][(w+x)(v

-+y+z)+v]

(c) f = [(w+x-)(w-+y)]

___________

z + (w-+y)z

-

3.19. (a)


16/42

- 3.16 -

3.19. (continued)

(b)

(c)

3.20. (a) f = xy-z + (xy-z+w)

______

+ wy-

= A + (A+w)

____

+ wy-

where A=xy-z

= A + (0+w)

____

+ wy-

by T3.12b

= A + w-+ wy

-

= xy-z + w

-+ wy

-

= xy-z + w

-+ y-

= w-+ y-(xz+1)

= w-+ y-


17/42

- 3.17 -

3.20. (continued)

(b) f = [(w-xy-z)

_____

z+x-z-]w-+ wyz

= [(w+x-+y+z

-)z+x

-z-]w-+ wyz

= (wz+x-z+yz+x

-z-)w-+ wyz

= w-x-z + w-yz + w-x-z- + wyz

= w-x-(z+z

-) + yz(w

-+w)

= w-x-+ yz

3.21. (a) f = w[(x+y)z+x-z-]

(b)

3.22. (a) f(w,x,y,z) = 3m(2,3,6,10,15) + dc(0,1,5,9,11,13,14)

= JM(4,7,8,12) + dc(0,1,5,9,11,13,14)


18/42

- 3.18 -

3.22. (continued)

(b)

w x y z f f-

0 0 0 0 - -

0 0 0 1 - -

0 0 1 0 1 0

0 0 1 1 1 0

0 1 0 0 0 1

0 1 0 1 - -

0 1 1 0 1 0

0 1 1 1 0 1

1 0 0 0 0 1

1 0 0 1 - -

1 0 1 0 1 0

1 0 1 1 - -

1 1 0 0 0 1

1 1 0 1 - -

1 1 1 0 - -

1 1 1 1 1 0

(c) f-(w,x,y,z) = 3m(4,7,8,12) + dc(0,1,5,9,11,13,14)

= JM(2,3,6,10,15) + dc(0,1,5,9,11,13,14)


19/42

- 3.19 -

3.23. (a) Using perfect induction:

x y z nand(y,z) nand[x,nand(y,z)] nand(x,y) nand[nand(x,y),z]

0 0 0 1 1 1 1

0 0 1 1 1 1 0

0 1 0 1 1 1 1

0 1 1 0 1 1 0

1 0 0 1 0 1 1

1 0 1 1 0 1 0

1 1 0 1 0 0 1

1 1 1 0 1 0 1

Since the columns for nand[x,nand(y,z)] and

nand[nand(x,y),z] are dissimilar, the nand-operation is

not associative.

(b) Using perfect induction:

x y z nor(y,z) nor[x,nor(y,z)] nor(x,y) nor[nor(x,y),z]

0 0 0 1 0 1 0

0 0 1 0 1 1 0

0 1 0 0 1 0 1

0 1 1 0 1 0 0

1 0 0 1 0 0 1

1 0 1 0 0 0 0

1 1 0 0 0 0 1

1 1 1 0 0 0 0

Since the columns for nor[x,nor(y,z)] and

nor[nor(x,y),z] are dissimilar, the nor-operation is

not associative.


20/42

- 3.20 -

3.24. (a) f = {[(u-v-)

___

(wx)

___

y]

__________

z}

_____________

= (u+v)(w-+x-)y + z

-

(b) f = {[x(xy)

___

]

______

[y(xy)

___

]

______

}

_______________

= x(x-+y-) + y(x-+y-)

= xy-+ x-y

(c) f = {[(x-+y+z

-)

______

+w-+z]

____________

+[(w+x-)

____

+y]

________

}

________________________

= (xy-z+w-+z)(w

-x+y)

= (w-+z)(w

-x+y)

(d) f = {[(v+w)

____

+(x-+y)

____

]

____________

+[(x-+y)

____

+z]

________

}

________________________

= (v-w-+xy-)(xy-+z)

(e) f = {[(u-+v-)

____

+(w+x)

____

+y-]z}

_________________

= (u-+v-)(w+x)y + z

-

3.25. (a)


21/42

- 3.21 -

3.25. (continued)

(b)


22/42

- 3.22 -

3.25. (continued)

(c)


23/42

- 3.23 -

3.25. (continued)

(d)


24/42

- 3.24 -

3.26. (a)


25/42

- 3.25 -

3.26. (continued)

(b)


26/42

- 3.26 -

3.26. (continued)

(c)


27/42

- 3.27 -

3.26. (continued)

(d)


28/42

- 3.28 -

3.27. (a)

(b)


29/42

- 3.29 -

3.27. (continued)

(c)


30/42

- 3.30 -

3.27. (continued)

(d)


31/42

- 3.31 -

3.28. (a)

(b)


32/42

- 3.32 -

3.28. (continued)

(c)


33/42

- 3.33 -

3.28. (continued)

(d)


34/42


35/42

- 3.35 -

3.29. (continued)

(b) Nor realization

f = [(w-+x)y+u

-+v](w+z)y


36/42

- 3.36 -

3.30. (x+y)r(x+z) = (x+y)

____

(x+z) + (x+y)(x+z)

____

= x-y-(x+z) + (x+y)x

-z-

= x-y-z + x

-yz-

= x-(y-z+yz

-)

= x-(yrz)

Alternatively, perfect induction could be used.

3.31. Using perfect induction

x y z yz xryz xry xrz (xry)(xrz)

0 0 0 0 0 0 0 0

0 0 1 0 0 0 1 0

0 1 0 0 0 1 0 0

0 1 1 1 1 1 1 1

1 0 0 0 1 1 1 1

1 0 1 0 1 1 0 0

1 1 0 0 1 0 1 0

1 1 1 1 0 0 0 0

Since the columns for xryz and (xry)(xrz) are dissimilar,

the exclusive-or-operation is not distributive over theand-operation.

3.32. (a) xu1 = x-.1-+ x.1

= x-.0 + x

= x

(b) xu0 = x-.0-+ x.0

= x-.1

= x-

(c) xux = x-x-+ xx

= x-+ x

= 1


37/42

- 3.37 -

3.32. (continued)

(d) xux-= x

-x + xx

-

= 0 + 0

= 0

(e) x-uy- = xy + x-y-

= xuy

(f) x-uy = xy

-+ x-y

= xry

= (xuy)

____

(g) (x+y)u(x+z) = (x+y)

____

(x+z)

____

+ (x+y)(x+z)

= x-y-x-z-+ x + yz

= x-y-z- + x + yz

= y-z-+ x + yz

= x + (yuz)

(h) xuyu(x+y) = xu[yu(x+y)]

= xu[y-(x+y)

____

+y(x+y)]

= xu(y-x-+xy+y)

= xu(x-y-+y)

= xu(x-+y)

= x-(x-+y)

____+ x(x

-+y)

= x-xy-

+ xx-+ xy

= xy

3.33. From the definition of the dual of a function, xry should

first be complemented by DeMorgan's law and then each of

the variables in the complemented expression should be

complemented. Applying DeMorgan's law gives

(xry)____

=(x-y+xy

-)

______=(x+y

-)(x-+y). Complementing each variable in

(x+y-)(x-+y) gives (x

-+y)(x+y

-). Therefore,

(xry)dual

= (x-+y)(x+y

-)

= (xry)

____

= xuy


38/42

- 3.38 -

3.34. (a)

According to statement 1, bn-1

should be connected to

gn-1

. According to statement 2, for n-1$k$1,

bk-1

=f(bk,gk-1

). This function is given by the truth

table

bk

gk-1

bk-1

0 0 0

0 1 1

1 0 1

1 1 0

which is described by bk-1

=bkrgk-1

. Thus, each of the

outputs bk-1

for n-1$k$1 is the exclusive-or of the

next higher order binary digit (bk) and the present

order Gray bit (gk-1

). The corresponding logic diagram

is


39/42

- 3.39 -

3.34. (continued)

(b) From part (a) we have

bn-1

= gn-1

bk-1

= bkrgk-1

for n-1$k$1

We now need expressions for gn-1 and gk-1 for n-1$k$1.

Clearly, gn-1

=bn-1

. For the remaining expression,

consider property (x) in Table 3.16. Letting X=gk-1

,

Y=bk, and Z=b

k-1, b

k-1=bkrgk-1

has the form XrY=Z.

Thus, X=YrZ or gk-1

=bkrbk-1

for n-1$k$1. A logic

diagram for an n-bit binary-to-Gray converter is:

3.35. Position: 7 6 5 4 3 2 1

Code group format: b4

b3

b2

p3

b1

p2

p1

c*1

= 0 if even parity over inputs b4, b

2, b

1, and p

1

c*1

= 1 if odd parity over inputs b4, b

2, b

1, and p

1


40/42

- 3.40 -

3.35. (continued)

Truth table for c*1:

b4

b2

b1

p1

c*1

0 0 0 0 0

0 0 0 1 1

0 0 1 0 1

0 0 1 1 0

0 1 0 0 1

0 1 0 1 0

0 1 1 0 0

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 0

1 0 1 1 1

1 1 0 0 0

1 1 0 1 1

1 1 1 0 1

1 1 1 1 0

c*1

= b-4b-2b-1p1+ b-4b-2b1p-1+ b-4b2b-1p-1+ b

-4b2b1p1+ b

4b-2b-1p-1

+ b4b-2b1p1

+ b4b2b-1p1+ b

4b2b1p-1

= b-4b-2(b-1p1+b1p-1) + b

-4b2(b-1p-1+b1p1) + b

4b-2(b-2p-1+b1p1)

+ b4b2(b-1p1+b

1p-1)

= b-4b-2(b1rp1) + b

-4b2(b1rp1)

______

+ b4b-2(b1rp1)

______

+ b4b2(b1rp1)

= (b1rp1)(b-4b-2+b4b2) + (b1rp1)

______

(b-4b2+b4b-2)

= (b1rp1)(b

4rb2)

______

+ (b1rp1)

______

(b4rb2)

Let b1rp1=X and b

4rb2=Y, then c*

1= XY

-+ X-Y = XrY

Therefore, c*1= (b

1rp1) r (b

4rb2)


41/42

- 3.41 -

3.35. (continued)

Using a similar procedure,

c*2

= (b1rp2) r (b

4rb3)

c*3

= (b2rp3) r (b

4rb3)

Each c*ican be realized with a network of the form:

To correct an appropriate bit, it is first necessary to

detect each of the seven combinations of c*3c*2c*1which can

be done with seven and-gates. To invert a bit if in error,

an exclusive-or-gate can be used. For example, to detect

c*3c*2c*1=011 and correct the corresponding bit, b

1, in

position 3, the following network can be used:

B1=b1if the output of the and-gate is 0 and B

1=b-1if the

output of the and-gate is 1.


42/42

3.35. (continued)

Logic Diagram:

Documents

Chapter 3 Problem Solutions -y 3.1. (a) Xy + Xy