Chapter 4
4.1. SOP form:
f
=
x
1
x
2
+
x
2
x
3
POS form:
f
= (
x
1
+
x
2
)(
x
2
+
x
3
)
4.2. SOP form:
f
=
x
1
x
2
+
x
1
x
3
+
x
2
x
3
POS form:
f
= (
x
1
+
x
3
)(
x
1
+
x
2
)(
x
2
+
x
3
)
4.3. SOP form:
f
=
x
1
x
2
x
3
x
4
+
x
1
x
2
x
3
x
4
+
x
2
x
3
x
4
POS form:
f
= (
x
1
+
x
4
)(
x
2
+
x
3
)(
x
2
+
x
3
+
x
4
)(
x
2
+
x
4
)(
x
1
+
x
3
)
4.4. SOP form:
f
=
x
2
x
3
+
x
2
x
4
+
x
2
x
3
x
4
POS form:
f
= (
x
2
+
x
3
)(
x
2
+
x
3
+
x
4
)(
x
2
+
x
4
)
4.5. SOP form:
f
=
x
3
x
5
+
x
3
x
4
+
x
2
x
4
x
5
+
x
1
x
3
x
4
x
5
+
x
1
x
2
x
4
x
5
POS form:
f
= (
x
3
+
x
4
+
x
5
)(
x
3
+
x
4
+
x
5
)(
x
2
+
x
3
+
x
4
)(
x
1
+
x
3
+
x
4
+
x
5
)(
x
1
+
x
2
+
x
4
+
x
5
)
4.6. SOP form:
f
=
x
2
x
3
+
x
1
x
5
+
x
1
x
3
+
x
3
x
4
+
x
2
x
5
POS form:
f
= (
x
1
+
x
2
+
x
3
)(
x
1
+
x
2
+
x
4
)(
x
3
+
x
4
+
x
5
)
4.7. SOP form:
f
=
x
3
x
4
x
5
+
x
3
x
4
x
5
+
x
1
x
4
x
5
+
x
1
x
2
x
4
+
x
3
x
4
x
5
+
x
2
x
3
x
4
+
x
2
x
3
x
4
x
5
POS form:
f
= (
x
3
+
x
4
+
x
5
)(
x
3
+
x
4
+
x
5
)(
x
1
+
x
2
+
x
3
+
x
4
+
x
5
)
4.8.
f
=
∑
m
(0
,
7)
f
=
∑
m
(1
,
6)
f
=
∑
m
(2
,
5)
f
=
∑
m
(0
,
1
,
6)
f
=
∑
m
(0
,
2
,
5)
etc.
4.9.
f
=
x
1
x
2
x
3
+
x
1
x
2
x
4
+
x
1
x
3
x
4
+
x
2
x
3
x
4
4.10. SOP form:
f
=
x
1
x
2
x
3
+
x
1
x
2
x
4
+
x
1
x
3
x
4
+
x
1
x
2
x
3
+
x
1
x
3
x
4
+
x
2
x
3
x
4
POS form:
f
= (
x
1
+
x
2
+
x
3
)(
x
1
+
x
2
+
x
4
)(
x
1
+
x
3
+
x
4
)(
x
2
+
x
3
+
x
4
)(
x
1
+
x
2
+
x
3
+
x
4
)
The POS form has lower cost.
4.11. The statement is false. As a counter example consider
f
(
x
1
, x
2
, x
3
) =
∑
m
(0
,
5
,
7)
.
Then, the minimum-cost SOP form
f
=
x
1
x
3
+
x
1
x
2
x
3
is unique.
But, there are two minimum-cost POS forms:
f
= (
x
1
+
x
3
)(
x
1
+
x
3
)(
x
1
+
x
2
)
and
f
= (
x
1
+
x
3
)(
x
1
+
x
3
)(
x
2
+
x
3
)
4-1

4.12. If each circuit is implemented separately:
f
=
x
1
x
4
+
x
1
x
2
x
3
+
x
1
x
2
x
4
Cost
= 15
g
=
x
1
x
3
x
4
+
x
2
x
3
x
4
+
x
1
x
3
x
4
+
x
1
x
2
x
4
Cost
= 21
In a combined circuit:
f
=
x
2
x
3
x
4
+
x
1
x
3
x
4
+
x
1
x
2
x
3
x
4
+
x
1
x
2
x
3
g
=
x
2
x
3
x
4
+
x
1
x
3
x
4
+
x
1
x
2
x
3
x
4
+
x
1
x
2
x
4
The first 3 product terms are shared, hence the total cost is 31.
4.13. If each circuit is implemented separately:
f
=
x
1
x
2
x
4
+
x
2
x
4
x
5
+
x
3
x
4
x
5
+
x
1
x
2
x
4
x
5
Cost
= 22
g
=
x
3
x
5
+
x
4
x
5
+
x
1
x
2
x
4
+
x
1
x
2
x
4
+
x
2
x
4
x
5
Cost
= 24
In a combined circuit:
f
=
x
1
x
2
x
4
+
x
2
x
4
x
5
+
x
3
x
4
x
5
+
x
1
x
2
x
4
x
5
g
=
x
1
x
2
x
4
+
x
2
x
4
x
5
+
x
3
x
4
x
5
+
x
1
x
2
x
4
x
5
+
x
3
x
5
The first 4 product terms are shared, hence the total cost is 31. Note that in this implementation
f
⊆
g
, thus
g
can be realized as
g
=
f
+
x
3
x
5
, in which case the total cost is lowered to 28.
4.14.
f
= (
x
3
↑
g
)
↑
((
g
↑
g
)
↑
x
4
)
where
g
= (
x
1
↑
(
x
2
↑
x
2
))
↑
((
x
1
↑
x
1
)
↑
x
2
)
4.15.
f
= (((
x
3
↓
x
3
)
↓
g
)
↓
((
g
↓
g
)
↓
(
x
4
↓
x
4
))
, where
g
= ((
x
1
↓
x
1
)
↓
x
2
)
↓
(
x
1
↓
(
x
2
↓
x
2
))
. Then,
f
=
f
↓
f
.
4.16.
f
= (
g
↑
k
)
↑
((
g
↑
g
)
↑
(
k
↑
k
))
, where
g
= (
x
1
↑
x
1
)
↑
(
x
2
↑
x
2
)
↑
(
x
5
↑
x
5
)
and
k
= (
x
3
↑
(
x
4
↑
x
4
))
↑
((
x
3
↑
x
3
)
↑
x
4
)
4.17.
f
= (
g
↓
k
)
↓
((
g
↓
g
)
↓
(
k
↓
k
))
, where
g
=
x
1
↓
x
2
↓
x
5
and
k
= ((
x
3
↓
x
3
)
↓
x
4
)
↓
(
x
3
↓
(
x
4
↓
x
4
))
. Then,
f
=
f
↓
f
.
4.18.
f
=
x
1
(
x
2
+
x
3
)(
x
4
+
x
5
) +
x
1
(
x
2
+
x
3
)(
x
4
+
x
5
)
4.19.
f
=
x
1
x
3
x
4
+
x
2
x
3
x
4
+
x
1
x
3
x
4
+
x
2
x
3
x
4
= (
x
1
+
x
2
)
x
3
x
4
+ (
x
1
+
x
2
)
x
3
x
4
This requires 2 OR and 2 AND gates.

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