Boolean algebra and Logic Simplification
Key point
The first two problems at S. Nos. 1 and 2 are on the Number of Boolean expressions for
a given number of variables.
The number of Boolean expressions for n variables is
Note that for n variable Boolean function one can have 2n Boolean inputs.
1. The number of Boolean function that can be generated by n variables is equal to:
(c)
(a)
(d)
(b)
[GATE 1990 : 1 Mark]
Ans. (b)
The number of district Boolean expressions for n variables is
Where, n is the number of variables.
2. The number of distinct Boolean expressions of 4 variables is
(a) 16
(c) 1024
(b) 256
(d) 65536
[GATE 2003 : 1 Mark]
Ans. (d)
We know that the number of Boolean expressions for n variables is
So, for
3. Two 2’s complement number having sign bits x and y are added and the sign bit of
the result is z. Then, the occurrence of overflow is indicated by the Boolean function.
(a)
(c) ̅ ̅
̅
(b) ̅ ̅ ̅
(d)
[GATE 1998 : 1 Mark]
Ans. (c)
This problem is based on 2’s complement addition/subtraction (covered in
number systems). Here we have to write the Boolean expression for the case
where overflow occurs.
In the 2’s complement arithmetic we had seen that if MSB (Sign bit) of operands
(Minuend & Subtrahend) is same, and the MSB of the result is different or vice
versa an overflow occurs.
i.e.
or, if
Overflow occurs
So, we can write
̅̅
̅ is the Boolean function for the overflow to occur.
Key points:
The problems in Q Nos. 3, 4, 5, 6, 7 and 8 are on the simplification of Boolean
expressions based on algebraic methods like using laws and theorems of Boolean
algebra. But this method becomes complex with increase in number of variables and
number of terms.
̅
4. The logical expression
(a)
̅
(b)
is equivalent to
(c)
(d)
̅
[GATE 1999 : 1 Mark]
Ans. (d)
̅
̅ )(
(
(
)
)
(
)
̅
5. The minimized form of the logical expression ( ̅ ̅ ̅
̅
̅
(a) ̅ ̅
(c) ̅
̅
̅
(b) ̅ ̅ ̅
(d)
̅
̅
̅
̅
̅ ) is
̅
̅
[GATE 1999 : 2 Marks]
Ans. (a)
̅ ̅̅
̅ ̅
̅
̅ ̅
Combining first two terms
̅ ̅( ̅
̅
)
̅̅
̅
̅
̅
̅ (̅
)
̅ (̅
)(̅
)
̅ (̅
)
̅
̅̅
̅
̅
̅
̅
̅̅
(̅
̅)
̅̅
(̅
)( ̅
̅̅
(̅
̅̅
̅
̅)
̅)
̅
6. The Boolean expression
̅
(a) ̅
̅
̅
(b) ̅
̅ is equivalent to
(c)
(d)
̅
̅
̅
̅
̅
̅
̅
[GATE 2004 : 2 Marks]
Ans. (d)
̅
̅)
(
̅(
̅)
̅
̅
̅ ̅
̅ ̅
̅
̅
7. The Boolean expression for the truth table shown is
A
0
0
0
0
1
1
1
1
(a)
(b)
B
0
0
1
1
0
0
1
1
(
(
C
0
1
0
1
0
1
0
1
F
0
0
0
1
0
0
1
0
)( ̅
̅ )( ̅
̅)
)
(c) ̅ (
(d) ̅ (
̅ )( ̅
)( ̅
)
̅)
[GATE 2005 : 2 Marks]
Ans. (a)
We observe that the expression needed in the result is of POS form but in truth
table the output is 0 for many terms but 1 only for 2 terms so we write the SOP
equation. Then convert it to POS
̅
̅
(̅
̅)
)( ̅
(
(
)
̅)
Since
̅̅̅̅̅̅̅
̅̅
̅̅
So, ̅̅̅̅̅̅ ̅̅̅̅̅̅̅̅̅̅̅̅
( ̅ ̅ )(
)
(
)
8. If X = 1 in the logic equation
̅ )+- * ̅
,
*̅ ( ̅
(a)
̅
(b)
̅(
)+
then
(c)
(d)
[GATE 2009 : 2 Marks]
Ans. (d)
Given logic equation is
*̅
,
(̅
̅ )+- * ̅
̅(
, so the first term is *
2nd term is *
̅(
)+
)+
+
̅
̅
Key Points:




Prime implicants :A product term in SOP which cannot be further simplified by
combination with other terms.
SOP (Sum of product) and POS (producer Sum) expressions.
K-Map has 2k cells where k is no. of variables.
Don’t care combinations: Represented as‘d’. d’s can be combined with 1’s to
simplify, where needed.
9. The K-map for a Boolean function is shown in figure. The number of essential prime
implicants for this function is
AB
00
01
11
10
00
1
1
0
1
01
0
0
0
1
11
1
0
0
0
10
1
0
0
1
CD
(a) 4
(b) 5
(c) 6
(d) 8
[GATE 1998 : 1 Mark]
Ans. (a)
AB
CD
00
00
01
1
1
11
01
10
1
1
11
1
10
1
1
On the K-Map retaining only minimum terms (corresponding to 1)
The number of prime implacants are
̅̅ ̅
̅̅
̅̅
̅̅
Four prime implicants
10. The number of products terms in the minimized sum-of-product expression obtained
through the following K-map is (where, “d” denotes don’t care states)
1
0
0
1
0
d
0
0
0
0
d
1
1
0
0
1
(a) 2
(b) 3
(c) 4
(d) 5
[GATE 2006 : 2 Marks]
Ans. (a)
AB
CD
00 01 11 10
00
1
0
0
1
01
0
d
0
0
11
0
0
d
1
10
1
0
0
1
To write the minimized expression (d terms do not simplify the expression)
̅̅
̅
So there are only two product terms
11. The Boolean expression
̅̅ ̅
̅ ̅
(a)
̅̅ ̅
̅
(b)
̅̅ ̅
̅
̅ ̅
̅
̅
̅ ̅
(c)
(d)
̅
̅
̅ ̅ can be minimized to
̅ ̅ ̅
̅ ̅
̅ ̅ ̅
̅̅
[GATE 2007 : 2 Marks]
Ans. (d)
CD
AB
00
01
10
11
1
00
01
1
11
1
10
1
The Boolean expression
̅ ̅̅
̅
̅
̅̅
̅̅
Min terms can be plotted on K-Map
The minimized expression is
̅
̅
̅̅
̅̅
Alternative method
̅ ̅̅
̅
̅
̅̅
̅̅
Combining first and third terms
(̅
̅
)̅̅
̅
̅
̅̅
̅
̅̅
̅̅
12. In the sum of products function (
̅
(a) ̅
̅ ̅ ̅
(b) ̅
)
∑(
), the prime implicates are
̅
(c) ̅ ̅ ̅
̅ ̅ ̅
(d) ̅ ̅ ̅
[GATE 2012 : 2 Marks]
Ans. (a)
X
YZ
00
01
11
10
1
1
0
1
(
1
1
)
∑(
̅
)
̅
So prime implicants are
̅ and ̅