"Correct Answer - Option 3 : g2 (h3 h2 h1 h
h3 h2 h1 h0 be the gray code representation of a number n and g1g2g3g0 be the gray code representation of the number (n + 1) modulo 16.
The truth table is as shown below.
Decimal
|
Binary
|
h3
|
h2
|
h1
|
h0
|
g3
|
g2
|
g1
|
g0
|
0
|
0000
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
0001
|
0
|
0
|
0
|
1
|
0
|
0
|
1
|
1
|
2
|
0010
|
0
|
0
|
1
|
1
|
0
|
0
|
1
|
0
|
3
|
0011
|
0
|
0
|
1
|
0
|
0
|
1
|
1
|
0
|
4
|
0100
|
0
|
1
|
1
|
0
|
0
|
1
|
1
|
1
|
5
|
0101
|
0
|
1
|
1
|
1
|
0
|
1
|
0
|
1
|
6
|
0110
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
0
|
7
|
0111
|
0
|
1
|
0
|
0
|
1
|
1
|
0
|
0
|
8
|
1000
|
1
|
1
|
0
|
0
|
1
|
1
|
0
|
1
|
9
|
1001
|
1
|
1
|
0
|
1
|
1
|
1
|
1
|
1
|
10
|
1010
|
1
|
1
|
1
|
1
|
1
|
1
|
1
|
0
|
11
|
1011
|
1
|
1
|
1
|
0
|
1
|
0
|
1
|
0
|
12
|
1100
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
1
|
13
|
1101
|
1
|
0
|
1
|
1
|
1
|
0
|
0
|
1
|
14
|
1110
|
1
|
0
|
0
|
1
|
1
|
0
|
0
|
0
|
15
|
1111
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
From the above truth table,
g3(h3 h2 h1 h0) = Σ(4, 9, 10, 11, 12, 13, 14, 15)
g2(h3 h2 h1 h0) = Σ(2, 4, 5, 6, 7, 12, 13, 15)
g1(h3 h2 h1 h0) = Σ(1, 2, 3, 6, 10, 13, 14, 15)
g0(h3 h2 h1 h0) = Σ(0, 1, 6, 7, 10, 11, 12, 13)
"