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Consider numbers represented in 4-bit Gray code. Let $ h_{3}h_{2}h_{1}h_{0}$  be the Gray code  representation of a number $n$ and let $ g_{3}g_{2}g_{1}g_{0}$ be the Gray code of $ (n+1)(modulo 16)$  value of the number. Which one of the following functions is correct?

  1. $ g_{0}(h_{3}h_{2}h_{1}h_{0})=\sum (1,2,3,6,10,13,14,15) $
  2. $ g_{1}(h_{3}h_{2}h_{1}h_{0})=\sum (4, 9,10,11,12,13,14,15) $
  3. $ g_{2}(h_{3}h_{2}h_{1}h_{0})=\sum (2, 4,5,6,7,12,13,15) $
  4. $ g_{3}(h_{3}h_{2}h_{1}h_{0})=\sum (0,1,6,7,10,11,12,13) $

7 Answers

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The question contain two functions H and G

H(n)-→ it gives the gray code value of n

G(n) → it gives the gray code value of (n+1)mod 16.so its output is between gray code values of 0-16.

so eg: n=5

H(5)=H(0101)=0111

G(5)=G(5+1 mod 16)=H(6)=H(0110)=0101

so G(n)=H(n+1)       where 0<= n <= 15

 

so if you check g3 g2 g1 g0  for which inputs they are 1 then you get the answer which is C

Answer:

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