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Problem: Determine the immediate successors of the following 9-tuple in the reflected Gray Code of order 9. $$111111111.$$

My Attempt: I am using the following algorithm to solve this problem:

Begin with the $n$ tuple $a_{n-1}a_{n-2}...a_0=00...0.$

While the $n-$ tuple $a_{n-1}a_{n-2}...a_0\neq 10...0,$ do the following:

  1. Computer $\sigma(a_{n-1}a_{n-2}...a_0)=a_{n-1}+a_{n-2}+...+a_0.$
  2. If $\sigma(a_{n-1}a_{n-2}...a_0)$ is even, change $a_0$ from $0$ to $1$ or $1$ to $0$.
  3. Else, determine $j$ such that $a_j=1$ and $a_i=0$ for all $i$ with $j>i$ and then change $a_{j+1}$ from $0$ to $1$ or $1$ to $0$.

We note that in step $(3)$ we may have $j=0$, that is, $a_0=1;$ in this case there is no $i$ with $i<j$, and we change $a_1$ as instructed in step $(3)$.

Using this information I computed $\sigma(111111111)=9$, which is odd. But since there is no $i$ with $i<j$, $a_1=0$ and thus the successor gray code is $111111101.$ I want to know whether this solution is correct or not.

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1 Answer 1

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Gray code word 111111111 corresponds to binary 101010101.

The binary successor is 101010110 which corresponds to your result Gray code word 111111101

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