the nth gray code

Question

the formula for calculating nth gray code is :

(n-1) XOR (floor((n-1)/2))  
(Source: wikipedia)

I encoded it as:

int gray(         


        

Answer 1

The Wikipedia entry you refer to explains the equation in a very circuitous manner.

However, it helps to start with this:

Therefore the coding is stable, in the sense that once a binary number appears in Gn it appears in the same position in all longer lists; so it makes sense to talk about the reflective Gray code value of a number: G(m) = the m-th reflecting Gray code, counting from 0.

In other words, Gn(m) & 2^n-1 is either Gn-1(m & 2^n-1) or ~Gn-1(m & 2^n-1). For example, G(3) & 1 is either G(1) or ~G(1). Now, we know that Gn(m) & 2^n-1 will be the reflected (bitwise inverse) if m is greater than 2^n-1.

In other words:

G(m, bits), k= 2^(bits - 1)
G(m, bits)= m>=k ? (k | ~G(m & (k - 1), bits - 1)) : G(m, bits - 1)
G(m, 1) = m

Working out the math in its entirety, you get (m ^ (m >> 1)) for the zero-based Gray code.