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I'm posting this as more of a theory/mathematic how-to followup to a stackoverflow question.

The non-iterative method for calculating graycode depends on Log2N bytes, to store position information for the next bit in the iteration sequence.
Specifically, the goal is to know the next bit to change without having to look at the current code.

However, for the 3 bit gray code, there's a iteration sequence 0,1,2,1,0,1,2,1 that can be represented with a much simpler function - maintain "0,1,2,1" in a register and rotate each time (as an example of a more general permutation).
To reduce the necessary state, this could be kept in two positions, starting 0,1 and a constant function of "xor, permute" applied: 0,3 => 1,0 => 2,1 => 3,2 => 0,3 (the bit to change being the first, and 3 handled as 1 since only 0,1,2 are valid)

Is it possible for graycodes to exist for higher values of N, such that the iteration function can be calculated with just a permutation operation?
If not, with permute/xor?
Is there a more general way to think of applying constant operations to define the state change?

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    @TonyK: I know, I like to think of them as (Hamiltonian) cycles in an n-cube. I was suggesting that perhaps frLich could improve the question. I did a little work on Gray codes, but it isn't computationally efficient and so not applicable to this question. Shame ;).2011-01-14

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The iteration function at stage $t$ is the highest power of $2$ that divides $t$ (this will result in a somewhat different code than yours, but still goes over all numbers, changing only one bit at a time). In general, since there are $n$ possible bits to flip, you will have to store at least $\log_2 n$ bits in memory (if you're not allowed to look at the actual counter).

Some processors have a machine instruction that calculates this "highest power of $2$ dividing a number", or perhaps something similar like "leading zero". This is an efficient way to implement Gray counting.

If you don't have such a machine instruction, another good choice is to unroll the increment loop for some small $2^K$, and use some smart iterative approach (using lookup tables) to handle the updates which are not hard-coded.

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    Your answer points to the confounding flood on Google. From the link, "calculating highest power of $2$" requires iteration on a mirror of the counter - whereas with this specific sequence, the answer is known in constant time without needing the counter. Thus the question of how to find similar sequences for larger N instead of using binary reflected gray code.2011-01-14
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Set $a_0=0$. To get $a_{n+1}$ from $a_n$, just change bit $r_n$, where $r_n$ is the number of trailing $1$-bits in the binary representation of $n$. So we get:

$(r_n) = \lbrace 0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,5,\dots \rbrace$

$(a_n) = \lbrace 0,1,3,2,6,7,5,4,12,13,15,14,10,11,9,8,24,25,27,26,30,31,29,28,20,\dots \rbrace$

This construction gives the standard Gray code.

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    Well then, I really don't know how to satis$f$y you.2011-01-17