Non-recursive Grey code algorithm understanding

This is task from algorithms book.

The thing is that I completely don't know where to start!

Trace the following non-recursive algorithm to generate the binary reflexive
Gray code of order 4. Start with the n-bit string of all 0’s. 
For i = 1, 2, ... 2^n-1, generate the i-th bit string by flipping bit b in the 
previous bit string, where b is the position of the least significant 1 in the 
binary representation of i.

So I know the Gray code for 1 bit should be 0 1, for 2 00 01 11 10 etc.

Many questions

1) Do I know that for n = 1 I can start of with 0 1?

2) How should I understand "start with the n-bit string of all 0's"?

3) "Previous bit string"? Which string is the "previous"? Previous means from lower n-bit? (for instance for n=2, previous is the one from n=1)?

4) How do I even convert 1-bit strings to 2-bit strings if the only operation there is to flip?

This confuses me a lot. The only "human" method I understand so far is: take sets from lower n-bit, duplicate them, invert the 2nd set, add 0's to every element in 1st set, add 1's do every elements in 2nd set. Done (example: 0 1 -> 0 1 | 0 1 -> 0 1 | 1 0 -> 00 01 | 11 10 -> 11 01 11 10 done.

Thanks for any help

The answer to all four your questions is that this algorithm does not start with lower values of n. All strings it generates have the same length, and the i-th (for i = 1, ..., 2ⁿ-1) string is generated from the (i-1)-th one.

Here is the fist few steps for n = 4:

Start with G₀ = 0000

To generate G₁, flip 0-th bit in G₀, as 0 is the position of the least significant 1 in the binary representation of 1 = 0001_b. G₁ = 0001.

To generate G₂, flip 1-st bit in G₁, as 1 is the position of the least significant 1 in the binary representation of 2 = 0010_b. G₂ = 0011.

To generate G₃, flip 0-th bit in G₂, as 0 is the position of the least significant 1 in the binary representation of 3 = 0011_b. G₃ = 0010.

To generate G₄, flip 2-nd bit in G₃, as 2 is the position of the least significant 1 in the binary representation of 4 = 0100_b. G₄ = 0110.

To generate G₅, flip 0-th bit in G₄, as 0 is the position of the least significant 1 in the binary representation of 5 = 0101_b. G₅ = 0111.