It is well known that all symmetric group can be generated using two generators

The two generators are:

1) $(1,2)$

2) $(1,2,3,\dots ,n)$

Question: Is there a deterministic algorithm to generate all permutations without repetition using only these two generators?

(Bonus 1: The algorithm generates the permutations in a cycle. Bonus 2: Not requiring the inverse of generator 2)

Edit: As point out by John, this is equivalent to a Hamiltonian path in the Cayley graph of $S_n$ with these two generators.

It is easy to generate all of them without repetition using $n-1$ generators, by the
Steinhaus-Johnson-Trotter algorithm.

It is easy to generate all of them, with repetition, using two generators.

However I was unable to find a way to generate all without repetition and using only two generators.

As this approach seems natural, I suspect someone should have worked on it but I was unable to find any references online.

Does anyone knows the status of this problem?

digraph(so the inverse of generator 2 is not used). This has been proved impossible for even $n$. For odd $n$ I did it by computer up to $n=11$ (that's a digraph of 39,916,800 vertices) but as far as I know the question remains open. – Brendan McKay Jun 17 '12 at 3:02