Well, I'm not really fast (average around 1 minute) but I've developed a really basic way of doing the Rubik's cube that requires only four algorithms*!
*Assuming you know how to get the first cross, with two of the algorithms being F'RFR' and FRUR'U'F'. If you can do Petrus up to the last layer, then you only need three algorithms (this time without FRUR'U'F'), each of the remaining algorithms being only four turns (with some extra logic).
Actually we can easily prove that we need at least 4 algorithms (not steps.. I don't know the proper English term for this) to solve the rubik's cube: that is,
Permutation of Edges, Permutation of Corners, Orientation of Edges, Orientation of Corners. and this also leads to 4 step solving. Think of the blind solving.
P.S. Fridrich Method is also 4 steps.
Ah, yes, I didn't mean that my method was only 4 steps. In fact, the method I developed is nearly identical (in both the number of steps and the goal of each step) as the fundamental method (which I think was 7 steps). The main interest I have in the method I developed is the re-usability of "algorithms" (I use algorithm to mean any series of turns) across multiple steps leads to being able to solve any cube with only 4 algorithms memorized. To go into more detail:
Step 1: Permutation and orientation of first-layer edges
Step 2: Permutation and orientation of first-layer corners (Alg. 1 for each corner)
Step 3: Permutation and orientation of middle-layer edges (Alg. 2 for each edge)
Step 4: Orientation of last-layer edges (FRUR'U'F')
Step 5: Permutation of last-layer edges (Alg. 2 x4)
Step 6: Permutation of last-layer corners (Alg. 3 x4)
Step 7: Orientation of last-layer corners (Alg. 1 x6 for each corner)
Another thing is that the last three steps create isolated cycles, which means that you can do steps 5, 6, and 7 in any order you like.
If you'd like, I could go into more detail, but it isn't really that amazing or practical--only interesting.
