quickest way to get beginner solving rubik's?(all instruction written+using only 3*3)

thanks for replies so far. i'm going to take a look at this "petrus method", and the page on "commutators" right after posting this reply. just in case they don't settle it, a few questions:

is "standard notation" the best notation/method of cube description you can imagine?

what would your "simplest guide" consist of? how would you keep the number of algorithms to a minimum, and make them really easy to use/learn/understand? how woudl you reduce the brain effort/time to learn to do the cube by yourself, to a minimum

can you think of ways to improve on this guide?:
http://www.learn2cube.com/

Let $C_i$ be a colour, where $i = 1, 2, ..., 6$, and where the $C_i$-coloured centre is opposite the $C_j$-coloured centre, where $j = i \pmod 3, i \ne j$. So for example $C_2$ is opposite $C_5$. Let $(C_p)$ be the centre piece with colour $C_p$, $(C_p, C_q)$ be a two-coloured piece for some $p \ne q \pmod 3$, and $(C_p, C_q, C_r)$ be a three-coloured piece (a corner piece) for some $p \ne q \ne r \pmod 3$.

Let $S_p$ be the side with centre $(C_p)$, let $(S_p, S_q)$ be the edge piece between sides $S_p$ and $S_q$, and let $(S_p, S_q, S_r)$ be the corner where the sides $S_p, S_q, S_r$ intersect. Then, if a cube is solved, each piece $(C_i, C_j)$ is in the position $(S_i, S_j)$ and each piece $(C_p, C_q, C_r)$ is in the position $(S_p, S_q, S_r)$.

Let also $[M_1, M_2, ..., M_i]$ be a sequence of rotations for some $M_i \in \{R, R*, L, L*, T, T*, U, U*, F, F*, B, B*\}$, where R, L, T, U, F and B represent the right, left, top, bottom, front and back sides respectively; and where a lack of asterisk resembles a clockwise rotation if that side were facing you, and an asterisk represents an anticlockwise rotation. Each instruction $M_i$ must be carried out in order, and the instruction sequence should be repeated until the desired outcome is reached.

1. Rotate $(C_1,C_n)$ piece to $(S_1, S_n)$ for all $n \ne 1 \pmod 3$. If a piece rotates to $S_1$ such that its $C_n$-coloured side is in contact with $C_1$, for some $n \ne 1$, then rotate the whole cube such that $(C_1, C_n)$ is on your right at the top of the cube, and then do $[R*,T,F*,T*]$.

Result: A cross of colour $C_1$ on the side $S_1$

2. Rotate the cube such that $S_1$ is on the bottom, and move the piece $(C_1, C_a, C_b)$ to $(S_4, S_a, S_b)$, then do $[R,T,R*,T*]$ until it is in the position $(S_1, S_a, S_b)$ and the colours are coordinated correctly. If no pieces $(C_1, C_a, C_b)$ are on the $S_4$ face, then $[R,T,R*,T*]$ will get them up there.

Result: Side $S_1$ completed with the first layer of sides $S_2, S_3, S_5, S_6$ completed (resembling an upside-down T-shape on them).

3. Move the piece $(C_p, C_q)$ for some $p, q \not \in \{1, 4\}$ such that the side $C_p$ facing you is in contact with $(C_p)$. If $S_q$ is on the right, do $[T,R,T*,R*,T*,F*,T,F]$. If it is on the left, do $[T*,L*,T,L,T,F,T*,F*]$. If a piece $(C_p, C_q)$ is in position $(S_p, S_q)$ but with the colours the wrong way round, then either of the above algorithms gets it out (depending on whether it is in the left or right.

Result: First two layers of $S_2, S_3, S_5, S_6$ complete.

4. If there is a $C_4$-coloured cross on the side $S_4$, stop. If there is a line of colour $C_4$ on the side $S_4$, then rotate the cube so that the line is on top and going from left-to-right, and do $[F,R,T,R*,T*,F*]$. If there is an L-shape, rotate it so that $S_4$ is facing up with the L-shape in the far-left corner, and then do $[F,R,T,R*,T*,F*]$ and follow the previous instructions. If there is only a dot, do $[F,R,T,R*,T*,F*]$ and then follow the previous instructions.

Result: $C_4$-coloured cross on the side $S_4$

5. Do $[T]$ until there are two pieces $(C_4, C_i), (C_4, C_j)$ such that the $C_i$ and $C_j$-coloured sides are in contact with $(C_i)$ and $(C_j)$, respectively. If the pieces are adjacent, i.e. if $i \ne j \pmod 3$, then rotate the cube such that $S_4$ is facing up and the two aforementioned pieces are to your right and at the far-side, respectively; then do $[R,T,R*,T,R,T,T,R*,T]$. If these pieces are opposite, i,e. if $i = j \pmod 3$, then rotate the cube so that $S_4$ is facing up and one of the aforementioned pieces is facing you, and then do $[R,T,R*,T,R,T,T,R*]$ and follow the previous instructions

Result: Correctly-oriented cross of colour $C_4$ on side $S_4$

6. Let $N$ be the number of pieces $(C_4, C_p, C_q)$ that are in the position $(S_4, S_p, S_q)$. If $N = 4$, then stop. If $0 < N < 4$, then rotate the cube such that $S_4$ is facing up and such that one such piece is to your near-right. Then do $[T,R,T*,L,T,R*,T*,L*]$ until $N = 4$. If $N = 0$, then do $[T,R,T*,L,T,R*,T*,L*]$ and repeat the above instructions.

Result: All corners in the cube in the correct place, but not necessarily correctly-oriented

7. Turn the cube over such that the side $S_1$ is facing down and such that an incorrectly-oriented pieces are to your near-right. Do $[R,T,R*,T*]$ until the side of the piece of colour $C_4$ is on the side $S_4$. Then, do $[ B ]$ until another incorrectly-oriented corner is to your near-right, and repeat until the side $S_4$ is complete.

Result: All corners correctly-oriented

8. Do $[ B ]$ and/or $[ T ]$ until the cube is solved.

Result: Solved cube

I've made the problem even more concise and readable

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