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

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    you have to write a guide for someone who wants to solve a rubik's cube. they own only a 3*3*3 cube.
    the whole guide must be done only with writing and diagrams. no video/animation.

    what's the simplest way? how would you make it as easy as possible for them to solve it? you want to get them solving it without needing a guide to look at, as quickly as possible.
    what notation/method of description? and which method of solving the cube?
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    There is a standard notation for cube operations - you'll find a description just about anywhere you search, for instance, here on Wikipedia. Diagrams may be helpful here.

    If the goal is to memorise a solution as quickly as possible, then I would suggest learning a strategy with the fewest algorithms. (On the other hand, if the goal is to solve a given cube as quickly as possible, then it would be necessary to learn lots of special-case algorithms.)
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    doesnt it depend on the cube when you start i.e. what colours are where etc... ?

    and maybe theres a book somewhere?
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    Use your head, is what I do.
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    Commutators>Algorithms

    Fact.
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    If by commutators you mean things like what are listed on this page, I don't see what the difference is. It's just another set of algorithms.
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    Depends whether the guy finds it easier memorizing strings of foolproof moves or working it out and memorizing relatively few moves.

    Petrus method is pretty good in regard to the latter.
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    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/
    • PS Helper
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    PS Helper
    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

    :cool:
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    (Original post by happyguy123)
    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/
    Petrus method only uses 2 algorithms, but does require quite a lot of just getting used to a cube.
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    i use a combination:
    fridrich f2l (intuitive) + petrus cross orientation and last slot at the same time + 2-look OLL + 1 PLL
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    (Original post by nuodai)
    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

    :cool:
    :holmes:

    That is amazing - I am very very impressed. I think the number of your posts I need to rep will take me to the end of the year easily :tongue:
    • PS Helper
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    I've made the problem even more concise and readable

    Code:
    ?!(i!1$3)(P(1:1,i:i))[M(1,i:1,i)?!C(1,i)[S(1,|)/.|%.|.]][S(4,|)]?!(i!1$
    3)(j!1!i$3)(P(1:1,i:i,j:j))[M(1ij:4ij)/|/.|.?!P(1:1,i:i,j:j)[/|/.|.]]?!
    (i!1$3)(j!1$3!i$3)(P(i:i,j:j))[?P(i:j,j:i)[|/|./.|.%.|%]M(ij:4j)?S(i,/)
    [|/|./.|.%.|%]?S(i,\)[|.\.|\|%|.%.]?P(i:j,j:i)[|/|./.|.%.|%]]?!(P(4:4,:
    ))[?(i=j$3!1$3)(P(4:4,i:)%P(4:4,j:))[M(4i:4i)S(i,\)%/|/.|.%.]?(i!j$3!1$
    3)(P(4:4,i:)%P(4:4,j:))[M(4i:4i)S(i,\)S(j,@)%/|/.|.%.]?!(i!1$3)(P(4:4,i
    :))[%/|/.|.%.]]?!(i!1$3)(P(4:4,i:i)):?!(i!1$3)(j!1$3)(P(i:i,j:j))[|]?(i
    =j$3)[S(i,%)/|/.|/||/.]?(i!j$3)[S(i,/)S(j,@)/|/.|/||/.|]]?!(i!1$3)(j!1$
    3!i$3)(M(4ij:4ij))[?(P(4:!4,i:!i,j:!j))[|/|.\|/.|.\.]S(i,%)S(j,/)|/|.\|
    /.|.\.]S(1,|)?!(i!1$3)(j!1$3!i$3)(P(4:4,i:i,j:j))[S(4ij,_/%)?!(P(4:4,i:
    i,j:j))[/|/.|.]_]?!(i)(j!i$3)(k!i$3!j$3)(P(i:i,j:j,k:k)&P(i:i,j:j))[__|]
    That just about does it. Anyone fancy decoding it?

    Spoiler:
    Show
    Presenting it like this is easier on the eye:
    Code:
    ?!(i!1$3)(P(1:1,i:i))[
      M(1,i:1,i)
      ?!C(1,i)[S(1,|)/.|%.|.]
    ]
    [S(4,|)]
    ?!(i!1$3)(j!1!i$3)(P(1:1,i:i,j:j))[
      M(1ij:4ij)/|/.|.
      ?!P(1:1,i:i,j:j)[/|/.|.]
    ]
    ?!(i!1$3)(j!1$3!i$3)(P(i:i,j:j))[
      ?P(i:j,j:i)[|/|./.|.%.|%]
      M(ij:4j)
      ?S(i,/)[|/|./.|.%.|%]
      ?S(i,\)[|.\.|\|%|.%.]
      ?P(i:j,j:i)[|/|./.|.%.|%]
    ]
    ?!(P(4:4,:))[
      ?(i=j$3!1$3)(P(4:4,i:)%P(4:4,j:))[M(4i:4i)S(i,\)%/|/.|.%.]
      ?(i!j$3!1$3)(P(4:4,i:)%P(4:4,j:))[M(4i:4i)S(i,\)S(j,@)%/|/.|.%.]
      ?!(i!1$3)(P(4:4,i:))[%/|/.|.%.]
    ]
    ?!(i!1$3)(P(4:4,i:i)):
      ?!(i!1$3)(j!1$3)(P(i:i,j:j))[|]
      ?(i=j$3)[S(i,%)/|/.|/||/.]
      ?(i!j$3)[S(i,/)S(j,@)/|/.|/||/.|]
    ]
    ?!(i!1$3)(j!1$3!i$3)(M(4ij:4ij))[
      ?(P(4:!4,i:!i,j:!j))[|/|.\|/.|.\.]
      S(i,%)S(j,/)|/|.\|/.|.\.
    ]
    S(1,|)
    ?!(i!1$3)(j!1$3!i$3)(P(4:4,i:i,j:j))[
      S(4ij,_/%)
      ?!(P(4:4,i:i,j:j))[/|/.|.]
      _
    ]
    ?!(i)(j!i$3)(k!i$3!j$3)(P(i:i,j:j,k:k)&P(i:i,j:j))[
      __|
    ]
    Spoiler:
    Show
    Code:
    ? - while the following is true, do the following instructions
    ! - not/not equal to/not true
    = - equal to
    $ - modulo
    ~ - or
    & - and
    P(a:p,b:q,c:r) - move and orient (a,b,c) to position (p,q,r)
    M(abc,pqr) - move (a,b,c) to (p,q,r) without necessarily orienting it
    S(abc,pqr) - rotate cube such that the side of colour a is in direction p, and so on
    [...] - do ...
    / /. - right clockwise, right anticlockwise
    \ \. - left
    | |. - top
    _ _. - bottom
    % %. - front
    @ @. - back
    Offline

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    (Original post by nuodai)
    I've made the problem even more concise and readable

    Code:
    ?!(i!1$3)(P(1:1,i:i))[M(1,i:1,i)?!C(1,i)[S(1,|)/.|%.|.]][S(4,|)]?!(i!1$
    3)(j!1!i$3)(P(1:1,i:i,j:j))[M(1ij:4ij)/|/.|.?!P(1:1,i:i,j:j)[/|/.|.]]?!
    (i!1$3)(j!1$3!i$3)(P(i:i,j:j))[?P(i:j,j:i)[|/|./.|.%.|%]M(ij:4j)?S(i,/)
    [|/|./.|.%.|%]?S(i,\)[|.\.|\|%|.%.]?P(i:j,j:i)[|/|./.|.%.|%]]?!(P(4:4,:
    ))[?(i=j$3!1$3)(P(4:4,i:)%P(4:4,j:))[M(4i:4i)S(i,\)%/|/.|.%.]?(i!j$3!1$
    3)(P(4:4,i:)%P(4:4,j:))[M(4i:4i)S(i,\)S(j,@)%/|/.|.%.]?!(i!1$3)(P(4:4,i
    :))[%/|/.|.%.]]?!(i!1$3)(P(4:4,i:i)):?!(i!1$3)(j!1$3)(P(i:i,j:j))[|]?(i
    =j$3)[S(i,%)/|/.|/||/.]?(i!j$3)[S(i,/)S(j,@)/|/.|/||/.|]]?!(i!1$3)(j!1$
    3!i$3)(M(4ij:4ij))[?(P(4:!4,i:!i,j:!j))[|/|.\|/.|.\.]S(i,%)S(j,/)|/|.\|
    /.|.\.]S(1,|)?!(i!1$3)(j!1$3!i$3)(P(4:4,i:i,j:j))[S(4ij,_/%)?!(P(4:4,i:
    i,j:j))[/|/.|.]_]?!(i)(j!i$3)(k!i$3!j$3)(P(i:i,j:j,k:k)&P(i:i,j:j))[__|]
    That just about does it. Anyone fancy decoding it?

    Spoiler:
    Show
    Presenting it like this is easier on the eye:
    Code:
    ?!(i!1$3)(P(1:1,i:i))[
      M(1,i:1,i)
      ?!C(1,i)[S(1,|)/.|%.|.]
    ]
    [S(4,|)]
    ?!(i!1$3)(j!1!i$3)(P(1:1,i:i,j:j))[
      M(1ij:4ij)/|/.|.
      ?!P(1:1,i:i,j:j)[/|/.|.]
    ]
    ?!(i!1$3)(j!1$3!i$3)(P(i:i,j:j))[
      ?P(i:j,j:i)[|/|./.|.%.|%]
      M(ij:4j)
      ?S(i,/)[|/|./.|.%.|%]
      ?S(i,\)[|.\.|\|%|.%.]
      ?P(i:j,j:i)[|/|./.|.%.|%]
    ]
    ?!(P(4:4,:))[
      ?(i=j$3!1$3)(P(4:4,i:)%P(4:4,j:))[M(4i:4i)S(i,\)%/|/.|.%.]
      ?(i!j$3!1$3)(P(4:4,i:)%P(4:4,j:))[M(4i:4i)S(i,\)S(j,@)%/|/.|.%.]
      ?!(i!1$3)(P(4:4,i:))[%/|/.|.%.]
    ]
    ?!(i!1$3)(P(4:4,i:i)):
      ?!(i!1$3)(j!1$3)(P(i:i,j:j))[|]
      ?(i=j$3)[S(i,%)/|/.|/||/.]
      ?(i!j$3)[S(i,/)S(j,@)/|/.|/||/.|]
    ]
    ?!(i!1$3)(j!1$3!i$3)(M(4ij:4ij))[
      ?(P(4:!4,i:!i,j:!j))[|/|.\|/.|.\.]
      S(i,%)S(j,/)|/|.\|/.|.\.
    ]
    S(1,|)
    ?!(i!1$3)(j!1$3!i$3)(P(4:4,i:i,j:j))[
      S(4ij,_/%)
      ?!(P(4:4,i:i,j:j))[/|/.|.]
      _
    ]
    ?!(i)(j!i$3)(k!i$3!j$3)(P(i:i,j:j,k:k)&P(i:i,j:j))[
      __|
    ]
    Spoiler:
    Show
    Code:
    ? - while the following is true, do the following instructions
    ! - not/not equal to/not true
    = - equal to
    $ - modulo
    ~ - or
    & - and
    P(a:p,b:q,c:r) - move and orient (a,b,c) to position (p,q,r)
    M(abc,pqr) - move (a,b,c) to (p,q,r) without necessarily orienting it
    S(abc,pqr) - rotate cube such that the side of colour a is in direction p, and so on
    [...] - do ...
    / /. - right clockwise, right anticlockwise
    \ \. - left
    | |. - top
    _ _. - bottom
    % %. - front
    @ @. - back
    I preferred how it was before .
 
 
 
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    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

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