- I actually figured this out last summer. Counting all mirrors and inverses I got

44 cases. It doesn't seem like a lot but there really aren't many edge

permutations for each corner orientation.

For all corners oriented (Fridrich PLL):

2 edge cycles

1 zig-zag

1 cross

1 lucky case

For 2 opposite corners twisted:

4 cases where 2 adjacent edges need to be exchanged

1 case where 2 oppostie egdes need to be exchanged (this also fixes zig-zag,

but you need to do an extra U after the sequence)

1 case with no edge permutation

For 2 adjacent corners twisted (x2 for 2 different cases)

4 cases where 2 adjacent edges need to be exchanged

1 case where 2 oppostie egdes need to be exchanged (this also fixes zig-zag,

but you need to do an extra U after the sequence)

1 case with no edge permutation

For 3 corners twisted (x2 for CW and CCW)

4 cases where 2 adjacent edges need to be exchanged

1 case where 2 oppostie egdes need to be exchanged (this also fixes zig-zag,

but you need to do an extra U after the sequence)

1 case with no edge permutation

For 4 corners twisted (all 4 facing same direction)

2 cases where 2 adjacent edges need to be exchanged

1 case where 2 oppostie egdes need to be exchanged (this also fixes zig-zag,

but you need to do an extra U after the sequence)

1 case with no edge permutation

the first one is half b/c the corner orientation is symmetrical both ways,

cancelling 2 of the cases.

For 4 corners twisted (the other case)

4 cases where 2 adjacent edges need to be exchanged

1 case where 2 oppostie egdes need to be exchanged (this also fixes zig-zag,

but you need to do an extra U after the sequence)

1 case with no edge permutation

Adding this up gives a total of 44 cases (minus the lucky case). It turns out that

these can all be solved using ONLY R, U, and L. So it's important to know

right and left handed finger tricks to take advantage of this.

The move average (based on the sequences I chose to use (Ron's solver

found more than one solution in most cases)) is about 11.1 moves.

Some of the sequences are same as the solutions to some OLL (Fridrich)

cases that affect edge permutation, so you should try some of those out first to

see if you can solve some cases without having to use Ron's solver.

I don't have these 44 cases online but it shouldn't take that long to find the

solutions. Allow Ron's cube solver to use only R, R', R2, L, L', L2, U, U', and

U2.

Andy

http://andyscubepage.tk - Hello Andy and all others,

I hope you won't mind me adding some comments.

It turns out that> these can all be solved using ONLY R, U, and L.

Actually they can be solved using ONLY R and U (two-generator), and

I think it is much easier to do it that way. I myself use only two-

gen algs with few exceptions.

You might know this, but the two-generator conservs the corners'

permutation so that all algorithms using R and U can be used or the

purpose of orientating corners and permuting edges. If some of you

don't believe it, or don't know why it is so, check my demonstration

on my website

http://www.geocities.com/portoseb/cube/demonstration.html

I think that this demonstration is the only one existing except for

the computer-generated ones that analysed all possible positions.

They should start giving cube problems in math contests ;)

Little more about these algs. Knowing the shortest 2-gen alg - the

Sune (7 moves) - one can generate algs consisting of two Sunes, done

one after the other in different positions of course. You can use

the AntiSune too, getting very many positions that can be solved in

a reasonable 14 moves. So actually you don't need to learn so many

algs (only if you're striving for optimality).

I think that all these algorithms are listed on Lars Petrus' page

http://lar5.com/cube/xMain.html

There are almost optimal ones too. The truly optimized ones are on

Bernard Helmstetter's page

http://www.ai.univ-paris8.fr/~bh/cube/

Sebastian - --- portoseb wrote:
> http://www.geocities.com/portoseb/cube/demonstration.html

You have shown that you can't swap two corners, but it is actually

> I think that this demonstration is the only one existing except for

> the computer-generated ones that analysed all possible positions.

much more restrictive. You cannot do any 3-cycles either. If you

could, then a conjugation of it would give you a 3-cycle in one face.

Combined with a turn of that face you would get a swap, which you

have shown is impossible.

I have done something very similar once, but I've never put it up on

my site. There is a proof in Singmaster's Notes on Rubik's Magic Cube

(5th ed.), which essentially boils down to much the same thing if I'm

not mistaken.

He labelled the positions (not the pieces) a-f. The five possible

pairing arrangements are then:

{ab,cd,ef}

{ac,de,bf}

{bd,af,ce}

{ad,be,cf}

{bc,ae,df}

The moves U and R permute these five arrangements. He then shows any

two different achievable corner permutations gives a different

permutation of these five sets, so there can be at most 5! corner

permutations. Therefore there are have 5! corner perms instead of the

6! you would normally expect with 6 corners, i.e. a factor 6 less.

It is easy to solve 3 corners in one face, and the other three will

then automatically be solved. Instead of the 3!=6 arrangements that

the 3 remaining corners normally could have, only the solved

arrangement is possible.

Jaap > He labelled the positions (not the pieces) a-f. The five possible

any

> pairing arrangements are then:

> {ab,cd,ef}

> {ac,de,bf}

> {bd,af,ce}

> {ad,be,cf}

> {bc,ae,df}

> The moves U and R permute these five arrangements. He then shows

> two different achievable corner permutations gives a different

the

> permutation of these five sets, so there can be at most 5! corner

> permutations. Therefore there are have 5! corner perms instead of

> 6! you would normally expect with 6 corners, i.e. a factor 6 less.

will

>

> It is easy to solve 3 corners in one face, and the other three

> then automatically be solved. Instead of the 3!=6 arrangements

that

> the 3 remaining corners normally could have, only the solved

This was the main idea of my demonstration too. But I just labeled

> arrangement is possible.

the pairs and watched how they moved around with R and U moves.

There was a very small number of possible positions, and a swap

would have meant that a forbidden arrangement would be possible. So

it is not possible to do a swap.

I used swaps rather than 3-cycles because it's much easier to see

them, and I use that in speedcubing.

Sebastian