## Re: corners orient and edge permutation

Expand Messages
• 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
Message 1 of 6 , Dec 7, 2004
• 0 Attachment
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 ... Actually they can be solved using ONLY R and U (two-generator),
Message 2 of 6 , Dec 8, 2004
• 0 Attachment
Hello Andy and all others,

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
• ... You have shown that you can t swap two corners, but it is actually much more restrictive. You cannot do any 3-cycles either. If you could, then a
Message 3 of 6 , Dec 9, 2004
• 0 Attachment
--- portoseb wrote:
> 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.

You have shown that you can't swap two corners, but it is actually
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}
{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
• ... any ... the ... will ... that ... This was the main idea of my demonstration too. But I just labeled the pairs and watched how they moved around with R and
Message 4 of 6 , Dec 10, 2004
• 0 Attachment
> 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}
> {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.

This was the main idea of my demonstration too. But I just labeled
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
Your message has been successfully submitted and would be delivered to recipients shortly.