Hey guys,
        It is now official: I can't make it to Nationals this year.  Though I
really want to, some financial issues are eating me alive this summer...  I
can't even get a loan to go, since I'm a full-time student (lame banks).  I'm
still cubing, but I simply don't have the transportation/funds.  I considered
hitchhiking/bus routes, but I also have classes for the summer at the
university, and it would take days for that type of transportation.
        I'm definately not walking out on the cubing thing.  "I'll be back."
   good luck everyone.  cuberz for life.
   -Brent Morgan

skeneegee <skeneegee@...> wrote:
           If anyone happens to be crossing the bay bridge towards Oakland at
night, check out what your cube looks like in the tunnel that goes
through Yerba Buena Island. Better yet, try to solve it(not if you're
driving of course).

--- In speedsolvingrubikscube@yahoogroups.com, d_funny007
<no_reply@...> wrote:
>
> Hi,
>
> Since I'm about to leave soon, I thought it would be a good idea to
> start a thread for people to exchange some last-minute info.
>
> Although most ppl are staying at the Coventry Motor Inn, me, Bob, and
> ChrisH worked out something over at the Adante Hotel.
>
> My cell num. is seven three four, 9 tree for, sx 3 sx too. (I am
> parinoid of internet bots/cralwers, lol.) Call me if you need a ride,
> I rented a car and plan to be there for 2 weeks. Say if you want to be
> picked up from either airport, I'll see what I can do, if you contact
> me a couple hours prior.
>
> Is there anything going on Thursday? Setup and going to see the
> Exploratorium would be good to do in the afternoon or something. I
> don't know if that would be a welcome idea though. I think that
> Thursday night there might be something at either one of the hotel
> lobbies or Clancy's place (contact me or him about that).
>
> I will be doing a lot of touristy stuff, and taking a lot of pics
> becasue I'm a photography freak. This time I brought my favorite
> camera and will be sure to take many pictures of you guys (some vid
> too).
>
> So far I got a good idea of who is and isn't comming. I do hear that
> Brent is having some difficulty with making arrangments, I fear.
> Perhaps if someone is driving from TX and can give him a lift...
>
> I know San Francisco pretty well guys, probably as well as any non-
> local that will be attending. Btw, the public transit system (BART) is
> one of the best in the country. It will take you from the airport to a
> few blocks from almost anything I can think of (most notably Berkeley
> U, Union Square, and the Wharf/Piers). Well except maybe the
> Exploratorium... I'm not sure how to get there using only BART and
> cable cars (busses perhaps?). I can help shuttle ppl form the Coventry
> to the Exploratorium though if that becomes a problem. I know that not
> all of you can afford a hefty series of cab fares.
>
> If you want to contact me, call me; I won't be reading this forum till
> after I come back. Good luck with getting things together, and have a
> safe trip.
>
>
> -Doug (almost packed!!)
>






:)
--Brent

---------------------------------
See the all-new, redesigned Yahoo.com.  Check it out.

[Non-text portions of this message have been removed]

No, in some cases (I think yours would apply), you should look
for "diagonal mirroring". Although the simple mirroring plus U
rotations *might* be enough/analgous/equivalent, but I have put
little thought into this as I am currently on vacation!

As a long time member of this fourm, I'd like to say that it is very
good to see another hardcore math/cs person like Bruce here! I've
been keeping up with his posts on this other fourm he uses too. Very
techincal stuff that I once wanted to see here, but after further
thought, it just wouldn't fit here. I was always the one rushing to
answer math questions, but I wasn't particularly patient in the past :
(.

I can try a verification of his computation when I get the chance. It
is most challenging :).


-Doug


--- In speedsolvingrubikscube@yahoogroups.com, "athefre"
<athefre@...> wrote:
>
> Thanks, 111 is better than 140, but not much.
>
> If you could reduce the number using mirrors and inverses, how much
> would it be?  If you don't mind.  I've been working hard for a
month
> trying to perfect everything so I can get to work on finding the
> algorithms for the idea I choose.
>
> Inverse = backwards
> Mirror = LUL'ULU2L' is the mirror of R'U'RU'R'U2R
>
> Correct?
>
> --- In speedsolvingrubikscube@yahoogroups.com, "Bruce Norskog"
> <brnorsk@> wrote:
> >
> > Hi,
> >
> > Yes, you're right. I considered rotations of the E layer, but not
> more
> > complicated adjustment moves like R2 E R2. If you allow that,
then
> the
> > middle multipliers in my table all become 1, and you can just
> multiply
> > the first and third number. With that, my 140 cases (excluding the
> > do-nothing case) get reduced to 111 cases. (I think I did the
> > arithmetic correctly.) Again, I haven't looked at using mirrors
and
> > inverses to reduce the number of algorithms further.
> >
> > Sorry, it looks like my table's formatting wasn't preserved, at
> least
> > if viewed from the Yahoo web site. You would think the Preview
> button
> > would actually show you what your post was going to look like,
> > wouldn't you? In Preview, it looked fine, but the actual post
> appears
> > to have all "redundant" space characters stripped out.
> >
> >  - Bruce
> > --- In speedsolvingrubikscube@yahoogroups.com, "athefre"
<athefre@>
> > wrote:
> > >
> > > Thanks.  All of what you said sounds right.  But there is one
> thing
> > > I'm not sure if you considered that I may have looked over in
> your
> > > post.
> > >
> > > What about the "empty spaces" available in E for the cases
where
> 2 E
> > > edges need to be placed?  Like, if you have an empty space at
FR
> and
> > > BR or you can have the spaces at FR and BL (although you could
do
> > > R2ER2 before the algorithm).
> > >
> > > If it really is 140 cases then that is WAY too many for me to
> make
> > > and learn.  I'm definitly going with my other option.
> > >
> > > --- In speedsolvingrubikscube@yahoogroups.com, "Bruce Norskog"
> > > <brnorsk@> wrote:
> > > >
> > > > Hi,
> > > >
> > > > From what I understand, you have 4 corner cubies in the U
layer
> to
> > > be
> > > > put into correct relative order (orientation doesn't matter).
> You
> > > have
> > > > 10 edges that can be permuted around without changing
> orientation.
> > > Of
> > > > those 10 edges, 4 are E-layer edges which can be considered
> > > > indistinguishable from each other. These E-layer edges are all
> > > > required to end up in the E layer. The other set of 6 edges
can
> also
> > > > be considered to be indistinguishable from each other. The U
> layer
> > > can
> > > > be rotated before (and after, if you want the corners
correctly
> > > placed
> > > > relative to the center) the algorithm. Likewise, the E layer
> can be
> > > > rotated before and after the algorithm. (Rotating after to
get
> the
> > > > E-layer centers back into correct position, if needed.)
> > > >
> > > > So to count the different cases you can have, consider the
> different
> > > > cases of where the E-layer edges can be, and count the cases
> for
> > > each
> > > > of the possible corner permutation situations (no swap, swap 2
> > > > adjacent, swap to diagonally opposite). First break down the
> edge
> > > > cases by how many might be in each layer. For each possible
> number
> > > of
> > > > E-layer edges in each of the layers, determine the number of
> cases
> > > > possible for each of the corner permutation situations.
> > > >
> > > > Then build a table of all the possibilities:
> > > >
> > > > (best viewed using fixed-width font)
> > > >
> > > > U-E-D      no swap      adj. swap    diag. swap
> > > > -----      -------      ---------    ----------
> > > > 4 0 0      1*1*1 =  1   1*1*1 =  1   1*1*1 =  1
> > > > 3 1 0      1*1*1 =  1   4*1*1 =  4   2*1*1 =  2
> > > > 3 0 1      1*1*2 =  2   4*1*2 =  8   2*1*2 =  4
> > > > 2 2 0      2*2*1 =  4   6*2*1 = 12   4*2*1 =  8
> > > > 2 1 1      2*1*2 =  4   6*1*2 = 12   4*1*2 =  8
> > > > 2 0 2      2*1*1 =  2   6*1*1 =  6   4*1*1 =  4
> > > > 1 3 0      1*1*1 =  1   4*1*1 =  4   2*1*1 =  2
> > > > 1 2 1      1*2*2 =  4   4*2*2 = 16   2*2*2 =  8
> > > > 1 1 2      1*1*1 =  1   4*1*1 =  4   2*1*1 =  2
> > > > 0 4 0      1*1*1 = (1)  1*1*1 =  1   1*1*1 =  1
> > > > 0 3 1      1*1*2 =  2   1*1*2 =  2   1*1*2 =  2
> > > > 0 2 2      1*2*1 =  2   1*2*1 =  2   1*2*1 =  2
> > > >                   ---          ---          ---
> > > >                    25           72           44
> > > >
> > > > So I get 25+72+44 = 141 cases. The 1 in parentheses in the
table
> > > > indicates the case where no algorithm needs to be performed.
So
> if
> > > you
> > > > don't count that case, then I get 140.
> > > >
> > > > I have not considered the diagonal symmetry in the above, but
> then,
> > > I
> > > > understand you were not asking for that to be taken into
> > > consideration.
> > > >
> > > > I just thought I would add my own comments about the edge
> > > orientation
> > > > issue.
> > > >
> > > > I agree with Doug in that the key in what you said was that F
> and B
> > > > moves flip four edges.
> > > >
> > > > From that I assume you mean, that to be oriented:
> > > >  - an edge cubie that belongs in the M or S slice, and is
> currently
> > > > located in one of those slices,  must have its U or D facelet
> > > aligned
> > > > with the U or D center
> > > >  - an edge cubie that belongs in the M or S slice, and is
> located in
> > > > the E slice, must have it U or D facelet aligned with the F
or
> B
> > > center.
> > > >  - an edge cubie that belongs in the E slice, and is located
in
> the
> > > E
> > > > slice, must have its F or B facelet aligned with the F or B
> center
> > > (or
> > > > equivalently, its R or L face aligned with the R or L center)
> > > >  - an edge cubie that belongs in the E slice, and is located
in
> the
> > > M
> > > > or S slice, must have its F or B face aligned with the U or D
> > > center.
> > > >
> > > > When an edge is in the inner slice that it belongs to, its
> usually
> > > > assumed that the edge would be oriented if each of its
facelets
> is
> > > > aligned with the same color center, or the center that is
> opposite
> > > > that center. (Someone could define edge orientation in a way
> such
> > > that
> > > > the above would not be the case, but I would say this is
rare.)
> But
> > > > when an edge is moved to a different inner slice than the one
it
> > > > belongs in, it is not generally as clear what it means to be
> > > oriented.
> > > >
> > > > Doug mentioned a way of defining edge orientation such that
> moving L
> > > > or R a quarter-turn flips four edges. There is yet another
way
> of
> > > > defining edge orientation that I have used in computer
analyses
> of
> > > the
> > > > cube. You can define edge orientation such that moving any of
> the
> > > > layers U, D, L, R, F, or B a quarter-turn flips all four
edges
> > > moved.
> > > > This is the most symmetrical way of defining edge
orientation.
> But
> > > > define edge orientation in the way that makes the most sense
> for
> > > your
> > > > situation. With your way, you can keep all edges oriented
> simply by
> > > > avoiding F, F', B, and B' moves (F2 and B2 okay, of course).
> > > >
> > > >  - Bruce
> > > > --- In speedsolvingrubikscube@yahoogroups.com, "athefre"
> <athefre@>
> > > > wrote:
> > > > >
> > > > > Yeah, it was supposed to say "DFL".
> > > > >
> > > > > I don't really understand or know anything about inverses
and
> > > mirrors
> > > > > and symmetry and all of that crazy stuff but hopefully this
> helps:
> > > > >
> > > > > -Add in the inverses the stuff like that but tell me how
many
> > > > > distinct cases there are with those included and without.
> > > > >
> > > > > -Don't count U adjustments.  I don't mind having to adjust
U
> > > before
> > > > > doing an algorithm.
> > > > >
> > > > > So far I'm thinking it's around 102.  If so, no way.  I'm
> going
> > > with
> > > > > my other option.  This is what I've been counting:
> > > > >
> > > > > Already permuted: 17 cases
> > > > > Diagonal swap: 18 cases (1 for E edges already in E)
> > > > > Adjacent swap: 69 cases (same as above)
> > > > >
> > > > > Is there a site that describes these kinds of things?
> > > > >
> > > > >
> > > > > --- In speedsolvingrubikscube@yahoogroups.com, d_funny007
> > > > > <no_reply@> wrote:
> > > > > >
> > > > > > That does help. Actually I use a different EO
definition...
> I
> > > treat
> > > > > > L and R as flipping 4 edges.
> > > > > >
> > > > > > Also, could you double check this: "The algorithm must not
> > > > > > mess up UFL, DL, DBL, DB, DBR, or DFR." It doesn't feel
> right.
> > > Are
> > > > > > you sure you don't mean 'DFL' there? Also what would you
> count
> > > as a
> > > > > > distinct case? I could group diagonally-symmetric cases
as
> one.
> > > I
> > > > > > could even group cases that use inverse algorithms
> together. If
> > > U
> > > > > > layer is not free for the first turn, than you could get
> what I
> > > > > like
> > > > > > to think of as a single case counted 4 times.
> > > > > >
> > > > > > This question sounds familiar, like I've already heard
> > > something
> > > > > > similar before, but it is definately a hard one and may
> take
> > > some
> > > > > > time.
> > > > > >
> > > > > >
> > > > > > -Doug
> > > > > >
> > > > > >
> > > > > >
> > > > > > --- In speedsolvingrubikscube@yahoogroups.com, "athefre"
> > > > > > <athefre@> wrote:
> > > > > > >
> > > > > > > I'm not too sure what you mean, but I'm using yellow on
> top,
> > > blue
> > > > > > on
> > > > > > > the right, orange in the front.  All of the yellow and
> white
> > > > > edges
> > > > > > > face the white or yellow center (it doesn't matter) and
> all
> > > of
> > > > > the
> > > > > > > blue and green edges are facing the blue or green
> centers.
> > > It's
> > > > > > like
> > > > > > > Petrus, the edges are oriented that way, and if you do
F
> or B
> > > it
> > > > > > > messes up 4 edges.
> > > > > > >
> > > > > > > Does that help.
> > > > > > >
> > > > > > > --- In speedsolvingrubikscube@yahoogroups.com, "Stefan
> > > Pochmann"
> > > > > > > <pochmann@> wrote:
> > > > > > > >
> > > > > > > > --- In
> speedsolvingrubikscube@yahoogroups.com, "athefre"
> > > > > > <athefre@>
> > > > > > > > wrote:
> > > > > > > > >
> > > > > > > > > All edges on the cube are already oriented before
> going
> > > to
> > > > > this
> > > > > > > > > step.
> > > > > > > >
> > > > > > > > There's no general definition for orientation so you
> need
> > > to
> > > > > > > provide
> > > > > > > > one.
> > > > > > > >
> > > > > > >
> > > > > >
> > > > >
> > > >
> > >
> >
>

Good tip, but failing that, do bring a print out of your e-mail
registration confirmation in order to exploit their half price
admission offer. I made a reminder a few weeks ago, but even I nearly
forgot about it when I was packing.

Anyhow, SanF is *amazing*. It's my first time here. However, it's
rather crowded downtown between 11am to 7pm for my taste. I guess I
don't look forward to a dense city as much as I would if I wasn't
from a dense city myself. Still the archetcture, landscape, and
diversity here is great. Just becareful of getting mugged (this
comming form a person who frequents downtowns of Toronto, Detroit,
and New York). I really didn't think it would be as bad, but I was
walking around at 1-2am and it got a bit scary. Do avoid Civc Center
area, as I was later warned by hotel front desk, lol.

Also if you are a big academic freak, then do visit BerkeleyU. It's
one of the top 3 when it comes to most branches of mathematics. The
nieghborhod is reminicent of AA, so it was right up my alley; felt at
home :). Hem, makes me want to plaster that campus with tournament
flyers, cuz I bet we could've gotten +20 audience that way.

Have a safe trip. BTW, your ears might pop riding the subway here...
BART takes you everywhere, it's very convienient. Well the
Exploratorium is requires a 15 minute cab ride though (form the
hotels I suspect you will likely be at).

Also, when/where will Mr. Cubesmith be?

Hem, Atante has several computers for internet access, but they
charge me a little... o_O


-Doug


--- In speedsolvingrubikscube@yahoogroups.com, Tyson Mao <tmao@...>
wrote:
>
> If you show up at the Exploratorium at 9 AM on Friday, you can get
in
> free with the rest of the Caltech crew.
>
> Tyson Mao
> Astrophysics '06
> California Institute of Technology
>
> On Jul 30, 2006, at 12:31 PM, d_funny007 wrote:
>
> > Hi,
> >
> >  Since I'm about to leave soon, I thought it would be a good idea
to
> >  start a thread for people to exchange some last-minute info.
> >
> >  Although most ppl are staying at the Coventry Motor Inn, me,
Bob, and
> >  ChrisH worked out something over at the Adante Hotel.
> >
> >  My cell num. is seven three four, 9 tree for, sx 3 sx too. (I am
> >  parinoid of internet bots/cralwers, lol.) Call me if you need a
ride,
> >  I rented a car and plan to be there for 2 weeks. Say if you want
to be
> >  picked up from either airport, I'll see what I can do, if you
contact
> >  me a couple hours prior.
> >
> >  Is there anything going on Thursday? Setup and going to see the
> >  Exploratorium would be good to do in the afternoon or something.
I
> >  don't know if that would be a welcome idea though. I think that
> >  Thursday night there might be something at either one of the
hotel
> >  lobbies or Clancy's place (contact me or him about that).
> >
> >  I will be doing a lot of touristy stuff, and taking a lot of pics
> >  becasue I'm a photography freak. This time I brought my favorite
> >  camera and will be sure to take many pictures of you guys (some
vid
> >  too).
> >
> >  So far I got a good idea of who is and isn't comming. I do hear
that
> >  Brent is having some difficulty with making arrangments, I fear.
> >  Perhaps if someone is driving from TX and can give him a lift...
> >
> >  I know San Francisco pretty well guys, probably as well as any
non-
> >  local that will be attending. Btw, the public transit system
(BART) is
> >  one of the best in the country. It will take you from the
airport to a
> >  few blocks from almost anything I can think of (most notably
Berkeley
> >  U, Union Square, and the Wharf/Piers). Well except maybe the
> >  Exploratorium... I'm not sure how to get there using only BART
and
> >  cable cars (busses perhaps?). I can help shuttle ppl form the
Coventry
> >  to the Exploratorium though if that becomes a problem. I know
that not
> >  all of you can afford a hefty series of cab fares.
> >
> >  If you want to contact me, call me; I won't be reading this
forum till
> >  after I come back. Good luck with getting things together, and
have a
> >  safe trip.
> >
> >  -Doug (almost packed!!)
> >
> >
> >
>

AM/PM? Which day??? Be specific man!

--- In speedsolvingrubikscube@yahoogroups.com, "Craig Bouchard"
<logitewty@...> wrote:
> I'll probably be there at 8:30 then :) I tend to be really early for
> stuff :)  I land in San Fran at 4:20 so if anyone wants to do
> something on thursday, I know people are doing stuff, just send an
> e-mail my way...

Thanks, I'm sure the number that he got is correct or at least very
close and I'm sure that the amount of cases is at least 100.  I can't
learn that many because the other steps in what I'm doing require
algorithms to be learned.

I'm glad you and Bruce are around.  It has made things a lot easier.

--- In speedsolvingrubikscube@yahoogroups.com, d_funny007
<no_reply@...> wrote:
>
> No, in some cases (I think yours would apply), you should look
> for "diagonal mirroring". Although the simple mirroring plus U
> rotations *might* be enough/analgous/equivalent, but I have put
> little thought into this as I am currently on vacation!
>
> As a long time member of this fourm, I'd like to say that it is
very
> good to see another hardcore math/cs person like Bruce here! I've
> been keeping up with his posts on this other fourm he uses too.
Very
> techincal stuff that I once wanted to see here, but after further
> thought, it just wouldn't fit here. I was always the one rushing to
> answer math questions, but I wasn't particularly patient in the
past :
> (.
>
> I can try a verification of his computation when I get the chance.
It
> is most challenging :).
>
>
> -Doug
>
>
> --- In speedsolvingrubikscube@yahoogroups.com, "athefre"
> <athefre@> wrote:
> >
> > Thanks, 111 is better than 140, but not much.
> >
> > If you could reduce the number using mirrors and inverses, how
much
> > would it be?  If you don't mind.  I've been working hard for a
> month
> > trying to perfect everything so I can get to work on finding the
> > algorithms for the idea I choose.
> >
> > Inverse = backwards
> > Mirror = LUL'ULU2L' is the mirror of R'U'RU'R'U2R
> >
> > Correct?
> >
> > --- In speedsolvingrubikscube@yahoogroups.com, "Bruce Norskog"
> > <brnorsk@> wrote:
> > >
> > > Hi,
> > >
> > > Yes, you're right. I considered rotations of the E layer, but
not
> > more
> > > complicated adjustment moves like R2 E R2. If you allow that,
> then
> > the
> > > middle multipliers in my table all become 1, and you can just
> > multiply
> > > the first and third number. With that, my 140 cases (excluding
the
> > > do-nothing case) get reduced to 111 cases. (I think I did the
> > > arithmetic correctly.) Again, I haven't looked at using mirrors
> and
> > > inverses to reduce the number of algorithms further.
> > >
> > > Sorry, it looks like my table's formatting wasn't preserved, at
> > least
> > > if viewed from the Yahoo web site. You would think the Preview
> > button
> > > would actually show you what your post was going to look like,
> > > wouldn't you? In Preview, it looked fine, but the actual post
> > appears
> > > to have all "redundant" space characters stripped out.
> > >
> > >  - Bruce
> > > --- In speedsolvingrubikscube@yahoogroups.com, "athefre"
> <athefre@>
> > > wrote:
> > > >
> > > > Thanks.  All of what you said sounds right.  But there is one
> > thing
> > > > I'm not sure if you considered that I may have looked over in
> > your
> > > > post.
> > > >
> > > > What about the "empty spaces" available in E for the cases
> where
> > 2 E
> > > > edges need to be placed?  Like, if you have an empty space at
> FR
> > and
> > > > BR or you can have the spaces at FR and BL (although you
could
> do
> > > > R2ER2 before the algorithm).
> > > >
> > > > If it really is 140 cases then that is WAY too many for me to
> > make
> > > > and learn.  I'm definitly going with my other option.
> > > >
> > > > --- In speedsolvingrubikscube@yahoogroups.com, "Bruce
Norskog"
> > > > <brnorsk@> wrote:
> > > > >
> > > > > Hi,
> > > > >
> > > > > From what I understand, you have 4 corner cubies in the U
> layer
> > to
> > > > be
> > > > > put into correct relative order (orientation doesn't
matter).
> > You
> > > > have
> > > > > 10 edges that can be permuted around without changing
> > orientation.
> > > > Of
> > > > > those 10 edges, 4 are E-layer edges which can be considered
> > > > > indistinguishable from each other. These E-layer edges are
all
> > > > > required to end up in the E layer. The other set of 6 edges
> can
> > also
> > > > > be considered to be indistinguishable from each other. The
U
> > layer
> > > > can
> > > > > be rotated before (and after, if you want the corners
> correctly
> > > > placed
> > > > > relative to the center) the algorithm. Likewise, the E
layer
> > can be
> > > > > rotated before and after the algorithm. (Rotating after to
> get
> > the
> > > > > E-layer centers back into correct position, if needed.)
> > > > >
> > > > > So to count the different cases you can have, consider the
> > different
> > > > > cases of where the E-layer edges can be, and count the
cases
> > for
> > > > each
> > > > > of the possible corner permutation situations (no swap,
swap 2
> > > > > adjacent, swap to diagonally opposite). First break down
the
> > edge
> > > > > cases by how many might be in each layer. For each possible
> > number
> > > > of
> > > > > E-layer edges in each of the layers, determine the number
of
> > cases
> > > > > possible for each of the corner permutation situations.
> > > > >
> > > > > Then build a table of all the possibilities:
> > > > >
> > > > > (best viewed using fixed-width font)
> > > > >
> > > > > U-E-D      no swap      adj. swap    diag. swap
> > > > > -----      -------      ---------    ----------
> > > > > 4 0 0      1*1*1 =  1   1*1*1 =  1   1*1*1 =  1
> > > > > 3 1 0      1*1*1 =  1   4*1*1 =  4   2*1*1 =  2
> > > > > 3 0 1      1*1*2 =  2   4*1*2 =  8   2*1*2 =  4
> > > > > 2 2 0      2*2*1 =  4   6*2*1 = 12   4*2*1 =  8
> > > > > 2 1 1      2*1*2 =  4   6*1*2 = 12   4*1*2 =  8
> > > > > 2 0 2      2*1*1 =  2   6*1*1 =  6   4*1*1 =  4
> > > > > 1 3 0      1*1*1 =  1   4*1*1 =  4   2*1*1 =  2
> > > > > 1 2 1      1*2*2 =  4   4*2*2 = 16   2*2*2 =  8
> > > > > 1 1 2      1*1*1 =  1   4*1*1 =  4   2*1*1 =  2
> > > > > 0 4 0      1*1*1 = (1)  1*1*1 =  1   1*1*1 =  1
> > > > > 0 3 1      1*1*2 =  2   1*1*2 =  2   1*1*2 =  2
> > > > > 0 2 2      1*2*1 =  2   1*2*1 =  2   1*2*1 =  2
> > > > >                   ---          ---          ---
> > > > >                    25           72           44
> > > > >
> > > > > So I get 25+72+44 = 141 cases. The 1 in parentheses in the
> table
> > > > > indicates the case where no algorithm needs to be
performed.
> So
> > if
> > > > you
> > > > > don't count that case, then I get 140.
> > > > >
> > > > > I have not considered the diagonal symmetry in the above,
but
> > then,
> > > > I
> > > > > understand you were not asking for that to be taken into
> > > > consideration.
> > > > >
> > > > > I just thought I would add my own comments about the edge
> > > > orientation
> > > > > issue.
> > > > >
> > > > > I agree with Doug in that the key in what you said was that
F
> > and B
> > > > > moves flip four edges.
> > > > >
> > > > > From that I assume you mean, that to be oriented:
> > > > >  - an edge cubie that belongs in the M or S slice, and is
> > currently
> > > > > located in one of those slices,  must have its U or D
facelet
> > > > aligned
> > > > > with the U or D center
> > > > >  - an edge cubie that belongs in the M or S slice, and is
> > located in
> > > > > the E slice, must have it U or D facelet aligned with the F
> or
> > B
> > > > center.
> > > > >  - an edge cubie that belongs in the E slice, and is
located
> in
> > the
> > > > E
> > > > > slice, must have its F or B facelet aligned with the F or B
> > center
> > > > (or
> > > > > equivalently, its R or L face aligned with the R or L
center)
> > > > >  - an edge cubie that belongs in the E slice, and is
located
> in
> > the
> > > > M
> > > > > or S slice, must have its F or B face aligned with the U or
D
> > > > center.
> > > > >
> > > > > When an edge is in the inner slice that it belongs to, its
> > usually
> > > > > assumed that the edge would be oriented if each of its
> facelets
> > is
> > > > > aligned with the same color center, or the center that is
> > opposite
> > > > > that center. (Someone could define edge orientation in a
way
> > such
> > > > that
> > > > > the above would not be the case, but I would say this is
> rare.)
> > But
> > > > > when an edge is moved to a different inner slice than the
one
> it
> > > > > belongs in, it is not generally as clear what it means to
be
> > > > oriented.
> > > > >
> > > > > Doug mentioned a way of defining edge orientation such that
> > moving L
> > > > > or R a quarter-turn flips four edges. There is yet another
> way
> > of
> > > > > defining edge orientation that I have used in computer
> analyses
> > of
> > > > the
> > > > > cube. You can define edge orientation such that moving any
of
> > the
> > > > > layers U, D, L, R, F, or B a quarter-turn flips all four
> edges
> > > > moved.
> > > > > This is the most symmetrical way of defining edge
> orientation.
> > But
> > > > > define edge orientation in the way that makes the most
sense
> > for
> > > > your
> > > > > situation. With your way, you can keep all edges oriented
> > simply by
> > > > > avoiding F, F', B, and B' moves (F2 and B2 okay, of course).
> > > > >
> > > > >  - Bruce
> > > > > --- In speedsolvingrubikscube@yahoogroups.com, "athefre"
> > <athefre@>
> > > > > wrote:
> > > > > >
> > > > > > Yeah, it was supposed to say "DFL".
> > > > > >
> > > > > > I don't really understand or know anything about inverses
> and
> > > > mirrors
> > > > > > and symmetry and all of that crazy stuff but hopefully
this
> > helps:
> > > > > >
> > > > > > -Add in the inverses the stuff like that but tell me how
> many
> > > > > > distinct cases there are with those included and without.
> > > > > >
> > > > > > -Don't count U adjustments.  I don't mind having to
adjust
> U
> > > > before
> > > > > > doing an algorithm.
> > > > > >
> > > > > > So far I'm thinking it's around 102.  If so, no way.  I'm
> > going
> > > > with
> > > > > > my other option.  This is what I've been counting:
> > > > > >
> > > > > > Already permuted: 17 cases
> > > > > > Diagonal swap: 18 cases (1 for E edges already in E)
> > > > > > Adjacent swap: 69 cases (same as above)
> > > > > >
> > > > > > Is there a site that describes these kinds of things?
> > > > > >
> > > > > >
> > > > > > --- In speedsolvingrubikscube@yahoogroups.com, d_funny007
> > > > > > <no_reply@> wrote:
> > > > > > >
> > > > > > > That does help. Actually I use a different EO
> definition...
> > I
> > > > treat
> > > > > > > L and R as flipping 4 edges.
> > > > > > >
> > > > > > > Also, could you double check this: "The algorithm must
not
> > > > > > > mess up UFL, DL, DBL, DB, DBR, or DFR." It doesn't feel
> > right.
> > > > Are
> > > > > > > you sure you don't mean 'DFL' there? Also what would
you
> > count
> > > > as a
> > > > > > > distinct case? I could group diagonally-symmetric cases
> as
> > one.
> > > > I
> > > > > > > could even group cases that use inverse algorithms
> > together. If
> > > > U
> > > > > > > layer is not free for the first turn, than you could
get
> > what I
> > > > > > like
> > > > > > > to think of as a single case counted 4 times.
> > > > > > >
> > > > > > > This question sounds familiar, like I've already heard
> > > > something
> > > > > > > similar before, but it is definately a hard one and may
> > take
> > > > some
> > > > > > > time.
> > > > > > >
> > > > > > >
> > > > > > > -Doug
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > --- In
speedsolvingrubikscube@yahoogroups.com, "athefre"
> > > > > > > <athefre@> wrote:
> > > > > > > >
> > > > > > > > I'm not too sure what you mean, but I'm using yellow
on
> > top,
> > > > blue
> > > > > > > on
> > > > > > > > the right, orange in the front.  All of the yellow
and
> > white
> > > > > > edges
> > > > > > > > face the white or yellow center (it doesn't matter)
and
> > all
> > > > of
> > > > > > the
> > > > > > > > blue and green edges are facing the blue or green
> > centers.
> > > > It's
> > > > > > > like
> > > > > > > > Petrus, the edges are oriented that way, and if you
do
> F
> > or B
> > > > it
> > > > > > > > messes up 4 edges.
> > > > > > > >
> > > > > > > > Does that help.
> > > > > > > >
> > > > > > > > --- In
speedsolvingrubikscube@yahoogroups.com, "Stefan
> > > > Pochmann"
> > > > > > > > <pochmann@> wrote:
> > > > > > > > >
> > > > > > > > > --- In
> > speedsolvingrubikscube@yahoogroups.com, "athefre"
> > > > > > > <athefre@>
> > > > > > > > > wrote:
> > > > > > > > > >
> > > > > > > > > > All edges on the cube are already oriented before
> > going
> > > > to
> > > > > > this
> > > > > > > > > > step.
> > > > > > > > >
> > > > > > > > > There's no general definition for orientation so
you
> > need
> > > > to
> > > > > > > > provide
> > > > > > > > > one.
> > > > > > > > >
> > > > > > > >
> > > > > > >
> > > > > >
> > > > >
> > > >
> > >
> >
>

I have a major problem on my hands right now.  I ordered cubes for the
exploratorium... except they were sent to Pasadena instead of San
Francisco.  They should arrive in Pasadena on Thursday morning.  If
you're driving up from Southern California and can bring cubes up for
me, please let me know immediately.

-Tyson

I land in San Fran at 4:20 PM on Thursday, will be at the hotel by
5:00 probably...e-mail me specifically if you want my cell number...I
wanna do something Thursday night...

Craig

--- In speedsolvingrubikscube@yahoogroups.com, d_funny007
<no_reply@...> wrote:
>
> AM/PM? Which day??? Be specific man!
>
> --- In speedsolvingrubikscube@yahoogroups.com, "Craig Bouchard"
> <logitewty@> wrote:
> > I'll probably be there at 8:30 then :) I tend to be really early for
> > stuff :)  I land in San Fran at 4:20 so if anyone wants to do
> > something on thursday, I know people are doing stuff, just send an
> > e-mail my way...
>

It's summer right now so there aren't that many people passing through
campus.  There will be a decent number of people from Cal that are
competing though.  At least there should be.

> Also if you are a big academic freak, then do visit BerkeleyU. It's
> one of the top 3 when it comes to most branches of mathematics. The
> nieghborhod is reminicent of AA, so it was right up my alley; felt at
> home :). Hem, makes me want to plaster that campus with tournament
> flyers, cuz I bet we could've gotten +20 audience that way.
>

Hi Doug,

The problem with the bright stickers the cubes come with is the edges
are cut at 90 degrees vertically. If they were cut at an angle then
they wouldn't catch fingertips like they do now.

I changed how I handled a cube so that my fingertips didn't catch the
edges and even with daily usage the stickers lasted a little over a
year before they began to look raggedy.

Cheers,

David J

--- In speedsolvingrubikscube@yahoogroups.com, d_funny007
<no_reply@...> wrote:
>
> Only the polypropylene stickers that come with most cubes will peel
> that easily. Polyvinyl chloride (PVC/polychloroethene) stickers from
> rubiks.com will not peel. (Although their corners do occasionally
> fold if very careless.) I do not recommend purchasing the
> polypropylene stickers from rubiks.com, they are basically laminated
> paper with a dye in between.
>
> So why do they sell both sticker types? The ones that last longer
> are muted in brightness and the ones that peel easily are quite
> bright and cheerful.
>
> The solution: make thicker PVC stickers!
>
> In general, due to their pricing and failure to make PVC stickers
> thick enough, I personally boycott them. Instead I opt to buy from
> cubesmith.com, which offers superior service and stickers of amazing
> quality and lifespan.
>
> Also, do note that cubesmith sets are somewhat "soft," and
> rubiks.com's PVC ones are rather "hard." This has trade-offs as well
> which is probably beyond the scope of this topic.
>
> The problem with painting your cube in anyway, is that it's going to
> end up looking terrible. Trust me, I've tried everyting. Besides, it
> somehow violates an intrinsic nature of the pop icon itself. And for
> the same reason I don't use tiled cubes. So unless you are going for
> a special look... I see no reason why any passionate cuber would
> want to do that to their cubes, let alone their main speedcube.
>
> On the other hand, is it really not alllowed in competitions? I was
> not aware of this. Plus, I'm sure that some one was using a painted
> cube at Nationals 04... o_O?? When was this rule added and why?
> (Just curious, doesn't really concern me.)
>
> Oh, I almost forgot to mention... rubiks.com has rather crappy
> service IMHO, but I've hear varing things from other buyers.
>
>
> -Doug
>
>
>
> --- In speedsolvingrubikscube@yahoogroups.com, kyuubree
> <no_reply@> wrote:
> >
> > Well my suggestion was not given with competitions in mind, but
> rather
> > a way to prevent sticker peeling.  Nail polish holds up very well.
> > Why aren't such cubes allowed, by the way?
> >
> > Marcus Stuhr
> > Wharton & CAS '09
> > University of Pennsylvania
> >
> > --- In speedsolvingrubikscube@yahoogroups.com, Tyson Mao <tmao@>
> wrote:
> > >
> > > No... I have to disagree.  If you intend to compete in
> competitions, do
> > > not use nail polish.  Such cubes are not allowed.
> > >
> > > Tyson Mao
> > > Astrophysics '06
> > > California Institute of Technology
> > >
> > > On Jul 26, 2006, at 10:12 PM, kyuubree wrote:
> > >
> > > > Do what I do. Peel off the stickers and repaint with nail
> polish:
> > > >
> > > > http://img206.imageshack.us/img206/7406/1001567jq8.jpg
> > > > http://img55.imageshack.us/img55/8977/1001570au1.jpg
> > > >
> > > >  --- In speedsolvingrubikscube@yahoogroups.com, kyle simmons
> > > >  <beamfreak314@> wrote:
> > > >  >
> > > >  > hi all.
> > > >  >
> > > >  > okay so im sure that just about everyone has the problem of
> their
> > > >  stickers wearing out, and end up buying those sticker-pack-
> things. i
> > > >  dont have the money or patience (lol, mostly patience), so i
> came up
> > > >  with a funny souloution, thats cheap and long lasting.
> > > >  >
> > > >  > i took a cheapy dollar store cube, and removed the coloured
> plastic
> > > >  squares (used instead of stickers), and sanded them. then i
> took what
> > > >  was left of the original stickers off my cube and then
> cleaned it. i
> > > >  sanded the cube and cleaned it again. this was to ensure a
> good bond
> > > >  between the cube an plastic bits. and then glued the plastic
> on the
> > > > cube.
> > > >  >
> > > >  > this brings up two obvoius issues, wich i noted right away.
> > > >  > 1. the cube was heavier, seemingly, alot heavier. probably
> > because i
> > > >  was still so used to my normal cube. i got used to this in
> about 30
> > > >  mins of solving.
> > > >  >
> > > >  > 2. the cube is bigger. true, but only by about, 1/8 of an
> inch,
> > wich
> > > >  i got fully used to in about an hour or so.
> > > >  >
> > > >  > this particullar mod cube is lubed with sewing machine oil.
> > > >  > (i experiment with different oils on my cubes and so far
> olive oil,
> > > >  and sewing machine oil are my favourites.)
> > > >  >
> > > >  > apologies if this is also on the net somewhere, or someone
> else
> > > >  posted a similar mod or suggestion.
> > > >  >
> > > >  > btw, bew record of 42 seconds. (yeah im a newbie)
> > > >  >
> > > >  > -kyle simmons
> > > >  >
> > > >  >
> > > >  >
> > > >  >
> > > >  > ---------------------------------
> > > >  > All new Yahoo! Mail -
> > > >  > ---------------------------------
> > > >  > Get a sneak peak at messages with a handy reading pane.
> > > >  >
> > > >  > [Non-text portions of this message have been removed]
> > > >  >
> > > >
> > > >
> > > >
> > >
> >
>

Hi Ron,

If your suggestion is to check for in general at least 18 (or x) moves
optimal solution, then this is something we could do.

That's what I meant.

I mislaid the URL for the WCA.

Concerning HTM and STM, Please reach a conclusion in favor of STM. :)

Thanks,

David J


--- In speedsolvingrubikscube@yahoogroups.com, "Ron van Bruchem"
<ron@...> wrote:
>
> Hi David,
>
> Thanks for the feedback.
> It is better to post your feedback on the WCA forum, because there
we can have a better overview of the discussions.
>
> I will put your suggestions on the TODO list for version 2007.
>
> For your suggestion of difficulty it is hard to define difficulty.
> If your suggestion is to check for in general at least 18 (or x)
moves optimal solution, then this is something we could do. Given a
majority.
> If your suggestion is to check for easy solutions in CFOP system,
then I think it is not a good idea. In that case you are influencing
randomness for a specific system.
>
> We have had several discussions about HTM and STM, but there was
never a conclusion. Therefore we kept it the way it is now.
>
> Have fun,
>
> Ron
>
>
>   ----- Original Message -----
>   From: d_j_salvia
>   To: speedsolvingrubikscube@yahoogroups.com
>   Sent: Monday, July 31, 2006 7:46 PM
>   Subject: [Speed cubing group] Re: WCA regulations version 2006 v2
is now official
>
>
>   Hi Ron,
>
>   I had no time to take part in the recent discussions, but I would like
>   to mention two things for the future.
>
>   1.) Have a standard for difficulty.
>   I suggest that once scrambles are randomly generated for a tournament
>   that they be run through a few cube solvers. This would allow you to
>   eliminate those scrambles which are too easy.
>
>   2.) Please reconsider and change the rule counting slice turns as two
>   moves. I request that this be changed to counting slice turns as one
>   move.
>
>   I've always seen each layer as equals, even though a slice turn
>   doesn't move corners. As it is now you are penalising corners first
>   solvers and others like me who use a lot of slice moves.
>
>   Cheers,
>
>   David J
>
>   --- In speedsolvingrubikscube@yahoogroups.com, "Ron" <ron@> wrote:
>   >
>   > Fellow cubers,
>   >
>   > The new version of the WCA regulations is now official.
>   > It is version 2006 v2, and available at
>   > http://www.speedcubing.com/events/regulations.html
>   >
>   > We kindly request all translators of former versions to update the
>   > regulations in other languages. If you are interested in
translating
>   > the WCA regulations to your language, then please let us know.
>   >
>   > Starting on July 21, 2006 all official WCA competitions must follow
>   > this new version of the regulations.
>   >
>   > Thank you all for your feedback.
>   >
>   > It is still possible to give more feedback.
>   > Errors will be corrected as soon as possible.
>   > Other feedback will be taken into account for the 2007 version.
This
>   > version is planned for December 2006.
>   >
>   > Happy cubing,
>   >
>   > Ron
>   >
>
>
>
>
>
> [Non-text portions of this message have been removed]
>

Hi Per,

Thanks for the info, as I am unfamiliar with Cube explorer.

I think that you don't have to start the night before the contest to
test scrambles, couldn't this be started months in advance.

You don't have to only use one computer to test scrambles.

If Cube Explorer can quickly come up with a ballpark figure for
difficulty then that might be adequate. Can it? I mean if the optimum
solution is 18 moves, would the "quick and dirty" solution be 20
moves? If the optimum solution is 12 moves would the quick and dirty
sollution be 15 or 16? So my question is how consistent is the program?

Perhaps Cube explorer could be used to determine wheter a particular
scramble would favor HTM or STM.

Cheers,

David J



--- In speedsolvingrubikscube@yahoogroups.com, "Per Kristen Fredlund"
<aspiring_to_love@...> wrote:
>
> Hi :-)
>
> On the practical side, it does take at least 30 minutes (depending
> on the computer) to find optimal solution for a given scramble with
> Cube Explorer. It quickly finds 20 or 19 move solution(s) but to
> really find the shortest proven solution takes a good while. So i
> don't think it's feasible to do this for a high number of scrambles.
> And then again why check this only for 3x3x3? We could check also
> for 4x4x4 or 5x5x5 or square-1 or megaminx ... I think it can be
> seen where im going with this. Im not worried about getting easy
> scrambles at competitions. I do want everyone to get same scrambles
> (possibly with random orientations). That's the most feasible option
> as i see it.
>
> (Has anyone really studied how optimal solution length compares
> version scramble length for a large number of random scrambles of
> different lenghths ?? Where do we stop getting "harder" scrambles by
> making them longer?)
>
> About move metric,this is only really interesting for fewest moves.
> And fewest movers have been dictated by Dan Harris (:-P) to follow
> htm for a good while already. I don't see any problem with that.
> Those who use inner slice turns heavily will feel punsihed by htm
> andvice versa those who are used to thinking in outer turns only
> will feel that those using inner slice turns a lot are favored by
> stm. So we can't satisfy everyone either way.
>
> I see 3 options :
> htm - because it's simply most common andmaybe the least
> controversial,mathematical basis
>
> stm - from a practical point of view maybe the most correct metric,
> a layer is a layer whether it's internal or external
>
> combined metric - a salomonic solution, noone should feel heavily
> favoured or dis-favoured
>
> The latter does most likely require a computer to determine the
> metric,and as discussed in another yahoo group it's not immune to
> ambiguity ;-)
>
> Cheers!
>
> -Per
>
> > --- In speedsolvingrubikscube@yahoogroups.com, "Ron van Bruchem"
> <ron@> wrote:
> >
> > Hi David,
> >
> > Thanks for the feedback.
> > It is better to post your feedback on the WCA forum, because there
> we can have a better overview of the discussions.
> >
> > I will put your suggestions on the TODO list for version 2007.
> >
> > For your suggestion of difficulty it is hard to define difficulty.
> > If your suggestion is to check for in general at least 18 (or x)
> moves optimal solution, then this is something we could do. Given a
> majority.
> > If your suggestion is to check for easy solutions in CFOP system,
> then I think it is not a good idea. In that case you are influencing
> randomness for a specific system.
> >
> > We have had several discussions about HTM and STM, but there was
> never a conclusion. Therefore we kept it the way it is now.
> >
> > Have fun,
> >
> > Ron
> >
> >
> >   ----- Original Message -----
> >   From: d_j_salvia
> >   To: speedsolvingrubikscube@yahoogroups.com
> >   Sent: Monday, July 31, 2006 7:46 PM
> >   Subject: [Speed cubing group] Re: WCA regulations version 2006
> v2 is now official
> >
> >
> >   Hi Ron,
> >
> >   I had no time to take part in the recent discussions, but I
> would like
> >   to mention two things for the future.
> >
> >   1.) Have a standard for difficulty.
> >   I suggest that once scrambles are randomly generated for a
> tournament
> >   that they be run through a few cube solvers. This would allow
> you to
> >   eliminate those scrambles which are too easy.
> >
> >   2.) Please reconsider and change the rule counting slice turns
> as two
> >   moves. I request that this be changed to counting slice turns as
> one
> >   move.
> >
> >   I've always seen each layer as equals, even though a slice turn
> >   doesn't move corners. As it is now you are penalising corners
> first
> >   solvers and others like me who use a lot of slice moves.
> >
> >   Cheers,
> >
> >   David J
> >
> >   --- In speedsolvingrubikscube@yahoogroups.com, "Ron" <ron@>
> wrote:
> >   >
> >   > Fellow cubers,
> >   >
> >   > The new version of the WCA regulations is now official.
> >   > It is version 2006 v2, and available at
> >   > http://www.speedcubing.com/events/regulations.html
> >   >
> >   > We kindly request all translators of former versions to update
> the
> >   > regulations in other languages. If you are interested in
> translating
> >   > the WCA regulations to your language, then please let us know.
> >   >
> >   > Starting on July 21, 2006 all official WCA competitions must
> follow
> >   > this new version of the regulations.
> >   >
> >   > Thank you all for your feedback.
> >   >
> >   > It is still possible to give more feedback.
> >   > Errors will be corrected as soon as possible.
> >   > Other feedback will be taken into account for the 2007
> version. This
> >   > version is planned for December 2006.
> >   >
> >   > Happy cubing,
> >   >
> >   > Ron
> >   >
> >
> >
> >
> >
> >
> > [Non-text portions of this message have been removed]
> >
>

I may have a solution for you, but it wouldn't get them there until
late Friday night.  With prelims starting on Friday, obviously that
isn't an ideal solution, but you've at least got this as an option.

Hopefully someone else has a solution to get them there earlier...


--Kirk


--- In speedsolvingrubikscube@yahoogroups.com, Tyson Mao <tmao@...>
wrote:
>
> I have a major problem on my hands right now.  I ordered cubes for
the
> exploratorium... except they were sent to Pasadena instead of San
> Francisco.  They should arrive in Pasadena on Thursday morning.  If
> you're driving up from Southern California and can bring cubes up
for
> me, please let me know immediately.
>
> -Tyson
>

Hi Kirk,

What's your solution?  If I can get half up on Thursday night and half
up on Friday night, I'd be just fine.

-Tyson

On Aug 2, 2006, at 1:37 PM, kirk83616 wrote:

> I may have a solution for you, but it wouldn't get them there until
>  late Friday night. With prelims starting on Friday, obviously that
>  isn't an ideal solution, but you've at least got this as an option.
>
>  Hopefully someone else has a solution to get them there earlier...
>
>  --Kirk
>
>  --- In speedsolvingrubikscube@yahoogroups.com, Tyson Mao <tmao@...>
>  wrote:
>  >
>  > I have a major problem on my hands right now. I ordered cubes for
>  the
>  > exploratorium... except they were sent to Pasadena instead of San
>  > Francisco. They should arrive in Pasadena on Thursday morning. If
>  > you're driving up from Southern California and can bring cubes up
>  for
>  > me, please let me know immediately.
>  >
>  > -Tyson
>  >
>
>
>

just sent you email with my cell #...

--- In speedsolvingrubikscube@yahoogroups.com, Tyson Mao <tmao@...>
wrote:
>
> Hi Kirk,
>
> What's your solution?  If I can get half up on Thursday night and
half
> up on Friday night, I'd be just fine.
>
> -Tyson
>
> On Aug 2, 2006, at 1:37 PM, kirk83616 wrote:
>
> > I may have a solution for you, but it wouldn't get them there
until
> >  late Friday night. With prelims starting on Friday, obviously
that
> >  isn't an ideal solution, but you've at least got this as an
option.
> >
> >  Hopefully someone else has a solution to get them there
earlier...
> >
> >  --Kirk
> >
> >  --- In speedsolvingrubikscube@yahoogroups.com, Tyson Mao <tmao@>
> >  wrote:
> >  >
> >  > I have a major problem on my hands right now. I ordered cubes
for
> >  the
> >  > exploratorium... except they were sent to Pasadena instead of
San
> >  > Francisco. They should arrive in Pasadena on Thursday
morning. If
> >  > you're driving up from Southern California and can bring
cubes up
> >  for
> >  > me, please let me know immediately.
> >  >
> >  > -Tyson
> >  >
> >
> >
> >
>

Hi,

This is about my 4x4x4 solver program that determines a solution for a
scrambled cube in no more than 79 slice turns.

Thanks to David for being willing to help. I haven't used Bittorrent
myself. In saying that you would host a tracker file, does that mean
you would store the files somewhere that would allow Bittorrent
to be transfer the files to others, or are you assuming they would be
stored by some other file hosting site (that your "tracker file" would
merely be referencing)?

I now have a version of the solver that may be usable by others. I
added the ability to specify facelet pattern and to specify the
location of the data files. I also eliminated a couple of large arrays
to reduce the amount of memory it uses. Its arrays now use up about
225 megabytes.

Anyways my plan at the moment is to make a few copies on DVD discs
tonight, and bring them to U.S. Nationals for people who will be there
and are interested in it. I may only have about a half-dozen copies
with me. If someone is there who is willing to host the files, that
would be great. Otherwise, I will probably figure out some way to make
them downloadable after returning from U.S. Nationals. If people out
there do not feel downloading is reasonable, I guess they can email me
and we can work something out. I also may create programs allowing the
user to generate the files, if I feel there is sufficient interest in
that alternative.

  - Bruce
--- In speedsolvingrubikscube@yahoogroups.com, "thomkirjava"
<snkenjoi@...> wrote:
>
> I'd be interested in a copy. I'll leave my server on and help seed if
> need be.
>
> ~Thom
>
> --- In speedsolvingrubikscube@yahoogroups.com, "David Barr"
> <david20708@> wrote:
> >
> > Another option is to share the file with Bittorrent.  1.4GB really
> > isn't that big a file by today's standards.  I downloaded a 14GB file
> > with Bittorrent, and it only took a couple days.  I'd be willing to
> > host a tracker for the file if you'd like.
> >
> > On 7/28/06, Bruce Norskog <brnorsk@> wrote:
> > >
> > >  Perhaps sometime after U.S. Nationals, I'll look into doing
this, if
> > >  people are interested in generating the files themselves after
> knowing
> > >  what the memory requirements are and how much runtime is involved.
> > >
> >
>

Hi everyone,

I'm leaving in about 6.5 hours to San Francisco.  I'll arrive tomorow
Thursday morning at 10:45am San Francisco time.

My cell number is nine 1 niine sevn, five sseven, 7 tree sevnn for.  I
also am worried about bots, so I'll follow Doug's example.

I'll be meeting Doug and we'll be doing a bit of sight seeing until
others arrive, and hopefully we can meet up with some other cubers who
are in the area.

See you all soon, and have a safe trip everyone!
Chris

Hi,

While optimal solvers (for 3x3x3) can take a long time to find the
solution for a deep position, they can determine if a position is
within 15 moves (HTM, aka FTM) of solved pretty fast, I believe. That
may be good enough. As they are, they won't automatically stop after a
certain depth, as far as I know. By the way, I believe Cube Explorer
only does HTM metric (but I don't have the latest version).  Mike
Reid's optimal solver can be compiled to do either QTM or HTM. It's
not clear to me if you can estimate the optimal length very well based
on a sub-optimal solver's solution length.

I think the issue of "scramble quality" is more important on the
2x2x2. With less than 4 million positions, I think easy scrambles are
likely to come up once in awhile, if there isn't something in place to
prevent them. There is almost a third of one percent chance of a
randomly chosen position to be depth five or less (HTM). I was in the
room when the current WR 2x2x2 solve occurred. I don't know what the
scramble was for that solve, though, although I assume I also had to
solve the same scramble.

- Bruce
--- In speedsolvingrubikscube@yahoogroups.com, "d_j_salvia"
<d_j_salvia@...> wrote:
>
> Hi Per,
>
> Thanks for the info, as I am unfamiliar with Cube explorer.
>
> I think that you don't have to start the night before the contest to
> test scrambles, couldn't this be started months in advance.
>
> You don't have to only use one computer to test scrambles.
>
> If Cube Explorer can quickly come up with a ballpark figure for
> difficulty then that might be adequate. Can it? I mean if the optimum
> solution is 18 moves, would the "quick and dirty" solution be 20
> moves? If the optimum solution is 12 moves would the quick and dirty
> sollution be 15 or 16? So my question is how consistent is the program?
>
> Perhaps Cube explorer could be used to determine wheter a particular
> scramble would favor HTM or STM.
>
> Cheers,
>
> David J
>
>
>
> --- In speedsolvingrubikscube@yahoogroups.com, "Per Kristen Fredlund"
> <aspiring_to_love@> wrote:
> >
> > Hi :-)
> >
> > On the practical side, it does take at least 30 minutes (depending
> > on the computer) to find optimal solution for a given scramble with
> > Cube Explorer. It quickly finds 20 or 19 move solution(s) but to
> > really find the shortest proven solution takes a good while. So i
> > don't think it's feasible to do this for a high number of scrambles.
> > And then again why check this only for 3x3x3? We could check also
> > for 4x4x4 or 5x5x5 or square-1 or megaminx ... I think it can be
> > seen where im going with this. Im not worried about getting easy
> > scrambles at competitions. I do want everyone to get same scrambles
> > (possibly with random orientations). That's the most feasible option
> > as i see it.
> >
> > (Has anyone really studied how optimal solution length compares
> > version scramble length for a large number of random scrambles of
> > different lenghths ?? Where do we stop getting "harder" scrambles by
> > making them longer?)
> >
> > About move metric,this is only really interesting for fewest moves.
> > And fewest movers have been dictated by Dan Harris (:-P) to follow
> > htm for a good while already. I don't see any problem with that.
> > Those who use inner slice turns heavily will feel punsihed by htm
> > andvice versa those who are used to thinking in outer turns only
> > will feel that those using inner slice turns a lot are favored by
> > stm. So we can't satisfy everyone either way.
> >
> > I see 3 options :
> > htm - because it's simply most common andmaybe the least
> > controversial,mathematical basis
> >
> > stm - from a practical point of view maybe the most correct metric,
> > a layer is a layer whether it's internal or external
> >
> > combined metric - a salomonic solution, noone should feel heavily
> > favoured or dis-favoured
> >
> > The latter does most likely require a computer to determine the
> > metric,and as discussed in another yahoo group it's not immune to
> > ambiguity ;-)
> >
> > Cheers!
> >
> > -Per
> >
> > > --- In speedsolvingrubikscube@yahoogroups.com, "Ron van Bruchem"
> > <ron@> wrote:
> > >
> > > Hi David,
> > >
> > > Thanks for the feedback.
> > > It is better to post your feedback on the WCA forum, because there
> > we can have a better overview of the discussions.
> > >
> > > I will put your suggestions on the TODO list for version 2007.
> > >
> > > For your suggestion of difficulty it is hard to define difficulty.
> > > If your suggestion is to check for in general at least 18 (or x)
> > moves optimal solution, then this is something we could do. Given a
> > majority.
> > > If your suggestion is to check for easy solutions in CFOP system,
> > then I think it is not a good idea. In that case you are influencing
> > randomness for a specific system.
> > >
> > > We have had several discussions about HTM and STM, but there was
> > never a conclusion. Therefore we kept it the way it is now.
> > >
> > > Have fun,
> > >
> > > Ron
> > >
> > >
> > >   ----- Original Message -----
> > >   From: d_j_salvia
> > >   To: speedsolvingrubikscube@yahoogroups.com
> > >   Sent: Monday, July 31, 2006 7:46 PM
> > >   Subject: [Speed cubing group] Re: WCA regulations version 2006
> > v2 is now official
> > >
> > >
> > >   Hi Ron,
> > >
> > >   I had no time to take part in the recent discussions, but I
> > would like
> > >   to mention two things for the future.
> > >
> > >   1.) Have a standard for difficulty.
> > >   I suggest that once scrambles are randomly generated for a
> > tournament
> > >   that they be run through a few cube solvers. This would allow
> > you to
> > >   eliminate those scrambles which are too easy.
> > >
> > >   2.) Please reconsider and change the rule counting slice turns
> > as two
> > >   moves. I request that this be changed to counting slice turns as
> > one
> > >   move.
> > >
> > >   I've always seen each layer as equals, even though a slice turn
> > >   doesn't move corners. As it is now you are penalising corners
> > first
> > >   solvers and others like me who use a lot of slice moves.
> > >
> > >   Cheers,
> > >
> > >   David J
> > >
> > >   --- In speedsolvingrubikscube@yahoogroups.com, "Ron" <ron@>
> > wrote:
> > >   >
> > >   > Fellow cubers,
> > >   >
> > >   > The new version of the WCA regulations is now official.
> > >   > It is version 2006 v2, and available at
> > >   > http://www.speedcubing.com/events/regulations.html
> > >   >
> > >   > We kindly request all translators of former versions to update
> > the
> > >   > regulations in other languages. If you are interested in
> > translating
> > >   > the WCA regulations to your language, then please let us know.
> > >   >
> > >   > Starting on July 21, 2006 all official WCA competitions must
> > follow
> > >   > this new version of the regulations.
> > >   >
> > >   > Thank you all for your feedback.
> > >   >
> > >   > It is still possible to give more feedback.
> > >   > Errors will be corrected as soon as possible.
> > >   > Other feedback will be taken into account for the 2007
> > version. This
> > >   > version is planned for December 2006.
> > >   >
> > >   > Happy cubing,
> > >   >
> > >   > Ron
> > >   >
> > >
> > >
> > >
> > >
> > >
> > > [Non-text portions of this message have been removed]
> > >
> >
>

Hey everyone,

      I am planning to attend US nationals on Saturday and Sunday, but
have never been to any cube related event before.  I was wondering if
it is generally possible to buy cubes from other cubers and/or
vendors.  I am specifically looking for a Square-1, a Rubik's Magic, a
Megaminx, a competition legal 2x2x2 cube, various Siamese/fused cubes,
and a Stackmat timer.  Would it be possible to purchase some or all of
these items at the event?  And if I register online, and then purchase
some cubes that I didn't have before, can sign up for additional
events during the registration time?

-Alexei Gousev

You may sign up for additional events at registration time.

-Tyson

On Aug 2, 2006, at 10:29 PM, agousev wrote:

> Hey everyone,
>
>  I am planning to attend US nationals on Saturday and Sunday, but
>  have never been to any cube related event before. I was wondering if
>  it is generally possible to buy cubes from other cubers and/or
>  vendors. I am specifically looking for a Square-1, a Rubik's Magic, a
>  Megaminx, a competition legal 2x2x2 cube, various Siamese/fused cubes,
>  and a Stackmat timer. Would it be possible to purchase some or all of
>  these items at the event? And if I register online, and then purchase
>  some cubes that I didn't have before, can sign up for additional
>  events during the registration time?
>
>  -Alexei Gousev
>
>
>

I'm getting to San Francisco on Friday around noon, so I'll get to the
Exploratorium by 1:30 or 2:00.  See you all there!

Tim Reynolds

Hey,

I want to wish everyone who is atteding the US Nationals the best of luck.
Unfortunately, I will
not be able to attend this weekend, but I hope to compete sometime soon. Have
fun!

Andy

http://andyscubepage.tk

WOOHOO!!!!!

I finally got my first sub-30 average fo one-handed solving. It also
is my first sub-29. :-) I'm really happy right now!! I'm gonna keep
practicing one-handed solving and see how good times I can get in the
future.

Average: 28.97 seconds
Individual Times: 29.31, 33.18, 25.37, 27.10, 31.74, 29.81, (34.43),
30.31, 26.09, 30.11, 26.71, (22.81)


/Gunnar Krig

How the heck do people do this one-handed? I tried it once and was
just so bad at it -- it's hard to not only grip the cube but to turn
faces at the same time.  Getting sub-30 times one-handed just seems
insane to me.  Great accomplishment!

Marcus Stuhr
Wharton & CAS '09
University of Pennsylvania

--- In speedsolvingrubikscube@yahoogroups.com, "Gunnar Krig"
<gunkr520@...> wrote:
>
> WOOHOO!!!!!
>
> I finally got my first sub-30 average fo one-handed solving. It also
> is my first sub-29. :-) I'm really happy right now!! I'm gonna keep
> practicing one-handed solving and see how good times I can get in the
> future.
>
> Average: 28.97 seconds
> Individual Times: 29.31, 33.18, 25.37, 27.10, 31.74, 29.81, (34.43),
> 30.31, 26.09, 30.11, 26.71, (22.81)
>
>
> /Gunnar Krig
>

Nice, Gunnar!
   One-handed cubing is a lot of fun

   my best avg is 26.07s, with a PB of 20.37s (almost sub-20 : ) )

   my next goal is to make a sub-26 avg, but that is very hard...still don't know
how Ryan does 24.xx avgs...haha

   Pedro

Gunnar Krig <gunkr520@...> escreveu:
           WOOHOO!!!!!

I finally got my first sub-30 average fo one-handed solving. It also
is my first sub-29. :-) I'm really happy right now!! I'm gonna keep
practicing one-handed solving and see how good times I can get in the
future.

Average: 28.97 seconds
Individual Times: 29.31, 33.18, 25.37, 27.10, 31.74, 29.81, (34.43),
30.31, 26.09, 30.11, 26.71, (22.81)

/Gunnar Krig






---------------------------------
  Novidade no Yahoo! Mail: receba alertas de novas mensagens no seu celular.
Registre seu aparelho agora!

[Non-text portions of this message have been removed]

I leave in 30 minutes.  I'll be in San Fran tonight around midnight.

You'll see me tomorrow at the comp, but if anybody feels compelled to
call me:

two hundred uno, double-8 nine, 4 sicks, sixtee niner

~ Bob

--- In speedsolvingrubikscube@yahoogroups.com, cmhardw <no_reply@...>
wrote:
>
> Hi everyone,
>
> I'm leaving in about 6.5 hours to San Francisco.  I'll arrive tomorow
> Thursday morning at 10:45am San Francisco time.
>
> My cell number is nine 1 niine sevn, five sseven, 7 tree sevnn for.  I
> also am worried about bots, so I'll follow Doug's example.
>
> I'll be meeting Doug and we'll be doing a bit of sight seeing until
> others arrive, and hopefully we can meet up with some other cubers who
> are in the area.
>
> See you all soon, and have a safe trip everyone!
> Chris
>

--- In speedsolvingrubikscube@yahoogroups.com, kyuubree <no_reply@...>
wrote:
>
> How the heck do people do this one-handed?

It's not that hard.

> I tried it once

And that's exactly your problem.

Stefan

Thanks, Bruce!

DJ


--- In speedsolvingrubikscube@yahoogroups.com, "Bruce Norskog"
<brnorsk@...> wrote:
>
> Hi,
>
> While optimal solvers (for 3x3x3) can take a long time to find the
> solution for a deep position, they can determine if a position is
> within 15 moves (HTM, aka FTM) of solved pretty fast, I believe. That
> may be good enough. As they are, they won't automatically stop after a
> certain depth, as far as I know. By the way, I believe Cube Explorer
> only does HTM metric (but I don't have the latest version).  Mike
> Reid's optimal solver can be compiled to do either QTM or HTM. It's
> not clear to me if you can estimate the optimal length very well based
> on a sub-optimal solver's solution length.
>
> I think the issue of "scramble quality" is more important on the
> 2x2x2. With less than 4 million positions, I think easy scrambles are
> likely to come up once in awhile, if there isn't something in place to
> prevent them. There is almost a third of one percent chance of a
> randomly chosen position to be depth five or less (HTM). I was in the
> room when the current WR 2x2x2 solve occurred. I don't know what the
> scramble was for that solve, though, although I assume I also had to
> solve the same scramble.
>
> - Bruce
> --- In speedsolvingrubikscube@yahoogroups.com, "d_j_salvia"
> <d_j_salvia@> wrote:
> >
> > Hi Per,
> >
> > Thanks for the info, as I am unfamiliar with Cube explorer.
> >
> > I think that you don't have to start the night before the contest to
> > test scrambles, couldn't this be started months in advance.
> >
> > You don't have to only use one computer to test scrambles.
> >
> > If Cube Explorer can quickly come up with a ballpark figure for
> > difficulty then that might be adequate. Can it? I mean if the optimum
> > solution is 18 moves, would the "quick and dirty" solution be 20
> > moves? If the optimum solution is 12 moves would the quick and dirty
> > sollution be 15 or 16? So my question is how consistent is the
program?
> >
> > Perhaps Cube explorer could be used to determine wheter a particular
> > scramble would favor HTM or STM.
> >
> > Cheers,
> >
> > David J
> >
> >
> >
> > --- In speedsolvingrubikscube@yahoogroups.com, "Per Kristen Fredlund"
> > <aspiring_to_love@> wrote:
> > >
> > > Hi :-)
> > >
> > > On the practical side, it does take at least 30 minutes (depending
> > > on the computer) to find optimal solution for a given scramble with
> > > Cube Explorer. It quickly finds 20 or 19 move solution(s) but to
> > > really find the shortest proven solution takes a good while. So i
> > > don't think it's feasible to do this for a high number of
scrambles.
> > > And then again why check this only for 3x3x3? We could check also
> > > for 4x4x4 or 5x5x5 or square-1 or megaminx ... I think it can be
> > > seen where im going with this. Im not worried about getting easy
> > > scrambles at competitions. I do want everyone to get same scrambles
> > > (possibly with random orientations). That's the most feasible
option
> > > as i see it.
> > >
> > > (Has anyone really studied how optimal solution length compares
> > > version scramble length for a large number of random scrambles of
> > > different lenghths ?? Where do we stop getting "harder"
scrambles by
> > > making them longer?)
> > >
> > > About move metric,this is only really interesting for fewest moves.
> > > And fewest movers have been dictated by Dan Harris (:-P) to follow
> > > htm for a good while already. I don't see any problem with that.
> > > Those who use inner slice turns heavily will feel punsihed by htm
> > > andvice versa those who are used to thinking in outer turns only
> > > will feel that those using inner slice turns a lot are favored by
> > > stm. So we can't satisfy everyone either way.
> > >
> > > I see 3 options :
> > > htm - because it's simply most common andmaybe the least
> > > controversial,mathematical basis
> > >
> > > stm - from a practical point of view maybe the most correct metric,
> > > a layer is a layer whether it's internal or external
> > >
> > > combined metric - a salomonic solution, noone should feel heavily
> > > favoured or dis-favoured
> > >
> > > The latter does most likely require a computer to determine the
> > > metric,and as discussed in another yahoo group it's not immune to
> > > ambiguity ;-)
> > >
> > > Cheers!
> > >
> > > -Per
> > >
> > > > --- In speedsolvingrubikscube@yahoogroups.com, "Ron van Bruchem"
> > > <ron@> wrote:
> > > >

Cubesmith.com ... stickers only cost $1.25 per set. Over thirty people
have told me to paint my cube. I didn't listen to a single person
because paint feels horrible, and it doesn't look nice... ever. No
matter what you do, a computer can do better than you at making solid
colors. Go Cubesmith.

-Tim

--- In speedsolvingrubikscube@yahoogroups.com, "Billy Gard"
<billygard@...> wrote:
>
> I painted my revenge cube. I took off the stickers and painte with
little jars of model paint, which has an acrylic-like toughness. I see
what you mean about it not looking very good. I guess if I had put
thin tape along all the edges it would have looked fine, as the paint
would form squares much like stickers, rather than covering pretty
much the whole face of the cubie.
>
> As for solving the pocket cube, I found from the "Jeff Conquers the
Cube in 45 Seconds" book about what he calls "correct edges". It is a
clever visual shortcut for corner placement (so that the four sides of
the layer are solid, i.e. correct edges). You have either no correct
edges, one correct edge, or all correct edges. If you have none, just
swap two diagonal corners. If you have one correct edge, swap the two
adjacent corners across from it. That's it. By the way, when doing the
second layer the same way, remember that in 2*2*2 speak, a double
parallel swap is equivalent in effect to a diagonal swap, and a 3-
cycle is equivalent to an adjacent swap.
>
> Billy
>
>
> [Non-text portions of this message have been removed]
>

Hi,

I've made it to San Francisco.  Unfortunately, my luggage did not.
I'll be just hanging out at my hotel tonight, waiting to see if my
luggage shows up. Fortunately, most critical things I carried on. I
may attempt to get to Exploratorium by 9AM.

  - Bruce

--- In speedsolvingrubikscube@yahoogroups.com, "Tim Reynolds"
<timothy.reynolds2@...> wrote:
>
> I'm getting to San Francisco on Friday around noon, so I'll get to the
> Exploratorium by 1:30 or 2:00.  See you all there!
>
> Tim Reynolds
>

Hey guys, I was wondering if it would be possible to create a cube
explorer like program that will solve cubes with undefined points in
the configuration, yet it can still identify isomorphic combinations!

That would be awsome.

Also, If you create a static reference, the cases are reduced
significantly, I just compiled 132 roux 2nd Block cases.  Yes, roux
is supposed to be intuitive.  But, there are just hard cases,
especially to do using the URM subset.  All the cases I generated
did this, =D very nice.

Anyway, Figure out what is isomorphic (any cube state with the same
moves whether inversed, reflected, or applied from a different angle)
Since we don't have a good hybrid of ACube and CubeExplorer, just go
over the tables you've generated and recognize the cases for
yourself.

Btw, the benifit to 2-gen F2B in roux is that you can identify the
CMLL permutation case while permuting the last C/E pair.

Also I'm curious what is your method for?  What do the steps involve?

I've worked with the Acube a lot lately, generating everything from
my new BLD algs for a new method, getting faster btw at execution.
Not too far from my goal.  Anyway, BLD algs, Roux F2B algs, a set of
24 algs that would permute the corners and orient edges on 4th c/e
pair insertion.

The roux algs are by far the biggest group, my BLD algs are the
deepest, and the C/E pair (Permute/Orient) algs.

Like I said have a reference point, such as in F2L algs, you align
the Corner above the C/E slot to recognize the case.

--- In speedsolvingrubikscube@yahoogroups.com, d_funny007
<no_reply@...> wrote:
>
> No, in some cases (I think yours would apply), you should look
> for "diagonal mirroring". Although the simple mirroring plus U
> rotations *might* be enough/analgous/equivalent, but I have put
> little thought into this as I am currently on vacation!
>
> As a long time member of this fourm, I'd like to say that it is
very
> good to see another hardcore math/cs person like Bruce here! I've
> been keeping up with his posts on this other fourm he uses too.
Very
> techincal stuff that I once wanted to see here, but after further
> thought, it just wouldn't fit here. I was always the one rushing
to
> answer math questions, but I wasn't particularly patient in the
past :
> (.
>
> I can try a verification of his computation when I get the chance.
It
> is most challenging :).
>
>
> -Doug
>
>
> --- In speedsolvingrubikscube@yahoogroups.com, "athefre"
> <athefre@> wrote:
> >
> > Thanks, 111 is better than 140, but not much.
> >
> > If you could reduce the number using mirrors and inverses, how
much
> > would it be?  If you don't mind.  I've been working hard for a
> month
> > trying to perfect everything so I can get to work on finding the
> > algorithms for the idea I choose.
> >
> > Inverse = backwards
> > Mirror = LUL'ULU2L' is the mirror of R'U'RU'R'U2R
> >
> > Correct?
> >
> > --- In speedsolvingrubikscube@yahoogroups.com, "Bruce Norskog"
> > <brnorsk@> wrote:
> > >
> > > Hi,
> > >
> > > Yes, you're right. I considered rotations of the E layer, but
not
> > more
> > > complicated adjustment moves like R2 E R2. If you allow that,
> then
> > the
> > > middle multipliers in my table all become 1, and you can just
> > multiply
> > > the first and third number. With that, my 140 cases (excluding
the
> > > do-nothing case) get reduced to 111 cases. (I think I did the
> > > arithmetic correctly.) Again, I haven't looked at using
mirrors
> and
> > > inverses to reduce the number of algorithms further.
> > >
> > > Sorry, it looks like my table's formatting wasn't preserved,
at
> > least
> > > if viewed from the Yahoo web site. You would think the Preview
> > button
> > > would actually show you what your post was going to look like,
> > > wouldn't you? In Preview, it looked fine, but the actual post
> > appears
> > > to have all "redundant" space characters stripped out.
> > >
> > >  - Bruce
> > > --- In speedsolvingrubikscube@yahoogroups.com, "athefre"
> <athefre@>
> > > wrote:
> > > >
> > > > Thanks.  All of what you said sounds right.  But there is
one
> > thing
> > > > I'm not sure if you considered that I may have looked over
in
> > your
> > > > post.
> > > >
> > > > What about the "empty spaces" available in E for the cases
> where
> > 2 E
> > > > edges need to be placed?  Like, if you have an empty space
at
> FR
> > and
> > > > BR or you can have the spaces at FR and BL (although you
could
> do
> > > > R2ER2 before the algorithm).
> > > >
> > > > If it really is 140 cases then that is WAY too many for me
to
> > make
> > > > and learn.  I'm definitly going with my other option.
> > > >
> > > > --- In speedsolvingrubikscube@yahoogroups.com, "Bruce
Norskog"
> > > > <brnorsk@> wrote:
> > > > >
> > > > > Hi,
> > > > >
> > > > > From what I understand, you have 4 corner cubies in the U
> layer
> > to
> > > > be
> > > > > put into correct relative order (orientation doesn't
matter).
> > You
> > > > have
> > > > > 10 edges that can be permuted around without changing
> > orientation.
> > > > Of
> > > > > those 10 edges, 4 are E-layer edges which can be considered
> > > > > indistinguishable from each other. These E-layer edges are
all
> > > > > required to end up in the E layer. The other set of 6
edges
> can
> > also
> > > > > be considered to be indistinguishable from each other. The
U
> > layer
> > > > can
> > > > > be rotated before (and after, if you want the corners
> correctly
> > > > placed
> > > > > relative to the center) the algorithm. Likewise, the E
layer
> > can be
> > > > > rotated before and after the algorithm. (Rotating after to
> get
> > the
> > > > > E-layer centers back into correct position, if needed.)
> > > > >
> > > > > So to count the different cases you can have, consider the
> > different
> > > > > cases of where the E-layer edges can be, and count the
cases
> > for
> > > > each
> > > > > of the possible corner permutation situations (no swap,
swap 2
> > > > > adjacent, swap to diagonally opposite). First break down
the
> > edge
> > > > > cases by how many might be in each layer. For each
possible
> > number
> > > > of
> > > > > E-layer edges in each of the layers, determine the number
of
> > cases
> > > > > possible for each of the corner permutation situations.
> > > > >
> > > > > Then build a table of all the possibilities:
> > > > >
> > > > > (best viewed using fixed-width font)
> > > > >
> > > > > U-E-D      no swap      adj. swap    diag. swap
> > > > > -----      -------      ---------    ----------
> > > > > 4 0 0      1*1*1 =  1   1*1*1 =  1   1*1*1 =  1
> > > > > 3 1 0      1*1*1 =  1   4*1*1 =  4   2*1*1 =  2
> > > > > 3 0 1      1*1*2 =  2   4*1*2 =  8   2*1*2 =  4
> > > > > 2 2 0      2*2*1 =  4   6*2*1 = 12   4*2*1 =  8
> > > > > 2 1 1      2*1*2 =  4   6*1*2 = 12   4*1*2 =  8
> > > > > 2 0 2      2*1*1 =  2   6*1*1 =  6   4*1*1 =  4
> > > > > 1 3 0      1*1*1 =  1   4*1*1 =  4   2*1*1 =  2
> > > > > 1 2 1      1*2*2 =  4   4*2*2 = 16   2*2*2 =  8
> > > > > 1 1 2      1*1*1 =  1   4*1*1 =  4   2*1*1 =  2
> > > > > 0 4 0      1*1*1 = (1)  1*1*1 =  1   1*1*1 =  1
> > > > > 0 3 1      1*1*2 =  2   1*1*2 =  2   1*1*2 =  2
> > > > > 0 2 2      1*2*1 =  2   1*2*1 =  2   1*2*1 =  2
> > > > >                   ---          ---          ---
> > > > >                    25           72           44
> > > > >
> > > > > So I get 25+72+44 = 141 cases. The 1 in parentheses in the
> table
> > > > > indicates the case where no algorithm needs to be
performed.
> So
> > if
> > > > you
> > > > > don't count that case, then I get 140.
> > > > >
> > > > > I have not considered the diagonal symmetry in the above,
but
> > then,
> > > > I
> > > > > understand you were not asking for that to be taken into
> > > > consideration.
> > > > >
> > > > > I just thought I would add my own comments about the edge
> > > > orientation
> > > > > issue.
> > > > >
> > > > > I agree with Doug in that the key in what you said was
that F
> > and B
> > > > > moves flip four edges.
> > > > >
> > > > > From that I assume you mean, that to be oriented:
> > > > >  - an edge cubie that belongs in the M or S slice, and is
> > currently
> > > > > located in one of those slices,  must have its U or D
facelet
> > > > aligned
> > > > > with the U or D center
> > > > >  - an edge cubie that belongs in the M or S slice, and is
> > located in
> > > > > the E slice, must have it U or D facelet aligned with the
F
> or
> > B
> > > > center.
> > > > >  - an edge cubie that belongs in the E slice, and is
located
> in
> > the
> > > > E
> > > > > slice, must have its F or B facelet aligned with the F or
B
> > center
> > > > (or
> > > > > equivalently, its R or L face aligned with the R or L
center)
> > > > >  - an edge cubie that belongs in the E slice, and is
located
> in
> > the
> > > > M
> > > > > or S slice, must have its F or B face aligned with the U
or D
> > > > center.
> > > > >
> > > > > When an edge is in the inner slice that it belongs to, its
> > usually
> > > > > assumed that the edge would be oriented if each of its
> facelets
> > is
> > > > > aligned with the same color center, or the center that is
> > opposite
> > > > > that center. (Someone could define edge orientation in a
way
> > such
> > > > that
> > > > > the above would not be the case, but I would say this is
> rare.)
> > But
> > > > > when an edge is moved to a different inner slice than the
one
> it
> > > > > belongs in, it is not generally as clear what it means to
be
> > > > oriented.
> > > > >
> > > > > Doug mentioned a way of defining edge orientation such
that
> > moving L
> > > > > or R a quarter-turn flips four edges. There is yet another
> way
> > of
> > > > > defining edge orientation that I have used in computer
> analyses
> > of
> > > > the
> > > > > cube. You can define edge orientation such that moving any
of
> > the
> > > > > layers U, D, L, R, F, or B a quarter-turn flips all four
> edges
> > > > moved.
> > > > > This is the most symmetrical way of defining edge
> orientation.
> > But
> > > > > define edge orientation in the way that makes the most
sense
> > for
> > > > your
> > > > > situation. With your way, you can keep all edges oriented
> > simply by
> > > > > avoiding F, F', B, and B' moves (F2 and B2 okay, of
course).
> > > > >
> > > > >  - Bruce
> > > > > --- In speedsolvingrubikscube@yahoogroups.com, "athefre"
> > <athefre@>
> > > > > wrote:
> > > > > >
> > > > > > Yeah, it was supposed to say "DFL".
> > > > > >
> > > > > > I don't really understand or know anything about
inverses
> and
> > > > mirrors
> > > > > > and symmetry and all of that crazy stuff but hopefully
this
> > helps:
> > > > > >
> > > > > > -Add in the inverses the stuff like that but tell me how
> many
> > > > > > distinct cases there are with those included and without.
> > > > > >
> > > > > > -Don't count U adjustments.  I don't mind having to
adjust
> U
> > > > before
> > > > > > doing an algorithm.
> > > > > >
> > > > > > So far I'm thinking it's around 102.  If so, no way.
I'm
> > going
> > > > with
> > > > > > my other option.  This is what I've been counting:
> > > > > >
> > > > > > Already permuted: 17 cases
> > > > > > Diagonal swap: 18 cases (1 for E edges already in E)
> > > > > > Adjacent swap: 69 cases (same as above)
> > > > > >
> > > > > > Is there a site that describes these kinds of things?
> > > > > >
> > > > > >
> > > > > > --- In speedsolvingrubikscube@yahoogroups.com,
d_funny007
> > > > > > <no_reply@> wrote:
> > > > > > >
> > > > > > > That does help. Actually I use a different EO
> definition...
> > I
> > > > treat
> > > > > > > L and R as flipping 4 edges.
> > > > > > >
> > > > > > > Also, could you double check this: "The algorithm must
not
> > > > > > > mess up UFL, DL, DBL, DB, DBR, or DFR." It doesn't
feel
> > right.
> > > > Are
> > > > > > > you sure you don't mean 'DFL' there? Also what would
you
> > count
> > > > as a
> > > > > > > distinct case? I could group diagonally-symmetric
cases
> as
> > one.
> > > > I
> > > > > > > could even group cases that use inverse algorithms
> > together. If
> > > > U
> > > > > > > layer is not free for the first turn, than you could
get
> > what I
> > > > > > like
> > > > > > > to think of as a single case counted 4 times.
> > > > > > >
> > > > > > > This question sounds familiar, like I've already heard
> > > > something
> > > > > > > similar before, but it is definately a hard one and
may
> > take
> > > > some
> > > > > > > time.
> > > > > > >
> > > > > > >
> > > > > > > -Doug
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > --- In
speedsolvingrubikscube@yahoogroups.com, "athefre"
> > > > > > > <athefre@> wrote:
> > > > > > > >
> > > > > > > > I'm not too sure what you mean, but I'm using yellow
on
> > top,
> > > > blue
> > > > > > > on
> > > > > > > > the right, orange in the front.  All of the yellow
and
> > white
> > > > > > edges
> > > > > > > > face the white or yellow center (it doesn't matter)
and
> > all
> > > > of
> > > > > > the
> > > > > > > > blue and green edges are facing the blue or green
> > centers.
> > > > It's
> > > > > > > like
> > > > > > > > Petrus, the edges are oriented that way, and if you
do
> F
> > or B
> > > > it
> > > > > > > > messes up 4 edges.
> > > > > > > >
> > > > > > > > Does that help.
> > > > > > > >
> > > > > > > > --- In
speedsolvingrubikscube@yahoogroups.com, "Stefan
> > > > Pochmann"
> > > > > > > > <pochmann@> wrote:
> > > > > > > > >
> > > > > > > > > --- In
> > speedsolvingrubikscube@yahoogroups.com, "athefre"
> > > > > > > <athefre@>
> > > > > > > > > wrote:
> > > > > > > > > >
> > > > > > > > > > All edges on the cube are already oriented
before
> > going
> > > > to
> > > > > > this
> > > > > > > > > > step.
> > > > > > > > >
> > > > > > > > > There's no general definition for orientation so
you
> > need
> > > > to
> > > > > > > > provide
> > > > > > > > > one.
> > > > > > > > >
> > > > > > > >
> > > > > > >
> > > > > >
> > > > >
> > > >
> > >
> >
>