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Re: Is there a way to figure the number of cases for a step?

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  • Daniel Beyer
    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
    Message 1 of 17 , Aug 3, 2006
      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.
      > > > > > > > > >
      > > > > > > > >
      > > > > > > >
      > > > > > >
      > > > > >
      > > > >
      > > >
      > >
      >
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