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Re: [Speed cubing group] Re: Please me with the Rubik's cube group elements order problem!

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  • Bart
    Where can I find a site on cube math? What s a corner cycle? What s parity?
    Message 1 of 29 , Jul 2, 2008
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      Where can I find a site on cube math? What's a corner cycle? What's parity?
    • a_ooms75
      ... Rubik s ... take ... it. ... with ... Could ... grateful. ... R F ... you ... here a site where you can fillin your 1260 alg
      Message 2 of 29 , Jul 2, 2008
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        --- In speedsolvingrubikscube@yahoogroups.com, "danix800"
        <danix800@...> wrote:
        >
        > I'm a big rubik's cube fan. It's well know that for a normal
        Rubik's
        > cube the cube group has its maximum element
        > order 1260, say, the maneuver UR'UF'D2 needs 1260 repeations to
        take
        > the cube back
        > to its original configuration. But the problem is, I can't prove
        it.
        > I've searched the web
        > extensively, no results! I guess the problem has something to do
        with
        > the cycles of the
        > edges or corners. And the cubie orientations are also important.
        Could
        > you please give me
        > some introductions or maybe whatever helps. I would be very
        grateful.
        > Also, some people around me said, for the picture cube, the
        > maximum order changes
        > to at least 1980, still, he can't prove it. Here's an example, F'
        R' F
        > D' U' F' L' D2 L' B L B2 R' D2 F',
        > it takes 990 times to recover the normal cube, but for the picture
        > cube, we need to recover
        > the orientations of the centerpieces, so it rises to 1980. Could
        you
        > say something about this?
        > Any help would be appreciated!
        >

        here a site where you can fillin your 1260 alg
        http://www.solvethecube.co.uk/tools.html and then go to order

        greets
        ~A.O.
      • mackymakisumi
        ... But I don t need the definition of an arbitrary correct orientation. A pure orientation of cubies is any equivalence class of legal sequences acting on
        Message 3 of 29 , Jul 2, 2008
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          --- In speedsolvingrubikscube@yahoogroups.com, "Stefan Pochmann"
          <stefan.pochmann@...> wrote:
          > Incomplete because you rely on definitions of orientations that you
          > omit. Nonobvious because it's not at all clear to me that your
          > statements are correct. And cumbersome compared to sticker cycles.

          But I don't need the definition of an arbitrary "correct" orientation.
          A pure orientation of cubies is any equivalence class of legal
          sequences acting on the Rubik's Cube by twisting/flipping any number
          of cubies in place, not relative to predefined orientation. And Bruce
          provided even less detail than me. Maybe the sticking point is the
          distinction between the cubie itself and the cubie position?

          More details:
          View the cube group elements as equivalence classes of legal sequences
          of moves on a Rubik's Cube, with the obvious equivalence relation
          (disregarding rotated centers), so that the cube group act on a
          Rubik's Cube in the obvious way. Any cube group element can be
          represented by a pure permutation of cubie _positions_ followed by a
          pure orientation of cubie _positions_. (That's clear enough. Don't
          make me write out the components of the semidirect product and exhibit
          an isomorphism.) By abuse of terminology, we will say that the cube
          group element _is_ a pure permutation followed by a pure orientation.
          Associate the permutation subgroup with a subgroup of S_12 X S_8 and
          write out the disjoint cycles in each direct product component. Take
          an n-cycle in this decomposition. Restrict the action of the cube
          group element to the cubies moved by this cycle. The restriction is
          now written as a pure permutation of n cubie positions followed by a
          pure orientation of the n cubie positions.

          Starting at any position on the cycle, number the positions 1 through
          n along the cycle. An n-cycle (1, ..., n) sends the cubie in position
          1 to position 2, etc. An orientation of n-cycle (1, ..., n) is a
          n-tuple (t_1, ..., t_n), where t_i is 0 if the corner in position i is
          left untwisted, 1 if twisted clockwise, and 2 if twisted
          counter-clockwise, where the cycle is written as a pure permutation
          followed by a pure orientation. The sum of orientations of the n-cycle
          is t_1 + ... + t_n. For example, a 3-cycle of edges (e_1, e_2, e_3,
          e_4) with orientation (1, 1, 1, 0) looks as follows, where "e_1 e_2
          e_3 e_4," for example, means that e_1 is in position 1, etc. We keep
          track of flipped edges using *.

          Starting configuration: e_1 e_2 e_3 e_4 (after renumbering, if necessary)
          Permute: e_4 e_1 e_2 e_3. Flip: e_4* e_1* e_2* e_3.
          Permute: e_3 e_4* e_1* e_2*. Flip: e_3* e_4 e_1 e_2*.
          Permute: e_2* e_3* e_4 e_1. Flip: e_2 e_3 e_4* e_1.
          Permute: e_1 e_2 e_3 e_4*. Flip: e_1* e_2* e_3* e_4*.

          Note that each cubie has been flipped three times, once each time it
          was in positions 1 through 3. The argument I provided below (a) and
          (b) should make more sense now.

          > In essence, you're checking every single piece of a cycle against an
          > unspecified artificial arbitrary orientation scheme, instead of
          > simply checking the cycle once as a whole for orientation in an
          > obvious natural way.

          An "obvious natural" way? You're such a Pochmann! ;-) Sticker cycles
          alone don't give you enough structure to figure out the maximum bound,
          though.

          -macky
        • d_funny007
          At first I didn t think that had anything to do with this proof, but I realize you can put exponents on sequences with that tool. It only suffices for
          Message 4 of 29 , Jul 2, 2008
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            At first I didn't think that had anything to do with this proof, but I
            realize you can put exponents on sequences with that tool.

            It only suffices for showing/proving that 1260 is a lower-bound. This
            is the existence claim/assertion Macky made. But to do this you have
            to also selectively try certain factors of it, by choosing from its
            prime factorization.

            Between Bruce and Macky they've got all the abstract math questions
            covered! I'm impressed. Good reads.


            > here a site where you can fillin your 1260 alg
            > http://www.solvethecube.co.uk/tools.html and then go to order
            >
            > greets
            > ~A.O.
            >
          • Stefan Pochmann
            ... Maybe, though I m not sure what you mean by that. But I think I understand you now (but whoa was reading your message hard). Earlier ... Only now do I
            Message 5 of 29 , Jul 2, 2008
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              --- In speedsolvingrubikscube@yahoogroups.com, "mackymakisumi"
              <mackymakisumi@...> wrote:
              >
              > Maybe the sticking point is the distinction between the cubie
              > itself and the cubie position?

              Maybe, though I'm not sure what you mean by that. But I think I
              understand you now (but whoa was reading your message hard). Earlier
              you wrote:

              > (a) An n-cycle of edges has order
              > n if an even number of edges are flipped
              > 2n if an odd number of edges are flipped

              Only now do I realize that you mean "are flipped" in the sense of
              "get flipped", i.e., the process rather than the result. I had
              misunderstood this, thinking about your orientation scheme for
              blindcubing where you determine which edges "are flipped".

              But I should've realized that that interpretation doesn't make sense
              here. Sloppy thinking by me, I apologize and walk away in shame.
              Although, I really recommend writing "get" instead of "are" to make
              it clearer. I believe I've come across this subtlety a few times
              myself when writing something and intentionally chose "get" to hint
              towards change rather than state.

              > View the cube group elements as equivalence classes of legal
              > sequences of moves on a Rubik's Cube, with the obvious equivalence
              > relation (disregarding rotated centers), so that the cube group act
              > on a Rubik's Cube in the obvious way.

              Not trying to nitpick or poke holes, but can you explain this a bit?
              It's not obvious to me at all what that relation is supposed to be. I
              pretty much understand the rest of your message, but this part has me
              totally stumped, and I'd like to understand it.

              > Sticker cycles alone don't give you enough structure to figure out
              > the maximum bound, though.

              Yeah, I only meant it for the lower level part of the proof, i.e.,
              the 2n/3n statement. And now that I understand what you mean, there's
              no big difference anymore. I was really acting noob, sorry again.

              Cheers!
              Stefan
            • Bob Burton
              you don t want to know about these things. possession of such knowledge may result in spending copious amounts of money on cheap plastic. ... What s parity?
              Message 6 of 29 , Jul 2, 2008
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                you don't want to know about these things. possession of such
                knowledge may result in spending copious amounts of money on cheap
                plastic.

                --- In speedsolvingrubikscube@yahoogroups.com, Bart <banaticus@...> wrote:
                >
                > Where can I find a site on cube math? What's a corner cycle?
                What's parity?
                >
              • Stefan Pochmann
                ... Earlier ... Alright, I think I just double-confused myself. I did have your blindcubing orientation scheme in mind, but actually only as part of the
                Message 7 of 29 , Jul 2, 2008
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                  --- In speedsolvingrubikscube@yahoogroups.com, "Stefan Pochmann"
                  <stefan.pochmann@...> wrote:
                  >
                  > --- In speedsolvingrubikscube@yahoogroups.com, "mackymakisumi"
                  > <mackymakisumi@> wrote:
                  > >
                  > > Maybe the sticking point is the distinction between the cubie
                  > > itself and the cubie position?
                  >
                  > Maybe, though I'm not sure what you mean by that. But I think I
                  > understand you now (but whoa was reading your message hard).
                  Earlier
                  > you wrote:
                  >
                  > > (a) An n-cycle of edges has order
                  > > n if an even number of edges are flipped
                  > > 2n if an odd number of edges are flipped
                  >
                  > Only now do I realize that you mean "are flipped" in the sense of
                  > "get flipped", i.e., the process rather than the result. I had
                  > misunderstood this, thinking about your orientation scheme for
                  > blindcubing where you determine which edges "are flipped".

                  Alright, I think I just double-confused myself. I did have your
                  blindcubing orientation scheme in mind, but actually only as part of
                  the picture. I did understand your "are flipped" as change rather
                  than state, but change *according to* a certain fixed orientation
                  scheme like you use for blindcubing, where you say an edge is or
                  isn't flipped if it moves from here to there. It just wasn't clear to
                  me that you meant it in a more general way where each piece "carries
                  its own orientation" with itself, so to speak. I'm not sure I've
                  thought about orientations this way before. I might've, but right now
                  I'm in a thoroughly confused state of mind.

                  Cheers!
                  Stefan
                • Bart
                  Last I checked, you could buy a Rubik s Cube for less than $20, including shipping to everywhere. If you wanted to put it together yourself, you can get them
                  Message 8 of 29 , Jul 2, 2008
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                    Last I checked, you could buy a Rubik's Cube for less than $20,
                    including shipping to everywhere. If you wanted to put it together
                    yourself, you can get them for less than half that price. I think
                    it's fairly easy to see the incredible price gulf between that and V
                    Cubes (at the previously quoted amount of $80 + shipping).

                    2008/7/2 Bob Burton <rubikscubewhiz@...>:
                    > you don't want to know about these things. possession of such
                    > knowledge may result in spending copious amounts of money on cheap
                    > plastic.
                    >
                    > --- In speedsolvingrubikscube@yahoogroups.com, Bart <banaticus@...> wrote:
                    >>
                    >> Where can I find a site on cube math? What's a corner cycle?
                    > What's parity?
                    >>
                    >
                    >
                    >
                    > ------------------------------------
                    >
                    > Yahoo! Groups Links
                    >
                    >
                    >
                    >
                  • Stefan Pochmann
                    Ok, I read that proof again, paying more attention. Improvement ideas below. ... edge ... You could shorten this part by saying that even the product-
                    Message 9 of 29 , Jul 2, 2008
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                      Ok, I read that proof again, paying more attention. Improvement ideas
                      below.

                      --- In speedsolvingrubikscube@yahoogroups.com, "mackymakisumi"
                      <mackymakisumi@...> wrote:
                      >
                      > A cycle with order divisible by 11 must be an 11-cycle of edges,
                      > leaving a 1-cycle. The total order for edges is maximized by giving
                      > both cycles an odd number of edge flips: lcm(22,2)=22. Since this
                      edge
                      > permutation is odd, the restriction on permutation parity forces an
                      > even corner permutation. We can easily list all distinct cycle types
                      > of A_8:
                      > Only even cycles: 1,1,1,1,1,1,1,1 3,1,1,1,1,1 5,1,1,1 7,1
                      > 3,3,1,1 5,3
                      > Two odd cycles: 2,2,1,1,1,1 4,2,1,1 6,2 4,4
                      > Two odd cycles and even cycle(s): 3,2,2,1
                      > Four odd cycles: 2,2,2,2.
                      >
                      > Ignoring the restriction on corner orientation, the total order for
                      > corners with cycle type l_1,...,l_k is maximized by multiplying each
                      > length by 3, according to (b). This at most lcm(3l_1, ...,
                      > 3l_k)=3lcm(l_1, ..., l_k), which takes the maximum 45 for cycle type
                      > 5,3. Thus these cube group elements have order at most lcm(22,45) =
                      > 2*3^2*5*11 = 990 < 1260.

                      You could shorten this part by saying that even the product-
                      maximizing partition of the eight corners, which is (8/e)^e=18.8,
                      isn't enough to reach the necessary (1260/22)/3=19.09.

                      For the product-maximizing partition:
                      http://mathforum.org/library/drmath/view/52441.html
                      http://www.jstor.org/pss/2690532

                      > Case 2: One factor of 3
                      > We must show that there cannot be four factors of 2. This means a
                      > flipped 8-cycle, which forces corners 7 and 5. The 3 must come from
                      > a 3-cycle of the edges, but again this violates the permutation
                      > parity.
                      > Done.

                      Aren't you already done at the point where you need 12 corners?

                      Cheers!
                      Stefan
                    • mackymakisumi
                      ... Nice! ... Haha, yeah. ... I m not sure how much you know about equivalence relation or group action, so I ll try to be as detailed as I can. Sorry if any
                      Message 10 of 29 , Jul 2, 2008
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                        --- In speedsolvingrubikscube@yahoogroups.com, "Stefan Pochmann"
                        <stefan.pochmann@...> wrote:
                        > You could shorten this part by saying that even the product-
                        > maximizing partition of the eight corners, which is (8/e)^e=18.8,
                        > isn't enough to reach the necessary (1260/22)/3=19.09.

                        Nice!

                        > > Case 2: One factor of 3
                        > > We must show that there cannot be four factors of 2. This means a
                        > > flipped 8-cycle, which forces corners 7 and 5. The 3 must come from
                        > > a 3-cycle of the edges, but again this violates the permutation
                        > > parity.
                        > > Done.
                        >
                        > Aren't you already done at the point where you need 12 corners?

                        Haha, yeah.

                        > > View the cube group elements as equivalence classes of legal
                        > > sequences of moves on a Rubik's Cube, with the obvious equivalence
                        > > relation (disregarding rotated centers), so that the cube group act
                        > > on a Rubik's Cube in the obvious way.
                        >
                        > Not trying to nitpick or poke holes, but can you explain this a bit?
                        > It's not obvious to me at all what that relation is supposed to be. I
                        > pretty much understand the rest of your message, but this part has me
                        > totally stumped, and I'd like to understand it.

                        I'm not sure how much you know about equivalence relation or group
                        action, so I'll try to be as detailed as I can. Sorry if any part
                        bores you.

                        We begin with the free group G on six letters, U, D, R, L, F, B. Its
                        elements are any finite products of these six letters and their
                        inverses (denoted U', etc, for convenience), and since I haven't
                        specified that these are actually moves on a Rubik's Cube, any two
                        distinct "words" (e.g. RRLL and LLRR; the identity, RRRR, R'R'R'R',
                        RRRRRRRR, etc) are considered distinct in this group ("a free group
                        has no relation"). The group operation is concatenation, or stringing
                        together the words one after another. The groups axioms are easily
                        checked.

                        Definition (based on Wiki http://en.wikipedia.org/wiki/Group_action):
                        If G is a group and X is a set, then a group action of G on X is a
                        function from X x G (all ordered pairs (x, g), where x is in X and g
                        is in G) into X, where x*g denotes the image of (x, g), that satisfies
                        the following two axioms:
                        1. x*(gh) = (x*g)*h for all g, h in G and x in X
                        2. x*e = x for every x in X (where e denotes the identity of G)
                        We say that G acts on X.
                        [Wiki uses left group action and writes x*g. Right group action works
                        better for us. Also, you usually use a dot instead of an *.]

                        Let X be the set of all legal configuration of the cube, with the
                        centers fixed. Then the free group G constructed above acts on X if we
                        assign the six letters their normal meanings as moves on a cube.
                        Intuitively, any finite sequence g of those six letters and their
                        inverses corresponds to a sequence of moves on the cube, and so we can
                        define x*g as the legal cube configuration that results when we apply
                        the moves corresponding to g to the legal configuration x. Axioms 1
                        holds as the group operation in G is concatenation (note that this
                        wouldn't have worked if we had used left group action instead) and
                        axiom 2 is obvious since the identity of G is the empty word,
                        corresponding to doing nothing.

                        Definition: An equivalence relation is a binary relation ~ on some set
                        G that satisfies the following for all a, b, and c in G.
                        Reflexivity: a ~ a for all a in G
                        Symmetry: If a ~ b, then b ~ a
                        Transitivity: If a ~ b and b ~ c, then a ~ c.
                        The equivalent class of a under ~, denoted [a], is the subset of G
                        whose elements b satisfy a ~ b.
                        When G is a group, ~ is said to be G-invariant if whenever g ~ u and h
                        ~ v, we have gh ~ uv.

                        Two examples of equivalence relations:
                        Equality on any set
                        Congruence modulo n on the integers.

                        It's easy to show that an equivalence relation on set G partitions G
                        into equivalent classes; that is, each element of G is in exactly one
                        equivalent class. For example, in the second example, there are n
                        congruence classes, [0] through [n-1], and every integer is congruent
                        to exactly one of 1 through n-1 mod n.

                        The idea is to introduce some equivalent relation on our free group G
                        so as to "collapse" it to the cube group. We want consider all
                        sequences with the same effect on the solved cube to be the same in
                        the cube group. So we use the action of G on X to define our
                        equivalence relation "~":

                        For g, h in G, g ~ h if and only if (solved)*g = (solved)*h, where
                        (solved) is the solved configuration in X.

                        It's easy to see that this is an equivalence relation. Moreover,

                        Claim 1: ~ is G-invariant.
                        Proof: This is easy to see for our example.

                        The set of equivalence classes, denoted by G/~, now has the desired
                        set structure of the cube group. It remains to place a group structure
                        by defining a group operation between two equivalent classes of G/~.
                        Again, we use the group action:

                        Claim 2: Let [g], [h] be equivalent classes in G/~ (so g, h are in G).
                        The binary operation [g][h] = [gh] makes G/~ into a group.
                        Proof: We first need to show that the operation is well defined. That
                        is, whatever elements of G we pick from the equivalent classes [g] and
                        [h], the operation returns the same equivalent class [gh]. But this is
                        precisely the G-invariance of ~; if g ~ u and h ~ v, meaning [g] = [u]
                        and [h] = [v], then gh ~ uv, so [gh] = [uv]. Closure, associativity,
                        and the existence of identity and inverse follow from the respective
                        properties of the group operation in G.

                        Introducing an equivalent relation is a standard way of "collapsing" a
                        group. The other, related, method uses quotient groups, which need a
                        bit more development. G-invariance is related to the concept of normal
                        subgroups and is the exact condition needed for us to be able to
                        induce the group structure on the smaller set G/~. It's a pretty
                        strong condition, but using action made it very easy in our example.

                        > Although, I really recommend writing "get" instead of "are" to make
                        > it clearer. I believe I've come across this subtlety a few times
                        > myself when writing something and intentionally chose "get" to hint
                        > towards change rather than state.

                        OK, I can see the source of confusion.

                        > And now that I understand what you mean, there's
                        > no big difference anymore. I was really acting noob, sorry again.

                        No problem! I know you're a very reasonable person, and it's cool that
                        you want to understand the little details.

                        -macky
                      • mackymakisumi
                        Right before Claim 2, skip Again, we use the group action. -macky
                        Message 11 of 29 , Jul 2, 2008
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                          Right before Claim 2, skip "Again, we use the group action."

                          -macky

                          > Again, we use the group action:
                          >
                          > Claim 2: Let [g], [h] be equivalent classes in G/~ (so g, h are in G).
                        • Tyson Mao
                          I really don t think it s a linear scale. ... [Non-text portions of this message have been removed]
                          Message 12 of 29 , Jul 2, 2008
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                            I really don't think it's a linear scale.

                            On Wed, Jul 2, 2008 at 5:00 PM, Bart <banaticus@...> wrote:

                            > Last I checked, you could buy a Rubik's Cube for less than $20,
                            > including shipping to everywhere. If you wanted to put it together
                            > yourself, you can get them for less than half that price. I think
                            > it's fairly easy to see the incredible price gulf between that and V
                            > Cubes (at the previously quoted amount of $80 + shipping).
                            >
                            > 2008/7/2 Bob Burton <rubikscubewhiz@... <rubikscubewhiz%40yahoo.com>
                            > >:
                            >
                            > > you don't want to know about these things. possession of such
                            > > knowledge may result in spending copious amounts of money on cheap
                            > > plastic.
                            > >
                            > > --- In speedsolvingrubikscube@yahoogroups.com<speedsolvingrubikscube%40yahoogroups.com>,
                            > Bart <banaticus@...> wrote:
                            > >>
                            > >> Where can I find a site on cube math? What's a corner cycle?
                            > > What's parity?
                            > >>
                            > >
                            > >
                            > >
                            > > ------------------------------------
                            > >
                            > > Yahoo! Groups Links
                            > >
                            > >
                            > >
                            > >
                            >
                            >


                            [Non-text portions of this message have been removed]
                          • Clancy Cochran
                            bart- yeah you re pretty much a dipshit, and no one cares about your extremely ignorant opinion on this matter. if you had even a clue as to the amount of
                            Message 13 of 29 , Jul 2, 2008
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                              bart- yeah you're pretty much a dipshit, and no one cares about your extremely ignorant opinion on this matter. if you had even a clue as to the amount of money and effort that goes into producing a toy on a worldwide scale (its not JUST the plastic) then you might be able to appreciate these for what they are -- probably the most complex twisty puzzles ever built and well worth the prices being asked. you're ignorant and uninformed, kindly quit trying to be a troll, it hardly works and will just make you look like a bigger dumbass (if that's even possible). if you think its unfair or not reasonable, produce 6x6 and 7x7 cubes for a price less than theirs of the same quality and i'll eat my hat. until then....cork it, asshat.


                              --- On Wed, 7/2/08, Bart <banaticus@...> wrote:
                              From: Bart <banaticus@...>
                              Subject: Re: [Speed cubing group] Re: Please me with the Rubik's cube group elements order pr
                              To: speedsolvingrubikscube@yahoogroups.com
                              Date: Wednesday, July 2, 2008, 4:00 PM



                              Last I checked, you could buy a Rubik's Cube for less than $20,

                              including shipping to everywhere. If you wanted to put it together

                              yourself, you can get them for less than half that price. I think

                              it's fairly easy to see the incredible price gulf between that and V

                              Cubes (at the previously quoted amount of $80 + shipping).



                              2008/7/2 Bob Burton <rubikscubewhiz@ yahoo.com>:

                              > you don't want to know about these things. possession of such

                              > knowledge may result in spending copious amounts of money on cheap

                              > plastic.

                              >

                              > --- In speedsolvingrubiksc ube@yahoogroups. com, Bart <banaticus@. ..> wrote:

                              >>

                              >> Where can I find a site on cube math? What's a corner cycle?

                              > What's parity?

                              >>

                              >

                              >

                              >

                              > ------------ --------- --------- ------

                              >

                              > Yahoo! Groups Links

                              >

                              >

                              >

                              >
                            • bladez740
                              ... wrote: i ll eat my hat. until then....cork it, asshat. ... Not the slayer hat!
                              Message 14 of 29 , Jul 2, 2008
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                                --- In speedsolvingrubikscube@yahoogroups.com, Clancy Cochran
                                <perscription_death@...> wrote:
                                i'll eat my hat. until then....cork it, asshat.
                                >

                                Not the slayer hat!
                              • Bart
                                Sure, I d moved on to asking about a different topic, until Bob threw that back up in my face again. If you d care to actually answer the question so that I
                                Message 15 of 29 , Jul 2, 2008
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                                  Sure, I'd moved on to asking about a different topic, until Bob threw
                                  that back up in my face again. If you'd care to actually answer the
                                  question so that I can learn more about Rubik's Cubes, I'd love to
                                  hear it.
                                • bryanlogancube
                                  Well, you can get a cube in the store for less than $10, and shipping on DIY s usually end up being more than $10 total, but those are facts, so I m sure
                                  Message 16 of 29 , Jul 2, 2008
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                                    Well, you can get a cube in the store for less than $10, and shipping
                                    on DIY's usually end up being more than $10 total, but those are
                                    facts, so I'm sure you're not too concerned about them.

                                    Anyways, if a Rubik's Cube is 3x3x3, wouldn't a 6x6x6 be the
                                    equivilant of 8 Rubik's cubes?

                                    Anyways, if you're trying to figure out corner cycle and parity, why
                                    not try Google?

                                    I'd suggest getting a new Yahoo ID and try not to be an idiot in the
                                    future.

                                    --- In speedsolvingrubikscube@yahoogroups.com, Bart <banaticus@...> wrote:
                                    >
                                    > Last I checked, you could buy a Rubik's Cube for less than $20,
                                    > including shipping to everywhere. If you wanted to put it together
                                    > yourself, you can get them for less than half that price. I think
                                    > it's fairly easy to see the incredible price gulf between that and V
                                    > Cubes (at the previously quoted amount of $80 + shipping).
                                    >
                                  • Bob Burton
                                    A simple Google search would yield the answer to each of your three quesitons. If you want a website about cube math, Google Rubik s Cube Math If you want
                                    Message 17 of 29 , Jul 2, 2008
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                                      A simple Google search would yield the answer to each of your three
                                      quesitons.

                                      If you want a website about "cube math," Google 'Rubik's Cube Math'
                                      If you want to know about a "corner cycle," Google 'Rubik's Cube
                                      corner cycle'
                                      If you want to know about "parity," Google 'Rubik's Cube parity'

                                      The first link for each of those three Google searches yields the
                                      answer to your question.

                                      ...however, I'd stay away from the Rubik's Cube math. You seem to
                                      have the mathematical ability and reasoning of a goldfish.

                                      Bob

                                      --- In speedsolvingrubikscube@yahoogroups.com, Bart <banaticus@...> wrote:
                                      >
                                      > Sure, I'd moved on to asking about a different topic, until Bob threw
                                      > that back up in my face again. If you'd care to actually answer the
                                      > question so that I can learn more about Rubik's Cubes, I'd love to
                                      > hear it.
                                      >
                                    • Bart
                                      Thanks. A guy asked what my thoughts were about the price. I answered. People jumped all over my case and I defended my reasoning. Then those people that, I
                                      Message 18 of 29 , Jul 3, 2008
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                                        Thanks. A guy asked what my thoughts were about the price. I
                                        answered. People jumped all over my case and I defended my reasoning.
                                        Then those people that, I presume, have a financial interest in
                                        seeing the company succeed started using profanity and slinging other
                                        insults. Whatever floats their boat, I guess. Maybe they should be
                                        upset with the guy who asked to hear people's opinion in the first
                                        place.
                                      • Stefan Pochmann
                                        ... reasoning. ... other ... Nobody minds you saying they cost too much for you. The problem started when you claimed a ridiculous x1000 markup and insulted
                                        Message 19 of 29 , Jul 3, 2008
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                                          --- In speedsolvingrubikscube@yahoogroups.com, Bart <banaticus@...>
                                          wrote:
                                          >
                                          > Thanks. A guy asked what my thoughts were about the price. I
                                          > answered. People jumped all over my case and I defended my
                                          reasoning.
                                          > Then those people that, I presume, have a financial interest in
                                          > seeing the company succeed started using profanity and slinging
                                          other
                                          > insults. Whatever floats their boat, I guess. Maybe they should be
                                          > upset with the guy who asked to hear people's opinion in the first
                                          > place.
                                          >

                                          Nobody minds you saying they cost too much for you. The problem
                                          started when you claimed a ridiculous x1000 markup and insulted the V-
                                          cubes people.

                                          Stefan
                                        • Stefan Pochmann
                                          ... And it even works when done right. Should be e^(8/e)=18.97. ... I know equivalence relations, but again misread your post. I missed that you actually went
                                          Message 20 of 29 , Jul 3, 2008
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                                            --- In speedsolvingrubikscube@yahoogroups.com, "mackymakisumi"
                                            <mackymakisumi@...> wrote:
                                            >
                                            > --- In speedsolvingrubikscube@yahoogroups.com, "Stefan Pochmann"
                                            > <stefan.pochmann@> wrote:
                                            > > You could shorten this part by saying that even the product-
                                            > > maximizing partition of the eight corners, which is (8/e)^e=18.8,
                                            > > isn't enough to reach the necessary (1260/22)/3=19.09.
                                            >
                                            > Nice!

                                            And it even works when done right. Should be e^(8/e)=18.97.


                                            > I'm not sure how much you know about equivalence relation or group
                                            > action, so I'll try to be as detailed as I can. Sorry if any part
                                            > bores you.

                                            I know equivalence relations, but again misread your post. I missed
                                            that you actually went down to the level of move sequences, as in the
                                            whole thread I was solely thinking in terms of their effects (cube
                                            group elements). I agree it's completely obvious when read correctly.
                                            And I'd actually say this obviousness is what made me misread your
                                            message. I need to be more careful.

                                            Group actions however still confuse me. I've read about them and
                                            similar stuff before but I have trouble grasping/remembering these
                                            things. I'll read your message again later when I have more time, and
                                            will hopefully understand it better then. Thanks for taking the time
                                            to explain, I imagine it's interesting/useful for others as well.

                                            Cheers!
                                            Stefan
                                          • bryanlogancube
                                            I don t think any of us have a financial interest in V-Cubes. We simply want them to succeed so they put out more cubes. But presumptions like this just make
                                            Message 21 of 29 , Jul 3, 2008
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                                              I don't think any of us have a financial interest in V-Cubes. We
                                              simply want them to succeed so they put out more cubes. But
                                              presumptions like this just make you look more ignorant.

                                              --- In speedsolvingrubikscube@yahoogroups.com, Bart <banaticus@...> wrote:
                                              >
                                              > Then those people that, I presume, have a financial interest in
                                              > seeing the company succeed started using profanity and slinging other
                                              > insults.
                                            • Alexander Goldberg
                                              V-Cubes profits would be the catalyst for 11x11x11 engineering. Our investment goes further than what we directly pay for.
                                              Message 22 of 29 , Jul 4, 2008
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                                                V-Cubes' profits would be the catalyst for 11x11x11 engineering.
                                                Our investment goes further than what we directly pay for.
                                              • Stefan Pochmann
                                                ... I built such a case yesterday and then today found your above message again. Here s an algorithm in four steps: x U L D L U L D L M U M U M U M U U
                                                Message 23 of 29 , Aug 5, 2008
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                                                  --- In speedsolvingrubikscube@yahoogroups.com, "Bruce Norskog"
                                                  <brnorsk@...> wrote:
                                                  >
                                                  > In this group of "permutable centers," the maximum order is 2520.
                                                  > This can be achieved by:
                                                  >
                                                  > corner cycles: 5*3, 3*3
                                                  > edge cycles: 7*2, 4*2, 1*1
                                                  > center cycles: 4, 1, 1
                                                  >
                                                  > LCM(15,9,14,8,1,4,1,1) = 2520

                                                  I built such a case yesterday and then today found your above message
                                                  again. Here's an algorithm in four steps:

                                                  x
                                                  U L D' L' U' L D L'
                                                  M' U M' U M' U M' U
                                                  U B' U' F2 M2 F2 M2 U B U'

                                                  The steps are small-effect algs with specific intentions:

                                                  1) Create several 4-cycles (and change edges+corners parity)
                                                  2) Create the misoriented corner 3-cycle and 5-cycle
                                                  3) Create the misoriented edge 4-cycle (in the L layer)
                                                  4) Create the edge 7-cycle

                                                  You can see steps 2-4 simply combine the 4-cycles created in step 1
                                                  in order to build the desired cycles.

                                                  Cheers!
                                                  Stefan
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