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Re: 5x5x5 help

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  • makimoto2000us
    ... apply steps 6 and 7 to ... Hi, Thanks for your interest in my method. Step3. (Middle edges.) and Step6. (Permute last U edges) may be the most difficult
    Message 1 of 23 , May 3 5:33 AM
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      --- In speedsolvingrubikscube@yahoogroups.com, hubexe <no_reply@y...>
      wrote:
      > in masayuki akimoto's solution to the professor cube, how do you
      apply steps 6 and 7 to
      > the 5x5x5?
      Hi,

      Thanks for your interest in my method.
      Step3. (Middle edges.) and Step6. (Permute last U edges)
      may be the most difficult parts in my method.
      I worry if you may not understand without pictures.


      Step6. (Permute last U edges)
      Before explanation, I would define each edge as El, Ec, Er.

      Ec :MD2M'
      El, Er :rl'D2r'l
      El,Ec, Er:rMl'D2r'M'l
      Those algorithms should not mix up center pieces.
      Use these algs effectively.

      If you have pair of El-Ec, Ec-Er, or El-Ec in same orientation, solve
      this pair first.
      El or Er left. You can apply Step 6-2 of 4x4x4.
      Ec left. MD2M' or S'U'SDS'USD'

      In the other case, solve either Ec or Er (El) pretending l, M, and r
      as one slice set, you can solve at least one edge either El, Ec, Er
      as same as 3x3x3 but you may need extra Ds to flip the rest of edges.
      If you solved Ec first, and if you could keep either Er or El which
      can be solved by rl'D2r'l, you can solve the last edge as same as
      4x4x4.

      If you solved Er or El first, you can also place Ec which can be
      solved by MD2M'. you can solve the last edge easily.


      I normally don't use but in case you have edge pair in different
      orientation that means Er-Ec or El-Ec, solve Ec with flipped Er or
      El.
      Then you can apply 7-2-2iv to solve Er and El simultanously.


      Step7 (Last Layer)
      7-1 (corners) Must be as same as smaller cubes.

      7-2-1,7-2-2
      You can use all the algorithms.

      7-2-3
      This is Chris's algorithm. You will have middle edges problem if you
      directly apply this for 5x5x5.
      You can fix the centers by E'L2Er2E'L2Er2.
      Or you can solve by combintaion of 7-2-2iii and 7-2-2-iv

      7-2-4
      You can directly apply them to 5x5x5.

      I hope this help you to finish 5x5x5 with my method.
      Good luck.

      Masayuki Akimoto
    • j_rueth
      Does anyone have a list of the 469 algs used in Watermans( i think) CF method? I d like to see them if possible... thanks jake
      Message 2 of 23 , May 3 6:08 AM
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        Does anyone have a list of the 469 algs used in Watermans( i think)
        CF method? I'd like to see them if possible... thanks

        jake
      • Wayne
        Where did you get the 469 figure from ? Wayne
        Message 3 of 23 , May 4 7:51 AM
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          Where did you get the 469 figure from ?

          Wayne


          --- In speedsolvingrubikscube@yahoogroups.com, j_rueth
          <no_reply@y...> wrote:
          > Does anyone have a list of the 469 algs used in Watermans( i think)
          > CF method? I'd like to see them if possible... thanks
          >
          > jake
        • j_rueth
          I remember someone saying that there is a corners first method uses 469 algs, i think dan gosbee was saying that this was the method he was using, or
          Message 4 of 23 , May 4 12:27 PM
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            I remember someone saying that there is a corners first method uses 469
            algs, i think dan gosbee was saying that this was the method he was using,
            or developed... I just wanted to see what this method is like

            :) jake

            --- In speedsolvingrubikscube@yahoogroups.com, "Wayne" <
            mylib_2000@y...> wrote:
            > Where did you get the 469 figure from ?
            >
            > Wayne
            >
            >
            > --- In speedsolvingrubikscube@yahoogroups.com, j_rueth
            > <no_reply@y...> wrote:
            > > Does anyone have a list of the 469 algs used in Watermans( i think)
            > > CF method? I'd like to see them if possible... thanks
            > >
            > > jake
          • Wayne
            His method may require that many but the Waterman system is probably around 150. Wayne ... 469 ... was using, ... think)
            Message 5 of 23 , May 4 1:19 PM
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              His method may require that many but the Waterman system is probably
              around 150.

              Wayne


              --- In speedsolvingrubikscube@yahoogroups.com, j_rueth
              <no_reply@y...> wrote:
              > I remember someone saying that there is a corners first method uses
              469
              > algs, i think dan gosbee was saying that this was the method he
              was using,
              > or developed... I just wanted to see what this method is like
              >
              > :) jake
              >
              > --- In speedsolvingrubikscube@yahoogroups.com, "Wayne" <
              > mylib_2000@y...> wrote:
              > > Where did you get the 469 figure from ?
              > >
              > > Wayne
              > >
              > >
              > > --- In speedsolvingrubikscube@yahoogroups.com, j_rueth
              > > <no_reply@y...> wrote:
              > > > Does anyone have a list of the 469 algs used in Watermans( i
              think)
              > > > CF method? I'd like to see them if possible... thanks
              > > >
              > > > jake
            • Kelley
              Dan Gosbee s CF method requires 463 algorithms. Kelley
              Message 6 of 23 , May 4 1:37 PM
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                Dan Gosbee's CF method requires 463 algorithms.

                Kelley
              • Gilles Roux
                ... I don t know any method requiring 469 algorithms , but if the method you re looking for belongs to the CF class of approaches, you could try the
                Message 7 of 23 , May 4 2:46 PM
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                  --- In speedsolvingrubikscube@yahoogroups.com, j_rueth <no_reply@y...>
                  wrote:
                  > Does anyone have a list of the 469 algs used in Watermans( i think)
                  > CF method? I'd like to see them if possible... thanks
                  >
                  > jake

                  I don't know any method requiring 469 "algorithms", but if the method
                  you're looking for belongs to the CF class of approaches, you could
                  try the cornersfirst Yahoo group.


                  Gilles.

                  PS: Josef Jelinek is building a page dealing with the method used by
                  Marc Waterman (work in progress). See
                  http://www.rubikscube.info/waterman/index.html
                • j_rueth
                  Thanks for the link Gilles, I ll check it out. I thought watermans solution was a lot bigger. Thanks jake
                  Message 8 of 23 , May 4 4:29 PM
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                    Thanks for the link Gilles, I'll check it out. I thought watermans
                    solution was a lot bigger. Thanks

                    jake
                  • cubacca1972
                    ... think) ... There are a total of 141 algs used in Waterman s method, provided that you don t run into any complications. This total includes a lot of
                    Message 9 of 23 , May 4 10:23 PM
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                      > Does anyone have a list of the 469 algs used in Watermans( i
                      think)
                      > CF method? I'd like to see them if possible... thanks
                      >
                      > jake

                      There are a total of 141 algs used in Waterman's method, provided
                      that you don't run into any complications. This total includes a
                      lot of mirrored algs.

                      Roughly, the method involves the following steps:

                      1. solve one face

                      2. orient and permute the remaining corners (1 of 43 algs)

                      Next, hold the solved face in your left hand, with the corners
                      solved in step 2 in the R face. The next steps are to solve the
                      right edge pieces (called "redges"), and orient all the middle edge
                      pieces (called "midges") in two algs. The M slice is referred to as
                      the "ring".

                      3. Solve 2 redges such that you leave 1 or 0 redges in the ring.
                      Do this by picking an alg which solves 1 redge in the ring and 1 in
                      the R face (1 of 6 algs), OR pick an alg which solves 2 redges in
                      the ring (1 of 6 algs)

                      4. Solve the last 2 redges, and orient all the midges in 1 alg. If
                      both redges are in the R face, then you have to pick 1 of 14 algs.
                      If one redge is in the R face, and one is in the ring, then pick one
                      of 48 algs.

                      5. solve the midges, 1 of 3 algs, which are trivial.

                      additional algs are included to handle situations where one or more
                      redges are solved by the end of step 2.

                      flip one redge and orient midges: 2 algs

                      solve one redge in the ring and orient midges: 16 algs (8 algs and
                      their mirrors)

                      if all redges are solved, orient midges: 3 algs

                      I think that the algs from step 3 are borrowed from the algs in step
                      4, but listed separately.

                      In addition to this set of algs, there are additional components to
                      the system designed to handle ugly situations such as when you have
                      3 or 4 redges cycled in the r face, or when you have all 4 redges in
                      the ring.

                      If you can master all of the bad situations in addition to the basic
                      (!) system, you should be able to solve the R face and orient the M
                      slice in 2 algs or processes, no matter what.

                      since you pretty much use only U, R, and M, you don't have to regrip
                      with your left hand too much (there are some algs with F moves).

                      Here are a few example algs from step 3 to show the basic idea. Try
                      doing the inverse of the algs on a solved cube to set these up.

                      To solve 2 redges in the ring, one at FU and one at FD (FU means
                      that the R facelet of the redge is on the F face as opposed to UF,
                      which puts the R facelet on the U face) , do this:

                      U2 (R) M U2 M' U M2 U (R)' U2

                      When you execute this alg, the redge at FU will get placed in the
                      hole at RU, and the redge at FD will get placed in the hole in the R
                      face which is turned up to the RU position with the turn (R). (R)'
                      is just the inverse of whatever R move you did in (R).

                      To solve 2 redges where one is in the ring at DF (R facelet of this
                      redge is in the D face), and one is in the correct hole in the R
                      face at the position RU but flipped, do this:

                      U2 (R) U' M U2 M2 U' (R)' U2

                      in this case, the move (R) means turn the R face so that the hole
                      that the redge in the ring belongs to gets moved to the RU position.

                      In any case its a neat but scary system.

                      Lucas
                    • Josef Jelinek
                      Only several more notes. To count exact number of algorithms in Waterman s method is quite difficult because each algorithm using (R) and (R) symbols is a
                      Message 10 of 23 , May 6 7:47 AM
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                        Only several more notes.

                        To count exact number of algorithms in Waterman's method
                        is quite difficult because each algorithm using (R) and (R)'
                        symbols is a shortcut for more algorithms and should
                        be learnt that way to get really fast in performing them...

                        The original Waterman's description of "bad luck cases"
                        and "complications" is too complicated and in fact can be
                        handled simpler with less algorithms with the same efficiency.
                        (about 4 + 2 algs)

                        The algorithms mentioned in the previous message (see below)
                        are not the most efficient (and this is true for more algs
                        in the Waterman's method). You can use the following
                        replacements (found by myself):

                        U M (R) U' M U M' (R)' U'
                        (instead of U2 (R) M U2 M' U M2 U (R)' U2)
                        - shorter, without half moves

                        U M' U' (R) U' M' U2 M2 U'
                        (instead of U2 (R) U' M U2 M2 U' (R)' U2)
                        - less half moves, only one (R) in alg., 1 more M turn,
                        there is a hole in RU and not the flipped edge (!)

                        In addition, I have found replacements for all algorithms
                        that use F turns (hard to perform quickly) and now
                        all algorithms use only U M R moves (!) and are of the
                        same length or even shorter (except for one or two algs.)

                        Josef

                        > There are a total of 141 algs used in Waterman's method, provided
                        > that you don't run into any complications. This total includes a
                        > lot of mirrored algs.
                        >
                        > Roughly, the method involves the following steps:
                        >
                        > 1. solve one face
                        >
                        > 2. orient and permute the remaining corners (1 of 43 algs)
                        >
                        > Next, hold the solved face in your left hand, with the corners
                        > solved in step 2 in the R face. The next steps are to solve the
                        > right edge pieces (called "redges"), and orient all the middle
                        edge
                        > pieces (called "midges") in two algs. The M slice is referred to
                        as
                        > the "ring".
                        >
                        > 3. Solve 2 redges such that you leave 1 or 0 redges in the ring.
                        > Do this by picking an alg which solves 1 redge in the ring and 1
                        in
                        > the R face (1 of 6 algs), OR pick an alg which solves 2 redges in
                        > the ring (1 of 6 algs)
                        >
                        > 4. Solve the last 2 redges, and orient all the midges in 1 alg.
                        If
                        > both redges are in the R face, then you have to pick 1 of 14
                        algs.
                        > If one redge is in the R face, and one is in the ring, then pick
                        one
                        > of 48 algs.
                        >
                        > 5. solve the midges, 1 of 3 algs, which are trivial.
                        >
                        > additional algs are included to handle situations where one or
                        more
                        > redges are solved by the end of step 2.
                        >
                        > flip one redge and orient midges: 2 algs
                        >
                        > solve one redge in the ring and orient midges: 16 algs (8 algs and
                        > their mirrors)
                        >
                        > if all redges are solved, orient midges: 3 algs
                        >
                        > I think that the algs from step 3 are borrowed from the algs in
                        step
                        > 4, but listed separately.
                        >
                        > In addition to this set of algs, there are additional components
                        to
                        > the system designed to handle ugly situations such as when you
                        have
                        > 3 or 4 redges cycled in the r face, or when you have all 4 redges
                        in
                        > the ring.
                        >
                        > If you can master all of the bad situations in addition to the
                        basic
                        > (!) system, you should be able to solve the R face and orient the
                        M
                        > slice in 2 algs or processes, no matter what.
                        >
                        > since you pretty much use only U, R, and M, you don't have to
                        regrip
                        > with your left hand too much (there are some algs with F moves).
                        >
                        > Here are a few example algs from step 3 to show the basic idea.
                        Try
                        > doing the inverse of the algs on a solved cube to set these up.
                        >
                        > To solve 2 redges in the ring, one at FU and one at FD (FU means
                        > that the R facelet of the redge is on the F face as opposed to UF,
                        > which puts the R facelet on the U face) , do this:
                        >
                        > U2 (R) M U2 M' U M2 U (R)' U2
                        >
                        > When you execute this alg, the redge at FU will get placed in the
                        > hole at RU, and the redge at FD will get placed in the hole in the
                        R
                        > face which is turned up to the RU position with the turn (R).
                        (R)'
                        > is just the inverse of whatever R move you did in (R).
                        >
                        > To solve 2 redges where one is in the ring at DF (R facelet of
                        this
                        > redge is in the D face), and one is in the correct hole in the R
                        > face at the position RU but flipped, do this:
                        >
                        > U2 (R) U' M U2 M2 U' (R)' U2
                        >
                        > in this case, the move (R) means turn the R face so that the hole
                        > that the redge in the ring belongs to gets moved to the RU
                        position.
                        >
                        > In any case its a neat but scary system.
                        >
                        > Lucas
                      • cubacca1972
                        ... Nicely done! I should have guessed that there was room for improvement with the original algs. On closer inspection of the algs in Waterman s method, I see
                        Message 11 of 23 , May 6 9:29 PM
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                          --- In speedsolvingrubikscube@yahoogroups.com, "Josef Jelinek"
                          <gloom@e...> wrote:
                          > Only several more notes.....
                          >
                          > The original Waterman's description of "bad luck cases"
                          > and "complications" is too complicated and in fact can be
                          > handled simpler with less algorithms with the same efficiency.
                          > (about 4 + 2 algs)
                          >
                          > The algorithms mentioned in the previous message (see below)
                          > are not the most efficient (and this is true for more algs
                          > in the Waterman's method). You can use the following
                          > replacements (found by myself):
                          >
                          > U M (R) U' M U M' (R)' U'
                          > (instead of U2 (R) M U2 M' U M2 U (R)' U2)
                          > - shorter, without half moves
                          >
                          > U M' U' (R) U' M' U2 M2 U'
                          > (instead of U2 (R) U' M U2 M2 U' (R)' U2)
                          > - less half moves, only one (R) in alg., 1 more M turn,
                          > there is a hole in RU and not the flipped edge (!)
                          >
                          > In addition, I have found replacements for all algorithms
                          > that use F turns (hard to perform quickly) and now
                          > all algorithms use only U M R moves (!) and are of the
                          > same length or even shorter (except for one or two algs.)
                          >
                          > Josef
                          >

                          Nicely done!

                          I should have guessed that there was room for improvement with the
                          original algs.

                          On closer inspection of the algs in Waterman's method, I see that
                          there are similar algs in table 4 which are similar to your second
                          example: U M' U' (R) U' M' U2 M2 U'.

                          Table 4, alg 17a: U M' U' (R) U M2 U2 M U

                          This makes me wonder why they published Table 2 with less efficient
                          algs borrowed from table 4 which accomplish the same thing!

                          How many algs have you found which are more efficient for the method?
                          I would love to see them.

                          I am sort of glad that I never had the guts to try to master the
                          original algs, given that I would have to spend a lot of time
                          unlearning the algs with more efficient alternatives.

                          Lucas
                        • Jason Baum
                          Hi guys, I need an alg to switch two opposite edge centers without affecting any other center pieces. I feel dumb because I can t figure this out on my own.
                          Message 12 of 23 , Oct 25, 2005
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                            Hi guys,

                            I need an alg to switch two opposite edge centers without affecting
                            any other center pieces. I feel dumb because I can't figure this out
                            on my own. =/ Help please!

                            -Jason
                          • Jason Baum
                            Nevermind, I figured it out. Probably not the most efficient way of doing it but at least it gets the job done. :P -Jason
                            Message 13 of 23 , Oct 25, 2005
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                              Nevermind, I figured it out. Probably not the most efficient way of
                              doing it but at least it gets the job done. :P

                              -Jason

                              --- In speedsolvingrubikscube@yahoogroups.com, "Jason Baum"
                              <speedrunningcuber@y...> wrote:
                              >
                              > Hi guys,
                              >
                              > I need an alg to switch two opposite edge centers without affecting
                              > any other center pieces. I feel dumb because I can't figure this out
                              > on my own. =/ Help please!
                              >
                              > -Jason
                              >
                            • Patrick Jameson
                              anyone have any tips for making a 5x5x5 loose. i have used silicon and it helped a little but it is still stiff. Patrick
                              Message 14 of 23 , Jun 10, 2007
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                                anyone have any tips for making a 5x5x5 loose. i have used silicon and
                                it helped a little but it is still stiff.

                                Patrick
                              • symbioticfear
                                Wearing it down is the best way, imo. Jonathan Choi
                                Message 15 of 23 , Jun 10, 2007
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                                  Wearing it down is the best way, imo.

                                  Jonathan Choi

                                  --- In speedsolvingrubikscube@yahoogroups.com, "Patrick Jameson"
                                  <poker19@...> wrote:
                                  >
                                  > anyone have any tips for making a 5x5x5 loose. i have used silicon and
                                  > it helped a little but it is still stiff.
                                  >
                                  > Patrick
                                  >
                                • stompey1
                                  Cube with it for maybe 2 or 3 months keeping the lubrication minimal, and then take it apart, claen it all out, lube it, and put it all back together. The lack
                                  Message 16 of 23 , Jun 10, 2007
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                                    Cube with it for maybe 2 or 3 months keeping the lubrication minimal,
                                    and then take it apart, claen it all out, lube it, and put it all back
                                    together. The lack of lube smooths out all the rough edges nicely.
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