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Re: [Speed cubing group] Re: wow... just........ wow..

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  • Arnaud van Galen
    I don t know anyone knows a 25 turn HTM scramble (without obvious cancellations) that result in the solved state again, but I do remember a challenge a year or
    Message 1 of 77 , Apr 1, 2007
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      I don't know anyone knows a 25 turn HTM scramble (without obvious cancellations) that result in the solved state again, but I do remember a challenge a year or more old that had a 13 moves one (if memory serves me correct) as a winner. I thought Stefan found it. There were also some longer ones naturally. I am not going to try to find a 25 moves one, but I think that if you search for that thread, you will see that one is extremely likely to exist.

      ----- Original Message -----
      From: d_funny007
      To: speedsolvingrubikscube@yahoogroups.com
      Sent: Saturday, March 31, 2007 8:03 PM
      Subject: [Speed cubing group] Re: wow... just........ wow..


      Intuitively this must be the case, and I think it's a assertion that
      everybody here is willing to believe. But this is much easier than
      proving the other things being discussed, all this takes is finding a
      single case. I wouldn't even call it a "proof," just a "example/counter-
      example".

      Just entertain me and see if anyone here can find a 25 turn HTM
      scramble that produces the identity/solved state. I'm having a hard
      time finding one. Oh and by "scramble," I am discounting the really
      trivial cancellations and such.

      > Whoa! There are positions not reachable in 25 turns? Proof please!
      >
      > Cheers!
      > Stefan
      >





      [Non-text portions of this message have been removed]
    • Stefan Pochmann
      ... about ... Those ... Ah, yes, I must ve made a computation mistake. I tried again and get 12 expected cubes for 13f as well now. I did it like this: First I
      Message 77 of 77 , Apr 12, 2007
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        --- In speedsolvingrubikscube@yahoogroups.com, h_kociemba
        <no_reply@...> wrote:
        >
        > > Which had to be expected. I think in a million random cubes,
        about
        > > three should be expected to be solvable in 13 moves or less.
        Those
        > > four you found with 13f, can you please find out their optimal
        > > solution lengths?
        >
        >
        > For 1 million cubes, the theoretical distribution for the *optimal*
        > maneuvers should look like this:
        >
        > 12f: ~1
        > 13f: ~12
        > 14f: ~160
        > 15f: ~2200
        > 16f: ~29000
        > 17f: ~260.000
        > 18f: ~690.000
        > 19f: ~30.000
        > 20f: probably less than 1

        Ah, yes, I must've made a computation mistake. I tried again and get
        12 expected cubes for 13f as well now. I did it like this:

        First I looked at the right column in this statistic:
        http://cubezzz.homelinux.org/drupal/?q=node/view/68

        Multiplied the last number twice more with 13 (roughly the branching
        factor). Multiply with a million, divide by 4.3e19.

        Yeah I know, probably there's a statistic somewhere out there about
        the overall distribution, but I don't know where.

        Cheers!
        Stefan
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