Thread: "Cubes, beer and Beethoven"

=3D'font-size:
10.0pt;font-family:Arial;color:navy'>Thanks David – I didn’t ex=
pect
a response so quickly (and certainly not before my question had come out). =

=3D'font-size:
10.0pt;font-family:Arial;color:navy'> 

=3D'font-size:
10.0pt;font-family:Arial;color:navy'>I will check the sites you refer me to=
and
let you know if they answer my question. I suspect I was not very clear in =
my
original posting, but by ‘3D version’ I meant the 4^3, not the =
3^3
and it was from this that I made guesses about the 4^4. Each of us when sol=
ving
probably creates his own lingo but I should be more careful when communicat=
ing:
by ‘centres’ I meant the 8 one-colour pieces on the interior of
each cubic face of the 4^4. Since you cannot distinguish orientations and
positions of these I wondered if their arrangement influenced the subsequen=
t
placing of the two-colour pieces, such that if it was wrong these last coul=
d
not be aligned. In my solution I never had to break and rearrange the one
colour pieces after I had placed them. As I mentioned, I did have to break =
and
rearrange the two-colour pieces before the three-colour pieces would align.=
The
corners, however, dropped into place first time. If there really are 3
independent issues, all at 50%, then I was presumably 3/8 lucky only to hit=
one
of them.

=3D'font-size:
10.0pt;font-family:Arial;color:navy'> 

=3D'font-size:
10.0pt;font-family:Arial;color:navy'>When solving the 3^4 for the first tim=
e using
something like Roice’s method you often come up against seemingly
impossible situations that are not impossible (and that is where the fun is=
).
It is useful for 4^4 solvers to know that in solving that puzzle you someti=
mes
meet seemingly impossible situations that really are that, at least without
breaking and reconstructing work already done. Because of familiarity with =
the
3D puzzle (Rubik’s Revenge) I wasted no time agonising when I found t=
he
three edge pieces inverted (I could easily revert four, leaving one inverte=
d,
but it appeared impossible to invert just one or three).<=
/font>

=3D'font-size:
10.0pt;font-family:Arial;color:navy'> 

=3D'font-size:
10.0pt;font-family:Arial;color:navy'>Thanks again,=

=3D'font-size:
10.0pt;font-family:Arial;color:navy'> 

=3D'font-size:
10.0pt;font-family:Arial;color:navy'>Guy

=3D'font-size:
10.0pt;font-family:Arial;color:navy'> 

'font-size:
12.0pt'> 

'font-size:
10.0pt'>Note that my response to Guy's message preceded his and, to further
confusestyle=3D'font-size:10.0pt;font-family:"Courier New"'>

matters, I accidentally deleted the attribut=
ion
(to Guy) of the portions I

quoted from his message.



The fault is mine, as Guy's message was held=
for
moderation (as is the first

message of any new subscriber).  I got
carried away and formed my response

before I approved his message for posting,
forgetting that the copy I was

looking at was that which came with the requ=
est
for a moderator's approval.  :-(



Guy, your posts will no longer be held for
moderation.



Regards,

  David V.

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From: "Roice Nelson" <roice@gravitation3d.com>
Date: Sun, 8 May 2005 23:07:09 -0500
Subject: RE: [MC4D] Cubes, beer and Beethoven

Hi Guy,

I had a few comments...

> I will check the sites you refer me to and
> let you know if they answer my question.
> I suspect I was not very clear in my original
> posting, but by �3D version� I meant the 4^3,
> not the 3^3 and it was from this that I made
> guesses about the 4^4. Each of us when solving
> probably creates his own lingo but I should be
> more careful when communicating: by �centres�
> I meant the 8 one-colour pieces on the interior
> of each cubic face of the 4^4. Since you cannot
> distinguish orientations and positions of these
> I wondered if their arrangement influenced the
> subsequent placing of the two-colour pieces,
> such that if it was wrong these last could
> not be aligned.

My memory had failed me as usual, but I dug up some emails from when I
had worked on the 4^4. I did have exactly what you are describing
here happen to me (a parity issue such that I could not line up all
the 2 color pieces after placing the centers, forcing me to undo work
on the centers).

I had the parity issue you had as well, forcing me to break and
rearrange 2 color pieces to fully align the 3 color ones. Very
thankfully, I did not have the issues with the corners! After working
through the other 2 problems, I distinctly remember hoping hard for
that :) Maybe some of other 4^4 solvers have had a corner parity
issue, allowing us to empirical verify the existence of all of these
using the whole group.

I think there could be something to your question about the solution
method affecting parity problems arising in the 4^4 (centers-out vs.
layer by layer). Say for instance there are multiple situations that
can lead to parity problems for 4 color corners, and that these depend
not just on the 3 color edges (if an even/odd number of twists was
used to solve them or whatever), but on both the 3 color and 2 color
pieces. In the centers-out approach, certain situations leading to 4
color problems might never show up since some work to get 2 color and
3 color pieces "in parity" has already been done, and so maybe such
issues could only arise in a layer-by-layer approach. This is
hypothetical, but my point is that the solution algorithm itself
places additional constraints on the state of the puzzle and may limit
what is possible.

In fact, I wonder if the fully unconstrained 4^4 has more than 3
contributing situations that can lead to parity problems (maybe 3! or
something). I am not familiar enough with group theory to back this
up mathematically, but others might be able to. As David suggested,
maybe the information could be gleaned from Eric's page on the number
of permutations. It would be cool to have a specific explanation of
the number of these problems and their occurrence percentage.

Take Care,

Roice

P.S. I had put marks in my 4^4 solution at the time, so you can check
out the parity problems I encountered on it if you like. Here are
descriptions for my marks...

a Centers Placed
b First parity problem
c Done Combining Face Pieces
d Done Combining Edge Pieces
e Second parity problem
f Fixed Parity Problem
g Faces Placed
h Edges Placed
i Finished

---

From: "Guy Padfield" <guy@guypadfield.com>
Date: Mon, 9 May 2005 19:54:47 +0200
Subject: RE: [MC4D] Cubes, beer and Beethoven

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Thanks Roice. Remigiusz is supposed to be producing a solution to the 4^4 a=
t
this very moment, though he keeps getting sidetracked by his mission to
unscramble a 2^4 in under 20 moves. I hope he will report back on any parit=
y
problems he encounters.

=20

I haven=92t had time yet to study David=92s and Eric=92s articles, though I=
did
read Eric=92s when I first found the site, before I had attempted any of th=
e
cubes. I first thought about parity issues, at a much simpler level, at
about the same time David was writing that article in 1980 (when I was a
schoolboy). Once I had solved the Rubik cube (there were no published
solutions then) I realised the box was wrong to boast there was only one
solution, since the face centres could be orientated any way without anyone
noticing. So I marked up my cube to yield only one correct solution and
rapidly made an interesting discovery. If you leave the orientation of the
centres to the end, there are two possibilities. 50% of the time their net
rotation is 0=BA (mod 360), in which case they are easily solved in pairs. =
50%
of the time the net rotation is 180=BA, in which case you are forced to bre=
ak
and remake the edges and corners of a face to align the remaining centre
correctly. By coincidence I chose the word =91parity=92 to describe my
discovery, calling the first situation =91even=91 and the second =91odd=91.

=20

Few people seem to have attempted the 5^4. The 5^3 has just one parity issu=
e
(providing you don=92t mark it up) =96 once you have aligned the edges corr=
ectly
the corners always fall into place. It would be interesting to know if the
same is true for the 5^4.=20

=20

What would happen if you worked outside in? The corners have only one
possible =91correct=92 alignment with respect to each other so completing t=
he
edge pieces inside them would presumably always work without rebreaking the
corners. I strongly suspect, without proof, it would then be possible to
align the faces without rebreaking the edges and the centres without
rebreaking the faces (because of the independence of the parity issues when
working in the reverse direction). However, I also suspect the required
sequences would be much longer, undoing the benefit gained.

=20

Roice indirectly revealed his 4^4 solution strategy in his last post
responding to my query. Would it breach protocol for me to publish my
slightly different strategy here, if that would give future solvers a hint
or two? It is far less economical in terms of moves but also, I think, far
easier to execute. Or would that upset people trying to solve it for
themselves?

=20

Guy

=20

_____=20=20

From: 4D_Cubing@yahoogroups.com [mailto:4D_Cubing@yahoogroups.com] On Behal=
f
Of Roice Nelson
Sent: 09 May 2005 06:07
To: 4D_Cubing@yahoogroups.com
Subject: RE: [MC4D] Cubes, beer and Beethoven

=20

Hi Guy,

I had a few comments...

> I will check the sites you refer me to and
> let you know if they answer my question.
> I suspect I was not very clear in my original
> posting, but by =913D version=92 I meant the 4^3,
> not the 3^3 and it was from this that I made
> guesses about the 4^4. Each of us when solving
> probably creates his own lingo but I should be
> more careful when communicating: by =91centres=92
> I meant the 8 one-colour pieces on the interior
> of each cubic face of the 4^4. Since you cannot
> distinguish orientations and positions of these
> I wondered if their arrangement influenced the
> subsequent placing of the two-colour pieces,
> such that if it was wrong these last could
> not be aligned.

My memory had failed me as usual, but I dug up some emails from when I
had worked on the 4^4. I did have exactly what you are describing
here happen to me (a parity issue such that I could not line up all
the 2 color pieces after placing the centers, forcing me to undo work
on the centers).

I had the parity issue you had as well, forcing me to break and
rearrange 2 color pieces to fully align the 3 color ones. Very
thankfully, I did not have the issues with the corners! After working
through the other 2 problems, I distinctly remember hoping hard for
that :) Maybe some of other 4^4 solvers have had a corner parity
issue, allowing us to empirical verify the existence of all of these
using the whole group.

I think there could be something to your question about the solution
method affecting parity problems arising in the 4^4 (centers-out vs.
layer by layer). Say for instance there are multiple situations that
can lead to parity problems for 4 color corners, and that these depend
not just on the 3 color edges (if an even/odd number of twists was
used to solve them or whatever), but on both the 3 color and 2 color
pieces. In the centers-out approach, certain situations leading to 4
color problems might never show up since some work to get 2 color and
3 color pieces "in parity" has already been done, and so maybe such
issues could only arise in a layer-by-layer approach. This is
hypothetical, but my point is that the solution algorithm itself
places additional constraints on the state of the puzzle and may limit
what is possible.

In fact, I wonder if the fully unconstrained 4^4 has more than 3
contributing situations that can lead to parity problems (maybe 3! or
something). I am not familiar enough with group theory to back this
up mathematically, but others might be able to. As David suggested,
maybe the information could be gleaned from Eric's page on the number
of permutations. It would be cool to have a specific explanation of
the number of these problems and their occurrence percentage.

Take Care,

Roice

P.S. I had put marks in my 4^4 solution at the time, so you can check
out the parity problems I encountered on it if you like. Here are
descriptions for my marks...

a Centers Placed
b First parity problem
c Done Combining Face Pieces
d Done Combining Edge Pieces
e Second parity problem
f Fixed Parity Problem
g Faces Placed
h Edges Placed
i Finished




_____=20=20

Yahoo! Groups Links

* To visit your group on the web, go to:
http://groups.yahoo.com/group/4D_Cubing/
=20=20
* To unsubscribe from this group, send an email to:
4D_Cubing-unsubscribe@yahoogroups.com
=20
=20=20
* Your use of Yahoo! Groups is subject to the Yahoo!
Terms of Service.=20


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