## Re: Periods (Devil's Algorithm)

Expand Messages
• While we re on this topic I thought I would post something related that I found. How could one construct an algorithm that would cycle through all possible
Message 1 of 21 , Nov 4, 2004
While we're on this topic I thought I would post something related
that I found.

How could one construct an algorithm that would cycle through all
possible positions of the cube (a "Devil's Algorithm" as I've seen
it written).

The algorithms would be good as long as it cycled though all
43252003274489856000 positions on the cube, regardless of whether it
repeated positions or not.

This is of course possible, even with a random string of notation
that is absolutely enormously larger than the number of possible
combinations to the cube.

However, the interesting question is whether or not it is possible
to have an algorithm of length 43252003274489855999 (staring from
the solved state) that cycles through all possible positions of the
cube (including having started at the solved state). Maybe it
doesn't even have to cycle through all the positions, but just have
the first 43252003274489855999 moves pass through the same number of
states (starting from the solved, or technically any, state.

Maybe you could look at this by drawing a graph with
43252003274489856000 nodes as a bipartite graph including all even
an odd positions, and also grouped together with various
characteristics (say positions that are all 2 moves away from each
other are drawn close together). Each position could only reach 18
others, 17 if you include the fact that you had to do a move to get
to that position. To be more precise, only quarter turns would be
allowed, since performing a double turn is by the purest sense
either doing R R or R' R' or some other equivalent

I honestly haven't put too much thought into this to be honest, but
it is an interesting question nonetheless. Would using graph theory
and trying to create some sort of Hamiltonian chain, or even a
Hamiltonian cycle, work as a way to do this?

Since that would take too much computing power, are there any
patterns we can use to come up with such an algorithm?

Also would this algorithm have any sort of
regularity/symmetry/interesting charateristics, or would it be
totally random in appearance?

There would of course be an infinite number of algorithms that would
achieve this affect, however how many are there with length
43252003274489855999, if any at all? If there aren't any of that
length, what is the shortest possible lentgh you can have to cycle
through all the positions of the cube? Can you do it without
repeating any positions at all?

Can you repeat positions often, though come up with an easy pattern
for writing out a Devil's algorithm?

I guess the holy grail of this would be any algorithm of exactly
length 43252003274489856000 that perfectly cycled through all the
possible positions of the cube. Such that, starting from any state,
if you performed this algorithm you would not only end at the same
state you started at, but have cycled through all possible positions
as well.

Just some questions I thought might spark some conversation.

Chris
• Hey Eivind :D Yes u are correct. (FRBL)*315 will yield a solved cube again (when u start from solved :- P ) I guess i was confusing with supercubing
Message 2 of 21 , Nov 5, 2004
Hey Eivind :D

Yes u are correct.

(FRBL)*315 will yield a solved cube again (when u start from solved :-
P ) I guess i was confusing with supercubing calculations. (FRBL)*315
will have the following effect on centers : [F'R'B'L'], all are
rotated counterclockwise. So if one wants to solve the centers also a
power of 1260 is needed. And of course any permutation of order 1260
will fix the centers, so this is the maximum order also in the
supergroup ;-)

-Per

PS! One can quite easily verify orders of permutation using for
instance the FMC companion !! Just check which is the lowest order
yielding a solved cube. Incidentally, for FRBL, if it's repeated 105
times u get a corner 3-twist on the D layer :-)

-------==================-------

> --- In speedsolvingrubikscube@yahoogroups.com, "Eivind Fonn"
<htkra1d@h...> wrote:
>
> If memory serves me right, I believe it was stated in the link you
> posted yesterday that FRBL has an order of ~300.
>
> --- In speedsolvingrubikscube@yahoogroups.com, "Per Kristen
Fredlund"
> <aspiring_to_love@y...> wrote:
> >
> > Hi :-)
> >
> > The idea for finding the maaximum order of a permutation is to
> > realise that a permutation is a combination of disjoint corner-
cycles
> > and edge-cycles. So what u do is to decompose the permutation
into
> > it's individual cycles. Then simply determine the LCM (least
common
> > multiple) of these cycles. Also u have to keep in mind what
> > combinations of cycles will be possible to yield an overall even
> > permutation. And also u have only 8 corners and 12 edges to play
> > with ;-) A formal proof might have to list all possible corner-
cycle
> > orders and edge-cycle orders?? Or is there an easier way to prove
> > this??
> >
> > An easier example than urs of such a permutation is the FRBL
> > permutation.
> >
> > Happy cubing :-)
> >
> > -Per
> >
> > > --- In speedsolvingrubikscube@yahoogroups.com, "joel_vn"
> > <joel_vn@y...> wrote:
> > >
> > > Hi Eivind,
> > >
> > > (Very nice subject, btw. Maybe I can use something like this
for my
> > > lecture as well)
> > >
> > > In Dutch, they are called 'orde', which looks like order, but I
am
> > > not sure if that's really the right word for it.. What matters
is:
> > I
I
> > > can't help you with that. But I recently found that the maximum
> > > order is 1260. An alg. with this order is: RU2D'BD'. It took me
> > > about 50 minutes to verify this. However, I don't know how to
prove
> > > that this really is the max. order.
> > >
> > > -Joel
> > >
> > > --- In speedsolvingrubikscube@yahoogroups.com, "Eivind Fonn"
> > > <htkra1d@h...> wrote:
> > > >
> > > > I'm sitting here at school and a guy asked me about
permutations
> > > > period (or maybe they are called orders?). I.e. how many
times you
> > > > have to perform a permutation to get back to the identity. I
> > > remember
> > > > having seen a post or a site showing all the possible
periods. Can
> > > > someone point me in the right direction? Or if not, what was
the
> > > > largest possible period, and is there a simple algorithm that
has
> > > this
> > > > period?
• Hi Chris, I recommend beginning with a sequence which will go through all possible combinations of two sides being turned one quarter turn. What algorithm
Message 3 of 21 , Nov 5, 2004
Hi Chris,

I recommend beginning with a sequence which will go through all
possible combinations of two sides being turned one quarter turn. What
algorithm accomplishing this has the fewest moves? Is this, going
through 15 positions, a small enough part of the problem to allow us
to hold a contest?

Regards,

David J

P.S. The new IBM supercomputer can perform 43252003274489856000

wrote:
>
> While we're on this topic I thought I would post something related
> that I found.
>
> How could one construct an algorithm that would cycle through all
> possible positions of the cube (a "Devil's Algorithm" as I've seen
> it written).
>
> The algorithms would be good as long as it cycled though all
> 43252003274489856000 positions on the cube, regardless of whether it
> repeated positions or not.
>
> This is of course possible, even with a random string of notation
> that is absolutely enormously larger than the number of possible
> combinations to the cube.
>
> However, the interesting question is whether or not it is possible
> to have an algorithm of length 43252003274489855999 (staring from
> the solved state) that cycles through all possible positions of the
> cube (including having started at the solved state). Maybe it
> doesn't even have to cycle through all the positions, but just have
> the first 43252003274489855999 moves pass through the same number of
> states (starting from the solved, or technically any, state.
>
> Maybe you could look at this by drawing a graph with
> 43252003274489856000 nodes as a bipartite graph including all even
> an odd positions, and also grouped together with various
> characteristics (say positions that are all 2 moves away from each
> other are drawn close together). Each position could only reach 18
> others, 17 if you include the fact that you had to do a move to get
> to that position. To be more precise, only quarter turns would be
> allowed, since performing a double turn is by the purest sense
> either doing R R or R' R' or some other equivalent
>
> I honestly haven't put too much thought into this to be honest, but
> it is an interesting question nonetheless. Would using graph theory
> and trying to create some sort of Hamiltonian chain, or even a
> Hamiltonian cycle, work as a way to do this?
>
> Since that would take too much computing power, are there any
> patterns we can use to come up with such an algorithm?
>
> Also would this algorithm have any sort of
> regularity/symmetry/interesting charateristics, or would it be
> totally random in appearance?
>
> There would of course be an infinite number of algorithms that would
> achieve this affect, however how many are there with length
> 43252003274489855999, if any at all? If there aren't any of that
> length, what is the shortest possible lentgh you can have to cycle
> through all the positions of the cube? Can you do it without
> repeating any positions at all?
>
> Can you repeat positions often, though come up with an easy pattern
> for writing out a Devil's algorithm?
>
> I guess the holy grail of this would be any algorithm of exactly
> length 43252003274489856000 that perfectly cycled through all the
> possible positions of the cube. Such that, starting from any state,
> if you performed this algorithm you would not only end at the same
> state you started at, but have cycled through all possible positions
> as well.
>
> Just some questions I thought might spark some conversation.
>
>
> Chris
• Hey David, that s a really cool puzzle! Just some opening thoughts. First it would help to know the numer of possible states that can be reached using only 2
Message 4 of 21 , Nov 5, 2004
Hey David, that's a really cool puzzle!

Just some opening thoughts.

First it would help to know the numer of possible states that can be
reached using only 2 quarter turns (assuming that repetition of
turning the same face twice is allowed).

I count 115 positions reachable by doing two quarter turns. 96 if
they are perpendicular faces, 12 if they are parallel faces, and 7
if you turn the same face twice in succession.

The critical thing to notice is that all the positions that need to
be reached are even permutations, so we have to go through an odd
permutation to reach each new position.

I guess an optimal algorithm in this puzzle would be one that
reached all 115 positions that need to be reached, but never
repeated any odd permutation. You could also have an algorithm that
achieved all 115 positions, but repeated some or multiple odd
permutations.

If an optimal length algorithm exists then it would have to be 230
moves long. Assuming you start from the solved state and that every
second move you achieve a distinct even permutation that is also one
of our 115. I don't think an optimal algorithm would exist though.

You could something like
R F4 U4 B4 D4 L4 R R F4 U4 B4 D4 L4 R L F4 U4 B4 D4 R4 L L F4 U4 B4
D4 R4

To cycle through all 38 positions unique to the RL axis, but to get
to a positions such as UR where a face in the UD axis is turned
first, you have to go through the solved state.

So my proposal for a shortest alg would be the following: any face
turn followed by a 4 means 4 clockwise turns, just written shorter.

1) cycle through all positions unqiue to the RL axis
R F4 U4 B4 D4 L4 R R F4 U4 B4 D4 L4 R L F4 U4 B4 D4 R4 L L F4 U4 B4
D4 R4

2) solve the cube again
L

3) cycle through all the positions unique to the UD axis
U F4 R4 B4 L4 D4 U U F4 R4 B4 L4 D4 U D F4 R4 B4 L4 U4 D D F4 R4 B4
L4 U4

4) solve the cube again
D

5) cycle through all the positions unique to the FB axis
F U4 R4 D4 L4 B4 F F U4 R4 D4 L4 B4 F B U4 R4 D4 L4 F4 B B U4 R4 D4
L4 F4

and you could optionally solve the cube again with B to create a
true cycle.

So my alg is 264 moves if you make it a true cycle by adding the
final F and is:

R F4 U4 B4 D4 L4 R R F4 U4 B4 D4 L4 R L F4 U4 B4 D4 R4 L L F4 U4 B4
D4 R4 L U F4 R4 B4 L4 D4 U U F4 R4 B4 L4 D4 U D F4 R4 B4 L4 U4 D D
F4 R4 B4 L4 U4 D F U4 R4 D4 L4 B4 F F U4 R4 D4 L4 B4 F B U4 R4 D4 L4
F4 B B U4 R4 D4 L4 F4 B

and repeates positions therefore 264-230=34 times.

I think, due to the nature of having to have a certain face having
been turned "first" for many of the positions, that this would be an
optimal alg.

Just my guess though, but I think it works. I doubt my reasoning
counts as an actual proof, but I think it works out.

Chris

--- In speedsolvingrubikscube@yahoogroups.com, "d_j_salvia"
<d_j_salvia@y...> wrote:
>
> Hi Chris,
>
> I recommend beginning with a sequence which will go through all
> possible combinations of two sides being turned one quarter turn.
What
> algorithm accomplishing this has the fewest moves? Is this, going
> through 15 positions, a small enough part of the problem to allow
us
> to hold a contest?
>
> Regards,
>
> David J
>
> P.S. The new IBM supercomputer can perform 43252003274489856000
> calculations in about a week.
• Hey Per, just take a quarter of your alg: R y :-) Stefan
Message 5 of 21 , Nov 5, 2004
Hey Per, just take a quarter of your alg: R y :-)

Stefan

--- In speedsolvingrubikscube@yahoogroups.com, "Per Kristen Fredlund"
<aspiring_to_love@y...> wrote:
>
> Hey Eivind :D
>
> Yes u are correct.
>
> (FRBL)*315 will yield a solved cube again (when u start from solved
:-
> P ) I guess i was confusing with supercubing calculations.
• ... Hi Chris, a while ago I asked about something similar, like a combination of your idea and the original thread question. In my opinion the real holy grail
Message 6 of 21 , Nov 5, 2004
wrote:
>
> I guess the holy grail of this would be any algorithm of exactly
> length 43252003274489856000 that perfectly cycled through all the
> possible positions of the cube. Such that, starting from any state,
> if you performed this algorithm you would not only end at the same
> state you started at, but have cycled through all possible positions
> as well.

Hi Chris,

a while ago I asked about something similar, like a combination of
your idea and the original thread question. In my opinion the real
holy grail would be: An algorithm of length 43252003274489856000/1260
which, applied 1260 times, cycles through all states.

Cheers!
Stefan
• ... That s not really going to help, since there s no linear time algorithm to solve the problem, since in general it s NP-complete (and I doubt our special
Message 7 of 21 , Nov 5, 2004
--- In speedsolvingrubikscube@yahoogroups.com, "d_j_salvia"
<d_j_salvia@y...> wrote:
>
> P.S. The new IBM supercomputer can perform 43252003274489856000
> calculations in about a week.

That's not really going to help, since there's no linear time
algorithm to solve the problem, since in general it's NP-complete (and
I doubt our special case is special enough to make it easier):
http://mathworld.wolfram.com/HamiltonianCircuit.html

Cheers!
Stefan
• Yeah yeah Stefan, tricking with cube turns won t make any difference :-P -Cubix PS! Then any of the following will do: Ry, Ry , Rx, Rx , Rz, Rz (or even the
Message 8 of 21 , Nov 5, 2004
Yeah yeah Stefan, tricking with cube turns won't make any
difference :-P

-Cubix

PS! Then any of the following will do:

Ry, Ry', Rx, Rx', Rz, Rz' (or even the same ones w R' in place of R.
Or for that matter any other face clockwise or counterclockwise. By
symmetry and or inversion they are all the same :D

zzzzzzzzzzzzzz ..........

Hehe, happy cubing :-)

> --- In speedsolvingrubikscube@yahoogroups.com, "Stefan Pochmann"
<pochmann@g...> wrote:
>
> Hey Per, just take a quarter of your alg: R y :-)
>
> Stefan
>
> --- In speedsolvingrubikscube@yahoogroups.com, "Per Kristen
Fredlund"
> <aspiring_to_love@y...> wrote:
> >
> > Hey Eivind :D
> >
> > Yes u are correct.
> >
> > (FRBL)*315 will yield a solved cube again (when u start from
solved
> :-
> > P ) I guess i was confusing with supercubing calculations.
• Hey! I know it can be proven that one algorithm wouldn t be enough. Two algorithms is needed. I have even seen published 2 short algs to do that. But some
Message 9 of 21 , Nov 5, 2004
Hey!

I know it can be proven that one algorithm wouldn't be enough. Two
algorithms is needed. I have even seen published 2 short algs to do
that. But some quick search on the net didn't succeed in finding
those algs, or anything else about it for that matter. I'm pretty
sure Richard Carr knows what im talking about. I think it was
published a long time ago actually, like 1982 or -83. But internet
didn't exist back then :-P Not like we know it now anyway ;-)

-Per

> --- In speedsolvingrubikscube@yahoogroups.com, "Stefan Pochmann"
<pochmann@g...> wrote:
>
> --- In speedsolvingrubikscube@yahoogroups.com, cmhardw
> wrote:
> >
> > I guess the holy grail of this would be any algorithm of exactly
> > length 43252003274489856000 that perfectly cycled through all
the
> > possible positions of the cube. Such that, starting from any
state,
> > if you performed this algorithm you would not only end at the
same
> > state you started at, but have cycled through all possible
positions
> > as well.
>
> Hi Chris,
>
> a while ago I asked about something similar, like a combination of
> your idea and the original thread question. In my opinion the real
> holy grail would be: An algorithm of length
43252003274489856000/1260
> which, applied 1260 times, cycles through all states.
>
> Cheers!
> Stefan
• I think you re talking about something different. We re talking about one algorithm that during (!) its execution comes across all states. Btw, I doubt an
Message 10 of 21 , Nov 5, 2004
one algorithm that during (!) its execution comes across all states.

Btw, I doubt an algorithm of 43252003274489856000/1260 turns has been
published ;-)

Cheers!
Stefan

--- In speedsolvingrubikscube@yahoogroups.com, "Per Kristen Fredlund"
<aspiring_to_love@y...> wrote:
>
> Hey!
>
> I know it can be proven that one algorithm wouldn't be enough. Two
> algorithms is needed. I have even seen published 2 short algs to do
> that. But some quick search on the net didn't succeed in finding
> those algs, or anything else about it for that matter. I'm pretty
> sure Richard Carr knows what im talking about. I think it was
> published a long time ago actually, like 1982 or -83. But internet
> didn't exist back then :-P Not like we know it now anyway ;-)
>
> -Per
>
> > --- In speedsolvingrubikscube@yahoogroups.com, "Stefan Pochmann"
> <pochmann@g...> wrote:
> >
> > --- In speedsolvingrubikscube@yahoogroups.com, cmhardw
> > wrote:
> > >
> > > I guess the holy grail of this would be any algorithm of exactly
> > > length 43252003274489856000 that perfectly cycled through all
> the
> > > possible positions of the cube. Such that, starting from any
> state,
> > > if you performed this algorithm you would not only end at the
> same
> > > state you started at, but have cycled through all possible
> positions
> > > as well.
> >
> > Hi Chris,
> >
> > a while ago I asked about something similar, like a combination of
> > your idea and the original thread question. In my opinion the real
> > holy grail would be: An algorithm of length
> 43252003274489856000/1260
> > which, applied 1260 times, cycles through all states.
> >
> > Cheers!
> > Stefan
• ... You may have been thinking about this list of all orders in the Cubic Circular: http://www.geocities.com/jaapsch/puzzles/cubic3.htm#p35 It does not list
Message 11 of 21 , Nov 5, 2004
--- "Eivind Fonn" wrote:
> I remember
> having seen a post or a site showing all the possible periods. Can
> someone point me in the right direction? Or if not, what was the
> largest possible period, and is there a simple algorithm that has this
> period?

Circular:
http://www.geocities.com/jaapsch/puzzles/cubic3.htm#p35
It does not list examples of each, only how many there are.
Jaap
• ... Guess why I added that as a P.S.? Putting aside the devil s algorithm, if all the possble positions can be reached in 25 moves then with proper guidance
Message 12 of 21 , Nov 5, 2004
--- In speedsolvingrubikscube@yahoogroups.com, "Stefan Pochmann"
<pochmann@g...> wrote:
>
> --- In speedsolvingrubikscube@yahoogroups.com, "d_j_salvia"
> <d_j_salvia@y...> wrote:
> >
> > P.S. The new IBM supercomputer can perform 43252003274489856000
> > calculations in about a week.
>
> That's not really going to help, since there's no linear time
> algorithm to solve the problem, since in general it's NP-complete (and
> I doubt our special case is special enough to make it easier):
> http://mathworld.wolfram.com/HamiltonianCircuit.html
>
> Cheers!
> Stefan

Guess why I added that as a P.S.?

Putting aside the devil's algorithm, if all the possble positions can
be reached in 25 moves then with proper guidance that computer could
cover all the possible positions in 6 months.

If the positions were stored in such a way that all positions one move
away from each other were linked then God's Algorithm would be found
as the shortest paths. Tha this would take more than one of these
computers and cost several million dollars is beside the point! :)

Regards,

David J
• That wasn t what I had seen before, but it is much better - so thanks!
Message 13 of 21 , Nov 5, 2004
That wasn't what I had seen before, but it is much better - so thanks!

wrote:
>
> --- "Eivind Fonn" wrote:
> > I remember
> > having seen a post or a site showing all the possible periods. Can
> > someone point me in the right direction? Or if not, what was the
> > largest possible period, and is there a simple algorithm that has this
> > period?
>
> Circular:
> http://www.geocities.com/jaapsch/puzzles/cubic3.htm#p35
> It does not list examples of each, only how many there are.
> Jaap
• x is parallell to R though so Rx and its twins have order 4. ;)
Message 14 of 21 , Nov 5, 2004
x is parallell to R though so Rx and its "twins" have order 4. ;)

--- In speedsolvingrubikscube@yahoogroups.com, "Per Kristen Fredlund"
<aspiring_to_love@y...> wrote:
>
> Yeah yeah Stefan, tricking with cube turns won't make any
> difference :-P
>
> -Cubix
>
> PS! Then any of the following will do:
>
> Ry, Ry', Rx, Rx', Rz, Rz' (or even the same ones w R' in place of R.
> Or for that matter any other face clockwise or counterclockwise. By
> symmetry and or inversion they are all the same :D
>
> zzzzzzzzzzzzzz ..........
>
> Hehe, happy cubing :-)
>
> > --- In speedsolvingrubikscube@yahoogroups.com, "Stefan Pochmann"
> <pochmann@g...> wrote:
> >
> > Hey Per, just take a quarter of your alg: R y :-)
> >
> > Stefan
> >
> > --- In speedsolvingrubikscube@yahoogroups.com, "Per Kristen
> Fredlund"
> > <aspiring_to_love@y...> wrote:
> > >
> > > Hey Eivind :D
> > >
> > > Yes u are correct.
> > >
> > > (FRBL)*315 will yield a solved cube again (when u start from
> solved
> > :-
> > > P ) I guess i was confusing with supercubing calculations.
• Hey Eivind!! Umm, yes i was a bit rushed with that post. One third of all those cases will indeed have order 4 ;-) Well observed :-) Happy cubing!! -Cubix ...
Message 15 of 21 , Nov 6, 2004
Hey Eivind!!

Umm, yes i was a bit rushed with that post. One third of all those
cases will indeed have order 4 ;-) Well observed :-)

Happy cubing!!

-Cubix

> --- In speedsolvingrubikscube@yahoogroups.com, "Eivind Fonn"
<htkra1d@h...> wrote:
>
> x is parallell to R though so Rx and its "twins" have order 4. ;)
>
> --- In speedsolvingrubikscube@yahoogroups.com, "Per Kristen
Fredlund"
> <aspiring_to_love@y...> wrote:
> >
> > Yeah yeah Stefan, tricking with cube turns won't make any
> > difference :-P
> >
> > -Cubix
> >
> > PS! Then any of the following will do:
> >
> > Ry, Ry', Rx, Rx', Rz, Rz' (or even the same ones w R' in place of
R.
> > Or for that matter any other face clockwise or counterclockwise.
By
> > symmetry and or inversion they are all the same :D
> >
> > zzzzzzzzzzzzzz ..........
> >
> > Hehe, happy cubing :-)
> >
> > > --- In speedsolvingrubikscube@yahoogroups.com, "Stefan
Pochmann"
> > <pochmann@g...> wrote:
> > >
> > > Hey Per, just take a quarter of your alg: R y :-)
> > >
> > > Stefan
> > >
> > > --- In speedsolvingrubikscube@yahoogroups.com, "Per Kristen
> > Fredlund"
> > > <aspiring_to_love@y...> wrote:
> > > >
> > > > Hey Eivind :D
> > > >
> > > > Yes u are correct.
> > > >
> > > > (FRBL)*315 will yield a solved cube again (when u start from
> > solved
> > > :-
> > > > P ) I guess i was confusing with supercubing calculations.
• ... Hey Chris, I had meant all positions involving two sides, but you are allowing for a side to be turned twice; that s OK with me. For clarity s sake I ll
Message 16 of 21 , Nov 8, 2004
wrote:
>
> Hey David, that's a really cool puzzle!
>
> Just some opening thoughts.
>
> First it would help to know the numer of possible states that can be
> reached using only 2 quarter turns (assuming that repetition of
> turning the same face twice is allowed).
>
> I count 115 positions reachable by doing two quarter turns. 96 if
> they are perpendicular faces, 12 if they are parallel faces, and 7
> if you turn the same face twice in succession.

Hey Chris,

I had meant all positions involving two sides, but you are allowing
for a side to be turned twice; that's OK with me.

For clarity's sake I'll lay them all the possibilities out.

Two quarter turns allowing the second turn to be the same side as
the first, including repeated positions: (6*2)*(6*2)=144 positions: if
same side twice isn't allowed (6*2)*(5*2)=120

12 are two turns on one side resulting in a double turn

RR, LL, FF, BB, UU, DD
R'R', L'L', F'F', B'B', U'U', D'D'

12 are two turns on one side resulting in a solved cube

RR', R'R, LL', L'L, FF', F'F, BB', B'B, UU', U'U, DD', D'D

6 * 16 = 96 are adjacent sides

RF, RF', R'F, R'F'
RB, RB', R'B, R'B'
RU, RU', R'U, R'U'
RD, RD', R'D, R'D'

LF, LF', L'F, L'F'
LB, LB', L'B, L'B'
LU, LU', L'U, L'U'
LD, LD', L'D, L'D'

FR, FR', F'R, F'R'
FL, FL', F'L, F'L'
FU, FU', F'U, F'U'
FD, FD', F'D, F'D'

BR, BR', B'R, B'R'
BL, BL', B'L, B'L'
BU, BU', B'U, B'U'
BD, BD', B'D, B'D'

UR, UR', U'R, U'R'
UL, UL', U'L, U'L'
UF, UF', U'F, U'F'
UB, UB', U'B, U'B'

DR, DR', D'R, D'R'
DL, DL', D'L, D'L'
DF, DF', D'F, D'F'
DB, DB', D'B, D'B'

6 * 4 = 24 opposite side moves of which 12 are duplicates opposite
one another

RL same as LR
RL' same as L'R
R'L same as L R'
R'L' same as L'R'

FB same as BF
FB' same as B'F
F'B same as B F'
F'B' same as B'F'

UD same as DU
UD' same as D'U
U'D same as DU'
U'D' same as D'U'

All positions minus duplicate and solved positions 144 - 24 = 120
All positions minus doubled turns minus duplicate positions
144 - 24 - 12 = 108

I'll get to the rest later.

My best,

David J
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