I don't see a problem in the number on the website. You can compare your no=
te with the calculation here:
http://www.superliminal.com/cube/permutations.html
and see which step is different.=20
Nan
--- In 4D_Cubing@yahoogroups.com, "charliemckiz@..."
te:
>
> Today I tried to calculate the number of MC4D permutations. My answer is =
a half of what posted in the website http://www.superliminal.com/cube/cube.=
htm . Does any one want to check the answer? I'm pretty sure about mine. An=
d I see no mistake in the website solution.??
>
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Content-Type: text/plain; charset=utf-8
Content-Transfer-Encoding: quoted-printable
=20
My solution is as follow
Rules that MC4D obey:
Position:=20
4-color pieces: The permutation of 4-color pieces is always
even
2/3-color pieces: The net permutation of 3-color and 2-color
is always even.
Orientation: Consider rotation of a piece is a permutation
of shading sides of that piece. So:
4-color pieces: The permutation of rotation of a 4-color
piece is always even. Besides, the net rotation of all 4-color pieces is 0.=
=20
3-color pieces: The net permutation of rotation of all
3-color pieces is even.
2-color pieces: The net permutation of rotation of all
2-color pieces is even.
Number of different kinds of pieces:
4-color pieces: 16
3-color pieces: 32
2-color pieces: 24
=C2=A0
Based on the rules above:
Permutation of 4-color pieces: 16! =C3=B72
Orientation of 4-color pieces: (4! =C3=B72)15=C3=973
Permutation of
2/3-color pieces: 32!=C3=9724! =C3=B72
Orientation of 3-color pieces: (3!)32 =C3=B72
Orientation of 2-color pieces: (2)24 =C3=B72
Total: 16! =C3=B72=C3=97(4! =C3=B72)15=C3=973=C3=9732!=C3=9724! =C3=B72=C3=
=97(3!)32 =C3=B72=C3=97(2)24 =C3=B72
________________________________
From: Brandon Enright
To: 4D_Cubing@yahoogroups.com=20
Cc: charliemckiz@rocketmail.com; bmenrigh@ucsd.edu=20
Sent: Monday, April 30, 2012 3:38 PM
Subject: Re: [MC4D] About the number of permutations of MC4D calculating
=20
=C2=A0=20
-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1
On Mon, 30 Apr 2012 15:31:05 -0000
"charliemckiz@rocketmail.com"
> Today I tried to calculate the number of MC4D permutations. My answer
> is a half of what posted in the website
> http://www.superliminal.com/cube/cube.htm . Does any one want to
> check the answer? I'm pretty sure about mine. And I see no mistake in
> the website solution.??
>=20
Well, I've aways taken the permutation calculation as gospel but I
decided to give it a go myself and I've arrived at the same result.
I find the sticker-based counting of permutations to be confusing so
I've done it piece-based. Here are my notes.
That is, there are:
Permutations:
* 8 centers with 192 "permutations" available. Fixing to one cancels
all re-orientation of the tesseract
* 24 face centers that can be in any permutation
* 32 3-color edges who's parity is tied to the parity of the face
centers
* 16 4-color corners that can only be in an even permutation
Orientations:
* 8 centers with 24 orientations available (only 1 visually distinct)
* 24 face pieces with 8 orientations available (only 2 visually
distinct ones)
* 32 3-color edges with 6 orientations available
* 16 4-color edges with 12 orientations available
The calculation is further complicated in that:
* The total "flip" of the 24 face pieces is even (the 24th orientation
is determined by the previous 23)
* The orientation of the 32nd 2-color edge can only be in half (3) of
the orientations.
* The first 15 corners can be in any orientation but the last corner
can only be in 1/3 (4) orientations.
Taking all of that into account there should be:
(24! * 2^23) * ((32! / 2) * (6^31 * 3)) * ((16! / 2) * (12^15 * 4))
Which matches nicely with the website.
The only "tricky" portion of the calculation is counting the
orientations for the last 3-color and the last 4-color piece. I can
create a macro to demonstrate how to put a single 3-color piece or a
single 4-color piece in these reduced orientations if you'd like.
Regards,
Brandon
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=20
--964895041-548286253-1335839367=:14949
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: quoted-printable
is always
even
nd 2-color
is always even.
mutation
of shading sides of that piece. So:
-color
piece is always even. Besides, the net rotation of all 4-color pieces is 0.=
all
3-color pieces is even.
all
2-color pieces is even.
/span>
5s New Roman";mso-bidi-font-family:
"Times New Roman""> =C3=97style=3D"mso-fareast-font-family:
"Times New Roman";mso-bidi-font-family:"Times New Roman"=
;;mso-fareast-language:
ZH-CN">3
tion of
2/3-color pieces: 32!ont-family:
"Times New Roman";mso-bidi-font-family:"Times New Roman"=
;"> =C3=9724! =C3=B72
mso-bidi-font-family:"Times New Roman"">Orientation of 3-color pi=
eces: (3!)
32 =C3=B72
mso-bidi-font-family:"Times New Roman"">Orientation of 2-color pi=
eces: (2)
24 =C3=B72
mso-bidi-font-family:"Times New Roman"">Total: 16! =
=C3=B72s New Roman";mso-bidi-font-family:
"Times New Roman"">=C3=97(4! =C3=B72) 15=
oman";mso-bidi-font-family:"Times New Roman"">
=C3=97ly:"Times New Roman";
mso-bidi-font-family:"Times New Roman";mso-fareast-language:ZH-CN=
">3"Times New Roman";mso-bidi-font-family:
"Times New Roman"">=C3=97-language:ZH-CN">32!nt-family:"Times New Roman";mso-bidi-font-family:
"Times New Roman""> =C3=9724! =C3=B72=C3=97(3!) 32 =C3=
=B72=C3=97(2) 24 =C3=B72tyle=3D"mso-fareast-font-family:"Times New Roman";mso-bidi-font-f=
amily:
"Times New Roman";mso-fareast-language:ZH-CN">>
=3D"font-weight:bold;">From: Brandon Enright <bmenrigh@ucsd.e=
du>
To: 4D_Cubing@y=
ahoogroups.com
Cc: cha=
rliemckiz@rocketmail.com; bmenrigh@ucsd.edu
ght: bold;">Sent: Monday, April 30, 2012 3:38 PM
yle=3D"font-weight: bold;">Subject: Re: [MC4D] About the number =
of permutations of MC4D calculating
=20=20=20=20=20=20
=20=20=20=20=20=20
Hash: SHA1
On Mon, 30 Apr 2012 15:31:05 -0000
"t=3D"_blank" href=3D"mailto:charliemckiz%40rocketmail.com">charliemckiz@roc=
ketmail.com" <ocketmail.com" target=3D"_blank" href=3D"mailto:charliemckiz%40rocketmail.c=
om">charliemckiz@rocketmail.com> wrote:
> Today I tried to calculate the number of MC4D permutations. My answer<=
br>
> is a half of what posted in the website
> http://www.superliminal.com/cube/cube.htm . Does any one want to
> check the answer? I'm pretty sure about mine. And I see no mistake in<=
br>
> the website solution.??
>
Well, I've aways taken the permutation calculation as gospel but I
decided to give it a go myself and I've arrived at the same result.
I find the sticker-based counting of permutations to be confusing so
I've done it piece-based. Here are my notes.
That is, there are:
Permutations:
* 8 centers with 192 "permutations" available. Fixing to one cancels
all re-orientation of the tesseract
* 24 face centers that can be in any permutation
* 32 3-color edges who's parity is tied to the parity of the face
centers
* 16 4-color corners that can only be in an even permutation
Orientations:
* 8 centers with 24 orientations available (only 1 visually distinct)
* 24 face pieces with 8 orientations available (only 2 visually
distinct ones)
* 32 3-color edges with 6 orientations available
* 16 4-color edges with 12 orientations available
The calculation is further complicated in that:
* The total "flip" of the 24 face pieces is even (the 24th orientation
is determined by the previous 23)
* The orientation of the 32nd 2-color edge can only be in half (3) of
the orientations.
* The first 15 corners can be in any orientation but the last corner
can only be in 1/3 (4) orientations.
Taking all of that into account there should be:
(24! * 2^23) * ((32! / 2) * (6^31 * 3)) * ((16! / 2) * (12^15 * 4))
Which matches nicely with the website.
The only "tricky" portion of the calculation is counting the
orientations for the last 3-color and the last 4-color piece. I can
create a macro to demonstrate how to put a single 3-color piece or a
single 4-color piece in these reduced orientations if you'd like.
Regards,
Brandon
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=20=20=20=20=20
<=
/div>
--964895041-548286253-1335839367=:14949--
From: "schuma" <mananself@gmail.com>
Date: Tue, 01 May 2012 04:48:55 -0000
Subject: Re: About the number of permutations of MC4D calculating
Charlie,
Your number is not half of the number on the website, but 3/4 of the number=
. And the difference is for the orientation of the last 4C piece.=20
When all the other pieces are solved, the last 4C piece has 4 possible orie=
ntations, but not 3. So the total number of orientations of 4C should be (4=
!/2)^15*4. Another way to get the same number is first counting all possibl=
e orientations for the last piece: (4!/2)^16, and then noting that one out =
of three is possible so /3: (4!/2)^16/3. I guess you may confused this two =
ways of counting so you wrote (4!/2)^15*3.=20
Nan
--- In 4D_Cubing@yahoogroups.com, Charlie Mckiz
>
>=20=20
> My solution is as follow
>=20
> Rules that MC4D obey:
> Position:=20
> 4-color pieces: The permutation of 4-color pieces is always
> even
> 2/3-color pieces: The net permutation of 3-color and 2-color
> is always even.
> Orientation: Consider rotation of a piece is a permutation
> of shading sides of that piece. So:
> 4-color pieces: The permutation of rotation of a 4-color
> piece is always even. Besides, the net rotation of all 4-color pieces is =
0.=20
> 3-color pieces: The net permutation of rotation of all
> 3-color pieces is even.
> 2-color pieces: The net permutation of rotation of all
> 2-color pieces is even.
> Number of different kinds of pieces:
> 4-color pieces: 16
> 3-color pieces: 32
> 2-color pieces: 24
> =C2=A0
> Based on the rules above:
> Permutation of 4-color pieces: 16! =C3=B72
> Orientation of 4-color pieces: (4! =C3=B72)15=C3=973
> Permutation of
> 2/3-color pieces: 32!=C3=9724! =C3=B72
> Orientation of 3-color pieces: (3!)32 =C3=B72
> Orientation of 2-color pieces: (2)24 =C3=B72
> Total: 16! =C3=B72=C3=97(4! =C3=B72)15=C3=973=C3=9732!=C3=9724! =C3=B72=
=C3=97(3!)32 =C3=B72=C3=97(2)24 =C3=B72
>=20
>=20
> ________________________________
> From: Brandon Enright
> To: 4D_Cubing@yahoogroups.com=20
> Cc: charliemckiz@...; bmenrigh@...=20
> Sent: Monday, April 30, 2012 3:38 PM
> Subject: Re: [MC4D] About the number of permutations of MC4D calculating
>=20=20
>=20
> =C2=A0=20
> -----BEGIN PGP SIGNED MESSAGE-----
> Hash: SHA1
>=20
> On Mon, 30 Apr 2012 15:31:05 -0000
> "charliemckiz@..."
>=20
> > Today I tried to calculate the number of MC4D permutations. My answer
> > is a half of what posted in the website
> > http://www.superliminal.com/cube/cube.htm . Does any one want to
> > check the answer? I'm pretty sure about mine. And I see no mistake in
> > the website solution.??
> >=20
>=20
> Well, I've aways taken the permutation calculation as gospel but I
> decided to give it a go myself and I've arrived at the same result.
>=20
> I find the sticker-based counting of permutations to be confusing so
> I've done it piece-based. Here are my notes.
>=20
> That is, there are:
>=20
> Permutations:
> * 8 centers with 192 "permutations" available. Fixing to one cancels
> all re-orientation of the tesseract
> * 24 face centers that can be in any permutation
> * 32 3-color edges who's parity is tied to the parity of the face
> centers
> * 16 4-color corners that can only be in an even permutation
>=20
> Orientations:
> * 8 centers with 24 orientations available (only 1 visually distinct)
> * 24 face pieces with 8 orientations available (only 2 visually
> distinct ones)
> * 32 3-color edges with 6 orientations available
> * 16 4-color edges with 12 orientations available
>=20
> The calculation is further complicated in that:
> * The total "flip" of the 24 face pieces is even (the 24th orientation
> is determined by the previous 23)
> * The orientation of the 32nd 2-color edge can only be in half (3) of
> the orientations.
> * The first 15 corners can be in any orientation but the last corner
> can only be in 1/3 (4) orientations.
>=20
> Taking all of that into account there should be:
> (24! * 2^23) * ((32! / 2) * (6^31 * 3)) * ((16! / 2) * (12^15 * 4))
>=20
> Which matches nicely with the website.
>=20
> The only "tricky" portion of the calculation is counting the
> orientations for the last 3-color and the last 4-color piece. I can
> create a macro to demonstrate how to put a single 3-color piece or a
> single 4-color piece in these reduced orientations if you'd like.
>=20
> Regards,
>=20
> Brandon
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>=20
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> =3DgpYl
> -----END PGP SIGNATURE-----
>
From: joelkarlsson97@gmail.com
Date: 07 May 2016 01:52:34 -0700
Subject: Re: About the number of permutations of MC4D calculating
From: joelkarlsson97@gmail.com
Date: Sat, 7 May 2016 19:09:38 -0500
Subject: Re: About the number of permutations of MC4D calculating
From: David Vanderschel <DvdS@Austin.RR.com>
Date: Sat, 7 May 2016 22:00:28 -0700
Subject: Re: [MC4D] Re: About the number of permutations of MC4D calculating
--------------9589B02661F778E389F3513E
Content-Type: text/plain; charset=utf-8; format=flowed
Content-Transfer-Encoding: 7bit
Very nice, Roice! Thanks for the milestone announcement. That is pretty
special.
I decided to try the same with first-time solvers of the 3^4 and put the
result here
from anywhere because I'm not sure that I want to maintain it. It's
interesting to note that after an initial gradual exponential period it
also appears to go linear. My wild-assed guess is that the exponential
part ended when it was almost guaranteed that anyone looking for a 4D
Rubik's cube would find this one. The linear portion then represents the
number of people in the world with both access and sufficient interest.
IOW, it is proportional to the birthrate of eventual MC4D solvers which
I would expect to be nearly linear. But like I said, it's a pretty
wild-assed guess. Please speak up if you have a different idea.
-Melinda
On 5/4/2016 5:06 PM, Roice Nelson roice3@gmail.com [4D_Cubing] wrote:
>
>
> Sorry, yahoo ate the image of the graph. I should have assumed, since
> that has been happening recently to others. I just uploaded the image
> to the group photos area, and hopefully this link will work.
>
> https://groups.yahoo.com/neo/groups/4D_Cubing/photos/albums/711088661
>
> seeya,
> Roice
>
>
> On Wed, May 4, 2016 at 4:15 PM, Roice Nelson
>
> Hi Hypercubers,
>
> We recently passed 50 solvers of the 3x3x3x3x3. Can you believe it?
>
> I made a graph of 2^5 and 3^5 solutions by date, and the trend
> looks close to linear in both cases. The 3^5 has a bit more data
> points, and one thing that popped out is how solutions tends to
> come in clusters of 3 or 4. Maybe this happens because of
> occasional publicity bursts, or maybe groups of friends discover
> MC5D together.
>
> Cheers,
> Roice
>
--------------9589B02661F778E389F3513E
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: quoted-printable
">
Very nice, Roice! Thanks for the milestone announcement. That is
pretty special.
I decided to try the same with first-time solvers of the 3^4 and put
the result here=
.
It's not linked from anywhere because I'm not sure that I want to
maintain it. It's interesting to note that after an initial gradual
exponential period it also appears to go linear. My wild-assed guess
is that the exponential part ended when it was almost guaranteed
that anyone looking for a 4D Rubik's cube would find this one. The
linear portion then represents the number of people in the world
with both access and sufficient interest. IOW, it is proportional to
the birthrate of eventual MC4D solvers which I would expect to be
nearly linear. But like I said, it's a pretty wild-assed guess.
Please speak up if you have a different idea.
-Melinda
l.com"
type=3D"cite">
uld
have assumed, since that has been happening recently to others.=C2=
=A0
I just uploaded the image to the group photos area, and
hopefully this link will work.
Nelson < href=3D"mailto:roice3@gmail.com" target=3D"_blank">roice3@gma=
il.com>
wrote:
.8ex;border-left:1px #ccc solid;padding-left:1ex">
n
you believe it?
the trend looks close to linear in both cases.=C2=A0 The 3^=
5
has a bit more data points, and one thing that popped
out is how solutions tends to come in clusters of 3 or
4.=C2=A0 Maybe this happens because of occasional publicity
bursts, or maybe groups of friends discover MC5D
together.
--------------9589B02661F778E389F3513E--
From: phamthihoa4444@gmail.com
Date: 08 May 2016 02:17:19 -0700
Subject: Re: About the number of permutations of MC4D calculating
From: phamthihoa4444@gmail.com
Date: 08 May 2016 05:50:02 -0700
Subject: Re: About the number of permutations of MC4D calculating
From: joelkarlsson97@gmail.com
Date: 14 May 2016 11:02:08 -0700
Subject: Re: About the number of permutations of MC4D calculating
From: joelkarlsson97@gmail.com
Date: Sat, 21 May 2016 22:31:24 -0700
Subject: Re: About the number of permutations of MC4D calculating
--------------07990B80C1E89EE64C413518
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OK, I've made the changes at the bottom of the permutations page
and able to verify it, I'd appreciate it.
Thanks Joel!
-Melinda
On 5/8/2016 5:50 AM, joelkarlsson97@gmail.com [4D_Cubing] wrote:
>
> I believe that the factors should be arranged like this:
> B: 6^31 * 12^15*4 * 16!
> S: 6^32/2 * 12^16/3 * 16!/2
>
> Like Pham pointed out, the middle factors (regarding orientation of
> corner pieces) are matching.
>
> The first factor is probably, as Pham said, about the orientation of
> central edge pieces (since it's a group of 32 pieces). David V has
> already proven that Smith is correct here since the orientation of the
> last of those pieces is limited by 1/2 (see:
> https://groups.yahoo.com/neo/groups/4D_Cubing/conversations/topics/2136)
>
>
> The third is indeed about corner permutation. It seems like Balandraud
> missed to account for the parity of the permuta tions of corner pieces
> and that Smith is correct in this regard as well.
>
> To conclude, Balandraud seems to have missed that the last central
> edge piece (3C piece) can have three different orientations and that
> the parity of permutations of corner pieces always are even. Thus, I
> suggest that Melinda changes the count at Superliminal to the value
> provided by Smith Wolfram|Alpha: Computational Knowledge Engine
>
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">
OK, I've made the changes at the bottom of the
page. If anyone is curious and able to verify it, I'd
appreciate it.
Thanks Joel!
-Melinda
I believe that the factors should be arranged like this:
Like Pham pointed out, the middle factors (regarding
orientation of corner pieces) are matching.
=C2=A0The first factor is probably, as Pham said, about
the=C2=A0orientation of central edge pieces (since it's a group o=
f
32 pieces). David V has already proven that Smith is correct
here since the orientation of the last of those pieces is
limited by 1/2 (see:=C2=A0llow"
target=3D"_blank"
href=3D"https://groups.yahoo.com/neo/groups/4D_Cubing/conversations/topics/=
2136">https://groups.yahoo.com/neo/groups/4D_Cubing/conversations/topics/21=
36)
The third is indeed about corner permutation. It seems like=C2=A0=
line-height: normal; word-spacing: normal;">Balandraud
missed to account for the parity of the permuta tions of
corner pieces and that Smith is correct in this regard as
well.
To conclude,=C2=A0 font-size: 12.8px; line-height: normal; word-spacing:
normal;">Balandraud seems to have missed that the last
central edge piece (3C piece) can have three different
orientations and that the parity of permutations of corner
pieces always are even. Thus, I suggest that Melinda changes
the count at Superliminal to the value provided by Smith=C2=A0<=
a
moz-do-not-send=3D"true" rel=3D"nofollow" target=3D"_blank"
href=3D"http://www.wolframalpha.com/input/?i=3D%2816%21%2F2%29+*+%2824%21*3=
2%21%2F2%29+*+64%21%2F2+*+%2896%21%2F24%5E24%29%5E2+*+64%21%2F%288%21%29%5E=
8+*+96%21%2F%2812%21%29%5E8+*+48%21%2F%286%21%29%5E8+*+12%5E16%2F3+*+6%5E32=
%2F2+*+3%5E64%2F3+*+2%5E24%2F2+*+%282%5E96%2F2%29%5E2">Wolfram|Alpha:
Computational Knowledge Engine)
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