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Well congratulations, Andrey, that's quite a puzzle story!
It sounds like the perfect puzzle for you. I've worked on many projects
that came out to around 100 hours, and I like that size. It's a big
investment, but you live with it for a matter of weeks or months and it
becomes like family. And then it's over and you can relax and enjoy the
memories.
As you know, I find reflection moves to be very interesting, so I am
particularly happy to know that this puzzle is difficult, solvable, and
solved. Would you like to tell us a little more about what you mean
about working with the whole structure, and what you had to keep in your
head that you couldn't write down? Perhaps the best way to ask this is
simply, what advice would you give to someone who might consider
attempting this feat? Clearly lots of macros and preparation. What else?
And these orbits you describe: Are they the lovely intertwined paths of
length 10 that run in straight lines through the 120 cell?
Lastly, what are your thoughts on short solutions? Your solutions always
seem to be very efficient, even when you are not trying. Were you happy
to break half a million twists or was it everything you could do just to
finish in a reasonable time?
Congratulations again,
-Melinda
On 1/30/2014 1:22 AM, andreyastrelin@yahoo.com wrote:
>
> I did it! It took 427287 twists and almost 90 hours by timer. 4C stage
> went without unexpected troubles (after preparing of large set of
> macros including 120deg rotation of the single piece).
>
> Probably it's a most difficult puzzle that I met. It required work
> with structure of the whole 120-cell, and I had to keep it in mind,
> because it's difficult to draw such object on 2D paper...
>
> This puzzle is unique because of multiple parity problems in the web
> of orbits of different dimensions. And I still don't know was it a
> luck that 3C parities were solved by combination of 3 symmetrical
> clusters of flips, or there is really not much freedom of possible
> combinations at that stage.
>
>
> If anybody is going to try this puzzle I can only wish him good
> luck. And a lot of patience...
>
>
> Andrey
>
>
>
>
--------------020008020900060402050801
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------=_NextPart_000_0016_01CF1DB6.30E437F0
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charset="UTF-8"
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Wow!
This is a big one.
Congratulations!
Ed
----- Original Message -----=20
From: andreyastrelin@yahoo.com=20
To: 4D_Cubing@yahoogroups.com=20
Sent: Thursday, January 30, 2014 10:22 AM
Subject: [MC4D] 120Z solved!!!
=20=20=20=20
I did it! It took 427287 twists and almost 90 hours by timer. 4C stage we=
nt without unexpected troubles (after preparing of large set of macros incl=
uding 120deg rotation of the single piece).
Probably it's a most difficult puzzle that I met. It required work with=
structure of the whole 120-cell, and I had to keep it in mind, because it'=
s difficult to draw such object on 2D paper...
This puzzle is unique because of multiple parity problems in the web of=
orbits of different dimensions. And I still don't know was it a luck that =
3C parities were solved by combination of 3 symmetrical clusters of flips, =
or there is really not much freedom of possible combinations at that stage.
If anybody is going to try this puzzle I can only wish him good luck. A=
nd a lot of patience...
Andrey
=20=20
------=_NextPart_000_0016_01CF1DB6.30E437F0
Content-Type: text/html;
charset="UTF-8"
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=EF=BB=BF
I did it! It took 427287 twists and almost 90 hours by timer. 4C stage=
went=20
without unexpected troubles (after preparing of large set of macros inclu=
ding=20
120deg rotation of the single piece).
Probably it's a most difficult puzzle that I met. It=20
required work with structure of the whole 120-cell, and I had t=
o=20
keep it in mind, because it's difficult to draw such object on 2D paper..=
.
This puzzle is unique because of multiple parity problems in th=
e web=20
of orbits of different dimensions. And I still don't know was it a luck t=
hat=20
3C parities were solved by combination of 3 symmetrical clusters of flips=
, or=20
there is really not much freedom of possible combinations at that stage.<=
/P>
If anybody is going to try this puzzle I can only wis=
h him=20
good luck. And a lot of patience...
Andrey
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--9nlWnTS9ocNmxdhsK3FKVLWiM6a5jwOaA8HvrLV--
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Wow! Sounds not easy at all. I will finishing a bigger MT soon. The "hyp {1=
0,3} v001" with about 15'000 twists (8'000 for you). Hence: much easier.
Have you all heard that they have found an Hamiltonian through Rubik's 3^3 =
?
http://bruce.cubing.net/ham333/rubikhamiltonexplanation.html
Regards
Ed
----- Original Message -----=20
From: andreyastrelin@yahoo.com=20
To: 4D_Cubing@yahoogroups.com=20
Sent: Thursday, January 30, 2014 5:12 PM
Subject: Re: [MC4D] 120Z solved!!!
=20=20=20=20
Melinda, thank you!
Let's begin with 2C. It's easy to find that simple commutator gives 3-loo=
p of 2C in one line, so one may try to solve puzzle line by line (like we d=
o with simple edge-rotating MT). But soon he finds that some line have odd =
permutation of pieces... and there begins the fun!
Flipping of one cell changes parity of permutations of all six lines pa=
ssing through it. So when you solve one line, you push disorder to five oth=
ers. And what we can try is to shrink disorder area from whole puzzle to sm=
aller ball... to one cell... to nothing.
Consider one plane (=3D sphere in 3D, =3D equator sphere). It contains =
6 lines, and every cell in plane belongs to two of them. So sum of parities=
of these lines is always zero. (First advice: draw a web of lines in one p=
lane and remember it. You will need it very often). Our first task will be =
to clear one plane.
You don't want to get lost in 120-cell surface, so I suggest to select =
"central" cell (for example, the cell that is in the center from the start)=
and remember it. You may give it some special color if you want. I didn't.=
Then you need to find "polar plane" for this cell (that is equator if you =
consider central cell as a pole). It will be our first plane to work.
Select any line on it. Each of 5 other lines in the plane intersects th=
is one, so you can solve parity disorders by single flips of cells on the s=
elected line. When you solve 5 lines, the sixth will have zero parity.
When the plane is solved, solve one of halves of puzzle (using commutat=
ors) by moving disorders to another half.=20
Select another plane that is close to first one. It intersects with uns=
olved ball by pentagon, and you can solve its sides one by one. So your uns=
olved area is shrinking... Soon you will see the single cell where s ix wro=
ng lines are intersecting. Flip this cell and finish 2C stage.
In 3C situation is similar, but the orbit of 3C piece is the plane. Eve=
ry cell is intersected by 15 planes, and its flipping changes parity of all=
these planes. So at first I suggest to select some half-line (5 cells in t=
he line going through the center) and move disorders in all planes to this =
area or close to it. It is not easy, geometry of the orbit is not intuitive=
ly clear, so you should carefully select paths for the pieces when move the=
m to their place. But finally you get most of 3C pieces in place, and only =
edges of the selected cells may be swapped (plus five other pairs in planes=
that are not intersected with the selected line). And now there it is the =
most difficult part.=20
You need to find some "clusters" flipping of which will not change pari=
ties of lines, but changes parities of some planes. Clusters that I found a=
re just & quot;layers" of 120-cell: 12 cells adjacent to one, or 20 cells o=
f the third layer. Flipping of the first cluster changes parity of 32 plane=
s and the second changes 24 planes. In the polar picture they look like 3rd=
and 4th layers for the first cluster and second and 4th layers for the sec=
ond one. To apply them to your situation you need to build the polar pictur=
e of disordered planes (that is, find polar cells for them) and decompose t=
he set of these cells to spherical clusters. In this stage I got picture wi=
th third order symmetry, so centers of clusters were limited to the symmetr=
y axis. After the second attempt I got zero parities of all planes and coul=
d finish 2C and 3C stages.
With 4C it's easy to build 3-cycle jumping diagonally across two cells,=
but it's difficult to work with it. So you have to find a path that moves =
a 4C to adjacent one (5 flips) - it will help you to build 3-cycle for piec=
es in one face . My macro for it was 232 flips long. Then build a path that=
returns 4C to its place rotated (8 flips long path exists) - you get rotat=
ion of two opposite pieces of one cell. And then you need a path that retur=
ns 4C to its place in wrong orientation. I found 17-flip path for it and go=
t macro for flipping of two opposite pieces. Commutator of flipping and rot=
ation gives 3-rotation of the single piece, but it's long (about 300 flips)=
.
And it's enough to finish the solve. Some problems will be in the end w=
hen you can't find a pair of opposite cells to flip, but it is solvable by =
extra moves.
Believe of not, but I consider my solve as short one :) I tried to use =
short macros when it was possible and leave long ones only for situations w=
hen it was difficult to apply short versions.=20=20
=20=20=20=20=20=20
Andrey
=20=20
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=EF=BB=BF
Melinda, thank you!
Let's begin with 2C. It's easy to find that simple commutator gives 3-=
loop=20
of 2C in one line, so one may try to solve puzzle line by line (like=
we=20
do with simple edge-rotating MT). But soon he finds that some l=
ine=20
have odd permutation of pieces... and there begins the fun!
Flipping of one cell changes parity of permutations of all six =
lines=20
passing through it. So when you solve one line, you push disorder to five=
=20
others. And what we can try is to shrink disorder area from whole pu=
zzle=20
to smaller ball... to one cell... to nothing.
Consider one plane (=3D sphere in 3D, =3D equator sphere).=
It=20
contains 6 lines, and every cell in plane belongs to two of them. So=
sum=20
of parities of these lines is always zero. (First advice: draw a web of l=
ines=20
in one plane and remember it. You will need it very often). Our firs=
t=20
task will be to clear one plane.
You don't want to get lost in 120-cell surface, so I suggest to=
=20
select "central" cell (for example, the cell that is in the center from t=
he=20
start) and remember it. You may give it some special color if you want. I=
=20
didn't. Then you need to find "polar plane" for this cell (that is equato=
r if=20
you consider central cell as a pole). It will be our first plane to work.=
Select any line on it. Each of 5 other lines in the plane inter=
sects=20
this one, so you can solve parity disorders by single flips of cells=
on=20
the selected line. When you solve 5 lines, the sixth will have =
zero=20
parity.
When the plane is solved, solve one of halves of puzzle (using=
=20
commutators) by moving disorders to another half.
Select another plane that is close to first one. It inters=
ects=20
with unsolved ball by pentagon, and you can solve its sides one=
by=20
one. So your unsolved area is shrinking... Soon you will see=20
the single cell where s ix wrong lines are intersecting. Flip t=
his=20
cell and finish 2C stage.
In 3C situation is similar, but the orbit of 3C piece =
;is=20
the plane. Every cell is intersected by 15 planes, and its flipping chang=
es=20
parity of all these planes. So at first I suggest to=20
select some half-line (5 cells in the line going through t=
he=20
center) and move disorders in all planes to this area or close to it=
. It=20
is not easy, geometry of the orbit is not intuitively clear, so you=
=20
should carefully select paths for the pieces when move them to their plac=
e.=20
But finally you get most of 3C pieces in place, and only edges of the sel=
ected=20
cells may be swapped (plus five other pairs in planes that are not inters=
ected=20
with the selected line). And now there it is the most difficult=20
part.
You need to find some "clusters" flipping of which will not cha=
nge=20
parities of lines, but changes parities of some planes. Clusters that I f=
ound=20
are just & quot;layers" of 120-cell: 12 cells adjacent to one, or 20 =
cells=20
of the third layer. Flipping of the first cluster changes parity of 32 pl=
anes=20
and the second changes 24 planes. In the polar picture they look like 3rd=
and=20
4th layers for the first cluster and second and 4th layers for the s=
econd=20
one. To apply them to your situation you need to build the polar pic=
ture=20
of disordered planes (that is, find polar cells for them) and decomp=
ose=20
the set of these cells to spherical clusters. In this stage I got=20
picture with third order symmetry, so centers of clusters were limit=
ed to=20
the symmetry axis. After the second attempt I got zero parities of all pl=
anes=20
and could finish 2C and 3C stages.
With 4C it's easy to build 3-cycle jumping diagonally across tw=
o=20
cells, but it's difficult to work with it. So you have to find a path tha=
t=20
moves a 4C to adjacent one (5 flips) - it will help you to buil=
d=20
3-cycle for pieces in one face . My macro for it was 232 flips =
long.=20
Then build a path that returns 4C to its place rotated (8 flips long path=
=20
exists) - you get rotation of two opposite pieces of one cell. And t=
hen=20
you need a path that returns 4C to its place in wrong orientation. I foun=
d=20
17-flip path for it and got macro for flipping of two opposite=20
pieces. Commutator of flipping and rotation gives 3-rotation of the=
=20
single piece, but it's long (about 300 flips).
And it's enough to finish the solve. Some problems will be=
in=20
the end when you can't find a pair of opposite cells to flip, but it=
is=20
solvable by extra moves.
Believe of not, but I consider my solve as short one :) I =
tried=20
to use short macros when it was possible and leave long ones only fo=
r=20
situations when it was difficult to apply short versions.
Andrey
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I'm wondering whether there might be another fun puzzle type here, or at
least a useful tool for solving this one. Imagine a 2-color version, or
even 1-color in which inverted stickers are highlighted. This would be
very much like a "Lights Out" puzzle because the goal would be to turn
off all the highlighted stickers.
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--------------090201030708070103030704
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I don't recall that message but I know that Lights Out has been
mentioned before. My suggested model is based on all pieces (cubies) of
an entire cell rather than on individual pieces. Call it "cell based" if
you like. That seems to me to be the more natural version of Lights Out
for 'Z' puzzles. I would expect it to be much harder than a piece-based
puzzle, but also much easier than the normal Z puzzles.
On 1/30/2014 3:52 PM, mananself@gmail.com wrote:
>
> The geometric lights out puzzle is what I suggested earlier in this
> thread... Later I tried playing the Cube and the icosahedron version
> of "Lights out" on a piece of paper. "Clicking" a vertex flip the
> state of adjacent edges. They were pretty trivial to solve. Maybe if
> we increase the number of edges that each "click" affects, the puzzle
> becomes harder.
>
> The real deal here in 120-cell is pretty hard to illustrate... Some
> configurations
>
> great Lights out puzzles.
>
--------------090201030708070103030704
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">
I don't recall that message but I know that Lights Out has been
mentioned before. My suggested model is based on all pieces (cubies)
of an entire cell rather than on individual pieces. Call it "cell
based" if you like. That seems to me to be the more natural version
of Lights Out for 'Z' puzzles. I would expect it to be much harder
than a piece-based puzzle, but also much easier than the normal Z
puzzles.
helvetica, clean,
sans-serif;background-color:transparent;font-style:normal;">The
geometric lights out puzzle is what I suggested earlier in this
thread... Later I tried playing the Cube and the icosahedron
version of "Lights out" on a piece of paper. "Clicking" a vertex
flip the state of adjacent edges.=C2=A0
were pretty trivial to solve. Maybe if we increase the number
of edges that each "click" affects, the puzzle becomes harder.pan>
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--------------000809000706030102020607
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Ah, thanks for the reference, Andrey. Yahoo groups changed their UI
recently and I can't figure out how to search within all messages. Do
you know how?
I think that your line-based suggestion is different from both Nan's
geometry-based and my cell-based ideas. It sounds to me like your idea
makes the most sense on an unsliced puzzle. IE strictly 1C pieces only.
Is that right? That might make it the easiest of the 3, though every
click would affect 1/3 of the whole puzzle, so perhaps. It could also be
quite lovely. I could even imagine creating a physical version of this.
-Melinda
On 1/30/2014 4:18 PM, andreyastrelin@yahoo.com wrote:
>
>
> Melinda,
>
>
> It was message 2911:
> https://groups.yahoo.com/neo/groups/4D_Cubing/conversations/messages/2911
>
> Yes, it may be possible to show 120 balls connected by green and
> red lines. Click in each ball changes colors of all lines passing
> through it. Goal is to make all lines green
>
> For orientation you can enable user to change colors of balls (say,
> from white to yellow and back) independently of lines colors.
>
> With 2^36 different situations this puzzle may be not very easy,
> and it can give good idea about 120-cell geometry.
>
>
> Andrey
>
>
>
>
--------------000809000706030102020607
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Melinda,
=C2=A0=C2=A0 It was message 2911: rel=3D"nofollow" target=3D"_blank"
href=3D"https://groups.yahoo.com/neo/groups/4D_Cubing/conversations/message=
s/2911">https://groups.yahoo.com/neo/groups/4D_Cubing/conversations/message=
s/2911
=C2=A0=C2=A0 Yes, it may be possible to show 120 balls connected b=
y green
and red lines. Click in each ball changes colors of all lines
passing through it. Goal is to make all lines green
=C2=A0 For orientation you can enable user to change colors of bal=
ls
(say, from white to yellow and back) independently of lines
colors.
=C2=A0 With 2^36 different situations=C2=A0this puzzle=C2=A0may be=
not very
easy, and=C2=A0it can give good idea about 120-cell geometry.
=C2=A0 Andrey=C2=A0
=20=20=20=20=20=20
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That's a wonderful puzzle story, Nan. Congratulations! In the middle, I=20
believe you said it's the hardest puzzle you ever attempted, right? It=20
certainly sounds like a bear! Is anybody else working on it or thinking=20
about it? I may feel as frightened of it as other people are of MC4D=20
when I describe it to them. I often don't have the heart to tell them of=20
all the other crazy puzzles we deal with. The human mind can take only=20
so much amazement at one time.
I don't think you should feel disappointed by having used computer=20
assistance, especially because you wrote the code. That code is an=20
extension of your mind and required you to understand what you wanted it=20
to do which tells me that with sufficient time and patience, you could=20
in theory have performed it all by hand. Don Hatch wrote a program to=20
solve the 2^d and 3^d cubes in all dimensions > 2 I believe. I therefore=20
believe that he has solved them all, just not in the same way as human=20
solutions. This is by the same extension that we consider macro=20
solutions to be completely valid too, just not comparable to no-macro=20
solutions.
I am particularly curious about that 3C stage and wonder whether it=20
might suggest a much smaller puzzle with the same sort of difficulty.=20
Can you think of a 3C-only analog puzzle that requires a similarly deep=20
dive to solve but perhaps not involving so many cells? For instance, how=20
does this step compare with, say, an edges-only version of the 24Z? Or=20
MC4DZ, or even a 3D Void Cube=20
Now that I look at that last puzzle I realize that it looks and turns=20
exactly like a subset of MC4D! I want one! It looks simple but I expect=20
that a Z version would be much more difficult if it were somehow=20
possible to build.
So what's next for you, Nan? Some time outside in our lovely pre-spring=20
weather perhaps?
-Melinda
On 3/7/2014 11:15 PM, mananself@gmail.com wrote:
>
>
> Finally, I've solved the 120-cell Mirror Z. It took me 357330 steps=20
> (happen to be an even number!). The game clock shows 38 hours and 45=20
> minutes.
>
> As mentioned earlier, I used some script to analyze the orbits. I=20
> found that after solving the 3C pieces, all 2C orbits are evenly=20
> permuted. So, as planned, I solved 3C first, then 2C, then 4C. 3C took=20
> about 19 hours and 85k moves; 2C about 2 hour and 24k moves; and 4C=20
> about 17.5 hours and 248k moves. 2C pieces were indeed easy after 3C=20
> were solved.
>
> I have to admit that this time I went too deep using computer. I used=20
> a parity fixing strategy found by the script for 3C orbits: flipping=20
> 19 cells changes the parity of seven 3C orbits which pass through one=20
> cell. This is the most powerful 3C parity algorithm I've found.=20
> Without computer I'd have a hard time finding such 19 cells. I used=20
> this algorithm several times towards the end of the 3C step. I'm=20
> certainly not proud of it. But it's hard not to look at the 19 cells=20
> when the script has already generated it.
>
> As before, I used a monitoring script to keep track of progress so=20
> that I know it when I screw up things. I also used Git to backup the=20
> macro and log files.
>
> Scripts and files are dumped into this github repo:
>
> https://github.com/nanma80/Z120
>
> I've never organized the script though...
>
> Andrey, thank you for making this puzzle! Also, if you didn't solve=20
> the puzzle, I wouldn't even know how delicate this puzzle was and I=20
> wouldn't even attempt it.
>
> Nan
>
>
>
>=20
--------------050009090404080801030209
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Content-Transfer-Encoding: quoted-printable
">
That's a wonderful puzzle story, Nan. Congratulations! In the
middle, I believe you said it's the hardest puzzle you ever
attempted, right? It certainly sounds like a bear! Is anybody else
working on it or thinking about it? I may feel as frightened of it
as other people are of MC4D when I describe it to them. I often
don't have the heart to tell them of all the other crazy puzzles we
deal with. The human mind can take only so much amazement at one
time.
I don't think you should feel disappointed by having used computer
assistance, especially because you wrote the code. That code is an
extension of your mind and required you to understand what you
wanted it to do which tells me that with sufficient time and
patience, you could in theory have performed it all by hand. Don
Hatch wrote a program to solve the 2^d and 3^d cubes in all
dimensions > 2 I believe. I therefore believe that he has solved
them all, just not in the same way as human solutions. This is by
the same extension that we consider macro solutions to be completely
valid too, just not comparable to no-macro solutions.
I am particularly curious about that 3C stage and wonder whether it
might suggest a much smaller puzzle with the same sort of
difficulty. Can you think of a 3C-only analog puzzle that requires a
similarly deep dive to solve but perhaps not involving so many
cells? For instance, how does this step compare with, say, an
edges-only version of the 24Z? Or MC4DZ, or even a
Z for that matter?
Now that I look at that last puzzle I realize that it looks and
turns exactly like a subset of MC4D! I want one! It looks simple but
I expect that a Z version would be much more difficult if it were
somehow possible to build.
So what's next for you, Nan? Some time outside in our lovely
pre-spring weather perhaps?
-Melinda
357330 steps (happen to be an even number!). The game clock
shows 38 hours and 45 minutes.
orbits. I found that after solving the 3C pieces, all 2C
orbits are evenly permuted. So, as planned, I solved 3C first,
then 2C, then 4C. 3C took about 19 hours and 85k moves; 2C
about 2 hour and 24k moves; and 4C about 17.5 hours and 248k
moves. 2C pieces were indeed easy after 3C were solved.
computer. I used a parity fixing strategy found by the script
for 3C orbits: flipping 19 cells changes the parity of seven
3C orbits which pass through one cell. This is the most
powerful 3C parity algorithm I've found. Without computer I'd
have a hard time finding such 19 cells. I used this algorithm
several times towards the end of the 3C step. I'm certainly
not proud of it. But it's hard not to look at the 19 cells
when the script has already generated it.
progress so that I know it when I screw up things. I also used
Git to backup the macro and log files.=C2=A0
didn't solve the puzzle, I wouldn't even know how delicate
this puzzle was and I wouldn't even attempt it.
=20=20=20=20=20=20
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Nan and Andrey,
Perhaps a 3C-only 24-cell will be more difficult without the 2C pieces=20
to help get the 3C pieces into the correct parity? I'm wondering if we=20
can find a puzzle that preserves the most difficult part of the 120Z but=20
in a much smaller puzzle. I'm particularly interested in whether this=20
can be done in 3D. The nature of the Z puzzles would make it hard to=20
imagine how to produce physical versions, but it would be ideal to find=20
the lowest dimension in which this puzzle can exist. Thoughts?
As an aside, I've been meaning to ask Andrey what he would think of=20
renaming the Z puzzles to X. I'm proposing this because the shape of the=20
letter 'X' looks like a map of how all parts of a cell transform through=20
the cell center, ending up at their antipodes. I know it's a little late=20
in the game for this but I keep thinking of it so I thought I'd throw it=20
out there. No pressure, though. I like 'Z' as well because I have a=20
Nissan 350Z. :-)
-Melinda
On 3/8/2014 11:26 AM, mananself@gmail.com wrote:
>
>
> Melinda and Andrey, Thanks.
>
> As I posted earlier here, when I solved 24Z, I solved 2C first. There,=20
> 2C orbits do have a complicated orbit parity situation. But because=20
> there are less orbits and less pieces, the parity can be solved=20
> intuitively. After solving 2C, 3C orbits are all evenly permuted. So=20
> no drama for 3C in 24Z, unlike in 120Z. Maybe it's because an=20
> octahedron has 6 (even number) pairs of edges, but a dodecahedron has=20
> 15 (odd number) pairs of edges.
>
> 3^4 Z is even cuter. I looked up my note, and I solved 3^4 Z with the=20
> order of 4C, 2C, 3C. It's because 4C pieces are in a unusual group.=20
> And there are parity situation in each step. But they are not very=20
> difficult. It's hard to compare that puzzle to 24Z and 120Z.
>
> Andrey, you mentioned that you would like to try vertex turning=20
> 24-cell and 16-cell. Are they just equivalent to cell-turning 24-cell=20
> and 8-cell due to duality?
>
> Nan
>
>
>=20
--------------050305090608050901080501
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Melinda and Andrey, Thanks.
As I posted earlier here, when I solved 24Z, I solved 2C
first. There, 2C orbits do have a complicated orbit parity
situation. But because there are less orbits and less pieces,
the parity can be solved intuitively. After solving 2C, 3C
orbits are all evenly permuted. So no drama for 3C in 24Z,
unlike in 120Z. Maybe it's because an octahedron has 6 (even
number) pairs of edges, but a dodecahedron has 15 (odd number)
pairs of edges.
3^4 Z is even cuter. I looked up my note, and I solved 3^4 Z
with the order of 4C, 2C, 3C. It's because 4C pieces are in a
unusual group. And there are parity situation in each step. But
they are not very difficult. It's hard to compare that puzzle to
24Z and 120Z.
Andrey, you mentioned that you would like to try vertex
turning 24-cell and 16-cell. Are they just equivalent to
cell-turning 24-cell and 8-cell due to duality?
Nan
=20=20=20=20=20=20=20=20
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Very very interesting indeed! Thanks a lot for the link.
Ed
----- Original Message -----=20
From: mananself@gmail.com=20
To: 4D_Cubing@yahoogroups.com=20
Sent: Tuesday, March 11, 2014 6:09 AM
Subject: Re: [MC4D] RE: 120Z solved!!!
=20=20=20=20
I always think the solving the orbit parity is like a Lights Out game. Th=
ese days I did some research and found some Lights Out puzzles on Jaap's pu=
zzle page. First I found this Orbix puzzle:
http://www.jaapsch.net/puzzles/orbix.htm
It has the shape of a dodecahedron. The type-1 version is a simple and ne=
at Lights Out on a dodecahedron.
Jaap made a very powerful game, Lights Out on a Graph.
http://www.jaapsch.net/puzzles/lograph.htm
One can draw any graph and thus defines a Lights Out puzzle. Players can =
also share the puzzle definitions by uploading and downloading. There are a=
lot of predefined puzzles already. And of course there isn't one that's eq=
uivalent to 3C of 120 cell. That's too crazy... But there is a hypercube in=
the folder "Les". I could spend a lot of time on this game.
Melinda asked me what's next for me. I was playing 120-cell Mirror Z on m=
y commute. Today I played Candy Crush. It was not easy...
Nan
---In 4D_Cubing@yahoogroups.com,
Nan and Andrey,
Perhaps a 3C-only 24-cell will be more difficult without the 2C pieces to=
help get the 3C pieces into the correct parity? I'm wondering if we can fi=
nd a puzzle that preserves the most difficult part of the 120Z but in a muc=
h smaller puzzle. I'm particularly interested in whether this can be done i=
n 3D. The nature of the Z puzzles would make it hard to imagine how to prod=
uce physical versions, but it would be ideal to find the lowest dimension i=
n which this puzzle can exist. Thoughts?
As an aside, I've been meaning to ask Andrey what he would think of renam=
ing the Z puzzles to X. I'm proposing this because the shape of the letter =
'X' looks like a map of how all parts of a cell transform through the cell =
center, ending up at their antipodes. I know it's a little late in the game=
for this but I keep thinking of it so I thought I'd throw it out there. No=
pressure, though. I like 'Z' as well because I have a Nissan 350Z. :-)
-Melinda
On 3/8/2014 11:26 AM, mananself@... wrote:
Melinda and Andrey, Thanks.
As I posted earlier here, when I solved 24Z, I solved 2C first. There=
, 2C orbits do have a complicated orbit parity situation. But because there=
are less orbits and less pieces, the parity can be solved intuitively. Aft=
er solving 2C, 3C orbits are all evenly permuted. So no drama for 3C in 24Z=
, unlike in 120Z. Maybe it's because an octahedron has 6 (even number) pair=
s of edges, but a dodecahedron has 15 (odd number) pairs of edges.
3^4 Z is even cuter. I looked up my note, and I solved 3^4 Z with the=
order of 4C, 2C, 3C. It's because 4C pieces are in a unusual group. And th=
ere are parity situation in each step. But they are not very difficult. It'=
s hard to compare that puzzle to 24Z and 120Z.
Andrey, you mentioned that you would like to try vertex turning 24-ce=
ll and 16-cell. Are they just equivalent to cell-turning 24-cell and 8-cell=
due to duality?
Nan
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href=3D"mailto:mananself@gmail.com">mananself@gmail.com
I always think the solving the orbit parity is like a Lights Out game.=
=20
These days I did some research and found some Lights Out puzzles on Jaap'=
s=20
puzzle page. First I found this Orbix puzzle: