--005045016447ff4d550479c199b8
Content-Type: text/plain; charset=UTF-8
Hello everybody! This is Chris again, and I thought some of you might like
to hear an update on how my length 4 solve of the {5}x{4} duoprism went.
I thought that with the length 5 down, I would take a bit of a break, but
the gap in the record table felt so tempting, and I had be thinking about
how I already have done most of the work needed to solve it by doing the
length 3 and 5 puzzles. Turns out that my algorithms for 3 cycling face and
edge pieces were perfectly usable in the length 4 case to build up all faces
and edges. Only thing that caught me a little off during this phase was the
reappearance of, what I guess could be called degenerate pieces. In the
length 2 and 4 puzzle, there are some pieces in the edge and face blocks
that don't have the same number of colors as the other pieces. As in, there
are some 1c pieces that are equivalent to corresponding length 5 2c pieces,
where the color that is part of the square torus is gone due to the shape of
the cuts. Likewise, there are 2c edge pieces sandwiched between two 3c edge
pieces. Turns out this isn't too much of a problem, and it actually kind of
makes it easier as you don't have to worry about an extra color matching up,
and the algorithms are exactly the same as the length 5 case.
The real problem with the length 4 is the possibility of parity issues. A
parity issue, for those unaware, occurs in even length puzzles precisely
because you don't have 'central' pieces to each block, so when you build
your centers, faces, and edges, you can't be sure that you are putting them
together in the correct permutation to be solvable or not when you reduce it
to its equivalent length 3 puzzle. Oh yeah, if you want to solve the puzzle
for yourself, you might want to opt out of the following discussion on the
parity issues present. In that case, you can skip the next paragraph.
For faces in this puzzle, it turns out that with a 3 cycle algorithm for
each of the kinds of face pieces, you can actually solve any potential
parity issue with just those macros alone. The faces between pentagonal
facets (2c(5,5)) seem to have no issues. The 2c(4,4) faces are pretty easy
to fix if you end up with a case where you need to flip one or swap two
blocks. This is because while those two cases have odd parity when you
consider the reduced length 3 puzzle, in terms of the individual pieces they
have even parity so can be fixed by the even 3 cycles (remember that 3 cycle
is an 'even' permutation). If you consider the 2c(5,4) faces, there are
basically two kinds of pieces. Pieces on a given diagonal within the face
can be cycled by algorithms, but you can never place a face piece outside of
its diagonal. You can discover this by playing around with the various
kinds of possible twists. So this case is similar to the 2c(4,4), except
you have to do each diagonal separately, whereas 2c(4,4) has all the pieces
identical. The real problem comes with edges. From the previous logic, it
can be seen that a single swap of two edges isn't too bad as that is
equivalent to swapping two pairs of pieces, which is even. So that can be
fixed with a 3 cycle macro of edge pieces. But if you end up with a a
single flipped 3c(5,4,4) edge, then bad luck for you: that is the only real
parity problem that can occur in the length 4 pentagonal duoprism that can't
be fixed by the macros you already have from length 5! I played around a
bunch before I finally stepped back to think about the problem and realized
what I just stated, which is that it is impossible to fix with just my 3
cycle. Once I did that, I spent a couple days of thinking about the problem
in my free time without actually touching the puzzle, because I didn't
really know what to do. When I thought that it might be necessary to break
everything up and rebuild, I had to then categorize the possible twists, to
determine exactly which twist I need to do to apply an odd permutation to
edge pieces so that I get the correct parity. Turns out there are only two
possibilities when you think about it: twisting a pentagonal facet on a
2c(5,4) face / 3c(5,4,4) edge (equivalent), or doing the same twist while
holding down '2' on your keyboard. I'm not sure sure what that is called,
so I'll just call it twist2. Now, the first twist can be thrown out because
it does something else that is odd: it also does an odd permutation (to be
precise, a single pair swap) of the 2c(5,5) face central piece. This
problem can only be fixed by the very same twist, which unfortunately means
that no matter how you twist this way, you will have either your edges in
odd parity, or your 2c(5,5) faces in odd parity. So the only possibility is
the twist2. Now that I had this knowledge in hand, I proceeded first to try
to find a better way to use this knowledge than just rebuilding from
scratch. I won't give the algorithm here, but turns out if you consider
just the 1c pieces, you can use this twist2 along with some other carefully
chosen twists to make a move sequence that changes parity of the edge
pieces, and keeps your 1c pieces intact. You have to remember though that
whatever your algorithm is, it 'must' have an odd number of these twist2s in
order to put the cube in a solvable orientation. Turns out that the
algorithm I used kept all edge blocks together, and only broke up a handful
of the 2c(4,4) face blocks. This is not as good as an algorithm that just
flips the edge I want, but it's much better than the alternative of
rebuilding everything! So I then proceeded to use my 2c(4,4) face 3 cycle
macros to rebuild the faces, then solved as a length 3 puzzle, and voila!
Victory!
Well, it was quite an adventure, but I'm glad I followed this pentagonal
duoprism road this far. There are still be length 6 or 7 puzzles left
undone, and I'm quite sure that my solves were pretty inefficient and can be
toppled in twist count too though :D. Not sure I have the time and effort
to take on anything above length 5 though, but they are awesome achievements
which I'm sure someone will one complete one day I'm quite sure.
Not sure what I'll attempt next, but one thing that is sorely missing from
my hypercubing experience, is the 5D cube. I have avoided that monster for
a long time! I like the 4D puzzles because the 3D projection that we can
work with in MagicCube 4D makes it possible to use innate 3D visualization
skills. Of course this is also why I have problems with really small 4D
puzzles like the simplex, because they are so small that the 4D-ness is
really important, whereas other puzzles are large enough that you can work
in locally 3D regions effectively. The 5D cube is another beast entirely
though, as you can't rely on 3D instincts to guide you at all it feels.
Nevertheless, one day I'll will make a serious push to solve that one too I
hope! Until then, I will enjoy this closer to 3D MagicCube 4D software as
much as I can :)
Chris
--005045016447ff4d550479c199b8
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
Hello everybody!=C2=A0 This is Chris again, and I thought some of you might=
like to hear an update on how my length 4 solve of the {5}x{4} duoprism we=
nt.
I thought that with the length 5 down, I would take a bit of a b=
reak, but the gap in the record table felt so tempting, and I had be thinki=
ng about how I already have done most of the work needed to solve it by doi=
ng the length 3 and 5 puzzles.=C2=A0 Turns out that my algorithms for 3 cyc=
ling face and edge pieces were perfectly usable in the length 4 case to bui=
ld up all faces and edges.=C2=A0 Only thing that caught me a little off dur=
ing this phase was the reappearance of, what I guess could be called degene=
rate pieces.=C2=A0 In the length 2 and 4 puzzle, there are some pieces in t=
he edge and face blocks that don't have the same number of colors as th=
e other pieces.=C2=A0 As in, there are some 1c pieces that are equivalent t=
o corresponding length 5 2c pieces, where the color that is part of the squ=
are torus is gone due to the shape of the cuts.=C2=A0 Likewise, there are 2=
c edge pieces sandwiched between two 3c edge pieces.=C2=A0 Turns out this i=
sn't too much of a problem, and it actually kind of makes it easier as =
you don't have to worry about an extra color matching up, and the algor=
ithms are exactly the same as the length 5 case.
The real problem with the length 4 is the possibility of parity issues.=
=C2=A0 A parity issue, for those unaware, occurs in even length puzzles pre=
cisely because you don't have 'central' pieces to each block, s=
o when you build your centers, faces, and edges, you can't be sure that=
you are putting them together in the correct permutation to be solvable or=
not when you reduce it to its equivalent length 3 puzzle.=C2=A0 Oh yeah, i=
f you want to solve the puzzle for yourself, you might want to opt out of t=
he following discussion on the parity issues present.=C2=A0 In that case, y=
ou can skip the next paragraph.
For faces in this puzzle, it turns out that with a 3 cycle algorithm fo=
r each of the kinds of face pieces, you can actually solve any potential pa=
rity issue with just those macros alone.=C2=A0 The faces between pentagonal=
facets (2c(5,5)) seem to have no issues.=C2=A0 The 2c(4,4) faces are prett=
y easy to fix if you end up with a case where you need to flip one or swap =
two blocks.=C2=A0 This is because while those two cases have odd parity whe=
n you consider the reduced length 3 puzzle, in terms of the individual piec=
es they have even parity so can be fixed by the even 3 cycles (remember tha=
t 3 cycle is an 'even' permutation).=C2=A0 If you consider the 2c(5=
,4) faces, there are basically two kinds of pieces.=C2=A0 Pieces on a given=
diagonal within the face can be cycled by algorithms, but you can never pl=
ace a face piece outside of its diagonal.=C2=A0 You can discover this by pl=
aying around with the various kinds of possible twists.=C2=A0 So this case =
is similar to the 2c(4,4), except you have to do each diagonal separately, =
whereas 2c(4,4) has all the pieces identical.=C2=A0 The real problem comes =
with edges.=C2=A0 From the previous logic, it can be seen that a single swa=
p of two edges isn't too bad as that is equivalent to swapping two pair=
s of pieces, which is even.=C2=A0 So that can be fixed with a 3 cycle macro=
of edge pieces.=C2=A0 But if you end up with a a single flipped 3c(5,4,4) =
edge, then bad luck for you:=C2=A0 that is the only real parity problem tha=
t can occur in the length 4 pentagonal duoprism that can't be fixed by =
the macros you already have from length 5!=C2=A0 I played around a bunch be=
fore I finally stepped back to think about the problem and realized what I =
just stated, which is that it is impossible to fix with just my 3 cycle.=C2=
=A0 Once I did that, I spent a couple days of thinking about the problem in=
my free time without actually touching the puzzle, because I didn't re=
ally know what to do.=C2=A0 When I thought that it might be necessary to br=
eak everything up and rebuild, I had to then categorize the possible twists=
, to determine exactly which twist I need to do to apply an odd permutation=
to edge pieces so that I get the correct parity.=C2=A0 Turns out there are=
only two possibilities when you think about it:=C2=A0 twisting a pentagona=
l facet on a 2c(5,4) face / 3c(5,4,4) edge (equivalent), or doing the same =
twist while holding down '2' on your keyboard.=C2=A0 I'm not su=
re sure what that is called, so I'll just call it twist2.=C2=A0 Now, th=
e first twist can be thrown out because it does something else that is odd:=
=C2=A0 it also does an odd permutation (to be precise, a single pair swap) =
of the 2c(5,5) face central piece.=C2=A0 This problem can only be fixed by =
the very same twist, which unfortunately means that no matter how you twist=
this way, you will have either your edges in odd parity, or your 2c(5,5) f=
aces in odd parity.=C2=A0 So the only possibility is the twist2.=C2=A0 Now =
that I had this knowledge in hand, I proceeded first to try to find a bette=
r way to use this knowledge than just rebuilding from scratch.=C2=A0 I won&=
#39;t give the algorithm here, but turns out if you consider just the 1c pi=
eces, you can use this twist2 along with some other carefully chosen twists=
to make a move sequence that changes parity of the edge pieces, and keeps =
your 1c pieces intact.=C2=A0 You have to remember though that whatever your=
algorithm is, it 'must' have an odd number of these twist2s in ord=
er to put the cube in a solvable orientation.=C2=A0 Turns out that the algo=
rithm I used kept all edge blocks together, and only broke up a handful of =
the 2c(4,4) face blocks.=C2=A0 This is not as good as an algorithm that jus=
t flips the edge I want, but it's much better than the alternative of r=
ebuilding everything!=C2=A0 So I then proceeded to use my 2c(4,4) face 3 cy=
cle macros to rebuild the faces, then solved as a length 3 puzzle, and voil=
a!=C2=A0 Victory!
Well, it was quite an adventure, but I'm glad I followed this penta=
gonal duoprism road this far.=C2=A0 There are still be length 6 or 7 puzzle=
s left undone, and I'm quite sure that my solves were pretty inefficien=
t and can be toppled in twist count too though :D.=C2=A0 Not sure I have th=
e time and effort to take on anything above length 5 though, but they are a=
wesome achievements which I'm sure someone will one complete one day I&=
#39;m quite sure.=C2=A0
Not sure what I'll attempt next, but one thing that is sorely missi=
ng from my hypercubing experience, is the 5D cube.=C2=A0 I have avoided tha=
t monster for a long time!=C2=A0 I like the 4D puzzles because the 3D proje=
ction that we can work with in MagicCube 4D makes it possible to use innate=
3D visualization skills.=C2=A0 Of course this is also why I have problems =
with really small 4D puzzles like the simplex, because they are so small th=
at the 4D-ness is really important, whereas other puzzles are large enough =
that you can work in locally 3D regions effectively.=C2=A0 The 5D cube is a=
nother beast entirely though, as you can't rely on 3D instincts to guid=
e you at all it feels.=C2=A0 Nevertheless, one day I'll will make a ser=
ious push to solve that one too I hope!=C2=A0 Until then, I will enjoy this=
closer to 3D MagicCube 4D software as much as I can :)
Chris
--005045016447ff4d550479c199b8--
Hi Chris,
Nice write-up there, and well done on suitably conquering the {5}x{4} duopr=
isms :). I did skip the paragraph on the parity though, I fully intend to =
get around to solving these myself and I'd rather figure it out for myself =
(not to mention trying to topple the twist counts ;) )
I am fully of the belief that if you enjoy the 4D puzzles a lot and can sol=
ve them, then 5D is certainly worth a go for the extra challenge. It is ju=
st the same principle as 4D, just taken a bit further and as such, more con=
fusing to start with. However, you may be pleased to know that I think tha=
t 3D visualisation skills (and 4D 'visualisation' as well as much as it exi=
sts) are still perfectly valid, as long as you put in enough effort to get =
your head around how the different axes behave. Just play around with it f=
or a while and you get the hang of it, that's what I did. Also, all the te=
chniques from 4D work too.
I'll be looking out for where your name crops up next for solving something=
around here.
Matthew
--- In 4D_Cubing@yahoogroups.com, Chris Locke
>
> Hello everybody! This is Chris again, and I thought some of you might li=
ke
> to hear an update on how my length 4 solve of the {5}x{4} duoprism went.
>=20
> I thought that with the length 5 down, I would take a bit of a break, but
> the gap in the record table felt so tempting, and I had be thinking about
> how I already have done most of the work needed to solve it by doing the
> length 3 and 5 puzzles. Turns out that my algorithms for 3 cycling face =
and
> edge pieces were perfectly usable in the length 4 case to build up all fa=
ces
> and edges. Only thing that caught me a little off during this phase was =
the
> reappearance of, what I guess could be called degenerate pieces. In the
> length 2 and 4 puzzle, there are some pieces in the edge and face blocks
> that don't have the same number of colors as the other pieces. As in, th=
ere
> are some 1c pieces that are equivalent to corresponding length 5 2c piece=
s,
> where the color that is part of the square torus is gone due to the shape=
of
> the cuts. Likewise, there are 2c edge pieces sandwiched between two 3c e=
dge
> pieces. Turns out this isn't too much of a problem, and it actually kind=
of
> makes it easier as you don't have to worry about an extra color matching =
up,
> and the algorithms are exactly the same as the length 5 case.
>=20
> The real problem with the length 4 is the possibility of parity issues. =
A
> parity issue, for those unaware, occurs in even length puzzles precisely
> because you don't have 'central' pieces to each block, so when you build
> your centers, faces, and edges, you can't be sure that you are putting th=
em
> together in the correct permutation to be solvable or not when you reduce=
it
> to its equivalent length 3 puzzle. Oh yeah, if you want to solve the puz=
zle
> for yourself, you might want to opt out of the following discussion on th=
e
> parity issues present. In that case, you can skip the next paragraph.
>=20
> For faces in this puzzle, it turns out that with a 3 cycle algorithm for
> each of the kinds of face pieces, you can actually solve any potential
> parity issue with just those macros alone. The faces between pentagonal
> facets (2c(5,5)) seem to have no issues. The 2c(4,4) faces are pretty ea=
sy
> to fix if you end up with a case where you need to flip one or swap two
> blocks. This is because while those two cases have odd parity when you
> consider the reduced length 3 puzzle, in terms of the individual pieces t=
hey
> have even parity so can be fixed by the even 3 cycles (remember that 3 cy=
cle
> is an 'even' permutation). If you consider the 2c(5,4) faces, there are
> basically two kinds of pieces. Pieces on a given diagonal within the fac=
e
> can be cycled by algorithms, but you can never place a face piece outside=
of
> its diagonal. You can discover this by playing around with the various
> kinds of possible twists. So this case is similar to the 2c(4,4), except
> you have to do each diagonal separately, whereas 2c(4,4) has all the piec=
es
> identical. The real problem comes with edges. From the previous logic, =
it
> can be seen that a single swap of two edges isn't too bad as that is
> equivalent to swapping two pairs of pieces, which is even. So that can b=
e
> fixed with a 3 cycle macro of edge pieces. But if you end up with a a
> single flipped 3c(5,4,4) edge, then bad luck for you: that is the only r=
eal
> parity problem that can occur in the length 4 pentagonal duoprism that ca=
n't
> be fixed by the macros you already have from length 5! I played around a
> bunch before I finally stepped back to think about the problem and realiz=
ed
> what I just stated, which is that it is impossible to fix with just my 3
> cycle. Once I did that, I spent a couple days of thinking about the prob=
lem
> in my free time without actually touching the puzzle, because I didn't
> really know what to do. When I thought that it might be necessary to bre=
ak
> everything up and rebuild, I had to then categorize the possible twists, =
to
> determine exactly which twist I need to do to apply an odd permutation to
> edge pieces so that I get the correct parity. Turns out there are only t=
wo
> possibilities when you think about it: twisting a pentagonal facet on a
> 2c(5,4) face / 3c(5,4,4) edge (equivalent), or doing the same twist while
> holding down '2' on your keyboard. I'm not sure sure what that is called=
,
> so I'll just call it twist2. Now, the first twist can be thrown out beca=
use
> it does something else that is odd: it also does an odd permutation (to =
be
> precise, a single pair swap) of the 2c(5,5) face central piece. This
> problem can only be fixed by the very same twist, which unfortunately mea=
ns
> that no matter how you twist this way, you will have either your edges in
> odd parity, or your 2c(5,5) faces in odd parity. So the only possibility=
is
> the twist2. Now that I had this knowledge in hand, I proceeded first to =
try
> to find a better way to use this knowledge than just rebuilding from
> scratch. I won't give the algorithm here, but turns out if you consider
> just the 1c pieces, you can use this twist2 along with some other careful=
ly
> chosen twists to make a move sequence that changes parity of the edge
> pieces, and keeps your 1c pieces intact. You have to remember though tha=
t
> whatever your algorithm is, it 'must' have an odd number of these twist2s=
in
> order to put the cube in a solvable orientation. Turns out that the
> algorithm I used kept all edge blocks together, and only broke up a handf=
ul
> of the 2c(4,4) face blocks. This is not as good as an algorithm that jus=
t
> flips the edge I want, but it's much better than the alternative of
> rebuilding everything! So I then proceeded to use my 2c(4,4) face 3 cycl=
e
> macros to rebuild the faces, then solved as a length 3 puzzle, and voila!
> Victory!
>=20
> Well, it was quite an adventure, but I'm glad I followed this pentagonal
> duoprism road this far. There are still be length 6 or 7 puzzles left
> undone, and I'm quite sure that my solves were pretty inefficient and can=
be
> toppled in twist count too though :D. Not sure I have the time and effor=
t
> to take on anything above length 5 though, but they are awesome achieveme=
nts
> which I'm sure someone will one complete one day I'm quite sure.
>=20
> Not sure what I'll attempt next, but one thing that is sorely missing fro=
m
> my hypercubing experience, is the 5D cube. I have avoided that monster f=
or
> a long time! I like the 4D puzzles because the 3D projection that we can
> work with in MagicCube 4D makes it possible to use innate 3D visualizatio=
n
> skills. Of course this is also why I have problems with really small 4D
> puzzles like the simplex, because they are so small that the 4D-ness is
> really important, whereas other puzzles are large enough that you can wor=
k
> in locally 3D regions effectively. The 5D cube is another beast entirely
> though, as you can't rely on 3D instincts to guide you at all it feels.
> Nevertheless, one day I'll will make a serious push to solve that one too=
I
> hope! Until then, I will enjoy this closer to 3D MagicCube 4D software a=
s
> much as I can :)
>=20
> Chris
>