From an October 2013 discussion at http://www.speedsolving.com/forum/showthread.php?44398:
What are the shortest possible PLL algs, in terms of how long they are
when written down? This should be more interesting than just finding
move-optimal stuff, and I think there's a lot of room for creativity.
The most common algs may not be the shortest!
To start with, the following notations are OK:
- [P,Q] is a commutator (that is, P Q P' Q')
- [P:Q] is a setup move/conjugate (that is, P Q P')
- (P)n is P repeated n times; parentheses are unnecessary if it's clear what P is
- Any face or slice moves, rotations, or lowercase moves.
Also, the algorithm can do the PLL on any face, and you do not have to
include any adjustment at the end, or rotations at the very start/end.
And the shortest algs we found:
A: L'[F,R'B2R]L, L'[R'd2R,U]L (thanks to Stefan Pochmann)
E: [[RU'L:D2],U2] (thanks to TDM)
F: R'URU'R2F'U'FU[R,F]R2
G: [RL:U2][F'UB':d2], [R'UL',d2][BF:U2], L'R'U2LR[FU'B,d2], [LU'R,d2]B'F'U2BF
H: (M2U)6
J: [[RU'L:d2],U], [[R'UL':d2],U]
N: (r'DrU2)5, (rDr'U2)5
R: R[U2R'U2,UR'F'R]R', R'[U2RU2,U'RBR']R
T: [R2D':F2][B2D:L2]
U: M2uMu2MuM2, B2UMU2M'UB2
V: [F'UBU'F:U][U2,B]
Y: F2[DR2:U][R'U'R:F2]
Z: (UF2)6M'U2M (thanks to Moritz Karl)
Speedsolving Posts: "Cross" Meta-Method for 4x4x4
Posted at http://www.speedsolving.com/forum/showthread.php?17783 on December 18, 2009:
More methods than your cube has room for.
The Theory
The idea behind this meta-method (and I call it that because it encompasses so many normal methods, as you'll see in a bit) is to divide the solve into a small number of steps, each of which can be done relatively efficiently/quicky and in many different ways. There are in fact only two steps, which are as follows:
- 1: Solve the centers and all 8 of the edge pieces on D (or L, if you prefer).
- 2: Solve the rest of the puzzle.
So, why are these two steps good? The key is that the first step is very simple to do, while still leaving a lot of freedom to the solver. In this step, we build a structure similar to the cross from Fridrich; there is a lot of freedom here, because until the entire step is completed you have the freedom to do most outer layer turns and a few Rw U Rw'-type triggers. It may seem constraining, but in practice it is quite easy to set things up without destroying what you have. There are many ways to do this, from blockbuilding to a more reduction-oriented approach.
After we have done the first step, what we are left with is set up quite well for speedsolving: not only have we paired up the centers (the pieces that typically require either a lot of freedom or a lot of moves), but we have also fixed the bottom edges so we can guarantee never to have to look there. We are now free to do U, Uw, and D moves, as almost all moves here will keep the orientation of the center blocks constant, and we can of course use any of the common F2L triggers on the outer layers. Even though there aren't a lot of layers that we can freely turn without breaking up the solved pieces, almost everything we can do has a useful effect on the cube and is also very suitable for speedcubing.
Step 1: The Cross
There are a lot of ways to do the cross step. Here are some of the ways I've tried, although they are by no means all of the ways to do this step:
K4 Style
The actual K4 method sets up some corners as well, so this is a variation on it. The first step is to set up two opposite centers, one of which should be your cross color. Next, using the M layer, you pair up 3 of the cross edges, placing them correctly on the left face (if you're right-handed). Finally, you finish the centers without disturbing the cross edges, and then pair up the last cross edge and insert it.
Reduction
First, simply solve all centers like in reduction. Then pair up the four cross edges (you can do this by pairing two edges at a time, or with a more 'freestyle' approach where you freely do double layer turns on one slice) and place them.
Columns
Start by doing only the D center, and place it on the bottom. Then set up the rest of the pieces by doing eight columns - each column has two center pieces and one edge piece. You can do the columns in pretty much any order, but it is important to realize that you can insert a column with Ru'R' type moves, not just rU2r'.
M-Slice
First, solve two centers and place them on R/L, but make sure they are not your cross color centers. Next, pair up (using the M slice) and solve the DL and DR cross edges. Finally, use the M slice to build the remaining three centers and two edge pairs in whatever order seems most convenient, making sure not to move the DL/DR edges. Notice that there's still a lot of freedom since B, U, and F are free, so you can do any kind of rUr' type triggers, and you shouldn't have to use too many slice moves.
Step 2: The Rest
Again, there are many ways to finish off the method; here are the ones I've played around with:
Traditional K4
First, place the corners of the first layer (note that it is a bit more efficient to do this during Step 1 if you can). Next, fill in the second and third layer by inserting the 8 edge pieces with commutators. Finally, solve the last layer with CLL and then ELL.
F2L-Style K4
Place the corners of the first layer, using F2L techniques to also place one edge piece (on average) per slot. Next, fill in the rest of the second and third layer with commutators, although you should only need about 4 commutators on average instead of 8. Finally, use CLL and ELL to finish the last layer.
Reduction (Yau)
First, pair up the edges; 3-2-2 pairing (start with a u, place three edges and do a u', then do 2-pairing for the rest) is suggested. Next, solve the F2L and LL with whatever variant of Fridrich you are most comfortable with.
Commutators
Solve the corners and then finish the rest of the edges two at a time with commutators.
Commutator Fridrich
Place the first layer's corners along with as many edges as you can, using F2L techniques. Then use OLL and PLL to try to place as many last layer edge pieces as you can. Finally, solve the remaining edges two at a time with commutators.
The Method(s) In Practice
The three well-known 4x4 methods built around this are K4 (with two major variants) and Yau. The standard version of K4 uses the "K4 style" cross and the "traditional K4" rest, although it has the slight optimization of doing the four first-layer corners during the cross step. There is also a variation of K4, sometimes used by Dan Cohen, which is a "K4 style" cross followed by an "F2L-style K4" rest. Finally, the Yau method uses a "K4 style" cross with a "reduction" rest.
Sub-minute averages have been achieved with all three of these methods, and Dan Cohen has even broken the 50-second barrier using Yau. Yau was also the method he used to achieve the current WR single of 36.46 seconds. It is clear that this meta-method has some serious potential; all that remains is to practice a lot :)
More methods than your cube has room for.
The Theory
The idea behind this meta-method (and I call it that because it encompasses so many normal methods, as you'll see in a bit) is to divide the solve into a small number of steps, each of which can be done relatively efficiently/quicky and in many different ways. There are in fact only two steps, which are as follows:
- 1: Solve the centers and all 8 of the edge pieces on D (or L, if you prefer).
- 2: Solve the rest of the puzzle.
So, why are these two steps good? The key is that the first step is very simple to do, while still leaving a lot of freedom to the solver. In this step, we build a structure similar to the cross from Fridrich; there is a lot of freedom here, because until the entire step is completed you have the freedom to do most outer layer turns and a few Rw U Rw'-type triggers. It may seem constraining, but in practice it is quite easy to set things up without destroying what you have. There are many ways to do this, from blockbuilding to a more reduction-oriented approach.
After we have done the first step, what we are left with is set up quite well for speedsolving: not only have we paired up the centers (the pieces that typically require either a lot of freedom or a lot of moves), but we have also fixed the bottom edges so we can guarantee never to have to look there. We are now free to do U, Uw, and D moves, as almost all moves here will keep the orientation of the center blocks constant, and we can of course use any of the common F2L triggers on the outer layers. Even though there aren't a lot of layers that we can freely turn without breaking up the solved pieces, almost everything we can do has a useful effect on the cube and is also very suitable for speedcubing.
Step 1: The Cross
There are a lot of ways to do the cross step. Here are some of the ways I've tried, although they are by no means all of the ways to do this step:
K4 Style
The actual K4 method sets up some corners as well, so this is a variation on it. The first step is to set up two opposite centers, one of which should be your cross color. Next, using the M layer, you pair up 3 of the cross edges, placing them correctly on the left face (if you're right-handed). Finally, you finish the centers without disturbing the cross edges, and then pair up the last cross edge and insert it.
Reduction
First, simply solve all centers like in reduction. Then pair up the four cross edges (you can do this by pairing two edges at a time, or with a more 'freestyle' approach where you freely do double layer turns on one slice) and place them.
Columns
Start by doing only the D center, and place it on the bottom. Then set up the rest of the pieces by doing eight columns - each column has two center pieces and one edge piece. You can do the columns in pretty much any order, but it is important to realize that you can insert a column with Ru'R' type moves, not just rU2r'.
M-Slice
First, solve two centers and place them on R/L, but make sure they are not your cross color centers. Next, pair up (using the M slice) and solve the DL and DR cross edges. Finally, use the M slice to build the remaining three centers and two edge pairs in whatever order seems most convenient, making sure not to move the DL/DR edges. Notice that there's still a lot of freedom since B, U, and F are free, so you can do any kind of rUr' type triggers, and you shouldn't have to use too many slice moves.
Step 2: The Rest
Again, there are many ways to finish off the method; here are the ones I've played around with:
Traditional K4
First, place the corners of the first layer (note that it is a bit more efficient to do this during Step 1 if you can). Next, fill in the second and third layer by inserting the 8 edge pieces with commutators. Finally, solve the last layer with CLL and then ELL.
F2L-Style K4
Place the corners of the first layer, using F2L techniques to also place one edge piece (on average) per slot. Next, fill in the rest of the second and third layer with commutators, although you should only need about 4 commutators on average instead of 8. Finally, use CLL and ELL to finish the last layer.
Reduction (Yau)
First, pair up the edges; 3-2-2 pairing (start with a u, place three edges and do a u', then do 2-pairing for the rest) is suggested. Next, solve the F2L and LL with whatever variant of Fridrich you are most comfortable with.
Commutators
Solve the corners and then finish the rest of the edges two at a time with commutators.
Commutator Fridrich
Place the first layer's corners along with as many edges as you can, using F2L techniques. Then use OLL and PLL to try to place as many last layer edge pieces as you can. Finally, solve the remaining edges two at a time with commutators.
The Method(s) In Practice
The three well-known 4x4 methods built around this are K4 (with two major variants) and Yau. The standard version of K4 uses the "K4 style" cross and the "traditional K4" rest, although it has the slight optimization of doing the four first-layer corners during the cross step. There is also a variation of K4, sometimes used by Dan Cohen, which is a "K4 style" cross followed by an "F2L-style K4" rest. Finally, the Yau method uses a "K4 style" cross with a "reduction" rest.
Sub-minute averages have been achieved with all three of these methods, and Dan Cohen has even broken the 50-second barrier using Yau. Yau was also the method he used to achieve the current WR single of 36.46 seconds. It is clear that this meta-method has some serious potential; all that remains is to practice a lot :)
Speedsolving Posts: Extension to the Tower of Hanoi
Posts from http://www.speedsolving.com/forum/showthread.php?34968, in January 2012.
I didn't know where else to put this, but I'm sure some people here will find this interesting. Hopefully this is actually new and not an obvious thing that I somehow couldn't find on Google.
So, you're probably all familiar with the Tower of Hanoi problem. You have three pegs, and n disks of different sizes on one of the pegs, arranged with the largest disks on the bottom. The goal is to move all the disks to a different peg; the rules are that (a) you can only move one disk at a time (from one peg to another peg), and (b) you can't place a disk on top of a smaller disk. The problem is not too difficult to solve if you are just trying to finish it, but the mathematically interesting thing is that it requires O(2^n) moves - roughly doubling for every new disk you add. The reason for this is that, to move n disks, you must first move the smallest n-1 disks to another peg, then move the biggest disk, and finally move the rest of the disks a second time.
For a while I've been casually trying to think of a way to extend this to create puzzles which require O(k^n) moves (for n disks, and for a k as large as we want). I think I've got an idea which will let us do this. Here is a rough picture of the idea, for a small case:
So, here's what's going on here. Choose some number j. Instead of 3 pegs you have 2j+1 pegs, and instead of disks that cover one peg, they cover j adjacent pegs. The trick is that the only way to move a disk is to move all of the smaller disks to j of the other j+1 pegs - and, even then, this will only allow you to move the remaining disk clockwise or counterclockwise by one peg.
Figuring out the number of moves is a little trickier than in the normal 3-peg version, but at least we know that we can only move the biggest disk if all of the other disks are in a pile. To move a pile of n disks by one space, we need to (1) move n-1 disks by j spaces, (2) move the biggest disk, and then (3) move n-1 disks by j spaces again. To move a pile of n disks by k spaces, though, we need to (1) move n-1 disks by j spaces, (2) repeat (move the biggest disk, then move n-1 disks one space) k-1 times, (3) move the biggest disk one last time, and then (4) move n-1 disks by j spaces again to finish. Only moves of 1 and j spaces seem to be important here, so if we write those as
and 
the recursion looks like this:
Or, just in terms of the 1-move numbers:
Anyway, it looks like the k constant (the number for which grows like 
) works out to 
. So that's pretty interesting, and we can definitely get k as large as we like by choosing a sufficiently high j. In fact, we can set k to be any even integer: for instance, let j=41, and then the sequence grows as fast as 
.
Of course, boards with 83 pegs and disks with 41 holes are not
particularly nice to play with, but seeing as the 5-disk puzzle takes
14,751 moves to just move the disks one space over, I don't think a lot
of people would complain.
Incidentally, if you don't like the idea of disks spanning multiple pegs, you can get the same effect by using 1-peg disks and adding the rule that you cannot put a disk onto a peg if a smaller disk is on that peg or one of the next j-1 pegs in either direction.
Alright guys, I made a sim. Try it here:
http://mzrg.com/js/hanoi/
>Very interesting, didn't read all of it though. Just curious, how did you come up with this idea?
I basically was thinking about how to expand the problem while keeping the restricted feel of the original one (only really one way to move stuff around) and I randomly thought of the idea of disks that would cover more than one peg, and of just barely having enough pegs to move those disks around. I initially had a wrong idea of how the math would turn out (since I didn't realize you could take shortcuts when moving a group of disks more than one space at a time) but the idea turned out to be what I wanted anyway.
PS: Wolfram Alpha gives the following closed form for
:
I didn't know where else to put this, but I'm sure some people here will find this interesting. Hopefully this is actually new and not an obvious thing that I somehow couldn't find on Google.
So, you're probably all familiar with the Tower of Hanoi problem. You have three pegs, and n disks of different sizes on one of the pegs, arranged with the largest disks on the bottom. The goal is to move all the disks to a different peg; the rules are that (a) you can only move one disk at a time (from one peg to another peg), and (b) you can't place a disk on top of a smaller disk. The problem is not too difficult to solve if you are just trying to finish it, but the mathematically interesting thing is that it requires O(2^n) moves - roughly doubling for every new disk you add. The reason for this is that, to move n disks, you must first move the smallest n-1 disks to another peg, then move the biggest disk, and finally move the rest of the disks a second time.
For a while I've been casually trying to think of a way to extend this to create puzzles which require O(k^n) moves (for n disks, and for a k as large as we want). I think I've got an idea which will let us do this. Here is a rough picture of the idea, for a small case:
So, here's what's going on here. Choose some number j. Instead of 3 pegs you have 2j+1 pegs, and instead of disks that cover one peg, they cover j adjacent pegs. The trick is that the only way to move a disk is to move all of the smaller disks to j of the other j+1 pegs - and, even then, this will only allow you to move the remaining disk clockwise or counterclockwise by one peg.
Figuring out the number of moves is a little trickier than in the normal 3-peg version, but at least we know that we can only move the biggest disk if all of the other disks are in a pile. To move a pile of n disks by one space, we need to (1) move n-1 disks by j spaces, (2) move the biggest disk, and then (3) move n-1 disks by j spaces again. To move a pile of n disks by k spaces, though, we need to (1) move n-1 disks by j spaces, (2) repeat (move the biggest disk, then move n-1 disks one space) k-1 times, (3) move the biggest disk one last time, and then (4) move n-1 disks by j spaces again to finish. Only moves of 1 and j spaces seem to be important here, so if we write those as
and the recursion looks like this:Or, just in terms of the 1-move numbers:
Anyway, it looks like the k constant (the number for which
grows like ) works out to . So that's pretty interesting, and we can definitely get k as large as we like by choosing a sufficiently high j. In fact, we can set k to be any even integer: for instance, let j=41, and then the sequence grows as fast as .
Of course, boards with 83 pegs and disks with 41 holes are not
particularly nice to play with, but seeing as the 5-disk puzzle takes
14,751 moves to just move the disks one space over, I don't think a lot
of people would complain.Incidentally, if you don't like the idea of disks spanning multiple pegs, you can get the same effect by using 1-peg disks and adding the rule that you cannot put a disk onto a peg if a smaller disk is on that peg or one of the next j-1 pegs in either direction.
Alright guys, I made a sim. Try it here:
http://mzrg.com/js/hanoi/
>Very interesting, didn't read all of it though. Just curious, how did you come up with this idea?
I basically was thinking about how to expand the problem while keeping the restricted feel of the original one (only really one way to move stuff around) and I randomly thought of the idea of disks that would cover more than one peg, and of just barely having enough pegs to move those disks around. I initially had a wrong idea of how the math would turn out (since I didn't realize you could take shortcuts when moving a group of disks more than one space at a time) but the idea turned out to be what I wanted anyway.
PS: Wolfram Alpha gives the following closed form for
:
Speedsolving Posts: Almost-2gen LL Algs
From http://www.speedsolving.com/forum/showthread.php?18539 (January 2010):
From any ZBLL we can easily get to PLL using only 2gen moves. All PLLs are a U, Z, or H perm away from either solved, R perm, or Y perm. Therefore, if we can do R and Y perm with at most 2 D-layer moves, we can do any ZBLL with at most 2 D-layer moves.
So, here is an R perm with exactly 2 D-layer moves:
R U2 R D R' U R D' R' U' R' U R U R' U
And here is a Y perm (but not an efficient one):
R' U' D (R' U' R U)3 D' U R (R U R' U R U2 R')2 U'
I imagine these algs could be very useful for OH - everyone loves 2gen moves, and keeping D turns to a minimum is very helpful. Unfortunately, I don't know of any way to generate optimal algs of this type. Does anyone have an idea?
P.S. You can also do every ZBLL with and at most two F
moves, although it's less useful IMO. You can prove this with the T perm
and this Y perm:
F (R U R' U')3 F' U (R U R' U R U2 R')2 U2
From http://www.speedsolving.com/forum/showthread.php?708&p=680961#post680961 (December 2011):
Dumping some optimal 2-F-move PLLs (there are probably a few optimal ones I missed but the optimal lengths should be accurate):
From any ZBLL we can easily get to PLL using only 2gen moves. All PLLs are a U, Z, or H perm away from either solved, R perm, or Y perm. Therefore, if we can do R and Y perm with at most 2 D-layer moves, we can do any ZBLL with at most 2 D-layer moves.
So, here is an R perm with exactly 2 D-layer moves:
R U2 R D R' U R D' R' U' R' U R U R' U
And here is a Y perm (but not an efficient one):
R' U' D (R' U' R U)3 D' U R (R U R' U R U2 R')2 U'
I imagine these algs could be very useful for OH - everyone loves 2gen moves, and keeping D turns to a minimum is very helpful. Unfortunately, I don't know of any way to generate optimal algs of this type. Does anyone have an idea?
P.S. You can also do every ZBLL with
F (R U R' U')3 F' U (R U R' U R U2 R')2 U2
From http://www.speedsolving.com/forum/showthread.php?708&p=680961#post680961 (December 2011):
Dumping some optimal 2-F-move PLLs (there are probably a few optimal ones I missed but the optimal lengths should be accurate):
A-perm a HTM-optimal: 11f R U' R F2 R' U R' U' R2 F2 R2 QTM-optimal: 15q R U' R F2 R' U R' U' R2 F2 R2 A-perm a HTM-optimal: 11f R2 F2 R2 U R U' R F2 R' U R' QTM-optimal: 15q R2 F2 R2 U R U' R F2 R' U R'
E-perm
HTM-optimal: 18f R' U2 R U' F' R' U R' U2 R U R' U R2 F R' U' R R' U R F' R2 U' R U' R' U2 R U' R F U R' U2 R R U R' U R' F U' R2 U' R2 U2 R' U' R F' R' U R2 R2 U F' U R' U' R U R' U' R U R' U' R F U' R2 R2 U F' R' U R U' R' U R U' R' U R U' F U' R2 R2 U' R F R' U R U2 R2 U R2 U F' R U' R U' R' R2 U2 F' U' R' U R U' R' U R U' R' U R F U2 R2 R2 U2 F' R' U' R U R' U' R U R' U' R U F U2 R2 F U R2 U' R U R' U R2 U' R' U R' U' R U' R F' F U' R U' R U R' U R U2 R' U R' U' R U2 R' F' F U' R U' R' U2 R U' R' U R' U2 R U R' U R F' F U' R' U R' U' R2 U R' U R U' R2 U R U' R F' F R U2 R' U R U' R U2 R' U' R U' R' U R' U F' F R' U R' U R U' R U R2 U' R U' R' U R2 U' F' F R' U R' U' R2 U R' U' R U' R2 U R U' R U F' F R' U' R U' R' U2 R U' R U R' U2 R U R' U F' QTM-optimal: 20q R2 U F' U R' U' R U R' U' R U R' U' R F U' R2 R2 U F' R' U R U' R' U R U' R' U R U' F U' R2 F U R2 U' R U R' U R2 U' R' U R' U' R U' R F' F U' R U' R U R' U R U2 R' U R' U' R U2 R' F' F U' R U' R' U2 R U' R' U R' U2 R U R' U R F' F U' R' U R' U' R2 U R' U R U' R2 U R U' R F' F R U2 R' U R U' R U2 R' U' R U' R' U R' U F' F R' U R' U R U' R U R2 U' R U' R' U R2 U' F' F R' U R' U' R2 U R' U' R U' R2 U R U' R U F' F R' U' R U' R' U2 R U' R U R' U2 R U R' U F'
F-perm HTM-optimal: 15f R2 U R2 U2 F2 U R2 U2 R2 U R2 U' R2 U2 F2 R2 U R2 U' R2 U2 R2 U' F2 U2 R2 U' R2 U' F2 F2 U R2 U R2 U2 F2 U R2 U2 R2 U R2 U' R2 F2 U2 R2 U R2 U' R2 U2 R2 U' F2 U2 R2 U' R2 R2 U' F2 U2 R2 U' R2 U2 R2 U R2 U F2 U2 R2 R2 U2 F2 U' R2 U' R2 U2 R2 U R2 U2 F2 U R2 QTM-optimal: 19q R U R' U' R' U R U F' U R' U R U2 F R U' R' R U R' F' U2 R' U' R U' F U' R' U' R U R U' R' R U R U F' U' R U R' U' R' U2 F U' R' U2 R' R U2 R U F' U2 R U R U' R' U F U' R' U' R' R' U' R U' R' U R U R2 F' R U R U' R' F U R R' U' F' R U R' U' R' F R2 U' R' U' R U R' U R
G-perm a HTM-optimal: 15f F2 U R U2 R' U' R' U2 R2 U R' F2 R' U R R' U' F2 U R U2 R U2 R' U F2 U2 R U' R' QTM-optimal: 18q R U R' U' R' U F R U R U' R' F' U R' U2 R
G-perm b HTM-optimal: 15f R' U' R F2 R U' R2 U2 R U R U2 R' U' F2 R U R' U2 F2 U' R U2 R' U2 R' U' F2 U R QTM-optimal: 18q R' U2 R U' F R U R' U' R' F' U' R U R U' R'
G-perm c
HTM-optimal: 14f R2 U2 R2 F2 U' R2 U R2 U F2 U2 R2 U' R2 QTM-optimal: 17q R2 U' R U' R U R' U R' F' R' U R U' F R'
G-perm d HTM-optimal: 14f R2 U R2 U2 F2 U' R2 U' R2 U F2 R2 U2 R2 QTM-optimal: 17q R F' U R' U' R F R U' R U' R' U R' U R2
J-perm a HTM-optimal: 13f R' F R' U' R2 U' R2 U R' F' R U R2 R2 U' R' F R U' R2 U R2 U R F' R QTM-optimal: 16q R' F R' U' R2 U' R2 U R' F' R U R2 R2 U' R' F R U' R2 U R2 U R F' R F R' U R U' F' U R' U2 R U' R' U2 R R' U2 R U R' U2 R U' F U R' U' R F' R' U' R U F R U' R' U' R' U R2 U R' F' R' U' R U' R' U F U R' U' R F' R' U R2 R2 U' R F R' U R U' F' U' R U R' U R F R U' R2 U' R U R U R' F' U' R' U R
J-perm b HTM-optimal: 11f R U' F U' R' U' R U F' U2 R' R U2 F U' R' U R U F' U R' QTM-optimal: 12q R U' F U' R' U' R U F' U2 R' R U2 F U' R' U R U F' U R'
N-perm a HTM-optimal: 15f R2 U R2 U' R2 U' F2 U' R2 U R2 U F2 U2 R2 R2 U2 F2 U' R2 U' R2 U F2 U R2 U R2 U' R2 QTM-optimal: 21q R U2 R2 U' R2 F' R U R' U' R' F R' U R U2 R' R U2 R' U' R F' R U R U' R' F R2 U R2 U2 R' R U R' U R U' F U' R' U' R U F' U2 R' U2 R U' R' R U R' U2 R U2 F U' R' U R U F' U R' U' R U' R'
N-perm b HTM-optimal: 15f R2 U F2 U2 R2 U R2 U2 R2 U' R2 U2 F2 U' R2 R2 U' R2 U2 F2 U' R2 U2 R2 U F2 U2 R2 U R2 QTM-optimal: 23q? R U' R' U R' U2 R2 U R2 F2 U R U R U' R' U' F2 F2 U R U R' U' R' U' F2 R2 U' R2 U2 R U' R U R' R' U2 R' U' R2 U F' U' R U R' U' R2 U2 R F R' U R R' U2 F U R2 U2 R' U' R U R' F' R U2 R' U' R' U R R' U' R U R U2 R' F R U' R' U R U2 R2 U' F' U2 R R' U' R F' R' U2 R2 U R U' R' U F U' R2 U R U2 R
R-perm a HTM-optimal: 13f R2 F2 U R U R' U' R' U' F2 R' U R' R U' R F2 U R U R U' R' U' F2 R2 QTM-optimal: 16q R2 F2 U R U R' U' R' U' F2 R' U R' R U' R F2 U R U R U' R' U' F2 R2
R-perm b
HTM-optimal: 13f R2 F R U R U' R' F' R U2 R' U2 R R' U2 R U2 R' F R U R' U' R' F' R2 QTM-optimal: 16q R2 F R U R U' R' F' R U2 R' U2 R R' U2 R U2 R' F R U R' U' R' F' R2
T-perm
HTM-optimal: 14f R U R' U' R' F R2 U' R' U' R U R' F' F R U' R' U R U R2 F' R U R U' R' QTM-optimal: 15q R U R' U' R' F R2 U' R' U' R U R' F' F R U' R' U R U R2 F' R U R U' R'
V-perm HTM-optimal: 15f R2 U2 R2 U R2 U2 F2 U R2 U R2 U2 F2 U' R2 R2 U2 R2 U' F2 U2 R2 U' R2 U' F2 U2 R2 U R2 R2 U F2 U2 R2 U' R2 U' F2 U2 R2 U' R2 U2 R2 R2 U' R2 U2 F2 U R2 U R2 U2 F2 U R2 U2 R2 QTM-optimal: 19q R U' R' U R F R' U' R2 U R U' R U R2 F' R' R F R2 U' R' U R' U' R2 U R F' R' U' R U R'
Y-perm
HTM-optimal: 14f R2 U' R2 U F2 U' R2 U' R2 U F2 R2 U R2 R2 U' R2 F2 U' R2 U R2 U F2 U' R2 U R2 QTM-optimal: 19q R2 U' R2 U' R2 U R' F' R U R2 U' R' F R R' F' R U R2 U' R' F R U' R2 U R2 U R2 R' U' R U' R U' F U' R' U' R U F' U2 R2 U R R' U' R2 U2 F U' R' U R U F' U R' U R' U R
GeoGebra Files for Twisty Puzzle Classification
In 2009 and 2010, I actually ended up making several more GeoGebra files for exploring the possible twisty puzzles. I don't think I ever actually published them, and they may not be 100% finished, but they definitely ought to be out in the wild. So here's a zip file with what I have: http://www.mzrg.com/filepost/GeoGebra_TPCP.zip
First, there are some 2D state diagram views. The files here are Tvv, Tee, Tev (that's all the T ones, since vertex and face turns are equivalent); Cee, Cvv, Cff, Cfv; Dff; Oee; and a partial Iee (there are so many Ie puzzles by themselves!). The display has one face of the polyhedron, and a 2D state diagram.
Second, and perhaps more interesting, are the files marked "Cview" etc. These display colored polyhedra with 6 sliders at once: two v, two f, and two e. I haven't actually done the work of classifying the puzzles, of course, but you can set up and view some pretty complicated things, and it's neat to play with the sliders and see how the puzzles change. The icosahedron one is the most complicated and does not completely work, but I haven't touched it in years and I don't remember what points and lines do what...
First, there are some 2D state diagram views. The files here are Tvv, Tee, Tev (that's all the T ones, since vertex and face turns are equivalent); Cee, Cvv, Cff, Cfv; Dff; Oee; and a partial Iee (there are so many Ie puzzles by themselves!). The display has one face of the polyhedron, and a 2D state diagram.
Second, and perhaps more interesting, are the files marked "Cview" etc. These display colored polyhedra with 6 sliders at once: two v, two f, and two e. I haven't actually done the work of classifying the puzzles, of course, but you can set up and view some pretty complicated things, and it's neat to play with the sliders and see how the puzzles change. The icosahedron one is the most complicated and does not completely work, but I haven't touched it in years and I don't remember what points and lines do what...
Speedsolving Posts: 2D State Diagrams
The posts here are from January 2009 and can be found in this topic: http://www.speedsolving.com/forum/showthread.php?8628
This is a topic on puzzle theory - specifically, it deals with the theory of symmetrical twisty puzzles. There's no real "theory" section in this forum but it doesn't belong in Off-Topic, so I've put it here so more people will see it. (This does have applications to speedcubing, in terms of solving gelatinbrain puzzles.)
So, as some of you may know, symmetrical twisty puzzles can be described by talking about the shape and the number of separate types of turns in each of the three basic categories (face turns, edge turns, and vertex turns) which are named around where the axis (which the pieces rotate around) points. A 3x3x3, for instance, is a cube with one type of turn (in the face turn category); in my categorization scheme that would make it type Cf (C for cube, f for face turn). The only other type-Cf puzzle is the 2x2x2, which has a deeper turn that just happens to line up with the turn on the opposite face. (The 4x4x4 and 5x5x5 have two face turn types, so they are type Cff. Similarly the 6x6x6 and 7x7x7 are type Cfff.) Note that I'm considering a 3x3 with slightly shallower or deeper cuts to be the same, because it has the same number and types of pieces and the exact same solution, so the only difference is in the shape of the pieces.
If we want to talk about symmetrical puzzles with only one type of turn, it's relatively simple, because we can describe the depth of the turn as a single number. But if we have two types of turns, we need two numbers: for instance, on a 5x5x5, if we imagine a turn of depth 0 to turn nothing and a turn of depth 1 to turn the entire cube, the two turns have depths 1/5 and 2/5. But it's hard to imagine exactly what numbers will give a certain puzzle, and which numbers will give a different one. So the question is this: how many possible puzzles are there of a given type, and, more importantly, how can we understand where one puzzle begins and another ends?
My answer to this is the 2D State Diagram. This particular one describes the Tvv-type polyhedra (that is, tetrahedra with two vertex turn types - note that a vertex turn and a face turn are the same thing on a tetrahedron). The two axes correspond to the two depths of turn, and each one goes from 0 (no part of the puzzle is turned) to 1 (the entire puzzle is turned).
So what are the lines in the diagram? Each one represents some kind of puzzle which only exists for a very specific set of turn depths. The diagonal line from the top-left to bottom-right represents the degenerate case where the two depths are the same (so, type Tv puzzles with only one type of cut). Each of those big spaces between the lines, however, is a specific puzzle, which remains the same anywhere in that space. There are also puzzles on the lines themselves, and each place where lines intersect is another puzzle, albeit a very specific one because even a small change in either of the two depths will make it different. The Pyraminx, for example, is located at the intersection of a horizontal or vertical 2/3 line, and the diagonal line from bottom left to top right.
Finally, I have prepared an interactive simulation (!) of this state diagram, so that you can play with it to see how it works out. It uses the GeoGebra geometry program, which I'm really fond of. You can download it from my website. To use it, just move the red point around inside the square (use the white pointer symbol to move a point around), and watch how the puzzle changes when you cross or move along lines.
I wrote up the state diagram for type Dff puzzles (dodecahedra with two different depths of face turns). The Gigaminx is from this category. Note that this category is MUCH more complicated than the Tvv type I showed in the first post - I'm not completely sure of the count, but I think there are a whopping 98 puzzles of this type!!! I'm really glad I made an interactive simulation, because there is no way I'd ever want to draw all of them out by hand. You can also see that there are 7 puzzles along the diagonal (these are the Df type puzzles).
In the image, I have marked a few lines as green. Those lines are distinctive because, although they separate different puzzles, the puzzles on the lines themselves are equivalent to one of the puzzles on one side of the line. For instance the Megaminx and Supernova, although they look like they might different puzzles, function exactly the same way.The full way to count the number of puzzles is:
- Any open space is a separate puzzle, although you have to make sure to only count puzzles on one side of the main diagonal (for everything here).
- Any line segment (i.e. a line between two intersections) is a separate puzzle, except (i) if the line segment is green; (ii) if the line segment is on the main diagonal; (iii) if the line segment is on the top or left. The second and third cases remove puzzles with essentially only one depth of turn, and the first removes a puzzle that looks like a different puzzle but isn't.
- Any intersection of lines is a separate puzzle, as long as (i) it isn't on the the top or left or the main diagonal, and (ii) it has at least two black lines crossing it.
The Tvv case gives 11+10+2 = 23 puzzles like this (I think). Note that, although I haven't added it in yet, the halfway lines and the bottom-left-to-top-right diagonal in the Tvv diagram should also be green. It's only important to know what lines are green and what are not when you are counting puzzles, though.
Finally, there is a GeoGebra simulation of this, although I have made the two-dimensional slider much larger to compensate for the complexity. That way you can really see what lines you are crossing.
For each polyhedron there are actually six types of second-order puzzles: the familiar vv, ee, and ff, but also fv, ev, and ef. The first three use only one type of cut, so they're symmetrical, but the other three use two different types of cuts and have much more complicated drawings. As an example I've made one for the Cfv group. There are around 50 puzzles in this group, most of which have probably never been seen before. I hope to eventually make an interactive simulation of all 5*6 categories of second-order regular twisty puzzles. I imagine there will be over 1000 puzzles in total!
This is a topic on puzzle theory - specifically, it deals with the theory of symmetrical twisty puzzles. There's no real "theory" section in this forum but it doesn't belong in Off-Topic, so I've put it here so more people will see it. (This does have applications to speedcubing, in terms of solving gelatinbrain puzzles.)
So, as some of you may know, symmetrical twisty puzzles can be described by talking about the shape and the number of separate types of turns in each of the three basic categories (face turns, edge turns, and vertex turns) which are named around where the axis (which the pieces rotate around) points. A 3x3x3, for instance, is a cube with one type of turn (in the face turn category); in my categorization scheme that would make it type Cf (C for cube, f for face turn). The only other type-Cf puzzle is the 2x2x2, which has a deeper turn that just happens to line up with the turn on the opposite face. (The 4x4x4 and 5x5x5 have two face turn types, so they are type Cff. Similarly the 6x6x6 and 7x7x7 are type Cfff.) Note that I'm considering a 3x3 with slightly shallower or deeper cuts to be the same, because it has the same number and types of pieces and the exact same solution, so the only difference is in the shape of the pieces.
If we want to talk about symmetrical puzzles with only one type of turn, it's relatively simple, because we can describe the depth of the turn as a single number. But if we have two types of turns, we need two numbers: for instance, on a 5x5x5, if we imagine a turn of depth 0 to turn nothing and a turn of depth 1 to turn the entire cube, the two turns have depths 1/5 and 2/5. But it's hard to imagine exactly what numbers will give a certain puzzle, and which numbers will give a different one. So the question is this: how many possible puzzles are there of a given type, and, more importantly, how can we understand where one puzzle begins and another ends?
My answer to this is the 2D State Diagram. This particular one describes the Tvv-type polyhedra (that is, tetrahedra with two vertex turn types - note that a vertex turn and a face turn are the same thing on a tetrahedron). The two axes correspond to the two depths of turn, and each one goes from 0 (no part of the puzzle is turned) to 1 (the entire puzzle is turned).
So what are the lines in the diagram? Each one represents some kind of puzzle which only exists for a very specific set of turn depths. The diagonal line from the top-left to bottom-right represents the degenerate case where the two depths are the same (so, type Tv puzzles with only one type of cut). Each of those big spaces between the lines, however, is a specific puzzle, which remains the same anywhere in that space. There are also puzzles on the lines themselves, and each place where lines intersect is another puzzle, albeit a very specific one because even a small change in either of the two depths will make it different. The Pyraminx, for example, is located at the intersection of a horizontal or vertical 2/3 line, and the diagonal line from bottom left to top right.
Finally, I have prepared an interactive simulation (!) of this state diagram, so that you can play with it to see how it works out. It uses the GeoGebra geometry program, which I'm really fond of. You can download it from my website. To use it, just move the red point around inside the square (use the white pointer symbol to move a point around), and watch how the puzzle changes when you cross or move along lines.
I wrote up the state diagram for type Dff puzzles (dodecahedra with two different depths of face turns). The Gigaminx is from this category. Note that this category is MUCH more complicated than the Tvv type I showed in the first post - I'm not completely sure of the count, but I think there are a whopping 98 puzzles of this type!!! I'm really glad I made an interactive simulation, because there is no way I'd ever want to draw all of them out by hand. You can also see that there are 7 puzzles along the diagonal (these are the Df type puzzles).
In the image, I have marked a few lines as green. Those lines are distinctive because, although they separate different puzzles, the puzzles on the lines themselves are equivalent to one of the puzzles on one side of the line. For instance the Megaminx and Supernova, although they look like they might different puzzles, function exactly the same way.The full way to count the number of puzzles is:
- Any open space is a separate puzzle, although you have to make sure to only count puzzles on one side of the main diagonal (for everything here).
- Any line segment (i.e. a line between two intersections) is a separate puzzle, except (i) if the line segment is green; (ii) if the line segment is on the main diagonal; (iii) if the line segment is on the top or left. The second and third cases remove puzzles with essentially only one depth of turn, and the first removes a puzzle that looks like a different puzzle but isn't.
- Any intersection of lines is a separate puzzle, as long as (i) it isn't on the the top or left or the main diagonal, and (ii) it has at least two black lines crossing it.
The Tvv case gives 11+10+2 = 23 puzzles like this (I think). Note that, although I haven't added it in yet, the halfway lines and the bottom-left-to-top-right diagonal in the Tvv diagram should also be green. It's only important to know what lines are green and what are not when you are counting puzzles, though.
Finally, there is a GeoGebra simulation of this, although I have made the two-dimensional slider much larger to compensate for the complexity. That way you can really see what lines you are crossing.
For each polyhedron there are actually six types of second-order puzzles: the familiar vv, ee, and ff, but also fv, ev, and ef. The first three use only one type of cut, so they're symmetrical, but the other three use two different types of cuts and have much more complicated drawings. As an example I've made one for the Cfv group. There are around 50 puzzles in this group, most of which have probably never been seen before. I hope to eventually make an interactive simulation of all 5*6 categories of second-order regular twisty puzzles. I imagine there will be over 1000 puzzles in total!
TPCP: Type Ce
(Old post from 2008)
These puzzles are rather complicated and there tend to be a lot of them, which tends to be the case with edge-turning types. They are also quite difficult to produce in real life, although one or two of them have actually been created. These are the first really complex puzzles we've seen. Note that these puzzles are related to type Oe.
These puzzles are rather complicated and there tend to be a lot of them, which tends to be the case with edge-turning types. They are also quite difficult to produce in real life, although one or two of them have actually been created. These are the first really complex puzzles we've seen. Note that these puzzles are related to type Oe.
 | Puzzle Ce1: Gelatinbrain Number: n/a Common Name: n/a Types of Pieces: 6 fixed centers, 12 trivial edges, 24 triangular centers in 4 orbits, 8 corners. Solution: n/a |
 | Puzzle Ce2: Gelatinbrain Number: 3.3.1 Common Name: Helicopter Cube, Bevel Cube Types of Pieces: 24 centers in 4 orbits, 8 corners. Solution: n/a |
 | Puzzle Ce3:Gelatinbrain Number: 3.3.2 Common Name: n/a Types of Pieces: 6 inner centers, 24 pentagonal centers, 24 inner triangular centers in 4 orbits, 48 outer triangular centers in 2 orbits, 12 edges, 8 corners. Solution: n/a |
 | Puzzle Ce4:Gelatinbrain Number: 3.3.3 Common Name: n/a Types of Pieces: 6 inner centers, 24 inner triangular centers, 48 outer triangular centers in 2 orbits, 12 edges, 8 corners. Solution: n/a |
 | Puzzle Ce5:Gelatinbrain Number: 3.3.4 Common Name: n/a Types of Pieces: 6 inner centers, 24 adjacent triangular centers, 24 diagonal triangular centers, 48 tetragonal centers in 2 orbits, 12 edges, 8 corners. Solution: n/a |
 | Puzzle Ce6:Gelatinbrain Number: 3.3.5 Common Name: n/a Types of Pieces: 6 inner centers, 24 inner triangular centers, 48 outer triangular centers in 2 orbits, 12 edges, 8 corners. Solution: n/a |
 | Puzzle Ce7: Gelatinbrain Number: 3.3.6 Common Name: n/a Types of Pieces: 6 inner centers, 24 pentagonal centers, 24 inner triangular centers, 48 outer triangular centers in 2 orbits, 12 edges, 8 corners. Solution: n/a |
 | Puzzle Ce8:Gelatinbrain Number: 3.3.7 Common Name: Little Chop Types of Pieces: 24 centers. Solution: n/a |