tag:blogger.com,1999:blog-33352713070010515872018-05-13T23:54:25.149-07:00Notes on Twisty PuzzlesMichael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.comBlogger25125tag:blogger.com,1999:blog-3335271307001051587.post-11645105760396787102014-06-29T17:13:00.001-07:002014-06-29T17:13:26.748-07:00Speedsolving Post: "Shortest" PLL Algs?From an October 2013 discussion at <a href="http://www.speedsolving.com/forum/showthread.php?44398">http://www.speedsolving.com/forum/showthread.php?44398</a>:<br /><br />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!<br /><br />To start with, the following notations are OK:<br />- [P,Q] is a commutator (that is, P Q P' Q')<br />- [P:Q] is a setup move/conjugate (that is, P Q P')<br />- (P)n is P repeated n times; parentheses are unnecessary if it's clear what P is<br />- Any face or slice moves, rotations, or lowercase moves.<br />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.<br /><br /><br /><br /><br /><br /><br />And the shortest algs we found:<br /><br /><br />A: L'[F,R'B2R]L, L'[R'd2R,U]L (thanks to Stefan Pochmann)<br />E: [[RU'L:D2],U2] (thanks to TDM)<br />F: R'URU'R2F'U'FU[R,F]R2<br />G: [RL:U2][F'UB':d2], [R'UL',d2][BF:U2], L'R'U2LR[FU'B,d2], [LU'R,d2]B'F'U2BF<br />H: (M2U)6<br />J: [[RU'L:d2],U], [[R'UL':d2],U]<br />N: (r'DrU2)5, (rDr'U2)5<br />R: R[U2R'U2,UR'F'R]R', R'[U2RU2,U'RBR']R<br />T: [R2D':F2][B2D:L2]<br />U: M2uMu2MuM2, B2UMU2M'UB2<br />V: [F'UBU'F:U][U2,B]<br />Y: F2[DR2:U][R'U'R:F2]<br />Z: (UF2)6M'U2M (thanks to Moritz Karl)Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-36180000408946019072014-06-29T17:07:00.000-07:002014-06-29T17:07:00.442-07:00Speedsolving Posts: "Cross" Meta-Method for 4x4x4Posted at <a href="http://www.speedsolving.com/forum/showthread.php?17783">http://www.speedsolving.com/forum/showthread.php?17783</a> on December 18, 2009:<br /><br /><i>More methods than your cube has room for.</i><br /><br /><span style="font-size: medium;">The Theory</span><br /><br />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:<br />- 1: Solve the centers and all 8 of the edge pieces on D (or L, if you prefer).<br />- 2: Solve the rest of the puzzle.<br /><br />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.<br /><br />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. <br /><br /><span style="font-size: medium;">Step 1: The Cross</span><br />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:<br /><br /><b>K4 Style</b><br />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.<br /><br /><b>Reduction</b><br />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.<br /><br /><b>Columns</b><br />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'.<br /><br /><b>M-Slice</b><br />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.<br /><br /><span style="font-size: medium;">Step 2: The Rest</span><br />Again, there are many ways to finish off the method; here are the ones I've played around with:<br /><br /><b>Traditional K4</b><br />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.<br /><br /><b>F2L-Style K4</b><br />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.<br /><br /><b>Reduction (Yau)</b><br />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.<br /><br /><b>Commutators</b><br />Solve the corners and then finish the rest of the edges two at a time with commutators. <br /><br /><b>Commutator Fridrich</b><br />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.<br /><br /><span style="font-size: medium;">The Method(s) In Practice</span><br />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.<br /><br />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 :) Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-32674730008999391522014-06-29T17:00:00.001-07:002014-06-29T17:00:59.148-07:00Speedsolving Posts: Extension to the Tower of HanoiPosts from <a href="http://www.speedsolving.com/forum/showthread.php?34968">http://www.speedsolving.com/forum/showthread.php?34968</a>, in January 2012.<br /><br />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.<br /><br />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.<br /><br />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:<br /><img alt="" border="0" src="http://i286.photobucket.com/albums/ll119/qqwref/extendedhanoi.png" /><br />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.<br /><br />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 <a href="https://www.blogger.com/null"><img alt="m_1(n)" src="http://www.speedsolving.com/%7Epatjk/speedsolving/forum/vlatex/img/4afa667ba3ebf9ae50d21e001a176c6b-1.gif" style="border: 0px; vertical-align: middle;" title="m_1(n)" /></a> and <a href="https://www.blogger.com/null"><img alt="m_j(n)" src="http://www.speedsolving.com/%7Epatjk/speedsolving/forum/vlatex/img/8e52c61e1a43228cc9d0bea30cb68048-1.gif" style="border: 0px; vertical-align: middle;" title="m_j(n)" /></a> the recursion looks like this:<br /><a href="https://www.blogger.com/null"><img alt="m_1(n) = 2 m_j(n-1) + 1" src="http://www.speedsolving.com/%7Epatjk/speedsolving/forum/vlatex/img/a354d23f23bbe1815dcd7e527899161c-1.gif" style="border: 0px; vertical-align: middle;" title="m_1(n) = 2 m_j(n-1) + 1" /></a><br /><a href="https://www.blogger.com/null"><img alt="m_j(n) = 2 m_j(n-1) + (j-1) m_1(n-1) + j" src="http://www.speedsolving.com/%7Epatjk/speedsolving/forum/vlatex/img/15c6db28a4056cc3cc0964aa20ab4507-1.gif" style="border: 0px; vertical-align: middle;" title="m_j(n) = 2 m_j(n-1) + (j-1) m_1(n-1) + j" /></a><br />Or, just in terms of the 1-move numbers:<br /><a href="https://www.blogger.com/null"><img alt="m_1(n) = 2 m_1(n-1) + (2j-2) m_1(n-2) + 2j-1" src="http://www.speedsolving.com/%7Epatjk/speedsolving/forum/vlatex/img/8a499f7daa3620e61bebfbc6b3ad5f08-1.gif" style="border: 0px; vertical-align: middle;" title="m_1(n) = 2 m_1(n-1) + (2j-2) m_1(n-2) + 2j-1" /></a><br /><br />Anyway, it looks like the k constant (the number for which <a href="https://www.blogger.com/null"><img alt="m_1(n)" src="http://www.speedsolving.com/%7Epatjk/speedsolving/forum/vlatex/img/4afa667ba3ebf9ae50d21e001a176c6b-1.gif" style="border: 0px; vertical-align: middle;" title="m_1(n)" /></a> grows like <a href="https://www.blogger.com/null"><img alt="k^n" src="http://www.speedsolving.com/%7Epatjk/speedsolving/forum/vlatex/img/aafabbf195eb3424ca815b9afe5caff1-1.gif" style="border: 0px; vertical-align: middle;" title="k^n" /></a>) works out to <a href="https://www.blogger.com/null"><img alt="1 + \sqrt{2j-1}" src="http://www.speedsolving.com/%7Epatjk/speedsolving/forum/vlatex/img/8f29c10d2aae6d07c1fea446b96d16a3-1.gif" style="border: 0px; vertical-align: middle;" title="1 + \sqrt{2j-1}" /></a>. So that's pretty interesting, and we can definitely get k as large as we like by choosing a sufficiently high j<a href="https://www.blogger.com/null"></a>. In fact, we can set k to be any even integer: for instance, let j=41, and then the sequence <a href="https://www.blogger.com/null"><img alt="m_1(n)" src="http://www.speedsolving.com/%7Epatjk/speedsolving/forum/vlatex/img/4afa667ba3ebf9ae50d21e001a176c6b-1.gif" style="border: 0px; vertical-align: middle;" title="m_1(n)" /></a> grows as fast as <a href="https://www.blogger.com/null"><img alt="10^n" src="http://www.speedsolving.com/%7Epatjk/speedsolving/forum/vlatex/img/d928cc1f8a6326dd8ff809d6d813ba3f-1.gif" style="border: 0px; vertical-align: middle;" title="10^n" /></a>. 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.<br /><br />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 <u>or one of the next j-1 pegs in either direction</u>.<br /><br /><hr /><br />Alright guys, I made a sim. Try it here:<br /><a href="http://mzrg.com/js/hanoi/" target="_blank">http://mzrg.com/js/hanoi/</a><br /><br />>Very interesting, didn't read all of it though. Just curious, how did you come up with this idea?<br /><br />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.<br /><br />PS: Wolfram Alpha gives the following closed form for <a href="https://www.blogger.com/null"><img alt="m_1" src="http://www.speedsolving.com/%7Epatjk/speedsolving/forum/vlatex/img/377b1a53b01e907138040867edc7cac2-1.gif" style="border: 0px; vertical-align: middle;" title="m_1" /></a>:<br /><a href="https://www.blogger.com/null"><img alt="m_1(n) = -1 + \frac{(1+\sqrt{2j-1})^n - (1-\sqrt{2j-1})^n}{\sqrt{2j-1}}" src="http://www.speedsolving.com/%7Epatjk/speedsolving/forum/vlatex/img/ee8eaa7229b815495812e91954e89a4b-1.gif" style="border: 0px; vertical-align: middle;" title="m_1(n) = -1 + \frac{(1+\sqrt{2j-1})^n - (1-\sqrt{2j-1})^n}{\sqrt{2j-1}}" /></a>Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-89486479364917732062014-06-29T16:40:00.000-07:002014-06-29T16:40:40.668-07:00Speedsolving Posts: Almost-2gen LL AlgsFrom <a href="http://www.speedsolving.com/forum/showthread.php?18539">http://www.speedsolving.com/forum/showthread.php?18539</a> (January 2010):<br /><br />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.<br /><br />So, here is an R perm with exactly 2 D-layer moves:<br />R U2 R D R' U R D' R' U' R' U R U R' U<br />And here is a Y perm (but not an efficient one):<br />R' U' D (R' U' R U)3 D' U R (R U R' U R U2 R')2 U'<br /><br />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?<br /><br />P.S. You can also do every ZBLL with <r> and at most two F moves, although it's less useful IMO. You can prove this with the T perm and this Y perm:<br />F (R U R' U')3 F' U (R U R' U R U2 R')2 U2 </r><br /><br />From <a href="http://www.speedsolving.com/forum/showthread.php?708&p=680961#post680961">http://www.speedsolving.com/forum/showthread.php?708&p=680961#post680961</a> (December 2011):<br /><br />Dumping some optimal 2-F-move PLLs (there are probably a few optimal ones I missed but the optimal lengths should be accurate): <br /><br /><pre><b>A-perm a</b><br />HTM-optimal: 11f<br /> R U' R F2 R' U R' U' R2 F2 R2<br /><br />QTM-optimal: 15q<br /> R U' R F2 R' U R' U' R2 F2 R2<br /><br /><b>A-perm a</b><br />HTM-optimal: 11f<br /> R2 F2 R2 U R U' R F2 R' U R'<br /><br />QTM-optimal: 15q<br /> R2 F2 R2 U R U' R F2 R' U R'<br /> </pre><pre><b>E-perm</b></pre><pre>HTM-optimal: 18f<br /> R' U2 R U' F' R' U R' U2 R U R' U R2 F R' U' R<br /> R' U R F' R2 U' R U' R' U2 R U' R F U R' U2 R<br /> R U R' U R' F U' R2 U' R2 U2 R' U' R F' R' U R2<br /> R2 U F' U R' U' R U R' U' R U R' U' R F U' R2<br /> R2 U F' R' U R U' R' U R U' R' U R U' F U' R2<br /> R2 U' R F R' U R U2 R2 U R2 U F' R U' R U' R'<br /> R2 U2 F' U' R' U R U' R' U R U' R' U R F U2 R2<br /> R2 U2 F' R' U' R U R' U' R U R' U' R U F U2 R2<br /> F U R2 U' R U R' U R2 U' R' U R' U' R U' R F'<br /> F U' R U' R U R' U R U2 R' U R' U' R U2 R' F'<br /> F U' R U' R' U2 R U' R' U R' U2 R U R' U R F'<br /> F U' R' U R' U' R2 U R' U R U' R2 U R U' R F'<br /> F R U2 R' U R U' R U2 R' U' R U' R' U R' U F'<br /> F R' U R' U R U' R U R2 U' R U' R' U R2 U' F'<br /> F R' U R' U' R2 U R' U' R U' R2 U R U' R U F'<br /> F R' U' R U' R' U2 R U' R U R' U2 R U R' U F'<br /><br />QTM-optimal: 20q<br /> R2 U F' U R' U' R U R' U' R U R' U' R F U' R2<br /> R2 U F' R' U R U' R' U R U' R' U R U' F U' R2<br /> F U R2 U' R U R' U R2 U' R' U R' U' R U' R F'<br /> F U' R U' R U R' U R U2 R' U R' U' R U2 R' F'<br /> F U' R U' R' U2 R U' R' U R' U2 R U R' U R F'<br /> F U' R' U R' U' R2 U R' U R U' R2 U R U' R F'<br /> F R U2 R' U R U' R U2 R' U' R U' R' U R' U F'<br /> F R' U R' U R U' R U R2 U' R U' R' U R2 U' F'<br /> F R' U R' U' R2 U R' U' R U' R2 U R U' R U F'<br /> F R' U' R U' R' U2 R U' R U R' U2 R U R' U F'<br /> </pre><pre><b>F-perm</b><br />HTM-optimal: 15f<br /> R2 U R2 U2 F2 U R2 U2 R2 U R2 U' R2 U2 F2<br /> R2 U R2 U' R2 U2 R2 U' F2 U2 R2 U' R2 U' F2<br /> F2 U R2 U R2 U2 F2 U R2 U2 R2 U R2 U' R2<br /> F2 U2 R2 U R2 U' R2 U2 R2 U' F2 U2 R2 U' R2<br /> R2 U' F2 U2 R2 U' R2 U2 R2 U R2 U F2 U2 R2<br /> R2 U2 F2 U' R2 U' R2 U2 R2 U R2 U2 F2 U R2<br /><br />QTM-optimal: 19q<br /> R U R' U' R' U R U F' U R' U R U2 F R U' R'<br /> R U R' F' U2 R' U' R U' F U' R' U' R U R U' R'<br /> R U R U F' U' R U R' U' R' U2 F U' R' U2 R'<br /> R U2 R U F' U2 R U R U' R' U F U' R' U' R'<br /> R' U' R U' R' U R U R2 F' R U R U' R' F U R<br /> R' U' F' R U R' U' R' F R2 U' R' U' R U R' U R<br /> </pre><pre><b>G-perm a</b><br />HTM-optimal: 15f<br /> F2 U R U2 R' U' R' U2 R2 U R' F2 R' U R<br /> R' U' F2 U R U2 R U2 R' U F2 U2 R U' R'<br /><br />QTM-optimal: 18q<br /> R U R' U' R' U F R U R U' R' F' U R' U2 R<br /> </pre><pre><b>G-perm b</b><br />HTM-optimal: 15f<br /> R' U' R F2 R U' R2 U2 R U R U2 R' U' F2<br /> R U R' U2 F2 U' R U2 R' U2 R' U' F2 U R<br /><br />QTM-optimal: 18q<br /> R' U2 R U' F R U R' U' R' F' U' R U R U' R'<br /> </pre><pre><b>G-perm c</b></pre><pre>HTM-optimal: 14f<br /> R2 U2 R2 F2 U' R2 U R2 U F2 U2 R2 U' R2<br /><br />QTM-optimal: 17q<br /> R2 U' R U' R U R' U R' F' R' U R U' F R'<br /> </pre><pre><b>G-perm d</b><br />HTM-optimal: 14f<br /> R2 U R2 U2 F2 U' R2 U' R2 U F2 R2 U2 R2<br /><br />QTM-optimal: 17q<br /> R F' U R' U' R F R U' R U' R' U R' U R2<br /> </pre><pre><b>J-perm a</b><br />HTM-optimal: 13f<br /> R' F R' U' R2 U' R2 U R' F' R U R2<br /> R2 U' R' F R U' R2 U R2 U R F' R<br /><br />QTM-optimal: 16q<br /> R' F R' U' R2 U' R2 U R' F' R U R2<br /> R2 U' R' F R U' R2 U R2 U R F' R<br /> F R' U R U' F' U R' U2 R U' R' U2 R<br /> R' U2 R U R' U2 R U' F U R' U' R F'<br /> R' U' R U F R U' R' U' R' U R2 U R' F'<br /> R' U' R U' R' U F U R' U' R F' R' U R2<br /> R2 U' R F R' U R U' F' U' R U R' U R<br /> F R U' R2 U' R U R U R' F' U' R' U R<br /> </pre><pre><b>J-perm b</b><br />HTM-optimal: 11f<br /> R U' F U' R' U' R U F' U2 R'<br /> R U2 F U' R' U R U F' U R'<br /><br />QTM-optimal: 12q<br /> R U' F U' R' U' R U F' U2 R'<br /> R U2 F U' R' U R U F' U R'<br /> </pre><pre><b>N-perm a</b><br />HTM-optimal: 15f<br /> R2 U R2 U' R2 U' F2 U' R2 U R2 U F2 U2 R2<br /> R2 U2 F2 U' R2 U' R2 U F2 U R2 U R2 U' R2<br /><br />QTM-optimal: 21q<br /> R U2 R2 U' R2 F' R U R' U' R' F R' U R U2 R'<br /> R U2 R' U' R F' R U R U' R' F R2 U R2 U2 R'<br /> R U R' U R U' F U' R' U' R U F' U2 R' U2 R U' R'<br /> R U R' U2 R U2 F U' R' U R U F' U R' U' R U' R'<br /> </pre><pre><b>N-perm b</b><br />HTM-optimal: 15f<br /> R2 U F2 U2 R2 U R2 U2 R2 U' R2 U2 F2 U' R2<br /> R2 U' R2 U2 F2 U' R2 U2 R2 U F2 U2 R2 U R2<br /><br />QTM-optimal: 23q?<br /> R U' R' U R' U2 R2 U R2 F2 U R U R U' R' U' F2<br /> F2 U R U R' U' R' U' F2 R2 U' R2 U2 R U' R U R'<br /> R' U2 R' U' R2 U F' U' R U R' U' R2 U2 R F R' U R<br /> R' U2 F U R2 U2 R' U' R U R' F' R U2 R' U' R' U R<br /> R' U' R U R U2 R' F R U' R' U R U2 R2 U' F' U2 R<br /> R' U' R F' R' U2 R2 U R U' R' U F U' R2 U R U2 R<br /> </pre><pre><b>R-perm a</b><br />HTM-optimal: 13f<br /> R2 F2 U R U R' U' R' U' F2 R' U R'<br /> R U' R F2 U R U R U' R' U' F2 R2<br /><br />QTM-optimal: 16q<br /> R2 F2 U R U R' U' R' U' F2 R' U R'<br /> R U' R F2 U R U R U' R' U' F2 R2<br /> </pre><pre><b>R-perm b</b></pre><pre>HTM-optimal: 13f<br /> R2 F R U R U' R' F' R U2 R' U2 R<br /> R' U2 R U2 R' F R U R' U' R' F' R2<br /><br />QTM-optimal: 16q<br /> R2 F R U R U' R' F' R U2 R' U2 R<br /> R' U2 R U2 R' F R U R' U' R' F' R2<br /> </pre><pre><b>T-perm</b></pre><pre>HTM-optimal: 14f<br /> R U R' U' R' F R2 U' R' U' R U R' F'<br /> F R U' R' U R U R2 F' R U R U' R'<br /><br />QTM-optimal: 15q<br /> R U R' U' R' F R2 U' R' U' R U R' F'<br /> F R U' R' U R U R2 F' R U R U' R'<br /> </pre><pre><b>V-perm</b><br />HTM-optimal: 15f<br /> R2 U2 R2 U R2 U2 F2 U R2 U R2 U2 F2 U' R2<br /> R2 U2 R2 U' F2 U2 R2 U' R2 U' F2 U2 R2 U R2<br /> R2 U F2 U2 R2 U' R2 U' F2 U2 R2 U' R2 U2 R2<br /> R2 U' R2 U2 F2 U R2 U R2 U2 F2 U R2 U2 R2<br /><br />QTM-optimal: 19q<br /> R U' R' U R F R' U' R2 U R U' R U R2 F' R'<br /> R F R2 U' R' U R' U' R2 U R F' R' U' R U R'<br /> </pre><pre><b>Y-perm</b></pre><pre>HTM-optimal: 14f<br /> R2 U' R2 U F2 U' R2 U' R2 U F2 R2 U R2<br /> R2 U' R2 F2 U' R2 U R2 U F2 U' R2 U R2<br /><br />QTM-optimal: 19q<br /> R2 U' R2 U' R2 U R' F' R U R2 U' R' F R<br /> R' F' R U R2 U' R' F R U' R2 U R2 U R2<br /> R' U' R U' R U' F U' R' U' R U F' U2 R2 U R<br /> R' U' R2 U2 F U' R' U R U F' U R' U R' U R</pre>Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-59904638348818218302014-06-29T14:35:00.003-07:002014-06-29T14:35:41.540-07:00GeoGebra Files for Twisty Puzzle ClassificationIn 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: <a href="http://www.mzrg.com/filepost/GeoGebra_TPCP.zip">http://www.mzrg.com/filepost/GeoGebra_TPCP.zip</a><br /><br />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.<br /><br />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...Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-73232797912368160912014-06-29T14:25:00.003-07:002014-06-29T14:35:58.060-07:00Speedsolving Posts: 2D State DiagramsThe posts here are from January 2009 and can be found in this topic: <a href="http://www.speedsolving.com/forum/showthread.php?8628">http://www.speedsolving.com/forum/showthread.php?8628</a><br /><br />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.)<br /><br />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.<br /><br />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?<br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="http://i286.photobucket.com/albums/ll119/qqwref/Tvv_diagram.png" /></div><br />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).<br /><br />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.<br /><br />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 <a href="http://www.geogebra.org/cms/">GeoGebra</a> geometry program, which I'm really fond of. You can download it from <a href="http://www.mzrg.com/filepost/Tvv.ggb">my website</a>. 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.<br /><br /><hr /><br />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).<br /><br /><div class="separator" style="clear: both; text-align: center;"><img border="0" src="http://i286.photobucket.com/albums/ll119/qqwref/Dff_diagram.png" /></div><br />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:<br />- 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).<br />- 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.<br />- 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.<br />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.<br /><br />Finally, there is a <a href="http://www.mzrg.com/filepost/Dff.ggb">GeoGebra simulation</a> 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. <br /><br /><hr /><br />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 <a href="http://www.mzrg.com/filepost/Cfv.ggb" target="_blank">Cfv</a> 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! Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-42369413870490022102014-06-29T14:17:00.000-07:002014-06-29T14:17:57.777-07:00TPCP: Type Ce(Old post from 2008)<br /><br />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.<br /><br /><table><tbody><tr><td><a href="http://2.bp.blogspot.com/_FwZfrOWBA0g/SHO3bvzy9nI/AAAAAAAAACE/C8RzUIhM1Ik/s1600-h/Ce_1_homemade.PNG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" src="http://2.bp.blogspot.com/_FwZfrOWBA0g/SHO3bvzy9nI/AAAAAAAAACE/C8RzUIhM1Ik/s320/Ce_1_homemade.PNG" id="BLOGGER_PHOTO_ID_5220718080370603634" style="cursor: pointer; float: left; margin: 0pt 10px 10px 0pt;" /></a></td><td><span style="font-size: medium;">Puzzle Ce1:</span><br />Gelatinbrain Number: n/a<br />Common Name: n/a<br />Types of Pieces: 6 fixed centers, 12 trivial edges, 24 triangular centers in 4 orbits, 8 corners.<br />Solution: n/a</td></tr><tr><td><a href="http://3.bp.blogspot.com/_FwZfrOWBA0g/SHO3bwympGI/AAAAAAAAACM/gZd_XWfJQxM/s1600-h/Ce_2.PNG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" src="http://3.bp.blogspot.com/_FwZfrOWBA0g/SHO3bwympGI/AAAAAAAAACM/gZd_XWfJQxM/s320/Ce_2.PNG" id="BLOGGER_PHOTO_ID_5220718080634037346" style="cursor: pointer; float: left; margin: 0pt 10px 10px 0pt;" /></a></td><td><span style="font-size: medium;">Puzzle Ce2:</span><br />Gelatinbrain Number: 3.3.1<br />Common Name: Helicopter Cube, Bevel Cube<br />Types of Pieces: 24 centers in 4 orbits, 8 corners.<br />Solution: n/a</td></tr><tr><td><a href="http://3.bp.blogspot.com/_FwZfrOWBA0g/SHO3cMddRvI/AAAAAAAAACU/qZPc4metFSs/s1600-h/Ce_3_homemade.PNG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" src="http://3.bp.blogspot.com/_FwZfrOWBA0g/SHO3cMddRvI/AAAAAAAAACU/qZPc4metFSs/s320/Ce_3_homemade.PNG" id="BLOGGER_PHOTO_ID_5220718088061535986" style="cursor: pointer; float: left; margin: 0pt 10px 10px 0pt;" /></a></td><td><span style="font-size: medium;">Puzzle Ce3:</span>Gelatinbrain Number: 3.3.2<br />Common Name: n/a<br />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.<br />Solution: n/a</td></tr><tr><td><br /></td></tr><tr><td><a href="http://3.bp.blogspot.com/_FwZfrOWBA0g/SHO3cIjNZTI/AAAAAAAAACc/NkzIgH0CaHg/s1600-h/Ce_4.PNG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" src="http://3.bp.blogspot.com/_FwZfrOWBA0g/SHO3cIjNZTI/AAAAAAAAACc/NkzIgH0CaHg/s320/Ce_4.PNG" id="BLOGGER_PHOTO_ID_5220718087011919154" style="cursor: pointer; float: left; margin: 0pt 10px 10px 0pt;" /></a></td><td><span style="font-size: medium;">Puzzle Ce4:</span>Gelatinbrain Number: 3.3.3<br />Common Name: n/a<br />Types of Pieces: 6 inner centers, 24 inner triangular centers, 48 outer triangular centers in 2 orbits, 12 edges, 8 corners.<br />Solution: n/a</td></tr><tr><td><a href="http://1.bp.blogspot.com/_FwZfrOWBA0g/SHO3cTVGlYI/AAAAAAAAACk/xpWQutYRzkQ/s1600-h/Ce_5_homemade.PNG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" src="http://1.bp.blogspot.com/_FwZfrOWBA0g/SHO3cTVGlYI/AAAAAAAAACk/xpWQutYRzkQ/s320/Ce_5_homemade.PNG" id="BLOGGER_PHOTO_ID_5220718089905542530" style="cursor: pointer; float: left; margin: 0pt 10px 10px 0pt;" /></a></td><td><span style="font-size: medium;">Puzzle Ce5:</span>Gelatinbrain Number: 3.3.4<br />Common Name: n/a<br />Types of Pieces: 6 inner centers, 24 adjacent triangular centers, 24 diagonal triangular centers, 48 tetragonal centers in 2 orbits, 12 edges, 8 corners.<br />Solution: n/a</td></tr><tr><td><a href="http://1.bp.blogspot.com/_FwZfrOWBA0g/SHO3ihGJvNI/AAAAAAAAACs/cBvlSMydOFo/s1600-h/Ce_6.PNG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" src="http://1.bp.blogspot.com/_FwZfrOWBA0g/SHO3ihGJvNI/AAAAAAAAACs/cBvlSMydOFo/s320/Ce_6.PNG" id="BLOGGER_PHOTO_ID_5220718196680146130" style="cursor: pointer; float: left; margin: 0pt 10px 10px 0pt;" /></a></td><td><span style="font-size: medium;">Puzzle Ce6:</span>Gelatinbrain Number: 3.3.5<br />Common Name: n/a<br />Types of Pieces: 6 inner centers, 24 inner triangular centers, 48 outer triangular centers in 2 orbits, 12 edges, 8 corners.<br />Solution: n/a</td></tr><tr><td><a href="http://4.bp.blogspot.com/_FwZfrOWBA0g/SHO3irj82BI/AAAAAAAAAC0/D3FMx9mK6LI/s1600-h/Ce_7_homemade.PNG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" src="http://4.bp.blogspot.com/_FwZfrOWBA0g/SHO3irj82BI/AAAAAAAAAC0/D3FMx9mK6LI/s320/Ce_7_homemade.PNG" id="BLOGGER_PHOTO_ID_5220718199489484818" style="cursor: pointer; float: left; margin: 0pt 10px 10px 0pt;" /></a></td><td><span style="font-size: medium;">Puzzle Ce7:</span><br />Gelatinbrain Number: 3.3.6<br />Common Name: n/a<br />Types of Pieces: 6 inner centers, 24 pentagonal centers, 24 inner triangular centers, 48 outer triangular centers in 2 orbits, 12 edges, 8 corners.<br />Solution: n/a</td></tr><tr><td><a href="http://3.bp.blogspot.com/_FwZfrOWBA0g/SHO3jCXlszI/AAAAAAAAAC8/zG3cNFeUwx4/s1600-h/Ce_8.PNG" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" src="http://3.bp.blogspot.com/_FwZfrOWBA0g/SHO3jCXlszI/AAAAAAAAAC8/zG3cNFeUwx4/s320/Ce_8.PNG" id="BLOGGER_PHOTO_ID_5220718205611651890" style="cursor: pointer; float: left; margin: 0pt 10px 10px 0pt;" /></a></td><td><span style="font-size: medium;">Puzzle Ce8:</span>Gelatinbrain Number: 3.3.7<br />Common Name: Little Chop<br />Types of Pieces: 24 centers.<br />Solution: n/a</td></tr></tbody></table>Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-33835688937561729042013-05-19T20:34:00.004-07:002013-05-19T20:35:22.411-07:00Speedsolving Posts: A Huge Cube MethodI really put a lot of thought into making a method for every part that would scale well - remain easy even with extremely large cubes. And when I say extremely large, this is the method I solved the 111x111x111 with. It's probably still useful on smaller huge cubes (like 20x20x20) but when you can see a whole face at a time, and count layers, some parts can be done quite a bit faster.<br /><br />From <a href="http://www.speedsolving.com/forum/showthread.php?42084">http://www.speedsolving.com/forum/showthread.php?42084</a>:<br /><br />First, some basics. I solved most faces in two steps: "cleaning" and finishing up. The idea of cleaning a face is to solve as many pieces as possible as quickly as possible. After that step, which typically solves 50% or more of the face, I can go through the longer step of putting each of the remaining pieces of that color into the center, one at a time or in blocks. As for notation, I'll be using something like SiGN (xR means turn the xth layer on the R face, or the first layer if x is absent), except that I'll also be using lowercase p and q as variables to better describe classes of commutators.<br /><br />1) Solve the edges. I do edges first for large computer cubes because reduction (centers first) scales badly, and because having solved edges actually lets me use them to keep track what part of the puzzle I'm working on. A lot of the time I do a pR' move or something to start working on a row of one center, and the incorrect edge reminds me which row it is. With so many layers it's impossible to do it by counting or visual inspection alone. Anyway, for the first 9 edges I did basically the same thing as with other cubes. For the last three edges I used moves like pR U2 pR' (for the 10th edge only), pR2 U2 pR U2 pR2, and pR' F U' R F' U pR. Instead of solving one orbit at a time like on my keyboard solves, I went for an edge at a time and solved pieces in the same position in groups. For instance on the last 3 edges there are only 6 possible positions for each piece, so with 109 pieces in each edge there are a lot of pieces in the same position which can be solved in the same way, all at once, by replacing a pR move with a whole bunch of slice moves along the same layer. This took about an hour in total.<br /><br />2) Clean the first center. I used white. Hold the white face on U, and then for each row (perpendicular to the F face), we do something like this: move white centers out of the way on the corresponding row on F using pU moves, then qR', then solve as many white centers into that layer as possible using only pU-axis moves, then qR. By "solve as many white centers ... as possible" I mean I basically look at that row on R, B, and L, and for every white center I see, move that layer the right amount. Note that these rows are vertical, which is something I did because of the way Iso is set up (when zoomed in, it's easy to scroll through a vertical row on F or R using just the up/down arrow keys). You could do any of this from a different angle if you wanted. A typical layer might be something like (2U 4U 7U ...) 3R' y (3U 6U 8U ...) y (U2 4U2 8U2 ...) y (2U' 3U' 4U' ...) y 3R. Each center on white corresponds to 5 stickers, and a white center on any of those 5 stickers means the center gets solved. I think roughly 60% of the center ends up solved.<br /><br />3) Finish the first center. Now I can hold the center on F and look at R. If I see a white center anywhere, I can solve it using pF qU' pF' qU. Four-move commutators are fast! However, I can't just do that randomly, but instead, I have to do it so that the center moves into a place on the white center that didn't have a solved center already. To check if there is room I do qU' qU, and if there isn't, I rotate the white center and repeat as necessary. You'll see this type of checking often as I continue. Note that this requires centers to be adjacent, and since we're solving from all five, we have to swap two adjacent centers. On the 5x5x5 the algorithm for this is 2R U2 2R' 2L' U2 2L U 3R U' 3R', which you then have to properly undo later.<br /><br />4) Clean the second center. I used yellow. This is the same idea as the first center - hold yellow on U (so, white on D) and proceed row by row. However, instead of using pR' moves to move a row onto the F face, I used pR U2 pR' moves. This does solve the layers in a different order, but that's okay. For the middle layer I used U pR U' pR'. Each center on yellow again corresponds to 5 stickers (any yellow piece in one of those 5 places means that piece gets solved), and I think roughly 67% of the center ends up solved after this.<br /><br />5) Finish the second center. Again I swap yellow with another center, so I can hold yellow on F, white on U, and the solved center on R. The commutator this time is pF2 qU' pF2 qU (or pF2 qU pF2 qU' for the mirror). Since I can't just do x rotations and keep going, I have to do some extra center swaps to bring each of the four non-white/yellow centers into the right place. There was also plenty of the qU' qU moves to check if there's room to put a piece.<br /><br />6) Clean the third center. I used orange (for visibility). Place orange on U and white/yellow on L/R. Now we can do pU2 moves but not pU moves. So I do something like the white center cleaning, where I do pU2 moves to move orange centers out of the way, then a qR' move to bring the row down, then do more pU2 moves to solve as many orange centers as possible, then a qR move. A typical row clean might look something like (2U2 4U2 7U2 ...) 3R' y2 (2U2 3U2 4U2 ...) y2 3R. Each center on orange corresponds to only 3 stickers, solving about 57% of the center.<br /><br />7) Solve the third center. This is just like what I do on yellow, with the pF2 qU' pF2 qU moves, since they keep U and D solved, and again I use plenty of qU' qU moves. In fact, there are fewer center swaps than last time, since the D center can stay where it is.<br /><br />8) Clean the fourth center. As a general rule you want this to be adjacent to the third center, so I chose green (since it was brighter). Unfortunately the cleaning setup I use only really lets us clean a center opposite of the third one, so we place orange on D and green on F, then swap F and U, and begin. The cleaning itself is just like the third center except, again, we use pR U2 pR' rather than pR'. Each center on green corresponds to 3 stickers, solving about 70% of the center.<br /><br />9) Solve the fourth center. Things got pretty messy here and it took a lot longer than I expected; most of that was probably because I can't use 4-move commutators any more and thus had to switch to 8-move ones. I held green on F and orange on D, and then did commutators like [pU2, qR U qR']. In fact I did all of the pU2 moves ahead of time, so that for each row I could immediately do a qR move for each piece I wanted to solve (using pU2 pu2 to check), and then solve many piece in that row at once. The problem, apart from the large number of qR moves, is that I couldn't stray too far from U, because I had to do U turns. So I ended up having to stay on the top half of F, and doing a lot of F and B moves to be able to solve things. Oh yeah, and once I got rid of all the green pieces on B, I had to swap U and B and do it again.<br /><br />10) Solve the last two centers. No cleaning here - I couldn't see any way to quickly solve most of the center. I held the two centers on F and R, and solved one vertical row of R at a time. The commutator looks something like [pU', R' qU' R]. Basically, I would do a pU' for every center in that row that I could solve, then do the R' qU' R, undo the pU moves, and finish the commutator. Then rotate F and go again - in at most 4 separate block commutators I would solve all pieces in that row. (With one caveat - if the x-center closer to D is unsolved, this commutator won't work on it. When that happened I had to do it individually.) After each finished row I did a qF2 move to get it out of the way and make the next row easy to see; since I had blue on R, I ended up with a whole bunch of green rows on that face. Since I had to be in view of the F-R edge to do R tuns, I only did the first half of the blue face before undoing all those green rows, turning that center by 180 degrees, and doing it again. <br /><br />And that's it! It might sound kind of complicated, but each step is pretty straightforward once you know what's going on, and I spent the vast majority of my time doing the same kind of thing over and over. The low move/piece count was mostly due to the cleaning stages, where I could solve a ton of pieces in one or two moves each. Those stages would have an absurdly low movecount in axial turn metric / snyder metric, with something like 400 moves solving 5000+ pieces. And in fact, you can do a cleaning stage on the same center more than once - just turn the center to a different orientation and continue. With four of these cleaning stages you should be able to solve every piece on the affected centers. I didn't do this because the returns do diminish, and so I don't think it's worth it to clean more than once for any individual center. Maybe for the fourth center it's worth it though! Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-55258524478953719562013-05-15T11:55:00.002-07:002013-05-15T11:55:38.283-07:00Speedsolving Posts: Parity ErrorsI'm going to be posting some of my longer posts on speedsolving.com here. I think I've written some very useful things down there, but it's a kinda big forum (and I've made a lot of posts!) so it's easy for stuff to get lost. The plan is to copy some of them here, with a link to the original post (for those who want to view the thread). Here goes...<br /><br /><hr />From <a href="http://www.speedsolving.com/forum/showthread.php?p=852756">http://www.speedsolving.com/forum/showthread.php?p=852756</a>:<br /><br />In speedcubing there are two things we think of as parity: "reduction parity" and mathematical parity. Some things we call parity fall into both categories, but others only fall into one of them, and you will often see people disagree on whether certain things count as parity, because they disagree that both of these two definitions are valid.<br /><br />Reduction parity occurs when you try to reduce the puzzle so it can be solved by a constrained set of moves, putting it into some subset of the positions. However, you can often reach a position which seems like it is in your subset, but which is actually not, and to solve the puzzle you have to briefly go outside your constrained set of moves to bring the puzzle back into the subset you want. Typically the number of positions you can encounter is some small multiple of the number of positions you expect. The obvious example is PLL parity in 4x4x4: all the centers and edges are properly paired, so you expect to be able to finish the puzzle with only outer layer turns, but this isn't quite possible. OLL parity falls under this definition too (so the reduced 4x4x4 has four times as many positions as you would expect). Square-1 parity also falls under this definition - your constrained set of moves are any moves that keep the puzzle in cubeshape. BLD parity is not of this type (the solver has not reduced the puzzle).<br /><br />Mathematical parity is based on the idea of the mathematical definition of an odd permutation. Basically, at least one orbit of pieces has an odd permutation, and thus cannot be solved with just 3-cycles. You see this type of parity crop up in blindfolded solves, because blindfolded solvers attempt to solve most or all of the puzzle with 3-cycles and thus an odd permutation is very noticeable. OLL parity on 4x4x4 and Square-1 parity can also be thought of as this way, as they originate from some kind of 2-cycle. PLL parity on 4x4x4 is not of this type (it can be solved with 3-cycles). Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-14325389123714619712009-03-26T23:32:00.000-07:002009-03-27T02:30:38.176-07:00Pixeling, Or, Putting Hearts on Your 7x7x7<img src="http://solvethecube.110mb.com/nxnimagecube.php?n=7&stickers=rrrrrrrrrgrgrrrgggggrrgggggrrrgggrrrrrgrrrrrrrrrrwwwwwwwwwrwrwwwrrrrrwwrrrrrwwwrrrwwwwwrwwwwwwwwwwgggggggggwgwgggwwwwwggwwwwwgggwwwgggggwgggggggggg&size=100" style="float: left;" /><br />Pixeling is a technique that I discovered to make certain types of pretty patterns on NxNxN cubes. It affects the centers, only, in a large series of 3-cycles. Specifically pixeling is used to generate patterns where a certain shape of centers is moved in two 3-cycles, from (say) the U->R->F faces and D->L->B faces, in a very efficient way. An example is the hearts picture at left, which can be created with just 11 block turns (i.e. turns of a group of adjacent slices all in the same direction)! If you ask me that's ridiculously efficient.<br /><br />Here's the motivation (with pictures below). You'll need to try these patterns out to understand where pixeling comes from. First we start with the basic center pattern on the 3x3:<br /><a href="http://www.randelshofer.ch/cube/rubik/?MR-MUMRMU-">M E' M' E</a>.<br />This just moves the centers around. We can also do this on the 5x5 (or any NxNxN cube, really):<br /><a href="http://www.randelshofer.ch/cube/professor/?WR-WUWRWU-">2-4r' 2-4u 2-4r 2-4u'</a>.<br />But on the 5x5 there are related patterns. The following show that you can move blocks of centers around (or a single center), and that the blocks don't even need to be contiguous:<br /><a href="http://www.randelshofer.ch/cube/professor/?VR-VUVRVU-">2-3r' 2-3u 2-3r 2-3u'</a><br /><a href="http://www.randelshofer.ch/cube/professor/?M1R-M1LM1UM1D-M1RM1L-M1U-M1D">(2R' 2L) (2U 2D') (2R 2L') (2U' 2D)</a>.<br /><img src="http://solvethecube.110mb.com/nxnimagecube.php?n=3&stickers=wwwwrwwwwggggwggggrrrrgrrrr&size=100" /><img src="http://solvethecube.110mb.com/nxnimagecube.php?n=5&stickers=wwwwwwrrrwwrrrwwrrrwwwwwwggggggwwwggwwwggwwwggggggrrrrrrgggrrgggrrgggrrrrrr&size=100" /><img src="http://solvethecube.110mb.com/nxnimagecube.php?n=5&stickers=wwwwwwwrrwwwrrwwwwwwwwwwwgggggggwwgggwwgggggggggggrrrrrrrggrrrggrrrrrrrrrrr&size=100" /><img src="http://solvethecube.110mb.com/nxnimagecube.php?n=5&stickers=wwwwwwrwrwwwwwwwrwrwwwwwwggggggwgwgggggggwgwggggggrrrrrrgrgrrrrrrrgrgrrrrrr&size=100" /><img src="http://solvethecube.110mb.com/nxnimagecube.php?n=5&stickers=wwwwwwwwrwwwwwwwwwwwwwwwwggggggggwggggggggggggggggrrrrrrrrgrrrrrrrrrrrrrrrr&size=100" /><br /><br /><br /><img src="http://solvethecube.110mb.com/nxnimagecube.php?n=5&stickers=wwwwwwrrrwwwrrwwwwrwwwwwwggggggwwwgggwwggggwggggggrrrrrrgggrrrggrrrrgrrrrrr&size=100" style="float: left;" />You can basically now make any pattern we want using a series of these block maneuvers, and the efficiency on those patterns is not bad at all. But we can do better. Try this maneuver on a 5x5, which creates the staircase pattern to the left:<br /><a href="http://www.randelshofer.ch/cube/professor/?M1R-M1D-MR-MUM1LM1UWRWU-">2R' 4U 3R' 3U 4R' 2U 2-4r 2-4u'</a><br />This is the most introductory example of pixeling. Do you see what is happening there? It's a bit tricky, but try to keep track. Every time we do a U turn, we are basically storing a line of three centers on the L-face. Together, those lines make the entire center (and the shape that we want to pixel in, which is inside it), which is done when we make the 2U move. Then we simply return the vertical slices to their original position and move all the horizontal slices back. This can also be done using the vertical R slices to store the center lines in the D face, of course, and some patterns turn out to be more efficient one way than the other, so it's important to try both ways.<br /><br /><img src="http://solvethecube.110mb.com/nxnimagecube.php?n=7&stickers=wwwwwwwwwwrwwwwwrrrwwwrrrrrwwwwrwwwwwwrwwwwwwwwwwggggggggggwgggggwwwgggwwwwwggggwggggggwggggggggggrrrrrrrrrrgrrrrrgggrrrgggggrrrrgrrrrrrgrrrrrrrrrr&size=100" style="float: left;" /><br />We're not limited to just R' turns, though - sometimes we want to use R slices to remove a center from our shape. To the left is an arrow shape for the 7x7 that can be created like this:<br /><a href="http://www.randelshofer.ch/cube/vcube7/?WR-MUNRNL-N3UN3RN3L-NUVD-MRWU-">2-6r' 4U 2R 2L' 3U 3R 3L' 2U 2-3d' 4R 2-6u'</a><br />Notice that before every time we do a turn of a horizontal slice in the U direction, we set up the R slices so that only the ones which contain the centers we want in our shape are misaligned. Also, this time we didn't do the U slices in order, but this is also fine; the pixeling maneuver still works.<br /><br /><img src="http://solvethecube.110mb.com/nxnimagecube.php?n=7&stickers=rrrrrrrrrgrgrrrgggggrrgggggrrrgggrrrrrgrrrrrrrrrrwwwwwwwwwrwrwwwrrrrrwwrrrrrwwwrrrwwwwwrwwwwwwwwwwgggggggggwgwgggwwwwwggwwwwwgggwwwgggggwgggggggggg&size=100" style="float: left;" />The end result is that there are many, many ways to pixel any given shape. It will take a bit of trial and error to do a shape efficiently, but the efficiency can be really amazing. If you're making a shape yourself, remember to try both directions (vertical and horizontal slices to set up the shape) and a few different orders of slices. If one of those cycles in the wrong order that you want, you can just invert the pattern. Finally, let's return to that heart pattern. Here's the best we can do for the heart pattern with blocks and with pixeling, respectively:<br />Blocks: <a href="http://www.randelshofer.ch/cube/vcube7/?NR-NLM2UNRNL-M2U-N3R-N3LV4UN3RN3L-V4U-MR-V4D-MRV4D">(2R' 2L 3-4u 2R 2L' 3-4u') (3R' 3L 2-5u 3R 3L' 2-5u') (4R' 3-6u 4R 3-6u')</a> (16 moves)<br />Pixeling: <a href="http://www.randelshofer.ch/cube/vcube7/?WR-V4UN3RN3L-NU-ND-MRVDNRNL-M2U-">(3-4u 2R' 2L 2-3d' 4R' 2D 2U 3R' 3L 2-5u' 2-6r)'</a> (11 moves)<br /><br />Happy patterning! Thanks go to Werner Randelshofer for the cube animation applets and to JoÃ«l van Noort for his NxN ImageCube (which I used to generate the images you see here).Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com1tag:blogger.com,1999:blog-3335271307001051587.post-8156542572918179872009-01-22T15:26:00.000-08:002009-01-22T22:07:12.301-08:00How to Solve 6x6+ (on hi-games)Let's start by saying that I don't really like the "grip shift" controls. Thus, the way I solve big cubes is designed to be somewhat fast (although I'm sure there are faster ways out there) while changing the grip as few times as possible. As it turns out, reduction is the best method for this.<br /><h2>Centers</h2>Obviously, centers require lots of face turns, but also lots of slice turns. So how can we solve them with a small number of grip shifts? As it turns out, we can do a pretty decent job using U, F, D, B, and any r and l slices. Only build and insert rows on the x axis (faces U, F, D, and B), and although this uses a lot of cube rotations it isn't too restrictive. Besides, you can only see U and F anyway. So then the question is what slices we need. Here's how I do it:<br /><span style="font-weight: bold;">6x6 and 7x7</span>: shift both right and left in once. Now you can do all of the usual moves (3r, 3R, 2r, 3l, 3L, 2l) with each hand.<br /><span style="font-weight: bold;">8x8 and 9x9</span>: shift right in twice and left in once. The left hand can do 2l, 3l, 3L and the right hand can do 4r, 3r, 4R.<br /><span style="font-weight: bold;">10x10</span>: shift right in three times and left in once. The left hand can do 2l, 3l, 3L and the right hand can do 5r, 4r, 5R. Note that 4R is actually not possible (except by doing 4r y2 3l'), so you should try to work around this when building centers.<br /><br />Note that in the cases where the right and left hands are asymmetrical, I really suggest pairing from the inside out, since slice moves are not always possible or convenient. The goal is to avoid cube rotations so you can spend all your time actually doing moves. As with other large cubes, by the way, I really suggest pairing up 1x(n-1) rows of centers, and putting them together as soon as they are done.<br /><h2>Edges</h2>My method for edges is worse, but it still works pretty acceptably. The way I do this is to use the D slice to do slice moves and pair up edges on the E slice, since I tend to use the other slices to move edges around. Currently I use WO for moving the D slice in (up) and SL for moving it out (down), but you can use anything you want. Then I pair up edges multislice-style: I start by pairing up the innermost pieces, and then move out. Note that it's also possible to just use the L slice (and pair up edges on M), and use only RUDFB moves to insert edges, but personally this doesn't feel natural to me so I don't use it.<br /><br />My method of pairing up 5x5-style edges is freeslice (the bigcubes.com) style, which means I pair up one edge group (of 3 pieces) at a time and don't fix centers until I have eight edges paired and out of the E slice. Since this normally requires U moves, I end up doing a LOT of z2 moves. As I said, this isn't the best way to pair up edges, but it works. Note that if you end up with a parity case you are better off temporarily moving the edges to the M slice, shifting R and L in by an appropriate amount, and doing the parity normally. This is possible because you don't need to do single-layer R and L turns.<br /><br />So I pair up edges in two steps on 6x6 and 7x7, three steps on 8x8 and 9x9, and four steps on 10x10. On the last step, I just use the standard 5x5 pairing (which is on M), because if I have the right and left grips on the far outside of the cube I can do slice moves AND R and L turns.<br /><h2>3x3</h2>The 3x3 stage is pretty standard, press space and just solve it normally. On even cubes there are parity issues to deal with. The way I do this is, for OLL parity, shift L and R in until the r and l slices turn half of the cube, and then the algorithm is:<br />r U2 x r U2 r U2 r' U2 l U2 r' U2 r U2 r' U2 r'<br />and for PLL parity, just shift R in so that the r slice is halfway, and then do:<br />z r2 z' r2 U2 r2 y2 L2 y2 U2 r2 z r2 z'.<br />There are a lot of cube rotations so it looks longer than it is, but it's still kinda slow.<br /><br />So, good luck with bigcubes! Make sure to give each one a try, because for a lot of them you'll get on the top 100 list if you solve it, no matter how long it takes.Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com1tag:blogger.com,1999:blog-3335271307001051587.post-89556087429015408852008-11-28T18:21:00.000-08:002008-11-28T18:30:20.288-08:00Centers Last (old)(Disclaimer: I wrote this guide in June 2007, on twistypuzzles.com. Here it is, with just a few edits. I haven't changed up the method much if at all since then.)<br /><br />This is a method for solving arbitrarily large cubes. I decided to make a topic because I want to keep a permanent reference for this method, and because I just recently improved it a bit and I thought I'd like to share that with you.<br /><br />From personal experience, it becomes faster than the reduction method for cubes of a certain size or greater, but that size is determined by a lot of factors (such as how fast you can make slice moves and how good you are at finding pieces for reduction). Suffice it to say, though, that if I ever attempt a truly huge cube (say 40x40x40) I'm not going to use reduction. This method does use about 30-50% more moves than a reduction strategy for a given size of cube, though, and although you have better recognition you will need more physical endurance to be able to complete a cube with this strategy.<br /><br />The basic method is just this:<br />1) Solve all of the edges and corners. They must be solved relative to the centers on an odd cube.<br />2) Solve all of the centers without disturbing the edges and corners (much).<br /><br />There are a lot of subtleties, though. Each step has many different ways to complete it, and there are a lot of small (but very useful) shortcuts. Throughout this guide, I'm going to use the r* notation; what this means is, on an NxNxN cube, to do a clockwise turn on all R-side slices from the second slice to the floor(N/2)th slice. For example, on a 6x6x6 cube r*' would mean to turn the 2nd and 3rd R slices counterclockwise. This is useful for generalized algorithms.<br /><br />My current step 1 is done by matching up pieces of edge groups. I'll mostly be using 3x3x3-like moves for this part, so think of everything as corners, edge groups, and center groups. Using only the outer layers, move an edge group with a lot of similar pieces to DF (but if you can't find one in 4-5 seconds don't bother), and choose a color pair of edges that you will place in that spot. Make sure to remember which color is on top. Now, looking at the R and L layers, find edge pieces which match that color pair and move them to UF or UB using R/U/L moves, then shoot them down to DF with the appropriate slice move. Once DF is completely grouped up, you can do something like R' L F to move it into the R or L layer; keep doing this, while never disturbing the previously grouped edge groups. When you have 7 of them, you will not be able to do this any more; then, move the M layer to the E layer (just do z), and group another edge group together using horizontal slices. You can use moves like RUR' to get the fifth unfinished edge group into E. When you are done, move the finished group into U or D. At this point, U and D should each have 4 finished edge groups. You can actually finish one more edge group in E quickly (remember, you can do things like R2 because centers are irrelevant) before you have to do moves like (vertical slice) R F' U R' F (vertical slice)' to finish off the remaining edges. Note that you can only be sure of solving 11 of 12 edges like this. Don't try to use last-two-tredges or parity algorithms yet, though.<br /><br />When you have paired up at least 11 edges, it's time to solve the edges and corners into a cage. Do it like the 3x3x3 step. Make sure it's solved relative to the centers if you're doing an odd cube. Here is a trick: when you get to the orient step (OLL for Fridrich, step 3 for Petrus) it's very helpful if the last edge group, the one which will almost certainly have parities in it, is one of the edges you are going to orient. If you're not color neutral, you can force this easily: for example, I solve Fridrich with a yellow last layer, so I can simply pair up all of the edge groups which don't contain yellow first, which will ensure that my parity edge is in the last layer when I get to it. So fix all of the orientation parities at the same time, then do OLL, then fix the permutation parity if you need to, then do PLL. The permutation parity is very easy, by the way: just do r*2 U2 r*2 U2 r*2.<br /><br />So edges and corners are all solved; now onto step 2. In this step, we use commutators to solve all of the centers. The goal is to solve one center completely, then the opposite center, and then any other center, an adjacent center, and the last two. But again, there are a lot of subtleties and tricks here. First of all, a lot of people tend to use commutators like r U' r' d r U r' d'. There's nothing wrong with that, but I much prefer commutators in the form of r U l' U' r' U l and r U' l' U r' U' l. These commutators are very fast to execute because all of the interior slices are on the same axis, and you can see what is going on by only looking at two faces (where the centers are coming from and where they are going), leaving you free to look for the next center on an adjacent face. You'll have to experiment with this algorithm to get a feel for it, but you should know a couple things: (a) you can vary the affected centers in the 3-cycle by varying which slices you use (and the r- and l-like slices don't even have to be the same depth, i.e. r U 3r U' r' U 3r works fine); (b) x-centers can only be solved by one of these, but others can be solved by both; (c) the first move of the commutator should move a center piece that you want to move to the U layer into a space in U that does not already have a U-color center in it; (d) it's possible to bring entire blocks up to U by using more than one slice in place of one (or both) of the component slices of the commutator, but the first move has to bring the block to a place in U that is mostly or fully unoccupied, and some blocks are simply too large to do this with; and (e) it's best to only use setup moves involving U and only try to solve one piece or block at a time, so that the edges and corners don't get further than an AUF away from solved.<br /><br />Now that we've gone over the small-scale details, let's concentrate on the large-scale ones. When you're solving the first center, pieces can come from all five centers, but you will notice that bringing them from the opposite center is time-consuming for a couple of reasons, notably 180-degree turns and slow recognition. There is a solution, however: the move r* F2 r*' l*' F2 l* will swap the F and D centers, completely on an even cube and mostly on an odd cube, although it messes up some edges so you will have to undo it after you solve the U center. So one way you could do the first center group (say white) is to bring every white center cubie from one face into the white face, then swap that face's center group (which contains no white cubies now) with the center group opposite white, and finally just bring all of the white centers (from the center group that used to be on bottom) onto the white face before reswapping the two center groups. Personally I like to hold the target face on U, and then bring all the target center pieces on F and R into the target face before doing y2 and doing it again. Once you have the first center, the opposite center is the same, but this time every center piece you want to bring in will be in an adjacent face, which makes it a little easier.<br /><br />Now, you have four centers left to do. Pick one and hold it on U, then transfer the target pieces from F to U, swap F and D, transfer the target pieces from F to U, swap F and D, do y2, transfer the target pieces from F to U. If you want to be fast, this should be mostly mechanical, as it's a lot of the same type of move done from more or less the same angle. Now pick an adjacent face and hold it on U, with the just-finished face on B. Transfer the target pieces from F to U, swap F and D, transfer, swap. Remember that if you do one swap, you must do a second one, because otherwise the edges will get messed up. Now hold the last two centers on F and U (the order is irrelevant) and finish the cube. You're done!<br /><br />By the way, if you want to use this on cubes smaller than 6x6x6, there are a couple of key differences. Depending on how good you are at reduction, it may be faster for you to just hold the edges on the E layer and pair them up normally. If that works for you, do it. Also, on the smaller cubes, you shouldn't bother with swapping centers, because there are not enough centers to swap to make it worthwhile. If you want to reduce the number of centers you have to move from one face to an opposite one, you can try to find a face that has as many centers as possible on that face and the adjacent ones, and start solving there. But unless you are very slow at the centers part of reduction, this will probably be slower for you for small cubes, so I don't recommend it for the small ones.Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com1tag:blogger.com,1999:blog-3335271307001051587.post-50784693042078223012008-11-28T12:55:00.000-08:002008-11-28T12:57:58.655-08:00Solving 4x4x4 and 5x5x5 SupercubesAlthough the 4x4 and 5x5 supercubes look much more difficult and impressive than the standard ones (even experienced bigcubers will wonder how you can find pieces so quickly!), they aren't much harder than the normal cubes: all they take is a slightly different method. The following tips and tricks were all gotten through experience and practice. Note that I'm going to describe the reduction method below, for a couple reasons: I think it's the fastest known way to do supercubes (and normal cubes for that matter), I'm most familiar with it, and you're probably most familiar with it. So if you're not a reduction user, then sorry, the only advice I can give you is to go practice :)<br /><br /><span style="font-size:130%;">Centers</span><br /><br />Centers are the only part that's really different about supercubes, although they are not the only part that you have to solve differently. In a supercube, every center has to be solved into the correct spot - they're not interchangeable anymore. Thus, instead of just solving blocks of centers, you have to actually pay attention to what goes where.<br /><br />In a 4x4, you only have x-centers, which have three colors in the standard Pochmann sticker design (which I really recommend, because just drawing arrows or cutting corners off the stickers makes it very hard to see what goes where). I always put 4x4 centers together by making two 1x2 blocks. This probably doesn't require any explanation at all, although you might find it helpful to know that if you pair up the two yellow centers across a red-yellow 'edge', the other stickers will be paired across an orange-yellow 'edge' (since orange is the opposite of yellow). This can make recognition and setup easier.<br /><br />If you go to solve centers, you might notice that the last center doesn't automatically solve itself anymore. That should happen 5/6 of the time, but fortunately there are algorithms to fix this quickly. If you want to swap Ufr and Ufl, you can use this A-perm-like algorithm:<br />x r' U r' d2 r U' r' d2 r2<br />If you want to do a diagonal swap, on the other hand, the algorithm I currently use comes from the Square-1 and is as follows:<br />r2 U D r2 U r2 U D r2 U r2 U D r2.<br />So your centers should be solved and you can now start on the edges step.<br /><br />In a 5x5, on the other hand, you have many different types of centers, and now the fixed center, although it cannot move, must be oriented along with the rest of them. Again I pretty much solve the centers with my normal method (1x3 blocks, with the first block containing the fixed center) although it takes a bit longer to find the exact pieces I am looking for. If you see any blocks already formed for the first center or two, that will be very helpful, so try to preserve them.<br /><br />On 5x5 you still have last center problems, but since PLLs are hard to recognize and turning a fixed center (only) by 180 degrees is slow, I prefer the following method. First I solve a 2x3 block on the second to last center, and position it so I can do rUr'-type triggers without messing it up. Then I intuitively solve all of the T-centers on U and F. It's important to solve them relative to the fixed center (not just relative to each other). There is one other case here, that of parity, where you have two edges swapped. The quick solution is just this:<br />r U2 r U2 r U2 r U2 r.<br />That algorithm should do an opposite swap of edges on U (a Usl-Usr swap to be exact), so now you can be certain to be able to solve the edges only. Finally, I have six T-centers left; I do an F turn to set them up to a more convenient location, and then finish off the centers with U turns and the Niklas commutator (which has two forms):<br />r U l' U' r' U l<br />l' U' r U l U' r'.<br />Sometimes it is actually more convenient to just solve the F centers with this commutator, and then finish off U with an A perm (which does exactly what you think it does). If you get an E or an H, good luck, you might as well just do two As. The A perms (for 5x5) are:<br />x r' U r' d2 r U' r' d2 r2<br />x r2 d2 r U r' d2 r U' r.<br /><br /><span style="font-size:130%;">Edges</span><br /><br />If you got the Eastsheen supercubes, you might notice that your edges and corners have more colors than they need to. The extra colors don't affect anything (although sometimes they can make recognition for PLL a little easier, on 3x3), and after a couple of dozen solves you'll learn to ignore them.<br /><br />For 4x4, you can do edges exactly as you would on a standard cube. There is no change at all no matter what method you use, so go full speed on this step.<br /><br />For 5x5, if you use AvG or any other kind of 2-pair edges method, do whatever you normally do. You'll notice that you never get parity - this is because we fixed it during the centers step. Convenient, isn't it? If the T-centers are solved, you can never get parity on the wings. Anyway, if you use the freeslice/bigcubes method, the biggest difference is that during the first 8 edges you can't do half turns of the faces (F2/R2/B2/L2 if you solve in the E slice, for instance) unless the slices are all solved. Just making them parallel isn't enough, because that will actually slightly mess up two centers. You will also notice that, during the last 4 edges, you can't do the m U2 m' U2 algorithm, because that also messes up centers a bit. (If you really want to do it, you should do the inverse algorithm twice.) You should also be careful during the last 2 edges: if you solve this in one step, be very careful about whether your algorithm messes up centers or not, because some will and some won't. If your algorithm does mess up centers, you can always just fix the edges with the r [flip edge] r' algorithm, which is completely center-safe.<br /><br /><span style="font-size:130%;">3x3x3</span><br /><br />The final step on both 4x4 and 5x5 is to solve a supercube 3x3. You don't need to learn any new tricks for this, you just have to be more careful. When you're putting together the cross, make sure that the centers are all oriented the right way; since this takes so much more thought than a normal cross I do not at all expect you to do this in one look. Feel free to do it one or two edges at a time. Now, if the centers on the first two layers are correctly oriented, you are basically home free: almost all F2L algs will keep them solved (the only one I know of that messes them up is the 2-gen edge insert algorithm, but you can always use an alternate algorithm), and most OLL algs will keep them solved as well. Some PLL algs do, and for certain ones (U perm, for instance) it might be helpful to learn alternate algorithms that don't mess up the centers at all, since you have to spend extra time fixing it. Here's one possible alternate U perm:<br />F2 U (M' U2 M U2 M' U2 M) U F2.<br />If you do end up with a messed up center, you should twist both it and the U center, because the U center hasn't been fixed to a specific orientation yet anyway. The following (intuitive) algorithm will flip the front center clockwise and the left one counterclockwise (note that here M and E mean every slice together):<br />M' U M E M' U' M E'.<br /><br />There are three parity algorithms that you should know. The first two are the standard 4x4 parity algorithms (so of course they only show up on even supercubes). For the orientation parity, you can still use the normal one although that will twist the R or L center by a quarter turn, so I strongly suggest that you use some kind of slice-turn-only parity. My favorite is this one:<br />2R U2 x 2R U2 2R U2 2R' U2 2L U2 2L' x' U2 2R U2 2R' U2 2R'.<br />For PLL parity, the normal algorithm will mess up centers pretty badly, so here is one that won't:<br />x' 2U' R F' U R' F 2U 2D F' R U' F R' 2D' x<br />Finally, at the very end on both 4x4 and 5x5 supercubes, half of the time the U center will be twisted by 180 degrees. There are two algorithms I use for this, either of which can be faster depending on the cube and your style:<br />(R L U2 R' L' U)2<br />(R' U' R U')5.<br /><br />And with that, you know everything you need to solve the 4x4 and 5x5 supercubes every time. Good luck, and good times!Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com3tag:blogger.com,1999:blog-3335271307001051587.post-38872739128913262292008-11-17T14:01:00.000-08:002008-11-28T12:58:59.313-08:00Advanced Counting: Burnside's LemmaSometimes the theory we have discussed so far is not enough to calculate the number of positions of a puzzle. Fortunately there is a theorem from group theory that will help us with this. It is called Burnside's Lemma (although he was not the first to discover or prove it):<br /><br /><span style="font-weight: bold;">Theorem: </span>Suppose we have a finite set X, and a finite group G of permutations of X. For each element g in G write |Fix(g)| for the number of elements of X that are not changed by g, and write |G| for the number of elements of G. Then the number of orbits of X is equal to 1/|G| Î£<sub>gâˆˆG</sub> |Fix(g)|.<br /><br />What does this mean, though? Well, when we are counting the number of positions of a puzzle, we want the number of positions regardless of orientation - you shouldn't be able to get from one position to another just by rotating the puzzle, but we also don't want to miss any. The way we do this is by setting X to the set of every possible position in every possible orientation and G the possible ways to rotate the cube. The number of elements in X is relatively easy to calculate (it's like finding the number of positions of a puzzle, except that you don't fix anything; for example the size of X for the 2x2 cube would be 8! 3^7), and the number of elements in G can mostly be looked up from a small table, since it is 12 for a puzzle with tetrahedral symmetry, 24 for puzzle with the symmetry of a cube or octahedron, and 60 for one with the symmetry of a dodecahedron or icosahedron. Now, the number of orbits is simply the number of positions of a puzzle regardless of orientation, which is exactly what we want - so as long as we can compute the Fix(g) values, we know exactly how many positions a puzzle has, even if the puzzle has repeated pieces.<br /><br />One really important thing to notice is that, for any cycle in a permutation in G, every piece on the cycle has to look the same. This is pretty important because it tells us that if we have, say, three of each piece, it is impossible for g to fix any positions if g contains a 4-cycle (or bigger), because there are no sets of 4 of the same piece. Also, if g contains only 3-cycles, and the number of duplicate pieces in each set is not a multiple of 3, g cannot fix any positions.<br /><br />I want to do two examples to show how this works. First I'll do the very simple example of a cube with three red faces and three blue faces which can be switched any way we want. We already saw that our basic approximation gave something like .83 when the actual answer was 2. Let's use Burnside's Lemma. The size of X is 6 choose 3 or 20, and the size of G is 24 because the puzzle is cubical. Let's calculate the |Fix(g)|'s: if g is the permutation that does nothing, 20 positions remain the same; if g is a rotation around a corner (there are 8 of these), 2 positions are fixed; if g is a 90-degree rotation around a face (there are 6 of these), there are no fixed positions; if g is a 180-degree rotation around a face (there are 3 of these), there are 4 fixed positions; and if g is a flip around an edge (there are 6 of these), there are no fixed positions. So the total number of positions is 1/24 (20 + 8*2 + 0 + 3*4 + 0) = 1/24 (48) = 2. It's exact!<br /><br />Now let's do a very hard example: the (non-jumbled) Little Chop or 24-cube, which has 24 pieces, with 4 each of 6 colors and no orientation. The size of X is 24!/(4!^6) and the size of G is 24 again. Calculate the values of |Fix(g)|: if g does nothing, 24!/(4!^6) positions remain the same; if g rotates around a corner, we create 8 3-cycles so no positions can be fixed; if g is a 90-degree rotation around a face, there are 6 4-cycles so 6! positions are fixed; if g is a 180-degree rotation around a face, there are 12 2-cycles so 12!/(2!^6) positions are fixed; and if g is a flip around an edge, there are again 12 2-cycles so 12!/(2!^6) positions are fixed. The total number of positions is 1/24 (24!/(4!^6) + 6*6! + 3*12!/(2!^6) + 6*12!/(2!^6)), which works out to 135,277,941,853,080. We can't really verify this, but if we have done our calculations right it is the correct value. Note that with our basic computations, fixing one of the 4 pieces of a specific color, we would get an approximate value of 1/4 (23!/(3! * 4!^5)) = 135,277,939,046,250, which is pretty close (but not exact).Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-49071200395946934462008-07-08T10:50:00.000-07:002008-12-08T21:22:29.525-08:00TPCP: Type CvThe Cv puzzles were some of the first simple twisty puzzles to be developed. One in particular, the Skewb, is probably the puzzle with the most distinct shape modifications. These puzzles are related to the Tv and Of puzzles.<br /><br /><table><tbody><tr></tr><tr><td><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_FwZfrOWBA0g/SHOqCkup_aI/AAAAAAAAABc/3Kz1l8w2bgM/s1600-h/Cv_1.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://1.bp.blogspot.com/_FwZfrOWBA0g/SHOqCkup_aI/AAAAAAAAABc/3Kz1l8w2bgM/s320/Cv_1.PNG" alt="" id="BLOGGER_PHOTO_ID_5220703354248363426" border="0" /></a><br /><br /></td><td><br /><span style="font-size:130%;">Puzzl</span><span style="font-size:130%;">e </span><span style="font-size:130%;">Cv1</span>:<br />Gelatinbrain Number: n/a<br />Common Name: n/a<br />Types of Pieces: 6 fixed centers, 8 trivial tips.<br />Solution: Turn the trivial tips to solve them.<br /></td></tr><tr><td><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_FwZfrOWBA0g/SHOqC1_v_VI/AAAAAAAAABk/RYgDvFN3xL0/s1600-h/Cv_2_homemade.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://4.bp.blogspot.com/_FwZfrOWBA0g/SHOqC1_v_VI/AAAAAAAAABk/RYgDvFN3xL0/s320/Cv_2_homemade.PNG" alt="" id="BLOGGER_PHOTO_ID_5220703358883462482" border="0" /></a><br /><br /></td><td><span style="font-size:130%;">Puzzle Cv</span><span style="font-size:130%;">2:<br /></span>Gelatinbrain Number: n/a<br />Common Name: n/a<br />Types of Pieces: 6 fixed centers, 8 trivial corners, 12 edges.<br />Solution: First position all trivial corners. Then the edges can all be solved intuitively by 4-move 3-cycle commutators of adjacent corners.</td></tr><tr><td><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_FwZfrOWBA0g/SHOqC8lzeVI/AAAAAAAAABs/-rLekYSsp0Q/s1600-h/Cv_3.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://4.bp.blogspot.com/_FwZfrOWBA0g/SHOqC8lzeVI/AAAAAAAAABs/-rLekYSsp0Q/s320/Cv_3.PNG" alt="" id="BLOGGER_PHOTO_ID_5220703360653687122" border="0" /></a><br /><br /></td><td><br /><span style="font-size:130%;">Puzzle Cv</span><span style="font-size:130%;">3:<br /></span>Gelatinbrain Number: 3.2.4<br />Common Name: Dino Cube<br />Types of Pieces: 12 edges.<br />Solution: Use basic block building to solve all edges.</td></tr><tr><td><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_FwZfrOWBA0g/SHOqCxTJewI/AAAAAAAAAB0/wF1hKA3sSyU/s1600-h/Cv_4.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://2.bp.blogspot.com/_FwZfrOWBA0g/SHOqCxTJewI/AAAAAAAAAB0/wF1hKA3sSyU/s320/Cv_4.PNG" alt="" id="BLOGGER_PHOTO_ID_5220703357622647554" border="0" /></a><br /><br /></td><td><br /><span style="font-size:130%;">Puzzle Cv</span><span style="font-size:130%;">4:<br /></span>Gelatinbrain Number: 3.2.2<br />Common Name: Master Skewb<br />Types of Pieces: 6 inner centers, 24 outer centers in 2 orbits, 12 edges, 8 corners.<br />Solution: Solve inner centers and corners like Cv5. Then use commutators of the type (2DLF) (URF URB' URF') (2DLF)' (URF URB' URF')' to solve the edges. Finally use commutators of the type (2DLF) (ULF DRF' ULF' DRF) (2DLF)' (ULF DRF' ULF' DRF)' to solve the outer centers.<br /></td></tr><tr><td><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_FwZfrOWBA0g/SHOqDBtvyWI/AAAAAAAAAB8/yEGVo9W9YlQ/s1600-h/Cv_5.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://1.bp.blogspot.com/_FwZfrOWBA0g/SHOqDBtvyWI/AAAAAAAAAB8/yEGVo9W9YlQ/s320/Cv_5.PNG" alt="" id="BLOGGER_PHOTO_ID_5220703362029177186" border="0" /></a><br /><br /></td><td><br /><span style="font-size:130%;">Puzzle Cv5:<br /></span>Gelatinbrain Number: 3.2.1<br />Common Name: Skewb<br />Types of Pieces: 6 centers, 8 corners.<br />Solution: Pair up the top center with its 4 corners intuitively. Then solve all centers using DLF' DRF DLF DRF'. Finally use (DLF' DRF DLF DRF')2 to orient all remaining corners.<br /></td></tr></tbody></table>Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-81506743438165793682008-06-30T02:07:00.000-07:002008-12-08T21:22:29.791-08:00TPCP: Type CfThe Cf puzzles are very simple to describe: they are just the very well-known 2x2x2 and 3x3x3 cube. There are, of course, much more efficient methods than those described here.<br /><br /><table><tbody><tr><td><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_FwZfrOWBA0g/SGii0V7ptaI/AAAAAAAAABM/QxrjkonFz7U/s1600-h/Cf_1.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://1.bp.blogspot.com/_FwZfrOWBA0g/SGii0V7ptaI/AAAAAAAAABM/QxrjkonFz7U/s320/Cf_1.PNG" alt="" id="BLOGGER_PHOTO_ID_5217599188432958882" border="0" /></a><br /><br /></td><td><br /><span style="font-size:130%;">Puzzle Cf1</span>:<br />Gelatinbrain Number: 3.1.2<br />Common Name: 3x3x3 Rubik's Cube<br />Types of Pieces: 6 fixed centers, 12 edges, 8 corners.<br />Solution: Solve all edges using U-layer turns and the 3-cycle U2 R' L F2 R L'. Then solve the permutation of corners with commutators of the type (L) (U' R U) (L)' (U' R U)'. Finally orient all corners with commutators of the type (R U2 R' F' U2 F) (D) (R U2 R' F' U2 F)' (D)'.<br /></td></tr><tr><td><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_FwZfrOWBA0g/SGijCbQoO4I/AAAAAAAAABU/nVcBQgCWn7s/s1600-h/Cf_2.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://4.bp.blogspot.com/_FwZfrOWBA0g/SGijCbQoO4I/AAAAAAAAABU/nVcBQgCWn7s/s320/Cf_2.PNG" alt="" id="BLOGGER_PHOTO_ID_5217599430381288322" border="0" /></a><br /><br /></td><td><br /><span style="font-size:130%;">Puzzle Cf</span><span style="font-size:130%;">2:<br /></span>Gelatinbrain Number: 3.1.1<br />Common Name: 2x2x2 Rubik's Cube<br />Types of Pieces: 8 corners.<br />Solution: Solve the D face intuitively. Then solve the U face using R U R' U R U2 R' U2 to twist three U-layer corners clockwise. Finally solve the puzzle using R U2 R' U' R U2 R' F R' F' R, a corner transposition.<br /></td></tr></tbody></table>Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-65766894779127976232008-06-29T21:15:00.000-07:002008-12-08T21:22:30.177-08:00TPCP: Type TeThese puzzles are tetrahedral with edge turns. Puzzles Te3 and Te4 in this series have extra moves if you remove the restriction that they must be tetrahedral, since in fact they can be considered as shape modifications of the 3x3x3 and 2x2x2 Rubik's Cube.<br /><table><tbody><tr><td><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_FwZfrOWBA0g/SGhenWTKCbI/AAAAAAAAAAs/slf-UK8J0FM/s1600-h/Te_1.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://2.bp.blogspot.com/_FwZfrOWBA0g/SGhenWTKCbI/AAAAAAAAAAs/slf-UK8J0FM/s320/Te_1.PNG" alt="" id="BLOGGER_PHOTO_ID_5217524198402558386" border="0" /></a><br /></td><td><br /><span style="font-size:130%;">Puzzle Te1</span>:<br />Gelatinbrain Number: n/a<br />Common Name: n/a<br />Types of Pieces: 4 fixed centers, 6 trivial edges, 12 centers in 3 orbits, 4 corners.<br />Solution: Solve the edges by twisting them in place. The centers can then be solved with four-move commutators of adjacent edges. Finally the corners can be solved with commutators of the form (DL) (DU DR DU DR) (DL)' (DU DR DU DR)'.<br /></td></tr><tr><td><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_FwZfrOWBA0g/SGhfj55vLyI/AAAAAAAAAA0/cHanclaefVg/s1600-h/Te_2.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://1.bp.blogspot.com/_FwZfrOWBA0g/SGhfj55vLyI/AAAAAAAAAA0/cHanclaefVg/s320/Te_2.PNG" alt="" id="BLOGGER_PHOTO_ID_5217525238751768354" border="0" /></a><br /><br /></td><td><br /><span style="font-size:130%;">Puzzle Te</span><span style="font-size:130%;">2:<br /></span>Gelatinbrain Number: 5.2.1<br />Common Name: Senior Pyraminx (?)<br />Types of Pieces: 6 trivial edges, 12 centers in 3 orbits, 4 corners.<br />Solution: Solve the edges by twisting them in place. The centers can then be solved with four-move commutators of adjacent edges. Finally the corners can be solved with commutators of the form (DL) (DU DR DU DR) (DL)' (DU DR DU DR)'.</td></tr><tr><td><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_FwZfrOWBA0g/SGhg_vsxLKI/AAAAAAAAAA8/4BUnv7WiVTs/s1600-h/Te_3_homemade.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://2.bp.blogspot.com/_FwZfrOWBA0g/SGhg_vsxLKI/AAAAAAAAAA8/4BUnv7WiVTs/s320/Te_3_homemade.PNG" alt="" id="BLOGGER_PHOTO_ID_5217526816560983202" border="0" /></a><br /><br /></td><td><br /><span style="font-size:130%;">Puzzle Te3:<br /></span>Gelatinbrain Number: 5.2.3<br />Common Name: Mastermorphix (half turns only)<br />Types of Pieces: 6 trivial edges, 4 corners, 4 triangular centers, 12 trapezoidal centers in 3 orbits.<br />Solution: Solve the edges by turning them in place. Solve one triangular center and one adjacent corner using four-move commutators of adjacent edge turns, then solve the rest of the triangular centers and corners (and the edges) by alternating the two turns which do not disturb the two that were solved before. Finally the trapezoidal centers can all be solved with commutators of the form (DL) (DU DR DU) (DL)' (DU DR DU)'.</td></tr><tr><td><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_FwZfrOWBA0g/SGhhfL5XIuI/AAAAAAAAABE/ogQG4BL9z30/s1600-h/Te_4.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://2.bp.blogspot.com/_FwZfrOWBA0g/SGhhfL5XIuI/AAAAAAAAABE/ogQG4BL9z30/s320/Te_4.PNG" alt="" id="BLOGGER_PHOTO_ID_5217527356705940194" border="0" /></a><br /></td><td><br /><span style="font-size:130%;">Puzzle Te4:<br /></span>Gelatinbrain Number: n/a<br />Common Name: Pyramorphix (half turns only)<br />Types of Pieces: 4 corners, 4 centers.<br />Solution: First pair up one corner with one center. Then there are only two possible moves which do not break up this pair; alternate between them until the puzzle is solved.<br /></td></tr></tbody></table>Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-13857625702077595582008-06-29T20:50:00.000-07:002008-12-08T21:22:30.655-08:00TPCP: Type TvThese puzzles are tetrahedral with vertex turns. Because the tetrahedron is the unique regular polyhedron to have a face opposite a vertex, all face-turning tetrahedral puzzles can also be considered to be vertex-turning, and for simplicity's sake they will thus fall under the Tv classification.<br /><table><tbody><tr><td><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_FwZfrOWBA0g/SGhYpV1HxxI/AAAAAAAAAAM/5cLEpknpRz8/s1600-h/Tv_1.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://3.bp.blogspot.com/_FwZfrOWBA0g/SGhYpV1HxxI/AAAAAAAAAAM/5cLEpknpRz8/s320/Tv_1.PNG" alt="Tv1" id="BLOGGER_PHOTO_ID_5217517635566552850" border="0" /></a><br /></td><td><br /><span style="font-size:130%;">Puzzle Tv1</span>:<br />Gelatinbrain Number: n/a<br />Common Name: n/a<br />Types of Pieces: 1 fixed center, 4 trivial tips.<br />Solution: Turn each trivial tip, in any order, to match the fixed center.<br /></td></tr><tr><td><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_FwZfrOWBA0g/SGhZOXJn5iI/AAAAAAAAAAU/uYtJFX3j8zY/s1600-h/Tv_2_homemade.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://3.bp.blogspot.com/_FwZfrOWBA0g/SGhZOXJn5iI/AAAAAAAAAAU/uYtJFX3j8zY/s320/Tv_2_homemade.PNG" alt="Tv2" id="BLOGGER_PHOTO_ID_5217518271576139298" border="0" /></a><br /></td><td><br /><span style="font-size:130%;">Puzzle Tv2:<br /></span>Gelatinbrain Number: 5.1.1<br />Common Name: n/a<br />Types of Pieces: 1 fixed center, 4 corners, 6 edges.<br />Solution: Turn each corner, in any order, to match the fixed center. Then all edges can be solved with simple 4-move commutators of adjacent vertex turns.<br /></td></tr><tr><td><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_FwZfrOWBA0g/SGhZVincwdI/AAAAAAAAAAc/tf_YpZie7XM/s1600-h/Tv_3.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://1.bp.blogspot.com/_FwZfrOWBA0g/SGhZVincwdI/AAAAAAAAAAc/tf_YpZie7XM/s320/Tv_3.PNG" alt="Tv3" id="BLOGGER_PHOTO_ID_5217518394913112530" border="0" /></a><br /></td><td><br /><span style="font-size:130%;">Puzzle Tv3:<br /></span>Gelatinbrain Number: 5.1.2<br />Common Name: Pyraminx<br />Types of Pieces: 4 corners, 6 edges.<br />Solution: Turn each corner so that they are solved relative to each other. Then all edges can be solved with simple 4-move commutators of adjacent vertex turns.<br /></td></tr><tr><td><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_FwZfrOWBA0g/SGhZYJp4P3I/AAAAAAAAAAk/xSjPwqNgyvI/s1600-h/Tv_4.PNG"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://3.bp.blogspot.com/_FwZfrOWBA0g/SGhZYJp4P3I/AAAAAAAAAAk/xSjPwqNgyvI/s320/Tv_4.PNG" alt="Tv4" id="BLOGGER_PHOTO_ID_5217518439752023922" border="0" /></a><br /></td><td><br /><span style="font-size:130%;">Puzzle Tv4:<br /></span>Gelatinbrain Number: 5.1.3<br />Common Name: Halpern-Meier Tetrahedron<br />Types of Pieces: 4 corners, 6 edges, 4 centers.<br />Solution: Turn each corner so that they are solved relative to each other. Then all edges can be solved with simple 4-move commutators of adjacent vertex turns. If centers are unsolved they must be in a double transposition, which can be solved with (UDR UDL UDR' UDL')3.<br /></td></tr></tbody></table>Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-30257523572131690562008-06-29T20:32:00.001-07:002008-06-29T21:43:44.963-07:00Twisty Polyhedron Categorization ProjectThere are a lot of twisty puzzles out there, and they can be categorized in a multitude of ways. This project, the TPCP, is an attempt to categorize all of the most basic twisty puzzles into a list, ideally giving each one an identification code, an image, the gelatinbrain number or common name if applicable, a list of the types of pieces and their orbits, and a basic solution. I do not expect to complete all of it but I hope to get a good chunk of it done.<br /><br />The most basic twisty puzzles, as I define them, satisfy the following characteristics:<br />- They are based around a regular polyhedron.<br />- There is exactly one type of axis around which turns can be made.<br />- All possible turn axes are represented, and they all have the same depth. Thus the puzzle is symmetrical and remains mathematically unchanged under a rotation.<br />- Each face of the solved puzzle has one unique solid color and the goal is to bring the puzzle into a state where each face has one solid color.<br />- After any turn, the puzzle is still in a regular-polyhedral form.<br /><br />The foundation for the TPCP is very mathematical. Each puzzle belongs to one type, which is described by a capital letter denoting the polyhedron (T, C, O, D, and I for tetrahedron, cube, octahedron, dodecahedron, and icosahedron, respectively) and a lowercase letter denoting the type of axis (f, e, and v for face, edge, and vertex turns, respectively). Thus for example the 3x3x3 cube belongs to type Cf. Within the types, each puzzle gets a natural number, starting with puzzle 1 for the shallowest cut and increasing for deeper cuts. Puzzles with different-depth cuts which function in the same way are considered to be the same. Even trivial puzzles will be considered, except for puzzles with a cut depth of 0 where turns do not move any pieces and thus no scrambling at all is possible. In this case the 3x3x3 cube would be Cf1.<br /><br />For standardization of images I have chosen to use an image size given by the downloaded Gelatinbrain applet: 312x312 pixels. Of course I have used the lossless .png format so the images should not take too much time to load. Images for puzzles which are not represented in the downloadable applet have been created by me.<br /><br />Note that in this project if a set of pieces is described as "trivial" the pieces cannot be permuted relative to each other but do have an orientation.<br /><br />Although it is not addressed in this project, it is possible to use this categorization system to create types for higher-order puzzles (which I call compound types), and I may investigate some of these series in the future. The type of, for example, a dodecahedral puzzle with two types of edge turn and one type of vertex turns would be Deev, and the Super-X would belong to type Cfv. Although the numbers within the types cannot be as unambiguous as in the TPCP puzzles, we can still assign each puzzle in a compound type a number, with the expectation that someone wishing to do research in this topic would look up a puzzle's number in an online list.<br /><br />With that said, let's start to classify some of the types.Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-39765236318828702642008-05-25T01:29:00.000-07:002008-05-25T02:48:20.996-07:00Crosslinked Forum/UWR ListNo theory in this article, it's just a proposal I've been thinking of but don't have the skills to program. I've never seen this in practice, but it was an idea I had a while ago and which I'm sure a lot of other people have had independently. It ought to be written up, though...<br /><br />The thing that inspired this was the realization that people look to the unofficial world record (UWR) pages on speedcubing.com as a reference, but that many fast people don't post their times at all, so they are clearly not representative. I've often heard the excuse that the official times are the only times that matter, but the fact is that most events and puzzles do not have any official counterpart and the only way to compare times on them is through an unofficial list of this sort.<br /><br />It seems to me that this would solve this problem, plus many additional ones which I'll talk about later. The basic idea is to have a forum where users enter their unofficial personal best times. The times would then be organized in a list in their profile, and there would also be a UWR section which would automatically sort and display the times that users have submitted. Ideally it would be very easy to submit times, so that people would grow used to submitting new PBs as often as they post in Accomplishment threads. A timer could also be added, and then users' single and rolling average-of-12 times could even be automatically updated when they finish a session. From a database point of view it would be a relatively simple addition to the forum system. The way I would do it would be to take a working forum and then add the databases of records and categories; then we would just have to make a way to submit records, display them in people's profiles, and display them in the UWR format. Other additions such as timers or a compare-users function would just involve different ways of interacting with this system.<br /><br />This doesn't sound like that much of an improvement over a forum like speedsolving.com, but it would solve a lot of problems. There wouldn't be many problems with fake times (like there apparently are on the UWR list) because the mods could simply delete or ban the offending account. Signatures wouldn't need to contain people's best times because they would be easily found in the user's profile. Nothing would have to be manually updated on the unofficial record list, so whenever someone updated their time or e-mail the list would update, and you'd never have duplicate entries. Similarly it would be easy to add new events: you could allow people to create their own events, with certain restrictions, and perhaps show all events with at least two people in them. Since the UWR list and forums would be on the same website, it would be a lot easier to compare people, and you could even have a function which takes two users and compares all of their personal records automatically. You'd also end up with an account for everyone who wants to be on the unofficial list, which means that the community could be more organized than ever; it wouldn't be much of a problem to make different forum sections for people from different countries or people who speak different languages, and perhaps eventually everyone could be using the same website as their speedcubing community (rather than having speedsolving.com for most English-speaking people and then a bunch of other separate websites for various other languages or countries). This would also be a very convenient centralized place for people to talk about their records, since it would all be on one site.<br /><br /><br />The question is, though, should everyone move from speedsolving or wherever to use this? I kind of think that we should agree on some kind of website that would satisfy all of our needs as a community for the next few years. Cubing is definitely not dying down; in fact it is picking up faster and faster, and I think it's about time we become very organized with the community. Speedsolving is a pretty good forum but it has a big problem with posting times: people can't really display much of their accomplishments because of small signatures, so you have to already know who people are in order to figure out who's amazingly fast. In fact some of the faster people I know don't even know what their best times are, which seems kind of odd given that so many other people want to know that. This tells me that more organization is needed. There's also a problem of lots of smaller, regional cubing forums; this means that the community isn't all in one place, so it can be tricky to find someone who isn't on speedsolving, and if something is posted in one place it won't necessarily reach everyone. So I figure we could use this kind of certainty.<br /><br />The last thing I think I should say is that, once we are sure we have a forum that will satisfy everything we want, moving to it won't be that hard. Initially you will have people using the old and new system concurrently, but eventually the old system will just become an archive of past cubing discussions. This seems to be the eventual fate of the speedcubing Yahoo group, which was replaced by speedsolving because the Yahoo group message system is much less convenient and organized than the forum system. I am pretty sure that, if a community website can be made which is clearly superior to the existing community forums, we will eventually have everyone in the new website, and it will be much easier to share records and ideas. It will definitely take a while to program and perfect this, but I think that if we use lookahead we can see that it will be well worth it in the long run.Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-80014732810682279112008-05-11T00:55:00.000-07:002008-05-11T01:03:35.540-07:00Number of Positions of Generalized Twisty PolyhedraI wrote this on 3/30/08.<br /><br />The goal of this article is to determine the number of positions of various twisty polyhedra. I will first show an algorithmic theory where we can find the number of positions in well-behaved twisty polyhedra (that is, puzzles which are symmetrical and have no move constraints based on the position) without difficulty, and then use it to find the number of positions in a few infinite families of theoretical twisty polyhedra, and a few examples of how to modify this theory for trickier puzzles.<br /><br /><span style="font-size:130%;">Theory of Positions</span><br /><br />Before we start determining the number of positions of a puzzle we must fix the orientation somehow. This is necessary since simply turning a puzzle around without doing moves on it will not change its position. The normal way to do this is to fix one specific piece, or a set of pieces which are fixed relative to one another. If this is impossible we can divide the number of positions we calculate later by the number of possible positions of a solved puzzle (most usually 12 for a puzzle with tetrahedral symmetry, 24 for a puzzle with cubical or octahedral symmetry, or 60 for a puzzle with icosahedral or dodecahedral symmetry) but it is not certain that this will give us the correct result. Note, however, that if we have a few choices for what piece to fix, you will get the same answer no matter which piece you choose.<br /><br />After this the first step to finding the number of positions of a twisty puzzle is to determine the orbits. An orbit is the set of positions that a given non-fixed piece can be in, and orbits containing the same positions are considered the same. In a 3x3x3 cube we will typically fix the orientation by forcing the centers to be in specific places; then there are two orbits, one for the 12 edges and one for the 8 centers. Note that there can be more than one orbit containing similar-looking pieces; for example on the 3x3x3 square group (where only half turns can be used) there are in fact three edge orbits and two corner orbits.<br /><br />In general we will proceed by finding the number of positions in each orbit and multiplying them together. Because of parity concerns, the best way to do this is to order the orbits in some way and consider them one by one, assuming that while considering one orbit all pieces in the previously considered orbits are solved. This way, when two orbits have the same parity, we see that one does not obey a parity constraint and the other does, so we correctly divide by 2 only once.<br /><br />So how many positions are in a single orbit? This is generally an easy problem, as it can be directly calculated with a little casework. First we consider the permutations. If there are n pieces, we start with n!; if there are n pieces each of k types, as in the centers of a higher-order Rubik's Cube, we start with (nk)!/n!^k. If the pieces of the orbit have to obey a (permutation) parity constraint then we divide by 2; note that this is only possible in the first type because if we have indistinguishable pieces we can always make a 2-cycle of same-colored pieces to cancel out a wrong parity. Next we consider the orientation of the pieces. If the pieces cannot be oriented differently when they are in a given position we do not have to consider orientation and we can just multiply by 1. Otherwise, if we let k be the number of possible orientations one piece can have if it is a given position and n be the number of pieces then we should multiply by k^n. If the pieces of the orbit obey an orientation parity, we divide by its order, which is generally 2 or 3 but can be higher.<br /><br />Let's work through an example to illustrate the technique: the 3x3x3 cube. As I already mentioned, if we fix the orientation by fixing the centers' positions, we have two orbits: edges and corners. First we'll consider edges: there are 12 pieces (12!) with no parity (since corners are unknown), and each has an orientation of order 2 (2^12) with an orientation parity (/2), so the total number of possibilities here is 12! * 2^11. Then we consider corners: there are 8 pieces (8!) with parity (/2, since edges are fixed), and each has an orientation of order 3 (3^8) with an orientation parity (/3), so the total number of possibilties is 8!/2 * 3^7. Thus the total number of positions is (12! * 2^11)(8!/2 * 3^7) = 43252003274489856000, as I'm sure you already know.<br /><br /><span style="font-size:130%;">Number of Positions of Infinite Families of Twisty Polyhedra</span><br /><br />The most important concept here is that of layers. It is difficult to put into words, but the number of layers of a puzzle is basically the number of parallel, twistable slices (not counting slices on the opposite side of the puzzle's center). Puzzles with one layer include the 2x2x2, 3x3x3, Megaminx, and Dino Cube; puzzles with two layers include the 4x4x4, 5x5x5, Gigaminx, and Lattice Cube. A family of puzzles is basically a set of puzzles that function similarly (have the same axes, the same fixed pieces, the same shape, and so on), but which differ in the number of layers, a constant which I will call n throughout the rest of the article. I don't count 'trivial tips' (twistable elements of a puzzle which do not affect any other elements, are not affected by any other elements, and can always be solved in one move) as layers, because they do not share any pieces with the rest of the puzzle; thus, although most families only have twistable puzzles for n >= 1, there are some families where the case for n = 0 has more than 1 possibility.<br /><br />There are many families of puzzles; however, most families have only had one puzzle constructed or even considered, and there are other families (such as stacked pucks and 2x2x(2n+1) cuboids) where adding one layer simply multiplies the number of combinations by a fixed amount. In this article I will only consider the number of combinations for families of which at least two puzzles have been constructed or designed. Of course, these are the most popular families to consider. All numbers I give have been calculated using the formula given here.<br /><br /><span style="font-weight: bold;">Odd NxNxN Cubes: </span>Just like the 3x3x3, we will hold the centers constant. Then our orbits are: 12 edges with 2 orientations each (12! * 2^11), 8 corners with 3 orientations each (8!/2 * 3^7), n-1 orbits of 24 edge wings ((24!)^(n-1)), and n^2-n different orbits of 24 centers, all of which are made up of 4 pieces each of 6 types ((24!/4!^6)^(n^2-n)). Thus the total number of positions is 12! * 8! * 24!^(n^2-1) * 4!^(-6n^2+6n) * 2^10 * 3^7. For this family the existing puzzles are n=1 (3x3x3, with 4.325 * 10^19 positions), n=2 (5x5x5, with 2.829 * 10^74 positions), and n=3 (7x7x7, with 1.950 * 10^160 positions).<br /><br /><span style="font-weight: bold;">Odd NxNxN Supercubes:</span> This is like the odd NxNxN cubes, except that there are much more parities since the center pieces have no duplicates, and also the centers have orientation. Again keep the centers fixed, then the orbits are 6 centers with 4 orientations each (1 * 4^6), 12 edges with 2 orientations each (12!/2 * 2^11), 8 corners with 3 orientations each (8!/2 * 3^7), n-1 orbits of 24 edge wings ((24!)^(n-1)), and n^2-n orbits of 24 centers ((24!/2)^(n^2-n)). So the total number of positions is 12! * 8! * 2^(-n^2+n+21) * 3^7 * 24!^(n^2-1). The puzzles that have been made so far are n=1 (3x3x3 Supercube, with 8.858 * 10^22 positions) and n=2 (5x5x5 Supercube, with 5.289 * 10^93 positions).<br /><br /><span style="font-weight: bold;">Even NxNxN Cubes:</span> We can't hold centers constant, because there are no centers, but if we fix one specific corner we will always be guaranteed a fixed orientation. So our orbits are: 7 corners with 3 orientations each (7! * 3^6), n-1 orbits of 24 edge wings ((24!)^(n-1)), and n^2-2n+1 orbits of 24 centers, each of 4 pieces in 6 colors ((24!/4!^6)^(n^2-2n+1)). The total number of positions here is 7! * 24!^(n^2-n) * 4!^(-6n^2+12n-6) * 3^6. The puzzles in this family that have been constructed are n=1 (2x2x2, with 3.674 * 10^6 positions), n=2 (4x4x4, with 7.401 * 10^45 positions), and n=3 (6x6x6, with 1.572 * 10^116 positions).<br /><br /><span style="font-weight: bold;">Even NxNxN Supercubes:</span><span style="font-weight: bold;"></span> This is very much like the odd NxNxN supercubes. Keep a corner fixed, and our orbits become 7 corners with 3 orientations each (7! * 3^6), n-1 orbits of 24 edge wings ((24!)^(n-1)), and n^2-2n+1 orbits of 24 centers, all with parity ((24!/2)^(n^2-2n+1)). So the total number of positions is 7! * 24!^(n^2-n) * 3^6 * 2^(-n^2+2n-1). The existing puzzles are n=1 (2x2x2, with 3.674 * 10^6 positions) and n=2 (4x4x4 Supercube, with 7.072 * 10^53 positions).<br /><br /><span style="font-weight: bold;">Megaminx Family:</span> These puzzles work just like the odd NxNxN cubes, except with larger numbers of pieces in each orbit. However, in these puzzles all turns create 5-cycles, so every orbit has a permutation parity. There are 30 edges with 2 orientations each (30!/2 * 2^29), 20 corners with 3 orientations each (20!/2 * 3^19), n-1 orbits of 60 edge wings ((60!/2)^(n-1)), and n^2-n orbits of 60 centers, in 12 colors of 5 pieces each ((60!/5!^12)^(n^2-n)). So the total number of positions is 30! * 20! * 60!^(n^2-1) * 5!^(-12n^2+12n) * 2^(28-n) * 3^19. The existing puzzles are n=1 (Megaminx, with 1.007 * 10^68 positions) and n=2 (Gigaminx, with 3.648 * 10^263 positions).<br /><br /><span style="font-weight: bold;">Pyraminx Family:</span> These puzzles are slightly more tricky to calculate since some of them have middle centers (where there are only 4 pieces instead of 12) and some do not. I am not including the trivial tips in the calculation because they are not an inherent part of the puzzle, but of course to put them back in we can just multiply by 3^4. In this puzzle we are keeping the corners in the same position, although their orientation can change. Note that, just as in the Megaminx family, all turns create even permutations, so every orbit has permutation parity. In this family our orbits are 4 corners with 3 orientations each (1 * 3^4), 6 middle edges with 2 orientations each (6!/2 * 2^5), n-1 orbits of 12 edge wings ((12!/2)^(n-1)), a total of floor((n-1)^2/3) orbits of 12 centers, in 3 centers for each of 4 colors ((12!/3!^4)^(floor((n-1)^2/3))), and ((n-1)^2 mod 3) orbits of 4 centers ((4!/2)^((n-1)^2 mod 3)). Thus the number of positions is 6! * 2^(5-n-((n-1)^2 mod 3)) * 3^4 * 12!^(n-1+floor((n-1)^2/3)) * 3!^(-4floor((n-1)^2/3)) * 4!^((n-1)^2 mod 3). The existing puzzles are n=1 (Tetraminx, with 9.331 * 10^5 positions), n=1 with tips (Pyraminx, with 7.558 * 10^7 positions), and n=2 with tips (Master Pyraminx, with 2.172 * 10^17 positions).<br /><br /><span style="font-weight: bold;">Magic Octahedron Family:</span>This family works just like the Pyraminx family, but now there are parities involved with some of the edges since a turn creates edge 4-cycles. Again, we want to fix the corners, and we're not going to fix the tips. The orbits are 6 corners with 4 orientations each (1 * 4^6), n-1 sets of 24 edge wings ((24!)^(n-1)), one orbit of middle edges (12!/2 * 2^11), a total of floor((n-1)^2/3) orbits of 24 centers, 3 each in 8 colors ((24!/3!^8)^(floor((n-1)^2/3)), and ((n-1)^2 mod 3) orbits of 8 centers ((8!/2)^((n-1)^2 mod 3)). So the number of positions is 2^(22-((n-1)^2 mod 3)) * 24!^(n-1+floor((n-1)^2/3)) * 12! * 3!^(-8floor((n-1)^2/3)) * 8!^((n-1)^2 mod 3). The existing puzzles are n=1 (Magic Jewel, with 2.009 * 10^15 positions), n=1 with tips (Magic Octahedron, with 8.229 * 10^18 positions), and n=2 with tips (Master Octahedron, with 1.029 * 10^47 positions).<br /><br /><span style="font-weight: bold;">Trajber's Octahedron Family:</span> Although the Trajber's Octahedron functions in an analogous way to an odd NxNxN cube, the pieces are colored differently, which means that a different method is required when solving it, and it has a different number of positions. Thus it makes sense to consider it a separate puzzle. Note that octahedra made from even NxNxN cubes are entirely different, in that they have many solutions and no piece has multiple stickers. We will hold the corners constant, although allowing them to rotate. Thus our orbits are 6 corners with 4 orientations each (1 * 4^6), 12 middle edges with two orientations each (12!/2 * 2^11), n-1 orbits of 24 edge wings ((24!)^(n-1)), n^2-n outer center orbits, with 3 pieces each of 8 colors ((24!/3!^8)^(n^2-n)), and 8 inner centers (8!/2). Thus the total number of positions is 12! * 8! * 2^21 * 24!^(n^2-1) * 3!^(-8n^2+8n). The existing puzzles are n=1 (Trajber's Octahedron, with 4.050 * 10^19 positions) and n=2 (5x5x5 Trajber's Octahedron, with 3.429 * 10^78 positions).<br /><br /><span style="font-weight: bold;">Dino Cube Family:</span> This is an interesting set of puzzles to think about, although it is easy to solve, since you basically just have a large number of orbits that are all solved like a dino cube. Note that a true n-layer Dino Cube will have 2n(n+1) stickers on each face, and also has two solutions - if you solve it to a mirrored color scheme you will not encounter any problems. This doesn't change the number of combinations, but it does make the puzzle slightly 'easier'. There are n(n+1)/2 orbits on this puzzle, each of which must have an even permutation because all turns create 3-cycles, and no pieces have orientation since they are in a given orientation whenever they are in a given position. We will fix one piece in the centermost orbit (11!/2) and then keep the other ones free ((12!/2)^((n^2+n)/2-1)), so the total number of positions is 11! * 12!^((n^2+n-2)/2) * 2^((-n^2-n)/2). The existing puzzles are n=1 (Dino Cube, with 1.996 * 10^7 positions) and n=2 (Lattice Cube, with 1.145 * 10^24 positions).<br /><br /><span style="font-size:130%;">Number of Positions of Some Other Interesting Twisty Polyhedra</span><br /><br />We can use these techniques for basically any twisty polyhedron, but in some specific cases we need to use a trick to find the correct answer, because if we are not careful we will miss an important detail about the puzzle. I'll illustrate each one with a puzzle. I will make a large table with the number of positions for each of a number of existing puzzles in another article, so if you are interested in the number of positions of a specific puzzle that isn't mentioned here it should be on that list.<br /><br /><span style="font-weight: bold;">Siamese Cube:</span> The Siamese Cube is composed of two separate puzzles which do not share pieces. Since they are the same, you can find the number of positions by finding the number of positions for one of those puzzles and squaring it. Fix the block and the number of positions is ((24)(120 * 3^5)(11!/2 * 2^10))^2 or 2.046 * 10^32.<br /><br /><span style="font-weight: bold;">Helicopter Cube:</span> The tricky thing about this cube is that (in its non-jumbleable form) it actually has four orbits of centers, which each contains one center of each color, and at least one orbit can have an odd permutation as long as the permutation of all of the orbits combined is even (since each turn flips parity on two orbits but doesn't change the overall parity). So if we fix a corner the number of positions ends up as (7! * 3^6)(6!)(6!)(6!)(6!/2) or 4.937 * 10^17.<br /><br /><span style="font-weight: bold;">Alexander's Star:</span> In the standard coloring scheme each piece has a duplicate. So we want to choose one pair of pieces and fix one. There are two identical pieces, though, so we can fix each position exactly two ways, so we have to divide the result by 2. But note that there still may be positions which look the same in more than one of these states (which we assume to be different), which would mean that the result we calculate is actually too small. Generally there are a very small number of these, not enough to influence the calculated result if we write it scientific notation, but it is important to realize that if we wrote out the number in full it would almost certainly be incorrect. The number of positions is approximately (29*(28!/2!^14) * 2^28)/2 or 7.243 * 10^34.<br /><br />Here is another way to look at this phenomenon: consider a "puzzle" with six tiles arranged like the faces of a cube, where you can switch any two tiles at any time. Suppose we color these tiles so there are three red ones and three blue ones. Now intuition tells us there are exactly 2 positions; our calculation, on the other hand, says that we can fix one of them in a specific orientation and get (5!/(2!*3!))/3*4 = 120/144 or 0.833... positions. Of course this is a contrived result, with many more identical positions than a real puzzle would have, but it does show how the calculated number is too small.Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com2tag:blogger.com,1999:blog-3335271307001051587.post-14127199207489405072008-05-10T01:15:00.000-07:002008-05-10T01:13:04.069-07:00Group TheoryGroup theory is a branch of theoretical mathematics that can be used to describe the way the functioning of the Rubik's Cube and various related puzzles. Although pure group theory can get extremely complicated (there is one proof, a classification of what are known as simple groups, which took an estimated tens of thousands of pages in its first incarnation) there are several simple applications of group theory to twisty puzzles.<br /><br />First, we should explain what a group is. A group is nothing more than a set of elements and some binary operation on those elements; so we could make a group out of the set of integers and the operation +. Generally people describe a group just as the set of elements, since the operation is usually obvious from context. A group also has to satisfy several properties:<br />- Closure: Whenever we apply the operation to two elements in our group, the result is an element in the group.<br />- Associativity: If a, b, and c are in the group, and * is our operation, then (a * b) * c and a * (b * c) are always the same thing. This means that we can write things like a * b * c * a * c without needing large numbers of parentheses, which is very convenient.<br />- Identity: There is some identity element, so that if we apply the operation to the identity element and any element a the result will just be a again.<br />- Inverse: Every element has an inverse in the group, so that if you apply the operation to the element and its inverse it becomes the identity.<br /><br />This is a pretty abstract concept, but fortunately it isn't too difficult to use. Basically, each puzzle has one group, with the elements being the puzzle's possible positions. Our operation is composition. Basically, suppose we have two elements in the group, A and B. Then A * B is what you get when you take the moves you would use to reach B from a solved cube, and apply them to A instead of a solved cube. This is a very simple and useful operation for several reasons. First, the identity is the solved state, which is easy to work with. Second, if we define a set of basic moves (for example F, R, U, B, L, and D, and their inverses, for a 3x3x3) we can get to every position from the identity using some sequence of these moves, which means that every element in the group can be written as some sequence like F * B * L inverse * D * D, and we can even describe positions by the way that they can be written like this. Third, the inverse of a position is also easy to find: group theory tells us that the inverse of A * B * ... * Z is Z inverse * .... * B inverse * A inverse. Since the inverse of basic moves are basic moves, we can very easily find the inverse of a certain position by just switching the order of the basic moves that bring you to that position and then inverting each one of them. Note that the inverse of the inverse of any position P is just P again.<br /><br />By the way, since the operation * is always implied, we often write a series of moves in what is called multiplicative notation, where we write the separate moves next to each other without a *, and use exponents (like in algebra) to indicate doing something multiple times or inverting it. This means that we would write something like F B L^-1 D^2 instead of F * B * L inverse * D * D. This is the origin of our modern notational system, which has been simplified so that the inverse sign (which normally looks like a superscript "-1") is changed to a ' and the square sign (a superscript "2") is changed to a normal 2. I believe it was David Singmaster who created the modern Rubik's Cube notation this way. As for my notation system, it works just like this, but includes many more basic moves so that sequences of moves can be written more succinctly.<br /><br />Along with the concept of a group comes the idea of a subgroup. A subgroup is what you get when you find a normal group and create another group with the same operation but only some of the elements. Because puzzle groups are only finitely large, it turns out that all you need to check to make sure a subgroup is valid is closure, so it's easy to find subgroups. Note that a subgroup always has to contain the identity (solved state). Since we always use the same operation, you can think of a subgroup as basically describing a sub-puzzle of a normal puzzle. A few interesting subgroups of the 3x3x3 are:<br />- The "two-generator" subgroup, all positions which can be solved using only R and U turns;<br />- The orientation-only subgroup, positions where all pieces are in place, but can have any orientation;<br />- The half turn subgroup or square group, positions that can be solved using only 180-degree turns.<br /><br />The concept of a subgroup is closely tied in with the idea of a generator. A generator for a group is a set of moves or algorithms in that group, so that you can reach any position in the group only using moves or algorithms from that set (or their inverses). * For a given set of moves, there is always some group that is generated by them, that is, some group that contains exactly the positions that you can reach using those moves (in some order and doing them as many times as you want). For any set of moves, say the set R, U, and F2, we write the group generated by them as <R,U,F2>. One particularly famous generated group is a subgroup which we have already discussed: the subgroup of 3x3x3 positions which can be solved using only R and U turns. This is known as the 2-generator group, since it is <R,U>, the group generated by moves of two adjacent faces. Playing with generators can create a lot of new and interesting challenges out of a normally non-challenging puzzle.<br /><br />Another useful result of group theory is the use of so-called "cycle notation". Imagine a group where, instead of a normal puzzle, you have a bunch of pieces labeled 1 through n, and you can put them in any order (permutation) you want. We can still use the composition operation for this group, and it works a lot like the puzzle groups we have looked at. The question is: how do we represent objects in this group? What we will do is represent a permutation by following where pieces go. Give each piece a spot, labeled 1 through n, which represents its original location; so we can write down a permutation by writing the list of numbers we see, following the locations from 1 to n. So if we have five pieces and the permutation leaves them looking like "3 4 2 1 5" we can use this to represent the permutation. There is a better way, though: instead of just looking at each piece, follow where pieces are actually sent. In this case, the piece that was at position 1 is now at position 4; the piece at position 4 is now at position 2; the piece at position 2 is now at position 3; and the piece at position 3 is now at position 1. We can encode all of this information by writing (1 4 2 3). This is what we call a cycle, and we can imagine making this permutation by simultaneously moving all of the pieces in it to the location where it will go. More specifically this is a 4-cycle, because it has 4 elements. Now what about the piece at position 5? It's still there, and although we could write this as a 1-cycle we usually do not, because that piece doesn't have to move at all. This is really useful, because (1 4 2 3) is a valid permutation for ANY number of pieces, as long as you have at least 4 of them! If we have more than one cycle at the same time, we can write them by writing them next to each other. So if 1 goes to 2 and 2 goes to 1, and 3 goes to 4 and 4 goes to 3, we will write this as (1 2)(3 4).<br /><br />We call a cycle with n elements an n-cycle. In puzzles we usually consider permutations on the pieces of the puzzle. Because of some more complicated math that will be discussed later, the basic unit that you can use to solve puzzles is the 3-cycle, which moves three pieces around. You may already know a few 3-cycles of pieces on the 3x3x3. Permutations of the pieces, however, is very important in the study of how to solve puzzles, as well as integral to blindfold solving.<br /><br />There are two last useful mathematical terms which are very useful to understand cubing. They are conjugation and commutation. In conjugation you take two move sequences, which we will call X and Y, and then you make the sequence X Y X^-1. The interesting thing about this is that we end up with the same cycles as in Y, just with different pieces being moved around. You can use this to make other algorithms out of a given one, because the pieces that are moved in the conjugated algorithm are actually the pieces that X puts in the spots Y moves around. Commutation is similar: for two move sequences X and Y the commutator is X Y X^-1 Y^-1. In theory this can make many of the cube's possible patterns, but in practice we mostly use commutators to create nice 3-cycle maneuvers, or to flip/twist two pieces on a puzzle. The way it works is this: have X be a single turn, and then choose Y so that it modifies just one piece on the layer. If Y flips or twists the piece, you will end up with one piece in X's layer flipped/twisted and another flipped/twisted the same amount in the opposite direction. If Y replaces the piece with some piece from another layer, you will end up with a 3-cycle. The really useful thing about commutators is that if you understand them well enough you can literally make up commutators quickly while you are solving a puzzle, and do any 3-cycle you want in just a few seconds.<br /><br />Well, those are most of the group theory topics you are likely to see. I've tried to explain them simply, but if you don't understand you can look them up on Wikipedia or look for a topic on the speedsolving.com forum. Again, good luck!Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-39326249282445316532008-04-20T01:56:00.000-07:002008-04-28T09:49:04.779-07:00Some Basic VocabularyThere are a lot of terms in the theory of cubing that won't be at all understandable to outsiders, or even that are used in a different way in cubing than in the outside world. I'm going to define some of them; since I have an imperfect memory this list is not complete, but just a list of some things which I think are commonly or easily misunderstood and which I'd like to explain.<br /><br /><span style="font-style: italic;">Algorithm</span>: This just means a sequence of moves, whether it is made up on the spot or memorized beforehand. The common abbreviation is "alg". Typically it is a sequence that accomplishes something specific (such as solving part of a cube), but people sometimes talk about scrambling algs. A way to solve a specific puzzle is not an algorithm, but a method (see below). Some methods have many algorithms to memorize, and some require less or no memorization.<br /><br /><span style="font-style: italic;">Bandaged</span>: A puzzle is bandaged if some of its pieces are connected to what would normally be adjacent pieces in a way that restricts the movement of the puzzle depending on where those pieces are, which often increases the puzzle's difficulty. A bandaging is a way of connecting pieces together to make a bandaged puzzle.<br /><br /><span style="font-style: italic;">Blindfold Solving</span>: This is a way of solving a puzzle by memorizing the positions of all of its pieces, then putting on a blindfold and solving the whole puzzle without looking at it. The typical abbreviation for blindfold or a blindfold solve is BLD. There are really two types of BLD: solving all of the pieces like a speedsolve, and tracking where they go while you memorize, then memorizing what moves to do; or solving a few pieces at a time while not affecting the others, so you can memorize where each piece is supposed to go, which is often easier.<br /><br /><span style="font-style: italic;">Block</span>: This can either mean a group of adjacent layers on one axis, or a cuboidal set of pieces. A block turn is a turn of some group of adjacent layers. Block building is a type of intuitive step of a method where you assemble a block of pieces.<br /><br /><span style="font-style: italic;">Center</span>: Any piece which belongs to only one face. Some puzzles have a lot of these on each face, in many different types. Since they have only one color, though, if a puzzle is not a supercube (see below) they can often be interchanged with each other without making the puzzle unsolved. Centers typically do not have orientation, except sometimes on a supercube.<br /><br /><span style="font-style: italic;">Corner</span>: Any piece which belongs to more than two faces. Corners usually have from two to five possible orientations. Note that trivial tips (see below) are generally not considered corners, though, since they don't affect other pieces and are thus not really part of the puzzle.<br /><br /><span style="font-style: italic;">Cube</span>: Although formally this means a cubical puzzle, the Rubik's Cube is so popular that people will often casually refer to any twisty puzzle as a cube.<br /><br /><span style="font-style: italic;">Cuber</span>: Anyone who is interested in solving the Rubik's Cube or other twisty puzzles. If they are also interested in solving as fast as they can, they are a speedcuber.<br /><br /><span style="font-style: italic;">Edge</span>: Any piece which belongs to two faces. Edges usually have two possible orientations.<br /><br /><span style="font-style: italic;">Face</span>: One face of the puzzle. A face is solved when all of its stickers match (on a normally-colored cube this would mean they were all the same color). However, solving a face on many puzzles (such as 3x3x3) is useless, since it might still be necessary to permute the pieces on that face in order to solve the puzzle; it is much better to solve in terms of layers or blocks.<br /><br /><span style="font-style: italic;">God's Algorithm</span>: In general this term means a method to always solve a puzzle in the fewest possible number of moves. However, it can be used in a few different ways: sometimes it means a way to find this solution for any given scramble (or a list of solutions for every scramble), sometimes it means a solution for a given position, and sometimes it is used to mean the maximum number of moves that an optimal solution for a scramble can take.<br /><br /><span style="font-style: italic;">Intuitive</span>: A step is intuitive if you solve it by figuring out what moves to do during the solve, rather than by using algorithms you know (which is called an 'algorithmic' step). Typically people think of commutators as intuitive, since you don't memorize the moves that are in them, but instead learn the concept. Intuitive steps in a method are good because they require less memorization, take fewer moves, and are generally more fun to execute, but it does take more practice to be fast at them.<br /><br /><span style="font-style: italic;">Lucky</span>: A solve is lucky if one or more of the steps in the method is skipped; solves can be very easy without being lucky. Note that if a solver uses a different algorithm than normal to deliberately skip a step in their method, most people would say that does not make the solve lucky.<br /><br /><span style="font-style: italic;">Method</span>: A method is a predetermined way of solving a puzzle in a series of steps. Typically methods are named after their creator(s) or the steps themselves. A method created by Stefan Pochmann might be referred to as the Pochmann method, Pochmann, or even poch; abbreviations are common if the name is long. For example, there is a family of methods called Corners First (CF) because the first step or first few steps solve the corners of the puzzle without regard to other pieces.<br /><br /><span style="font-style: italic;">Metric</span>: A specific and well-defined way to count the length of an algorithm or a puzzle solution. The most common metrics are htm or ftm (half turn or face turn metric, where every move of one face, or block containing a 1st slice, is counted as one turn), stm or btm (slice turn or block turn metric, where every move of one face, block, or slice is counted as one turn; some would say stm does not count blocks as one move but as separate slices), and qtm (like htm, except that only turns of the smallest valid angle increment (90 degrees on 3x3x3, for example) are counted as one move, and larger turns may be more than one move).<br /><br /><span style="font-style: italic;">Orbit</span>: This is the set of all pieces that can be moved to a specific location on the puzzle without rotating the puzzle. Sometimes there is more than one orbit of pieces that look the same.<br /><br /><span style="font-style: italic;">Orientation</span>: The way that a piece is positioned, independent of where it is on the puzzle; you can define a 'correct' orientation for each piece and location. On a Rubik's Cube, for instance, corners have three possible orientations. When you do moves to give a set of pieces the correct orientation, you are orienting them (not orientating). Pieces with the wrong orientation are typically said to be flipped (in the case of edges) or twisted (in the case of corners and centers) and algorithms to solve them are thus called flippers or twisters.<br /><br /><span style="font-style: italic;">Parity</span>: As I'll discuss later, pieces can normally be solved in cycles of three pieces, but if pieces cannot be solved like this then there is a (permutation) parity. There can also be orientation parities, if only one piece in an orbit is wrongly oriented, but these do not occur on all puzzles. The algorithm to fix any parity without disturbing other pieces is typically very long.<br /><br /><span style="font-style: italic;">Permutation</span>: Where a piece is on the puzzle, independent of the way it is positioned; if a piece is in the correct permutation it might still be misoriented (twisted or flipped). When you do moves to give a set of pieces the correct permutation, you are permuting them (not permutating).<span style="font-style: italic;"></span><br /><br /><span style="font-style: italic;">Slice</span>: One of the layers on an axis, not counting the outermost layer. A slice turn (or sometimes just slice) is a turn of one of these layers only.<br /><br /><span style="font-style: italic;">Supercube</span>: A puzzle where some or all pieces are marked so that the position and orientation of every piece must be correct for the puzzle to be solved.<br /><br /><span style="font-style: italic;">Trivial tip</span>: A piece on the puzzle which can always be brought to its solved position in one move without disturbing any other pieces; they are called trivial because they are so easy to solve that they are often not even mentioned in methods. They are also known as tips. A pyraminx has four of these.<br /><br /><span style="font-style: italic;">Twisty puzzle</span>: Basically a sequential move puzzle where the goal is to turn various parts of it until it reaches one of its solved states. A pretty good list can be found at the twistypuzzles.com website. These are often simply referred to as puzzles, or even cubes.Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-46643155552818967052008-04-20T00:07:00.000-07:002008-04-28T09:49:21.608-07:00NotationIn this post I'll describe my generalized notation for twisty puzzles. I might stray from this notation at times but I think this does a pretty decent job of describing moves for puzzles. There are a few levels of notation built on each other, and although it might seem overcomplicated most applications of this (such as writing down specific algorithms) only use simple features of it: for example, the well-known Sune algorithm for the 3x3x3 is still written R U R' U R U2 R'.<br /><br />Not all of this will necessary make sense or seem useful to you now. However, if I use some notation that you are not familiar with, it would be helpful to come back to this post as a reference.<br /><br />First you give each face of your puzzle a letter (or more than one if you are careful about it). Fortunately, for algorithms on dodecahedral and icosahedral puzzles, you usually don't have to label all of the faces. The standard naming for cubical puzzles, which you have probably seen before if you are into speedsolving, is to call the faces U (for the upper face), D (for the down face), F (for the front face), B (for the back face), R (for the right face), and L (for the left face). Typically if you see three faces in a shot of a cube they are assumed to be U, F, and R. I'll assume that you know what faces I am using when I talk about cubical puzzles, but if I am discussing a different puzzle I'll probably describe how I am naming faces or at least make it so that you can determine the meanings from context.<br /><br />Now you can label the axes that you turn pieces around. There are three types of axes: faces, edges, and corners. Check out the puzzles at gelatinbrain if you aren't entirely sure what this means; the most common puzzles with various axes are the Rubik's Cube (face), Bevel/Helicopter Cube (edge), and Pyraminx (corner). Face axes are labeled by their face, edge axes are labeled by the two faces they touch, and corner axes are labeled by the three faces they touch; the face's names can be in any order, but they should not have spaces between them. This means that if you see an axis label (say UF on a 3x3x3) you should be able to quickly determine what type of axis it belongs to, so this notation can still handle puzzles with more than one type of notation.<br /><br />Finally, we get to the turns themselves. Turns are all around an axis, so you should modify the following basic types of turns based on the axis. For the examples I'll use the axis UR. We're going to suppose there is more than one slice around the axis (like in big cubes) for the purpose of the notation. Label the slices with integers starting at 1 for the outermost slice and ascending as you move deeper into the puzzle. There are five types of moves; all moves are clockwise turns as seen from a vantage point outside the puzzle.<br />- 3UR is a clockwise turn around the UR axis, turning the 3rd slice only (a slice turn). If the number is 1 (1UR) then it is allowed to just omit the 1, since this is a very common type of move.<br />- 3ur is a clockwise turn around the UR axis, turning all slices up to the 3rd slice (a block turn). If the number is 2 (2ur) then it is allowed to omit the 2, since this is the smallest block turn that can't be just written as a slice, and is also very common.<br />- 2-4ur is a clockwise turn around the UR axis, turning slices from the 2nd to the 4th (a block turn). This isn't a common type of move, but it's useful for writing patterns in block turn metric.<br />- *ur is a clockwise turn around the UR axis, turning basically half of the slices on the UR axis (and its opposite axis) rounded down. On a 4x4x4 or 5x5x5 cube this would be 2 slices; on a 6x6x6 or 7x7x7 it would be 3. Note that since it is lowercase it should be a block turn. This is useful for writing algorithms in a more general form.<br />- cUR is a clockwise cube rotation around the UR axis. Note that it's unambiguous because having lowercase and uppercase letters together is unique. Observe that sometimes we might need cube rotations around axes that you can't turn; for example on a 3x3x3 cube sometimes a rotation like cURF is useful.<br /><br />Separate moves should always be written with spaces separating them. A group of moves can be enclosed with parentheses (i.e. ( and )). You can modify a move or a group of moves by putting a number after it (to indicate it is to be done that many times) or a ' (to indicate that its inverse is to be performed instead). Thus combinations such as 3UR' and r2 are possible. If we want to use both of these, either R2' or R'2 is acceptable, but personally I will tend to use R2' because I like it more.<br /><br />Rarely, in generalized patterns or commutators, variables are needed to show that any slice or any amount of turn can be used. Typically any letters that are not in use to describe axes are acceptable. As in the notation we have already seen, if a variable is placed before the axis name (such as nr) it describes a slice number, and if a variable is placed after (such as Rx) it describes a number of times a turn should be done.<br /><br /><br />Finally, I'd like to say that it's true that this notation differs a bit from the conventional notation, especially when it comes to large NxNxN puzzles (bigcubes, that is), but I think that it is better generalizable to other puzzles and more consistent than the standard notation. Moves such as (rm') or Rw, as written in standard notation, are difficult to expand to different puzzles and often unwieldy. I hope my notation is reasonable enough to be understood by most visitors here.Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0tag:blogger.com,1999:blog-3335271307001051587.post-10384200270961621952008-04-16T09:49:00.000-07:002008-04-28T09:49:28.320-07:00IntroductionThis is a blog about twisty puzzles themselves. I aim to explain the way they work and the way they are solved so that the reader should eventually gain a deep understanding of the way these puzzles work, a thing which is sadly lacking in many of today's speedcubers. I also hope to put some of the methods I use or have discovered on here. Some of the material here might be original research and some may not; it's safe to assume that I've created anything that I explicitly credit to me, but I don't necessarily claim authorship of all that I write here.<br /><br />The first couple posts will probably concern themselves with the mathematics of the cube, to get a feel for why things work as they do. Although I'm considering majoring in Math I'm not going to be too formal or theoretical, so anyone with an early high school understanding of mathematics (or better) should be fine. Then after that I'll start to go into the more interesting stuff, or perhaps some more complicated mathematics. Don't think of this as a kind of textbook where every section builds on the one before it: the material in these opening sections should be enough to understand all or most of the rest. If you are confused about something you can always ask me, or the friendly folks on speedsolving.com.<br /><br />So what's with this name "Notes on Twisty Puzzles"? It is based off of the title of an early Rubik's Cube treatise by David Singmaster which he called "Notes on Rubik's Magic Cube" and which was a great leap ahead in understanding how the standard 3x3x3 cube works. Despite not having read his books about the Cube I do recommend that you read them if you have a chance, as I've heard they are excellent.<br /><br />Finally, all opinions expressed herein, and all writing, are mine unless stated otherwise. Opinions can and do change, so if you have a problem with anything here feel free to tell me about it.<br /><br />Have fun, and good luck!Michael Gottliebhttp://www.blogger.com/profile/06863626487601055059noreply@blogger.com0