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You are not logged in. #1 20120529 23:13:27
Rubik's cube algorithm with period 5?Can you design a Rubik's Cube algorithm that gives you back the cube you started from when repeated 5 times,if it exists? If it doesn't existcan you prove it? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #2 20120530 04:09:15
Re: Rubik's cube algorithm with period 5?How about? Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #3 20120530 05:04:04
Re: Rubik's cube algorithm with period 5?Nice one,Bob! How did you find it? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #4 20120530 06:12:57
Re: Rubik's cube algorithm with period 5?I have a book I would strongly recommend called 'Handbook of Cubik Math' by Frey and Singmaster. You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #5 20120530 06:21:41
Re: Rubik's cube algorithm with period 5?I think I get why it can produce those periods. I think it can produce any period that is a divisor of 30 i.e. 1,2,3,5,6,10,15 and 30. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #6 20120530 06:27:05
Re: Rubik's cube algorithm with period 5?Btw,I have found that (R' U)**9 has period 7. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #7 20120530 06:36:16
Re: Rubik's cube algorithm with period 5?hi Stefy, So this is the number of distinct 'moves' that can be found. These form a 'group' with order N. The order of the only possible sub groups must divide N so, for example, no cycle of order 41 can be found. But 7 divides N so I suspect that means a 7 cycle can be found. unfortunately, knowing that does not help you to find one. I'll have a look through the book and see if any are given. (I seem to remember I did have one years ago, because I tried to make a cube with no colours, just days, dates, and months so you could use it as a calendar. Each day you would advance to the next date by doing the right cycles on the cublets. The tricky bit was to find a way to hide the unneeded dates around the back and sides so the front showed just the right things. I'll post again when / if I find a 7 cycle. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #8 20120530 06:47:22
Re: Rubik's cube algorithm with period 5?Arha. You got one before I had typed my long post. Well done! You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #9 20120530 07:12:54
Re: Rubik's cube algorithm with period 5?Change the 9 to 7 and you get a 9period. But which question did you predict? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #10 20120530 16:52:20
Re: Rubik's cube algorithm with period 5?The move R'U has a 9 cycle and a 7 cycle. So the number of repeats can be used to 'eliminate' one cycle. Look at the cublets and work out where each one moves. You should be able to find some 3cycles and 2cycles. By repeating the move three times you are left with a simple pair of 2cycles. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #11 20120530 21:18:59
Re: Rubik's cube algorithm with period 5?I just saw the long post. I missed it when I posted my reply. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #12 20120530 22:06:22
Re: Rubik's cube algorithm with period 5?The date cube was going to have a 12cycle of months; a 7cycle of days of the week; a 10cycle of unit digits; and a 4cycle of blank, 1, 2, 3. These last two would come together to make days in the month such as 28 or 7 or 31. We've got one of those mantlepiece ones where you have to advance the cards manually. I wasn't planning this would cycle through the whole year automatically; as you observe that would not be possible. But my version would take more skill because you'd have to know all the cycles and when to apply them. You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #13 20120530 22:08:52
Re: Rubik's cube algorithm with period 5?What about leap years? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #14 20120530 22:11:19
Re: Rubik's cube algorithm with period 5?You just rotate to make the 8 into a 9 as any other month. Then you have to go twice for 9 into 0 onto 1 and move the 2 onto 3 onto blank. You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #15 20120530 22:18:31
Re: Rubik's cube algorithm with period 5?So you have center pieces 1,2,3 and 3 blanks,what do you have on the edges and corners? Last edited by anonimnystefy (20120530 22:24:49) The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #16 20120531 04:36:43
Re: Rubik's cube algorithm with period 5?Hi Bob The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #17 20120531 06:57:51
Re: Rubik's cube algorithm with period 5?hi Stefy
You must be joking!! You'll get repetitive strain injury.
I'll check it out and post again later. You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #18 20120531 07:01:07
Re: Rubik's cube algorithm with period 5?I hope you didn't think I would actually do the algorithm. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #19 20120531 07:59:18
Re: Rubik's cube algorithm with period 5?RL'U You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #20 20120531 08:06:42
Re: Rubik's cube algorithm with period 5?Now that is some nice analysing! I must have done a move wrong because I repeated the algorithm 200 times and the cubelets didn't go back to their place. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #21 20120616 03:07:18
Re: Rubik's cube algorithm with period 5?Hi everyone, this is my first post on this forum! I tend to spend time on writing up my posts, so they are often times lengthy if I have enough to say on the subject!
Actually, there is a simple way, but it can be tedious. Before I explain this, I need to provide a little background information. 1) Apply a 3cycle commutator to a solved cube which affects slots A, B, and C. 2) Apply the same commutator to, for example, slots C, D, and E, so that they overlap. , we generate a 5cycle of either edges or corners (which ever version of the commutator we chose). You probably will need to apply setup moves (conjugates) before the 3cycle commutator to affect different pieces than what the commutator affects by itself. You may also do a cyclic shift of the 3cycle commutator you have to affect different positions. A cyclic shift of the corner commutator I gave, for example, is: R U' L' U R' U' L U (original) => U' L' U R' U' L U R, where you conjugate with the inverse of the first move (or the inverse of the last move) to not add more moves to the original commutator, but you affect different pieces (yay!) A more useful commutator to generate more corner positions might be: R' F' R U R2 U' R' F R U R2 U' = [R' F' R U, R2]. Also, you can overlap two pieces from the previous 3cycle commutator as well. (You are not just limited to overlapping one piece of two 3cycle commutators). Just make sure that the very last 3cycle you do also involves the first piece you affected (which is you have left alone after the first application of the 3cycle to the solved cube) to "close" the cycle. So once you have a commutator which swaps only three corners or only three edges, you can overlap each 3cycle to obtain a higher order cycle. If you have an even permutation, this is all you will need to do. If you have an odd permutation (remember that both the corners and the edges cycle types will be odd permutations if one is), you need to do a 2cycle PLL first, and then overlap that 2cycle with a 3cycle commutator (and overlap that 4cycle "marriage" with more 3cycle commutators). And actually, I have modeled all of the 2cycle PLL algorithms in the speedsolving wiki as commutators and conjugates. A commutator, if you haven't caught on, is A B A' B' = [A,B] and a conjugate C of some algorithm X is C X C' = [C:X]. So you can see the decompositions of the 2cycle PLL algorithms as commutators and conjugates (just so that you won't feel that these PLL algorithms are gibberish) if you look up the following: Link>>Look up "Entire Set of Speedsolving Wiki 2Cycle PLL Algorithms Decomposed" in Google. It's the first result. So that's all you really need to make any cycle class type with edges and with corners. Now, you can actually shorten (optimize) the length of your products of commutators by inputting the entire sequence into an optimal 3x3x3 solver. I recommend the following: >>Link: Look up "cube explorer 5.0" in Google. It should be the first result. **A way that might be easier is to make a high order cycle by just repeating the 2cycle PLL over and over, sharing just one piece from the previous application of the algorithm at a time. Of course, set up moves (and/or you can do a cyclic shift of the algorithm) in between each application. This of course will scramble the edges if you do the corners first, or scramble the corners if you do the edgesunless you do what you ought to do if you choose to just use a 2cycle PLL: make sure you have the 2cycle PLL affect the same two corners (or same two edges if you are permuting the corners first) every application by proper application of setup moves. Just keep in mind that whether you use the 2cycle PLLs once for an odd permutation purpose only OR if you use it as your only base algorithm, you take note of which two edges it affects (if you choose to do corners first) or viceversa.
This is incorrect. We must represent the sequence R L' U as: (urf)(ufl,rub,rbd,rdf)(dbl,fdl,ulb)(ur,br,dr,fr,uf,ul,bl,dl,fl,ub) Now, in order to compute the order of the Rubik's Cube Group, which is 43 252 003 274 489 856 000, the number of permutations which corners contribute is "only" . From the formula at a link I previously mentioned for calculating the number of different cycle types (first search result in Google for "Cycle classes of permutations"), we see that the number of possible 12cycles with 24 objects (stickers) is Note that the sum of all of the number of possible cycle classes must add up to the total number of permutations which that object contributes to the diameter of its group, and clearly only the total of one cycle class is more than , which is a contradiction to the order of the Rubik's Cube Group. Q.E.D. I could also argue (without mathematics) that your notion is false by the following example Consider the 3cycle corner commutator R U' L' U R' U' L U = [R, U' L' U]. [R, U' L' U]3 = the identity. Also consider the 3cycle corner commutator L' D R D' L D R' D' = [L', D R D']. [L', D R D']3 = the identity. [R, U' L' U] [L', D R D'] = [L', D R D'] [R, U' L' U], and therefore, by the results above, ([R, U' L' U] [L', D R D'])3 = ([L', D R D'] [R, U' L' U])3 = the identity. Now what if we were to consider your argument. Then we would assume that [R, U' L' U] is a 3cycle because it takes 3 times of repeating it to achieve the identity. If we twist two of the three corners it affects (with two more commutators following it): [R, U' L' U] (F' U' B' U F U' B U) (U R' D' R U' R' D R) //2corner twist But when we repeat this 3 times, it shouldn't be solved because it's a 9cycle now. But ([R, U' L' U] (F' U' B' U F U' B U) (U R' D' R U' R' D R) )3 = the identity? (we can do the same for the commutator [L', D R D'] and achieve the same result). From this, we at least have to conclude that a disjoint 3cycle which doesn't share an orientation with any piece outside of that cycle will stay a 3cycle. To verify this, let's let the two original 3cycle corner commutators share an orientation (let's twist one corner which is in [R, U' L' U] and one that is in [L', D R D']. It shouldn't matter which corner from each we choose. [R, U' L' U] [L', D R D'] (D R U R' D' R U' R') (R' F L F' R F L' F') //2corner twist of the two front right corners. ([R, U' L' U] [L', D R D'] (D R U R' D' R U' R') (R' F L F' R F L' F') )3 the identity, but ([R, U' L' U] [L', D R D'] (D R U R' D' R U' R') (R' F L F' R F L' F') )9 is. But by my "proof", that algorithm, even though it takes 9 times to achieve the identity, it is two disjoint 3cycles which share orientation with each other, not a 9cycle.
Actually there is. Bruce Norskog found what is commonly called "The Devil's Algorithm", which is what you are asking about, earlier this year for the entire 2x2x2 cube group and the entire 3x3x3 cube group. Last edited by cmowla (20120616 06:10:50) #22 20120616 04:32:58
Re: Rubik's cube algorithm with period 5?hi cmowla You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #23 20120616 06:07:38
Re: Rubik's cube algorithm with period 5?Thanks for the warm welcome! So yes, we only differ a little. I can see why you thought what you did, by that definition. In fact, the movement of the "wing edges" (as well as the movable inner center pieces) of the 4x4x4 cube and larger cubes are strictly permutations and so they match that definition perfectly with no exception (too bad the 4x4x4 cube and larger cubes as a whole do not resemble groups though): they are much easier to track than the corners and (central) edges of the 3x3x3. Last edited by cmowla (20120616 06:14:32) 