3x3x3 Rubik's cube
The 3x3x3 Rubik's cube is probably the most famous and best-selling combinatorial puzzle of all time. Its solving principles can be even used elsewhere - they apply to the Pyraminx or Megaminx, for example. Six independent faces, each of which has 3x3 = 9 squares, can be rotated. The aim of the puzzle is to obtain exactly one color on each face.
On this page you will find:
- notation
- intuitive puzzle solving method
- a bit of theory - commutators
- world record videos
- and much more
Notation
Enabled JavaScript in your web browser is necessary to display the content on this site correctly. Consequently, animated Rubik's cube simulators, that serve as an interactive text addition, are possible to see. In case you don't see other puzzle simulators on other pages for some reason, try to read brief instructions on how to run java applets. Java is free (and so is JavaScript).
The 3x3x3 Rubik's cube consists of three little cubie sets. We distinguish:
Corners | Centers | Edges | ||
It is obvious that the centers (center pieces) are one-colored, the edges (edge pieces) are two-colored and the corners (corner pieces) are three-colored. Two terms are closely related to the colors of individual pieces: orientation and permutation. Significance of these terms is shown on next simulators.
orientation of two edges | permutation of three edges |
On the left-hand side cube, both white-green and white-blue edges are initially oriented incorrectly. In other words, these two edges are at the right places, but they are both flipped. On the right-hand side cube, both white-orange, white-red and white-green edges are initially permuted wrongly (click and hold the mouse button in a close cube surrounding and see for yourself by dragging, or tapping in case of touch device). Buttons below the simulators control a cube solving process. By left mouse button (a finger in case of touch devices) you can usually rotate with selected layer (face) of a cube.
Out of curiosity, let's now calculate the number of different combinations for a 3x3x3 Rubik's cube (you may calmly skip this part up to the pair of cube images with letters, because it has no effect on the solve itself). We'll start with centers. They are, believe it or not, mutually fixed - they form an axial cross of a cube itself. That means we can not arbitrarily permute them. Since they are one-colored, they have only one orientation. It implies that there is only one way how centers can be mutually combined. That's simply because they are fixed (which is not always true - for instance, when calculating the number of combinations for the NxNxN Rubik's cube, it is considered that centers can be combined by 24 ways in a so-called fixed corner model. However, in such a case a number for corner combinations will be 24x smaller than it is stated in the calculation on this page, since we work with a so-called fixed centers model here).
Regarding edges, the situation is somewhat different. There are 12 of them on a cube in total. Thus we can theoretically place the first edge at any of 12 locations. Then, the second edge can be placed at any of 11 remaining locations (twelfth location is already occupied by the first edge). The third edge can be placed at any of 10 remaining locations (eleventh location is occupied
by the second edge and twelfth location is occupied by the first edge). If we continued in this way further up to the twelfth edge, we would come to a number 12·11·10·...·2·1 = 12! (twelve factorial). And that is a number of permutations for edges.
Since each edge is two-colored, it has two orientations. Hence we can place the first edge at a cube by 2 different ways. The second edge can be placed at a cube by two possible ways again. The same applies to the other edges as well. A number of orientations for edges is then 2·2·2·...·2·2 = 2^{12}
A number of possible combinations for edges is then 12!·2^{12}, theoretically.
Corners are left. To calculate their number of combinations, we will proceed analogically with a number of combinations for edges. There are 8 corners on a cube in total, thus the first corner can be placed at any of 8 locations. Then, the second corner can be placed
at any of 7 remaining locations (eighth location is occupied by the first corner). If we continued further up to the eighth corner, we would come to a number 8·7·6·...·2·1 = 8! (eight factorial). By that we obtained a number of permutations for corners.
Since each corner is three-colored, it has three orientations. Therefore we can place the first corner at a cube by 3 different ways. The second corner can be placed at a cube by three possible ways again. The same applies to the other corners as well. A number of orientations for corners is then 3·3·3·...·3·3 = 3^{8}
A number of possible combinations for corners is then 8!·3^{8}, theoretically.
Mathematical field called a group theory, however, tells us two interesting things. It says something like this: "if you want to permute (swap) only two corners on a 3x3x3 Rubik's cube, you won't succeed. Two edges will be always permuted along with two corners, too. The same applies to edges as well" (if you want to swap only two edges, two corners will be permuted too - for an explanation see the article about a parity problem). And it is adding: "if you want to orient all 8 corners independently on each other, you won't succeed. Only 7 of them will be independent, the eight one will be dependent on one of these 7 corners. The same applies to edges" (11 out of 12 may be independent). What does that mean?
On a 3x3x3 Rubik's cube we can theoretically calculate a number of possible combinations: 12!·2^{12} for edges and 8!·3^{8} for corners. Nevertheless, as a result of a group theory, it comes down to a number 8!·3^{8}·1/3 for corners (7 corners can be oriented by all three their ways, but the eight one must have only one out of its three possible orientations) and 12!/2 (because it is not possible to swap only two edges - the details are mentioned in a parity problem article) ·2^{12}·1/2 for edges (11 edges can be oriented by all two their ways, but the twelfth one must have only one out of its two possible orientations).
A little more practical view on the matter can be considered if we imagine solved 3x3x3 cube. If we orient only one corner in its location by 120° clockwise manually, the puzzle will be unsolvable by legal moves. If we, from a solved cube state, orient only one corner by 240° clockwise (i.e. by 120° counter-clockwise), we will again reach a configuration which can not be solved by legal moves. Hence only 1 out of 3 these states/configurations is solvable (that one, when the cube is solved) and the same applies to the other corner permutations (8! in total) as well. Therefore, it has been multiplied by a factor of 1/3 (i.e. it has been divided by 3) in a derivation above.
Analogically, if we orient only one edge by 180°, the puzzle will be unsolvable by legal moves. The same applies to all edge permutations (let's say we don't know the number of those yet). Hence only 1 out of 2 these states/configurations is solvable (that one, when the cube is solved) and as a result, it has been multiplied by a factor of 1/2 (i.e. it has been divided by 2) in a derivation above.
A swap of two corners and simultaneously two edges can be achieved using a T-algorithm, which is used in a blindfolded solving, for example. From above it comes that it is always possible to swap two corners (see 8! of corner permutations). If we did such a swap by hand after a T-algorithm execution, it would end up in a configuration which requires to swap only two edges. That case cannot be solved by legal moves for all 12! edge permutations (only 1 out of 2 described situations is solvable - that one, in which two corners are swapped along with two edges), thus it has been multiplied by a factor of 1/2 (i.e. it has been divided by 2) in a derivation above - if interested, have a look at already linked article about a parity problem to see more.
Finally we get the real number of combinations for a 3x3x3 Rubik's cube, which is equal to [1] · [(12!/2)·(2^{11})] · [(8!·3^{7})] = 43 252 003 274 489 856 000, or roughly 4.3·10^{19}. Our task is to find only one state in this enormously big number - that one in which the cube is solved. I'd like to compare it with something, but finding a needle in a haystack seems not adequate to me in this case. Maybe hundreds, thousands or millions of haystacks.
P.S. a comparison exists after all - see an article about why the Rubik's cube is so hard.
Prior to solving the cube, it is useful to agree on a notation first. Six layers (faces, if you like) can be rotated and following two pictures show us how.
A few words regarding the pictures: what you see is conventional English notation. Letters are derived from locations: Up, Down, Right, Left, Front and Back. By an apostrophe (comma above the letter), for instance in mathematics in case of functions, an inverse is denoted. In case of a Rubik's cube we will understand an inverse (opposite) move by that. If a letter is followed by a digit, it specifies how many times we have to make a move. As an example, U2 means we have to execute a U move twice in a row. R2' move = R2 = 2·R' = 2·R.
Each corner can be described using the letters R, L, U, D, F and B. The same applies to centers and edges. By URF, for instance, an upper-right-front corner is meant. Similarly, by DB a lower-back edge is meant. Sometimes the same edge can be viewed in two ways: say RF edge on the picture above, which is a blue-yellow one. If we describe the same edge as FR, then it becomes yellow-blue. Therefore, an order of letters is associated with colored stickers on edges. We treat corners analogically.
Before we get into the actual puzzle solving, a Rubik's cube simulator solving system will be outlined. First we will start by white face. It can be recognized as the one which contains a white center piece. As we already know, centers are mutually fixed, hence a white one is unambiguous identifier of a white face. The same applies to other centers. You may notice gray cubies on the simulators. Those ones are not important for a solving, they appear on a cube because of better clarity (so you don't have to pay attention to them at all).
Additional notation |
x: | y: | z: | ||
S: | E: | M: | ||
r: | u: | f: | ||
Intuitive puzzle solving method
Intuitive Rubik's cube method described below uses four solving steps, nevertheless, they can be reduced to only two of them. If we first solve the edges (by the same manner as described in steps no. 1-3), then it remains to solve all the corners (by the same manner as described in step 4).
This method is not suitable for speedsolving. Still, I am personally able to achieve sub 50 seconds on average with it. On the contrary, in case of speedsolving I achieved 18.94 s (average of 12 solves), but I had to memorize 15 algorithms - see more on a page about the corners-first method.
Intuitive puzzle solving method - step 1
The aim is to solve a layer without one corner. Something like this (see a simulator below):
How do we achieve that? One way is to solve edges first, followed by corners (once again I remind you gray cubies are not important) - as it is shown on a subsequent simulator.
Solving order is following: white-green, white-red, white-blue and white-orange edge.
That was easy. Solving of three out of four corners will be simple, too. See a next simulator.
Solving order: white-green-red, white-green-orange and white-blue-orange corner.
Let's denote a location to which a white-blue-red corner belongs as the working corner. You may be asking yourself: "why not to solve a whole, complete layer first"? The answer you will find in the next step of a method.
Intuitive puzzle solving method - step 2
Now we will solve, using the working corner, any three edges which belong to the second layer. At the end you should come to something like what can be seen on a simulator below.
If we solved a whole layer in the previous step, we couldn't now solve ¾ of the second layer in a way so that the first layer wouldn't be scrambled again - that's the purpose of already mentioned working corner (for the sake of completeness it should be said there are plenty of ways how to solve the second layer without messing up the first one, however, we will be using the working corner).
Since centers are fixed, the second layer is de facto composed of just four edges. Those can appear either in the upper layer (an ideal case) or the middle layer. If they are in the upper layer, we directly solve them by means of the working corner (see a red-green and orange-green edges on a simulator below); if they are in the middle layer and not solved (that means either oriented or permuted wrongly), we will place them to the upper layer first (using the working corner) and then apply a solution from the previous case - a blue-orange edge on a simulator is being solved that way.
Solving order: green-red, green-orange and blue-orange edge.
Intuitive puzzle solving method - step 3
At this point we will solve all 5 remaining edges, so the puzzle will look like this:
We have been using the working corner in the previous step and now a time for the working edge has come. With it, as an analogical element to the working corner, we will solve everything we need. The working edge on a simulator initially occupies a location to which a blue-red edge belongs. Therefore it is a location above the working corner location. The idea is to solve - i.e. orient and permute correctly - 4 edges that belong to the third (upper) layer. In conjunction with the group theory paragraph, the working edge will be automatically solved as well.
On the left-hand side simulator we can principally insert an edge that is at the working edge location to the upper layer by two ways: F' and R (if you are not familiar with the letters, have a look at the notation section). If we execute an F' move, a blue-yellow edge (which is initially placed at the position of the working edge) will be oriented wrongly. But if we execute an R move instead, it will be oriented correctly. Hence an orientation is not a problem anymore (we will use the latter mentioned possibility, i.e. an R move) and thus we need to figure out its permutation. In order to correctly permute edges in the upper layer, we will use only a rotation of that layer, which means U, U' or U2 move. It will be explained on the left-hand side cube in more detail. We see that a green-yellow edge is solved. That's because it is oriented and permuted correctly. That being said, it represents some kind of reference point for us. Blue-yellow edge belongs to directly opposite location in the upper layer, therefore a move [1] rotates the upper layer in a way that after a move [2] these two edges are opposite to each other, thus solved (after performing a U move). Next, it is principally possible to make a U move, however, a simulator goes by a U' move [3] - another edge is about to get to the working edge location (a U move would lead to fast solution of edges in the upper layer, therefore a U' move has been chosen by a reason of exercise). [4] - red-yellow edge is at the working edge location, and two layers except for the working edge and working corner locations are solved again [5] - permutation (thus solving) of a blue-yellow and green-yellow edges [6] - permutation (thus solving) of a red-yellow edge [7] - blue-red edge has to be placed at the working edge location, since all the other edges belonging to the upper layer are already correctly oriented as well as permuted (thus solved) in there [8] - all edges in the upper layer are permuted (after U2 move), but an orange-yellow one is oriented incorrectly [9] - setup move for a swap of the red-blue and orange-yellow edges (this move doesn't need to be executed, but because of a future solving order, I personally find it convenient / more illustrative to mention it on a simulator) [10 - 16] - a swap of the orange-yellow and blue-red edges, while all the other edges keep the same orientation as well as permutation [17] - setup move to an inverse swap of the orange-yellow and red-blue edges (the same as in case of [9] move) [18 - 24] a re-swap of the orange-yellow and red-blue edges, only this time with the difference that an orange-yellow edge is oriented correctly |
Intuitive puzzle solving method - step 4
If you understood edge-solving conception, a solving of corners won't be anything new for you. As you guess correctly, the goal of this step will be obtaining of the following, somewhat humdrum ;-), cube appearance.
We will solve maybe as many as possible corners in the upper layer with the help of the working corner. While doing so, it is convenient to obey two recommendations:
1) try not to place a corner belonging to the lower layer to the working corner location - the reason is obvious: corners in the upper layer are being solved by means of the working corner. In other words, the working corner is being placed to the upper layer and that layer cannot be solved when you place there a corner belonging to the lower layer. (The point doesn't apply strictly to all situations, e.g. for only two misoriented corners - then it is less time-consuming to solve this situation another way, see bellow).
2) if possible, keep an orientation of the working corner in a way so that a color of the upper layer would be visible on the working corner when looking from right or left, i.e. not from bellow. In other words, a color of center piece in the upper layer should not be the same as a color of working corner when viewed from below. (In the most simply way: there is a yellow center piece in the upper layer on the first simulator of this step, and there is a white color on the working corner when looking from below.) The reason is following: more difficult solution to a corner placed at the working corner location.
What to do if it is not possible to obey point 1 or 2 during the solve (there is no such situation on the following simulator) will be discussed next.
There is a blue-orange-yellow corner at the working corner location. Move [1] only rotates the cube because of a better view. The working corner belongs to the upper layer, between the blue and orange centers. [2] is a setup move, it is done in order to be able to solve the working corner [3 - 5] - solving of the working corner, another corner from the upper layer must replace it [6] - at the working corner location (currently still situated in the upper layer) is a red-green-yellow corner [7 - 8] - insertion of a new working corner into its initial location [9] - re-solving of "two layers" (obviously without the working corner) [10] - "solving" of the upper layer in the meaning of a permutation of edges [11] - cube rotation because of a better view, red-green-yellow corner is at the working corner location [12] - upper layer rotation in order to be able to solve the working corner [13 - 15] - solving of the working corner, another corner from the upper layer must replace it [16] - at the working corner location (currently still situated in the upper layer) is a blue-red-white corner [17 - 18] - insertion of a new working corner into its initial location [19] - re-solving of two layers (the working corner is solved by luck) [20] - "solving" of the upper layer in the meaning of a permutation of edges |
By outlined procedure it is always doable to solve the Rubik's cube. Sometimes it takes a lot of time because you can't meet either point 1 or 2 mentioned above. In that case you need to use some finesse. The trick is to seemingly scramble already solved cubies (corners in the upper layer) in such a way that the cube would be solvable at further solving of the working corners (see below).
Hmm. Perhaps I am explaining it wrong. I would like to know via an e-mail what you think about this passage. Can it be understood? What to write better? Where have you possibly lost yourself?
Back to simulators. The previous one ended-up with two misoriented corners. As already noted, this situation is always solvable by placing a misoriented corner from the upper layer to the working corner location and a reverse process of giving the working corner back must be done "skillfully" - in a way so that the cube would be solved in a final effect. It has been also mentioned (see point 1) that this way (i.e. an orientation of just two corners) is heavy time-consuming. Now let's introduce faster technique. The good news is that we already used that in the past, therefore it won't be new for us.
We can probably agree that if we execute U and U' moves on a solved Rubik's cube in a row, it will be still solved. That's all we need to know.
There are two unsolved (misoriented) corners in the upper layer. What if we orient correctly one of them without changing anything else in the upper layer? Well, we would get only one incorrectly oriented corner in that layer, but the rest of a cube would be scrambled. But what if we take that second misoriented corner in the upper layer and do the inverse moves to those which properly oriented
the previous corner? Two things would happen: first, previously scrambled cube would get back to its initial state again (analogy of U and U' in a row) and second, according to a group theory paragraph, the second corner in the upper layer would get solved (i.e. it would be oriented correctly). That's because a case in which only one corner is oriented incorrectly can not occur on a Rubik's cube. [1 - 6] - orientation of a blue-red-yellow corner in a way so that the rest of the upper layer wouldn't be changed - lower two layers can be scrambled freely [7] - setup move; since an orientation process of the first corner has been started at the RUF location, so an orientation process of the seconds corner has to be started at this location [8 - 13] - orientation of a green-yellow-orange corner = executing of the inverse moves to [1 - 6] moves [14] - solving of the upper layer = an inverse move to [7] move |
Let's recap (as general and logically easiest as possible, because there is a plenty of possibilities/combinations for permutations and orientations of several unsolved corners) regarding the two points mentioned above: if you have a corner belonging to the lower layer at the working corner location, just place it somewhere to the upper layer (typically by 3 moves without counting some "preparatory" moves), replace it by another unsolved corner (by 1 move) and put it back to the lower layer (by 3 moves which are inverse to those previous three). That's enough for solving the cube. If you want to speed up certain processes and orient only a corner in the lower layer (along with some other corner, for example from the upper layer), be inspired by the last illustrated (until now) simulator (eventually by next simulator on the right-hand side below (see the moves 68-69)).
To solve few last corners, the real experts use conjugations in this intuitive method (a number of moves required to solve a cube will decrease significantly by that). However, understanding of their theory (see below) is not obligatory to beginners by any means.
Still confused? Don't worry - two example solves are prepared for you. Those include also a solution to situations which haven't been discussed in this tutorial through the simulators. Moreover, you can always ask the author ;-). I will explain, consult, write up, rewrite.
Example solves
At the end you can see two typical example solves for a Rubik's cube, using an intuitive method. Letters below the simulators stand for a scrambling algorithm applied to originally solved cubes.
D2 R2 B2 L' R2 B2 R2 D' R2 F U B R2 D' B L' F' U2 F2 U' D2 R D' B U2 | D F2 L U L D F2 B2 U' D F B' D R2 F2 U' F2 B' U D F2 D' F2 U D2 |
A bit of theory - commutators
A commutator (more precisely a commutator of a permutation) is a powerful weapon for solving the combinatorial puzzles more or less intuitively. Mathematically it can be described as A B A' B' and with an advantage it is used to influence only a few pieces of the puzzle. Both A and B may comprise some move sequences, an inverse sequence is denoted by an apostrophe.
Commutator orienting two corners: | A = F D2 F' R' D2 R B = U2' A' = R' D2' R F D2' F' B' = U2 |
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Commutator permuting three corners: | A_{1} = R' D' R B_{1} = U' A'_{1} = R' D R B'_{1} = U auxiliary move C = U2' A_{2} = L D' L' B_{2} = U' A'_{2} = L D L' B'_{2} = U auxiliary move C' = U2 |
How do the commutators work? Quite simply. Let's explain their principle on two simulators above.
On first of them, a blue-red-yellow corner is being solved in the first part (A) and the puzzle is being scrambled while doing so (note that only two lower layers have been scrambled, the upper one remained unchanged - except for a solved corner). Naturally, it is better if the puzzle is being scrambled "suitably". That means in a way that the second part (B) could be done so that it wouldn't affect the scrambled puzzle part (i.e. two lower layers). The third part (A') influences the second unsolved corner (green-orange-yellow one) by an inverse manner to initially unsolved first corner. About the fourth part (B') can be said that it is putting the puzzle back to its original state before an algorithm execution, furthermore the chosen corners are solved as well.
On second simulator, two commutators are used. By a first part of first of them (A_{1}), a blue-red-yellow corner is getting out of the upper layer, whereas the rest of the layer is unchanged. In a second part of the commutator (B_{1}) we rotate the upper layer in a way so that it would be possible to place earlier mentioned corner at the correct location. Third part (A'_{1}) solves that corner by an inverse manner to that one which got a corner out of the upper layer. Fourth part (B'_{1}) is "aligning" the upper layer to its initial state with the difference that a blue-red-yellow corner is solved now. The second commutator on this simulator solves remaining corners analogically.
The 4^{th} step of a Rubik's cube intuitive solving method is a beautiful example of the application of commutators.
If the positions of 3 pieces are mutually exchanging in a commutator, you can cycle these three pieces in an inverse direction by a double execution of that commutator (the same applies to two pieces in case of their orientation). However, execution of the same commutator in an inverse order has the same effect as well. Since the commutator has a form of A B A' B', an inverse order is B A B' A' - thus instead of A move we will start by B move and the rest is similar for both sequences. Therefore, it is not necessary at all to memorize original, as well as inverse, order of moves in a commutator. You only need to remember (or derive) the A and B moves.
In order to have the narration complete, there must be a link to the conjugations. Those can be, for example, utilized in a blindfolded solving of a Rubik's cube. Both commutators and conjugations are widely used for intuitive solving of combinatorial puzzles.
World record videos
As a football has FIFA and athletics has IAAF, also the 3x3x3 Rubik's cube has some sort of board that organizes the competitions worldwide. It is WCA - World Cube Association. Thus it can be officially competed in a solving of the Rubik's cube, even in several events. Usually it is competed in two formats: single fastest solve of the puzzle and average solve. As an average, five consecutive times of one round are taken, the best and the worst time is not considered and from the remaining three times an arithmetic mean is calculated.
event: 3x3x3 single solve | ||
name: Patrick Ponce (USA) | ||
result: 4.69 s | ||
scramble: B2 R2 U F2 D2 B U' R U' L F L F' L2 R D R' | ||
solution: x' y2 r' U D' R' F' U D' // cross R' U R2 // F2L 1 U' R' U2 L' U L // F2L 2 U' R U' R' U L U L' // F2L 3 U U' U' F' R U R' U' R' F R // F2L 4 U' R U R' U R U2' R' U2 // OLLCP |
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cube brand: Boron Valk M | ||
solving method: CFO(P) | ||
personal opinion on used method: see below | ||
competition: Rally In The Valley; 2. 9. 2017; USA |
event: 3x3x3 average solve | ||
name: Feliks Zemdegs (Australia) | ||
result: 5.97 s | ||
scrambles: available upon request | ||
solutions: available upon request | ||
cube brand: GAN 356 Air SM | ||
solving methods: CFOP, ZBLL | ||
personal opinion on used methods: see below | ||
competition: Latin America Cubing Tour - Chía; 28-29. 6. 2017; Columbia |
Apart from two-handed competing, it is also possible to try to solve the cube only one-handed.
event: 3x3x3 one-handed, single solve | ||
name: Feliks Zemdegs (Australia) | ||
result: 6.88 s | ||
scramble: available upon request | ||
solution: available upon request |
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cube brand: MoYu AoLong v1 | ||
solving method: CFO(P) | ||
personal opinion on used method: see below | ||
competition: Canberra Autumn; 9-10. 5. 2015; Australia |
event: 3x3x3 one-handed, average solve | ||
name: Max Park (USA) | ||
result: 10.31 s | ||
scrambles: available upon request | ||
solutions: available upon request |
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cube brand: GAN 357 Ultimate | ||
solving method: CFOP | ||
personal opinion on used method: see below | ||
competition: World Rubik's Cube Championship; 13-16. 7. 2017; France |
Believe it or not, it may be also competed in a solving without using hands. The average solve has different format - an arithmetic mean of all three times is calculated (therefore the competitor doesn't have 5 attempts, only 3 of them).
event: 3x3x3 with feet, single solve | ||
name: Jakub Kipa (Poland) | ||
result: 20.57 s | ||
scramble: available upon request | ||
solution: available upon request | ||
cube brand: Sheng En F-II | ||
solving method: CFO(P) | ||
personal opinion on used method: see below | ||
competition: Radomsko Cube Theory; 27-28. 6. 2015; Poland |
event: 3x3x3 with feet, average solve | ||
name: Jakub Kipa (Poland) | ||
result: 28.16 s | ||
scrambles: available upon request | ||
solutions: available upon request | ||
cube brand: Sheng En F-II | ||
solving method: CFOP | ||
personal opinion on used method: see below | ||
competition: Mielec Cube Day; 11-12. 6. 2016; Poland |
A little bit distinct event from WCA concept is a solving of a cube by the fewest moves. In July 2010, it has been proven that the 3x3x3 Rubik's cube is optimally solvable from any state/configuration in 20 moves or less. Competitors have a time limit of 1 hour to find the shortest solution. Since 2014 a new average format has been introduced, in which a competitor's arithmetic mean of all three attempts is calculated (therefore there are not 5 attempts, only 3 of them).
American Tim Wong broke the mythical threshold of 20 moves as the first one (see below). Later, Marcel Peters from Germany and Vladislav Ushakov from Belorussia followed him (with the same result of 19 moves). Marcel Peters set the world record for an average solve with a result of 24.33 moves (see below) as the first one. Later, Baiqiang Dong from China followed him.
event: 3x3x3 fewest moves, single solve | ||
name: Tim Wong (USA) | ||
result: 19 moves | ||
scramble: available upon request | ||
solution: available upon request | ||
solving method: NISS | ||
personal opinion on used method: see below | ||
competition: Irvine Fall; 10. 11. 2015; USA |
event: 3x3x3 fewest moves, average solve | ||
name: Marcel Peters (Germany) | ||
result: 24.33 moves | ||
scrambles: available upon request | ||
solutions: available upon request | ||
solving methods: NISS, blockbuilding, inserting of several pieces at once | ||
personal opinion on used methods: see below | ||
competition: Schwandorf Open; 28-29. 5. 2016; Germany |
In addition, the 3x3x3 Rubik's cube can be solved while blindfolded. World record videos can be seen in a blindfolded solving section. If you find a so-called speedcubing interesting, check out an article about where to buy a Rubik's cube, what cube is the best and how to solve it faster. You can also read an article about what cube you should get for a speedcubing. Last but not least, have a look at the solving methods that are used by the fastest speedcubers.
Personal opinion on methods used in the world records
3x3x3 Rubik's cube:
World record holder for a single solve is Patrick Ponce, world record holder for an average solve is Feliks Zemdegs. Both are the top world-class speedcubers. They have extremely fast finger tricks, know many of algorithms and orient themselves incredibly quickly during the solve (they look ahead).
CFOP, as their used method, is ideal for them. Primarily it offers algorithmic approach, in which the intuitive moves are not required (those are usually executed slower). Unlike the methods based on a sequential addition of blocks, CFOP has a disadvantage in a higher average move count.
Both Patrick and Feliks know several solving methods and techniques (i.e. more algorithms for given cases). For all of them, let's mention at least OLS (OLL + Last lot) and OLLCP (OLL + Corner permutation). In OLS, last corner-edge pair belonging to the first two layers is being solved along with an orientation of the last layer (symbolically expressed as F2L 4 + OLL in case of CFOP). In OLLCP, after solving of two layers, the remaining layer is oriented in one algorithm, while corners are already solved after that. Thus it might seem, at first sight, that these speedcubers skip one step of a method far more often.
Personally, I don't think CFOP is a method of the future - to solve first two layers more effectively is already practically impossible, hence a solution could be in solving of the last layer more effectively (or solving of the last layer + few pieces in two previous layers). Anyway, far greater promise I see in a combination of corners first methods and block buildings, i.e. Roux method (or some variant of it). Alternatively in a new method which is, for now, waiting to be discovered.
3x3x3 Rubik's cube one-handed:
As a matter of fact, it is principally speedcubing with the difference that a cuber has much more time to plan during the solving process, because the moves are usually slowly executable when using one hand in comparison with two hands.
Unlike a two-handed solving, I would be careful with recommendation of Roux method since M moves are generally hard to execute in just one hand. Laying the cube down on a table with subsequent execution of M move can be an alternative. I find CFOP as suitable method, especially with some procedures that will reduce the number of moves. One of them is ZBLL where the last layer, after an orientation of edges, is solved by just one algorithm. Another convenient method seems ZZ to me, mainly because it allows to solve F2L only using the moves R, U and L. That is useful for one-handed solving in particular. The duo COLL + EPLL won't maybe reduce a move count compared to OLL + PLL, but those are usually easily (thus fast) executable, therefore the time required to recognize a given case could be reduced. The same applies to OLLCP + EPLL and WV + PLL, respectively.
Feliks' world record is controversial. Instead of getting the right scramble, he got a mis-scramble (eighth move should be R', but a scrambler did R instead). Since WCA, after an investigation, found that a cube has been scrambled in the spirit of the rules (except for that incident) and no disproportionate advantage has been given to Feliks in comparison with the other competitors, the record has been recognized despite a fact his cube has been scrambled differently compared to the rest of competitors' cubes.
It was sufficient to solve corners (by COLL technique) after solving of first two layers, because Feliks got the edges (which were initially oriented correctly) solved themselves by luck. He wouldn't probably reach that with the right scramble.
3x3x3 Rubik's cube with feet:
Quite frankly, I admit I have no experience with this kind of solving whatsoever. The question is whether to use rather methods with a higher move count, or a lower move count but with harder-to-execute algorithms.
Since a cube-manipulation with feet won't be probably pleasant and especially fast in any way, I would rather choose a speedcubing variant. Another possible aspect may be a sort of "finger memory". Lower move count methods offer a higher degree of freedom than algorithm-based methods. I can imagine that the moves memorized by hand-fingers are better adapted to feet-fingers (algorithmic methods) than in a case when one hasn't been memorizing any algorithms (lower move count methods).
3x3x3 Rubik's cube - fewest moves:
It is a specific event. That being said, it indicates the used solving methods will be specific as well.
Unwritten rule is that it is being started with a building of a block, to which are gradually added another blocks. I find Heise, Petrus or Snyder method promising. If you choose CFOP, a combination with ZBLL may be profitable. Or you can directly use ZB - good luck, you will need it ;-). For the hardcore ones there is also a possibility to solve the last layer in one algorithm (out of 1211). Although Roux method has a low move count, M move is taken as two moves in accordance with WCA notation (R L' + puzzle rotation x'). From this perspective, Roux is probably not the best choice.
As it can be seen, the more optimal algorithms one knows, the better for a fewest moves solving. Variants or combinations of above-mentioned methods are, in my opinion, the best starting point for a fewest moves solving.
However, if you can not find a promising beginning, you might want to try NISS technique - Normal Inverse Scramble Switch. If you can find a promising beginning for this inverse scramble, you simply solve it and at the end invert the whole solution, which will solve original scramble. It sounds pretty wild, nevertheless, NISS enables you to find a good start of a solution. True experts can also use a lot of other techniques over the solve, e.g. commutators, conjugations, a so-called pre-scramble, inserting of several pieces at once, etc.
The page was graphically improved by Michael Feather, Jiříček and Conrad Rider.