Edge recovery formula of the fifth layer of the fifth-order Rubik's cube

The process of this scheme is as follows: first floor-second floor, third floor-fourth floor, fifth floor. Before reading the solution, please take a look at the following code about rotating all edges: All above are 90 degrees. If you add "2", such as "L2", it means that L turns to 180 degrees. For each face, the solution is represented by the following code: edge (ed) wing (w) angle (co).

Cross (Cr) Point (P) Center (C) First, restore the first floor. When solving the first layer, we must turn the "first face" and the "first circle" right at the same time. The solution is not difficult, based on the experience of the third-order Rubik's cube, it is easy to solve. Second, restore the second and third floors 2. 1. Restore the fork (Cr) on the second floor. If the available Cr cannot be found on the fourth floor, the available Cr can be transferred to the fourth floor using formulas (2-5) and (3- 1). Equation 2- 1-F2 u 'f22.2. The wing (w) of the second layer is restored as a third-order Rubik's cube I..

If you can't find the available W on the fifth floor, you can use formulas (2-2) and (3- 1) to transfer the available W to the fifth floor. Please imagine mirror shooting. Equation 2-2-u f u' f' l f' l f 2.3. Restore the edge of the third layer.

It is similar to (2-2), but it is a third-order Rubik's Cube II.

If you can't find an available Ed on the fifth floor, you can use Formula (2-3) to transfer the available Ed to the fifth floor. Please imagine mirror shooting. Equation 2-3-uuff u 'u 'f 'f'll f 'l 'l 'l 'ff 2.4. Restore the point of the second layer (P)

If you can't find the available P on the fifth floor, you can use formulas (2-4) and (3- 1) to transfer the available P to the fifth floor. Please imagine mirror shooting. Equation 2-4-F u 'f u 'l u l 2.5. Restoring the bifurcation (Cr) of the third layer is similar to (2-4).

If the available Cr cannot be found on the fifth floor, the available Cr can be transferred to the fifth floor by formulas (2-5) and (3- 1). Imagine the formula 2-5-ff u 'f 'f 'u 'u 'l 'uuli iii. The fourth and fifth floors are restored to 3. 1. Restore the fork (Cr) on the fifth floor. The goal of this step is to turn it into a small cross 3x3 in the center of the fifth floor.

Only one formula is needed for this step, and there may be any small picture below. If it is not the case of any small picture in the figure below, just rotate the U or U layer slightly and meet one of the following conditions. Equation 3-1-r' r' u' u' f' f' uuffrr 3.2. The direction of the fifth floor edge (ed). The goal of this step is to become a 5x5 cross on the fifth floor.

Only one formula is needed for this step, and there may be any small picture below. The formula is similar to (3- 1), but it is regarded as a third-order Rubik's cube. I. Formula 3-2-R 'u 'f 'u 'f R3.3. The position of the fifth floor corner (co) in this step, don't worry about the directions of the four corners on the fifth floor. Take this article as an example, that is, whether the blue face is on the top or not, as long as the small squares belonging to the four corners are in the right position. Formula 3-3-l r' u' r u' r u' r u' r u' r formula 3-4 3-4-r 'l u 'r u 'l u 'l Other cases can be solved by combining formulas (3-3) and (3-4). 3.4. The direction formula of the angle (Co) of the fifth floor is 3-5-R 'u 'r 'u2Rau2 and 3-6-RauR 'uRau2. Other situations can be solved by combining formulas (3-5) and (3-6). 3.5. The formula for the position of the edge (Ed) of the fifth floor is 3-7-(3-5) → u '→ (3-6) → u formula 3-8-(3-6) → u → (3-5) → u' In other cases, the formula (3-7) can be used. 3.6. Restore the wings (W) on the fourth and fifth floors, first turn the W on the fifth floor better, and then turn the W on the fourth floor (because there is 8 W on the fifth floor and only 4 W on the fourth floor). If there are only two W's left to switch, then going to U will turn into three W's to switch. Formula 3-9-l r' u' r u' r u' r u' r u' r u' r formula 3-9-3-9-r 'l u 'r u 'l u 'l 'l' 3.7. Restore the remaining fork (Cr) formula on the fourth floor 3- 10-l r. E r E' L E L' Figure 3 When using the formula in this step, it is often necessary to temporarily rotate each face to match the position in the formula. Remember to write down the process of temporary rotation so that it can be restored to its original state when the formula is perfect. Because the formula can only exchange three Cr's, but if there are only two Cr's to be exchanged in the end, as shown in Figure 3, you can borrow 1 Cr on the U surface as the third block, as follows: b 'r2→ (3-10) → r2b.3.8. The remaining points (p) of the fourth and fifth floors are recovered as formula 3- 1 1-l r' u' r u' r r formula 3-1/ -r' l u' l u' l u' l' l' 438+02-r' l d' l' d d d l' formula 3-12-r u' r' r' f f f' formula 3-655 is not necessary.

When using the formula in this step, it is usually necessary to temporarily rotate each surface to match the position in the formula. Remember to write down the process of temporary rotation so that it can be restored to its original state when the formula is perfect. Because the formula can only be interchanged with more than three P's, if only two P's are interchanged in the end, as shown in Figure 3-3, you can borrow 1 P on the F surface as the third block, as follows: f' → (3-12) → f.