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3 out of the last 5

Positioning of 3 out of the last 5 corners

The 'workbench' is the place where we have been accumulating the side effects of the rotations we performed to solve the cube. Now that the cubies outside the workbench are finished, it would become particularly challenging to place the last 5 corners.

However, again, a tough problem (like for the 'Tutorial.Y.The Trap' ) is matched by a very simple solution: just change the old workbench and use this new workbench instead.


The change of the 'workbench' will not affect the 5 cubies already sorted on the White face: OWB, WB, RW, RWG, RB, if we care to always use the same 'sledgehammer' (or its reverse)

To be more precise, the cubies will be affected. But at the end, every thing will be automatically sorted correctly by the cube itself.


The overall strategy is quite straightforward: swap the corners in the rdf and ruf positions by employing the 'sledgehammers' technique (R F' R' F)

Look the cubie on rdf position. Bring the cubie (on the 3rd layer) located in its destination into position ruf. Run the sledgehammer (R F' R' F) (an odd number of times) until the cubie is into its correct place and with the correct orientation.

On the other hand, if the cubie on the 3rd layer is in the right position but has the wrong orientation. Again, run the sledgehammer (R F' R' F) (an even number of times) until the cubie reach the correct orientation.


There can be 3 cases:

In this case, the final result is the cube fully solved. It is not impossible, but it is rare.

The result of this 2 cases ('easy' and 'standard') is what normally happen.

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The 'sledgehammer' (R F' R' F) would be sufficient to solve this step. In some cases, however, it may be necessary to execute it more than three times in a row. In such cases, it is advisable to use its 'reverse' (F' R F R') Due to the cyclicity property, 5 iterations of the first sledgehammer are equivalent to a single rotation of its reverse. (Refer to 'Tutorial.Sledgehammer.Cycle' for more information.)