After completing a 2x2x3 block, assuming the edges are oriented, one can reduce the cube from a state of
Using a similar approach to CPLS or Seth (see menu above), place the DBR corner in place. For these purposes, the orientation of hte corner does not matter. This should always be achieved in two or less moves, with only a 1/3 chance of it being 2 moves.
From this point, use [CPLS recognition] to decifer the current permutation of your corners. There should be a 1/6 chance that the cube is already in a
For the other 5/6 of the time, you can reduce the cube with a fast
D B D2 L' F' R B2 R2 D F B D L' R U F L' F R D' U R2 L2 U' F2 Scramble WCA, then x2 to inspect/solve R2 B' R' U F' D2 R D' L R2 U2 L' U R' U' L' U2 R U2 BR->FR U' L' U R2 U' L Algs: Good U' D' R U2 R' D BL->FR L' U R U' R' L FL->FR U L U' R U L' BR->FR U' L' U R2 U' L AdjL U D R' U2 R D' AdjB It's already
Once you have mastered this technique, feel free to try placing the DFR corner rather than the DBR corner and make algs for that. Please read up on CPLS before asking any questions; the recognition from this system is key in understanding any of this.