Two example 3x3 blindfolded solves
Follow two complete Old Pochmann memorisation examples, including short cycles, cycle breaks, flipped pieces and parity.
A short corner cycle and two flipped edges
R2 L2 F2 R B' L F L2 U2 R2 U' B2 D F2 D L2 B2 R2 U' BShow the scramble as notation images
Corner memorisation
The sticker in the A buffer position belongs at S. Continue tracing: S goes to U, U to M, M to W and W to Q. Q returns to the A buffer piece.
This corner cycle is unusually short. The other two corners are already solved, so no cycle break is required.
Parity is required
The corner memo contains five swaps: S, U, M, W and Q. Because that number is odd, perform the parity algorithm after executing the corner memo and before beginning the edges.
Main edge cycle
Start at B and trace B → S → I → N → O → L → K → P → A → E. E returns to the buffer piece.
Two edge pieces have not appeared in this cycle. They are not solved: inspection shows that both are flipped in place.
A twisted buffer and a corner cycle break
R2 B D L' D B2 R' L2 U2 F B2 R2 F L2 D2 B' L2 BShow the scramble as notation images
Corner cycle from a twisted buffer
The corner buffer contains the correct piece, but that piece is twisted. It cannot begin a normal buffer cycle, so start a new cycle immediately. Choosing P is convenient, making P the first memo letter.
Trace P → O → M → G → N → K. K is on the same physical corner as P, so this closes the cycle and accounts for the twist.
Second corner cycle
Two corner pieces remain untouched. Start a cycle break at F, then trace F → S → I. I completes this second cycle.
Parity is required again
The complete corner memo has nine letters, so the number of corner swaps is odd. Execute the parity algorithm between the corner and edge phases.
Edge memorisation
Starting from B, trace B → T → K → I → A → P → W → G → F → R → V → D. D returns to the buffer.
Every unsolved edge is included in this cycle, so no cycle break or additional flipped-edge memo is needed.