1. R' U' F U' L2 F2 R2 D L2 F2 D' U2 R F L' F2 R2 F U R F' R2 U R' U' F
| Ім'я | Середнє | Найкраще | Опис | |
|---|---|---|---|---|
| 1 | Юра Рябов | 23 | Solution: F2 D' B2 U' B2 U F2 D' B2 U' R2 F2 L2 B2 R' U L2 U L F' U' L' F'(23) Explanation: (F L U F) // EO(4/4) (L') // DR-4E4C(1/5) (U' L2 U' R) // DR, 2c5(4/9) F2 [U' L2 U' L2 U R2 D' L2 D'] // HTR(10/19) R2 F2 L2 B2 // LS(4/23) [] = D' B2 U' B2 U F2 D' B2 U'(+0/23) Other: (L' F2 R' F) U' F2 R2 L2 U' F // 4b5 in 10 (F L U F' L U' R D' L2 D) // 2c4 in 10 (F L U F) U2 R U' R D R' U // 2c4 in 11 (R' D2 L) B (F' B2 D R2 U2 D2 F) // 4a2 in 11 (R' U' F) L2 D' R L B2 R2 L U' // 4a3 in 11 Very unpleasant scramble in a sense that 3-4 movers on inverse have a lot to check, but all resulting short drs are bad subsets, didn't have time to check any 5movers because of that |
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| 2 | Микита Гриценко | 23 | solution: R L B2 L D2 F R2 D2 B2 L2 R' D2 R F2 U2 D2 R' B2 D B' F R2 U' (23) (U R2 F' B D') // EO (5/5) R L B2 L D2 F // DR 2c3 (6/11) (B2 R D2 U2 F2 R' D2 R) // HTR (8/19) (L2 B2 D2 R2) // finish (4/23) |