1. R' U' F R2 B2 L2 D' F2 D2 U B2 D B U' B F R' U' L2 B2 L2 F D' R' U' F
| Ім'я | Середнє | Найкраще | Опис | |
|---|---|---|---|---|
| 1 | Микита Гриценко | 22 | L2 B2 D B2 R2 U' B2 L2 U2 B2 L2 U2 R' D2 U2 L' D2 U B L2 F R2 (R2 F' L2 B' ) // EO (4/4) (U' D2 L U2 D2 R) // DR 4b2 (6/10) L2 B2 D B2 R2 U' // HTR (6/16) B2 L2 U2 B2 L2 U2 // finish (6/22) |
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| 2 | Юра Рябов | 23 | Solution: U2 L2 F' B' D2 F' B' U' R2 U B2 U' L2 R F2 U' B2 U' F2 R' B' L2 F(23) Explanation: (F' L2 B) // EO(3/3) (R) // DR-4E4C(1/4) (F2 U B2 U F2 R') // DR, 2c3(6/10) (L2 U B2 U' R2 U) // HTR(6/16) (B F D2 B F L2 U2) // solved(7/23) Other: L2 B2 R (F) L2 D B2 D' B // 2c4 in 9, for 22 you need M2 extension of htr i had, sad (R2 F' L2 B') (U' D2 L U2 D2 R) // 4b2 4e in 10 with tons of variations and double slice potential, but only 22 optimal for which you need a wierd 6move htr->slice Pretty bad for the scramble, balancing 2 variations of 2c3 in 10, like 8 variations of 4b2 in 10 and 2c4 in 9 is not fun. Missed a simple 21 from (F' L2 B D' R) rzp i somehow didn't find |
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| 3 | Тимур Остапець | 35 | Solution: R2 F2 R' B2 F' U' F B' L2 D' R U2 R' D R U R' U R U' F R F' U' R' U R' U' R' U R F2 U R2 F2(35) R2 F2 R' //EO(3/3) B2 F' U' F B' L2 # //2x2x2(6/9) (F2 R2 U' F2) // 2x2x3(4/13) R U' R' U R U' // f2l-1(6/19) F R F' U' R' U R' U' R' U R // 3c(11/30) # = D' R U2 R' D R U2 R'(8-3//35) |