群論與魔方：Dan Knights攻略解析(下)

 定理2： 對一個處於初始狀態的魔方進行F、B、R、L、U、D的任意複合後，8個角塊的「扭轉數總和」在「模3加法」下等於0，12個邊塊的「翻轉數總和」在「模2加法」下等於0，而6個中心塊則維持不動。

第四步：使上面的四個邊塊正常

FRUR−1U−1F−1     (9)

FRUR−1U−1F−1 = (fru1, flu0)(bru2, blu0)(fu1, ru1, bu0)     (10)

第五步：使上面的四個邊塊歸位

RUR−1URU2R−1     (11)

RUR−1URU2R−1 = (fru1, blu1)(bru0, flu1)(ru0, bu0, lu0)     (12)

第六步：使上面的四個角塊歸位

URU−1L−1UR−1U−1L     (13)

URU−1L−1UR−1U−1L = (flu1, bru1, blu1)     (14)

第七步：使上面的四個角塊正常

R−1D−1RD         (3)

R−1D−1RD = (fru1, frd1)(brd1, bld0)(fr1, rd1, bd0)         (4)

(brd1, bld0)(brd1, bld0)(brd1, bld0)(brd1, bld0)(brd1, bld0)(brd1, bld0) = I
(fr1, rd1, bd0)(fr1, rd1, bd0)(fr1, rd1, bd0)(fr1, rd1, bd0)(fr1, rd1, bd0)(fr1, rd1, bd0) = I

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