# 群論與魔方：超級魔方的性質與操作(下)

## 絕對奇偶性和相對奇偶性

「超級魔方」與「普通魔方」的還原過程基本相同，在還原的最初幾步(即《群論與魔方：更高階魔方的還原攻略》所列舉的第I至第IV步)中，我們的目標是搞定六個面的中心區和12條邊的邊區，暫時無須理會各個中心區、邊區和角塊的相互位置。在此一階段，「非正中夾心層」旋轉和外表面旋轉對還原起著很不同的作用，現以下圖為例以作說明：

(f21, u12, u21, u32, u23, b12, d21)

(f21, u23, b12, d21)     (1)

(f21, u12, b12, d21)     (2)

R-Parity(Ci) = 0     (3)

 定理19： 設P1和P2為配成一對的「配對中心塊」軌道，M1和M2為P1、P2所在「非正中夾心層」的「十字中心塊」軌道，那麼P1和P2的「相對奇偶性」相等，且都等於M1和M2的「相對奇偶性」在模2算術下之和。

R-Parity(P1) = R-Parity(P2) = R-Parity(M1) +2 R-Parity(M2)     (4)

A-Parity(Ei) = R-Parity(Mi)     (5)

A-Parity(C) = A-Parity(E0) = A-Parity(Ci) = A-Parity(C0)     (6)

## 一些新公式

Ux−1 R Uz−1 R−1 Ux R Uz R−1 = (X, Y, Z)     (7)

Ux−1 R−1 Uz−1 R Ux R−1 Uz R = (X, Y, Z)     (8)

U3−1 R−1 U1−1 R U3 R−1 U1 R = (f31, r33, f11)

U1−1 R−1 U5−1 U2−1 R U1 R−1 U5 U2 R = (f15, r11, f55)(f12, r41, f25)

U1−1 U4−1 R U3−1 R−1 U1 U4 R U3 R−1 = (f13, r31, f31)(f43, r34, f34)

L−1 R F−1 B U−1 D Ln D−1 U B−1 F R−1 L U−n = (u3n)(ln)     (9)

(U R L U2 R−1 L−1)2 = (u2)     (10)

(UU1U2 RR1R2 LL1L2 U2U12U22 R−1R1−1R2−1 L−1L1−1L2−1)2     (11)

(UU1 RR1 LL1 U2U12 R−1R1−1 L−1L1−1)2     (12)

(A, B, C, D)(E, F, G, H) = (A, B)(A, C)(A, D)(E, F)(E, G)(E, H)
(A, B, C, D)(A, B, C, D) = (A, C)(B, D)

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