引言
四阶魔方（4×4×4 Rubik’s Cube）是比经典三阶更复杂的组合谜题。它没有固定的中心块，存在”中心块交换奇偶性”和”边块翻转”等特殊现象。本文将用Java实现四阶魔方的核心模拟与求解算法，重点讲解群论置换表示、降阶法还原策略以及Kociemba两阶段算法在四阶上的适配思路。
核心算法原理
1. 四阶魔方的状态表示
四阶魔方共 $16 \times 6 = 96$ 个贴纸（或 $8$ 角块 + $24$ 边块 + $24$ 中心块）。采用紧凑编码：
- 角块：8个，每个3方向，状态空间 $8! \times 3^7$
- 边块：24个，每个2方向，但四阶边块成对出现，记为12对
- 中心块：24个，分6组各4个，仅需记录相对位置
2. 降阶法（Reduction Method）
四阶求解的标准策略是降阶为三阶：
- 中心配对：将4个同颜色中心块聚到一起
- 边配对：将成对的边块拼合为”复合边”
- 三阶求解：用标准CFOP或Kociemba算法解决
3. Kociemba两阶段算法
Phase 1：将魔方转到 $G1$ 子群（所有边方向正确，UD层角块在UD层）
Phase 2：在 $G1$ 内用预计算表求解
Java核心实现
package rubik;
import java.util.*;
/**
* 四阶魔方模拟器与求解器
* 核心算法：降阶法 + Kociemba两阶段适配
*/
public class Rubik4x4 {
// 六个面的颜色常量
public static final int U = 0, D = 1, F = 2, B = 3, L = 4, R = 5;
public static final String[] COLOR_NAMES = {"白","黄","绿","蓝","橙","红"};
// 每个面4x4的贴纸状态
private int[][][] state; // [face][row][col]
// 角块定义：(面,行,列) 四阶角块位置与三阶一致
private static final int[][][] CORNERS = {
{{U,0,0},{B,0,3},{L,0,0}}, {{U,0,3},{B,0,0},{R,0,3}},
{{U,3,0},{F,0,0},{L,0,3}}, {{U,3,3},{F,0,3},{R,0,0}},
{{D,0,0},{F,3,0},{L,3,3}}, {{D,0,3},{F,3,3},{R,3,0}},
{{D,3,0},{B,3,3},{L,3,0}}, {{D,3,3},{B,3,0},{R,3,3}}
};
public Rubik4x4() {
state = new int[6][4][4];
reset();
}
/**
* 重置为还原状态
*/
public void reset() {
for (int f = 0; f < 6; f++) {
for (int r = 0; r < 4; r++) {
for (int c = 0; c < 4; c++) {
state[f][r][c] = f;
}
}
}
}
/**
* 核心操作：旋转指定面（90度顺时针/逆时针）
* 同时带动相邻四行的边块循环
*/
public void rotateFace(int face, boolean clockwise) {
int[][] temp = new int[4][4];
// 面内旋转
for (int r = 0; r < 4; r++) {
for (int c = 0; c < 4; c++) {
if (clockwise) {
temp[c][3 - r] = state[face][r][c];
} else {
temp[3 - c][r] = state[face][r][c];
}
}
}
for (int r = 0; r < 4; r++) {
System.arraycopy(temp[r], 0, state[face][r], 0, 4);
}
// 边块循环（四阶特有：每次影响相邻面的整行/列）
cycleAdjacentRows(face, clockwise);
}
/**
* 相邻面边块循环：四阶每次转动影响相邻面的4个贴纸
*/
private void cycleAdjacentRows(int face, boolean cw) {
int[][] cycles = getAdjacentCycle(face);
int[] buffer = new int[4];
if (cw) {
// 保存第一组
for (int i = 0; i < 4; i++) buffer[i] = getSticker(cycles[0][i]);
// 循环移位
for (int g = 0; g < 3; g++) {
for (int i = 0; i < 4; i++) {
setSticker(cycles[g][i], getSticker(cycles[g+1][i]));
}
}
for (int i = 0; i < 4; i++) setSticker(cycles[3][i], buffer[i]);
} else {
for (int i = 0; i < 4; i++) buffer[i] = getSticker(cycles[3][i]);
for (int g = 3; g > 0; g--) {
for (int i = 0; i < 4; i++) {
setSticker(cycles[g][i], getSticker(cycles[g-1][i]));
}
}
for (int i = 0; i < 4; i++) setSticker(cycles[0][i], buffer[i]);
}
}
/**
* 获取某面转动时相邻面的边块位置
* 每个面有4组相邻边，每组4个位置
*/
private int[][] getAdjacentCycle(int face) {
// 每组4个int[3] = {face, row, col}
switch (face) {
case U: return new int[][]{
{B,0,3},{B,0,2},{B,0,1},{B,0,0},
{R,0,3},{R,0,2},{R,0,1},{R,0,0},
{F,0,3},{F,0,2},{F,0,1},{F,0,0},
{L,0,3},{L,0,2},{L,0,1},{L,0,0}
};
case D: return new int[][]{
{F,3,0},{F,3,1},{F,3,2},{F,3,3},
{R,3,0},{R,3,1},{R,3,2},{R,3,3},
{B,3,0},{B,3,1},{B,3,2},{B,3,3},
{L,3,0},{L,3,1},{L,3,2},{L,3,3}
};
case F: return new int[][]{
{U,3,0},{U,3,1},{U,3,2},{U,3,3},
{R,0,0},{R,1,0},{R,2,0},{R,3,0},
{D,0,3},{D,0,2},{D,0,1},{D,0,0},
{L,3,3},{L,2,3},{L,1,3},{L,0,3}
};
case B: return new int[][]{
{U,0,3},{U,0,2},{U,0,1},{U,0,0},
{L,0,0},{L,1,0},{L,2,0},{L,3,0},
{D,3,3},{D,3,2},{D,3,1},{D,3,0},
{R,3,3},{R,2,3},{R,1,3},{R,0,3}
};
case L: return new int[][]{
{U,0,0},{U,1,0},{U,2,0},{U,3,0},
{F,0,0},{F,1,0},{F,2,0},{F,3,0},
{D,0,0},{D,1,0},{D,2,0},{D,3,0},
{B,3,3},{B,2,3},{B,1,3},{B,0,3}
};
case R: return new int[][]{
{U,3,3},{U,2,3},{U,1,3},{U,0,3},
{B,0,3},{B,1,3},{B,2,3},{B,3,3},
{D,3,3},{D,2,3},{D,1,3},{D,0,3},
{F,3,3},{F,2,3},{F,1,3},{F,0,3}
};
default: return new int[4][3];
}
}
private int getSticker(int[] pos) { return state[pos[0]][pos[1]][pos[2]]; }
private void setSticker(int[] pos, int val) { state[pos[0]][pos[1]][pos[2]] = val; }
/**
* 随机打乱：执行n次随机转动
*/
public void scramble(int n) {
Random rand = new Random();
String[] moves = {"U","D","F","B","L","R","Uw","Dw","Fw","Bw","Lw","Rw"};
for (int i = 0; i < n; i++) {
String move = moves[rand.nextInt(moves.length)];
boolean cw = rand.nextBoolean();
applyMove(move, cw);
}
}
/**
* 应用标准记号转动
* 大写字母=外层，加w=双层（同时转内层）
*/
public void applyMove(String move, boolean clockwise) {
switch (move) {
case "U": rotateFace(U, clockwise); break;
case "D": rotateFace(D, clockwise); break;
case "F": rotateFace(F, clockwise); break;
case "B": rotateFace(B, clockwise); break;
case "L": rotateFace(L, clockwise); break;
case "R": rotateFace(R, clockwise); break;
// 双层转动：外层+内层
case "Uw": rotateFace(U, clockwise); rotateInnerLayer(U, clockwise); break;
case "Dw": rotateFace(D, clockwise); rotateInnerLayer(D, clockwise); break;
case "Fw": rotateFace(F, clockwise); rotateInnerLayer(F, clockwise); break;
case "Bw": rotateFace(B, clockwise); rotateInnerLayer(B, clockwise); break;
case "Lw": rotateFace(L, clockwise); rotateInnerLayer(L, clockwise); break;
case "Rw": rotateFace(R, clockwise); rotateInnerLayer(R, clockwise); break;
}
}
/**
* 转动内层（Slice move）
*/
private void rotateInnerLayer(int face, boolean cw) {
// 简化为交换相邻面中间两行
// 实际实现需根据面的方向精确处理
}
/**
* 检查是否已还原
*/
public boolean isSolved() {
for (int f = 0; f < 6; f++) {
for (int r = 0; r < 4; r++) {
for (int c = 0; c < 4; c++) {
if (state[f][r][c] != f) return false;
}
}
}
return true;
}
/**
* 打印魔方展开图
*/
public void print() {
// 展开图上半部分（U面）
for (int r = 0; r < 4; r++) {
System.out.print(" ");
for (int c = 0; c < 4; c++) System.out.print(COLOR_NAMES[state[U][r][c]] + " ");
System.out.println();
}
// 中间环（L F R B）
for (int r = 0; r < 4; r++) {
for (int c = 0; c < 4; c++) System.out.print(COLOR_NAMES[state[L][r][c]] + " ");
System.out.print(" ");
for (int c = 0; c < 4; c++) System.out.print(COLOR_NAMES[state[F][r][c]] + " ");
System.out.print(" ");
for (int c = 0; c < 4; c++) System.out.print(COLOR_NAMES[state[R][r][c]] + " ");
System.out.print(" ");
for (int c = 0; c < 4; c++) System.out.print(COLOR_NAMES[state[B][r][c]] + " ");
System.out.println();
}
// D面
for (int r = 0; r < 4; r++) {
System.out.print(" ");
for (int c = 0; c < 4; c++) System.out.print(COLOR_NAMES[state[D][r][c]] + " ");
System.out.println();
}
}
/**
* 降阶法求解：中心配对 -> 边配对 -> 三阶求解
* 实际求解需要大量预计算表，这里展示框架
*/
public List<String> solveReduction() {
List<String> solution = new ArrayList<>();
// Step 1: 中心配对（将4个同色中心聚到一起）
solution.addAll(solveCenters());
// Step 2: 边配对（将成对边块合并）
solution.addAll(solveEdges());
// Step 3: 当作三阶求解（使用Kociemba或CFOP）
solution.addAll(solveAs3x3());
return solution;
}
private List<String> solveCenters() {
// 中心配对算法：通过交换将同色中心归位
List<String> moves = new ArrayList<>();
// 简化实现：直接返回空，实际需复杂的中心交换算法
return moves;
}
private List<String> solveEdges() {
// 边配对算法：将分散的边块配对为复合边
List<String> moves = new ArrayList<>();
return moves;
}
private List<String> solveAs3x3() {
// 调用三阶求解器（需额外实现）
List<String> moves = new ArrayList<>();
return moves;
}
public static void main(String[] args) {
Rubik4x4 cube = new Rubik4x4();
System.out.println("初始状态：");
cube.print();
System.out.println("\n打乱后：");
cube.scramble(20);
cube.print();
System.out.println("\n已还原？" + cube.isSolved());
}
}
复杂度分析
| 组件 | 状态空间 | 说明 |
|---|---|---|
| 角块 | $8! \times 3^7 \approx 8.8 \times 10^7$ | 与三阶相同 |
| 边块（配对后） | $12! \times 2^{11} \approx 2.0 \times 10^9$ | 12对复合边 |
| 中心 | $\frac{24!}{(4!)^6} \approx 2.8 \times 10^{13}$ | 仅相对位置 |
| 总状态空间 | $\approx 7.4 \times 10^{45}$ | 远超三阶的 $4.3 \times 10^{19}$ |
扩展思考
- God’s Number：四阶魔方的God’s Number（最坏还原步数）为77步（HTM度量）
- 中心奇偶性：四阶可能存在”单中心交换”现象，需特殊OLL parity公式修正
- 并行求解：可用IDA*并行搜索Phase 1和Phase 2的解
总结
四阶魔方展示了群论在组合谜题中的深度应用。降阶法将高维问题分解为可管理的子问题，而Kociemba算法的两阶段策略则是子群搜索的经典范例。Java中的面-行-列三维数组结构清晰表达了魔方的几何关系。
