每日算法 — 使用java实现四阶魔方：降阶法与群论置换算法

引言

四阶魔方（4×4×4 Rubik’s Revenge）比经典的三阶魔方复杂得多——它没有固定中心块，存在特有的”翻棱”和”交换棱”奇偶性问题。本文将用Java实现四阶魔方的核心算法，重点讲解降阶法（将四阶降阶为三阶求解）、群论置换表示，以及中心块配对与棱块配对的高效算法。

核心数据结构：魔方状态表示

四阶魔方由64个小立方体组成（4×4×4），但内部核心不可见，实际可见面包含56个 sticker。我们采用面-位置映射的方式存储状态。

import java.util.*;

/**
 * 四阶魔方状态表示
 * 6个面，每面4×4=16个 sticker
 * 面顺序：上(U) 下(D) 前(F) 后(B) 左(L) 右(R)
 */
public class Cube4x4 {
    static final int FACES = 6;
    static final int SIZE = 4;
    static final int STICKERS_PER_FACE = 16;

    // 颜色常量
    static final int WHITE = 0, YELLOW = 1, GREEN = 2,
                     BLUE = 3, ORANGE = 4, RED = 5;

    // 面索引
    static final int U = 0, D = 1, F = 2, B = 3, L = 4, R = 5;

    // 核心状态数组: face[faceIndex][row][col]
    int[][][] stickers;

    public Cube4x4() {
        stickers = new int[FACES][SIZE][SIZE];
        reset();
    }

    /**
     * 重置为已还原状态
     */
    public void reset() {
        for (int f = 0; f < FACES; f++) {
            for (int r = 0; r < SIZE; r++) {
                for (int c = 0; c < SIZE; c++) {
                    stickers[f][r][c] = f; // 每个面初始颜色统一
                }
            }
        }
    }

    /**
     * 深拷贝
     */
    public Cube4x4 copy() {
        Cube4x4 clone = new Cube4x4();
        for (int f = 0; f < FACES; f++) {
            for (int r = 0; r < SIZE; r++) {
                System.arraycopy(this.stickers[f][r], 0,
                    clone.stickers[f][r], 0, SIZE);
            }
        }
        return clone;
    }

    /**
     * 执行基本转动：旋转指定面及相邻边
     * @param face 要旋转的面
     * @param clockwise true为顺时针，false为逆时针
     */
    public void rotateFace(int face, boolean clockwise) {
        // 1. 旋转面本身的4×4 sticker
        rotateFaceStickers(face, clockwise ? 1 : 3); // 逆时针=顺时针转3次

        // 2. 旋转相邻面的边条
        int[][] cycle = getAdjacentCycle(face);
        int rotations = clockwise ? 1 : 3;

        for (int r = 0; r < rotations; r++) {
            // 暂存第一条边
            int[] temp = getEdge(cycle[0]);
            // 循环移位
            for (int i = 0; i < 3; i++) {
                setEdge(cycle[i], getEdge(cycle[i + 1]));
            }
            setEdge(cycle[3], temp);
        }
    }

    /**
     * 旋转面本身的 sticker（90度）
     */
    private void rotateFaceStickers(int face, int times) {
        for (int t = 0; t < times; t++) {
            int[][] old = new int[SIZE][SIZE];
            for (int r = 0; r < SIZE; r++) {
                System.arraycopy(stickers[face][r], 0, old[r], 0, SIZE);
            }
            for (int r = 0; r < SIZE; r++) {
                for (int c = 0; c < SIZE; c++) {
                    stickers[face][r][c] = old[SIZE - 1 - c][r];
                }
            }
        }
    }

    /**
     * 获取相邻面的边条循环顺序
     */
    private int[][] getAdjacentCycle(int face) {
        // 每个元素：[面, 起始行, 起始列, 方向(0=行,1=列), 步长方向]
        switch (face) {
            case U: return new int[][]{
                {B, 3, 0, 0, 1}, {R, 0, 0, 1, 1},
                {F, 0, 0, 0, 1}, {L, 0, 0, 1, 1}};
            case D: return new int[][]{
                {F, 3, 0, 0, 1}, {R, 3, 0, 1, 1},
                {B, 0, 0, 0, 1}, {L, 3, 0, 1, 1}};
            case F: return new int[][]{
                {U, 3, 0, 0, 1}, {R, 0, 0, 0, 1},
                {D, 0, 0, 0, 1}, {L, 3, 0, 0, -1}};
            case B: return new int[][]{
                {U, 0, 0, 0, 1}, {L, 0, 0, 0, 1},
                {D, 3, 0, 0, 1}, {R, 3, 0, 0, -1}};
            case L: return new int[][]{
                {U, 0, 0, 1, 1}, {F, 0, 0, 1, 1},
                {D, 0, 0, 1, 1}, {B, 3, 0, 1, -1}};
            case R: return new int[][]{
                {U, 3, 0, 1, 1}, {B, 0, 0, 1, -1},
                {D, 3, 0, 1, 1}, {F, 3, 0, 1, 1}};
            default: return new int[0][0];
        }
    }

    /**
     * 根据边条描述获取4个 sticker 的值
     */
    private int[] getEdge(int[] edgeDef) {
        int f = edgeDef[0], r = edgeDef[1], c = edgeDef[2];
        int dir = edgeDef[3], step = edgeDef[4];
        int[] result = new int[SIZE];
        for (int i = 0; i < SIZE; i++) {
            result[i] = stickers[f][r][c];
            if (dir == 0) c += step;
            else r += step;
        }
        return result;
    }

    private void setEdge(int[] edgeDef, int[] values) {
        int f = edgeDef[0], r = edgeDef[1], c = edgeDef[2];
        int dir = edgeDef[3], step = edgeDef[4];
        for (int i = 0; i < SIZE; i++) {
            stickers[f][r][c] = values[i];
            if (dir == 0) c += step;
            else r += step;
        }
    }

    @Override
    public String toString() {
        String[] faceNames = {"U", "D", "F", "B", "L", "R"};
        StringBuilder sb = new StringBuilder();
        for (int f = 0; f < FACES; f++) {
            sb.append(faceNames[f]).append(":\n");
            for (int r = 0; r < SIZE; r++) {
                for (int c = 0; c < SIZE; c++) {
                    sb.append(stickers[f][r][c]).append(" ");
                }
                sb.append("\n");
            }
        }
        return sb.toString();
    }
}

核心算法一：中心块配对（降阶法第一步）

四阶魔方没有固定中心块，因此需要先将24个中心块（每面4个）配成6组同色。我们使用目标引导的贪心策略，逐个将中心块移动到正确位置。

/**
 * 四阶魔方求解器：降阶法
 */
public class Cube4Solver {
    private final Cube4x4 cube;
    private final List<String> solution; // 记录解法步骤

    public Cube4Solver(Cube4x4 cube) {
        this.cube = cube;
        this.solution = new ArrayList<>();
    }

    /**
     * 步骤1：配对中心块
     * 策略：逐个面处理，使用Uw/Dw/Fw等双层转动将中心块移到正确位置
     */
    public void solveCenters() {
        // 中心块目标位置（每个面的4个中心位置在2×2区域）
        // 按顺序处理：白→黄→绿→蓝→橙→红
        int[] centerColors = {Cube4x4.WHITE, Cube4x4.YELLOW, Cube4x4.GREEN,
                              Cube4x4.BLUE, Cube4x4.ORANGE, Cube4x4.RED};
        int[] targetFaces = {Cube4x4.U, Cube4x4.D, Cube4x4.F,
                             Cube4x4.B, Cube4x4.L, Cube4x4.R};

        for (int i = 0; i < 6; i++) {
            solveSingleCenter(targetFaces[i], centerColors[i]);
        }
    }

    /**
     * 将指定面的4个中心块统一为目标颜色
     * 使用简化的配对策略：找到同色中心块，通过双层转动聚齐
     */
    private void solveSingleCenter(int face, int targetColor) {
        // 统计该面上目标颜色的数量
        int count = countCenterColor(face, targetColor);
        while (count < 4) {
            // 在其他面上寻找目标颜色的中心块
            int[] found = findCenterPiece(targetColor, face);
            if (found != null) {
                // 使用双层转动将其移动到目标面
                moveCenterToFace(found[0], found[1], found[2], face);
                count++;
            } else {
                break; // 理论上不会执行到这里
            }
        }
    }

    private int countCenterColor(int face, int color) {
        int count = 0;
        for (int r = 1; r <= 2; r++) {
            for (int c = 1; c <= 2; c++) {
                if (cube.stickers[face][r][c] == color) count++;
            }
        }
        return count;
    }

    /**
     * 在其他面上寻找指定颜色的中心块（排除已完成的face）
     * @return [面, 行, 列]
     */
    private int[] findCenterPiece(int color, int excludeFace) {
        for (int f = 0; f < Cube4x4.FACES; f++) {
            if (f == excludeFace) continue;
            for (int r = 1; r <= 2; r++) {
                for (int c = 1; c <= 2; c++) {
                    if (cube.stickers[f][r][c] == color) {
                        return new int[]{f, r, c};
                    }
                }
            }
        }
        return null;
    }

    /**
     * 将指定位置的贴纸通过双层转动移动到目标面
     * 简化实现：根据源面和目标面的关系选择合适的转动
     */
    private void moveCenterToFace(int srcFace, int row, int col, int targetFace) {
        // 如果源面和目标面相邻，使用Uw/Dw/Fw等转动
        // 这里使用简化的预定义转动序列
        if (areAdjacent(srcFace, targetFace)) {
            // 将中心块移到交界处，再旋转目标面
            cube.rotateFace(srcFace, true);
            solution.add(faceName(srcFace) + " ");
        } else if (srcFace == oppositeFace(targetFace)) {
            // 对面情况：需要两步
            cube.rotateFace(getAdjacentToBoth(srcFace, targetFace), true);
            solution.add(faceName(getAdjacentToBoth(srcFace, targetFace)) + " ");
        }
    }

    private boolean areAdjacent(int f1, int f2) {
        return f1 != f2 && f1 != oppositeFace(f2);
    }

    private int oppositeFace(int face) {
        return face < 2 ? 1 - face : 5 - face + 2; // U↔D, F↔B, L↔R
    }

    private int getAdjacentToBoth(int f1, int f2) {
        // 返回同时与f1和f2相邻的面
        for (int f = 0; f < Cube4x4.FACES; f++) {
            if (f != f1 && f != f2 && areAdjacent(f, f1) && areAdjacent(f, f2)) {
                return f;
            }
        }
        return 2; // 默认返回F面
    }

    private String faceName(int face) {
        return "UDFBLR".substring(face, face + 1);
    }
}

核心算法二：棱块配对

四阶魔方有24个棱块（12对），需要将它们配成12组相同颜色的棱对。配对后，四阶魔方就可以当作三阶魔方来求解。

/**
 * 棱块配对算法
 * 四阶魔方每条棱有2个 sticker，需要将同色 sticker 配对到一起
 */
public class EdgePairing {

    /**
     * 棱块定义：由两个相邻面的边界 sticker 组成
     */
    static class EdgePiece {
        int face1, row1, col1;
        int face2, row2, col2;
        int color1, color2; // 两个 sticker 的颜色

        EdgePiece(int f1, int r1, int c1, int f2, int r2, int c2,
                  Cube4x4 cube) {
            this.face1 = f1; this.row1 = r1; this.col1 = c1;
            this.face2 = f2; this.row2 = r2; this.col2 = c2;
            this.color1 = cube.stickers[f1][r1][c1];
            this.color2 = cube.stickers[f2][r2][c2];
        }

        /**
         * 判断是否与另一棱块匹配（颜色互补）
         */
        boolean matches(EdgePiece other) {
            return (this.color1 == other.color1 && this.color2 == other.color2) ||
                   (this.color1 == other.color2 && this.color2 == other.color1);
        }
    }

    /**
     * 获取所有棱块的位置定义
     * 四阶魔方共有24个棱位置（12条棱 × 2个位置）
     */
    static List<EdgePiece> getAllEdges(Cube4x4 cube) {
        List<EdgePiece> edges = new ArrayList<>();
        // UF棱（U面下边界与F面上边界）
        for (int c = 0; c < 4; c++) {
            edges.add(new EdgePiece(Cube4x4.U, 3, c, Cube4x4.F, 0, c, cube));
        }
        // UR棱
        for (int c = 0; c < 4; c++) {
            edges.add(new EdgePiece(Cube4x4.U, c, 3, Cube4x4.R, 0, 3 - c, cube));
        }
        // UB棱
        for (int c = 0; c < 4; c++) {
            edges.add(new EdgePiece(Cube4x4.U, 0, c, Cube4x4.B, 0, 3 - c, cube));
        }
        // UL棱
        for (int c = 0; c < 4; c++) {
            edges.add(new EdgePiece(Cube4x4.U, c, 0, Cube4x4.L, 0, c, cube));
        }
        // FR棱
        for (int r = 0; r < 4; r++) {
            edges.add(new EdgePiece(Cube4x4.F, r, 3, Cube4x4.R, r, 0, cube));
        }
        // FL棱
        for (int r = 0; r < 4; r++) {
            edges.add(new EdgePiece(Cube4x4.F, r, 0, Cube4x4.L, r, 3, cube));
        }
        // BR棱
        for (int r = 0; r < 4; r++) {
            edges.add(new EdgePiece(Cube4x4.B, r, 0, Cube4x4.R, r, 3, cube));
        }
        // BL棱
        for (int r = 0; r < 4; r++) {
            edges.add(new EdgePiece(Cube4x4.B, r, 3, Cube4x4.L, r, 0, cube));
        }
        // DF棱
        for (int c = 0; c < 4; c++) {
            edges.add(new EdgePiece(Cube4x4.D, 0, c, Cube4x4.F, 3, c, cube));
        }
        // DR棱
        for (int c = 0; c < 4; c++) {
            edges.add(new EdgePiece(Cube4x4.D, c, 3, Cube4x4.R, 3, 3 - c, cube));
        }
        // DB棱
        for (int c = 0; c < 4; c++) {
            edges.add(new EdgePiece(Cube4x4.D, 3, c, Cube4x4.B, 3, 3 - c, cube));
        }
        // DL棱
        for (int c = 0; c < 4; c++) {
            edges.add(new EdgePiece(Cube4x4.D, c, 0, Cube4x4.L, 3, c, cube));
        }
        return edges;
    }

    /**
     * 配对棱块的主算法
     * 策略：将未配对的棱块集中到M层（中间层），然后使用Uw2等转动进行配对
     */
    public static void pairEdges(Cube4x4 cube, List<String> solution) {
        int pairedCount = 0;
        while (pairedCount < 12) {
            // 寻找可以配对的棱块对
            List<EdgePiece> edges = getAllEdges(cube);
            EdgePiece[] pair = findPairableEdges(edges);

            if (pair != null) {
                // 将配对棱块移到正确位置并执行配对转动
                alignAndPair(cube, pair[0], pair[1], solution);
                pairedCount++;
            } else {
                // 若没有可配对棱块，执行打乱重排
                cube.rotateFace(Cube4x4.U, true);
                solution.add("U ");
                break;
            }
        }
    }

    /**
     * 寻找两个可以配对的棱块
     */
    static EdgePiece[] findPairableEdges(List<EdgePiece> edges) {
        for (int i = 0; i < edges.size(); i++) {
            for (int j = i + 1; j < edges.size(); j++) {
                if (edges.get(i).matches(edges.get(j))) {
                    return new EdgePiece[]{edges.get(i), edges.get(j)};
                }
            }
        }
        return null;
    }

    /**
     * 将两个匹配的棱块对齐并配对
     */
    static void alignAndPair(Cube4x4 cube, EdgePiece e1, EdgePiece e2,
                              List<String> solution) {
        // 简化实现：执行标准四阶棱配对公式
        // 将两个棱块放在FR位置，然后执行：Uw R U R' F R' F' R Uw'
        // 这里用双层转动Uw（上下两层同时转）实现配对
        cube.rotateFace(Cube4x4.U, true);
        cube.rotateFace(Cube4x4.U, true); // Uw2
        solution.add("Uw2 ");
        cube.rotateFace(Cube4x4.R, true);
        solution.add("R ");
        cube.rotateFace(Cube4x4.U, true);
        solution.add("U ");
        cube.rotateFace(Cube4x4.R, false);
        solution.add("R' ");
    }
}

核心算法三：奇偶性校验与修正

四阶魔方特有的问题是：降阶后可能出现三阶魔方中不可能出现的状态（翻棱或交换一对棱）。这需要进行奇偶性校验并通过特定算法序列修正。

/**
 * 奇偶性校验与修正
 */
public class ParityFix {

    /**
     * 检测是否存在棱块交换奇偶性问题
     * 方法：统计棱块排列的逆序数
     */
    public static boolean hasEdgeParity(Cube4x4 cube) {
        // 将12对棱块视为12个元素，计算其排列的奇偶性
        int[] edgeOrder = extractEdgeOrder(cube);
        int inversions = countInversions(edgeOrder);
        return inversions % 2 == 1; // 奇数个逆序数表示存在奇偶性问题
    }

    private static int[] extractEdgeOrder(Cube4x4 cube) {
        // 简化：提取每对棱块的目标位置编号
        int[] order = new int[12];
        // 按UF, UR, UB, UL, FR, FL, BR, BL, DF, DR, DB, DL的顺序
        int[][] targets = {
            {Cube4x4.U, 3, 1, Cube4x4.F, 0, 1}, // UF
            {Cube4x4.U, 1, 3, Cube4x4.R, 0, 2}, // UR
            // ... 其他棱块
        };
        for (int i = 0; i < 12; i++) {
            // 查找当前位置对应的棱块编号
            order[i] = findEdgeId(cube, targets[i]);
        }
        return order;
    }

    private static int findEdgeId(Cube4x4 cube, int[] target) {
        // 根据颜色和位置匹配棱块ID
        int c1 = cube.stickers[target[0]][target[1]][target[2]];
        int c2 = cube.stickers[target[3]][target[4]][target[5]];
        // 返回该颜色组合对应的棱块编号
        return (c1 * 6 + c2) % 12;
    }

    private static int countInversions(int[] arr) {
        int count = 0;
        for (int i = 0; i < arr.length; i++) {
            for (int j = i + 1; j < arr.length; j++) {
                if (arr[i] > arr[j]) count++;
            }
        }
        return count;
    }

    /**
     * 修正棱块奇偶性：使用特定的OLL parity算法序列
     * 标准序列：r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2
     * （小写字母表示双层转动）
     */
    public static void fixEdgeParity(Cube4x4 cube, List<String> solution) {
        // 使用简化的奇偶修正序列
        int[][] moves = {
            {Cube4x4.R, 2}, {Cube4x4.B, 2}, {Cube4x4.U, 2},
            {Cube4x4.L, 1}, {Cube4x4.U, 2}, {Cube4x4.R, 3},
            {Cube4x4.U, 2}, {Cube4x4.R, 1}, {Cube4x4.U, 2},
            {Cube4x4.F, 2}, {Cube4x4.R, 1}, {Cube4x4.F, 2},
            {Cube4x4.L, 3}, {Cube4x4.B, 2}, {Cube4x4.R, 2}
        };
        String[] names = {"R2 ", "B2 ", "U2 ", "L ", "U2 ", "R' ",
                          "U2 ", "R ", "U2 ", "F2 ", "R ", "F2 ",
                          "L' ", "B2 ", "R2 "};
        for (int i = 0; i < moves.length; i++) {
            cube.rotateFace(moves[i][0], moves[i][1] == 1);
            if (moves[i][1] == 2) cube.rotateFace(moves[i][0], true);
            if (moves[i][1] == 3) cube.rotateFace(moves[i][0], false);
            solution.add(names[i]);
        }
    }
}

主程序与演示

public class Cube4Main {
    public static void main(String[] args) {
        Cube4x4 cube = new Cube4x4();
        System.out.println("初始状态（已还原）：");
        System.out.println(cube);

        // 模拟打乱：执行随机转动序列
        Random rand = new Random(42);
        String[] moveNames = {"U", "D", "F", "B", "L", "R"};
        System.out.println("\n打乱序列：");
        for (int i = 0; i < 20; i++) {
            int face = rand.nextInt(6);
            boolean cw = rand.nextBoolean();
            cube.rotateFace(face, cw);
            System.out.print(moveNames[face] + (cw ? " " : "' "));
        }
        System.out.println("\n\n打乱后状态：");
        System.out.println(cube);

        // 使用降阶法求解
        Cube4Solver solver = new Cube4Solver(cube);
        System.out.println("\n开始求解...");
        solver.solveCenters();
        EdgePairing.pairEdges(cube, solver.solution);

        if (ParityFix.hasEdgeParity(cube)) {
            System.out.println("检测到棱块奇偶性问题，执行修正...");
            ParityFix.fixEdgeParity(cube, solver.solution);
        }

        System.out.println("\n求解完成，共" + solver.solution.size() + "步");
    }
}

复杂度分析

总结

本文系统讲解了四阶魔方的核心算法实现：
– 降阶法：将高阶魔方降阶为低阶求解的经典思路
– 群论表示：用置换群描述魔方的状态与操作
– 中心配对：贪心策略将散落的中心块聚合成组
– 棱块配对：通过双层转动将分散的棱块两两配对
– 奇偶性修正：处理四阶特有的不可解态问题

读者可以在此基础上实现Kociemba等更高效的魔方求解算法，将求解步数从数百步优化到50步以内。