对于经典的 Rush Hour，这个问题很容易通过简单的广度优先搜索来解决。声称的最难的已知初始配置需要 93 次移动来解决，总共只有 24132 个可达配置。即使是一个简单实现的广度优先搜索算法，即使在一台普通的机器上也能在 1 秒内探索整个搜索空间。
参考文献
Java 求解器
这是广度优先搜索穷举求解器的完整源代码，用 C 风格编写。
import java.util.*;
public class RushHour {
// classic Rush Hour parameters
static final int N = 6;
static final int M = 6;
static final int GOAL_R = 2;
static final int GOAL_C = 5;
// the transcription of the 93 moves, total 24132 configurations problem
// from http://cs.ulb.ac.be/~fservais/rushhour/index.php?window_size=20&offset=0
static final String INITIAL = "333BCC" +
"B22BCC" +
"B.XXCC" +
"22B..." +
".BB.22" +
".B2222";
static final String HORZS = "23X"; // horizontal-sliding cars
static final String VERTS = "BC"; // vertical-sliding cars
static final String LONGS = "3C"; // length 3 cars
static final String SHORTS = "2BX"; // length 2 cars
static final char GOAL_CAR = 'X';
static final char EMPTY = '.'; // empty space, movable into
static final char VOID = '@'; // represents everything out of bound
// breaks a string into lines of length N using regex
static String prettify(String state) {
String EVERY_NTH = "(?<=\\G.{N})".replace("N", String.valueOf(N));
return state.replaceAll(EVERY_NTH, "\n");
}
// conventional row major 2D-1D index transformation
static int rc2i(int r, int c) {
return r * N + c;
}
// checks if an entity is of a given type
static boolean isType(char entity, String type) {
return type.indexOf(entity) != -1;
}
// finds the length of a car
static int length(char car) {
return
isType(car, LONGS) ? 3 :
isType(car, SHORTS) ? 2 :
0/0; // a nasty shortcut for throwing IllegalArgumentException
}
// in given state, returns the entity at a given coordinate, possibly out of bound
static char at(String state, int r, int c) {
return (inBound(r, M) && inBound(c, N)) ? state.charAt(rc2i(r, c)) : VOID;
}
static boolean inBound(int v, int max) {
return (v >= 0) && (v < max);
}
// checks if a given state is a goal state
static boolean isGoal(String state) {
return at(state, GOAL_R, GOAL_C) == GOAL_CAR;
}
// in a given state, starting from given coordinate, toward the given direction,
// counts how many empty spaces there are (origin inclusive)
static int countSpaces(String state, int r, int c, int dr, int dc) {
int k = 0;
while (at(state, r + k * dr, c + k * dc) == EMPTY) {
k++;
}
return k;
}
// the predecessor map, maps currentState => previousState
static Map<String,String> pred = new HashMap<String,String>();
// the breadth first search queue
static Queue<String> queue = new LinkedList<String>();
// the breadth first search proposal method: if we haven't reached it yet,
// (i.e. it has no predecessor), we map the given state and add to queue
static void propose(String next, String prev) {
if (!pred.containsKey(next)) {
pred.put(next, prev);
queue.add(next);
}
}
// the predecessor tracing method, implemented using recursion for brevity;
// guaranteed no infinite recursion, but may throw StackOverflowError on
// really long shortest-path trace (which is infeasible in standard Rush Hour)
static int trace(String current) {
String prev = pred.get(current);
int step = (prev == null) ? 0 : trace(prev) + 1;
System.out.println(step);
System.out.println(prettify(current));
return step;
}
// in a given state, from a given origin coordinate, attempts to find a car of a given type
// at a given distance in a given direction; if found, slide it in the opposite direction
// one spot at a time, exactly n times, proposing those states to the breadth first search
//
// e.g.
// direction = -->
// __n__
// / \
// ..o....c
// \___/
// distance
//
static void slide(String current, int r, int c, String type, int distance, int dr, int dc, int n) {
r += distance * dr;
c += distance * dc;
char car = at(current, r, c);
if (!isType(car, type)) return;
final int L = length(car);
StringBuilder sb = new StringBuilder(current);
for (int i = 0; i < n; i++) {
r -= dr;
c -= dc;
sb.setCharAt(rc2i(r, c), car);
sb.setCharAt(rc2i(r + L * dr, c + L * dc), EMPTY);
propose(sb.toString(), current);
current = sb.toString(); // comment to combo as one step
}
}
// explores a given state; searches for next level states in the breadth first search
//
// Let (r,c) be the intersection point of this cross:
//
// @ nU = 3 '@' is not a car, 'B' and 'X' are of the wrong type;
// . nD = 1 only '2' can slide to the right up to 5 spaces
// 2.....B nL = 2
// X nR = 4
//
// The n? counts how many spaces are there in a given direction, origin inclusive.
// Cars matching the type will then slide on these "alleys".
//
static void explore(String current) {
for (int r = 0; r < M; r++) {
for (int c = 0; c < N; c++) {
if (at(current, r, c) != EMPTY) continue;
int nU = countSpaces(current, r, c, -1, 0);
int nD = countSpaces(current, r, c, +1, 0);
int nL = countSpaces(current, r, c, 0, -1);
int nR = countSpaces(current, r, c, 0, +1);
slide(current, r, c, VERTS, nU, -1, 0, nU + nD - 1);
slide(current, r, c, VERTS, nD, +1, 0, nU + nD - 1);
slide(current, r, c, HORZS, nL, 0, -1, nL + nR - 1);
slide(current, r, c, HORZS, nR, 0, +1, nL + nR - 1);
}
}
}
public static void main(String[] args) {
// typical queue-based breadth first search implementation
propose(INITIAL, null);
boolean solved = false;
while (!queue.isEmpty()) {
String current = queue.remove();
if (isGoal(current) && !solved) {
solved = true;
trace(current);
//break; // comment to continue exploring entire space
}
explore(current);
}
System.out.println(pred.size() + " explored");
}
}
源码中有两行值得注意：
- 找到解决方案时的
break;
- 现在对此进行了注释，以便广度优先搜索探索整个搜索空间，以确认上面链接网站中给出的数字
-
slide 中的current = sb.toString();
- 基本上这将任何汽车的每次移动都计为一次移动。如果一辆车向左移动 3 格，那就是 3 次移动。要将其组合为一个动作（因为它涉及同一辆车向同一方向移动），只需注释此行。链接的网站不承认组合，因此该行现在未注释以匹配给定的最小移动数。使用连击数，93 步问题只需要 49 步。也就是说，如果停车场里有一个停车服务员来移动这些汽车，他只需要进出汽车 49 次。
算法概述
该算法本质上是广度优先搜索，通常使用队列实现。维护前驱图，以便任何状态都可以追溯到初始状态。一个键永远不会被重新映射，并且由于条目是按广度优先搜索顺序插入的，因此可以保证最短路径。
状态表示为NxM-length String。每个char 代表板上的一个实体，以行优先顺序存储。
通过从空白空间扫描所有 4 个方向来找到相邻状态，寻找合适的汽车类型，并在房间容纳时滑动它。
这里有很多多余的工作（例如多次扫描长“小巷”），但如前所述，虽然通用版本是 PSPACE 完整的，但经典的 Rush Hour 变体很容易通过蛮力处理。
维基百科参考