8 puzzle with A* algorithm

本文介绍了一个使用A*搜索算法解决8谜题问题的实现案例。通过定义Board类来表示8谜题的状态,并利用Solver类进行求解。文章详细展示了如何计算启发式函数(如汉明距离和曼哈顿距离),以及如何使用优先队列来高效寻找最优路径。

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The main idea is from:

http://www.cs.princeton.edu/courses/archive/spr10/cos226/assignments/8puzzle.html

A* is used to find the best solution. The detail of A* could be found:

http://web.mit.edu/eranki/www/tutorials/search/

#include <iostream>
#include <vector>
#include <algorithm>
#include <list>
#include <queue>
#include <unordered_map>
#include <string>
#include <sstream>
#include <stdexcept>
#include <memory>
#include <Windows.h>


using namespace std;

class Board;
class Solver;

class Board
{
	friend class Solver;
	friend class compare_node;
public:
	typedef enum
	{
		TL_SPACE,
		TL_1,
		TL_2,
		TL_3,
		TL_4,
		TL_5,
		TL_6,
		TL_7,
		TL_8,
	}TILE;

	Board()
	{
		for (int i = 0; i != 8; tiles.push_back(TILE(++i)));
		tiles.push_back(TL_SPACE);
	}
	Board(int t[3][3])
	{
		for (int i = 0; i != 3; ++i)
			for (int j = 0; j != 3; ++j)
				tiles.push_back(TILE(t[i][j]));
	}

	int hamming()
	{
		int count = 0;
		for (int i = 0; i != tiles.size(); ++i)
			if (tiles[i] != TILE((i + 1)) && tiles[i] != TL_SPACE)
				++count;
		return count;
	}

	int manhattan()
	{
		int count = 0;
		int num;
		for (int i = 0; i != tiles.size(); ++i)
		{
			if (tiles[i] == TL_SPACE)
				continue;
			num = (tiles[i] + 8) % 9;
			count = count + (abs(num / 3 - i / 3) + abs(num % 3 - i % 3));
		}
		return count;
	}

	bool equals(const Board &b) const
	{
		bool ret = true;
		for (int i = 0; i != b.tiles.size(); ++i)
		{
			if (this->tiles[i] != b.tiles[i])
			{
				ret = false;
				break;
			}
		}
		return ret;
	}

	void neighbors(vector<Board> &nghbrs)
	{
		nghbrs.clear();
		int pos = 0;
		for (; pos != tiles.size() && tiles[pos] != TL_SPACE; ++pos);
		int row = pos / 3;
		int col = pos % 3;
		if (row > 0)
		{
			nghbrs.push_back(*this);
			nghbrs.back().swap(pos, (row - 1) * 3 + col);
		}
		if (row < 2)
		{
			nghbrs.push_back(*this);
			nghbrs.back().swap(pos, (row + 1) * 3 + col);
		}
		if (col > 0)
		{
			nghbrs.push_back(*this);
			nghbrs.back().swap(pos, row * 3 + col - 1);
		}
		if (col < 2)
		{
			nghbrs.push_back(*this);
			nghbrs.back().swap(pos, row * 3 + col + 1);
		}
	}

	string toString()
	{
		string s;
		ostringstream convert;
		for (int i = 0; i != tiles.size(); ++i)
			convert << static_cast<int>(tiles[i]) << " ";
		s = convert.str();
		return s;
	}

private:
	vector<TILE> tiles;
	Board *parent = nullptr;
	int f = 0;
	int g = 0;
	int h = 0;

	void swap(size_t pos1, size_t pos2)
	{
		if (pos1 >= tiles.size() || pos2 >= tiles.size())
			throw runtime_error("position not match");
		std::swap(tiles[pos1], tiles[pos2]);
	}
};

// function used to compared board in heap
class compare_node
{
public:
	bool operator()(Board *&l, Board *&r)
	{
		return l->f > r->f;
	}
};

// vector for board pointers
class bvector : public vector<Board*>
{
public:
	bvector()
	{
		make_heap(this->begin(), this->end(), compare_node());
	}

	void push_node(Board *b)
	{
		this->push_back(b);
		push_heap(this->begin(), this->end(), compare_node());
	}

	void pop_node()
	{
		pop_heap(this->begin(), this->end(), compare_node());
		this->pop_back();
	}
};

class Solver
{
	friend class Board;
public:
	Solver() = default;
	Solver(Board *p) : initial(p) {}


	void init()
	{
		if (initial == nullptr)
		{
			cerr << "Solver not initialized" << endl;
			return;
		}

		if (!isSolvable())
		{
			cerr << "No solution existed" << endl;
			return;
		}


		initial->g = 0;
		initial->h = initial->manhattan() + initial->hamming();
		initial->f = initial->g + initial->h;

		open_ref[initial->toString()] = initial;
		open_set.push_node(initial);


		while (!open_set.empty())
		{

			Board *closed_node = open_set.front();

			open_ref.erase(closed_node->toString());
			closed_ref[closed_node->toString()] = closed_node;
			open_set.pop_node();

			if (closed_node->equals(*end))
			{
				Board *p = closed_node;
				while (p->parent != nullptr)
				{
					solution.push_back(p->toString());
					p = p->parent;
				}
				solution.push_back(initial->toString());
				reverse(solution.begin(), solution.end());
				return;
			}

			vector<Board> neighbors;

			closed_node->neighbors(neighbors);

			for (auto iter = neighbors.begin(); iter != neighbors.end(); ++iter)
			{
				Board *new_node = new Board(*iter);
				new_node->g = 0;
				new_node->h = new_node->manhattan() + new_node->hamming();
				new_node->f = new_node->f + new_node->h;

				// ignore the neighbor which is already evaluated.
				if (closed_ref.find(new_node->toString()) != closed_ref.end())
				{
					delete new_node;
					continue;
				}

				if (open_ref.find(new_node->toString()) != open_ref.end())
				{
					delete new_node;
					continue;
				}

				new_node->parent = closed_node;


				open_set.push_node(new_node);


				open_ref[new_node->toString()] = new_node;


			}

		}
		return;
	}


	vector<string> getSolution()
	{
		return solution;
	}

	// http://www.geeksforgeeks.org/check-instance-8-puzzle-solvable/
	// It is not possible to solve an instance of 8 puzzle if the number of inversions is odd
	// in the initial state
	bool isSolvable()
	{
		int count = 0;
		for (int i = 0; i != initial->tiles.size(); ++i)
			for (int j = i + 1; j != initial->tiles.size(); ++j)
				if (initial->tiles[i] != Board::TILE::TL_SPACE &&
					initial->tiles[j] != Board::TILE::TL_SPACE &&
					static_cast<int>(initial->tiles[i]) > static_cast<int>(initial->tiles[j]))
					++count;
		return (count % 2 == 0);
	}

	~Solver()
	{
		delete initial;
		delete end;
		for (auto iter = open_ref.begin(); iter != open_ref.end(); ++iter)
			delete iter->second;
		for (auto iter = closed_ref.begin(); iter != closed_ref.end(); ++iter)
			delete iter->second;
	}

private:
	Board *initial = nullptr;
	Board *end = new Board();

	bvector open_set;
	unordered_map<string, Board*> open_ref;
	unordered_map<string, Board*> closed_ref;
	vector<string> solution;
};

int main()
{
	int t1[3][3] = { { 1, 2, 3 },{ 4, 5, 6 },{ 8, 7, 0 } };
	int t2[3][3] = { { 8, 1, 3 },{ 4, 0, 2 },{ 7, 6, 5 } };

	Board *b1 = new Board(t1);
	Board *b2 = new Board(t2);
	
	Solver s1(b1);
	Solver s2(b2);
	
	s1.init(); 
	
	s2.init();
	auto ret2 = s2.getSolution();
	cout << "The solution to " << b1->toString() << endl;
	for (auto str : ret2)
		cout << str << endl;
	system("PAUSE");
	return 0;
}


include using namespace std; struct node{ int nodesun[4][4]; int pre; //上一步在队列中的位置 int flag ; //步数标识,表示当前的步数为有效的 int value; //与目标的差距 int x,y; //空格坐标 }queue[1000]; //移动方向数组 int zx[4]={-1,0,1,0}; int zy[4]={0,-1,0,1}; //当前步数 int top; int desti[4][4];//目标状态 int detect(struct node *p)//检查是否找到 {int i,j; for(i=1;i<4;i++) for(j=1;jnodesun[i][j]!=desti[i][j]) return 0; return 1; } //打印 void printlj() {int tempt; int i,j; tempt=top; while(tempt!=0) { for(i=1;i<4;i++) for(j=1;j<4;j++) {cout<<queue[tempt].nodesun[i][j]; if(j==3) cout<<" "<<endl; } tempt=queue[tempt].pre; } } //现在状态与目标状态有多少个不同位置 int VALUE(struct node *p) {int count=0; int i,j; for(i=1;i<4;i++) for(j=1;jnodesun[i][j]!=desti[i][j]) count++; return count; } void main() { //初始化 int i,j,m,n,f; int min=10; int temp,find=0,minnumber; top=1; for(i=1;i<4;i++) for(j=1;j<4;j++) {cout<<"请输入第"<<i<<"行"<<"第"<<j<<"列的值"<>temp; queue[1].nodesun[i][j]=temp; } cout<<"请输入初始状态的空格的位置(行)"<>temp; queue[1].x=temp; cout<<"请输入初始状态的空格的位置(列)"<>temp; queue[1].y=temp; queue[1].value=VALUE(&queue[1]); queue[1].pre=0; //上一步在队列中的位置 queue[1].flag=0; //目标状态 for(i=1;i<4;i++) for(j=1;j<4;j++) {cout<<"请输入目标状态第"<<i<<"行"<<"第"<<j<<"列的值"<>temp; desti[i][j]=temp; } //根据估价函数 while(!find&&top>0) { for(i=1;i<=top;i++) //////////////////////////////////////////// //min为上一图中与目标图有多少个元素不相同,queue[i]为当前图与目标图有多少个元素不相同通过这两个数的比较,就可以得出当前图较之上一图向目标图接近同时把当前的i记录下来进行下一步比较 {if(queue[i].value<min&&queue[i].flag==0) {minnumber=i;// min=queue[i].value; //还有多少不同的位数 } } queue[minnumber].flag=1; //表示此位有效 ////////////////////////////////////// // for(f=0;f=1&&i=1&&j<=3) {top++; ///////////////////////////////////////////// //位置交换 queue[top]=queue[minnumber]; queue[top].nodesun[m][n]=queue[minnumber].nodesun[i][j]; queue[top].nodesun[i][j]=0; /////////////////////////////////////// //空格移动方向 queue[top].x=i; queue[top].y=j; /////////////////////////////////////// queue[top].pre=minnumber; //上一步在队列中的位置 queue[top].value=VALUE(&queue[top]); //有多少位与目标不同 queue[top].flag=0; //标识位初始化 if(detect(&queue[top])) //检查是否为目标 {printlj(); //打印 find=1; //设找到标识位 break; } } } } }
尽管提供的引用并未涉及 A* 算法或 8 数码问题的具体内容,但可以基于专业知识提供详细的解答。 ### A* Algorithm Implementation for the 8-Puzzle Problem #### Overview of the 8-Puzzle Problem The 8-puzzle is a sliding puzzle that consists of a frame of numbered square tiles in random order with one tile missing. The goal is to rearrange the tiles into their correct positions by making sliding moves that use the empty space. To solve this problem using the **A*** algorithm, it's necessary to define an appropriate heuristic function and implement the search process efficiently[^2]. --- #### Key Concepts Behind A* 1. **Heuristic Function**: This estimates the cost from the current state to the goal state. - Common heuristics include Manhattan distance (sum of horizontal and vertical distances between each tile’s current position and its target position). 2. **Evaluation Function**: Combines both actual path cost \( g(n) \) and estimated remaining cost \( h(n) \). - Formula: \( f(n) = g(n) + h(n) \), where: - \( g(n) \): Cost to reach node \( n \) from start. - \( h(n) \): Estimated cost from \( n \) to goal. 3. **Priority Queue**: Used to select nodes based on minimal evaluation value \( f(n) \). --- #### Python Code Example Below is a complete implementation of the A* algorithm tailored for solving the 8-puzzle: ```python import heapq from copy import deepcopy def manhattan_distance(state): """Calculate the total Manhattan distance as the heuristic.""" goal_state = [[0, 1, 2], [3, 4, 5], [6, 7, 8]] dist = 0 for i in range(3): for j in range(3): if state[i][j] != 0: # Ignore blank tile val = state[i][j] target_x, target_y = divmod(val, 3) dist += abs(i - target_x) + abs(j - target_y) return dist def get_neighbors(state): """Generate all possible next states by moving the blank tile.""" neighbors = [] zero_i, zero_j = [(i, j) for i in range(3) for j in range(3) if state[i][j] == 0][0] directions = [ (-1, 0), (1, 0), (0, -1), (0, 1) # Up, Down, Left, Right ] for di, dj in directions: ni, nj = zero_i + di, zero_j + dj if 0 <= ni < 3 and 0 <= nj < 3: new_state = deepcopy(state) new_state[zero_i][zero_j], new_state[ni][nj] = new_state[ni][nj], new_state[zero_i][zero_j] neighbors.append(new_state) return neighbors def astar_8puzzle(initial_state, goal_state=[[0, 1, 2], [3, 4, 5], [6, 7, 8]]): """Solve the 8-puzzle using A* algorithm.""" open_set = [] # Priority queue closed_set = set() initial_tuple = tuple(map(tuple, initial_state)) # Convert list-of-lists to hashable type heapq.heappush(open_set, (manhattan_distance(initial_state), 0, initial_tuple)) came_from = {initial_tuple: None} g_score = {initial_tuple: 0} while open_set: _, current_g, current_tup = heapq.heappop(open_set) current_state = [list(row) for row in current_tup] if current_state == goal_state: break closed_set.add(current_tup) for neighbor in get_neighbors(current_state): neighbor_tup = tuple(map(tuple, neighbor)) tentative_g = current_g + 1 if neighbor_tup in closed_set or (neighbor_tup in g_score and tentative_g >= g_score[neighbor_tup]): continue came_from[neighbor_tup] = current_tup g_score[neighbor_tup] = tentative_g priority = tentative_g + manhattan_distance(neighbor) heapq.heappush(open_set, (priority, tentative_g, neighbor_tup)) # Reconstruct Path path = [] current_node = tuple(map(tuple, goal_state)) while current_node: path.append([list(row) for row in current_node]) current_node = came_from[current_node] return path[::-1] # Test Case if __name__ == "__main__": initial_state = [[1, 2, 3], [4, 0, 5], [6, 7, 8]] solution_path = astar_8puzzle(initial_state) print(f"Solved! Steps taken: {len(solution_path)-1}") for step in solution_path: print(step) ``` --- #### Explanation of the Code 1. **Manhattan Distance Calculation**: - Measures how far away misplaced tiles are from their final destinations[^3]. 2. **Neighbor Generation**: - Explores valid movements of the blank tile within bounds. 3. **A* Search Logic**: - Uses `heapq` for efficient access to minimum-cost nodes during exploration. - Tracks visited states via `closed_set`. 4. **Path Reconstruction**: - Backtracks through parent pointers stored in `came_from` dictionary after reaching the goal. --- ### Performance Considerations For larger puzzles like the 15-tile variant, memory usage may become prohibitive due to exponential growth in potential states. Optimizations such as iterative deepening or bidirectional searches can mitigate these issues[^4].
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