曼哈顿距离启发式的 A* 算法 [英] A* Algorithm with Manhattan Distance Heuristic
问题描述
我一直在用 C 语言开发一个 15 拼图解算器.我的代码使用了大量内存，但我遇到了一些问题.
I've been working on a 15-puzzle solver in C. And I had some issues with the huge amounts of memory that my code uses.
我不会发布我的代码，因为它太长了......我已经实现了我正在使用的大部分库，它可能只会给你带来困惑.让我们从基础开始.
I won't be posting my code because it's too long... I've implemented most of the libraries I'm using and it will probably just bring confusion to you. Let's start on the basics.
我现在使用的东西是:(所有这些都是用 C 实现的)
The things that I'm using right now are: (All of them implemented in C)
- 斐波那契堆:
/* Struct for the Fibonacci Heap */
typedef struct _fiboheap {
int size; // Number of nodes in the heap
node min; // Pointer to the minimun element in the Heap
funCompare compare; // Function to compare within nodes in the heap
freeFunction freeFun;// Function for freeing nodes in the heap
} *fiboheap;
/* Struct of a Fibonacci Heap Node */
typedef struct _node {
node parent; // Pointer to the node parent
node child; // Pointer to the child node
node left; // Pointer to the right sibling
node right; // Pointer to the left sibling
int degree; // Degree of the node
int mark; // Mark
void *key; // Value of the node (Here is the element)
} *node;
- 哈希表
唯一我自己没有实现的东西，我使用的是uthash.
The only thing I haven't implemented myself, I'm using uthash.
- 15-Puzzle States 代表
这是一个有趣的话题.在解释这里的情况之前，让我们先想一想 15-puzzle 问题......众所周知，15-puzzle 有 16 个瓷砖(我们'将空白瓷砖重新计算为编号为 0 的瓷砖).那么，有多少种可能的状态有 15 谜题呢?嗯，这是阶乘(16)状态.所以，这是很多状态.这意味着我们希望我们的状态尽可能小……如果我们的初始状态离解太远，我们可能会探索太多的状态，以至于我们的程序内存会爆炸.
This is an interesting topic. Before explaining the situation here, let's just think a little bit about the 15-puzzle problem... As we know, 15-Puzzle has 16 tiles (We're counting the blank tile as the tile with the number 0). So, how many possible states has the 15-puzzle problem? Well, it's factorial(16) states. So, that's a lot of states. That means that we want our states to be as small as possible... If our initial state is too far away from the solution, we may explore so much states that our program memory will just explode.
所以...我的 15 拼图表示包括使用位和逻辑运算符:
So... my 15-puzzle representation consist of using bits and logical operators:
/* Struct for the 15-puzzle States Representation */
typedef struct _state {
unsigned int quad_1; /* This int represent the first 8 numbers */
unsigned int quad_2; /* This int represent the last 8 numbers */
unsigned short zero; /* This is the position of the zero */
} *state;
所以我正在做的是使用逻辑运算符来移动和更改数字，使用最小的空间.
So what I'm doing is using logical operators to move and change the numbers, using the minimum space.
注意这个结构体的大小是12个字节(因为它有三个整数)
Note that the size of this struct is 12 bytes (Because it has three integers)
- 曼哈顿距离
这就是著名的曼哈顿距离启发式算法.这里基本上就是计算每个号码当前位置到目标状态下号码位置的距离之和.
This is just the well known Manhattan Distance Heuristic. Where you basically calculate the sum of the distances of each number current position to the number position in the goal state.
- A* 实现和与 A* 一起使用的节点结构
让我们从节点开始.
typedef struct nodo_struct {
nodo parent; // Pointer to the parent of the node
state estado; // State associated with the node
unsigned int g : 8; // Cost of the node
action a; // Action by which we arrived to this node
// If null, it means that this is the initial node
} *nodo;
注意这些节点与斐波那契堆中的节点无关.
Note that these nodes had nothing to do with the nodes in fibonacci heap.
现在这个话题的主要原因......我目前使用的A*伪代码.
And now the main reason of this topic... The pseudocode of A* I'm currently using.
a_star(state initial_state) {
q = new_fibo_heap; // Sorted by (Cost g) + (Manhattan Distance)
// It will have nodes which contain a pointer to the state
q.push(make_root_node(initial_state));
closed = new_hash_table();
while (!q.empty()) {
n = q.pop();
if ((n->state ∉ closed) || (n->g < dist(n->state))) {
/* The dist used above is stored in the hash table as well */
closed.insert(n->state);
dist(n->state) = n->g; // Update the hash table
if (is_goal(n->state)) {
return extract_solution(n); // Go through parents, return the path
}
for <a,s> ∈ successors(n->state) {
// 'a' is the action, It can be 'l', 'r', 'u', 'd'
// 's' is the state that is a successor of n
if (manhattan(s) < Infinity) {
q.push(make_node(n,a,s));
// Assuming that manhattan is safe
}
}
}
}
return NULL;
}
所以我还不能回答的问题是...
So the questions I haven't been able to answer yet are...
你如何有效地管理内存?你能重用状态或节点吗?这会带来什么后果?
我的意思是，如果你看一下伪代码.它不考虑重用状态或节点.它只是不断地分配节点和状态，即使它们之前已经计算过.
I mean, If you take a look at the pseudocode. It's not considering re-using states or nodes. It just keeps allocating over and over nodes and states, even though they've been calculated before.
我一直在思考这个问题……每次运行求解器时，它都可以非常快速地扩展数百万个节点.正如我们所知，当您找到另一条成本低于前一条路径的路径时，A* 可能会重新探索节点.这意味着......如果我们探索 100 万个节点，它将是 2400 万字节(哇)......考虑到每个节点都有一个状态，每个节点将是 14 个字节，这些是 1400 万字节...
I've been thinking a lot about this... Each time you run the solver, it can expand millions of nodes real quick. And as we know, A* may re-explore nodes when you find another path with a cost less than the previous one. That means... If we explore 1 million nodes, it will be 24 millions of bytes (Wow)... And considering that each node has a state, that would be 14 bytes per node, those are 14 millions of bytes...
最后，我需要的是一种重用/释放空间的方法，这样当我执行这个求解器时我的计算机就不会崩溃.
(PD:抱歉长篇幅)
推荐答案
你这样做是为了作业还是为了好玩?
Are you doing this for an assignment or for fun?
如果是为了好玩，那就不要使用 A*，使用 IDA*.IDA* 会更快，并且几乎不使用内存.此外，您可以使用额外的内存来构建更好的启发式方法，并获得更好的性能.(如果您在 google 上搜索模式数据库"，您应该会找到足够的信息.这些信息是由 Jonathan Schaeffer 和 Joe Culberson 发明的，但 Rich Korf 和 Ariel Felner 进行了重要的详细研究.)
If it's for fun, then don't use A*, use IDA*. IDA* will be faster and it uses almost no memory. Furthermore, you can use the extra memory to build better heuristics, and get even better performance. (You should find sufficient information if you google "pattern database". These were invented by Jonathan Schaeffer and Joe Culberson, but studied in significant detail by Rich Korf and Ariel Felner.)
IDA* 有一些缺点，并不适用于所有领域，但它对于滑动瓷砖拼图来说几乎是完美的.
IDA* has some drawbacks and doesn't work in every domain, but it is just about perfect for the sliding-tile puzzle.
另一种可以提供帮助的算法是广度优先启发式搜索.这篇论文和其他论文讨论了如何避免完全存储封闭列表.
Another algorithm which could help is Breadth-First Heuristic Search. This and other papers discuss how you can avoid storing the closed list entirely.
基本上，很多聪明人之前已经解决了这个问题，并且他们已经发布了他们的方法/结果，以便您可以向他们学习.
Basically, a lot of smart people have tackled this problem before and they've published their methods/results so you can learn from them.
这里有一些提高 A* 的技巧:
Here are some tips to improve your A*:
我发现斐波那契堆没有太大的速度增益，因此您可能想要使用更简单的数据结构.(尽管自从我上次尝试后，可用的实现可能有所改进.)
I have found that there isn't much of a speed gain from Fibonacci heaps, so you might want to use a simpler data structure. (Although available implementations might have improved since I tried this last.)
节点的 f-cost 会以 2 为增量跳跃.因此，您可以对 f-cost 进行分桶，而只需担心对同一 f-cost 层中的项目进行排序.FIFO 队列实际上工作得很好.
The f-cost of a node will jump in increments of 2. Thus, you can bucket your f-costs and only worry about sorting items in the same f-cost layer. A FIFO queue actually works pretty well.
您可以使用本文中的想法 将 15-puzzle 表示转换为排列，完整表示需要大约 43 位.但是，扩展状态会变得更加昂贵，因为您必须转换为不同的表示来生成移动.
You can use the ideas in this paper to convert the 15-puzzle representation into a permutation, which will take about 43 bits for the full representation. But, it becomes more expensive to expand states because you have to convert into a different representation to generate moves.
避免完全使用禁止操作符来存储封闭列表.(参见之前的广度优先启发式搜索论文或这篇关于 Best-First Frontier search 的论文 了解更多详情.)
Avoid storing the closed list entirely using forbidden operators. (See the previous Breadth-First Heuristic Search paper or this paper on Best-First Frontier search for more details.)
希望这些要点能够解决您的问题.如果您需要，我很乐意提供更多链接或说明.
Hopefully these points will address your issues. I'm happy to provide more links or clarification if you need them.
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