用C++的A*算法求解n个难题

Question

我正在C++中实现C++，以解决n难题.

我试图在这链接中实现伪代码。

总成本(F=H+G)计算是“成本取决于错误放置的瓷砖数量(启发式)+从初始状态(G)开始的步骤”。给出了AStar函数的算法。

问题是，我有一个无限循环的情况。我怎么才能解决这个问题？

PS:如果需要的话，我可以给出AStar中使用的其他函数的实现。

任何帮助都将不胜感激。

void AStar(const int size, int** puzzle)
{
int moveCount = 0;                                                                  // initialize G(n)
int**goalState = GoalState(size);                                                   // initialize  and assign goal state
int openListIndex = 0;                                                              // initialize open list index
vector<node> openList;                                                              // initialize open list
vector<node> closedList;                                                            // initialize closed list

node startNode;                                                                     // initialize start node
startNode.puzzleArray = puzzle;                                                     // assign start node's state
startNode.cost = moveCount + Heuristics(goalState,puzzle,size);                     // assign start node's cost

node goalNode;                                                                      // initialize goal node
goalNode.puzzleArray = goalState;                                                   // assign goal node's state

openList.push_back(startNode);                                                      // push start node to the open list

while (!openList.empty())                                                           // loop while open list is not empty
{
    node currentNode = CalculateLowestCost(&openList, &closedList);                 // initialize current node which has the lowest cost, pop it from open list, push it to the closed list
    int** currentState = currentNode.puzzleArray;                                   // initialize and assign current state array
    /*********************************************************************************************/
    if (GoalCheck(goalState, currentState, size)) break;                            // GOAL CHECK//
    /*********************************************************************************************/
    vector<char> successorDirectionList = CalculateSuccessor(size, currentState);   // initialize a char vector for the directions of the successors

    int**successor;                                                                 // initialize successor state
    node successorNode;                                                             // initialize successor node
    moveCount++;                                                                    // advance G(n)

    for (;!successorDirectionList.empty();)                                         // loop over the successor list
    {
        char direction = successorDirectionList.back();                             // take a direction from the list
        successorDirectionList.pop_back();                                          // remove that direction from the list
        successor = MoveBlank(currentState, size, direction);                       // assign successor state

        successorNode.puzzleArray = successor;                                      // assign successor node's state
        successorNode.cost = moveCount + Heuristics(goalState,currentState,size);   // assign successor node's cost

        //vector<node> stateCheckList = openList;                                   // copy the open list for the checking the nodes in that list

        bool flagOpen = false;
        bool flagClosed = false;
        int locationOpen = -1;
        int locationClosed = -1;

        for (int i=0; i<openList.size(); i++)
        {
            int** existing = openList[i].puzzleArray;
            int existingCost = openList[i].cost;

            if (StateCheck(successor, existing, size))
            {
                locationOpen = i;
                if (successorNode.cost > existingCost)
                {
                    flagOpen = true;
                    break;
                }
            }
        }
        if (flagOpen) continue;

        int** existingInOpen;
        if(locationOpen != -1) 
        {
            existingInOpen = openList[locationOpen].puzzleArray;
            openList.erase(openList.begin()+locationOpen);
        }

        for (int i=0; i<closedList.size(); i++)
        {
            int** existing = closedList[i].puzzleArray;
            int existingCost = closedList[i].cost;

            if (StateCheck(successor, existing, size))
            {
                locationClosed = i;
                if (successorNode.cost > existingCost)
                {
                    flagClosed = true;
                    break;
                }
            }
        }
        if (flagClosed) continue;

        int**existingInClosed;
        if(locationClosed != -1)
        {
            existingInClosed = closedList[locationClosed].puzzleArray;
            closedList.erase(closedList.begin()+locationClosed);
        }

        openList.push_back(successorNode);
    }    
}

}

Answer 1

由于循环的可能性，也就是将您带回到您已经访问过的状态的一系列移动，所以检查重复状态是很重要的(显然，树搜索不是问题)。我不太明白你在做什么检查这个，但这很可能是问题所在。在编写Haskell实现(details 这里和这里)时，我遇到了类似的问题，问题归结为处理探索集的问题。做得对，一切都正常。(要解决4x4的难题仍然是一件偶然的事情，特别是如果你从状态空间中的目标开始很长一段时间，但这主要是因为A*的缺陷和我们处理可能发生的动作的天真方式。)

Answer 2

是否从公开列表中删除所选择的状态？这是一个非常简单和编写良好的C#代码，也许它可以帮助您：http://geekbrothers.org/index.php/categories/computer/12-solve-8-puzzle-with-a和A*自动避免循环，因为它采取了您要考虑的步骤

Answer 3

我还实现了一个n字谜(深度优先和*)，它进入无限循环。这是因为下面的循环:上，左，下，右，上，左.我真的不知道是否有办法防止这样的事情(我能做的最大就是防止像左-右-左)这样的循环.通过记忆它最后的动作)。

不知道这是否是导致你循环的原因，最好的方法是在每次迭代中打印板，看看到底发生了什么;)