Question

I am implementing A* algorithm in C++ to solve the n-puzzle problem.
I tried to implement the pseudocode in this link.
Total cost(F=H+G) calculation is “cost depends on the number of misplaced tiles (Heuristics) + steps from initial state (G)”. The algorithm of the AStar function given below.

The problem is, I am having an infinite loop situation. How can I solve this?

PS: I can give the implementations of the other functions used in AStar, if requested.

Any help would be appreciated.

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

Because of the possibility of loops, i.e. a sequence of moves that takes you back to a state you’ve already visited, it’s important to check for duplicate states (not a problem with tree search, obviously). I can’t quite follow what you’re doing with your checking for this, but that’s likely to be where the problem lies. I had a similar sort of problem to you when writing a Haskell implementation (details here and here), and it came down to a problem of handling the explored set. Get that right, and everything works. (Getting solutions for the 4×4 puzzle remains a bit of a hit-and-miss affair, especially if you start from a state a long way from the goal in state space, but that’s mostly down to the deficiencies of A* and the naïve way we’re dealing with possible moves.)