// javascript-astar
// http://github.com/bgrins/javascript-astar
// Freely distributable under the MIT License.
// Implements the astar search algorithm in javascript using a binary heap.

var astar = {
    init: function(grid) {
        for(var x = 0, xl = grid.length; x < xl; x++) {
            for(var y = 0, yl = grid[x].length; y < yl; y++) {
                var node = grid[x][y];
                node.f = 0;
                node.g = 0;
                node.h = 0;
                node.visited = false;
                node.closed = false;
                node.parent = null;
            }
        }
    },
    heap: function() {
        return new BinaryHeap(function(node) {
            return node.f;
        });
    },
    search: function(grid, start, end, heuristic) {
        astar.init(grid);
        heuristic = heuristic || astar.manhattan;

        var openHeap = astar.heap();

        openHeap.push(start);

        while(openHeap.size() > 0) {

            // Grab the lowest f(x) to process next.  Heap keeps this sorted for us.
            var currentNode = openHeap.pop();

            // End case -- result has been found, return the traced path.
            if(currentNode === end) {
                var curr = currentNode;
                var ret = [];
                while(curr.parent) {
                    ret.push(curr);
                    curr = curr.parent;
                }
                return ret.reverse();
            }

            // Normal case -- move currentNode from open to closed, process each of its neighbors.
            currentNode.closed = true;

            var neighbors = astar.neighbors(grid, currentNode);
            for(var i=0, il = neighbors.length; i < il; i++) {
                var neighbor = neighbors[i];

            var getCoords = new String(neighbor);
            var finalCoords = getCoords.split(",");
            var finalX = finalCoords[0];
            var finalY = finalCoords[1];

                if(neighbor.closed || !jQuery.inArray(board[finalY][finalX], blocked_tiles)) {
                    // Not a valid node to process, skip to next neighbor.


                    continue;
                }

                // The g score is the shortest distance from start to current node.
                // We need to check if the path we have arrived at this neighbor is the shortest one we have seen yet.
                // 1 is the distance from a node to it's neighbor - this could be variable for weighted paths.
                var gScore = currentNode.g + 1;
                var beenVisited = neighbor.visited;

                if(!beenVisited || gScore < neighbor.g) {

                    // Found an optimal (so far) path to this node.  Take score for node to see how good it is.
                    neighbor.visited = true;
                    neighbor.parent = currentNode;
                    neighbor.h = neighbor.h || heuristic(neighbor.pos, end.pos);
                    neighbor.g = gScore;
                    neighbor.f = neighbor.g + neighbor.h;

                    if (!beenVisited) {
                        // Pushing to heap will put it in proper place based on the 'f' value.
                        openHeap.push(neighbor);
                    }
                    else {
                        // Already seen the node, but since it has been rescored we need to reorder it in the heap
                        openHeap.rescoreElement(neighbor);
                    }
                }
            }
        }

        // No result was found - empty array signifies failure to find path.
        return [];
    },
    manhattan: function(pos0, pos1) {
        // See list of heuristics: http://theory.stanford.edu/~amitp/GameProgramming/Heuristics.html

        var d1 = Math.abs (pos1.x - pos0.x);
        var d2 = Math.abs (pos1.y - pos0.y);
        return d1 + d2;
    },
    neighbors: function(grid, node) {
        var ret = [];
        var x = node.x;
        var y = node.y;

        if(grid[x-1] && grid[x-1][y]) {
            ret.push(grid[x-1][y]);
        }
        if(grid[x+1] && grid[x+1][y]) {
            ret.push(grid[x+1][y]);
        }
        if(grid[x] && grid[x][y-1]) {
            ret.push(grid[x][y-1]);
        }
        if(grid[x] && grid[x][y+1]) {
            ret.push(grid[x][y+1]);
        }
        return ret;
    }
};