astar-list.js

/* 	astar-list.js http://github.com/bgrins/javascript-astar
	MIT License
	
	** You should not use this implementation (it is quite slower than the heap implementation) **
	
	Implements the astar search algorithm in javascript
	Based off the original blog post http://www.briangrinstead.com/blog/astar-search-algorithm-in-javascript
	It has since been replaced with astar.js which uses a Binary Heap and is quite faster, but I am leaving
	it here since it is more strictly following pseudocode for the Astar search
	**Requires graph.js**
*/

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

        var openList   = [];
        openList.push(start);

        while(openList.length > 0) {

        	// Grab the lowest f(x) to process next
            var lowInd = 0;
            for(var i=0; i<openList.length; i++) {
			    if(openList[i].f < openList[lowInd].f) { lowInd = i; }
		    }
		    var currentNode = openList[lowInd];

		    // 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
		    openList.remove(lowInd);
		    currentNode.closed = true;

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

			    if(neighbor.closed || neighbor.isWall()) {
				    // not a valid node to process, skip to next neighbor
				    continue;
			    }

			    // 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
			    var gScore = currentNode.g + 1; // 1 is the distance from a node to it's neighbor
			    var gScoreIsBest = false;

			    if(!neighbor.visited) {
				    // This the the first time we have arrived at this node, it must be the best
				    // Also, we need to take the h (heuristic) score since we haven't done so yet

				    gScoreIsBest = true;
				    neighbor.h = heuristic(neighbor.pos, end.pos);
				    neighbor.visited = true;
				    openList.push(neighbor);
			    }
			    else if(gScore < neighbor.g) {
				    // We have already seen the node, but last time it had a worse g (distance from start)
				    gScoreIsBest = true;
			    }

			    if(gScoreIsBest) {
				    // Found an optimal (so far) path to this node.  Store info on how we got here and
				    //  just how good it really is...
				    neighbor.parent = currentNode;
				    neighbor.g = gScore;
				    neighbor.f = neighbor.g + neighbor.h;
				    neighbor.debug = "F: " + neighbor.f + "<br />G: " + neighbor.g + "<br />H: " + neighbor.h;
				}
		    }
        }

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