A * Отслеживание в гексагональной сетке

Может ли кто-нибудь указать мне на простой пример, который реализует алгоритм поиска путей * в сетке гексагональной в JS). Я заставил его работать над квадратной сеткой, однако все мои попытки заставить его работать на шестиугольной сетке потерпели неудачу.

Вот как выглядит моя сетка:

Я использую ту же технику, чтобы как рисовать сетку, так и генерировать координаты, как показано в этой теме.

Здесь данные сетчатых коордов вместе с начальными, конечными коордами:

        [0, 0] , [0, 1],  [0, 2],
    [1, 0],  [1, 1],  [1, 2],  [1, 3],
 [2, 0],  [2, 1],  [2, 2],  [2, 3],  [2, 4],
    [3, 0],  [3, 1], [3, 2],  [3, 3], 
        [4, 0], [4, 1],  [4, 2]


start_point: [0,2]
end_point: [4.0]

После обновления расчета расстояния в манхэттен до:

var dx = pos1[0] - pos0[0];
    var dy = pos1[1] - pos0[1];

    var dist;
    if ( Math.sign(dx) == Math.sign(dy) ){
        dist = Math.abs (dx + dy);
    }else{
        dist = Math.max(Math.abs(dx), Math.abs(dy))
    }

return dist;

Получаю этот результат:

а также способ вычисления кратчайшего пути:

if (!Array.prototype.remove) {
    Array.prototype.remove = function(from, to) {
        var rest = this.slice((to || from) + 1 || this.length);
        this.length = from < 0 ? this.length + from : from;
        return this.push.apply(this, rest);
    };
}

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].content = false;
                grid[x][y].visited = false;
                grid[x][y].closed = false;
                grid[x][y].debug = "";
                grid[x][y].parent = null;
				console.log([grid[x][y].coords[0],grid[x][y].coords[1]])
            }
        }
    },
    search: function(grid, start, end, heuristic) {
        this.init(grid);
        heuristic = heuristic || this.manhattan;

        var openList = [];
		
		//// find the start and end points in the grid ////
		start = grid[start.pos[0]][start.pos[1]];
		end =  grid[end.pos[0]][end.pos[1]];
		
		console.log( start, end )
		
        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 = this.neighbors(grid, currentNode);
            for(var i=0; i<neighbors.length;i++) {
                var neighbor = neighbors[i];

                if( neighbor.closed || neighbor.content == 2 ) { // 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 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.coords, end.coords);
                    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) { //// heuristics : use manhattan distances  ////
        var dx = pos1[0] - pos0[0];
        var dy = pos1[1] - pos0[1];
		
        return  Math.abs (dx + dy);
    },
    neighbors: function(grid, node) {
        var ret = [];
        var x = node.coords[0];
        var y = node.coords[1];
		
        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;
    }
};