a * search algorithm added

src/algorithms/search/a-star searching/A Star Searching Algorithm.js

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+var astar = {
+    init: function(grid) {
+        for(var x = , xl = grid.length; x < xl; x++) {
+            for(var y = , yl = grid[x].length; y < yl; y++) {
+                var node = grid[x][y];
+                node.f = ;
+                node.g = ;
+                node.h = ;
+                node.cost = 1;
+                node.visited = false;
+                node.closed = false;
+                node.parent = null;
+            }
+        }
+    },
+    heap: function() {
+        return new BinaryHeap(function(node) {
+            return node.f;
+        });
+    },
+    search: function(grid, start, end, diagonal, heuristic) {
+        astar.init(grid);
+        heuristic = heuristic || astar.manhattan;
+        diagonal = !!diagonal;
+
+        var openHeap = astar.heap();
+
+        openHeap.push(start);
+
+        while(openHeap.size() > ) {
+
+            // 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;
+
+            // Find all neighbors for the current node. Optionally find diagonal neighbors as well (false by default).
+            var neighbors = astar.neighbors(grid, currentNode, diagonal);
+
+            for(var i=, il = neighbors.length; i < il; i++) {
+                var neighbor = neighbors[i];
+
+                if(neighbor.closed || neighbor.isWall()) {
+                    // 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.
+                var gScore = currentNode.g + neighbor.cost;
+                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, diagonals) {
+        var ret = [];
+        var x = node.x;
+        var y = node.y;
+
+        // West
+        if(grid[x-1] && grid[x-1][y]) {
+            ret.push(grid[x-1][y]);
+        }
+
+        // East
+        if(grid[x+1] && grid[x+1][y]) {
+            ret.push(grid[x+1][y]);
+        }
+
+        // South
+        if(grid[x] && grid[x][y-1]) {
+            ret.push(grid[x][y-1]);
+        }
+
+        // North
+        if(grid[x] && grid[x][y+1]) {
+            ret.push(grid[x][y+1]);
+        }
+
+        if (diagonals) {
+
+            // Southwest
+            if(grid[x-1] && grid[x-1][y-1]) {
+                ret.push(grid[x-1][y-1]);
+            }
+
+            // Southeast
+            if(grid[x+1] && grid[x+1][y-1]) {
+                ret.push(grid[x+1][y-1]);
+            }
+
+            // Northwest
+            if(grid[x-1] && grid[x-1][y+1]) {
+                ret.push(grid[x-1][y+1]);
+            }
+
+            // Northeast
+            if(grid[x+1] && grid[x+1][y+1]) {
+                ret.push(grid[x+1][y+1]);
+            }
+
+        }
+
+        return ret;
+    }
+};

src/algorithms/search/a-star searching/README.md

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+This is the actual implementation of the algorithm. I will do my best to explain what is going on, but feel free to just look at the source of the example, or just download astar.js.
+
+There are three functions that we keep track of for nodes that we look at:
+
+g(x): The total cost of getting to that node (pretty straightforward). If we reach a node for the first time or reach a node in less time than it currently took, then update the g(x) to the cost to reach this node.
+h(x): The estimated time to reach the finish from the current node. This is also called a heuristic. We online need to update this if it is not set already, since the distance to the finish will not change even if the path we took to arrive at a node changes. Note: There are many different ways to guess how far you are from the end, I use the Manhattan distance in this implementation.
+f(x): Simply g(x) + h(x). The lower the f(x), the better. Think about it like this: the best node is one that takes the least total amount of time to arrive at and to get to the end. So, a node that took only 1 step to arrive at and 5 to get to the end is more ideal than one that took 10 to arrive and and only 1 to get to the end.
+
+
+
+
+
+PsuedoCode:
+push startNode onto openList
+while(openList is not empty) {
+ currentNode = find lowest f in openList
+ if currentNode is final, return the successful path
+ push currentNode onto closedList and remove from openList
+ foreach neighbor of currentNode {
+     if neighbor is not in openList {
+            save g, h, and f then save the current parent
+            add neighbor to openList
+     }
+     if neighbor is in openList but the current g is better than previous g {
+             save g and f, then save the current parent
+     }
+ }