Shortest path A* f(n) = g(n) + h(n)

When you use A*, it chooses the best node which is the closest to the goal right? ( using f(n) = g(n) + h(n)) ( using Manhattan distance for h(n) )

But what if there are walls between the start and goal. I can't explain it in words but I'll show a picture.

_{(source: uploadir.com)}

If A* chooses the node that is closest to the goal, why is the path not the one encircled in red? But the one encircled in green. I really don't understand A* especially when there are impassable cells/tiles/nodes/etc. (walls). Also, you can see this picture I made like in this video http://www.youtube.com/watch?v=DINCL5cd_w0 (Path Finding Algorithms (A*, Dijkstra, Bi-Directional BFS)) at 1:20

A* also considers the length of the current path as well as the distance to the goal. All possible paths are expanded, but with a priority on those which are:

1. Closest to the goal
2. Shortest

The path cost, f(n) is equal to the cost at the previous step, g(n), plus a factor based on distance to the goal, h(n). So for every extra grid space the path goes through, the cost of the path increases. This will effectively create a balance between short paths, and paths which move directly to the goal.

So by the time your example paths have joined up again, it is the purple path which is shortest, and so that will be the path which is expanded first, and will eventually reach the goal first.

There is a Udacity course: Programming a Robotic Car which has a good section on the A* (and similar) algorithms.