# Talk:A* search algorithm

## moving into a barrier position

How does a path   move into   a barrier position?

I see that none of the programming examples show a path (so far) that "moves into" a barrier,
else we'd be seeing a total cost for a path that exceeds   100.

Does there need to be a cost when "moving into" a barrier, since no path has (apparently) done that?

How would one show a path that   moves into   a barrier?   -- Gerard Schildberger (talk) 16:42, 29 January 2017 (UTC)

While A* does not evaluate every possible move, it does internally check the cost of moving into a barrier square. For this reason, the cost of moving into a barrier square is required. However, there is always a lower cost alternative while still moving in the correct general direction (according to the heuristic), it should never actually be part of the maze solution.
As you pointed out, the method for showing a path that moves into a barrier is left undefined. However, it should not be part of the a* solution so any method would be fine.--TimSC (talk) 16:58, 29 January 2017 (UTC)
So if a "move into" a barrier is never shown (because there is always a lower cost solution), why have a cost (for moving into a barrier) at all?   -- Gerard Schildberger (talk) 18:48, 29 January 2017 (UTC)
In terms of understanding, there are two approaches where the "move into barrier" may lead:
1. The barrier is a soft restriction. This means one may consider the barrier to find the path, and it will be considered when there are no paths without barrier move. In terms of coding, this places the barrier move in the queue with the least priority by assigning a huge cost. Then, at the first time a path successfully is found, the high cost will indicate a barrier was crossed.
2. The barrier is a hard restriction. This means when there are no valid paths, the algorithm will return an empty path. In terms of coding, when the move puts a barrier, the move will not be added to the queue. Then, as the priority queue is popped, if there aren't any valid paths, the queue will empty and the A* shall raise the problem.

Also (above), you mentioned that:   A* doesn't evaluate every possible move.

However, in the Wikipedia article (link), it states:

A* is an informed search algorithm, or a best-first search, meaning that it solves problems by searching among all possible paths to the solution (goal) for the one that incurs the smallest cost (least distance travelled (sic), shortest time, etc.), and among these paths it first considers the ones that appear to lead most quickly to the solution.   ...     (underscoring and italics added by me).   -- Gerard Schildberger (talk) 18:59, 29 January 2017 (UTC)
A* does not need to evaluate every possible move if two things happen:
1. h(x) is always lower than the actual remaining cost.
2. h(x_1) < h(x_2) + d(x_1, x_2).

## grid orientation

It would seem that some programming examples are using a non-standard orientation of the
grid display,   with the   (0,0)   origin point in the   top-left   of the display area,   with positive
values for   X   (columns)   going downward,   instead of   going upward.

For me, it doesn't make me no never-mind no-how, but it took a wee bit of fixin' for my
programming example   (not yet posted)   to match the existing displayed grids.   -- Gerard Schildberger (talk) 16:42, 29 January 2017 (UTC)

In my experience, matrix representation typically has 0,0 in the upper left hand corner. Opengl typically puts 0,0 in the lower left corner or the center, depending on context. Presumably we also have other standards which might be relevant here. But unless we state which standard we're working with... "the nice thing about standards is that there's so many to choose from". --Rdm (talk) 16:09, 17 January 2022 (UTC)

## A more interesting example

A barrier is created by setting a square's value effectively to infinity. How this is achieved by the algorithm should be implementation dependant. Some languages support the concept intrinsically, certainly 100 should not be a magic number. Would it not be more interesting if the squares had values other than 1 or infinity. Say randomly assigned?--Nigel Galloway (talk) 10:33, 30 January 2017 (UTC)

How is this algorithm related to solving a Hidato Puzzle? or the others for that matter. Does anyone intend to produce a solution using this algorithm?--Nigel Galloway (talk) 10:33, 30 January 2017 (UTC)