The AStar library implements the classic A* path-finding algorithm. It uses a min priority queue for managing potential paths, ordered by each
path's known and estimated cost. The AStar class delegates map-related functionality to a `MapOracle`

protocol to determine valid positions
as well as the cost of using a position. Example of a `MapOracle`

can be found in the `AStarTests.swift`

file.

The AStar API is very basic. There is just the static `find`

method. Here is an example of it being used:

```
let mapData = MapData(data: [
[.🌊, .🌲, .🌲, .🌲, .🌲, .🌲, .🌲, .🌲],
[.🌊, .🌲, .🌲, .🌲, .🌲, .🌲, .🌲, .🌲],
[.🌲, .🌲, .🌲, .🌲, .🗻, .🌲, .🌲, .🌲],
[.🌲, .🌲, .🗻, .🗻, .🗻, .🗻, .🗻, .🌲],
[.🌲, .🌲, .🗻, .🌲, .🌲, .🗻, .🌊, .🌊],
[.🌲, .🌲, .🗻, .🌲, .🗻, .🌲, .🌲, .🌊],
[.🌊, .🌲, .🗻, .🌲, .🌲, .🌲, .🗻, .🗻],
[.🌊, .🌲, .🌲, .🌲, .🌲, .🌲, .🗻, .🌲]
])
let start = Coord2D(x: 4, y: 0)
let end = Coord2D(x: 4, y: 4)
func distanceToEnd(position: Coord2D) -> Int { abs(position.x - end.x) + abs(position.y - end.y) }
let path = AStar.find(mapOracle: mapOracle, considerDiagonalPaths: true,
heuristicCostCalulator: distancToEnd,
start: start, end: end)
```

You supply something that implements the `MapOracle`

protocol like the `MapData`

above. You decide if diagonal paths are acceptable,
and provide a way to estimate the cost of moving from a given point on the map to the end point (the *heuristic cost*). The start and end
points complete the `find`

request.

You get back an optional array of `Coord2D`

values. If `nil`

then there was no path to be found. Otherwise, the array will have the map
coordinates of the path that was found, starting at `start`

and ending with `end`

.

Here is the visual representation of the map with the found path. The starting position appears as a red flag (🚩) and the end position is a checkered flag (🏁). The path in between these two points contains an adventurer (🏃).

```
let image = mapData.asString(path: path!)
print(image)
🌊🌲🌲🌲🚩🌲🌲🌲
🌊🌲🌲🌲🌲🏃🌲🌲
🌲🌲🌲🌲🗻🌲🏃🌲
🌲🌲🗻🗻🗻🗻🗻🏃
🌲🌲🗻🌲🏁🗻🏃🌊
🌲🌲🗻🌲🗻🏃🌲🌊
🌊🌲🗻🌲🌲🌲🗻🗻
🌊🌲🌲🌲🌲🌲🗻🌲
```

The map contains three different terrain elements, each with their own cost for travelling into their square:

- 🌲 tree (1)
- 🌊 water (2)
- 🗻 boulder (∞)

The algorithm minimizes the cost of traveling over terrain elements while at the same time trying to keep to the shortest path to the goal. For comparison, here is what the algorithm found when constrained to not use diagonal moves:

```
🌊🌲🌲🌲🚩🌲🌲🌲
🌊🌲🌲🏃🏃🌲🌲🌲
🌲🏃🏃🏃🗻🌲🌲🌲
🌲🏃🗻🗻🗻🗻🗻🌲
🌲🏃🗻🏃🏁🗻🌊🌊
🌲🏃🗻🏃🗻🌲🌲🌊
🌊🏃🗻🏃🌲🌲🗻🗻
🌊🏃🏃🏃🌲🌲🗻🌲
```

There is another path to the right that is also 16 moves, but it goes over two 🌊 positions which increases the total cost of the trip by 2. Thus the algorithm chose the one shown above.