- references
- goal of this workshop:
- basic understanding of bfs, dfs and astar algorithms
- the ability to apply theory in practice
val frontier = stack
frontier push initialNode
val explored = set
while (frontier.nonEmpty) {
val current = frontier.pop
if (current is goal) return Some(current)
successors(current)
.filter(not in explored)
.foreach(child => {
explored add child
frontier push child
})
}
None
note also that you have some structure to track path
case class Node[T](state: T, parent: Option[Node[T]] = Option.empty)
- just take dfs and use queue instead of stack
- always finds the shortest solution path where one exists
- why?
- try to find path from Cracow to Warsaw with fewest stops (by train)
- dfs: keep going in the same direction and backtrack when a dead end
- route could go through Poznan
- bfs: check all of the stations one stop away from Cracow
- then, two stops away from Cracow and so on
- when you do find Warsaw, you will know you have found the route with the fewest stops
- all of the stations that are fewer stops away from Cracow were checked
- why?
- one important aspect of A* is
f = g + h
- G - distance between the current node and the start node
- H - heuristic (estimated distance from the current node to the end node)
- F - total cost of the node
- we can look at all our nodes and ask: "is this the best node I can pick to move forward?"
- and slightly modified Node class
case class AStarNode[T]( state: T, parent: AStarNode[T] = null, cost: Double = 0.0, heuristic: Double = 0.0 )
- code
val frontier = priorityQueue // by total cost frontier enqueue initial val explored = map // state, cost explored put (initial, 0.0) while (frontier.nonEmpty) { val current = frontier.dequeue() if (current is goal) return Some(current) for (child <- successors(current)) { val newCost = currentNode.cost + 1 if (child is not explored || explored(child) > newCost) { explored put (child, newCost) frontier enqueue child } } } None
explored(child) > newCost
- check if path to that square is better, using G cost as the measure
- a lower G cost means that this is a better path
- task 1
- you have a maze (n x m) with start location (startX, startY), goal location (goalX, goalY) and possible walls
- find path from start to goal
- note that the path could not exist or there could be many solutions
- example
and the solution
"S, , , ,X,X,X,X,X,X," "X,X,X, ,X,X,X,X,X,X," "X,X,X, ,X,X,X,X,X,X," " , , , ,X,X,X,X,X,X," " ,X,X,X,X,X,X,X,X,X," " ,X,X,X,X,X,X,X,X,X," " ,X,X,X,X,X,X,X,X,X," " , , , , ,X,X,X,X,X," "X,X,X,X, , , , , , ," "X,X,X,X,X,X,G, , , ,"
"S,*,*,*,X,X,X,X,X,X" "X,X,X,*,X,X,X,X,X,X" "X,X,X,*,X,X,X,X,X,X" "*,*,*,*,X,X,X,X,X,X" "*,X,X,X,X,X,X,X,X,X" "*,X,X,X,X,X,X,X,X,X" "*,X,X,X,X,X,X,X,X,X" "*,*,*,*,*,X,X,X,X,X" "X,X,X,X,*,*,*, , , " "X,X,X,X,X,X,G, , , "
- task 2
- three policemen and three thieves are on the west bank of a river
- they all must cross to the opposite bank of the river
- they have a riverboat that can hold two people
- there may never be more thieves than policemen
- riverboat must have at least one person on board to cross the river
- return the sequence of crossings