scala-algorithms-bfs-dfs-astar-workshop

• references
  • https://www.manning.com/books/classic-computer-science-problems-in-java
  • https://medium.com/@nicholas.w.swift/easy-a-star-pathfinding-7e6689c7f7b2

preface

• goal of this workshop:
  • basic understanding of bfs, dfs and astar algorithms
  • the ability to apply theory in practice

dfs

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)

bfs

• 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

astar

• 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

practice

• 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

    "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, , , ,"
    

    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, , , "
    

• 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