

500,000 clones get you out of a 10x10 maze - dxbydt
https://github.com/krishnanraman/Parallel-Pathfinding

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mkelley
So ... I'm not familiar with path-finding algorithms, but is this the most
efficient way to solve the problem?

~~~
dxbydt
Not by a long shot. An astar will get you out of a 10x10 maze in 10 steps
since this particular maze has no barriers... I asked my wife if she were
trapped in a maze how she would get out. She thought for a while and said...
if I could clone myself I would fill up the whole maze with me...and one of me
would be at the maze door in no time!

I thought that was a pretty radical and creative algorithm... so I coded it up

~~~
carb
So it's more of a 10x10 room than a 10x10 maze?

~~~
dxbydt
adding barriers to make a room into a maze is trivial. So if you were forking
my scala code, you see 4 filters -

    
    
          val res = Seq(x-1,x,x+1).map( a => Seq(y-1,y,y+1).map( b=> (a,b))).flatten
          .filterNot( ab => ab._1 == x && ab._2 == y)  // don't include me
          .filterNot( ab=> ab._1 < leftTop._1 || ab._1 > rightBottom._1) // don't include points outside the maze
          .filterNot( ab=> ab._2 < leftTop._2 || ab._2 > rightBottom._2)
          .filterNot( ab=> path.contains(ab)) // don't revisit points along your path
    
    

Now if you add 1 more filter, literally 1 more line of code, to eliminate
paths that are prohibited because they intersect with a barrier, you are good
to go. a room is simply a maze with zero barriers.

