There are 100 doors in a row that are all initially closed.

You make 100 passes by the doors.

The first time through, visit every door and  toggle  the door  (if the door is closed,  open it;   if it is open,  close it).

The second time, only visit every 2nd door   (door #2, #4, #6, ...),   and toggle it.

The third time, visit every 3rd door   (door #3, #6, #9, ...), etc,   until you only visit the 100th door.



Task

Answer the question:   what state are the doors in after the last pass?   Which are open, which are closed?


Alternate:
As noted in this page's   discussion page,   the only doors that remain open are those whose numbers are perfect squares.

Opening only those doors is an   optimization   that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.




import java.util.ArrayList
 
class Door
	:state
 
	def initialize
		@state=false
	end
 
	def closed?; !@state; end
	def open?; @state; end
 
	def close; @state=false; end
	def open; @state=true; end
 
	def toggle
		if closed?
			open
		else
			close
		end
	end
 
	def toString; Boolean.toString(@state); end
end
 
doors=ArrayList.new
1.upto(100) do
    doors.add(Door.new)
end 
 
1.upto(100) do |multiplier|
    index = 0
    doors.each do |door|
        Door(door).toggle if (index+1)%multiplier == 0
        index += 1
    end
end
 
i = 0
doors.each do |door| 
    puts "Door #{i+1} is #{door}."
    i+=1
end
 