How to resolve the algorithm 100 doors step by step in the Oz programming language

Published on 12 May 2024 09:40 PM
#Oz

How to resolve the algorithm 100 doors step by step in the Oz programming language

Table of Contents

Problem Statement

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.

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.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm 100 doors step by step in the Oz programming language

Source code in the oz programming language

declare
  NumDoors = 100
  NumPasses = 100

  fun {NewDoor} closed end

  fun {Toggle Door}
     case Door of closed then open
     [] open then closed
     end
  end

  fun {Pass Doors I}
     {List.mapInd Doors
      fun {$ Index Door}
         if Index mod I == 0 then {Toggle Door}
         else Door
         end
      end}
  end
  
  Doors0 = {MakeList NumDoors}
  {ForAll Doors0 NewDoor}

  DoorsN = {FoldL {List.number 1 NumPasses 1} Pass Doors0}
in
  %% print open doors
  {List.forAllInd DoorsN
   proc {$ Index Door}
      if Door == open then
	 {System.showInfo "Door "#Index#" is open."}
      end
   end
  }

  

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