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

Published on 12 May 2024 09:40 PM

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

Source code in the free programming language

program OneHundredIsOpen;

const
  DoorCount = 100;

var
  IsOpen: array[1..DoorCount] of boolean;
  Door, Jump: integer;

begin
  // Close all doors
  for Door := 1 to DoorCount do
    IsOpen[Door] := False;
  // Iterations
  for Jump := 1 to DoorCount do
  begin
    Door := Jump;
    repeat
      IsOpen[Door] := not IsOpen[Door];
      Door := Door + Jump;
    until Door > DoorCount;
  end;
  // Show final status
  for Door := 1 to DoorCount do
  begin
    Write(Door, ' ');
    if IsOpen[Door] then
      WriteLn('open')
    else
      WriteLn('closed');
  end;
  // Wait for 
  Readln;
end.

  

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