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

Published on 12 May 2024 09:40 PM
#Ada

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

Source code in the ada programming language

with Ada.Text_Io; use Ada.Text_Io;
 
 procedure Doors is
    type Door_State is (Closed, Open);
    type Door_List is array(Positive range 1..100) of Door_State;
    The_Doors : Door_List := (others => Closed);
 begin
    for I in 1..100 loop
       for J in The_Doors'range loop
          if J mod I = 0 then
             if The_Doors(J) = Closed then
                 The_Doors(J) := Open;
             else
                The_Doors(J) := Closed;
             end if;
          end if;
       end loop;
    end loop;
    for I in The_Doors'range loop
       Put_Line(Integer'Image(I) & " is " & Door_State'Image(The_Doors(I)));
    end loop;
 end Doors;


with Ada.Text_Io; use Ada.Text_Io;
 with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
 
 procedure Doors_Optimized is
    Num : Float;
 begin
    for I in 1..100 loop
       Num := Sqrt(Float(I));
       Put(Integer'Image(I) & " is ");
       if Float'Floor(Num) = Num then
          Put_Line("Opened");
       else
          Put_Line("Closed");
       end if;
    end loop;
 end Doors_Optimized;

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