How to resolve the algorithm 100 doors step by step in the Seed7 programming language

Published on 12 May 2024 09:40 PM
#Seed7

How to resolve the algorithm 100 doors step by step in the Seed7 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 Seed7 programming language

Source code in the seed7 programming language

$ include "seed7_05.s7i";
 
const proc: main is func
  local
    var array boolean: doorOpen is 100 times FALSE;
    var integer: pass is 0;
    var integer: index is 0;
    var array[boolean] string: closedOrOpen is [boolean] ("closed", "open");
  begin
    for pass range 1 to 100 do
      for key index range doorOpen do
        if index rem pass = 0 then
          doorOpen[index] := not doorOpen[index];
        end if;
      end for;
    end for;
    for key index range doorOpen do
      write(index lpad 3 <& " is " <& closedOrOpen[doorOpen[index]] rpad 7);
      if index rem 5 = 0 then
        writeln;
      end if;
    end for;
  end func;

$ include "seed7_05.s7i";

const proc: main is func
  local
    var integer: index is 0;
    var integer: number is 0;
    var array[boolean] string: closedOrOpen is [boolean] ("closed", "open");
  begin
    for index range 1 to 100 do
      number := sqrt(index);
      write(index lpad 3 <& " is " <& closedOrOpen[number**2 = index] rpad 7);
      if index rem 5 = 0 then
        writeln;
      end if;
    end for;
  end func;

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