How to resolve the algorithm 100 doors step by step in the MATLAB / Octave programming language

Published on 12 May 2024 09:40 PM
#Matlab

How to resolve the algorithm 100 doors step by step in the MATLAB / Octave 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 MATLAB / Octave programming language

Source code in the matlab programming language

a = false(1,100);
for b=1:100
  for i = b:b:100
    a(i) = ~a(i);
  end
end
a

for x=1:100;
  if sqrt(x) == floor(sqrt(x))
    a(i)=1;
  end
end
a

doors   = zeros(1,100); // 0: closed 1: open
for i = 1:100
    doors(i:i:100) = 1-doors(i:i:100)
end
doors

a = zeros(100,1);
for counter = 1:sqrt(100);
  a(counter^2) = 1;
end
a

function [doors,opened,closed] = hundredDoors()

    %Initialize the doors, make them booleans for easy vectorization
    doors = logical( (1:1:100) );
    
    %Go through the flipping process, ignore the 1 case because the doors
    %array is already initialized to all open
    for initialPosition = (2:100)
        doors(initialPosition:initialPosition:100) = not( doors(initialPosition:initialPosition:100) );
    end
    
    opened = find(doors); %Stores the numbers of the open doors
    closed = find( not(doors) ); %Stores the numbers of the closed doors
    
end

doors((1:10).^2) = 1;

doors

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