Write a function that produces a plot for the Cantor Set

The Cantor Set is an image that looks like this: http://library.thinkquest.org/2647/media/cantor.jpg

I am asked to write a function that takes the number of iterations(n) as input and produces the desired plot. I am told that the number of rows in each Matrix of n iterations is equal to the number of line segments.

I first started off doing this problem by finding a relationship between the number of iterations (n) and the number of rows (rn). Each matrix of n interation has the same number of columns but different number of rows, and I found the relationship between rn and n to be: rn=(2^n)

I then wrote out the matrices for n=0,1, and 2 iterations and got the following: When n=0, M equals [0 1]. When n=1, M equals [0 1/3; 2/3 1]. When n=2, M equals [0 1/9; 2/9 1/3; 2/3 7/9; 8/9 1] So the relationship between Mn and Mn-1 is the top half of the matrix Mn (first rn/2) is equal to the previous matrix Mn-1 times 1/3. And the bottom half of the matrix Mn is equal to the top half plus 2/3. I have found the relationships but I am not quite certain on how to write the script to graph the plot. I have started my function with the first iteration of Mprev=[0 1] and I am not sure how to add the additional rows for the other iterations. Please help! Thanks!!

[Walter Roberson]

Are you asking how to store an "irregular array" in MATLAB? Or are you asking for an assistance with your algorithm?

http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

Answer 1

[sixwwwwww]

Dear Jerry, here is the code which plotting Cantor Set:

n = input('Input number of iterations:');
a = cell(1,n);
for j = 1:n
  a{j} = linspace(0, 1, 2^j);
end
m = n;
for j = 1:n
  b = a{j};
  l = ones(1, length(b)) * m;
  plot(b, l, 'ro-'), hold on
  m = m - 1;
end
xlim([-0.1 1.1]), ylim([0 n+1])

I hope it helps. Good luck!

[Jerry]

Thanks so much! However, how would you rewrite the code to not show the whole line on the iterations but have a gap? Such as http://library.thinkquest.org/2647/media/cantor.jpg

[sixwwwwww]

Following is the code which produces the exact output as you desire:

n = input('Input number of iteratios:');
a = cell(1,n);
for j = 1:n
    a{j} = linspace(0, 1, 2^j);
end
m = n;
for j = 1:n
    b_old = a{j};
    b = reshape(b_old, 2, []);
    l = ones(1, 2) * m;
    for k = 1:size(b, 2)
        plot(b(:, k), l, 'ro-'), hold on
    end
    m = m - 1;
end
xlim([-0.1 1.1]), ylim([0 n+1])

Good luck!

[Rodrigo Osuna Orozco]

This is NOT the cantor set. This sets points at intervals of 1/(2^n - 1), for instance at 1/7 for the third iteration, instead of at 1/9.

See the function below.

[Tejas Kotwal]

Agree with Rodrigo. To fix replace the 2^j by 3^j + 1

Answer 2

[Yotam Stern]

here's a simple recursive funtion that plots both the cator set and the cantor function :) it just takes out the middle third in each iteration and activate itself on the other two thirds. i know it's a bit late but i've only seen your post now,

function cantr=cantor(l,a)

if nargin <1
    tic
    l=0;
    N=1E7;
    cantr=ones(1,N);
else
    cantr=a;
    l=l+1;
end
if l==14
    return
end
n=length(cantr);
cantr(ceil(n/3)+1:2*ceil(n/3))=0;
cantr(1:ceil(n/3))=cantor(l,cantr(1:ceil(n/3)));
cantr(2*ceil(n/3)+1:end)=cantor(l,cantr(2*ceil(n/3)+1:end));
if l==0
    subplot (2,1,1)
    plot(cantr);
    dy=1/sum(cantr);
    imcan=zeros(1,length(cantr));
    for i=2:length(cantr)
        if cantr(i)
            imcan(i)=imcan(i-1)+dy;
        else
            imcan(i)=imcan(i-1);
        end
    end
    subplot (2,1,2)
    plot (imcan)
    toc
end