Write a function that produces a plot for the Cantor Set

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Jerry
Jerry 2013-10-21
评论： Tejas Kotwal 2020-5-2
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!!

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sixwwwwww
sixwwwwww 2013-10-21
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!
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Rodrigo Osuna Orozco
Rodrigo Osuna Orozco 2019-3-15
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.

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Yotam Stern
Yotam Stern 2014-3-6
编辑：Yotam Stern 2014-3-6
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

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