Collatz Sequence Plotting Issue
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I have written code for the collatz problem, and performed some plotting for the number of steps versus the integer values. But now I am trying to plot the values of i versus my xi values *for i = 1 : 300.
Here is my code:
function [ns, seq] = collatz(n)
% first number insequence is n
seq(1) = n;
% position an index on the next element of sequence
i = 2;
% Repeat until you find a 1
while seq(i-1) ~= 1
% mod after division to find an even/odd number
if mod(seq(i-1), 2) == 0
% Step if even
seq(i) = seq(i-1)/2;
else
% Step taken if odd
seq(i) = 3*seq(i-1) + 1;
end
% index increment
i = i+1;
end
% Find length of the sequence
ns = length(seq);
Running Program:
>> [n, seq] = collatz(300)
n =
17
seq =
300 150 75 226 113 340 170 85 256 128 64 32 16 8 4 2 1
Then I wrote this code for the max values of xi with the largest number of steps for i = 1 : 300.
% For max number of steps for i = 1 : 300
clc; clear
% Sequence for max steps for
for i = 1 : 300
% Perform the function and get the number of elements
[n(i), seq{i}] = collatz(i);
end
% Find max number of steps found
[max_steps, ix] = max(n)
% Display the appropriate indexed sequence
seq{ix}
% Running Function:
max_steps =
128
ix =
231
How would I plot i on the x-axis versus my values of xi on the y - axis?
4 Comments
Walter Roberson
on 23 Apr 2012
Please clarify. There is no "x" variable in your code. You have an "ix" variable, but that is obtained as the position the maximum length occurs over i=1:n and is thus a one-time thing not something that can be plotted against "i". Unless, that is, you want to modify the code slightly so that "ix" are the index of the "running maximum" -- e.g., if at "i" = 58 the maximum so-far had been at "i = 43" then ix(58) would be 43 ??
Answers (1)
Thomas
on 23 Apr 2012
You could plot the number of steps required to get to 1 as shown in the Wolfram mathworld link
You could do it simply by
plot(n,'MarkerSize',10,'Marker','.','LineStyle','none');
% Create xlabel
xlabel('Number');
% Create ylabel
ylabel('Steps to reach 1');
EDIT If the largest upto a number is needed
for i=1:300
newmax(i)=max(n(1:i))
end
plot(newmax)
2 Comments
Thomas
on 23 Apr 2012
R you trying to see what the largest value at each of the numbers from 1:300 are i.e. if n=1 2 8 3 6 9 17
at 1 Largest =1
2, largest=2
3, largest =8
4,largest =8
5,largest=8
6 largest=9 and so on..?
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