Problem 21. Return the 3n+1 sequence for n

What's wrong in this code?

the assignment c = [c,a]; must be out of the else clause, but still inside of the while block.

how can i simplify??

Try this！:D
c=n;
while n>1
if mod(n,2)
n =3*n+1;
else
n =n/2;
end
c=[c n];
end

Thats right> construct c thru concat operator [..] instead of indexing it.
Index "i" doesn't care because it don't care the size of the sequence.

why this will go wrong?

very easy

How to reduce it's size?

This solution does not work for n = 3. Test Suite should include n = 3.

Hi all. I have no idea why there are many outputs of ans (the last of which is the right one), can anyone help to explain?

How can I reduce the size? Too many loops, yet I have to define the single commands :/

I don't think you should hard code the lists you should be constructing. Regex is fine, but you're limiting yourself by making such a narrow solution.
I personally don't like the regex solution because I'm trying to find interesting things about the language and I don't want to run through and find the shortest non regexp.

Would some kind soul enlighten me please?

it works fine on matlab when i assign 'c' a value

disp(c) contains the correct answer,unlike c. c is always equal to 1(last value of the iteration process)

This is a poor solution

Very clever

Nice

different approach but it used different functions frquenctly

this function collatz(N) produces specific values for N=10000, try this :
----------------------
N=collatz(1000);
plot(N);
-----------------------
% more interesting figure, derivate N
-----------------------
dN=diff(N);
figure,plot(dN)
-----------------------
dN looks like a sample of voice signal.

I really think the test set should go to much higher arguments. This sort of thing should not be allowed!

This is way more efficient than the recursive algorithms.

How?

Nice indeed.
I am surprised that it is more efficient to solve this with a recursive function than a while loop.
This solution has size 43:

while n(end) > 1
n = [n mod(n(end),2)*(n(end)*2.5+1) + n(end)/2];
end

Is the loop overhead greater than the recursion overhead? Or is the weak point having to go to n(end) at each iteration?

Strange I can not delete my first comment. I wanted to format it a bit nicer :)

Good solution, David!:)