for loop and while loop help with Collatz conjecture

Below is code for Collatz conjecture, that given an x value will print out the max number(mx) during the run for that x value and number of iterations(N) it takes to get to 1.

function [N, mx] = prob3_6(x)
% function [N, mx] = prob3_6(x)
% takes a positive integer x and finds number of iterations and maximum
% value along the way
X = x;
N = 0; % initialize interations to zero 
mx = x; % initialize maximum value
while X ~= 1
  if mod(X,2) % X is odd
      X = 3*X+1;
  else
      X = X/2;
  end
  N = N+1;
  %disp(X)
  if X > mx
      mx = X;
  end
end

I need to run a whole array of x through here like 2:1:100 and get how many iterations(N) it took for each value of x, and the max number of each value of x. I figured for loop around the while loop, like this

function [N, mx] = prob3_6(x)
X = x;
N = 0; % initialize interations to zero 
mx = x; % initialize maximum value
for k = 1:length(x)
while X ~= 1
  if mod(X,2) % X is odd
      X = 3*X+1;
  else
      X = X/2;
  end
  N = N+1;
  %disp(X)
  if X > mx
      mx = X;
  end
end
end

now where would I have to put k or maybe N or mx to get what I need, been trying different combos not sure how it should go.

Answer 1

[Geoff Hayes]

Luke - if x is now an array of elements, then wouldn't you do something like

 function [N, mx] = prob3_6(x)

 for k=1:length(x)
     X = x(k);
     N = 0; % initialize interations to zero 
     mx = x; % initialize maximum value
     while X ~= 1
         % etc.
     end
 end

It is really no different than what you had originally when x was a scalar. You just need to realize that it is an array and iterate over each element of it.

What I have omitted (since this is homework) is how you store the mx for each element in x. You will want to create an mx array which is the same dimension as x and then update the kth element of it.