Collatz Conjecture using for loop

Consider the following algorithm

• start with positive integer
• if the number is even, divide by 2 if not multiply it by 3 and add one
• repeat the process until the number 1 is obatined

We ae required to create a function that takes two inputs 'n'=positive integer and 'max_steps' and returns the number of steps rquired to reach 1. The function is meant to be such that if the number of steps reaches the value 'max_steps'(without the algorithm reaching 1) it returns Nan. The code has to use a for loop and at least 1 if statement

N='Nan';
steps = 0;
    while n~=1;
        if mod(n,2) == 0; 
            n = n/2;
        else
            n = 3*n + 1;
        end
        steps=steps+1;
    end
    if steps>max_steps
        disp=(N);
    end 

Most of the code is correct expect i cant seem to get 'Nan' when the number of steps reaches the value 'max_steps' and dont know i can implement a for loop

Any help will appreciated

Answer 1

[John D'Errico]

Your code is close to a working algorithm.

There are some issues truly good code would worry about. Could you ever have an overflow issue? I'm not talking about overflowing the realmax value for a double, but instead, the largest possible integer a double can represent exactly and unambiguously. This is flintmax in MATLAB.

flintmax

ans = 9.0072e+15

And flintmax might not be that terribly a huge number in this context. flintmax is 2^53-1. So good code would worry about that. Perhaps you can gain a little headroom, by using uint64 to represent your integers.

First, this must be a function. Displaying a result in NOT the same thing as returning a value from a function. So your code should be wrapped in a function, like this:

function steps = collatz(n,max_steps)

end

Now I'll put your while loop in there.

function steps = collatz(n,max_steps)
  steps = 0;
  while n~=1
    if mod(n,2) == 0; 
      n = n/2;
    else
      n = 3*n + 1;
    end
    steps=steps+1;
  end
end

But this never worries about how to stop that loop when there is a problem. So we need to do more. One idea is to use the break command, when max_steps is exceeded. Another idea is to change the value of N to a NaN, and then to put that into the test for the while loop.

function steps = collatz(n,max_steps)
  steps = 0;
  while n~=1 && ~isnan(n)
    % the collatz iteration
    if mod(n,2) == 0; 
      n = n/2;
    else
      n = 3*n + 1;
    end
    
    steps=steps+1; % increment the steps counter.
    
    if (steps >= max_steps) && (n~=1) % have we gone too far?
      n = NaN;
    end
  end
end

As you can see, this version will overwrite the value of n is we have hit the step counter limit. And then the while loop will fail.

The alternative is to use break.

function steps = collatz(n,max_steps)
  steps = 0;
  while n~=1
    % the collatz iteration
    if mod(n,2) == 0; 
      n = n/2;
    else
      n = 3*n + 1;
    end
    
    steps=steps+1; % increment the steps counter.
    
    if (steps >= max_steps) && (n~=1) % have we gone too far?
      n = NaN;
      break
    end
  end
end

And of course, there are surely other ways you might write it.

[B]

thanks alot