How can I set up the variables for this question?

Take any natural number n. If n is even, divide it by 2 to get n/2. If n is odd, multiply it by 3 and add 1 to obtain 3n+1. (This is called the Hailstone sequence).
Set the new number to n and repeat this process until n=1.
How many steps does it take for 71 to get to 1?
I'm a total idiot in Matlab in the real sense of the word. I have problem translating theories into practice. I need a leg up on understanding how I can apply concepts to solve this problem.

2 comentarios

cwc
cwc el 5 de Sept. de 2013
What is the best way to set up all the parameters? Is there some kind of mental process I must adopt? I can see all abstract reasoning very well but come problem solving mind just go blank!

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Image Analyst
Image Analyst el 5 de Sept. de 2013
Editada: Image Analyst el 5 de Sept. de 2013
I think you only need 2 variables, n and the number of steps you took. Use "n" because that's what they told you to use. Then pick a descriptive name for the other variable you need, like "numberOfSteps". Hints to get you started.
n = int32(42); % whatever.
numberOfSteps = int32(0);
while n ~= 1
if n is even, then % You do this
n = n/2
else
n = 3*n+1
end
numberOfSteps = ....... % You do this.
end
There's still some stuff in there for you to do, so good luck. By the way, that code is not that robust since there's no failsafe to prevent infinite loops, but I'll take your word on it that eventually you'll hit 1 as required.

1 comentario

cwc
cwc el 5 de Sept. de 2013
I shall work on it. Admittedly, I knew it involves loops and logical operators Just having problems condensing all the reasoning into syntax.

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Más respuestas (1)

cwc
cwc el 6 de Sept. de 2013
Editada: Image Analyst el 6 de Sept. de 2013
function numberofsteps = naturalnumber(n)
numberofsteps = 0;
while n~=1;
if mod(n,2) == 0;
n = n/2;
else
n = 3*n + 1;
end
numberofsteps = numberofsteps + 1;
end

2 comentarios

Image Analyst
Image Analyst el 6 de Sept. de 2013
Looks like it should work.
Walter Roberson
Walter Roberson el 6 de Sept. de 2013
Good thing that infinity is not a "natural number" ;-)
Still, I think the code will start misbehaving at about 2^51

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cwc
el 5 de Sept. de 2013

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