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This is problem 5.31 from Introduction to the Theory of Computation, 3rd Edition by Michael Sipser.

Let $$ f(x) = \begin{cases} 3x+1, & \text{for odd $x$} \\ x/2, & \text{for even $x$} \end{cases} $$ for any natural number $x$. If you start with an integer $x$ and iterate $f$, you obtain a sequence, $x$, $f(x)$, $f(f(x))$, $\dots$ . Stop if you ever hit 1. For example, if $x = 17$, you get the sequence 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. Extensive computer tests have shown that every starting point between 1 and a large positive integer gives a sequence that ends in 1. But the question of whether all positive starting points end up at 1 is unsolved; it is called the $3x + 1$ problem.

Suppose that $A_{TM}$ = $\{\langle M,w\rangle$∣ $M$ is a Turing Machine and $M$ accepts $w\}$ (Acceptance Turing machine) were decidable by a Turing Machine $H$. Use $H$ to describe a Turing Machine that is guaranteed to state the answer to the $3x + 1$ problem.

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    Are you talking about the Collatz Conjecture? – 2017-02-24
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    "Suppose that ATM were decidable by a TM H. Use H to describe a TM that is guaranteed to state the answer to the 3x + 1 problem" Could you rephrase that sentence, I don't understand the acronymns – 2017-02-24
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    It is, but note that the question is not to prove the conjecture. It is about Turing Machines and decidability. But what do you mean by $ATM$? – 2017-02-24
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    $A_{TM}$={⟨M,w⟩∣M on input w accepts}. – 2017-02-24

1 Answers 1

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I suppose the question is: if we can solve the halting problem, can we figure out the answer to the $3x+1$ (Collatz) problem?

Well, if we have a program $H$ that solves the halting problem (that is, it can decide whther some Turing machine $M$ with input $I$ halts or not), then do the following:

Create a Turing machine $F$ that takes a number $x$ and iterates through the $f(x)$ sequence, and stops once it reaches $1$. It is easy to show such a Turing machine $F$ exists .. here is its pseudocode:

F (input $i$)

$Begin$

$\quad$ While $i \not = 1$

$\quad \quad $ Case $i$ is even: $i = i/2$

$\quad \quad $ Case $i$ is odd: $i = 3*i+1$

$End$

(So this F routine will halt whenever the sequence ends with a 1 ... otherwise it will go on forever)

Now create a new Turing machine $C$ that starts with a given $i$, and that calls halting program $H$ on $F$ and $i$. If $H$ says that $F$ does not halt on $i$, then stop. If $H$ says that $F$ does halt $i$, then increase $i$ by $1$, and repeat the process. Again, it is easy to show that such a Turing machine $C$ exists, assuming $H$ exists .. here is its pseudocode:

C (input $i$)

$Begin$

$\quad$ While $H(F,i)$

$\quad \quad$ $i=i+1$

$End$

(so this routine $C$ will only stop if for some $i$, $H$ finds that $F$ does not halt on $i$. Effectively, $C$ looks for a counterexample to the Collatz conjecture. If there is one, then $C$ will eventually run into it. If there isn't, then $C$ will run forever)

Finally, call $H$ on $C$ and $1$. If $H$ says that $C$ with $1$ will not halt, then apparently the solution to the $3x+1$ is that the sequence will always end up with $1$ for any $x$ (that is, there is no counterexample to the Collatz conjecture ... so the Collatz Conjecture is true). If $H$ says that $C$ with $1$ does halt, then apparently the solution is that for some $x$ the sequence does not end up with $1$ (i.e. there is a counterexample to the Collatz conjecture). So, either way, you have solved the $3x+1$ Collatz problem.

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    I have tried something similar but got confused in that it is undecidable that C will halt or not if every i ends to 1. – 2017-02-24
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    @TjRocks If the sequence gets to $1$ for every $i$, then that means that $C$ with input $1$ will not halt. But that means that $H$ with input $C$ and $1$ will say that, and so then you know. – 2017-02-24
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    Now create a new Turing machine C that starts with a given i, and that calls halting program H on F and i. **If H says that F does not halt on i, then stop**. If H says that F does halt i, then increase i by 1, and repeat the process. Can you elaborate the line a bit like how we can say now it we not stop in F so stop C on i? – 2017-02-24
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    @TjRocks C goes through a loop where it keeps increasing $i$ and basically looks for a counterexample to the claim that the F routine will always halt (i.e. That the sequence ends up in a 1 for any x). So it will jump outside this loop once it finds such an $i$, otherwise it just keeps increasing $i$. I'll add some pseudocode to my post to make this more clear. – 2017-02-24