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Question: Has it been proven that the following decision problem is algorithmically decidable?

$$ P_\text{Collatz} := \text{Given $n \in \mathbb{N}_+$, does the Collatz-Iteration of $n$ eventually reach $1$?} $$

Obviously, we strongly suspect that the problem is decidable by the algorithm that outputs "yes" for any input. It gets more interesting if we allow negative inputs, but I guess we still strongly suspect decidability.

Be that as it may, Conway proved[1] that there is a "Collatz-like" function $g$ such that the following decision problem is not decidable:

$$ P_\text{g} :=\text{Given $n \in \mathbb{N}_+$, does the $g$-iteration of $n$ eventually reach $1$? } $$

So, is it known whether the specialization $g=\text{Collatz}$ is decidable?

[1]: See S.A. Kurtz and J. Simon, The Undecidability of the Generalized Collatz Problem, in Theory and Applications of Models of Computation: 4th International Conference, TAMC 2007, Shanghai, China, May 22-25, 2007, Proceedings (Lecture Notes in Computer Science, 4484).

nonuser
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Vincent
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    Can someone explain the downvotes? My question is substantially weaker than the Collatz conjecture itself. It's very possible that there is an easy way to see that the map is decidable while the Collatz conjecture is still hopeless. – Vincent Nov 21 '25 at 02:17
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    I think Collatz posts are just downvote magnets because they're so often low-quality. This one seems good! Maybe making the title more specific would help. – Karl Nov 21 '25 at 07:30
  • Is your question the same as "Is the original Collatz Conjecture undecidable ?" I think the answer is still "open". But, I have no references. – nonuser Nov 24 '25 at 00:56
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    @nonuser No, it's not the same question. The question you mention is a bit ambiguous. I think you mean "is the collatz conjecture independent of ZFC", which is a completely different question as mine. If you mean "is there an algorithm that computes whether the collatz conjecture (as a whole) is true", the the answer to that question is yes - we just don't know which algorithm it is. (it's either the one outputting "yes" for every input, or the one outputting "no" of every input). When using the word "undecidable" you have to be very precise when explaining what you mean. – Vincent Nov 24 '25 at 07:57
  • @Vincent I mean Turing-undecidability for the Collatz decision problem. – nonuser Nov 24 '25 at 09:00
  • @Vincent I am not sure I understand the question what is the difference between Collatz and g? I am asking so that I understand why doesn't theorem 1.4 here https://people.cs.uchicago.edu/~simon/RES/collatz.pdf answer it in the negative. – H.C Manu Nov 25 '25 at 16:10
  • @H.CManu The PDF looks like work-in-progress rather than the actual article, so some of the phrasing might not be formally precise. My understanding of Theorem 1.4 is that there is no single algorithm which takes a description (i.e. the values of $p$ and all the $a_i$, $b_i$) of a Collatz function $g$ and determines if that specific $g$ reaches $1$ for all possible starting points. This doesn't tell us much about any specific $g$. – Peter Košinár Nov 30 '25 at 01:29

1 Answers1

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The answer to your question is weakly related to the answer to the question below :

“Does there exist an algorithm to detect a (possible) loop that goes to infinity under the finite Collatz iterations ?”

There is no paper currently posted to arXiv.org that answers above question.

If there is a counterexample that leads to the cyclic loop, then there must be an algorithm that reaches the number of iterations in that loop and finds it in finite time, and the Turing machine theoretically halts in finite time - the output is always algorithmically computable.

In short, your question can be reduced to:

If it can be proven that there are no noncyclic counterexamples that go to infinity, the answer to your question is positive — Yes, the decision problem is always algorithmically decidable. Otherwise, the status of the question remains open.

Because no one has proven it to date. Proving the existence of, or even directly constructing, an algorithm to detect a (possible) counterexample that goes to infinity may be as difficult as solving the original Collatz problem.

Consider the generalized Collatz map

$$ T(n) = \begin{cases} \dfrac{n}{2}, & \text{if } n \text{ is even}, \\[6pt] 5n + 1, & \text{if } n \text{ is odd}. \end{cases} $$

This map grows much faster than the classical $3n+1$ function, and for many starting values it is still unknown whether the corresponding trajectory eventually enters a cycle or diverges to infinity.

A concrete example is the starting value

$$ n = 7. $$

Computations show that the orbit of $7$ under the $5n+1$ rule becomes extremely large. After several thousand iterations it already exceeds numbers far beyond $10^{1000}$, and no cycle or decrease has been observed. Nevertheless, there is no proof that the sequence

$$ 7,\; T(7),\; T^{2}(7),\; T^{3}(7),\;\ldots $$

either falls into a periodic loop or diverges to infinity. Thus, it remains unknown whether

$$ T^{k}(7) \longrightarrow \infty $$

or not.

This makes $n = 7$ a standard example of a natural number whose behavior under a Collatz-type iteration is still unknown. More generally, maps of the form

$$ T(n) = \begin{cases} \dfrac{n}{2}, & n \text{ even}, \\[6pt] an + 1, & n \text{ odd}, \end{cases} \qquad (a > 3) $$

are known to have dynamics complicated enough to encode arbitrary computations, which explains why proving convergence or divergence for specific initial values is extremely difficult.

Игорь
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