<p>As others have stated, counting the factors would most likely require factoring n. However, trial division can bound the number of factors. You know, for instance, that $N$ has at most $n$ factors, since no factor can be less than 2. By testing if $N$ is divisible by 2, you also know that $N$ has at most $\log_3(N)$ factors, etc. The downside is that each reduction in size is progressively harder - you have to test up to $N^{1/p}$ to rule out $N$ containing more than $p$ factors.</p>

CC BY-SA 3.0
2012-08-14T02:29:11.200Z
2012-08-14T03:07:08.953Z
2012-08-14T03:07:08.953Z
Foo Barrigno
<p>This isn't my answer, but Terrence Tao gave <a href="https://mathoverflow.net/questions/3820/how-hard-is-it-to-compute-the-number-of-prime-factors-of-a-given-integer/10062#10062">a beautiful answer</a> to this question on MathOverflow.</p>

<p>Here are the first few lines of his answer. To read the complete answer, follow the link.</p>

<blockquote>
  <p>There is a folklore observation that if one was able to quickly count the number of prime factors of an integer n, then one would likely be able to quickly factor n completely. So the counting-prime-factors problem is believed to have comparable difficulty to factoring itself.</p>
</blockquote>

<p>(I wasn't sure whether this should be an answer or a comment. But it is really an answer, although it isn't written by me. I've made the answer Community Wiki so that it may be upvoted or accepted without unnecessarily giving me reputation.)</p>

CC BY-SA 3.0
2012-08-11T03:16:12.603Z
In my opinion, a pointer to an answer like this deserves reputation points (so it should not be community wiki), but I understand that different people have different views.
Tsuyoshi Ito
CC BY-SA 3.0
2012-08-11T04:05:35.403Z
But this is not a formal reduction....
arnab
CC BY-SA 3.0
2012-08-11T12:59:26.750Z
@arnab: No, it is not.  That is why he wrote “then one would _likely_ be able to quickly factor n completely.”
Tsuyoshi Ito
CC BY-SA 3.0
2012-08-11T02:42:47.223Z
2012-08-11T02:42:47.223Z
2017-04-13T12:57:54.530Z
Robin Kothari
<blockquote>
  <p>Given an integer $N$ of length $n$ bits, how hard is it to output the number of prime factors (or alternatively number of factors) of $N$?</p>
</blockquote>

<p>If we knew the prime factorization of $N$, then this would be easy. However, if we knew the number of prime factors, or the number of general factors, it is not clear how we'd find the actual prime factorization.</p>

<p>Is this problem studied? Are there known algorithms that solve this question without finding the prime factorization?</p>

<p>This question is motivated by curiosity and partially by a <a href="https://math.stackexchange.com/q/58426/13668">math.SE question</a>.</p>

CC BY-SA 3.0
2011-08-19T08:43:57.080Z
If the number of prime factors is large, that would imply that N has a small factor which can be found easily. On the other hand, if the number of prime factors of N is small, say 2, then it is similar to the problem of factoring a product of two primes, and knowing that the number of factors is 2 doesn't seem to help. See [this question by Omid](http://cstheory.stackexchange.com/questions/7550/reducing-factoring-prime-products-to-factoring-integer-products-in-average-case) about their average hardness.
Kaveh
CC BY-SA 3.0
2011-08-19T08:54:12.273Z
One more thing, since division is in uniform $\mathsf{TC^0}$, the problem of counting all factors (not just prime factors) is in $\mathsf{\#TC^0}$ and therefore is also in $\mathsf{P}$ (and is probably also complete for $\mathsf{\#TC^0}$ under $\mathsf{AC^0}$ reductions).
Kaveh
CC BY-SA 3.0
2011-08-19T15:36:42.153Z
Kaveh, if you could expand your above comment into an answer, that would be great. I don't exactly see how division in $\text{TC}^0$ gets you to counting factors in $\#\text{TC}^0$ without also implying that factoring is in $\text{TC}^0$. This misunderstanding is likely due to my own failures but a more detailed answer would help.
Derrick Stolee
CC BY-SA 3.0
2011-08-19T18:11:24.483Z
I don't see why this would imply that factoring is in $\mathsf{\#TC^0}$. Here is why counting the number of factors (all of them) is in $\mathsf{\#TC^0}$: $Div(x,y)$ is in $\mathsf{TC^0}$, therefore $\#x\leq y: Div(x,y)$ is in $\mathsf{\#TC^0}$ by definition (Complexity Zoo does not have an entry for $\mathsf{\#TC^0}$). Since $\mathsf{TC^0} \subseteq \mathsf{L}$ we get that this is in $\mathsf{\#L}\subseteq\mathsf{DET}\subseteq \mathsf{NC^2}\subseteq \mathsf{P}$. But I agree that there is something strange going on here, since this would imply that $Primes \in \mathsf{NC^2}$ which is not
Kaveh
CC BY-SA 3.0
2011-08-19T18:15:53.710Z
known AFAIK! and this is too easy. But I don't see where the argument breaks. ps: I guess I know see, my definition of $\mathsf{\#TC^0}$ is no good (it is the same as $\mathsf{\#P}$) and that is the problem.
Kaveh
CC BY-SA 3.0
2011-08-20T21:24:42.403Z
@Kaveh I don't see how your argument breaks down based on your definition of $\# \mathsf{TC}^0$. Since $Div(x,y) \in \mathsf{TC}^0 \subseteq \mathsf{L}$, can't we conclude that $\# \{x \leq y: Div(x,y)\} \in \# \mathsf{L} \subseteq \mathsf{NC}^2$? So I think there must be an error somewhere other than the definition of $\# \mathsf{TC}^0$.
Artem Kaznatcheev
CC BY-SA 3.0
2011-08-20T22:16:47.860Z
@Artem, $\mathsf{\#L}$ is defined as the number of accepting paths of an $\mathsf{NL}$ machine, and an $\mathsf{NL}$ machine can use only logarithmic (in $|y|$) amount of space for guessing $x$. We are guessing too many bits if we use the definition I wrote, an $\mathsf{AC^0}$ computation with polynomially many guesses would capture $\mathsf{NP}$, similarly counting the number of $x$s of polynomial size that an $\mathsf{AC^0}$ machine accepts on them will give $\mathsf{\#P}$ (guess the computation also and verify that is really an accepting computation).
Kaveh
<cc.complexity-theory><counting-complexity><nt.number-theory><factoring>
How hard is it to count the number of factors of an integer?
Artem Kaznatcheev
# How hard is it to count the number of factors of an integer?

(Asked by: username_0 on 2011-08-19)

> Given an integer $N$ of length $n$ bits, how hard is it to output the number of prime factors (or alternatively number of factors) of $N$?

If we knew the prime factorization of $N$, then this would be easy. However, if we knew the number of prime factors, or the number of general factors, it is not clear how we'd find the actual prime factorization.

Is this problem studied? Are there known algorithms that solve this question without finding the prime factorization?

This question is motivated by curiosity and partially by a [math.SE question](https://math.stackexchange.com/q/58426/13668).

**Tags:** cc.complexity-theory, counting-complexity, nt.number-theory, factoring
> **Comments:**
> **username_1**: If the number of prime factors is large, that would imply that N has a small factor which can be found easily. On the other hand, if the number of prime factors of N is small, say 2, then it is similar to the problem of factoring a product of two primes, and knowing that the number of factors is 2 doesn't seem to help. See [this question by Omid](http://cstheory.stackexchange.com/questions/7550/reducing-factoring-prime-products-to-factoring-integer-products-in-average-case) about their average hardness.  
> **username_1**: One more thing, since division is in uniform $\mathsf{TC^0}$, the problem of counting all factors (not just prime factors) is in $\mathsf{\#TC^0}$ and therefore is also in $\mathsf{P}$ (and is probably also complete for $\mathsf{\#TC^0}$ under $\mathsf{AC^0}$ reductions).  
> **username_2**: Kaveh, if you could expand your above comment into an answer, that would be great. I don't exactly see how division in $\text{TC}^0$ gets you to counting factors in $\#\text{TC}^0$ without also implying that factoring is in $\text{TC}^0$. This misunderstanding is likely due to my own failures but a more detailed answer would help.  
> **username_1**: I don't see why this would imply that factoring is in $\mathsf{\#TC^0}$. Here is why counting the number of factors (all of them) is in $\mathsf{\#TC^0}$: $Div(x,y)$ is in $\mathsf{TC^0}$, therefore $\#x\leq y: Div(x,y)$ is in $\mathsf{\#TC^0}$ by definition (Complexity Zoo does not have an entry for $\mathsf{\#TC^0}$). Since $\mathsf{TC^0} \subseteq \mathsf{L}$ we get that this is in $\mathsf{\#L}\subseteq\mathsf{DET}\subseteq \mathsf{NC^2}\subseteq \mathsf{P}$. But I agree that there is something strange going on here, since this would imply that $Primes \in \mathsf{NC^2}$ which is not  
> **username_1**: known AFAIK! and this is too easy. But I don't see where the argument breaks. ps: I guess I know see, my definition of $\mathsf{\#TC^0}$ is no good (it is the same as $\mathsf{\#P}$) and that is the problem.  
> **username_0**: @Kaveh I don't see how your argument breaks down based on your definition of $\# \mathsf{TC}^0$. Since $Div(x,y) \in \mathsf{TC}^0 \subseteq \mathsf{L}$, can't we conclude that $\# \{x \leq y: Div(x,y)\} \in \# \mathsf{L} \subseteq \mathsf{NC}^2$? So I think there must be an error somewhere other than the definition of $\# \mathsf{TC}^0$.  
> **username_1**: @Artem, $\mathsf{\#L}$ is defined as the number of accepting paths of an $\mathsf{NL}$ machine, and an $\mathsf{NL}$ machine can use only logarithmic (in $|y|$) amount of space for guessing $x$. We are guessing too many bits if we use the definition I wrote, an $\mathsf{AC^0}$ computation with polynomially many guesses would capture $\mathsf{NP}$, similarly counting the number of $x$s of polynomial size that an $\mathsf{AC^0}$ machine accepts on them will give $\mathsf{\#P}$ (guess the computation also and verify that is really an accepting computation).  

---

## 2 answers

### Answer by username_4 on 2012-08-11. Score: 16 | Accepted answer

This isn't my answer, but Terrence Tao gave [a beautiful answer](https://mathoverflow.net/questions/3820/how-hard-is-it-to-compute-the-number-of-prime-factors-of-a-given-integer/10062#10062) to this question on MathOverflow.

Here are the first few lines of his answer. To read the complete answer, follow the link.

> There is a folklore observation that if one was able to quickly count the number of prime factors of an integer n, then one would likely be able to quickly factor n completely. So the counting-prime-factors problem is believed to have comparable difficulty to factoring itself.

(I wasn't sure whether this should be an answer or a comment. But it is really an answer, although it isn't written by me. I've made the answer Community Wiki so that it may be upvoted or accepted without unnecessarily giving me reputation.)

> **Comments:**
> **username_5**: In my opinion, a pointer to an answer like this deserves reputation points (so it should not be community wiki), but I understand that different people have different views.  
> **username_6**: But this is not a formal reduction....  
> **username_5**: @arnab: No, it is not.  That is why he wrote “then one would _likely_ be able to quickly factor n completely.”  

---

### Answer by username_3 on 2012-08-14. Score: 1

As others have stated, counting the factors would most likely require factoring n. However, trial division can bound the number of factors. You know, for instance, that $N$ has at most $n$ factors, since no factor can be less than 2. By testing if $N$ is divisible by 2, you also know that $N$ has at most $\log\_3(N)$ factors, etc. The downside is that each reduction in size is progressively harder - you have to test up to $N^{1/p}$ to rule out $N$ containing more than $p$ factors.


---

