# Is it a semiprime?

Surprisingly, I don't think we have a question for determining if a number is semiprime.

A semiprime is a natural number that is the product of two (not necessarily distinct) prime numbers.

Simple enough, but a remarkably important concept.

Given a positive integer, determine if it is a semiprime. Your output can be in any form so long as it gives the same output for any truthy or falsey value. You may also assume your input is reasonably small enough that performance or overflow aren't an issue.

Test cases:

input -> output
1     -> false
2     -> false
3     -> false
4     -> true
6     -> true
8     -> false
30    -> false   (5 * 3 * 2), note it must be EXACTLY 2 (non-distinct) primes
49    -> true    (7 * 7)      still technically 2 primes
95    -> true
25195908475657893494027183240048398571429282126204032027777137836043662020707595556264018525880784406918290641249515082189298559149176184502808489120072844992687392807287776735971418347270261896375014971824691165077613379859095700097330459748808428401797429100642458691817195118746121515172654632282216869987549182422433637259085141865462043576798423387184774447920739934236584823824281198163815010674810451660377306056201619676256133844143603833904414952634432190114657544454178424020924616515723350778707749817125772467962926386356373289912154831438167899885040445364023527381951378636564391212010397122822120720357
-> true, and go call someone, you just cracked RSA-2048


This is , so standard rules apply!

• @WheatWizard There's a slightly difference in that that one asks for 3 primes (not a big difference for almost all languages) and it's for distinct primes only (fairly different for some languages). I can discuss it with you in chat if you'd like to continue a more detailed discussion. – HyperNeutrino Aug 7 '17 at 18:30
• @WheatWizard You raise a good point, but similarly, we already have a bunch of many types of questions, and although, in contrast to what you express, most of them do add significant contribution to their area, this question has enough of a difference that I would believe that it warrants a separate question/post. – HyperNeutrino Aug 7 '17 at 18:33
• @hyperneutrino looking at the answers on both challenges, it looks like the difference is a single number in the source code, 2 vs 3. I would hardly call that a big difference. – Sriotchilism O'Zaic Aug 7 '17 at 18:37
• @WheatWizard There is also "distinct" vs "not distinct"... – HyperNeutrino Aug 7 '17 at 18:40
• @LordFarquaad Just because its a duplicate doesn't mean it is bad. In my mind being a duplicate is a good thing, it means that you are asking a thing that the community finds interesting enough to have already asked about. – Sriotchilism O'Zaic Aug 7 '17 at 18:50

# Brachylog, 2 bytes

Basically a port from Fatalize's answer to the Sphenic number challenge.

ḋĊ


Try it online!

### How?

ḋĊ - implicitly takes input
ḋ  - prime factorisation (with duplicates included)
Ċ - is a couple

• Right language for the job indeed :P – HyperNeutrino Aug 7 '17 at 18:55
• @Uriel Ċ is actually a built-in list of two variables; being a declarative language the output is, by default, a test for satisfaction (e.g. ḋ on its own would output true. for non-negative integers). – Jonathan Allan Aug 7 '17 at 19:10
• How is this 2 bytes? – harold Aug 8 '17 at 20:19
• @harold I just updated to make "bytes" in the header link to Brachylog's code-page. A hex dump would be c6 eb. – Jonathan Allan Aug 8 '17 at 20:21

# APL(NARS), 7 chars, 14 bytes

{2=≢π⍵}


test:

  f←{2=≢π⍵}
f¨1 2 3 4 6 8 30 49 95
0 0 0 1 1 0 0 1 1


# Japt, 6 5 bytes

k Ê¥2


Test it online

## Explanation

Does pretty much the same as most of the other answers: k gets the array of prime factors, Ê gets its length and ¥ checks for equality with 2.

# Pyke, 5 bytes

Pl02q


Try it here!

# Braingolf, 8 bytes

pl2e1:0|


Try it online!

# Java 8, 69 61 bytes

n->{int r=1,c=2;for(;r++<n;)for(;n%r<1;n/=r)c--;return c==0;}


-8 bytes thanks to @Nevay.

Try it here.

• You can remove the else statement (which could be else++r;) to save 8 bytes n->{int r=1,c=2;for(;r++<n;)for(;n%r<1;n/=r)c--;return c==0;}. – Nevay Aug 9 '17 at 13:26

# PHP, 75 bytes

<?php for($c=$t=2;$n=&$argv[1]>1;$t++)while($n%$t<1){$c--;$n/=$t;}echo+!$c;  Try it online! This script is called from the command line with the number being tested as its parameter. It prints out a 1 if the number was a semi-prime, else a 0. Explanation <?php  Work through each integer, $t, from 2 up, to see if it's a factor. We need to keep going, dividing by the factors until $n (an alias for $argv[1], the first CLI parameter) reaches 1. (And as there is a 2 to hand, we take the chance to initialize the counter, $c, to 2.) for ($c = $t = 2;$n = &$argv[1] > 1;$t++)


When we find a factor, we keep using it for as long as we can…

    while ($n %$t < 1) {


…each time deccrementing the factor counter…

        $c--;  …and recalculating our number. $n /= $t; }  When we reach the end, the counter can only be 0 if there were 2 factors. We ‘not’ the number, but this converts it to a boolean, so we use + to make it into 1 (the number was a semi-prime) or 0 (it wasn’t). echo +!$c;


# Python 3, 54 bytes

lambda n:0<sum((n%x<1)+(x**3==n)for x in range(2,n))<3


Try it online!

The previous verson had some rounding issues on large cube numbers (125,343,etc)
This calculates the amount of divisors (not only primes), if it has 1 or 2 it returns True.
The only exception is when a number has more than two prime factors but only two divisors. In this case it is a perfect cube of a prime (its divisors are its cube root and its cube root squared). x**3==n will cover this case, adding one to the cube root entry pushes the sum up to a count of 3 and stops the false-positive. thanks Jonathan Allan for writing with this beautiful explanation

• This claims 8 is semiprime – xnor Aug 7 '17 at 19:47
• n**(1/3)%1>0<sum... should work. – Dennis Aug 7 '17 at 21:20
• @xnor fixed it. – Rod Aug 7 '17 at 22:55
• Made a tiny edit (e.g. 6 cubed has many more divisors) – Jonathan Allan Aug 8 '17 at 19:04

# Perl 6, 43 bytes

{my \f=first $_%%*,2..$_;?f&&is-prime $_/f}  Try it online! f is the smallest factor greater than 1 of the input argument $_, or Nil if $_ is 1. The return value of the function is true if f is true (ie, not Nil) AND the input argument divided by the factor is prime. If $_ itself is prime, then f will be equal to $_, and $_ / f is 1, which is not prime, so the formula works in that case as well.

# J, 6 bytes

5 bytes will work as a one-off:

   2=#q: 8
0
2=#q: 9
1


I believe I need six when I define the function:

   semiprime =. 2=#@q:
(,. semiprime) 1 + i. 20
1 0
2 0
3 0
4 1
5 0
6 1
7 0
8 0
9 1
10 1
11 0
12 0
13 0
14 1
15 1
16 0
17 0
18 0
19 0
20 0


# Python 2, 90 bytes

def g(x,i=2):
while x%i:i+=1
return i
def f(n,l=0):
while 1%n:l+=1;n/=g(n)
return l==2


f takes an integer n greater than or equal to 1, returns boolean.

Try it online!

Test cases:

>>> f(1)
False
>>> f(2)
False
>>> f(3)
False
>>> f(4)
True
>>> f(6)
True
>>> f(8)
False
>>> f(30)
False
>>> f(49)
True
>>> f(95)
True


# Retina, 45 bytes

.+
$* ^(11+)(\1)+$
$1;1$#2$* A\b(11+)\1+\b ;  Try it online! Link includes test cases. Explanation: .+$*


Convert to unary.

^(11+)(\1)+1;1$#2$*


Try to find two factors.

A\b(11+)\1+\b


Ensure both factors are prime.

;


Ensure two factors were found.

# Ruby, 56 48 bytes

->x{r=c=2;0while x%r<1?(x/=r;c-=1):x>=r+=1;c==0}


Try it online!

### How it works:

->x{                    # Lambda function
r=c=2;              # Starting from r=2, c=2
0 while             # Repeat (0 counts as a nop)
x%r<1? (        # If x mod r == 0
x/=r:       # Divide x by r
c-=1        # decrease c
):              # else
x>=r+=1     # increase r, terminate if r>x
);
c==0                # True if we found 2 factors
}


Thanks Value Ink for the idea that saved 8 bytes.

• Why not just have c start at 0 and count up, instead of making it an array that you add all the factors to? That way you take out the need to use size at the end – Value Ink Aug 8 '17 at 1:20
• You are right, it's because I wrote the factorization function for another challenge and I reused it here. – G B Aug 8 '17 at 6:10

# Mathematica 32 bytes

Thanks to ngenesis for 1 byte saved

Tr@FactorInteger[#][[;;,2]]==2&

• Save one byte by using ;; instead of All. – ngenisis Aug 7 '17 at 21:30

# R, 67 bytes

c=0;n=scan();for(p in(1:n)[-1:-2]-1)while(n%%p<1){c=c+1;n=n/p};c==2


Try it online!

# Python with SymPy 1.1.1,  57  44 bytes

-13 bytes thanks to alephalpha (use 1.1.1's primeomega)

from sympy import*
lambda n:primeomega(n)==2


Try it online!

• lambda n:primeomega(n)==2 – alephalpha Aug 8 '17 at 3:41
• Ah that reminds me to upgrade from 1.0, thanks! – Jonathan Allan Aug 8 '17 at 4:01

# Gaia, 4 bytes

ḍl2=


4 bytes seems to be a common length, I wonder why... :P

Try it online!

### Explanation

ḍ     Prime factors
l    Length
2=  Equals 2?

• 4 bytes seems to be a common length, I wonder why... - Probably because all answers are prime factors, length, is equal to 2? – Mr. Xcoder Aug 8 '17 at 8:23
• @MrXcoder Yep, exactly – Business Cat Aug 8 '17 at 11:22
• 4 of which are mine BTW >_> – Mr. Xcoder Aug 8 '17 at 11:23
• 4 is also the first semiprime. Spooky! – Neil Aug 8 '17 at 11:47

# Pari/GP, 17 bytes

n->bigomega(n)==2


Try it online!

# Ruby, 35+8 = 43 bytes

Uses the -rprime flag to unlock the prime_division function.

->n{n.prime_division.sum(&:pop)==2}


Try it online!

# Mathematica, 31 29 bytes

Tr[Last/@FactorInteger@#]==2&


# C#, 112 Bytes

n=>{var r=Enumerable.Range(2,n);var l=r.Where(i=>r.All(x=>r.All(y=>y*x!=i)));return l.Any(x=>l.Any(y=>y*x==n));}


With formatting applied:

n =>
{
var r = Enumerable.Range (2, n);
var l = r.Where (i => r.All (x => r.All (y => y * x != i)));
return l.Any (x => l.Any (y => y * x == n));
}


And as test program:

using System;
using System.Linq;

namespace S
{
class P
{
static void Main ()
{
var f = new Func<int, bool> (
n =>
{
var r = Enumerable.Range (2, n);
var l = r.Where (i => r.All (x => r.All (y => y * x != i)));
return l.Any (x => l.Any (y => y * x == n));
}
);

for (var i = 0; i < 100; i++)
Console.WriteLine (\$"{i} -> {f (i)}");
}
}
}


Which has the output:

0 -> False
1 -> False
2 -> False
3 -> False
4 -> True
5 -> False
6 -> True
7 -> False
8 -> False
9 -> True
10 -> True
11 -> False
12 -> False
13 -> False
14 -> True
15 -> True
16 -> False
17 -> False
18 -> False
19 -> False
20 -> False
21 -> True
22 -> True
23 -> False
24 -> False
25 -> True
26 -> True
27 -> False
28 -> False
29 -> False
30 -> False
31 -> False
32 -> False
33 -> True
34 -> True
35 -> True
36 -> False
37 -> False
38 -> True
39 -> True
40 -> False
41 -> False
42 -> False
43 -> False
44 -> False
45 -> False
46 -> True
47 -> False
48 -> False
49 -> True
50 -> False
51 -> True
52 -> False
53 -> False
54 -> False
55 -> True
56 -> False
57 -> True
58 -> True
59 -> False
60 -> False
61 -> False
62 -> True
63 -> False
64 -> False
65 -> True
66 -> False
67 -> False
68 -> False
69 -> True
70 -> False
71 -> False
72 -> False
73 -> False
74 -> True
75 -> False
76 -> False
77 -> True
78 -> False
79 -> False
80 -> False
81 -> False
82 -> True
83 -> False
84 -> False
85 -> True
86 -> True
87 -> True
88 -> False
89 -> False
90 -> False
91 -> True
92 -> False
93 -> True
94 -> True
95 -> True
96 -> False
97 -> False
98 -> False
99 -> False


# Mathematica, 16 bytes

PrimeOmega@#==2&


PrimeOmega counts the number of prime factors, counting multiplicity.

• Dang, there's a builtin? – JungHwan Min Aug 8 '17 at 1:01
• @JungHwanMin If only there was SemiprimeQ – ngenisis Aug 8 '17 at 1:17
• Nice. I didn't know about PrimeOmega – DavidC Aug 8 '17 at 6:00

## JavaScript (ES6), 47 bytes

Returns a boolean.

n=>(k=1)==(d=n=>++k<n?n%k?d(n):d(n/k--)+1:0)(n)


### Demo

let f =

n=>(k=1)==(d=n=>++k<n?n%k?d(n):d(n/k--)+1:0)(n)

console.log(
[...Array(200).keys()].filter(f).join,
)

# Python 2, 75 65 bytes

lambda n:g(n)==2
g=lambda n,i=2:n/i and[g(n,i+1),1+g(n/i)][n%i<1]


Try it online!

All credit to xnor's answer for the original prime factorization code.

# Python 2, 67 bytes

lambda k:f(k)==2
f=lambda n,k=2:n/k and(f(n,k+1),1+f(n/k,k))[n%k<1]


Try it online!

-10 bytes thanks to @JonathanAllan!

Credit for the Prime factorization algorithm goes to Dennis (in the initial version)

• Did you use the code from Dennis' answer? If so, you should give credit. – totallyhuman Aug 7 '17 at 19:26
• @totallyhuman Oh yeah, sorry. I used it in 2 different answers today and I gave him credit in one of them, but I have forgotten to do that here once more. Thanks for spotting that! – Mr. Xcoder Aug 7 '17 at 19:28
• 67 bytes – Jonathan Allan Aug 7 '17 at 20:25
• @JonathanAllan Wow, thanks a lot! – Mr. Xcoder Aug 7 '17 at 20:27
• 55 bytes – Halvard Hummel Aug 8 '17 at 9:05

# Dyalog APL, 18 bytes

⎕CY'dfns'
2=≢3pco⎕


Try it online!

How?

⎕CY'dfns' - import pco

3pco⎕ - run pco on input with left argument 3 (prime factors)

2=≢ - length = 2?

# MATL, 5 bytes

Yfn2=


Try it online!

# Explanation

• Yf - Prime factors.

• n - Length.

• 2= - Is equal to 2?

# Husk, 4 bytes

Look ma no Unicode!

=2Lp


Try it online!

### How?

=2Lp - a one input function
p - prime factorisation (with duplicates included)
L  - length
=2   - equals 2?


# 05AB1E, 4 bytes

Òg2Q


Try it online!

How?

Ò       prime factors list (with duplicates)
g      length
Q    equal to
2     2


# Actually, 4 bytes

ol2=


Try it online!