31
\$\begingroup\$

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!

\$\endgroup\$
14
  • 4
    \$\begingroup\$ @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. \$\endgroup\$
    – hyper-neutrino
    Aug 7, 2017 at 18:30
  • 2
    \$\begingroup\$ @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. \$\endgroup\$
    – hyper-neutrino
    Aug 7, 2017 at 18:33
  • 2
    \$\begingroup\$ @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. \$\endgroup\$
    – Wheat Wizard
    Aug 7, 2017 at 18:37
  • 3
    \$\begingroup\$ @WheatWizard For example, if you compare the Jelly answer on that one: ÆEḟ0⁼7B¤ to the Jelly answer on this one: ÆfL=2, you'll notice that there is a significant difference; namely, that one has to check for distinctness. \$\endgroup\$
    – hyper-neutrino
    Aug 7, 2017 at 18:41
  • 3
    \$\begingroup\$ @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. \$\endgroup\$
    – Wheat Wizard
    Aug 7, 2017 at 18:50

37 Answers 37

20
\$\begingroup\$

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
\$\endgroup\$
4
  • 1
    \$\begingroup\$ Right language for the job indeed :P \$\endgroup\$
    – hyper-neutrino
    Aug 7, 2017 at 18:55
  • 2
    \$\begingroup\$ @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. on its own would output true. for non-negative integers). \$\endgroup\$ Aug 7, 2017 at 19:10
  • \$\begingroup\$ How is this 2 bytes? \$\endgroup\$
    – harold
    Aug 8, 2017 at 20:19
  • 1
    \$\begingroup\$ @harold I just updated to make "bytes" in the header link to Brachylog's code-page. A hex dump would be c6 eb. \$\endgroup\$ Aug 8, 2017 at 20:21
16
\$\begingroup\$

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?
\$\endgroup\$
0
8
\$\begingroup\$

Pyth, 4 bytes

q2lP

Test suite.


How?

q2lPQ     - Q is implicit input.

q2        - Is equal to 2?
    lPQ   - The length of the prime factors of the input.
\$\endgroup\$
1
  • \$\begingroup\$ Darn it, shorter builtins! :( \$\endgroup\$
    – hyper-neutrino
    Aug 7, 2017 at 18:28
8
\$\begingroup\$

Mathematica, 16 bytes

PrimeOmega@#==2&

PrimeOmega counts the number of prime factors, counting multiplicity.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Dang, there's a builtin? \$\endgroup\$ Aug 8, 2017 at 1:01
  • 1
    \$\begingroup\$ @JungHwanMin If only there was SemiprimeQ \$\endgroup\$
    – ngenisis
    Aug 8, 2017 at 1:17
  • \$\begingroup\$ Nice. I didn't know about PrimeOmega \$\endgroup\$
    – DavidC
    Aug 8, 2017 at 6:00
8
\$\begingroup\$

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

\$\endgroup\$
4
  • \$\begingroup\$ This claims 8 is semiprime \$\endgroup\$
    – xnor
    Aug 7, 2017 at 19:47
  • \$\begingroup\$ n**(1/3)%1>0<sum... should work. \$\endgroup\$
    – Dennis
    Aug 7, 2017 at 21:20
  • 1
    \$\begingroup\$ @xnor fixed it. \$\endgroup\$
    – Rod
    Aug 7, 2017 at 22:55
  • \$\begingroup\$ Made a tiny edit (e.g. 6 cubed has many more divisors) \$\endgroup\$ Aug 8, 2017 at 19:04
6
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\begingroup\$ 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 \$\endgroup\$
    – Value Ink
    Aug 8, 2017 at 1:20
  • \$\begingroup\$ You are right, it's because I wrote the factorization function for another challenge and I reused it here. \$\endgroup\$
    – G B
    Aug 8, 2017 at 6:10
5
\$\begingroup\$

Mathematica, 31 29 bytes

Tr[Last/@FactorInteger@#]==2&
\$\endgroup\$
4
\$\begingroup\$

Neim, 4 bytes

𝐏𝐥δ𝔼

Try it online!

\$\endgroup\$
7
  • \$\begingroup\$ Can you explain how this is 4 bytes?... I am totally confused. \$\endgroup\$
    – Mr. Xcoder
    Aug 7, 2017 at 18:39
  • \$\begingroup\$ Lol I just had this \$\endgroup\$
    – hyper-neutrino
    Aug 7, 2017 at 18:39
  • \$\begingroup\$ @Mr.Xcoder Neim has a custom code-page \$\endgroup\$ Aug 7, 2017 at 18:39
  • \$\begingroup\$ @Mr.Xcoder Using the Neim codepage, this is 𝐏, 𝐥, δ, and 𝔼 as single-bytes. \$\endgroup\$
    – hyper-neutrino
    Aug 7, 2017 at 18:39
  • \$\begingroup\$ @HyperNeutrino I just obfuscated the 2 a little bit, and now this is the only answer without 2 :) \$\endgroup\$
    – Okx
    Aug 7, 2017 at 18:40
4
\$\begingroup\$

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)

\$\endgroup\$
7
  • \$\begingroup\$ Did you use the code from Dennis' answer? If so, you should give credit. \$\endgroup\$ Aug 7, 2017 at 19:26
  • 2
    \$\begingroup\$ @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! \$\endgroup\$
    – Mr. Xcoder
    Aug 7, 2017 at 19:28
  • 1
    \$\begingroup\$ 67 bytes \$\endgroup\$ Aug 7, 2017 at 20:25
  • \$\begingroup\$ @JonathanAllan Wow, thanks a lot! \$\endgroup\$
    – Mr. Xcoder
    Aug 7, 2017 at 20:27
  • \$\begingroup\$ 55 bytes \$\endgroup\$ Aug 8, 2017 at 9:05
4
\$\begingroup\$

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`, `
)

\$\endgroup\$
4
\$\begingroup\$

Mathematica 32 bytes

Thanks to ngenesis for 1 byte saved

Tr@FactorInteger[#][[;;,2]]==2&
\$\endgroup\$
1
  • 1
    \$\begingroup\$ Save one byte by using ;; instead of All. \$\endgroup\$
    – ngenisis
    Aug 7, 2017 at 21:30
3
\$\begingroup\$

Jelly, 5 bytes

ÆfL=2

Try it online!

Explanation

ÆfL=2  Main link
Æf     Prime factors
  L    Length
   =   Equals
    2  2
\$\endgroup\$
3
\$\begingroup\$

Actually, 4 bytes

ol2=

Try it online!

\$\endgroup\$
3
\$\begingroup\$

05AB1E, 4 bytes

Òg2Q

Try it online!

How?

Ò       prime factors list (with duplicates)
 g      length
   Q    equal to
  2     2
\$\endgroup\$
3
\$\begingroup\$

MATL, 5 bytes

Yfn2=

Try it online!

Explanation

  • Yf - Prime factors.

  • n - Length.

  • 2= - Is equal to 2?

\$\endgroup\$
0
3
\$\begingroup\$

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?

\$\endgroup\$
3
\$\begingroup\$

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?
\$\endgroup\$
4
  • \$\begingroup\$ 4 bytes seems to be a common length, I wonder why... - Probably because all answers are prime factors, length, is equal to 2? \$\endgroup\$
    – Mr. Xcoder
    Aug 8, 2017 at 8:23
  • \$\begingroup\$ @MrXcoder Yep, exactly \$\endgroup\$ Aug 8, 2017 at 11:22
  • \$\begingroup\$ 4 of which are mine BTW >_> \$\endgroup\$
    – Mr. Xcoder
    Aug 8, 2017 at 11:23
  • \$\begingroup\$ 4 is also the first semiprime. Spooky! \$\endgroup\$
    – Neil
    Aug 8, 2017 at 11:47
3
\$\begingroup\$

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!

\$\endgroup\$
2
  • \$\begingroup\$ lambda n:primeomega(n)==2 \$\endgroup\$
    – alephalpha
    Aug 8, 2017 at 3:41
  • \$\begingroup\$ Ah that reminds me to upgrade from 1.0, thanks! \$\endgroup\$ Aug 8, 2017 at 4:01
3
\$\begingroup\$

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!

\$\endgroup\$
3
\$\begingroup\$

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.

\$\endgroup\$
4
  • 2
    \$\begingroup\$ 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;}. \$\endgroup\$
    – Nevay
    Aug 9, 2017 at 13:26
  • \$\begingroup\$ This ties with my regex solution at 61 bytes. \$\endgroup\$
    – Deadcode
    Feb 23 at 2:44
  • 1
    \$\begingroup\$ @Deadcode Actually, yours can be 10 bytes shorter by changing new String(new char[n]) to "x".repeat(n) (Java 11+). :) \$\endgroup\$ Feb 23 at 8:21
  • \$\begingroup\$ Ah, nice! It even works in TIO. Hmm... I know I saw this method before, as I had already upvoted this. Can't think of any reason I wouldn't have just included both, unless it didn't work on TIO back then... but TIO already stopped being updated by then, so I don't know. \$\endgroup\$
    – Deadcode
    Feb 23 at 16:35
3
\$\begingroup\$

J-uby + prime, 34 bytes

:prime_division|:*&:pop|:sum|:==&2

Try it online!

\$\endgroup\$
3
\$\begingroup\$

Regex (ECMAScript or better), 24 23 bytes

(?=((x(x*))\2+)\1+$)\3^

-1 byte thanks to Grimmy

Try it online! - ECMAScript
Try it online! - Python
Try it online! - Ruby

Takes its input in unary, as a string of x characters whose length represents the number.

This regex is based on the 18 byte "Is it prime?" regex (?=(x(x*))\1+$)\2^, which unlike its semiprime version above, is longer than the shortest known, the 16 byte ^(?!(xx+)\1+$)xx. But like ^(?=(xx+?)\1*$)\1$ (also 18 bytes), it has the interesting attribute of implicitly rejecting 0 and 1 as prime numbers, whereas the 16 byte regex must explicitly reject them (the xx at the end).

                  # tail = N = input number; there's no need to anchor here,
                  # because a start anchor is used at the end of this regex
(?=
    (             # \1 = largest proper divisor of N that is >=2
        (x(x*))   # \2 = largest proper divisor of \1; \3 = \2 - 1
        \2+       # Assert that \2 is a proper divisor of \1
    )
    \1+$          # Assert that \1 is a proper divisor of N
)
\3^               # Assert that \3==0, and thus \2==1, meaning the largest proper
                  # divisor of the largest proper divisor of N is 1, meaning N is
                  # semiprime

Regex (Perl / Java / PCRE / Pythonregex / .NET), 23 bytes

^(?>((x(x*))\2+)\1+$)\3

Try it online! - Perl
Try it online! - Java
Try it online! - PCRE
Try it online! - Python import regex
Try it online! - .NET

This is a direct port of the "ECMAScript or better" regex above, using an atomic group instead of an atomic lookahead. It is much faster on most regex engines, as the ^ start-anchor test is done first instead of last.

Regex (Perl / Java / PCRE / Pythonregex), 23 bytes

^((x+)\2*(?=\2$)){2}+x$

Try it online! - Perl
Try it online! - Java
Try it online! - PCRE
Try it online! - Python import regex

This is a port of the now-obsoleted 24 byte "ECMAScript or better" regex, using a possessive quantifier instead of an atomic lookahead to disable backtracking.

^                   # tail = input number
(
    (x+)            # \2 = largest strict divisor of tail
                    #    = tail / {smallest prime factor of tail}
    \2*(?=\2$)      # Assert that \2 is a strict divisor of tail; tail = \2
){2}+               # Execute the loop exactly 2 times, with backtracking disabled
x$                  # assert tail == 1

\$Expressions\ with\ interactive\ input\$

R, 59 58 54 bytes

grepl('^((.+)\\2*(?=\\2$)){2}+.$',strrep(1,scan()),,1)

Try it online! - test cases only

\$Anonymous\ functions\$

R, 63 62 58 53 bytes

\(n)grepl('^((.+)\\2*(?=\\2$)){2}+.$',strrep(1,n),,1)

Attempt This Online!

-1 byte thanks to Giuseppe
-4 bytes by using grepl() instead of sum(grep()) or any(grep())
-5 bytes by using a new anonymous function syntax introduced in R v4.1.0

The old 58 byte R v3.5.2 function: Try it online!

Ruby, 37 36 bytes

->n{?x*n=~/(?=((x(x*))\2+)\1+$)\3^/}

Try it online!
Try it online! - test cases only

PHP, 62 bytes

fn($n)=>preg_match('/^((.+)\2*(?=\2$)){2}+.$/',str_pad('',$n))

Try it online!

JavaScript (ES6), 46 45 bytes

n=>/(?=((.(.*))\2+)\1+$)\3^/.test(Array(n+1))

Try it online!

Java 8, 61 bytes

n->new String(new char[n]).matches("((.+)\\2*(?=\\2$)){2}+.")

Try it online!
Try it online! - test cases only

Java 11, 51 bytes

n->"x".repeat(n).matches("((x+)\\2*(?=\\2$)){2}+x")

Try it online!

\$Full\ programs\$

Retina, 30 29 bytes

.+
$*
^(?>((1(1*))\2+)\1+$)\3

Try it online! - test cases only

PHP, 64 bytes

<?=preg_match('/^((.+)\2*(?=\2$)){2}+.$/',str_pad('',$argv[1]));

Try it online!

\$\endgroup\$
2
  • 1
    \$\begingroup\$ I'll forgive you for missing this conversation. 23 bytes \$\endgroup\$
    – H.PWiz
    Jun 2 at 19:00
  • \$\begingroup\$ @H.PWiz Oh! Thanks. I came up with (?=(x(x*))\1+$)\2^ independently on 2021-03-06 (basing it on my earlier finding of ^(?>(x(x*))\1+$)\2 on 2018-12-07), and didn't realize Grimmy had already come up with it on 2019-04-21. Cool that a method that's longer for primes can be shorter for semiprimes. \$\endgroup\$
    – Deadcode
    Jun 4 at 19:43
2
\$\begingroup\$

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!

\$\endgroup\$
2
\$\begingroup\$

Vyxal, 3 bytes

ǐḢ₃

Try it Online!

\$\endgroup\$
2
\$\begingroup\$

Factor + math.primes.factors, 22 bytes

[ factors length 2 = ]

Try it online!

\$\endgroup\$
1
\$\begingroup\$

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.

\$\endgroup\$
0
1
\$\begingroup\$

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)}");
            Console.ReadLine ();
        }
    }
}

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
\$\endgroup\$
1
\$\begingroup\$

Pari/GP, 17 bytes

n->bigomega(n)==2

Try it online!

\$\endgroup\$
1
\$\begingroup\$

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.

\$\endgroup\$
1
\$\begingroup\$

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
\$\endgroup\$

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