Semiprime

Semiprime
You are encouraged to solve this task according to the task description, using any language you may know.

Semiprime numbers are natural numbers that are products of exactly two (possibly equal) prime numbers. Example: 1679 = 23 × 73 (This particular number was chosen as the length of the Arecibo message).

Write a function determining whether a given number is semiprime.

Haskell

Library: Data.Numbers.Primes

<lang Haskell>isSemiprime :: Int -> Bool isSemiprime n = (length factors) == 2 && (product factors) == n ||

               (length factors) == 1 && (head factors) ^ 2 == n
                   where factors = primeFactors n</lang>

Alternative (and faster) implementation using pattern matching: <lang Haskell>isSemiprime :: Int -> Bool isSemiprime n = case (primeFactors n) of

                  [f1, f2] -> f1 * f2 == n
                  otherwise -> False</lang>

Perl 6

Here is a naive, grossly inefficient implementation. <lang perl6>sub is-semiprime (Int $n --> Bool) {

   return False if $n.is-prime;
   return .is-prime given 
       $n div first $n %% *,
           grep &is-prime, 2 .. *;

}

use Test; my @primes = grep &is-prime, 2 .. 100; for ^5 {

   ok not is-semiprime(@primes.pick);
   ok is-semiprime([*] @primes.roll(2));
   ok not is-semiprime([*] @primes.roll(3));
   ok not is-semiprime([*] @primes.roll(4));

}</lang>

Output:

ok 1 - 
ok 2 - 
ok 3 - 
ok 4 - 
ok 5 - 
ok 6 - 
ok 7 - 
ok 8 - 
ok 9 - 
ok 10 - 
ok 11 - 
ok 12 - 
ok 13 - 
ok 14 - 
ok 15 - 
ok 16 - 
ok 17 - 
ok 18 - 
ok 19 - 
ok 20 -

Python

This imports Prime decomposition#Python <lang python>from prime_decomposition import decompose

def semiprime(n):

   d = decompose(n)
   try:
       return d.next() * d.next() == n
   except:
       return False</lang>

Output:

From Idle: <lang python>>>> semiprime(1679) True >>> [n for n in range(1,101) if semiprime(n)] [4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 33, 34, 35, 38, 39, 46, 49, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95] >>> </lang>

Racket

The first implementation considers all pairs of factors multiplying up to the given number and determines if any of them is a pair of primes. <lang Racket>#lang racket (require math)

(define (pair-factorize n)

 "Return all two-number factorizations of a number"
 (let ([up-limit (integer-sqrt n)])
   (map (λ (x) (list x (/ n x)))

(filter (λ (x) (<= x up-limit)) (divisors n)))))

(define (semiprime n)

 "Determine if a number is semiprime i.e. a product of two primes.

Check if any pair of complete factors consists of primes."

 (for/or ((pair (pair-factorize n)))
   (for/and ((el pair))
     (prime? el))))</lang>

The alternative implementation operates directly on the list of prime factors and their multiplicities. It is approximately 1.6 times faster than the first one (according to some simple tests of mine). <lang Racket>#lang racket (require math)

(define (semiprime n)

 "Alternative implementation.

Check if there are two prime factors whose product is the argument or if there is a single prime factor whose square is the argument"

 (let ([prime-factors (factorize n)])
   (or (and (= (length prime-factors) 1)

(= (expt (caar prime-factors) (cadar prime-factors)) n)) (and (= (length prime-factors) 2) (= (foldl (λ (x y) (* (car x) y)) 1 prime-factors) n)))))</lang>

REXX

<lang rexx>/* REXX ---------------------------------------------------------------

• 20.02.2014 Walter Pachl relying on prime decomposition
• --------------------------------------------------------------------*/

Call test 4 Call test 9 Call test 10 Call test 12 Call test 1679 Exit

test: Parse Arg z If isprime(z) Then Say z 'is semiprime' fl

             Else Say z 'is NOT semiprime' fl

Return

isprime:

 Parse Arg z
 fl=factr(z)
 Return words(fl)=2

/*----------------------------------FACTR subroutine-----------------*/ factr: procedure; parse arg x 1 z,list /*sets X&Z to arg1, LIST=. */ if x==1 then return /*handle the special case of X=1.*/ j=2; call .factr /*factor for the only even prime.*/ j=3; call .factr /*factor for the 1st odd prime.*/ j=5; call .factr /*factor for the 2nd odd prime.*/ j=7; call .factr /*factor for the 3rd odd prime.*/ j=11; call .factr /*factor for the 4th odd prime.*/ j=13; call .factr /*factor for the 5th odd prime.*/ j=17; call .factr /*factor for the 6th odd prime.*/

                                   /* [?]   could be optimized more.*/
                                   /* [?]   J in loop starts at 17+2*/
    do y=0  by 2;     j=j+2+y//4   /*insure J isn't divisible by 3. */
    if right(j,1)==5  then iterate /*fast check for divisible by 5. */
    if j*j>z          then leave   /*are we higher than the v of Z ?*/
    if j>Z            then leave   /*are we higher than value of Z ?*/
    call .factr                    /*invoke .FACTR for some factors.*/
    end   /*y*/                    /* [?]  only tests up to the v X.*/
                                   /* [?]  LIST has a leading blank.*/

if z==1 then return list /*if residual=unity, don't append*/

             return list z         /*return list,  append residual. */

/*-------------------------------.FACTR internal subroutine----------*/ .factr: do while z//j==0 /*keep dividing until we can't. */

        list=list j                /*add number to the list  (J).   */
        z=z%j                      /*% (percent)  is integer divide.*/
        end   /*while z··· */      /*  //   ?---remainder integer ÷.*/

return /*finished, now return to invoker*/</lang> Output

4 is semiprime  2 2
9 is semiprime  3 3
10 is semiprime  2 5
12 is NOT semiprime  2 2 3
1679 is semiprime  23 73

Ruby

<lang ruby>require 'prime'

1. 75.prime_division # Returns the factorization.75 divides by 3 once and by 5 twice => [[3, 1], [5, 2]]

class Integer

 def semi_prime?
   prime_division.map( &:last ).inject( &:+ ) == 2
 end

end

p 1679.semi_prime? # true p ( 1..100 ).select( &:semi_prime? )

1. [4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 33, 34, 35, 38, 39, 46, 49, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95]

</lang>