Prime counting function

Question

Introduction

The Prime Counting Function, also known as the Pi function \$\pi(x)\$, returns the amount of primes less than or equal to x.

Challenge

Your program will take an integer x which you can assume to be positive, and output a single integer equal to the amount of primes less than or equal to x. This is a code-golf challenge, so the winner will be the program with the fewest bytes.

You may use any language of your choosing provided that it existed before this challenge went up, but if the language has a built-in prime-counting function or a primality checking function (such as Mathematica), that function cannot be used in your code.

Example Inputs

Input:
1
Output:
0

Input:
2
Output:
1

Input:
5
Output:
3

A000720 - OEIS

Answer 1

05AB1E, 3 bytes

!fg

This assumes that factorization built-ins are allowed. Try it online!

How it works

!    Compute the factorial of the input.
 f   Determine its unique prime factors.
  g  Get the length of the resulting list.

Answer 2

Python 2, 45 bytes

f=lambda n,k=1,p=1:n/k and p%k+f(n,k+1,p*k*k)

Uses the Wilson's Theorem prime generator. The product p tracks (k-1)!^2, and p%k is 1 for primes and 0 for nonprimes.

Answer 3

MATL, 11, 10, 8, 5 bytes

:pYFn

Try it online!

I wrote a version that had a really cool explanation of how MATL's matrices work:

:YF!s1=1

But it's no longer relevant. Check out the revision history if you want to see it.

New explanation:

:p      % Compute factorial(input)
  YF    % Get the exponenents of prime factorization
    n   % Get the length of the array

Three bytes saved thanks to Dennis's genius solution

Answer 4

Jelly, 8 5 bytes

3 bytes saved thanks to @Dennis!

RÆESL

Try it online!

Port of DJMcMayhem's MATL answer (former version) refined by Dennis.

R          Range of input argument
 ÆE        List of lists of exponents of prime-factor decomposition
   S       Vectorized sum. This right-pads inner lists with zeros
    L      Length of result

Answer 5

Perl, 33 bytes

Includes +1 for -p

Give the input number on STDIN

primecount.pl

#!/usr/bin/perl -p
$_=1x$_;$_=s%(?!(11+)\1+$)%%eg-2

Gives the wrong result for 0 but that's OK, the op asked for support for positive integers only.

Answer 6

MediaWiki templates with ParserFunctions, 220 + 19 = 239 bytes

{{#ifexpr:{{{2}}}+1={{{1}}}|0|{{#ifexpr:{{{3}}}={{{2}}}|{{P|{{{1}}}|{{#expr:{{{2}}}+1}}|2}}|{{#ifexpr:{{{2}}} mod {{{3}}}=0|{{#expr:1+{{P|{{{1}}}|{{#expr:{{{2}}}+1}}|2}}|{{P|{{{1}}}|{{{2}}}|{{#expr:{{{2}}}+1}}}}}}}}}}}}

To call the template:

{{{P|{{{n}}}|2|2}}}

Arranged in Lisp style:

{{#ifexpr:{{{2}}} + 1 = {{{1}}}|0|
    {{#ifexpr:{{{3}}} = {{{2}}} |
        {{P|{{{1}}}|{{#expr:{{{2}}} + 1}}|2}} |
            {{#ifexpr:{{{2}}} mod {{{3}}} = 0 |
                {{#expr:1 + {{P|{{{1}}}|{{#expr:{{{2}}} + 1}}|2}} |
                {{P|{{{1}}}|{{{2}}}|{{#expr:{{{2}}} + 1}}}}}}}}}}}}

Just a basic primality test from 2 to n. The numbers with three braces around them are the variables, where {{{1}}} is n, {{{2}}} is the number being tested, {{{3}}} is the factor to check.

Answer 7

Retina, 31 bytes

Byte count assumes ISO 8859-1 encoding. Convert input to unary, generate the range from 1 to n, each on its own line. Match the primes.

.*
$*
\B
¶$`
m`^(?!(..+)\1+$)..

Try it online - Input much larger than 2800 either times out or runs out of memory.

References:

Martin's range generator

Martin's prime checker

Answer 8

Jelly, 13 11 10 9 8 7 6 bytes

Using no built-in prime functions whatsoever
-1 byte thanks to @miles (use a table)
-1 byte thanks to @Dennis (convert from unary to count up the divisors)

ḍþḅ1ċ2

TryItOnline
Or see the first 100 terms of the series n=[1,100], also at TryItOnline

How?

ḍþḅ1ċ2 - Main link: n
 þ     - table or outer product, n implicitly becomes [1,2,3,...n]
ḍ      - divides
  ḅ1   - Convert from unary: number of numbers in [1,2,3,...,n] that divide x
                             (numbers greater than x do not divide x)
    ċ2 - count 2s: count the numbers in [1,2,3,...,n] with exactly 2 divisors
                   (only primes have 2 divisors: 1 and themselves)

Answer 9

JavaScript (ES6), 45 43 bytes

f=(n,x=n)=>n>1&&(--x<2)+(n%x?f(n,x):f(n-1))

A modification of my 36 35 33-byte primality function (1 byte saved by @Neil, 2 by @Arnauld):

f=(n,x=n)=>n>1&--x<2||n%x&&f(n,x)

(I can't post this anywhere because Is this number a prime? only accepts full programs...)

Test snippet

f=(n,x=n)=>n>1&&(--x<2)+(n%x?f(n,x):f(n-1))

<input type="number" oninput="console.log('f('+this.value+') is '+f(this.value))" value=2>

Answer 10

Jelly, 3 bytes

!Æv

Try it online!

How it works

!Æv  Main link. Argument: n

!    Compute the factorial of n.
 Æv  Count the number of distinct prime factors.

Answer 11

C# 5.0 78 77

int F(int n){int z=n;if(n<2)return 0;for(;n%--z!=0;);return(2>z?1:0)+F(n-1);}

Ungolfed

int F(int n)
{
    var z = n;
    if (n < 2) return 0;
    for (; n % --z != 0;) ;
    return F(n - 1) + (2 > z ? 1 : 0);
}

Answer 12

PowerShell v2+, 98 bytes

param($n)if($j='001'[$n]){}else{for($i=1;$i-lt$n){for(;'1'*++$i-match'^(?!(..+)\1+$)..'){$j++}}}$j

Caution: This is slow for large input.

Basically the unary-based lookup from Is this number a prime?, coupled with a for loop and a $j++ counter. A little additional logic on the front to account for edge cases input 1 and 2, due to how the fenceposting works in the for loops.

Answer 13

05AB1E, 5 bytes

Assumes that prime factorization builtins are allowed.

Code:

LÒ1ùg

Explanation:

L      # Get the range [1, ..., input]
 Ò     # Prime factorize each with duplicates
  1ù   # Keep the elements with length 1
    g  # Get the length of the resulting array

Uses the CP-1252 encoding. Try it online!

Answer 14

CJam, 7 bytes

rim!mF,

Try it online! Uses a factorization function.

Explanation:

ri      | read input as integer
  m!    | take the factorial
    mF  | factorize with exponents (one element per prime)
      , | find length

Answer 15

Jelly, 6 bytes

Ḷ!²%RS

This uses only basic arithmetic and Wilson's theorem. Try it online! or verify all test cases.

How it works

Ḷ!²%RS  Main link. Argument: n

Ḷ       Unlength; yield [0, ..., n - 1].
 !      Factorial; yield [0!, ..., (n - 1)!].
  ²     Square; yield [0!², ..., (n - 1)!²].
    R   Range; yield [1, ..., n].
   %    Modulus; yield [0!² % 1, ..., (n - 1)!² % n].
        By a corollary to Wilson's theorem, (k - 1)!² % k yields 1 if k is prime
        and 0 if k is 1 or composite.
     S  Sum; add the resulting Booleans.

Answer 16

APL (Dyalog Unicode), 13 bytes^SBCS

2+.=0+.=⍳∘.|⍳

Try it online!

⍳ ɩndices 1…N
 ⧽ ∘.| remainder-table (using those two as axes)
⍳ ɩndices 1…N

0+.= the sum of the elements equal to zero (i.e. how many dividers does each have)

2+.= the sum of the elements equal to two (i.e. how many primes are there)

Answer 17

Python 3, 40 bytes

f=lambda n:1if n<1else(2**n%n==2)+f(n-1)

An odd integer k is prime if an only if 2**(k-1) is congruent to 1 mod k. Thus, we just check for this condition and add 1 for the case of k=2.

Answer 18

Pyke, 8 6 bytes

SmPs}l

Try it here!

Thanks to Maltysen for new algorithm

SmP    -    map(factorise, input)
   s   -   sum(^)
    }  -  uniquify(^)
     l - len(^)

Answer 19

Pyth - 7 6 bytes

Since others are using prime factorization functions...

l{sPMS

Test Suite.

Answer 20

Bash + coreutils, 30

seq $1|factor|egrep -c :.\\S+$

Ideone.

---

Bash + coreutils + BSD-games package, 22

primes 1 $[$1+1]|wc -l

This shorter answer requires that you have the bsdgames package installed: sudo apt install bsdgames.

Answer 21

C#, 157 bytes

n=>{int c=0,i=1,j;bool f;for(;i<=n;i++){if(i==1);else if(i<=3)c++;else if(i%2==0|i%3==0);else{j=5;f=1>0;while(j*j<=i)if(i%j++==0)f=1<0;c+=f?1:0;}}return c;};

Full program with test cases:

using System;

class a
{
    static void Main()
    {
        Func<int, int> s = n =>
            {
                int c = 0, i = 1, j;
                bool f;
                for (; i <= n; i++)
                {
                    if (i == 1) ;
                    else if (i <= 3) c++;
                    else if (i % 2 == 0 | i % 3 == 0) ;
                    else
                    {
                        j = 5;
                        f = 1 > 0;
                        while (j * j <= i)
                            if (i % j++ == 0)
                                f = 1 < 0;
                        c += f ? 1 : 0;
                    }
                }
                return c;
            };

        Console.WriteLine("1 -> 0 : " + (s(1) == 0 ? "OK" : "FAIL"));
        Console.WriteLine("2 -> 1 : " + (s(2) == 1 ? "OK" : "FAIL"));
        Console.WriteLine("5 -> 3 : " + (s(5) == 3 ? "OK" : "FAIL"));
        Console.WriteLine("10 -> 4 : " + (s(10) == 4 ? "OK" : "FAIL"));
        Console.WriteLine("100 -> 25 : " + (s(100) == 25 ? "OK" : "FAIL"));
        Console.WriteLine("1,000 -> 168 : " + (s(1000) == 168 ? "OK" : "FAIL"));
        Console.WriteLine("10,000 -> 1,229 : " + (s(10000) == 1229 ? "OK" : "FAIL"));
        Console.WriteLine("100,000 -> 9,592 : " + (s(100000) == 9592 ? "OK" : "FAIL"));
        Console.WriteLine("1,000,000 -> 78,498 : " + (s(1000000) == 78498 ? "OK" : "FAIL"));
    }
}

Starts to take awhile once you go above 1 million.

Answer 22

Jelly, 3 bytes

ÆRL

Jelly has a built-in prime counting function, ÆC and a prime checking function, ÆP, this instead uses a built-in prime generating function, ÆR and takes the length L.

I guess this is about as borderline as using prime factorisation built-ins, which would also take 3 bytes with !Æv (! factorial, Æv count prime factors)

Answer 23

Matlab, 60 bytes

Continuing my attachment to one-line Matlab functions. Without using a factorisation built-in:

f=@(x) nnz(arrayfun(@(x) x-2==nnz(mod(x,[1:1:x])),[1:1:x]));

Given that a prime y has only two factors in [1,y]: we count the numbers in the range [1,x] which have only two factors.

Using factorisation allows for significant shortening (down to 46 bytes).

g=@(x) size(unique(factor(factorial(x))),2);

Conclusion: Need to look into them golfing languages :D

Answer 24

MATL, 9 bytes

This avoids prime-factor decomposition. Its complexity is O(n²).

:t!\~s2=s

Try it online!

:     % Range [1 2 ... n] (row vector)
t!    % Duplicate and transpose into a column vector
\     % Modulo with broadcast. Gives matrix in which entry (i,j) is i modulo j, with
      % i, j in [1 2 ... n]. A value 0 in entry (i,j) means i is divisible by j
~     % Negate. Now 1 means i is divisible by j
s     % Sum of each column. Gives row vector with the number of divisors of each j
2=    % Compare each entry with 2. A true result corresponds to a prime
s     % Sum

Answer 25

PHP, 96 92 bytes

for($j=$argv[1]-1;$j>0;$j--){$p=1;for($i=2;$i<$j;$i++)if(is_int($j/$i))$p=0;$t+=$p;}echo $t;

Saved 4 bytes thanks to Roman Gräf

Test online

Ungolfed testing code:

$argv[1] = 5;

for($j=$argv[1]-1;$j>0;$j--) {
    $p=1;
    for($i=2;$i<$j;$i++) {
        if(is_int($j/$i)) {
            $p=0;
        }
    }
    $t+=$p;
}
echo $t;

Test online

Answer 26

Actually, 10 bytes

This was the shortest solution I found that didn't run into interpreter bugs on TIO. Golfing suggestions welcome. Try it online!

;╗r`P╜>`░l

Ungolfing

         Implicit input n.
;╗       Duplicate n and save a copy of n to register 0.
r        Push range [0..(n-1)].
`...`░   Push values of the range where the following function returns a truthy value.
  P        Push the a-th prime
  ╜        Push n from register 0.
  >        Check if n > the a-th prime.
l        Push len(the_resulting_list).
         Implicit return.

Answer 27

Java 8, 95 84 bytes

z->{int c=0,n=2,i,x;for(;n<=z;c+=x>1?1:0)for(x=n++,i=2;i<x;x=x%i++<1?0:x);return c;}

Explanation:

Try it online.

z->{                // Method with integer parameter and integer return-type
  int c=0,          //  Start the counter at 0
      n=2,          //  Starting prime is 2
      i,x;          //  Two other temp integers
  for(;n<=z;        //  Loop (1) as long as `n` is smaller than or equal to the input `z`
       c+=x>1?1:0)  //    and increase the counter if we've came across a prime
                    //    (if `x` is larger than 0, it means the current `n` is a prime)
     for(x=n++,i=2;i<x;x=x%i++<1?0:x);
                    //   Determine if the next integer in line is a prime by setting `x`
                    //   (and increase `n` by 1 afterwards)
                    //  End of loop (1) (implicit / single-line body)
  return c;         //  Return the resulting counter
}                   // End of method

Answer 28

Perl 6, 31 bytes

{+grep {$_%%none 2..^$_},2..$_}

Try it online!

Answer 29

(pure) BASH, 138 bytes

Look, Mom!
No preloaded prime tables, no prime related builtins, not even a division or plultimication!

a=2
l=0
while((a<=$1));do if((b[a]))
then((c=b[a]));else((c=a,l++));fi;((d=a+c))
while((b[d]));do((d+=c));done
((b[d]=c,a++));done;echo $l

Eating the pudding:

$ for i in 1 2 3 11033 ; do bash fsoe-primecounter-in-bash $i ; done
0
1
2
1337

Ok, probably looping while looking at candidates modulo all possible divisors would have been shorter in "pure BASH" too but we've seen those loops at least 1001 times now.

This solution builds its own prime table on the fly and so is bearably fast even in "pure BASH".

Answer 30

APL NARS, 150 bytes

∇r←g n;h;i;k
i←0⋄h←1+⍳n-1⋄→B
A:k←i⊃h⋄h←k∪(0≠k∣h)/h
B:→A×⍳(⍴h)≥i+←1
r←⍴h
∇

this would be the "Crivello di Eratostene" or something as that.

If h=2..n

In the first iteration of the loop eliminate each multiple of 2 in h except 2

in the second iteration of the loop eliminate each multiple of 3 in h except 3

...

return the number of final element of h.(in h primes return ⍴h)

  g¨1 2 5
0  1  3 
  g 20000
2262

As note negative for the question: There would be a range limit for the arg of the function; something as 1..20000 and each answer would indicate the range the argument has to be for to be safe to pass to the function (without one seg fault in the code, without one memory insufficient for the program, without one incorrect value return as result )

Answer 31

Pyth, 6 bytes

l{P.!Q

Explanation follows:

   .!Q Factorial of input
  P    List primes of factorial with multiplicity
 {     Remove duplicates formed by said multiplicity
l      Length of deduplicated list

Answer 32

Add++, 17 bytes

L,RB*f0b]@bLA1=!*

Try it online!

Answer 33

Kotlin, 85 bytes

{n:Int->var r=0
for(i in 2..n){var t=1
for(p in 2..i/2)if(i%p==0){t--
break}
r+=t}
r}

Try it online!

Try lambda for first time as saved 16 bytes.

Kotlin, 101 bytes

fun c(n:Int):Int{var r=0
for(i in 2..n){var t=1
for(p in 2..i/2)if(i%p==0){t--
break}
r+=t}
return r}

Try it online!

Answer 34

JavaScript (ES6), 50+2 46+2 43 bytes

Saved 3 5 bytes thanks to Neil:

f=n=>n&&eval(`for(z=n;n%--z;);1==z`)+f(n-1)

eval can access the n parameter.
The eval(...) checks if n is prime.

---

Previous solutions:
Byte count should be +2 because I forgot to name the function f= (needed for recursion)

46+2 bytes (Saved 3 bytes thanks to ETHproductions):

n=>n&&eval(`for(z=n=${n};n%--z;);1==z`)+f(n-1)

50+2 bytes:

n=>n&&eval(`for(z=${n};${n}%--z&&z;);1==z`)+f(n-1)

Answer 35

Java 7,102 bytes

Brute force

int f(int n){int i=2,j=2,c=1,t=0;for(;i<=n;j=2,c+=t==1?1:0,i++)for(;j<i;t=i%j++==0?j=i+1:1);return c;}

Ungolfed

int f(int n){
int i=2,j=2,c=1,t=0;
for(;i<=n;j=2,c+=t==1?1:0,i++)
    for(;j<i;)
        t=i%j++==0?j=i+1:1;
    return c;
 }

Answer 36

q 35 bytes

{sum 1=sum'[0=2_' a mod\: a:til x]}

Answer 37

Actually, 10 bytes

If my first Actually answer is disallowed for using a prime-generating function, here is a backup answer using Wilson's theorem. Golfing suggestions welcome. Try it online!

R`;D!²%`MΣ

Try it online

         Implicit input n.
R        Push range [1..n]
`...`M   Map the following function over the range. Variable k.
  ;        Duplicate k.
  D        Decrement one of the copies of k.
  !²       Push ((k-1)!)².
  %        Push ((k-1)!)² % k. This returns 1 if k is prime, else 0.
Σ        Sums the result of the map, adding all the 1s that represent primes, 
          giving the total number of primes less than n.
         Implicit return.

Answer 38

Racket 60 bytes

(require math)(λ(n)(length(filter prime?(range 1(+ 1 n)))))

Testing:

(require math)
(define f
    (λ(n) (length (filter prime? (range 1 (+ 1 n)))))
)

(f 1)
(f 2)
(f 3)
(f 5)

Output:

0
1
2
3

Answer 39

PARI/GP, 15 bytes

n->#factor(n!)~

Take the factorial and count its unique prime factors.

Answer 40

Actually, 3 bytes

This uses a prime factorization function, which may or may not be allowed after the OP clarifies. Golfing suggestions welcome. Try it online!

!yl

Ungolfing

     Implicit input n.
!    Push n factorial.
 y   Push a list of all of the positive prime factors of n! (every prime < n)
  l  Take the length of that list of primes.
     Implicit return.

Answer 41

Pyt, 3 bytes

řṗƩ

Explanation:

ř         Push [1,2,...,input]
 ṗ        Elementwise isPrime(k)
  Ʃ       Sum of array (autoconverts booleans to 0/1)

Try it online!