Minimum primes

From Rosetta Code
Minimum primes is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Task


Given three lists:

  • Numbers1 = [5,45,23,21,67]
  • Numbers2 = [43,22,78,46,38]
  • Numbers3 = [9,98,12,54,53]


then:

  1. Select the maximum (max) of Numbers[n], Numbers2[n] and Numbers3[n], where n <= 5 (one based).
  2. For each value of max, find the least prime, minPrime, such that minPrime >= max
  3. Add minPrime to a new list (Primes)
  4. Show Primes on this page.



ALGOL 68[edit]

Translation of: Wren

Can handle the possibility of the maximum elements being negative, 0, 1 or 2.

BEGIN # show the minimum prime >= the maximum elements of three lists #
    PR read "primes.incl.a68" PR
    []INT numbers1 = (  5, 45, 23, 21, 67 );
    []INT numbers2 = ( 43, 22, 78, 46, 38 );
    []INT numbers3 = (  9, 98, 12, 54, 53 );
    [ 1 : UPB numbers1 ]INT prime list;
    INT max element := numbers1[ 1 ];
    FOR i TO UPB numbers1 DO
        INT m := numbers1[ i ];
        IF numbers2[ i ] > m THEN m := numbers2[ i ] FI;
        IF numbers3[ i ] > m THEN m := numbers3[ i ] FI;
        IF m > max element THEN max element := m FI;
        prime list[ i ] := m
    OD;
    # construct a sieve of primes big enough for the maximum element #
    []BOOL prime = PRIMESIEVE ( max element * 2 );
    # replace the elements of prime list wih the smallest prime >= the element #
    FOR i TO UPB prime list DO
        INT m := prime list[ i ];
        # find the next prime >= m #
        IF   m <= 2 THEN m := 2 
        ELSE
            IF NOT ODD m THEN m +:= 1 FI;
            WHILE NOT prime[ m ] DO m +:= 2 OD
        FI;
        prime list[ i ] := m
    OD;
    print( ( "[" ) );
    FOR i TO UPB prime list DO print( ( " ", whole( prime list[ i ], 0 ) ) ) OD;
    print( ( " ]" ) )
END
Output:
[ 43 101 79 59 67 ]

AWK[edit]

# syntax: GAWK -f MINIMUM_PRIMES.AWK
BEGIN {
    n1 = split("5,45,23,21,67",numbers1,",")
    n2 = split("43,22,78,46,38",numbers2,",")
    n3 = split("9,98,12,54,53",numbers3,",")
    if (n1 != n2 || n1 != n3) {
      print("error: arrays must be same length")
      exit(1)
    }
    for (i=1; i<=n1; i++) {
      m = max(max(numbers1[i],numbers2[i]),numbers3[i])
      if (m % 2 == 0) { m++ }
      while (!is_prime(m)) { m += 2 }
      primes[i] = m
      printf("%d ",primes[i])
    }
    printf("\n")
    exit(0)
}
function is_prime(x,  i) {
    if (x <= 1) {
      return(0)
    }
    for (i=2; i<=int(sqrt(x)); i++) {
      if (x % i == 0) {
        return(0)
      }
    }
    return(1)
}
function max(x,y) { return((x > y) ? x : y) }
Output:
43 101 79 59 67


BASIC256[edit]

Translation of: FreeBASIC
dim Num1 = { 5,45,23,21,67}
dim Num2 = {43,22,78,46,38}
dim Num3 = { 9,98,12,54,53}

print "The minimum prime numbers of three lists"
print "[ ";
for n = 0 to 4
	maxi = max(Num1[n], max(Num2[n], Num3[n]))
	if maxi mod 2 = 0 then maxi += 1
	while not isPrime(maxi)
		maxi += 2
	end while
	print maxi; " ";
next n
print "]"
end

function max(a, b)
	if a > b then return a else return b
end function

function isPrime(v)
	if v < 2 then return False
	if v mod 2 = 0 then return v = 2
	if v mod 3 = 0 then return v = 3
	d = 5
	while d * d <= v
		if v mod d = 0 then return False else d += 2
	end while
	return True
end function


C[edit]

Translation of: Wren
#include <stdio.h>

#define TRUE 1
#define FALSE 0

int isPrime(int n) {
    int d;
    if (n < 2) return FALSE;
    if (n%2 == 0) return n == 2;
    if (n%3 == 0) return n == 3;
    d = 5;
    while (d*d <= n) {
        if (!(n%d)) return FALSE;
        d += 2;
        if (!(n%d)) return FALSE;
        d += 4;
    }
    return TRUE;
}

int max(int a, int b) {
    if (a > b) return a;
    return b;
}

int main() {
    int n, m;
    int numbers1[5] = { 5, 45, 23, 21, 67};
    int numbers2[5] = {43, 22, 78, 46, 38};
    int numbers3[5] = { 9, 98, 12, 54, 53};
    int primes[5]   = {};
    for (n = 0; n < 5; ++n) {
        m = max(max(numbers1[n], numbers2[n]), numbers3[n]);
        if (!(m % 2)) m++;
        while (!isPrime(m)) m += 2;
        primes[n] = m;
        printf("%d ", primes[n]);
    }
    printf("\n");
    return 0;
}
Output:
43 101 79 59 67 


C#[edit]

Translation of: Ring
...solution #1.
using System;
using System.Linq;
using static System.Console;

class Program {

  static int nxtPrime(int x) {
    int j = 2; do {
        if (x % j == 0) { j = 2; x++; }
        else j += j < 3 ? 1 : 2;
    } while (j * j <= x); return x; }

  static void Main(string[] args) {
    WriteLine("working...");
    int[] Num1 = new int[]{  5, 45, 23, 21, 67 },
          Num2 = new int[]{ 43, 22, 78, 46, 38 },
          Num3 = new int[]{  9, 98, 12, 54, 53 };
    int n = Num1.Length; int[] Nums = new int[n];
    for (int i = 0; i < n; i++)
      Nums[i] = nxtPrime(new int[]{ Num1[i], Num2[i], Num3[i] }.Max());
    WriteLine("The minimum prime numbers of three lists = [{0}]", string.Join(",", Nums));
    Write("done..."); } }
Output:

Same as Ring.

F#[edit]

This task uses Extensible Prime Generator (F#)

// Minimum primes. Nigel Galloway: October 29th., 2021
let N1,N2,N3=[5;45;23;21;67],[43;22;78;46;38],[9;98;12;54;53]
let fN g=primes32()|>Seq.find((<=)g)
printfn "%A" (List.zip3 N1 N2 N3|>List.map(fun(n,g,l)->fN(max (max n g) l)))
Output:
[43; 101; 79; 59; 67]

Factor[edit]

USING: math math.order math.primes prettyprint sequences ;

{ 5 45 23 21 67 } { 43 22 78 46 38 } { 9 98 12 54 53 }
[ max max 1 - next-prime ] 3map .
Output:
{ 43 101 79 59 67 }


FreeBASIC[edit]

#define MAX(a, b) iif((a) > (b), (a), (b))

Function isPrime(Byval ValorEval As Integer) As Boolean
    If ValorEval < 2 Then Return False
    If ValorEval Mod 2 = 0 Then Return ValorEval = 2
    If ValorEval Mod 3 = 0 Then Return ValorEval = 3
    Dim d As Integer = 5
    While d * d <= ValorEval
        If ValorEval Mod d = 0 Then Return False Else d += 2
    Wend 
    Return True
End Function

Dim As Integer Num1(5) = { 5,45,23,21,67}
Dim As Integer Num2(5) = {43,22,78,46,38}
Dim As Integer Num3(5) = { 9,98,12,54,53}

Print "The minimum prime numbers of three lists..."
Print "[";
For n As Integer = 0 To 4
    Dim As Integer maxi = MAX(num1(n), MAX(num2(n), num3(n)))
    If (maxi Mod 2 = 0) Then maxi += 1
    While Not isPrime(maxi)
        maxi += 2
    Wend
    Print maxi; ", ";
Next n
Print !"\b\b ]"
Sleep
Output:
[ 43, 101, 79, 59, 67 ]


Go[edit]

Translation of: Wren
Library: Go-rcu
package main

import (
    "fmt"
    "rcu"
)

func main() {
    numbers1 := [5]int{5, 45, 23, 21, 67}
    numbers2 := [5]int{43, 22, 78, 46, 38}
    numbers3 := [5]int{9, 98, 12, 54, 53}
    primes := [5]int{}
    for n := 0; n < 5; n++ {
        max := rcu.Max(rcu.Max(numbers1[n], numbers2[n]), numbers3[n])
        if max % 2 == 0 {
            max++
        }
        for !rcu.IsPrime(max) {
            max += 2
        }
        primes[n] = max
    }
    fmt.Println(primes)
}
Output:
[43 101 79 59 67]

jq[edit]

Works with: jq

Works with gojq, the Go implementation of jq

This entry uses `is_prime` as defined, for example, at Erdős-primes#jq.

Two solutions are presented following these preliminaries:

include "is_prime";  # reminder

def Numbers1: [5,45,23,21,67];
def Numbers2: [43,22,78,46,38];
def Numbers3: [9,98,12,54,53];

# Generate primes in range(m;n) provided m>=2
def primes(m; n):
  if m%2 == 0 then primes(m+1;n)
  else range(m; n; 2) | select(is_prime)
  end;

Explicit Iteration

[range(0;5)
 | [Numbers1[.], Numbers2[.], Numbers3[.]] | max
 | first(primes(.; infinite))]

Functional

[Numbers1, Numbers2, Numbers3]
 | transpose
 | [map(max | first(primes(.; infinite)))]
Output:
[43,101,79,59,67]

Julia[edit]

using Primes

println(nextprime.(maximum(hcat([5,45,23,21,67], [43,22,78,46,38], [9,98,12,54,53]), dims=2)))
Output:
[43; 101; 79; 59; 67;;]

Mathematica / Wolfram Language[edit]

minPrime[x_List] :=
 If[PrimeQ@Max@x, Max@x, NextPrime@Max@x]
 
 MapThread[
 minPrime@{##} &, {{5., 45, 23, 21, 67}, {43, 22, 78, 46, 38}, {9, 98,
    12, 54, 53}}]
Output:

{43,101,79,59,67}

Nim[edit]

const
  Numbers1 = [ 5, 45, 23, 21, 67]
  Numbers2 = [43, 22, 78, 46, 38]
  Numbers3 = [ 9, 98, 12, 54, 53]

var numbers: array[0..Numbers1.high, int]

template isEven(n: int): bool = (n and 1) == 0

func isPrime(n: Positive): bool =
  if n < 2: return false
  if n.isEven: return n == 2
  if n mod 3 == 0: return n == 3
  var k = 5
  var delta = 2
  while k * k <= n:
    if n mod k == 0: return false
    inc k, delta
    delta = 6 - delta
  result = true

func minPrime(n: int): int =
  if n == 2: return 2
  result = if n.isEven: n + 1 else: n
  while not result.isPrime():
    inc result, 2

for i in 0..numbers.high:
  let m = max(max(Numbers1[i], Numbers2[i]), Numbers3[i])
  numbers[i] = minPrime(m)

echo numbers
Output:
[43, 101, 79, 59, 67]

Perl[edit]

Library: ntheory
#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Minimum_primes
use warnings;
use ntheory qw( next_prime );
use List::Util qw( max );

my @Numbers1 = (5,45,23,21,67);
my @Numbers2 = (43,22,78,46,38);
my @Numbers3 = (9,98,12,54,53);

my @Primes = map {
  next_prime( max( $Numbers1[$_], $Numbers2[$_], $Numbers3[$_] ) - 1 )
  } 0 .. 4;

print "@Primes\n";
Output:
43 101 79 59 67

Phix[edit]

with javascript_semantics
function nextprime(sequence s)
    sequence res = repeat(0,length(s[1]))
    for i=1 to length(res) do
        res[i] = get_prime(length(get_primes_le(maxsq(vslice(s,i))-1))+1)
    end for
    return res
end function
printf(1,"%v\n",{nextprime({{ 5, 45, 23, 21, 67},
                            {43, 22, 78, 46, 38},
                            { 9, 98, 12, 54, 53}})})
Output:
{43,101,79,59,67}

Raku[edit]

Seems kind of pointless to specify a maximum of 5 terms when there are only 5 elements in each list but... ¯\_(ツ)_/¯

say ([Zmax] <5 45 23 21 67>, <43 22 78 46 38>, <9 98 12 54 53>)».&next-prime[^5];

sub next-prime { ($^m..*).first: &is-prime }
Output:
(43 101 79 59 67)

Ring[edit]

Solution #1[edit]

? "working..."

Num1 = [ 5,45,23,21,67]
Num2 = [43,22,78,46,38]
Num3 = [ 9,98,12,54,53]
n = len(Num1)
Nums = list(n)
 
for i = 1 to n
    Nums[i] = nxtPrime(max([Num1[i], Num2[i], Num3[i]]))
next

? "The minimum prime numbers of three lists = " + fmtArray(Nums)
put "done..."

func fmtArray(ar)
    rv = ar[1]
    for n = 2 to len(ar) rv += "," + ar[n] next
    return "[" + rv + "]"

func nxtPrime(x)
    j = 2
    while true
        if x % j = 0 j = 2 x++
        else j++ ok
        if j * j > x exit ok
    end return string(x)
Output:
working...
The minimum prime numbers of three lists = [43,101,79,59,67]
done...

Solution #2[edit]

load "stdlib.ring"
see "working..." + nl

Primes = []
Numbers1 = [5,45,23,21,67]
Numbers2 = [43,22,78,46,38]
Numbers3 = [9,98,12,54,53]

for n = 1 to len(Numbers1)
    Temp = []
    add(Temp,Numbers1[n])
    add(Temp,Numbers2[n])
    add(Temp,Numbers3[n])
    max = max(Temp)
    max--
    while true 
          max++
          if isprime(max) 
             exit
          ok
    end
    add(Primes,max) 
next

see "Minimum primes = "
see showArray(Primes)
see nl + "done..." + nl

func showArray(array)
     txt = ""
     see "["
     for n = 1 to len(array)
         txt = txt + array[n] + ","
     next
     txt = left(txt,len(txt)-1)
     txt = txt + "]"
     see txt
Output:
working...
Minimum primes = [43,101,79,59,67]
done...

Ruby[edit]

require "prime"
numbers1 = [ 5, 45, 23, 21, 67]
numbers2 = [43, 22, 78, 46, 38]
numbers3 = [ 9, 98, 12, 54, 53]

p [numbers1, numbers2, numbers3].transpose.map{|ar| (ar.max..).find(&:prime?) }
Output:
[43, 101, 79, 59, 67]

Sidef[edit]

var lists = [
    [ 5,45,23,21,67],
    [43,22,78,46,38],
    [ 9,98,12,54,53],
]

say lists.zip.map { next_prime(.max - 1) }
Output:
[43, 101, 79, 59, 67]

Wren[edit]

Library: Wren-math
import "./math" for Int

var numbers1 = [ 5, 45, 23, 21, 67]
var numbers2 = [43, 22, 78, 46, 38]
var numbers3 = [ 9, 98, 12, 54, 53]
var primes   = List.filled(5, 0)
for (n in 0..4) {
    var max = numbers1[n].max(numbers2[n]).max(numbers3[n])
    if (max % 2 == 0) max = max + 1
    while(!Int.isPrime(max)) max = max + 2
    primes[n] = max
}
System.print(primes)
Output:
[43, 101, 79, 59, 67]

XPL0[edit]

func IsPrime(N);        \Return 'true' if N is a prime number
int  N, I;
[if N <= 1 then return false;
for I:= 2 to sqrt(N) do
    if rem(N/I) = 0 then return false;
return true;
];

int Numbers1, Numbers2, Numbers3, N, Max;
[Numbers1:= [5,45,23,21,67];
 Numbers2:= [43,22,78,46,38];
 Numbers3:= [9,98,12,54,53];
 for N:= 0 to 4 do
        [Max:= Numbers1(N);
        if Numbers2(N) > Max then Max:= Numbers2(N);
        if Numbers3(N) > Max then Max:= Numbers3(N);
        while not IsPrime(Max) do Max:= Max+1;
        IntOut(0, Max);  ChOut(0, ^ );
        ];
]
Output:
43 101 79 59 67