Minimum primes
Appearance
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:
- Select the maximum (max) of Numbers[n], Numbers2[n] and Numbers3[n], where n <= 5 (one based).
- For each value of max, find the least prime, minPrime, such that minPrime >= max
- Add minPrime to a new list (Primes)
- Show Primes on this page.
ALGOL 68
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 ]
Arturo
lists: [
[ 5 45 23 21 67]
[43 22 78 46 38]
[ 9 98 12 54 53]
]
print map 0..dec size first lists 'i ->
first select.first (max map lists 'l -> l\[i])..∞ => prime?
- Output:
43 101 79 59 67
AWK
# 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
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
#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#
...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.
C++
#include <algorithm>
#include <cstdint>
#include <iostream>
#include <iterator>
#include <vector>
uint32_t next_prime(uint32_t number) {
uint32_t divisor = 2;
while ( divisor * divisor <= number ) {
if ( number % divisor == 0 ) {
number++;
divisor = 2;
} else {
divisor++;
}
}
return number;
}
int main() {
const std::vector<uint32_t> numbers1{ 5, 45, 23, 21, 67 };
const std::vector<uint32_t> numbers2{ 43, 22, 78, 46, 38 };
const std::vector<uint32_t> numbers3{ 9, 98, 12, 54, 53 };
std::vector<uint32_t> primes{};
for ( uint32_t n = 0; n < 5; ++n ) {
const uint32_t max = std::max({ numbers1[n], numbers2[n], numbers3[n] });
primes.emplace_back(next_prime(max));
}
std::copy(primes.begin(), primes.end(), std::ostream_iterator<uint32_t>(std::cout, " "));
}
- Output:
43 101 79 59 67
Delphi
function IsPrime(N: int64): boolean;
{Fast, optimised prime test}
var I,Stop: int64;
begin
if (N = 2) or (N=3) then Result:=true
else if (n <= 1) or ((n mod 2) = 0) or ((n mod 3) = 0) then Result:= false
else
begin
I:=5;
Stop:=Trunc(sqrt(N+0.0));
Result:=False;
while I<=Stop do
begin
if ((N mod I) = 0) or ((N mod (I + 2)) = 0) then exit;
Inc(I,6);
end;
Result:=True;
end;
end;
type TIntArray = array of integer;
type TNumList = array [0..2, 0..4] of integer;
const NumLists: TNumList = (
(5,45,23,21,67),
(43,22,78,46,38),
(9,98,12,98,53));
procedure GetColPrimes(NumList: TNumList; var ColPrimes: TIntArray);
{Get the Maxium value, find next prime and store result in array}
var X,Y,I,M: integer;
var Highest: integer;
begin
for X:=0 to High(NumLists[0]) do
begin
Highest:=0;
for Y:=0 to High(NumList) do
if NumLists[Y,X]>Highest then Highest:=NumList[Y,X];
SetLength(ColPrimes,Length(ColPrimes)+1);
ColPrimes[High(ColPrimes)]:=Highest;
end;
for I:=0 to High(ColPrimes) do
begin
M:=ColPrimes[I];
if (M mod 2)=0 then Inc(M);
while not IsPrime(M) do Inc(M,2);
ColPrimes[I]:=M;
end;
end;
procedure ShowColumnPrimes(Memo: TMemo);
{Show min value for columns in NumLists}
var ColPrimes: TIntArray;
var I: integer;
var S: string;
begin
GetColPrimes(NumLists,ColPrimes);
S:='[';
for I:=0 to High(ColPrimes) do
begin
if I<>0 then S:=S+' ';
S:=S+IntToStr(ColPrimes[I]);
end;
S:=S+']';
Memo.Lines.Add(S);
end;
- Output:
[43 101 79 101 67]
EasyLang
func nxtprim p .
j = 2
while j * j <= p
if p mod j = 0
j = 1
p += 1
.
j += 1
.
return p
.
num1[] = [ 5 45 23 21 67 ]
num2[] = [ 43 22 78 46 38 ]
num3[] = [ 9 98 12 54 53 ]
n = len num1[]
len nums[] n
for i to n
nums[i] = nxtprim higher higher num1[i] num2[i] num3[i]
.
print nums[]
- Output:
[ 43 101 79 59 67 ]
F#
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
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
#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
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]
J
] numbers =. 3 5 $ 5 45 23 21 67 43 22 78 46 38 9 98 12 54 53
5 45 23 21 67
43 22 78 46 38
9 98 12 54 53
4 p: <: >./ numbers
43 101 79 59 67
Java
import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.stream.IntStream;
public final class MinimumPrimes {
public static void main(String[] args) {
List<Integer> numbers1 = List.of( 5, 45, 23, 21, 67 );
List<Integer> numbers2 = List.of( 43, 22, 78, 46, 38 );
List<Integer> numbers3 = List.of( 9, 98, 12, 54, 53 );
List<Integer> primes = new ArrayList<Integer>();
IntStream.range(0, 5).forEach( i -> {
final int max = List.of( numbers1.get(i), numbers2.get(i), numbers3.get(i) ).stream()
.max(Comparator.naturalOrder()).get();
primes.addLast(nextPrime(max));
});
System.out.println(primes);
}
private static int nextPrime(int number) {
int divisor = 2;
while ( divisor * divisor <= number ) {
if ( number % divisor == 0 ) {
number += 1;
divisor = 2;
} else {
divisor += 1;
}
}
return number;
}
}
- Output:
[43, 101, 79, 59, 67]
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
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
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
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
#!/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
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}
Quackery
transpose
is defined at Matrix transposition#Quackery.
isprime
is defined at Primality by trial division#Quackery.
' [ 5 45 23 21 67 ]
' [ 43 22 78 46 38 ]
' [ 9 98 12 54 53 ]
3 pack transpose
[] swap witheach
[ unpack max max 1 -
[ 1+ dup isprime until ]
join ]
echo
- Output:
[ 43 101 79 59 67 ]
Raku
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
Solution #1
? "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
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...
RPL
By the letter
≪ {5 45 23 21 67} {43 22 78 46 38} {9 98 12 54 53} → numbers1 numbers2 numbers3
≪ numbers1 numbers2 numbers3 3 ≪ MAX MAX ≫ DOLIST
{ } → max primes
≪ 1 max SIZE FOR j
max j GET
IF DUP ISPRIME? NOT THEN NEXTPRIME END
'primes' SWAP STO+
NEXT primes
≫ ≫ ≫ 'TASK' STO
- Output:
1: {43 101 79 59 67}
Idiomatic
No need for local variables.
≪ {5 45 23 21 67} {43 22 78 46 38} {9 98 12 54 53}
3 ≪ MAX MAX ≫ DOLIST
≪ IF DUP ISPRIME? NOT THEN NEXTPRIME END ≫ MAP
≫ 'TASK' STO
Ruby
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
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
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
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