Pandigital prime: Difference between revisions

=={{header|ALGOL 68}}==

Uses the observations in the Factor sample - the prime we are looking for can only have 7 or 4 digits.

# As noted in the Factor sample, only 7 and 4 digit primes need be #

# considered: 1 is not prime, all 2, 3, 5, 6, 8 and 9 digit #

find pd prime( 1, "pandigital" );

find pd prime( 0, "pandigital0" )

{{out}}

<pre>

max pandigital0 prime: 76540231

</pre>

 

=={{header|C#|CSharp}}==

class Program {

if (x % (i += 2) == 0) goto nxt;

}

nxt: ;

}

fun('0');

}

{{out|Output @ Tio.run}}

 

begin

{{out}}

<pre>

</pre>

 

=={{header|Factor}}==

{{works with|Factor|0.99 2021-06-02}}

math.primes math.ranges present sequences sequences.cords ;

 

[ reverse 0 [ 10^ * + ] reduce-index prime? ] find-last nip

"The largest pandigital decimal prime is: " print

{{out}}

<pre>

{{trans|Wren}}

{{libheader|Go-rcu}}

 

import (

}

}

 

{{out}}

76,540,231

</pre>

 

=={{header|jq}}==

 

See e.g. [[Erd%C5%91s-primes#jq]] for a suitable implementation of `is_prime`.

# drawing from the digits in the input array,

# in descending numerical order

| select(is_prime);

 

{{out}}

<pre>

 

=={{header|Julia}}==

 

function pandigitals(firstdig, lastdig)

println("Max pandigital prime over [$firstdigit, 9] is ", pandigitals(firstdigit, 9))

end

<pre>

Max pandigital prime over [1, 9] is 7652413

</pre>

 

=={{header|Perl}}==

{{libheader|ntheory}}

 

use strict; # https://rosettacode.org/wiki/Pandigital_prime

is_prime($n) and exit ! print "$n\n";

} $digits;

{{out}}

<pre>7652413</pre>

Slightly different approach for optional portion of task.

 

use warnings;

use ntheory <forperm is_prime vecmax>;

} @{[0..$c]};

}

{{out}}

<pre>76540231</pre>

 

=={{header|Phix}}==

<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>

<span style="color: #004080;">sequence</span> <span style="color: #000000;">avail</span>

<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>

<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>

{{out}}

With full inner workings (the second "1" is really "01", a failing pandigital0), obviously removing the "?n" on the fourth line above will reduce the output to just four lines.<br>

76540231

Largest decimal pandigital0 prime with 8 digits:76,540,231

</pre>

 

=={{header|Raku}}==

{{out}}

<pre>7652413

 

Essentially, the CPU time was displayed as using 0.00 seconds &nbsp; (rounded to two fractional decimal digits).

pand = reverse(123456789) /*get a big 9-digit pandigital number. */

gp= 0 /*indicate that primes not generated. */

end /*k*/ /* [↓] a prime (J) has been found. */

#= #+1; @.#= j; sq.#= j*j; !.j= 1 /*bump #Ps; P──►@.assign P; P^2; P flag*/

{{out|output|text=&nbsp; when using the internal default input:}}

<pre>

 

=={{header|Ring}}==

see "The largest " + z + "..7 pandigital prime is "

for n = hi to 0 step -18

strn = string(n)

pandig = true

exit

ok

next

if pandig and isprime(n)

exit

ok

next

next

func isprime(n)

i = 5

while i * i < n

end

{{out}}

<pre>

</pre>

 

=={{header|Ruby}}==

Using the observations from the Factor code:

def find_pan(ar) = ar.permutation(ar.size).find{|perm| perm.join.to_i.prime? }.join.to_i

puts find_pan(digits)

digits << 0

{{out}}

<pre>7652413

</pre>

 

 

}

}

}

 

for (start in 1..0) {

var outer = false

System.print("The largest pandigital decimal prime which uses all the digits %(start)..n once is:")

for (n in [7, 4]) {

for (i in perms.count - 1..0) {

if (perms[i][-1] % 2 == 0 || perms[i][-1] == 5 || (start == 0 && perms[i][0] == "0")) continue

if (outer) break

}

 

{{out}}

The largest pandigital decimal prime which uses all the digits 0..n once is:

76,540,231

</pre>