RogerBW's Blog

The Weekly Challenge 154: Missing Padovan 03 March 2022

I’ve been doing the Weekly Challenges. The latest involved permutations and primality testing. (Note that this is open until 6 March 2022.)

Task 1: Missing Permutation

You are given possible permutations of the string 'PERL'.

Write a script to find any permutations missing from the list.

My basic approach is to make a list of all possible permutations (of the first argument in the list, rather than making it specific to one string), then delete from it those in the input list.

A fairly sharp dividing line separates the languages with some sort of built-in permutation engine (Python, Ruby, Raku) or readily-available module (Perl, Rust) and those without (the latter being PostScript, JavaScript, Kotlin and Lua). For those last I wrote a non-recursive implementation of Heap's Algorithm (dodging about a bit for Lua with its 1-based pseudo-arrays). For the others it was rather easier. The cleanest code is probably Python with itertools:

def missingpermutations(lst):
  perms=set("".join(x) for x in permutations(lst[0]))
  for x in lst:
    perms.discard(x)
  return list(perms)

But just for fun here's a permutor in PostScript:

/permute { % [array] {proc} permute runs proc on each permutation of array
    7 dict begin
    /subproc exch def
    /a exch def
    /n a length def
    /c [ n { 0 } repeat ] def
    mark a subproc cleartomark
    /i 0 def
    {
        i n ge {
            exit
        } if
        c i get i lt {
            i 2 mod 0 eq {
                0 i permute.swap
            } {
                c i get i permute.swap
            } ifelse
            mark a subproc cleartomark
            c i get 1 add c exch i exch put
            /i 0 def
        } {
            c i 0 put
            /i i 1 add def
        } ifelse
    } loop
    end
} bind def

/permute.swap {
    /bi exch def
    /ai exch def
    a ai get
    a bi get
    a exch ai exch put
    a exch bi exch put
} bind def

Task 2: Padovan Prime

A Padovan Prime is a Padovan Number that's also prime.

In number theory, the Padovan sequence is the sequence of integers P(n) defined by the initial values.

P(0) = P(1) = P(2) = 1

and then followed by

P(n) = P(n-2) + P(n-3)

Write a script to compute first 10 distinct Padovan Primes.

Opinions differ; Wikipedia quotes this definition, while the OEIS starts its system with (1, 0, 0). But 0 and 1 are canonically non-prime anyway.

With Raku I was able to use its lazy sequence generator:

sub padovanprime($ct) {
    my $pp=SetHash.new;
    for (1, 1, 1, -> $a, $b, $c { $a + $b } ... *) -> $padovan {
        if (isprime($padovan)) {
            $pp{$padovan}=True;
            if $pp.elems >= $ct {
                last;
            }
        }
    }
    return [$pp.keys.sort];
}

But, sadly, not to rely on the built-in and quite fast is-prime function. This will reliably return false if the number is known not to be prime, but returns true for either a prime or a number with unknown primality. (Perl's Math::Prime::Util at least has a three-state return for (no, unknown, yes).) With numbers of this size it probably won't matter, but as a result I imported the primality-testing code I'd written for other languages.

(Normally I'd use a sieve of Eratosthenes as in previous Challenges, but since I don't know in advance how large the numbers will be I fall back on lightly-optimised trial division.)

sub isprime($candidate) {

Check the easy cases first.

    if (!is-prime($candidate)) {
        return False;
    } elsif ($candidate==2) {
        return True;
    } elsif ($candidate==3) {
        return True;
    } elsif ($candidate % 2 == 0) {
        return False;
    } elsif ($candidate % 3 == 0) {
        return False;
    }

Failing that, start checking divisors of the form 6n±1 until we either exceed the square root of the candidate or find an even divisor.

    my $anchor=0;
    my $limit=floor(sqrt($candidate));
    while (True) {
        $anchor+=6;
        for ($anchor-1,$anchor+1) -> $t {
            if ($t > $limit) {
                return True;
            }
            if ($candidate % $t == 0) {
                return False;
            }
        }
    }
}

Although this is basically the same code I'd written for other languages, it was desperately slow in Raku, reliably taking about 58s to complete the test cases on my unloaded reference machine. Optimised compiled Rust took 0.2s; Kotlin took 0.45s even with Java runtime startup; Javascript under Node, 0.8s; Ruby, 1.4s; Rust unoptimised including compilation time 1.5s; Lua, 1.9s; Perl, 3.2s; Python, 6.7s, PostScript, 7.0s.

(Yeah, one might start to think that Rust is actually not a terrible scripting language…)

Full code on github.

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