RogerBW's Blog

Perl Weekly Challenge 105: Name Root 23 March 2021

I’ve been doing the Perl Weekly Challenges. The latest involved fractional exponentiation and sweating with the oldies. (Note that this is open until 28 March 2021.)

TASK #1 › Nth root

You are given positive numbers $N and $k.

Write a script to find out the $Nth root of $k. For more information, please take a look at the wiki page.

Well, all of my chosen languages have some sort of fractional exponentiation capability so return $k^(1/$N) would have been a valid answer. But this is meant to be at least some sort of challenge, so I implemented Newtonian iteration. In Raku:

sub nroot($n,$a) {
  my $xk=2;
  while (1) {
    my $xk1=(($n-1)*$xk+$a/($xk ** ($n-1)))/$n;
    if ($xk1==$xk) {
      last;
    }
    $xk=$xk1;
  }
  return $xk;
}

and the others look basically the same. The only fiddliness was in Rust, where there's no exponentiation operator, but rather library function calls.

        let xk1=((n-1.0)*xk+a/(xk.powf(n-1.0)))/n;

TASK #2 › The Name Game

You are given a $name.

Write a script to display the lyrics to the Shirley Ellis song The Name Game. Please checkout the wiki page for more information.

I admit I didn't bother with trying to work out where syllable stresses might lie; instead, I make a "tail" part that's the name with any initial consonants removed. (I assume that an initial Y will be a vowel sound, favouring Yves over Yanni; I suppose the next level of sophistication would treat Y as a consonant if and only if a vowel follows it.) If there was no initial consonant, drop the tail into lower case.

Then it's a matter of interpolating name and tail into the template. In Perl:

sub ng {
  my $name=shift;
  (my $tail=$name) =~ s/^[bcdfghjklmnpqrstvwxz]*//i;
  if ($tail eq $name) {
    $tail=lc($tail);
  }
  return "$name, $name, bo-b$tail\nBonana-fanna fo-f$tail\nFee fi mo-m$tail\n$name!";
}

Python has grown string interpolation in recent versions, so I didn't have to use its special formatting language.

    return f"{name}, {name}, bo-b{tail}\nBonana-fanna fo-f{tail}\nFee fi mo-m{tail}\n{name}!"

Rust is above all that sort of thing (though I do think that having to separate the variable list from the template is prone to producing errors, just like good old sprintf; there is a system of named parameters which might help but is rather verbose).

    return format!("{}, {}, bo-b{}\nBonana-fanna fo-f{}\nFee fi mo-m{}\n{}!",name,name,tail,tail,tail,name);

And of course hope that none of your users is called "Buck".

Full code on github.


  1. Posted by John P at 08:53pm on 23 March 2021

    Have you seen these?

    https://azgaar.github.io/Fantasy-Map-Generator https://watabou.itch.io/medieval-fantasy-city-generator

    I'm impressed by the results they get. Somebody has been doing a lot of work.

  2. Posted by RogerBW at 05:14pm on 29 March 2021

    Very nifty stuff – one of these days I'll have to take one and expand it to do e.g. interstellar political boundaries.

    Looking at other people's answers to these challenges:

    (Task 1) A surprising number of people simply used the built-in exponentiation, returning $k ** (1/$N) or something like it. On the one hand, yes, that's obviously faster to implement and run than my version. On the other, well, what's the point? It's in the language already. Of the people who shared my approach, several of them used a fixed number of iterations rather than checking for a convergence.

    (Task 2) Nobody had much to say about this one. Stem, interpolate. One solution allowed Y to be configured as a consonant.

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