RogerBW's Blog

The Weekly Challenge 157: Pythagoras in Brazil 24 March 2022

I’ve been doing the Weekly Challenges. The latest involved a variety of means and more number theory. (Note that this is open until 27 March 2022.)

Task 1: Pythagorean Means

You are given a set of integers.

Write a script to compute all three Pythagorean Means i.e Arithmetic Mean, Geometric Mean and Harmonic Mean of the given set of integers.

I ended up breaking this down into lots of separate functions. Each mean can be calculated on a reduce-like principle, but they have quite different functions inside them, so each one happens separately (particularly since if I ever need to to this in the real world I'll probably want only one of those means); then the unstated rounding to one decimal place is its own function again.

As for "reduce" itself, the idea of repeatedly applying a function to members of a list, the only languages I'm using that don't have something like it either built-in (most of them) or in a core module (Perl, via List::Util) are PostScript and Lua. And by coincidence, just before this came out, I'd been working on my iterables library for PostScript…

Some of the languages have built-in sum and even product functions which might be faster for large lists, but it seemed cleaner to lay everything out explicitly.

The overall function being tested: calculate, round and return all the means as a hashmap. Kotlin:

fun pythagoreanmeans(s: List<Int>): Map<String,Double> {
    return hashMapOf(
        "AM" to round1(arithmeticmean(s)),
        "GM" to round1(geometricmean(s)),
        "HM" to round1(harmonicmean(s))
    )
}

Round to one decimal place. (Raku, Python and Ruby have built-ins to do this.)

fun round1(x: Double): Double {
    return floor(10.0*x+0.5)/10.0;
}

The arithmetic mean: take the sum of the list values, divide by the length.

fun arithmeticmean(s: List<Int>): Double {
    return s.reduce { acc, x -> acc + x }.toDouble() / s.size
}

The geometric mean: take the product of the list values, exponentiate to (1/length).

fun geometricmean(s: List<Int>): Double {
    return s.reduce { acc, x -> acc * x }.toDouble().pow(1.0/s.size)
}

The harmonic mean: take the reciprocal of each entry, sum them, and divide into the length. (In some languages I initialised the reduce accumulator with 0 and added 1/x rather than inverting the whole series first.)

fun harmonicmean(s: List<Int>): Double {
    return s.size/(s.map {1.0/it}.reduce { acc, x -> acc + x })
}

Task 2: Brazilian Number

You are given a number $n > 3.

Write a script to find out if the given number is a Brazilian Number.

A positive integer number N has at least one natural number B where 1 < B < N-1 where the representation of N in base B has same digits.

Which for me breaks down to: for each possible number base (from 2 up to n-2), calculate the digits of the conversion into that base until one of them differs from the first one. Raku:

sub brazilian($n) {

Loop over number bases:

    for (2..$n-2) -> $b {

Set up the working copy of the parameter, the boolean value for whether a valid Brazilian form has been found, and the first digit.

        my $nn = $n;
        my $braz = True;
        my $digit = $nn % $b;
        while ($nn > 0) {

Compare the next digit with the one we've already established. If it's the same, continue. (This is slightly inefficient as we end up doing the first modulus twice, but I thought the code was cleaner.)

            if ($digit == $nn % $b) {
                $nn div= $b;
            } else {

Otherwise, this isn't a Brazilian form and we can stop here.

                $braz = False;
                last;
            }
        }

If we found a Brazilian form, return it; we don't need to look for more.

        if ($braz) {
            return True;
        }
    }

If we tried all the bases, it's not a Brazilian number.

    return False;
}

Full code on github.

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