# 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): Map { 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): 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): 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): 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. Comments on this post are now closed. If you have particular grounds for adding a late comment, comment on a more recent post quoting the URL of this one. 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