Exercise 1.24 of the course "Functional Programming in Erlang", by Simon Thompson

ex.erl

-module(ex).
-export([perimeter/1, area/1, enclose/1, bitsR/1, bitsTR/1]).

% Shapes are represented with these tuples:
%
%   {circle, {X, Y}, R}
%   {rectangle, {X, Y}, H, W}
%   {triangle, {X, Y}, A, B, C}
%
% These tuples are not able to express the orientation of the rectangles and
% triangles, but that is in line with the lesson that showcases area/1.

perimeter({circle, _, R}) ->
    2*math:pi()*R;
perimeter({rectangle, _, H, W}) ->
    2*(H + W);
perimeter({triangle, _, A, B, C}) ->
    A + B + C.

area({circle, _, R}) ->
    math:pi()*R*R;
area({rectangle, _, H, W}) ->
    H*W;
% This is called Heron's formula, yields the area given the sides.
area({triangle, _, A, B, C}) ->
    S = (A + B + C)/2,
    math:sqrt(S*(S - A)*(S - B)*(S - C)).

% A circle is minimally enclosed in a square whose side is the circle diameter.
enclose({circle, C, R}) ->
    Diameter = 2*R,
    {rectangle, C, Diameter, Diameter};
% A rectangle is minimally enclosed in itself.
enclose({rectangle, C, H, W}) ->
    {rectangle, C, H, W};
% The smallest rectangle enclosing a triangle has the largest side as its base
% and the triangle altitude corresponding to that side as its height. Since
% area = b*h/2, we can leverage the area function.
enclose({triangle, Center, A, B, C}) ->
    Base = lists:max([A, B, C]),
    Height = 2*area({triangle, Center, A, B, C})/Base,
    {rectangle, Center, Base, Height}.

bitsR(0) ->
    0;
bitsR(N) ->
    bitsR(N div 2) + N rem 2.

bitsTR(N) ->
    bitsTR(N, 0).

bitsTR(0, S) ->
    S;
bitsTR(N, S) ->
    bitsTR(N div 2, S + (N rem 2)).