Erlang nested list comprehensions - an anagram example

listcomp.erl

-module(listcomp).
-export([perms/1]).

% generates all possible letter permutations given a string of letters

perms( [] ) -> [[]];
perms(Phrase) ->
  [ [H|Rest] ||
    H <- Phrase,
    Rest <- perms(Phrase -- [H])].

% Explanation with "word" as Phrase (for my own understanding, some real code, some pseudo):

% At the deepest level of recursion, perms([]) will return [[]].

% Up one level - perms("d") means H == "d" and Rest == perms("d" -- "d")
% (which is the same as % perms([]) ), so perms("d") == [ [$d|[]] ] == ["d"].

% Step up one level, using "rd" as Phrase. perms("rd") will mean that for
% one iteration, H == "r" and Rest == ["d"], which results in "rd". For the
% next iteration, H == "d" and Rest == ["r"], which results in "dr". So
% perms("rd") == [ "rd", "dr" ].

% Step up yet another level, using perms("ord"). For the first iteration,
% H == "o" and Rest == [ "rd", "dr" ], so "o" is mapped to "ord", "odr". After
% the first iteration, we have:
% [ "ord", "odr"
% For the next iteration, H == "r" and Rest == [ "od", "do" ]. After the
% second iteration, we have:
% [ "ord", "odr", "rod", "rdo"
% For the third iteration, H == "d" and Rest == [ "or", "ro" ]. After the third
% iteration, we have:
% [ "ord", "odr", "rod", "rdo", "dor", "dro" ]

% Step up one more level, using perms("word"). For the first iteration, H == "w" and
% Rest == [ "ord", "odr", "rod", "rdo", "dor", "dro" ]. After the first iteration, we
% have:
% [ "word", "wodr", "wrod", "wrdo", "wdor", "wdro"
% and so on.