Euler task 20

euler20.erl

-module(euler20).
-export([solve/1]).

%% the solution for the problem
solve (N) -> sum_digits(factorial_rec(N), 0).


%% Tail recursive implementation
%% of the factorial function
%% Question: I think that it's much faster than
%% the second implementation. Is it correct?
factorial_rec (0) -> 1;
factorial_rec (1) -> 1;
factorial_rec (N) -> factorial_rec(N - 1) * N.


%% yet another implementation
%% I was using list:seq that reminds me of
%% ranges in other languages
factorial_lst (N) ->
    lists:foldl(fun(Curr, Prod) -> Curr * Prod end, 1, lists:seq(1, N)).

%% Finds a summ of the digits
sum_digits (N, 0) ->
    sum_digits(N div 10, N rem 10);

sum_digits (0, Sum) -> Sum;

sum_digits (N, Sum) ->
    sum_digits(N div 10, Sum + (N rem 10)).

In fact, the following is not tail-recursive:

factorial_rec (N) -> factorial_rec(N - 1) * N.

The second approach is going to be faster if Erlang supports lazy lists (or, at least, enumerators) and doesn't really allocate space for each item in list:seq.

First is not tail recursive, yes. But it will be faster