Euler 35

euler35.erl

-module(euler35).
-export([is_circular_pr/1, solve/0]).

is_circular_pr(Number) -> % takes a prime number as input, does not check the number if it itself is prime
    Check = lists:any(fun(X) -> (X - 48) rem 2 =:= 0 orelse (X - 48) == 5 end, integer_to_list(Number)),
    Length = length(integer_to_list(Number)),
    if
	Length =:= 1 -> is_prime(Number);
	Length > 1 -> 
	    case Check of
		true -> false;
		false -> is_circular_pr(integer_to_list(Number), length(integer_to_list(Number)) - 1, false)
	    end
    end.

is_circular_pr(_New_number, 0, Bool) ->
    Bool;
is_circular_pr([H | Tail], Count, _Bool) ->
    New_number = lists:append(Tail, [H]),
    Is_it_prime = is_prime(list_to_integer(New_number)),
    case Is_it_prime of
	true -> is_circular_pr(New_number, Count - 1, true);
	false -> false
    end.

% reused from another problem
is_prime(X) ->
    is_prime(X, 2).

is_prime(1, _) ->
    false;
is_prime(2, _) ->
    true;
is_prime(3, _) ->
    true;
is_prime(X, Last_divisor) when Last_divisor * Last_divisor > X ->
    true;
is_prime(X, Last_divisor) ->
    if 
	X rem Last_divisor == 0 ->
	    false;
	Last_divisor == 2 ->
	    is_prime(X, Last_divisor + 1);
	true ->
	    is_prime(X, Last_divisor + 2)
    end.

% reused
primes(Below) ->
    [X || X <- [2 | lists:seq(3, Below, 2)],
	   is_prime(X)].

solve() ->
    solve(primes(1000000), 0). % this step (generating primes) takes the most of the time needed to find the answer

solve([], Cnt) ->
    Cnt;
solve([H | T], Cnt) ->
    Check = is_circular_pr(H),
    case Check of
	true ->
	    solve(T, Cnt + 1);
	false ->
	    solve(T, Cnt)
    end.