Problem Definition: Euler Problem 2

In order to solve this problem in python, I have used the yield keyword to implements a generator to provide a sequence of Fibonacci’s numbers. I think I can do a better work in using list comprehension in the main routine to filter and add up the even numbers, but the current version is basically good enough.

“”“

Euler 2

Each new term in the Fibonacci sequence is generated by adding the previous two terms.
By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

Find the sum of all the even-valued terms in the sequence which
do not exceed one million.


”“”
def gen_fib_seq():
a = 1
b = 2
yield a
yield b
try:
 while 1:
   c = a + b
   a, b = b, c
   yield c
except StopIteration:
 pass


if name == “main”:
b = 0
onemillion = 1000000
for x in gen_fib_seq():
 if x % 2 == 0:
   if b + x < onemillion:
     b = b + x
   else:
     break
print b

The erlang turns out to be even simpler than I thought. First of all, because every object or operation can be abstracted as a process, I do not need special keyword ‘yield’ to create a special object – generator. Furthermore the guard clause of function acc remove the need to put in any ‘if’ or ‘break’ to do, for example, the even number testing as in the python version.

-module(p2).
-export([main/0]).

start() –> spawn(fun() –> fib(0, 1) end).

fib(N, M) –>
receive
 {Pid, next} –>
   Pid ! {self(), N + M},
   fib(M, N+ M)
end.

get_next_fib(Pid) –>
Pid ! {self(), next},
receive
 {, N} –>
   %% io:format(“~p~n”, [N])
   N
end.

%% accumulator
acc(N, Pid, Acc) when (N rem 2) =:= 1 –>
 N1 = get_next_fib(Pid),
 acc(N1, Pid, Acc);
acc(N, Pid, Acc) when N + Acc < 1000000 –>
 N1 = get_next_fib(Pid),
 acc(N1, Pid, Acc + N);
acc(, _, Acc) –>
 Acc.


%%%
%%% Program main()
%%%
main() –>
 Pid = start(),
 Acc = acc(1, Pid, 0),
 Acc.