From: Ruby Student <ruby.student@...>
Date: 2008-11-28T00:11:58+09:00
Subject: Re: Ayuda con Ejercicio

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On Sun, Nov 23, 2008 at 3:26 AM, Todd Benson <caduceass@gmail.com> wrote:

> On Thu, Nov 20, 2008 at 11:21 AM,  <tirado.carlos@gmail.com> wrote:
> > On Nov 19, 3:53 pm, Fernando Villafuerte <fernand...@hemac.com.mx>
> > wrote:
> >> lo que no se como hacer... es que el
> >> resultado consiste solamente en la suma de los resultados
> >> pero solo los que son n=FAmeros pares...
> >
> > Solo te hace falta agregar una variable para la suma, incrementandola
> > cuando el numero es par (al dividirlo en 2 el residuo es cero):
> >
> >> a =3D 1
> >> b =3D 2
> >> stop =3D 4_000_000
> >
> > suma =3D 0
> >
> >> (1...500).each do
> >>   s =3D a + b
> >>   a =3D b
> >>   b =3D s
> >
> > suma =3D suma + b if 0 =3D=3D b.modulo(2)
> >
> >>   break if a > stop
> >> end
> >
> > puts =3D suma
>
> For fun (no guarantee of accuracy, but works for 4_000_000)...
>
> include Math
> R =3D Math.sqrt(5) #radical
> P =3D (1 + R) / 2 #phi
> my_number =3D 4_000_000
> #inverting my_number to see approximately where n sits for F(n)
> max_inverted =3D (log((my_number - 0.5) * R) / log(P)).to_i
> #adding up F(n) for 0 to n
> even_sum =3D (0..max_inverted).inject do |s, i|
>  #next line is for direct F(n)
>  #the inversion above is the opposite of this
>  fib =3D (((P**i) - (1-P)**i) / R).floor
>  #see if it's even
>  fib % 2 =3D=3D 0 ? (s + fib) : s
> end
> p even_sum
>
>
> ...It would be interesting to see what ceilings (like 4_000_000) this fai=
ls
> for.
>
> cheers,
> Todd
>
>
Fernando,
El ejercicio del cual hablas es parte del web site: www.eulerorg.net, donde
se ofrece un largo numero de otros ejercicios interesantes. La siguiente is
mi solucion, aunque no la mejor, pero es otra manera.
Que te diviertas!

=3Dbegin
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 four million.
http://projecteuler.net
=3Dend

MAXSUM =3D 4000000

def fibo(previous_fibo, current_fibo,new_fibo,total)

  if new_fibo.remainder(2) =3D=3D 0
    if new_fibo <=3D MAXSUM
        total =3D total + new_fibo
    else
       puts "Found it. It is: #{total}"
       exit
    end
  end

  new_fibo =3D previous_fibo + current_fibo
  previous_fibo =3D current_fibo
  current_fibo =3D new_fibo
  fibo(previous_fibo, new_fibo, new_fibo, total)

end

fibo(1,2,2,0)


--=20
Ruby Student

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