missingfaktor <[email protected]> wrote:
> Text of the problem: 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, ... . By considering the
> terms in the Fibonacci sequence whose values do not exceed four million,
> find the sum of the even-valued terms.
>
> My solution:
>
> : million ( n -- n' ) 1000000 * ;
> : fibonacci ( n -- seq ) 0 1 rot '[ dup _ < ] [ swap over + dup ] produce
> 2nip ; ! Where has "tuck" gone?!
> 4 million fibonacci [ even? ] filter sum
>
> Any comments on how it can be simplified or made more idiomatic?
>
> --
> Cheers,
> missingfaktor <http://twitter.com/#!/missingfaktor>.
>
Well, first of all, your fibonacci word overshoots: 4000000 fibonacci last
. -> 5702887
Second, you could make it more readable like this:
> USE: locals
: next-fib ( a b -- b c ) [ + ] keep swap ; inline
>
:: fibs-until ( n -- seq ) 0 1 [ dup n < ] [ [ next-fib ] keep ] produce
> 2nip ;
>
Or you could go imperative route with:
> :: euler-2 ( n -- sum )
> 0 :> sum!
> 0 1
> [ dup even? [ [ sum + sum! ] keep ] when next-fib dup n < ] loop
> 2drop
> sum
> ;
This saves some time and memory, as Factor isn't lazy and this doesn't work
as well as Haskell's
> let fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
infixr 9 !>
> infixr 9 !
> (!) = flip (.)
> (!>) = flip ($)
let euler2 = fibs !> takeWhile (<4000000) ! filter even ! sum
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