The code that Danny wrote to create fib-stream works great with his
peek-fibs function, but I really don't understand stream-first.
I wrote:
; function to use the fib-stream to generate a list of all fibs < n
(define (fib-stream-less-than-n n)
(let ([fib-val (stream-first fib-stream)])
(if (>= fib-val n) '() (cons fib-val (fib-less-than-n n)))))
I get:
(fib-stream-less-than-n 500)
'(0 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377)
Note the two leading 0s. This perplexed me, until I read about
stream-first then I realized that I really didn't understand anything. I
believe the function above should return
'(0 0 0 0 0 0 0 ...)
Can someone tell me what is going on?
Thanks,
-Joe
On Mon, Feb 13, 2012 at 9:10 PM, Danny Yoo <d...@cs.wpi.edu> wrote:
> On Mon, Feb 13, 2012 at 11:52 PM, Joe Gilray <jgil...@gmail.com> wrote:
> > Warning: extreme newbie question ahead.
> >
> > I wrote the following fibonacci function:
> >
> > ; function that returns the next fibonacci number each time it is called
> > ; invoke as (fib)
> > (define fib
> > (let ([n0 -1] [n1 1])
> > (lambda ()
> > (let ([next (+ n0 n1)])
> > (set! n0 n1)
> > (set! n1 next))
> > n1)))
>
>
> One thing you can do is turn fib into a "sequence", and then from a
> sequence into a stream that knows how to remember its previous values.
>
> Here's what it might look like:
>
> ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
> #lang racket
> (require racket/sequence)
>
> (define fib
> (let ([n0 -1] [n1 1])
> (lambda ()
> (let ([next (+ n0 n1)])
> (set! n0 n1)
> (set! n1 next))
> n1)))
>
> (define fib-stream
> ;; Here's a sequence of the function:
> (let ([fib-sequence (in-producer fib 'donttellmecauseithurts)])
>
> ;; Let's wrap it and turn it into a stream that remembers...
> (sequence->stream fib-sequence)))
>
>
> ;; Ok, we've got a stream. Let's look at its first few elements.
> (define (peek-fibs n)
> (for ([elt fib-stream]
> [i (in-range n)])
> (displayln elt)))
>
> (peek-fibs 10)
> (printf "-----\n")
> (peek-fibs 20)
> ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
>
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