thanks. Very subtle difference... definitely not a good idea to try
to double an existing variable as an accumulator too...
I reworked it to:
(define (pi4 accuracy)
(let helper ((k 0) (r 0))
(let ((this (* (/ (expt -1 k) (expt 4 k))
(+ (/ 2 (+ (* 4 k) 1))
(/ 2 (+ (* 4 k) 2))
(/ 1 (+ (* 4 k) 3))))))
(if (< (abs this) accuracy)
(+ this r)
(helper (+ k 1) (+ this r))))))
and just to clear up, this is not someone else's pi... :p
On Wed, Feb 15, 2012 at 5:02 PM, Matthias Felleisen
<matth...@ccs.neu.edu> wrote:
>
> If you systematically apply cps and then accumulator transformations, you get
> this:
>
> (define (pi3 accuracy)
> (let helper ((k 0))
> (let ((this (formula k)))
> (if (< (abs this) accuracy)
> this
> (+ this (helper (+ k 1)))))))
>
> (pi3 .1)
>
> (define (pi3-cps accuracy)
> (let helper ((k 0) (c (lambda (x) x)))
> (let ((this (formula k)))
> (if (< (abs this) accuracy)
> (c this)
> (helper (+ k 1) (lambda (v) (c (+ this v))))))))
>
> (pi3-cps .1)
>
> (define (pi3-accu accuracy)
> (let helper ((k 0) (c 0))
> (let ((this (formula k)))
> (if (< (abs this) accuracy)
> (+ c this)
> (helper (+ k 1) (+ c this))))))
>
> (pi3-accu .1)
>
> Now when you compare the result of that with yours:
>
> (define (pi4 accuracy)
> (let helper ((k 1) (this (formula 0)))
> (if (< (abs this) accuracy)
> this
> (helper (+ k 1) (+ this (formula k))))))
>
> you immediately see why yours fails.
>
> ;; ---
>
> I would worry about accuracy because you are switching the direction of
> addition when you apply the accu transform, but perhaps you know why this
> doesn't matter for your context.
>
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