I am confused of the relative speed.
The two reverses in the explict algorithm is insignificant,and we also
ignore additions.Then the explict algorithm has 2*n
multiplications,and Mike's algorithm has 3*n multiplication + n
divisions.But it's speed is about 10x the speed of the explict
algorithm.What's wrong?

2008/3/10, Roger Hui <[EMAIL PROTECTED]>:
> > Where it came from before that, I know not.
>
>
> I first saw the algorithm in the APL Quote-Quad magazine
>  in the 1970's.  I'll see if I can find it in my (incomplete)
>  collection of Quote-Quads.
>
>
>
>
>  ----- Original Message -----
>  From: Mike Powell <[EMAIL PROTECTED]>
>  Date: Saturday, March 8, 2008 19:09
>  Subject: Re: [Jprogramming] Is there a tacit version EMA?
>  To: Programming forum <[email protected]>
>
>  > Xu,
>  >
>  > Your EMA question is related to algorithms for linear
>  > recurrence
>  > relations. There is a much loved algorithm (probably because
>  > it's
>  > usually found to be lean and mean) that I first encountered at
>  > IP
>  > Sharp in a newsletter quiz. We used it all over the place,
>  > particularly wherever a present value or discounted cash flow
>  > was in
>  > play. Where it came from before that, I know not. It goes like this.
>  >
>  > If you have a linear recurrence relation of the form:
>  >
>  > r[0] is m[0] times a[0]
>  > r[i] is a[i] + m[i] times r[i-
>  > 1]          (i >0)
>  >
>  > then this can be calculated in vector languages in the following way:
>  >
>  > a recur m;t
>  > t is timesscan m
>  > r is t times plusscan a div t
>  >
>  > I'd be most interested to see a J algorithm for this,
>  > particularly a
>  > tacit one. I note that we have a reflexive piece (t) and that
>  > there
>  > may be an opportunity for an inverse (* and %).
>  >
>  > As far as the EMA is concerned, I suggest Xu's original
>  > requirement is
>  > solved with:
>  >
>  > (y[0] , k times 1 drop y) recur (length y) replicate 1 - k
>
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