Thanks everybody for their help!
I solved the problem (almost) but only for my specific f(x,y), which
is enough for me for now. Paul Mooser's hint about accumulating state
was the key I needed. By introducing another accumulator I got my
color compositing to blend correctly from front to back.
T
I'm not sure how you're compositing your colors, but the function
you're using is not associative or commutative (I checked at lunch,
mmm), so I don't think you can reverse order.
It's:
a + b = (b1 + a1 (1 - b2), a2)
Your function might need to be something like
a + b = (b1*(1 - a2) + a1 (1 -
For a general non-commutative operation, I'm not sure you can do this
without keeping extra state around. You can clearly cheat by
accumulating state as you run reduce, and then as the last reduce step
perform the actual computation, but that isn't any better than the
original problem you're tryin
On Jun 4, 3:00 pm, CuppoJava wrote:
> Hey guys.
> Thanks for the help. I have to clarify my question a bit.
>
> f(x,y) and a0 are given and do not assume any properties.
> Find g(x,y) and b0, such that for *any* list of numbers v,
>
> (reduce f a0 v) = (reduce g b0 (reverse v))
This is not alway
Hey guys.
Thanks for the help. I have to clarify my question a bit.
f(x,y) and a0 are given and do not assume any properties.
Find g(x,y) and b0, such that for *any* list of numbers v,
(reduce f a0 v) = (reduce g b0 (reverse v))
--
In case it helps at all, my specif
On Thu, Jun 4, 2009 at 12:27 PM, Mark Reid wrote:
>
> Hi again,
>
> I misinterpreted the question first time around but here's another
> attempt.
>
> Given f, a0 and v, let b0 = (reduce f a0 v) = f( f( f( f(a0, v[1]), v
> [2]), ...), v[n])
>
> Now define the function g to ignore its second argume
On Jun 4, 6:23 am, CuppoJava wrote:
> Hey guys,
> I'm really stuck on this math question, and I'm wondering if you guys
> know of any links that may help me.
>
> Given: f(x,y), a0, a list of numbers v.
> Find: g(x,y) and b0 such that:
>
> (reduce f a0 v) = (reduce g b0 (reverse v))
>
> Thanks fo
On Jun 4, 2009, at 7:23, CuppoJava wrote:
> I'm really stuck on this math question, and I'm wondering if you guys
> know of any links that may help me.
>
> Given: f(x,y), a0, a list of numbers v.
> Find: g(x,y) and b0 such that:
>
> (reduce f a0 v) = (reduce g b0 (reverse v))
If that's for any
Hi again,
I misinterpreted the question first time around but here's another
attempt.
Given f, a0 and v, let b0 = (reduce f a0 v) = f( f( f( f(a0, v[1]), v
[2]), ...), v[n])
Now define the function g to ignore its second argument and return its
first. That is, g(x, y) = x.
Then (reduce g b0 (r
On Jun 3, 2009, at 11:23 PM, CuppoJava wrote:
>
> Hey guys,
> I'm really stuck on this math question, and I'm wondering if you guys
> know of any links that may help me.
>
> Given: f(x,y), a0, a list of numbers v.
> Find: g(x,y) and b0 such that:
>
> (reduce f a0 v) = (reduce g b0 (reverse v))
>
On Thu, Jun 4, 2009 at 8:37 AM, Mark Reid wrote:
>
> Hi,
>
> Maybe I'm missing something but doesn't + fit the bill (or any
> symmetric function)?
>
> (== (reduce + 3 [1 2 3]) (reduce + 3 [3 2 1])) ;; => true
>
> In this case f = g = + and a0 = b0 for any choice of a0 and v.
I think he can't cho
Hi,
Maybe I'm missing something but doesn't + fit the bill (or any
symmetric function)?
(== (reduce + 3 [1 2 3]) (reduce + 3 [3 2 1])) ;; => true
In this case f = g = + and a0 = b0 for any choice of a0 and v.
Regards,
Mark.
On Jun 4, 3:23 pm, CuppoJava wrote:
> Hey guys,
> I'm really stuck
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