On Jun 4, 3:00 pm, CuppoJava <patrickli_2...@hotmail.com> 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 always possible. Proof: assume, for some f and a, that we have the requisite g and b. Suppose a is such that (f a x) = x for any x. Case one: v = [v1]. It follows that (f a v1) = (g b v1) = v1. Case two: v = [v1 v2]. Then (f (f a v1) v2) = (f v1 v2) = (g (g b v2) v1) = (g v2 v1). Case three: v = [v1 v2 v3]. Then (f (f (f a v1) v2) v3) = (g (g (g b v3) v2) v1) (f (f v1 v2) v3) = (g (g v3 v2) v1) = (g (f v2 v3) v1) = (f v1 (f v2 v3)), for any v1, v2, v3. So f is associative. Thus, if f is not associative, either (f a x) != x for some x, or no such g, b exist. OTOH, if f is associative and we have (f a x) = x, then g = (flip f) works with b = a. I haven't been able to prove the existence or nonexistence of such g, b in other cases of f, a yet. I suspect f is required to be associative regardless of the value of a. > > ------------------------------ > > In case it helps at all, my specific problem is like this: > > v is a list of number pairs. > v = [ [1 1] [0.5 1] [0.1 1] [0.3 3] ] > a0 = [0 0] > > and f = > (fn [[c2 a2] [c a]] > [(+ c (* (- 1 a) c2)) a2]) This makes things much simpler, yes. I haven't analysed this function mathematically, so I can't say whether it's possible under the given constraints. But it might be, which is probably better than a "no". And it will be easier to prove things about this specific function than about all functions. > > It's the formula I'm using for blending translucent pixels on top of > each other. > Currently I'm drawing back to front and accumulating the color with > the above function. > > But for optimization purposes, it's much better to render front to > back and take advantage of an early exit condition. But for that, I > need a inverse-reduce function g(x,y), which is how this question > popped into my head. Does it have to be (reduce f a0 v) = (reduce g b0 (reverse v))? It might be simpler if it were (reduce f (cons a0 v)) = (reduce g (reverse (cons b0 v))). Does the early exit condition depend on f/g, or might it be possible to have g simply collect values of (reverse v), and then later pass them to f? What little I know of the subject makes me suspect the answers are both "yes it does", but it can't hurt to ask. > > Thanks so much for the help > -Patrick --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en -~----------~----~----~----~------~----~------~--~---
