Hello, I'm wondering how works faust internal evaluation with recurrence formulas and if there is a way to optimize it in the context where all the terms of the recurrence are asked. Let me explain by an example:
Let's have this pattern matching function: f(0,x) = x; f(n,x) = alpha*f(n-1,x); where alpha is a complicated expression, long to evaluate. Now, if I ask: process = par(i,M,f(i,x)); each output will be an increasing call of f(n,x) from i=0 to M-1. Each term in this parallel composition will be computed from recurrence formula, Thus, f(M-1,x) = alpha*f(M-2,x) = ..... = alpha^(M-1) f(0,X) I suppose faust evaluation compute alpha^(M-1) f(0,X). However, as I asked process= par(i,M,f(i,x)); faust already computed f(M-2,x) when i=M-1 and it could be used to compute straightly f(M-1,x) as f(M-1,x) = alpha f(M-2,x) In the light of this, I'm wondering if there is a way to tell faust in the code to use the previously computed expression f(i-1,x) instead of going all the way down the recurrence (up to f(0,x)) to compute the term f(i,x). Thank you for your help and suggestions Pierre ------------------------------------------------------------------------------ Check out the vibrant tech community on one of the world's most engaging tech sites, Slashdot.org! http://sdm.link/slashdot _______________________________________________ Faudiostream-users mailing list Faudiostream-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/faudiostream-users
