The more the merrier :-) I like the idea. The quad-tree decomposition I used has the drawback that pick-up notes could render it useless. Polynomial approximation minimizes that effect. > 3) Measure the (signed) distance of each successive note > above/below middle C. Form a vector of these, (v(1),v(2),...,v(n)). > Think of this as a function for the moment: v(t), t=1,2,...,n. In order > to make the result independent of the key, subtract the (weighted) mean > value from each v(i), and call the resulting vector V. This is not clear to me. I might be wrong, but by using intervals instead of notes, you are making it already key-independent. So I don't see the use of subtracting a mean. Anyway, even if you have to, I don't see how subtracting a mean would render it key-independent. > (f) Obvious problems: one has to be able to compare tunes in > different time signatures. This might come down to re-scaling the > polynomial. I don't think it is necessary to handle differences in time signature. > Finally: I suppose anyone suggesting an algorithm should have to > implement it---fair is fair---but while I could probably write the routine > from step two on (because it's pretty easy), the task of writing a parser > to get to step one is a bit much for me. Sorry. I had to write such a parser to implement the quad-tree decomposition. It's not entirely fit to reuse for something else, but I planned to make it more application-independent anyway, so this might be a good occasion. Would an object-oriented Perl module be of any use to you ? Brgds, Mark -- ------------------------------------------------------------------------------ Mark Vandenbroeck Email [EMAIL PROTECTED] Customer Support Phone +32-2-719.12.11 Oracle Belgium +32-2-719.59.51 (Support) Vuurberg 80 Fax +32-2-725.48.74 B-1831 Diegem +32-2-719.13.14 (Support) Belgium Web http://www.be.oracle.com/~mvandenb ------------------------------------------------------------------------------
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