posted 08 May 2011 20:14            

quote:

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eg. the "c" is switched off, everytime the active !P value *would* result in a "c", the actual output is the next activated note, "d" for example.

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Would this do it?

code:

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 60 nn + ((1 - !C) * 2 nn)

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When the !C switch was on (1), then you would hear 60 nn. When the !C switch is off (0), you would hear 62 nn.

posted 09 May 2011 09:07            

}

You know what i mean? this incomplete sequence with loads of brackets, added semitones, times 12, within an array, times 8 for all the sliders(!p1, !p2 ...) is this the only way to do it?
Thanks very much!

posted 11 May 2011 20:13            

quote:

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if eg. the "c" is switched off, everytime the active !P value *would* result in a "c", the actual output is the next activated note, "d" for example.

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So if we split up the task, the first thing you are doing is taking an arbitrary pitch and quantizing it to a pitch from the diatonic scale, correct? You have an index of:

((!p1*12) + (!OctaveRange*12)) mod: 12) rounded

Since you are extracting the pitch class using mod 12, you could remove the octave from the index calculation and use:

(!p1 * 12) of: #(0 0 2 2 4 5 5 7 7 9 9 11)

That would take a !p1 in the range of (0,1) and map it to intervals from a diatonic scale (white keys of a keyboard).

Next we can create an Array that includes your checkboxes. As I understand it, each element of the Array has to include all possible intervals (since if any checkbox is turned off, it has to play the next pitch higher) on that stage of the sequence. Is that correct? So the first element would be:

{!c true: 0 false: (!d true: 2 false: (!e true: 4 false: (!f true: 5 false: (!g true: 7 false: (!a true: 9 false: (!b true: 11 false: (0)))))))}

Is this what you had in mind as the behavior?