Code licensing/verifying Winner gets sparse Vector Exctraction source.

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1

Hi, long time member first time contributor.
Anywhoo, I have a simple ( oh my god you wouldn’t believe how) vector extraction routine,
Simply put I want to start a challange with this simple routine but useful if you put glowing rings on each of your joint’s to track motion in 3dimensions with just a camera. So if you want to try it out here it is…
And in the epidemiology of the challange the prize is in the wrapping,… Make for all of us a small reducable language for python’s code_ prototyping, say licensing a script to an individual for his use and if he forward’s it to a third party the code is traceable back to him… for example.

<code>
import types;
funct=types.FunctionType(types.CodeType(…,…))
</code>
Here is the prize in action…

2

sorry here’s the code for rgbToHue: b=‘870000870100870200660300640100860000740000640200830100448301007d05007c05006a0100830000017c050064030019640400197c05006404001964040019187049006405007d06007402007c04006402001b6404006406006407006703007c05006403001964050019198802008801008800006703006405006406006404006703007c0500640300196405001919198802008801008800006703006406006404006405006703007c050064030019640500191919187c06001b171483010053’; c=‘000122000a002001’; b2=‘6700007c00005d2b007d01007400008802008801008800006703007c0100198301007028006400007c010066020091020071060053’; c2=‘0900’; types.FunctionType(types.CodeType(3,0,6,9,67,bytes([int(b[a:a+2],16) for a in range(0,len(b),2)]),(None,types.CodeType(1,0,2,6,19,bytes([int(b2[a:a+2]) for a in range(0,len(b2),2)]),(1,),(‘float’,),(‘.0’,‘a’),‘’,‘’,5,bytes([int(c2[a:a+2]) for a in range(0,len(c2),2)])),3,-1,0,1,2,4),(‘len’,‘range’,‘float’,‘sort’,‘int’),(‘colors’,‘min’,‘max’,‘m’,‘a’,‘c’,‘diff’),‘rgbhue.py’,‘rgbhue’,4,bytes([int(c[a:a+2],16) for a in range(0,len(c),2)])),{‘len’:len,‘float’:float,‘range’:range,‘int’:int})([126,0,0],0,256) And… Here is