[ARCHIVE!] Any rookie question, so as not to clutter up the forum. Professionals, don't pass by. Can't go anywhere without you - 4. - page 77
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The complexity of the problem increases rapidly as N increases, because it requires calculating and enumerating N*(N-1)*(N-2)/6 circles (56 in the case of N=8)
Thanks for the option, I will try to pack it all into a more or less elegant form.
Oh, where did you get that N! / ((N-3)!*3! ) came from?
Thanks for the option, I'll try to package it all up in a more or less nifty way.
Well it is not the number of circles, but the number of distance triples - for a given circle position.
In short, the problem does not get any easier as it is numerical.
Well it's not the number of circles, but the number of distance triples - for a given circle position.
If it's not a secret, what is the application to the subject matter of the forum?
What is meant by "application"?
What is meant by the word "application"?
I see. It remains to come up with a GA to find the optimum circle. Analytically differentiating the modulus is kind of ugly.