[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

New comment
 
alsu:

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.
 
Mathemat:
Oh, where did you get that N! / ((N-3)!*3! ) came from?
the number of combinations of N by 3, aka the number of 3-element subsets of the N-element set
 
Elenn:

Thanks for the option, I'll try to package it all up in a more or less nifty way.
If it's no secret, what's the application to the forum theme?
 

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.

 
Mathemat:
Well it's not the number of circles, but the number of distance triples - for a given circle position.
We are not looking for distances from all points to a given circle, but first choose three points and look for the best circle for them. Thus, the number of circles equals the number of ways to choose three points.
 
alsu:
If it's not a secret, what is the application to the subject matter of the forum?

What is meant by "application"?
 
alsu: We are not looking for distances from all points to a given circle, but we are choosing three points first and looking for the optimum circle for them.
I see. Now we need to invent a GA to find the optimal circle. Analytically differentiating the modulus is somehow not nice.
 
Elenn:

What is meant by the word "application"?
Well, in the abstract, we have solved the problem, but in practice, what do dots and circles mean when applied to forex?
 
Mathemat:
I see. It remains to come up with a GA to find the optimum circle. Analytically differentiating the modulus is kind of ugly.
Yeah, you get all sorts of bad algorithms like simplex method, but in this case even more complicated, because there are roots inside. But if instead of simple sum take sum of squares, then modules disappear. Although there will be just as much fuss with the roots.) Instead of GA I would use some of the quasi-Newtonian methods...
 
And what is the least squared distance - by analogy with your d = |r - sqrt((xA-x0)^2 + (yA-y0)^2)|? There's a difficulty here too. The moduli will disappear, but the roots are unlikely...
1...707172737475767778798081828384...632
New comment