Ah I see. Yeah this probably depends a lot on the way you are representing
the bits and how you split the objectives. If each objective is an
ASCIIString of bits and they're held separately (i.e. objectives is a
Vector{ASCIIString}) what I have provided should work because the isless
function works on strings. It probably isn't the most efficient way to
compare them however.
On Friday, April 24, 2015 at 1:34:25 PM UTC-3, Ronan Chagas wrote:
>
> Hi Duane Wilson,
>
> Em sexta-feira, 24 de abril de 2015 11:52:08 UTC-3, Duane Wilson escreveu:
>>
>> Actually I was able to get the go ahead from my boss to create a gist of
>> the code I've produced. I've modified it a little bit, as some of the
>> things in there are really relevant.
>>
>> http://nbviewer.ipython.org/gist/dwil/5cc31d1fea141740cf96
>>
>> Any comments for optimizing this a little be more would be appreciated :)
>>
>>
> Thanks very much, it will help me and, if I can contribute, I will tell
> you :)
>
> The problem is that for each generation of this evolutionary algorithm I
> will need to check if n candidate points belong to the Pareto frontier,
> where n is the number of bits in the string.
> Thus, I need a really fast algorithm for this kind of operation. As I am
> doing now (I will post the code in github), it works fine until the
> frontier has a large number of elements.
> Just one example, suppose that I'm using n = 16 and I have 6,000 elements
> in the frontier. If I generate 1000 generations, then I will need to check
> if 16,000 points are in the frontier.
> It will lead to 96,000,000 comparisons considering that no point will be
> added to the frontier.
>
> Thanks,
> Ronan
>
>
