Algorithm Optimisation Championship. - page 8
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in general, like this, and of course there will be a time-keeping counter. outline:
How do you know the number of function parameters?
Give me a test function and we'll practice.
How do you know the number of function parameters?
Give me a test function and we'll practice.
The number of FF parameters is between 100 and 500. You should be guided by this roughly the scale of the tasks in the championship.
Examples of FF:
And if you add me a third time to the list, it will be 8! ))
Thank you for your vigilance :)
I will not be aware of the FF at the championship.
Once the championship has started and the participants have published their algorithms, let's start looking at FF options from the participants. Eventually we will create a "mix" of features (this is quite easy to do) and start testing. No one will know in advance what the algorithms will have to solve.
The examples above are smooth functions (it's not hard to get to a sharp peak on a smooth slope) , fairly simple.
We will make some parts of the FFs discrete, stepped. This will make "life" much more difficult for both GA-like (stochastic) and deterministic-based methods.
The examples above are smooth functions (it's not hard to get to a sharp peak on a smooth slope) , fairly simple.
We will make some parts of the FFs discrete, stepped. It will considerably complicate the "life" of both GA-like (stochastic) and deterministic methods.
What is shown on the picture - are these examples of surfaces in question?
Are the peaks the maximum values of the parameters being searched for?
So, for a limited number of "probing", it is necessary to get as close as possible to the peak of each vertex?
The height of each vertex does not go outside the cube. It means they are between maximal and minimal (on the plane) values. That is - inside the range.
Conclusion: There is a range of numerical values. Inside it the "peak" values are hidden. Each value must be found, or approached.
The number of "glances" of the algorithm at the "surface" is limited.
For the total number of "looks", you have to "see" the whole "surface" and reproduce its analogue with values of your "research" results.
We need an algorithm that finds the "peak" values themselves, or their nearest "analogues", as efficiently as possible.
Help me to find out, what in the picture of my problem representation is wrong?
Yes, these are the simplest examples of ffs (the second one is more complicated, because it has flat areas with nothing to hold on to).
You will need to find the global maximum, i.e. the 1st point. And of course within the given parameter limits.
Will there be negative values in the range too? The global maximum, is it the highest point of the whole surface?
The global maximum is the maximum value of the FF, and points with this value can be more than 1.
The FF value area is all numerical values that the machine can process.
The global maximum is the maximum value of the FF, and points with this value can be more than 1.
The FF value area is all numerical values that the machine can process.
So - the FF value area is not just a range with two boundaries, between which there is only emptiness and lonely peaks of vertices. It's a full surface with a relief that needs to be probed all the way through?
Does the FF feed the "surface topography curves" into the algorithm?
So, the algorithm has to access the FF a huge number of times to get a minimal "idea" of the "surface" topography.
Until now, I imagined it in a two-dimensional array space, which simply stores some values to be found in a limited number of tries, but judging from the pictures, the search space is actually three-dimensional...
In other words, the number of values to be searched is several orders of magnitude higher. So, the more calls to FF (surface views) to make a "relief map", the more accurately the surface vertices will be found. But the number of references should be reduced by competition rules... Something I understand... :)
So, if you access the surface (FF) a maximum number of times - you can create a perfect copy of the relief?
But then, the fewer times, the worse the result?