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I have seen a tutorial which said that you have to try different models and see which fits best on your data.

Can this be considered brute force? I have searched this on google and the closest answer that came up was on Quora (https://www.quora.com/If-today-s-Machine-Learning-is-considered-brute-force-how-do-you-envision-the-Machine-Learning-of-the-future).

Can someone please explain this question in relation to this link in as a beginner friendly manner as possible.

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I think that the answers in quora miss the point/do not understand the question. Those answers respond to the question "whether or not machine learning algorithms use brute force in order to reach their conclusion". If this was indeed your question, then those answers seem pretty fine.

It seems however that you are not asking this. If we consider a machine learning problem as "find the best model for a job" (a meta approach where we already know that we will use machine learning) and we approach it by trying different models and see which one works best, this is indeed a form of brute-forcing.

This is made more apparent if we consider that for many problems, there are characteristics of the data that can help us have an idea about which algorithms would perform the best before we even try anything. As an example, if the data are linearly separable, SVM is a good candidate. Since there are such ways to chose a model, if we simply try everything, it is brute-force.

In addition, another aspect of the machine learning process that is essentially brute force is hyperparameter tuning. I haven't seen anybody not trying a lot of different values for each parameter and keeping the best ones, because in the end we rarely have an idea about which value would work best, and even when we do, nobody wants to risk not trying something that could potentially lead to better results.

So, to sum up, the process of creating a machine learning model (not the algorithms themselves) has aspects of brute forcing that either complement or completely replace other methods.

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