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I am an enthusiast of programming and machine learning. Only a few months back I started learning about machine learning programming. Like many who don't have a quantitative science background I also started learning about ML by tinkering with the algorithms and datasets in the widely used ML package(caret R).

A while back I read a blog in which the author talks about usage of linear regression in ML. If I am remembering correct he talked about how all machine learning in the end uses some kind of "linear regression"(not sure whether he used this exact term) even for linear or non-linear problems. That time I didn't understood what he meant by that.

My understanding of using machine learning for non-linear data is to use a non linear algorithm to separate the data.

This was my thinking

Let's say to classify linear data we used linear equation $y=mx+c$ and for non linear data we use non-linear equation say $y=sin(x)$

enter image description here

This image is taken from sikit learn website of support vector machine. In SVM we used different kernels for ML purpose. So my initial thinking was linear kernel separates the data using a linear function and RBF kernel uses a non-linear function to separate the data.

But then I saw this blog where the author talks about Neural networks.

To classify the non linear problem in left subplot, the neural network transforms the data in such a way that in the end we can use simple linear separation to the transformed data in the right sub-plot

enter image description here

My question is whether all machine learning algorithms in the end uses a linear separation to classifiction(linear /non-linear dataset)?

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The answer is No. user20160 has a perfect answer, I will add 3 examples with visualization to illustrate the idea. Note, these plots may not be helpful for you to see if the "final decision" is in linear form but give you some sense about tree, boosting and KNN.

We will start with decision trees. With many splits, it is a non-linear decision boundary. And we cannot think all the previous splits are "feature transformations" and there are a final decision line at the end.

Another example is the boosting model, which aggregates many "weak classifiers" and the final decision boundary is not linear. You can think about it is a complicated code/algorithm to make the final prediction.

Finally, think about K Nearest Neighbors (KNN). It is also not a linear decision function at the end layer. in addition, there are no "feature transformations" in KNN.

Here are three visualizations in 2D space (Tree, Boosting and KNN from top to bottom). The ground truth is 2 spirals represent two classes, and the left subplot is the predictions from the model and the right subplot is the decision boundaries from the model.

Tree decision boundary

Boosting decision boundary

KNN decision boundary


EDIT: @ssdecontrol's answer in this post gives another perspective.

It depends on how we define the "transformation".

Any function that partitions the data into two pieces can be transformed into a linear model of this form, with an intercept and a single input (an indicator of which "side" of the partition the data point is on). It is important to take note of the difference between a decision function and a decision boundary.

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  • $\begingroup$ I don't want to critic, but the boosting seems a bit rough, no? Is it not possible to get a smoother result with different parameters? Sorry to be pernickety, because I find the all explanation very good. $\endgroup$ – YCR Jun 1 '16 at 15:56
  • $\begingroup$ @YCR I think that is the point of boosting where you have a rough decision boundary. The roughness is caused by aggregating many weak classifiers (in this example,they are trees). But I agree with you that the second example is not a good model, and it is overfitting :) $\endgroup$ – Haitao Du Jun 1 '16 at 16:53
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    $\begingroup$ (+1) Great visualization (I also use spirals a lot in my experimentations). A suggestion: plot the decision boundaries as image, and perhaps add probabiliity levels (if you are using probabilistic outputs) with contour. $\endgroup$ – Firebug Jun 24 '16 at 14:22
  • $\begingroup$ @Firebug great suggestion! these plot are generated in a grid and only can tell you the final label. Contour is much better. $\endgroup$ – Haitao Du Jun 24 '16 at 14:27
  • $\begingroup$ Look at my answer here: stats.stackexchange.com/a/218578/60613 $\endgroup$ – Firebug Jun 24 '16 at 14:45
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Some algorithms use a hyperplane (i.e. linear function) to separate the data. A prominent example is logistic regression. Others use a hyperplane to separate the data after a nonlinear transformation (e.g. neural networks and support vector machines with nonlinear kernels). In this case, the decision boundary is nonlinear in the original data space, but linear in the feature space into which the data are mapped. In the case of SVMs, the kernel formulation defines this mapping implicitly. Other algorithms use multiple splitting hyperplanes in local regions of data space (e.g. decision trees). In this case, the decision boundary is piecewise linear (but nonlinear overall).

However, other algorithms have nonlinear decision boundaries, and are not formulated in terms of hyperplanes. A prominent example is k nearest neighbors classification. Ensemble classifiers (e.g. produced by boosting or bagging other classifiers) are generally nonlinear.

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  • $\begingroup$ Neural Network is not a good example of a model class that uses hyperplanes after nonlinear transformation. The output layer may be (in many cases) a sigmoid activation, given you assume previous layers as a non-linear transformation to a feature space. $\endgroup$ – Cagdas Ozgenc Jun 22 '16 at 13:53
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    $\begingroup$ @CagdasOzgenc Let's consider the case of binary classification and a network w/ sigmoidal output, as you're suggesting. This is equivalent to logistic regression on the activations of the previous layer (using softmax outputs would be equivalent to multinomial logistic regression). So, the decision boundary is a hyperplane in feature space. The picture in the original question shows a nice example of this. $\endgroup$ – user20160 Jun 22 '16 at 21:06
  • $\begingroup$ I understand that when f(Ax) = 0 and f is one-to-one you can simply do f^-1 o f(Ax) = f^-1 (0) => Ax = 0 (or some constant c). So in case of sigmoid, you get a linear decision boundary. Basically are we talking about cases when f is not invertable? $\endgroup$ – Cagdas Ozgenc Jun 23 '16 at 11:08
  • $\begingroup$ Is f the output neuron's activation function and x the output of the previous layer? Not sure I understand what you're asking. $\endgroup$ – user20160 Jun 23 '16 at 11:25
  • $\begingroup$ x is a vector coming from neurons of previous layer, and f is output activation function. $\endgroup$ – Cagdas Ozgenc Jun 23 '16 at 11:29

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