In recent years, Convolutional Neural Networks (CNNs) have become the state-of-the-art for object recognition in computer vision. Typically, a CNN consists of several convolutional layers, followed by two fully-connected layers. An intuition behind this is that the convolutional layers learn a better representation of the input data, and the fully connected layers then learn to classify this representation based into a set of labels.

However, before CNNs started to dominate, Support Vector Machines (SVMs) were the state-of-the-art. So it seems sensible to say that an SVM is still a stronger classifier than a two-layer fully-connected neural network. Therefore, I am wondering why state-of-the-art CNNs tend to use the fully connected layers for classification rather than an SVM? In this way, you would have the best of both worlds: a strong feature representation, and a strong classifier, rather than a strong feature representation but only a weak classifier...

Any ideas?


3 Answers 3


What is an SVM, anyway?

I think the answer for most purposes is “the solution to the following optimization problem”: $$ \begin{split} \operatorname*{arg\,min}_{f \in \mathcal H} \frac{1}{n} \sum_{i=1}^n \ell_\mathit{hinge}(f(x_i), y_i) \, + \lambda \lVert f \rVert_{\mathcal H}^2 \\ \ell_\mathit{hinge}(t, y) = \max(0, 1 - t y) ,\end{split} \tag{SVM} $$ where $\mathcal H$ is a reproducing kernel Hilbert space, $y$ is a label in $\{-1, 1\}$, and $t = f(x) \in \mathbb R$ is a “decision value”; our final prediction will be $\operatorname{sign}(t)$. In the simplest case, $\mathcal H$ could be the space of affine functions $f(x) = w \cdot x + b$, and $\lVert f \rVert_{\mathcal H}^2 = \lVert w \rVert^2 + b^2$. (Handling of the offset $b$ varies depending on exactly what you’re doing, but that’s not important for our purposes.) In the ‘90s through the early ‘10s, there was a lot of work on solving this particular optimization problem in various smart ways, and indeed that’s what LIBSVM / LIBLINEAR / SVMlight / ThunderSVM / ... do. But I don’t think that any of these particular algorithms are fundamental to “being an SVM,” really.

Now, how do we train a deep network? Well, we try to solve something like, say, $$ \begin{split} \operatorname*{arg\,min}_{f \in \mathcal F} \frac1n \sum_{i=1}^n \ell_\mathit{CE}(f(x_i), y) + R(f) \\ \ell_\mathit{CE}(p, y) = - y \log(p) - (1-y) \log(1 - p) ,\end{split} \tag{$\star$} $$ where now $\mathcal F$ is the set of deep nets we consider, which output probabilities $p = f(x) \in [0, 1]$. The explicit regularizer $R(f)$ might be an L2 penalty on the weights in the network, or we might just use $R(f) = 0$. Although we could solve (SVM) up to machine precision if we really wanted, we usually can’t do that for $(\star)$ when $\mathcal F$ is more than one layer; instead we use stochastic gradient descent to attempt at an approximate solution.

If we take $\mathcal F$ as a reproducing kernel Hilbert space and $R(f) = \lambda \lVert f \rVert_{\mathcal F}^2$, then $(\star)$ becomes very similar to (SVM), just with cross-entropy loss instead of hinge loss: this is also called kernel logistic regression. My understanding is that the reason SVMs took off in a way kernel logistic regression didn’t is largely due to a slight computational advantage of the former (more amenable to these fancy algorithms), and/or historical accident; there isn’t really a huge difference between the two as a whole, as far as I know. (There is sometimes a big difference between an SVM with a fancy kernel and a plain linear logistic regression, but that’s comparing apples to oranges.)

So, what does a deep network using an SVM to classify look like? Well, that could mean some other things, but I think the most natural interpretation is just using $\ell_\mathit{hinge}$ in $(\star)$.

One minor issue is that $\ell_\mathit{hinge}$ isn’t differentiable at $\hat y = y$; we could instead use $\ell_\mathit{hinge}^2$, if we want. (Doing this in (SVM) is sometimes called “L2-SVM” or similar names.) Or we can just ignore the non-differentiability; the ReLU activation isn’t differentiable at 0 either, and this usually doesn’t matter. This can be justified via subgradients, although note that the correctness here is actually quite subtle when dealing with deep networks.

An ICML workshop paper – Tang, Deep Learning using Linear Support Vector Machines, ICML 2013 workshop Challenges in Representation Learning – found using $\ell_\mathit{hinge}^2$ gave small but consistent improvements over $\ell_\mathit{CE}$ on the problems they considered. I’m sure others have tried (squared) hinge loss since in deep networks, but it certainly hasn’t taken off widely.

(You have to modify both $\ell_\mathit{CE}$ as I’ve written it and $\ell_\mathit{hinge}$ to support multi-class classification, but in the one-vs-rest scheme used by Tang, both are easy to do.)

Another thing that’s sometimes done is to train CNNs in the typical way, but then take the output of a late layer as "features" and train a separate SVM on that. This was common in early days of transfer learning with deep features, but is I think less common now.

Something like this is also done sometimes in other contexts, e.g. in meta-learning by Lee et al., Meta-Learning with Differentiable Convex Optimization, CVPR 2019, who actually solved (SVM) on deep network features and backpropped through the whole thing. (They didn't, but you can even do this with a nonlinear kernel in $\mathcal H$; this is also done in some other "deep kernels" contexts.) It’s a very cool approach – one that I've also worked on – and in certain domains this type of approach makes a ton of sense, but there are some pitfalls, and I don’t think it’s very applicable to a typical "plain classification" problem.

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    $\begingroup$ Pegasos algorithm : hinge loss differentiable via sub gradient $\endgroup$
    – user318514
    Commented Oct 4, 2021 at 17:58
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    $\begingroup$ @Germania Yes, that's true; I just added a note about subgradients. (The hinge loss still isn't differentiable, but you can apply subgradient methods to optimize, which in practice generally looks like "just ignore that it's not differentiable" anyway.) $\endgroup$
    – Danica
    Commented Oct 4, 2021 at 22:08

Most of the theory behind the support vector machine assumes you are constructing a maximal margin classifier following a fixed transformation of the input space (via a kernel). The theory is less applicable if the fixed transformation has been learned from the data (as would be the case for the lower levels of the deep neural network). This means that there probably isn't much reason to expect a linear SVM to be better than a conventional (regularised) neural layer.


As far I can see, there are at least couple differences:

  1. CNNs are designed to work with image data, while SVM is a more generic classifier;
  2. CNNs extract features while SVM simply maps its input to some high dimensional space where (hopefully) the differences between the classes can be revealed;
  3. Similar to 2., CNNs are deep architectures while SVMs are shallow;
  4. Learning objectives are different: SVMs look to maximize the margin, while CNNs are not (would love to know more)

This being said, SVMs can work as good as CNNs provided good features are used with a good kernel function.

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    $\begingroup$ I think you may have misunderstood the question; it's about using an "SVM layer" at the end of the CNN. $\endgroup$
    – Danica
    Commented Aug 20, 2015 at 15:51
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    $\begingroup$ I understand the difference between a CNN and an SVM, but as @Dougal says, I'm asking more about the final layer of a CNN. Typically, this is a fully-connected neural network, but I'm not sure why SVMs aren't used here given that they tend to be stronger than a two-layer neural network. $\endgroup$ Commented Aug 20, 2015 at 15:58
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    $\begingroup$ @Karnivaurus Sorry for misreading your question. The idea is not new. Typically the last layer is thrown away and the output of the last layer is used as features in other classification algorithms. Why it is not done consistently and everywhere? The features of the last layer are typically so discriminative that there is no need of a sophisticated black box as SVM, a simple Logistic Regression does the job. This is my vision of things. $\endgroup$ Commented Aug 20, 2015 at 16:02

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