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I've been learning machine learning through Andrew Ng's Coursera. I've completed Andrew's homework on SVM, but it felt wishy washed and I'm having hard time taking it from 0 to finish.

Say you calculated the SVM "model" using svmTrain provided by Andrew Ng. And now you want to see what the predicted output (using Andrew's svmPredict) is for all the training samples you used to calculate the error. Easy enough:

model = svmTrain(X, y, C, @(x1, x2) gaussianKernel(x1, x2, sigma));
predictions = svmPredict(model, Xval);
err = mean(double(predictions ~= yval));

All that make sense. What trips me up is what svmPredict is doing, specifically,

elseif strfind(func2str(model.kernelFunction), 'gaussianKernel')
    % Vectorized RBF Kernel
    % This is equivalent to computing the kernel on every pair of examples
    X1 = sum(X.^2, 2);
    X2 = sum(model.X.^2, 2)';
    K = bsxfun(@plus, X1, bsxfun(@plus, X2, - 2 * X * model.X'));
    K = model.kernelFunction(1, 0) .^ K;
    K = bsxfun(@times, model.y', K);
    K = bsxfun(@times, model.alphas', K);
    p = sum(K, 2);

What equation/formula is it trying to solve here Gaussian Kernel (RBF)? What does the equation even look like?

Also, if I give you a new input $\vec u$, how would you find its predicted output?

The predicted output for the linear kernel looks pretty readable:

p = X * model.w + model.b;

Which is just solving $y=X\vec w+b$, with each example being a row in $X$.

Any help would be sincerely appreciated.

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In kernel SVM you map your data points into a possibly infinite dimension Hilbert space $\mathcal H$. It turns out that $w$, the normal vector to the separating hyperplane (which completely characterizes the decision boundary), has the form $$ w = \sum_{i=1}^n \alpha_i y_i \phi(x_i) $$ where $\phi$ is the map from input data space to feature space and the $\alpha_i$ are from the dual formation (and if $i$ is not a support vector then $\alpha_i = 0$). One of the things that makes SVM practical is that $w$ is a finite linear combination of the $\{\phi(x_i)\}$ which is due to the representer theorem.

We classify a point $x$ via $$ \text{sgn}(b + \langle w, x\rangle) = \text{sgn}\left(b + \sum_{i=1}^n \alpha_i y_i K(x_i, x)\right) $$ where $K$ is the kernel function. This is just determining which side of the hyperplane $x$ is on, and note how we never actually need any $\phi(x)$, we instead only need inner products which are provided by $K$.

There is no way around computing this, but for certain kernels there may be tricks to compute it faster (note how the code for "other Non-linear kernels" is exactly computing what I've described here). In particular, for $$ K(x, x') = \exp\left(-\gamma ||x - x'||^2\right) $$ we have $$ ||x - x'||^2 = ||x||^2 + ||x'||^2 - 2 x^T x' $$ so you can work with vectors rather than doing a double for loop which may be much faster. I don't know Matlab, but in their code, it looks like X1 computes the squared norms of each test point, and X2 computes the squared norm of each training point. Then they compute the sum of these minus 2 times the inner products of the data points, which is exactly the (or at least a) vectorized way to compute all pairwise $||x - x'||^2$. I'm guessing the part with model.kernelFunction(1, 0) is how you go from $||x - x'||^2$ to $\exp\left(-\hat \gamma ||x-x'||^2\right)$, and then the rest is just element-wise multiplication by $\vec \alpha$ and $\vec y$, before summing to get the single signed distance per test point.

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  • $\begingroup$ Thanks @Chaconne! Making some sense now. :) According to this lecture on Kernel methods, the $sgn(...)$ you have above is the $g(x)$ in the Lagrangian. You're saying that $g(x)$ is the final predicted output? What if I give you a new input $\vec u$? How would you find its predicted output? $\endgroup$ – Aziz Javed Aug 22 '17 at 20:09
  • $\begingroup$ @AzizJaved the function $g(x) = \text{sgn}\left(b + \sum_{i=1}^n \alpha_i y_i K(x_i, x)\right)$ tells you your decision for any $x$ in your input space. For a new input $\vec u$ you'd just compute $g(\vec u)$. $\endgroup$ – jld Aug 22 '17 at 20:17
  • $\begingroup$ Ah, since you can't directly compute the weight $\vec w$, you have to use the supporting vectors and dot product ALL of them with your new input $\vec u$, sum them up, and just take the sign of the answer. Is that correct?! #tooExcited #imGettingIt!! $\endgroup$ – Aziz Javed Aug 22 '17 at 20:22
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    $\begingroup$ @AzizJaved yeah exactly! We just look at the sign of $w^T \vec u + b$. In the general case we never actually see $w$, we only interact with it via dot products, and these dot products are simply evaluations of our nice kernel function so this is easy (which is exactly what the kernel trick is all about) $\endgroup$ – jld Aug 22 '17 at 20:26
  • $\begingroup$ and for linear kernels though, we know what the weight $\vec w$ is, we can just do $g(\vec x)=sgn(\vec w \bullet \vec x +b)$ - the $sgn$ function is still there. $\endgroup$ – Aziz Javed Aug 22 '17 at 20:36
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Thanks for the question, I too was looking for an explanation how Predict works. We need to use original data set to transform the test data set

So, we need to use the original training set to transform and the equation is here that's $K(x_i, x)$, where $x_i$ is the test data point and $x$ is the original data set

$$ g(x) = \text{sgn}\left(b + \sum_{i=1}^n \alpha_i y_i K(x_i, x)\right)$$

Curtsy Andrew Ng, ML Octave code, and the implementation goes like this, where Xi is the test data and X is the original data

def gaussianKernel(Xi, X, sigma):
    """
    Returns a gaussian kernel between Xi and X
    Returns:
        K: array
    """
    K = np.zeros((Xi.shape[0], X.shape[0]))
    for i, x1 in enumerate(Xi):
        for j, x2 in enumerate(X):
            x1 = x1.ravel()
            x2 = x2.ravel()
            K[i, j] = np.exp(-np.sum(np.square(x1 - x2)) / (2 * (sigma**2)))
    return K

And the function predict is

def predict_sgd(Xi,X,kernelFunction,w,b,sigma):
    K=kernelFunction(Xi, X, sigma)
    return [ 1 if np.dot(w, K[i])+b > 1 else 0
                       for i in range(len(K))]
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