2. Support Vector Machines
Elegant combination of statistical learning theory and machine learning – Vapnik
Good empirical results
Non-trivial implementation
Can be slow and memory intensive
Binary classifier
Much current work
Do Quadric review first, (including each layer just linear divider, will help see polynomial kernel issues

3. SVM Comparisons
In order to have a natural way to compare and gain intuition on SVMs, we will first do a brief review of:
Quadric/Higher Order Machines
Radial Basis Function Networks
Note that BP follows similar strategy
Non-linear map of input features into new feature space which is now linearly separable
But, BP learns the non-linear mapping
Perceptron Slide – Higher Order
(BP slide 8-10) – Multi-layer learning (Rosenblatt's simple perceptron) BP-slide 3 – linearly separable for top layer
IBL Slide RBF

4. SVM Overview
Non-linear mapping from input space into a higher dimensional feature space
Non adaptive – User defined by Kernel function
Linear decision surface (hyper-plane) sufficient in the high dimensional feature space
Note that this is the same as we do with standard MLP/BP
Avoid complexity of high dimensional feature space with kernel functions which allow computations to take place in the input space, while giving the power of being in the feature space
Get improved generalization by placing hyper-plane at the maximum margin

6. Maximum Margin and Support Vectors

7. Standard (Primal) Perceptron Algorithm
Target minus Output not used. Just add (or subtract) a portion (multiplied by learning rate) of the current pattern to the weight vector
If weight vector starts at 0 then learning rate can just be 1
R could also be 1 for this discussion

8. Dual and Primal Equivalence
Note that the final weight vector is a linear combination of the training patterns
The basic decision function (primal and dual) is
How do we obtain the coefficients αi

9. Dual Perceptron Training Algorithm
Assume initial 0 weight vector

10. Dual vs. Primal Form
Gram Matrix: all (xi·xj) pairs – Done once and stored (can be large)
αi: One for each pattern in the training set. Incremented each time it is misclassified, which would have led to a weight change in primal form
Magnitude of αi is an indicator of effect of pattern on weights (embedding strength)
Note that patterns on borders have large αi while easy patterns never effect the weights
Could have trained with just the subset of patterns with αi > 0 (support vectors) and ignored the others
Can train in dual. How about execution? Either way (dual could be efficient if support vectors are few)
Would if transformed feature space is still not linearly separable? αi would keep growing. Could do early stopping or bound the αi with some maximum C, thus allowing outliers.

11. Feature Space and Kernel Functions
Since most problems require a non-linear decision surface, we do a non-linear map Φ(x) = (Φ1(x),Φ2(x), …,ΦN(x)) from input space to feature space
Feature space can be of very high (even infinite) dimensionality
By choosing a proper kernel function/feature space, the high dimensionality can be avoided in computation but effectively used for the decision surface to solve complex problems - "Kernel Trick"
A Kernel is appropriate if the matrix of all K(xi, xj) is positive semi-definite (has non-negative eigenvalues). Even when this is not satisfied many kernels still work in practice (sigmoid).
Phi

12. Basic Kernel Execution
Primal:
Dual:
Gaussian kernel example, learned reduced weighted nearest neighbor style
F is a space made from inner products of the original feature space X
just show f(x) for output rather than sign, can always set to sign but f(x) has more info
Kernel version:
Now we see the real advantage of working in the dual form
Note intuition of execution: Gaussian (and other) Kernel similar to reduced weighted K-nearest neighbor

13. replace middle two layers with just k(x,xi)

14. Polynomial Kernels For greater dimensionality we can do
Walk through with two vectors x1 x2 and z1 z2 (next slide)
For greater dimensionality we can do

15. Polynomial Kernel Example
Assume a simple 2-d feature vector x: x1, x2
Note that a new instance x will be paired with training vectors xi from the training set using K(x, xi). We'll call these x z for this example.
Note that in the input space x we are getting the 2nd order terms: x12, x22, and x1x2

16. Polynomial Kernel Example
Following is the 3rd degree polynomial kernel

17. Polynomial Kernel Example
Note that for the 2nd degree polynomial with two variables we get the 2nd order terms x12, x22, and x1x2
Compare with quadric. We also add bias weight with SVM.
For the 2nd degree polynomial with three variables we would get the 2nd order terms x12, x22, x32, x1x2, x1x3, and x2x3
Note that we only get the dth degree terms. However, with some kernel creation/manipulation we could also include the lower order terms
Only a subset of the quadric terms, but we can get the others.

18. SVM Execution Assume novel instance x = <.4, .8>
Assume training set vectors (with bias = 0)
.5, .3 y= -1 α=1
-.2, .7 y= 1 α=2
What is the output for this case?
Show higher-order and kernel computation

19. SVM Execution Assume novel instance x = <.4, .8>
Assume training set vectors (with bias = 0)
.5, .3 y= -1 α=1
-.2, .7 y= 1 α=2
1·-1·(<.5,.3>·<.4,.8>)2 + 2·1·(<-.2,.7>·<.4,.8>)2 = = .2672
This is Kernel version, what about higher order version?
Kernel version

20. SVM Execution Assume novel instance x = <.4, .8>
Assume training set vectors (with bias = 0)
.5, .3 y= -1 α=1
-.2, .7 y= 1 α=2
1·-1·(<.5,.3>·<.4,.8>)2 + 2·1·(<-.2,.7>·<.4,.8>)2 = = .2672
1·-1·(.52 · ·.5·.3·.4· ·.82) = =
2·1·((-.2)2 · ·-.2·.7·.4· ·.82) = 2·( ) = .4608
Showed each term separate for Phi space (higher order)

21. Kernel Trick
So are we getting the same power as the Quadric machine without having to directly calculate the 2nd order terms?

22. Kernel Trick
So are we getting the same power as the Quadric machine without having to directly calculate the 2nd order terms?
No. With SVM we weight the scalar result of the kernel, which has constant coefficients in the 2nd order equation!
With Quadric we can have a separate learned weight (coefficient) for each term
Although do get to individually weight each support vector
Assume that the 3rd term above was really irrelevant. How would Quadric/SVM deal with that?
Quadric would weight the term as 0. SVM can't – it comes with the entire equation. It can weight the equation (but not individual terms within it). But combinations of weightings of the different equations still lets us separate lots of things (ala RBF – are there enough support vectors to work from to get our desired classification)

23. Kernel Trick Realities
Polynomial Kernel - all monomials of degree 2
x1x3y1y3 + x3x3y3y (all 2nd order terms)
K(x,y) = <Φ (x)·Φ (y)> = … + (x1x3)(y1y3) + (x3x3)(y3y3) + ...
Lot of stuff represented with just one <x·y>2
However, in a full higher order solution we would would like adaptive coefficients for each of these higher order terms (i.e. -2x1 + 3·x1x2 + …)
SVM does a weighted (embedding coefficients) sum of them all with individual constant internal coefficients
Thus, not as powerful as a higher order system with arbitrary weighting
The more desirable arbitrary weighting can be done in an MLP because learning is done in the layers between inputs and hidden nodes
SVM feature to higher order feature is a fixed mapping. No learning at that level.
Of course, individual weighting requires a theoretically exponential increase in terms/hidden nodes for which we need to find weights as the polynomial degree increases. Also need learning algorithms which can actually find these most salient higher-order features.
Do Quadric review first, will help see polynomial kernel issues
Remember, we don't learn weights (per se) even for the combined kernel values (just embedding strengths), but the strengths are based on set coefficients.

25. SVM vs RBF Comparison SVM commonly uses a Gaussian kernel
Kernel is a distance metric (ala K nearest neighbor)
How does the SVM differ from RBF?

26. SVM vs RBF Comparison SVM commonly uses a Gaussian kernel
How does the SVM differ from RBF?
SVM will automatically discover which training instances to use as prototypes during the quadratic optimization
SVM works only in the kernel space while RBF calculates values in the much larger exploded space
Both weight the different prototypes
RBF uses a perceptron style learning algorithm to create weights between prototypes and output classes
RBF supports multiple output classes and nodes have a vote for each output class, whereas SVM support vectors can only vote for two classes: 1 or -1
SVM will create a maximum margin hyperplane decision surface
Since internal feature coefficients are constants in the Gaussian distance kernel (for both SVM and RBF), SVM will suffer from fixed/irrelevant features just like RBF/k-nearest neighbor
Same Kernel Trick weakness – no learning in the kernel map from input to higher order feature space
Discuss irrelevant feature weakness of K-nn and the similar issue with SVM not being able to weight individual terms

27. Choosing a Kernel
Can start from a desired feature space and try to construct a kernel
More often one starts from a reasonable kernel and may not analyze the feature space
Some kernels are a better fit for certain problems, domain knowledge can be helpful
Common kernels:
Polynomial
Gaussian
Sigmoidal
Application specific

28. Maximum Margin Maximum margin can lead to overfit due to noise
Problem may not be linearly separable even in transformed feature space
Soft Margin is a common solution, allows slack variables
αi constrained to be >= 0 and less than C. The C allows outliers.
How to pick C? Can try different values for the particular application to see which works best.

29. Soft Margins

30. Quadratic Optimization
Optimizing the margin in the higher order feature space is convex and thus there is one guaranteed solution at the minimum (or maximum)
SVM Optimizes the dual representation (avoiding the higher order feature space) with variations on the following
Maximizing Σαi tends towards larger α subject to Σαiyi = 0 and α ≤ C (which both tend towards smaller α)
Without this term, α = 0 could suffice
2nd term minimizes number of support vectors since
Two positive (or negative) instances that are similar (high Kernel result) would increase size of term. Note that both (or either) instances usually not needed.
Two non-matched instances which are similar should have larger α since they are likely support vectors at the decision boundary (negative term helps maximize)
The optimization is quadratic in the αi terms and linear in the constraints – can drop C maximum for non soft margin
While quite solvable, requires complex code and usually done with a numerical methods software package – Quadratic programming
Differs from straight training set accuracy due to the alpha sub j term

31. Execution
Typically use dual form which can take advantage of Kernel efficiency
If the number of support vectors is small then dual is fast
In cases of low dimensional feature spaces, could derive weights from αi and use normal primal execution
Can get speed-up by dropping support vectors with embedding coefficients below some threshold

32. Standard SVM Approach
Select a 2 class training set, a kernel function (calculate the Gram Matrix), and choose the C value (soft margin parameter)
Pass these to a Quadratic optimization package which will return an α for each training pattern based on a variation of the following (non-bias version)
Patterns with non-zero α are the support vectors for the maximum margin SVM classifier.
Execute by using the support vectors
key slide

33. A Simple On-Line Alternative
Stochastic on-line gradient ascent
Can be effective
This version assumes no bias
Sensitive to learning rate
Stopping criteria tests whether it is an appropriate solution – can just go until little change is occurring or can test optimization conditions directly
Can be quite slow and usually quadratic programming is used to get an exact solution
Newton and conjugate gradient techniques also used – Can work well since it is a guaranteed convex surface – bowl shaped
Could skip these next 2 slides

34. Maintains a margin of 1 (typical in standard SVM implementation) which can always be done by scaling  or equivalently w and b
This is done with the (1 - actual) term below, which can update even when correct, as it tries to make the distance of support vectors to the decision surface be exactly 1
If parenthetical term < 0 (i.e. current instance is correct and beyond margin), then don’t update 

35. Large Training Sets
Big problem since the Gram matrix (all (xi·xj) pairs) is O(n2) for n data patterns
106 data patterns require 1012 memory items
Can’t keep them in memory
Also makes for a huge inner loop in dual training
Key insight: most of the data patterns will not be support vectors so they are not needed

36. Chunking
Start with a reasonably sized subset of the Data set (one that fits in memory and does not take too long during training)
Train on this subset and just keep the support vectors or the m patterns with the highest αi values
Grab another subset, add the current support vectors to it and continue training
Note that this training may allow previous support vectors to be dropped as better ones are discovered
Repeat until all data is used and no new support vectors are added or some other stopping criteria is fulfilled

37. SVM Issues Excellent empirical and theoretical potential
Multi-class problems not handled naturally. Basic model classifies into just two classes. Can do one model for each class (class i is 1 and all else 0) and then decide between conflicting models using confidence, etc.
How to choose kernel – main learning parameter other than margin penalty C. Kernel choice will include other parameters to be defined (degree of polynomials, variance of Gaussians, etc.)
Speed and Size: both training and testing, how to handle very large training sets (millions of patterns and/or support vectors) not yet solved
Adaptive Kernels: trained during learning?