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Then by the Cauchy-Schwartz inequality, we have
 
Then by the Cauchy-Schwartz inequality, we have
  
|kgamz|.
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<math>k\gamma_z(c^*) \leq \langle c^*,c_k \rangle \leq \|c^*\| \cdot \|c_k\| \leq \sqrt{k} \psi</math>.
 
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.. |kgamz| image:: tex
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  :alt: tex: k\gamma_z(c^*) \leq \langle c^*,c_k \rangle \leq \|c^*\| \cdot \|c_k\| \leq \sqrt{k} \psi
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It follows, then, that the number of required iterations of the Perceptron algorithm has a finite upper bound, i.e.
 
It follows, then, that the number of required iterations of the Perceptron algorithm has a finite upper bound, i.e.
  
|kbound|
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<math>k\leq \left( \frac{\psi}{\gamma_z(c^*)}\right)^2</math>
 
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.. |kbound| image:: tex
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  :alt: tex: k\leq \left( \frac{\psi}{\gamma_z(c^*)}\right)^2
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.. |zdef| image:: tex
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  :alt: tex: z=(x,y)\in\mathbb{Z}^m
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.. |z| image:: tex
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  :alt: tex: z
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.. |xiyiinz| image:: tex
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  :alt: tex: (x_i,y_i)\in z
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.. |gammaitildeDef| image:: tex
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  :alt: tex: \tilde{\gamma}_i = y_i\langle x_i,c \rangle
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.. |gammaztildeDef| image:: tex
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  :alt: tex: \tilde{\gamma}_z = \min_{(x_i,y_i)\in z} \tilde{\gamma}_i
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.. |gammaidef| image:: tex
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  :alt: tex: \gamma_i = \frac{\tilde{\gamma}_i(c)}{\|c\|}
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.. |gammazdef| image:: tex
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  :alt: tex: \gamma_z = \frac{\tilde{\gamma}_z(c)}{\|c\|}
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.. |psidef| image:: tex
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  :alt: tex: \psi = \max_{x_i\in x} \|\phi(x_i)\|
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.. |feature| image:: tex
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  :alt: tex: \phi(x_i)
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.. |xi| image:: tex
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  :alt: tex: x_i
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.. |stagekdef| image:: tex
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  :alt: tex: c_k = c_{k-1} + y_ix_i
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.. |ck| image:: tex
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  :alt: tex: c_k
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.. |k| image:: tex
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  :alt: tex: k
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.. |c*| image:: tex
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  :alt: tex: c^*
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.. |cc_kinnerprod| image:: tex
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  :alt: tex: \langle c^*,c_k \rangle = \langle c^*,c_{k-1} \rangle + y_i \langle c^*,x_i \rangle \geq \langle c^*,c_{k-1}\rangle + \gamma_z(c^*) \geq \ldots \geq k\gamma_z(c^*)
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.. |c0| image:: tex
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  :alt: tex: c_0=0
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.. |cknorm| image:: tex
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  :alt: tex: \|c_k\|^2 = \|c_{k-1}\|^2 + 2y_i\langle c_{k-1},x_i \rangle + \|x_i\|^2 \leq \|c_{k-1}\|^2 + \psi^2 \leq \ldots \leq k\psi^2.
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Revision as of 13:15, 28 March 2008

(See Lecture 10_Old Kiwi)

The following theorem, due to Novikoff (1962), proves the convergence of a perceptron using linearly-separable samples. This proof was taken from Learning Kernel Classifiers, Theory and Algorithms By Ralf Herbrich

Consider the following definitions:

A training set $ z=(x,y)\in\mathbb{Z}^m $

"Functional margin" on example $ (x_i,y_i)\in z $ is defined as $ \tilde{\gamma}_i = y_i\langle x_i,c \rangle $;

"Functional margin" on training set $ z $ is defined as $ \tilde{\gamma}_z = \min_{(x_i,y_i)\in z} \tilde{\gamma}_i $;

"Geometrical margin" on example $ (x_i,y_i)\in z $ is defined as $ \gamma_i = \frac{\tilde{\gamma}_i(c)}{\|c\|} $;

"Geometrical margin" on training set $ z $ is defined as $ \gamma_z = \frac{\tilde{\gamma}_z(c)}{\|c\|} $.

Let $ \psi = \max_{x_i\in x} \|\phi(x_i)\| $, where $ \phi(x_i) $ is the feature vector for $ x_i $.

Now let $ c_k $ be a final solution vector after $ k $ steps. Then the last update of the Perceptron algorithm has

$ c_k = c_{k-1} + y_ix_i $.

For any vector, $ c^* $, we have

$ \langle c^*,c_k \rangle = \langle c^*,c_{k-1} \rangle + y_i \langle c^*,x_i \rangle \geq \langle c^*,c_{k-1}\rangle + \gamma_z(c^*) \geq \ldots \geq k\gamma_z(c^*) $,

where the inequalities follow from repeated applications up to step 0 where we assume $ c_0=0 $. Similarly, by the algorithm definition,

$ \|c_k\|^2 = \|c_{k-1}\|^2 + 2y_i\langle c_{k-1},x_i \rangle + \|x_i\|^2 \leq \|c_{k-1}\|^2 + \psi^2 \leq \ldots \leq k\psi^2. $

Then by the Cauchy-Schwartz inequality, we have

$ k\gamma_z(c^*) \leq \langle c^*,c_k \rangle \leq \|c^*\| \cdot \|c_k\| \leq \sqrt{k} \psi $.

It follows, then, that the number of required iterations of the Perceptron algorithm has a finite upper bound, i.e.

$ k\leq \left( \frac{\psi}{\gamma_z(c^*)}\right)^2 $

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