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In my machine learning class I have been provided a weight vector that has the property that it is generously feasible ?

Formally, what does generously feasible mean? I can't seem to find a definition?

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If this is from Geoffrey Hinton's slides, he defines them in slide 23. In fact, this is the only real Google hit I could get for those words –  Ganesh Oct 15 '12 at 5:25
    
Can you provide the link? It doesn't show in my search results? –  Andrew Tomazos Oct 15 '12 at 6:34
    
tiny.cc/7lj7lw –  Ganesh Oct 15 '12 at 6:43
    
I've removed algebra tag, since we don't use algebra tag anymore, see meta for details. –  Martin Sleziak Oct 19 '12 at 18:03
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2 Answers 2

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If the weight vector in the current iteration is in the region between the hyperplane and the magnitude of input vector, i.e. $\vec{w^t_x} \: \epsilon \: [ \langle \vec{w_{x}},\vec{x} \rangle , |\vec{x}| ]$, where $\langle \vec{w_{x}},\vec{x} \rangle$ is the hyperplane, then, since the perceptron adds $\vec{x}$ or $-\vec{x}$ to the weights each iteration, the weight vector will oscillate around the hyperplane. Hence for the algorithm to terminate with a solution, it should be allowed to accept a solution in this feasible space, hence called the "generously feasible" space.

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So consider “generously feasible” weight vectors that lie within the feasible region by a margin at least as great as the length of the input vector that defines each constraint plane.

Every time the perceptron makes a mistake, the squared distance to all of these generously feasible weight vectors is always decreased by at least the squared length of the update vector.

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