Timeline for Is the value of $\alpha$ the same for all support vectors (SV) in the dual and what is the reason for it if they do or don't?
Current License: CC BY-SA 3.0
8 events
when toggle format | what | by | license | comment | |
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Mar 20, 2014 at 20:08 | vote | accept | Charlie Parker | ||
Mar 20, 2014 at 19:54 | answer | added | Marc Claesen | timeline score: 3 | |
Mar 20, 2014 at 19:46 | history | edited | Charlie Parker | CC BY-SA 3.0 |
made my question better english
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Mar 20, 2014 at 19:41 | comment | added | Charlie Parker | Don't worry, I think I also wrote the question a little weirdly, let me write it clearer, I agree its not the best wording I could have chosen. You should try look for a clarification before writing an angry comment and down voting though :P I was so sad for a second hehe But I understand the misunderstanding. I'll improve my question. | |
Mar 20, 2014 at 19:39 | history | edited | user603 | CC BY-SA 3.0 |
deleted 1 characters in body
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Mar 20, 2014 at 19:39 | comment | added | user603 | thanks for the clarification. I see now: I had read the equations a bit too fast and clearly not paid attention to the surrounding text. | |
Mar 20, 2014 at 19:36 | comment | added | Charlie Parker | @user603 in the case of no offset, for support vectors we have $y^{(i)} \theta \cdot x^{(i)} = 1$ (I said no slack and then later asked whats going on with slack). If $\theta = \sum^n_{i=1} \alpha_i y^{i} x^{(i)}$ then the constrain becomes: $y^{(i)} \sum^n_{i=1} \alpha_i y^{i} x^{(i)} \cdot x^{(i)} = 1$. Which is what I wrote. For support vectors we can change the inequality to an equality because its a support vector. There is nothing wrong about what I wrote. | |
Mar 19, 2014 at 21:51 | history | asked | Charlie Parker | CC BY-SA 3.0 |