Apr
9
awarded Student
Apr
9
comment On the stopping criterion of coordinate descent method
@elexhobby: Yes, it certainly does. However, my question is when the algorithm should stop to get an $\epsilon$-accurate solution (If we do not know the optimal value of course!) ;)
Apr
9
comment On the stopping criterion of coordinate descent method
Hi @RB. Let's consider only the objective function for linear SVM, that I stated above: $$\min f(\mathbf{x}) = \frac{1}{2}\mathbf{x}^\top K \mathbf{x} - C\mathbf{1}^\top\mathbf{x},\quad \mbox{s.t. } 0\le x_i\le C \quad i=1,\ldots,m.$$
Apr
8
revised On the stopping criterion of coordinate descent method for linear SVM with $\ell_1$-regularization
typo
Apr
8
revised On the stopping criterion of coordinate descent method for linear SVM with $\ell_1$-regularization
Update
Apr
8
revised On the stopping criterion of coordinate descent method for linear SVM with $\ell_1$-regularization
added 98 characters in body
Apr
8
asked On the stopping criterion of coordinate descent method for linear SVM with $\ell_1$-regularization
Apr
8
asked On the stopping criterion of coordinate descent method
Apr
4
comment prove that $\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}+(\sqrt{3}-1)(|a-1|+|b-1|+|c-1|)\ge 3\sqrt{3}$ if $a+b+c=3$
No problem, @Jonas12. If my answer resolves your problem, then please accept it. Thanks.
Apr
4
accepted SVM without offset
Apr
4
comment SVM without offset
This is exactly what I was looking forward. I read it once somewhere but then couldn't find it again. Thank you so much, @Dikran! (P/s: It would be nice if you edit the answer and replace it by this comment).
Apr
4
comment Proving inequality $\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\lt \sqrt{\frac{3}{2}}$
@daniel: Yeah I think it should be fine now. (Btw I will definitely post my solution, using Mixing Variables method, but maybe on next Friday.)
Apr
3
comment Proving inequality $\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\lt \sqrt{\frac{3}{2}}$
Hi @daniel. Do you realize that I have suggested a small modification to your solution to make it become correct? If you agree with my suggestion, then just modify it that way, and there's no need to delete the answer. If you don't, or if I was not clear enough, then just ask me for clarification. Best regards.
Apr
3
comment SVM without offset
I understand now. We can either use a rotation or a translation. What you have suggested is a translation, which is much simpler than a rotation. In this case, we have to solve another problem than "SVM-without-offset", haven't we? (I mean we cannot use the "SVMs-without-bias" solver).
Apr
3
comment SVM without offset
Hi @Dikran. Thanks for the answer. Could you please be more specific about "adding an extra input feature..."? My thinking: Suppose that we have already an "SVMs-without-bias" solver $F(X,y,C)$ (that returns $w$ and $\xi$). Now for an instance $(X,y,C)$, where the data $(X,y)$ are linearly separable by a hyperplane that does not pass through the origin, if one applies a suitable rotation to the data to obtain $X',y'$ (i.e. we go to another space, the same idea as "feature space" in kernels), then the above solver can be used to classify the new (rotated) features: $F(X',y',C)$.
Apr
3
asked SVM without offset
Apr
3
comment Proving inequality $\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\lt \sqrt{\frac{3}{2}}$
We can, however, assume that $a=\min(a,b,c)$ for example. In this case, the problem becomes maximizing $w(x,y)$ subject to $0<x\le 1/3$ and $x\le y \le 1$. The feasible region has been changed a bit, but probably the arguments in your solution can still be applied. With this (necessary) modification, your solution becomes correct. (P/s: Of course a nice analytic solution exists, maybe I will post it when I have time.)
Apr
3
comment Proving inequality $\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\lt \sqrt{\frac{3}{2}}$
As I said the inequality is not symmetric. It is only cyclic, not symmetric (clearly $f(a,b,c)\neq f(b,a,c)$), thus if we assume $a\le b\le c$ then generality is lost. Your arguments above even prove that. Indeed, according to your arguments, if $a\le b\le c$ then the maximum value of $f(a,b,c)$ is only $1.115$, i.e. $f(a,b,c)$ will never reach its true maximum value $1.179$, which can be achieved only in the case $b\le a\le c$ (or $c\le b\le a$ or $a\le c\le b$). ...(continued below)...
Apr
2
awarded Commentator
Apr
2
comment Proving inequality $\frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\lt \sqrt{\frac{3}{2}}$
One more remark: you found the maximum at $y=0^+$ and $x\approx 0.1547$, but these values clearly violate the constraint $x\leq y \leq 1/2$. By the way, you might be interested in the closed-form values of the numerical values that you found. Here they are: $$x=\frac{2+2\sqrt{3}}{3+2\sqrt{3}} \approx 0.845, \quad y =\frac{1}{3+2\sqrt{3}} \approx 0.1547$$ and the maximum value is $$\frac{2\sqrt{3}-2}{\sqrt{2\sqrt{3}}}+\sqrt{-1+\frac{2}{\sqrt{3}}} \approx 1.179.$$ These values were totally found by hand (+ a pen and a scrap paper :D).
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