Khue

 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 methodHi @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 $\ell_1$-regularizationtypo Apr 8 revised On the stopping criterion of coordinate descent method for linear SVM with $\ell_1$-regularizationUpdate Apr 8 revised On the stopping criterion of coordinate descent method for linear SVM with $\ell_1$-regularizationadded 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 offsetThis 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 offsetI 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 offsetHi @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