Theorem 1Suppose \(\mathbf X\) is full rank and hence \(\mathbf \Sigma_{\mathbf X \mathbf X}\) is invertible. Further assume \(\mathbf \Sigma \doteq \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} \mathbf \Sigma _{\mathbf X \mathbf Y}\) is full rank with \(n\) distinct eigenvalues \(\lambda_1 > \cdots > \lambda_n > 0\). Then \(\eqref{eq:nn_main_global}\) has no spurious local minima, except for equivalent versions of global minimum due to invertible transformations.

**Proof.** The partial derivatives of \(f\) are [this post provides a useful trick to do this kind of calculation] \[\begin{align*}
\frac{\partial}{\partial \mathbf A} f & = - \left(\mathbf Y - \mathbf A \mathbf B \mathbf X\right) \mathbf X^\top \mathbf B^\top, \newline
\frac{\partial}{\partial \mathbf B} f & = - \mathbf A^\top \left(\mathbf Y - \mathbf A \mathbf B \mathbf X\right) \mathbf X^\top.
\end{align*}\] Setting the derivatives to zero, we obtain that critical points of \(f\) are characterized by \[\begin{align} \label{eq:crit_ch1}
\mathbf A^\top \mathbf \Sigma_{\mathbf Y\mathbf X} & = \mathbf A^\top \mathbf A \mathbf B \mathbf \Sigma_{\mathbf X \mathbf X}
\end{align}\] and \[\begin{align} \label{eq:crit_ch2}
\mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf B^\top & = \mathbf A \mathbf B \mathbf \Sigma_{\mathbf X \mathbf X} \mathbf B^\top.
\end{align}\] Now we claim that at any critical point of \(f\), \[\begin{align} \label{eq:crit_ch3}
\mathbf A \mathbf B = \mathcal P_{\mathbf A} \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1}
\end{align}\] with \(\mathbf A\) satisfying \[\begin{align} \label{eq:crit_ch4}
\mathcal P_{\mathbf A} \mathbf \Sigma = \mathcal P_{\mathbf A} \mathbf \Sigma \mathcal P_{\mathbf A} = \mathbf \Sigma \mathcal P_{\mathbf A},
\end{align}\] where \(\mathcal P\) is the ortho-projector projecting onto the column span of the sub-indexed matrix. To see this, from \(\eqref{eq:crit_ch1}\) we obtain \[\begin{align*}
\mathbf B = \left(\mathbf A^\top \mathbf A\right)^{\dagger} \mathbf A^\top \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} + \left(\mathbf I - \mathbf A^{\dagger} \mathbf A\right) \mathbf L = \mathbf A^{\dagger} \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} + \left(\mathbf I - \mathbf A^{\dagger} \mathbf A\right) \mathbf L,
\end{align*}\] where \(\dagger\) denotes the Moore-Penrose pseudo-inverse and \(\mathbf L\) is any \(p \times n\) matrix. So \[\begin{align*}
\mathbf A \mathbf B = \mathbf A \mathbf A^{\dagger} \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} + \mathbf A \left(\mathbf I - \mathbf A^{\dagger} \mathbf A\right) \mathbf L = \mathcal P_{\mathbf A} \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1},
\end{align*}\] where we have used the fact that \(\mathbf X = \mathbf X \mathbf X^{\dagger} \mathbf X\) for any matrix \(\mathbf X\). Now by \(\eqref{eq:crit_ch2}\), \[\begin{align*}
\mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf B^\top \mathbf A^\top = \mathbf A \mathbf B \mathbf \Sigma_{\mathbf X \mathbf X} \mathbf B^\top \mathbf A^\top,
\end{align*}\] substituting the analytic expression obtained above for \(\mathbf A \mathbf B\) yields \[\begin{align*}
\mathcal P_{\mathbf A} \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} \mathbf \Sigma_{\mathbf X \mathbf Y} \mathcal P_{\mathbf A} = \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} \mathbf \Sigma_{\mathbf X \mathbf Y} \mathcal P_{\mathbf A} \Longleftrightarrow \mathcal P_{\mathbf A} \mathbf \Sigma \mathcal P_{\mathbf A} = \mathbf \Sigma \mathcal P_{\mathbf A}.
\end{align*}\] Note also \(\mathbf \Sigma \mathcal P_{\mathbf A} = \mathcal P_{\mathbf A} \mathbf \Sigma\) since \(\mathcal P_{\mathbf A}\mathbf \Sigma \mathcal P_{\mathbf A}\) is symmetric.

Next we show the claimed about global minimum. To this end, suppose at a critical point \(\operatorname{rank}\left(\mathbf A\right) = r \leq p\). Assuming the eigendecomposition of \(\mathbf \Sigma\) as \(\mathbf \Sigma = \mathbf U \mathbf \Lambda \mathbf U^\top\), then we have \[\begin{align*} \mathcal P_{\mathbf U^\top \mathbf A} = \mathbf U^\top \mathbf A \left(\mathbf A^\top \mathbf U \mathbf U^\top \mathbf A\right)^{\dagger} \mathbf A^\top \mathbf U = \mathbf U^\top \mathbf A\left(\mathbf A^\top \mathbf A\right)^{\dagger} \mathbf A^\top \mathbf U = \mathbf U^\top \mathcal P_{\mathbf A} \mathbf U, \end{align*}\] or \(\mathcal P_{\mathbf A} = \mathbf U \mathcal P_{\mathbf U^\top \mathbf A} \mathbf U^\top\). Now by \(\eqref{eq:crit_ch4}\) we obtain \[\begin{align*} \mathbf U \mathcal P_{\mathbf U^\top \mathbf A} \mathbf U^\top \mathbf U \mathbf \Lambda \mathbf U^\top = \mathcal P_{\mathbf A} \mathbf \Sigma = \mathbf \Sigma \mathcal P_{\mathbf A} = \mathbf U \mathbf \Lambda \mathbf U^\top \mathbf U \mathcal P_{\mathbf U^\top \mathbf A} \mathbf U^\top, \end{align*}\] which yields \[\begin{align} \mathcal P_{\mathbf U^\top \mathbf A} \mathbf \Lambda = \mathbf \Lambda \mathcal P_{\mathbf U^\top \mathbf A}. \end{align}\] It is easy to from the above \(\mathcal P_{\mathbf U^\top \mathbf A}\) must be diagonal as \(\mathbf \Lambda\) is diagonal with distinct values by our assumption. Then \(\mathcal P_{\mathbf U^\top \mathbf A} = \mathbf I_{\mathcal J}\) for an index set \(\mathcal J = \left\{\ell_1, \cdots, \ell_r\right\} \in [n]\) with \(1 \leq \ell_1 < \ell_2 < \cdots < \ell_r \leq n\) and \[\begin{align} \label{eq:diagonal_constraint} \mathcal P_{\mathbf A} = \mathbf U \mathcal P_{\mathbf U^\top \mathbf A} \mathbf U^\top = \mathbf U \mathbf I_{\mathcal J} \mathbf U^\top = \mathbf U_{\mathcal J} \mathbf U_{\mathcal J}^\top. \end{align}\] Hence column space is identical to the column space of \(\mathbf U_{\mathcal J}\) and hence at any critical point of \(f\), \(\mathbf A\) can be written in the form \[\begin{align} \label{eq:form_a_final} \mathbf A = \left[\mathbf U_{\mathcal J}, \mathbf 0_{n \times \left(p-r\right)}\right] \mathbf C \end{align}\] for some invertible \(\mathbf C\). So by \(\eqref{eq:crit_ch1}\) again \(\mathbf B\) will take the form \[\begin{align} \mathbf B = \mathbf A^{\dagger} \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} + \left(\mathbf I - \mathbf A^{\dagger} \mathbf A\right) \mathbf L \end{align}\] for some \(\mathbf L\) constrained by \(\eqref{eq:crit_ch2}\). Note that from \(\eqref{eq:form_a_final}\) we have \[\begin{align*} \mathbf A^{\dagger} = \mathbf C^{-1} \left[\mathbf U_{\mathcal J}^\top; \mathbf 0\right], \end{align*}\] where we have used Matlab notation for matrix concatenation. So \[\begin{align} \mathbf B & = \mathbf C^{-1} \begin{bmatrix} \mathbf U^\top_{\mathcal J} \newline \mathbf 0 \end {bmatrix} \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} + \left(\mathbf I - \begin{bmatrix} \mathbf U^\top_{\mathcal J} \newline \mathbf 0 \end {bmatrix} \begin{bmatrix} \mathbf U_{\mathcal J}, \mathbf 0 \end{bmatrix} \right) \mathbf L \newline & = \mathbf C^{-1}\begin{bmatrix} \mathbf U^\top_{\mathcal J} \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} \newline \mathbf 0 \end{bmatrix} + \mathbf C^{-1} \begin{bmatrix} \mathbf 0 & \newline & \mathbf I_{p - r} \end{bmatrix} \mathbf C \mathbf L \newline & = \mathbf C^{-1} \begin{bmatrix} \mathbf U^\top_{\mathcal J} \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} \newline \text{last $p-r$ rows of $\mathbf C \mathbf L$} \end{bmatrix}. \end{align}\] We will need to discuss two cases separately.

When \(r < p\), we can perturb the last \(p-r\) rows of \(\mathbf B\) with arbitrarily small noise to make \(\widehat{\mathbf B}\) full rank, and \(\mathbf A \mathbf B = \mathbf A \widehat{\mathbf B}\). Since when \(\widehat{\mathbf B}\) is full rank, \(f\left(\mathbf A, \widehat{\mathbf B}\right)\) is strictly convex in \(\mathbf A\), for any \(\varepsilon \in \left(0, 1\right)\), we can make \(\overline{\mathbf A} = \left(1-\varepsilon\right)\mathbf A + \varepsilon \widehat{\mathbf A}\left(\widehat{\mathbf B}\right)\) such that \[\begin{align*} f\left(\overline{A}, \widehat{\mathbf B}\right) < f\left(\mathbf A, \widehat{\mathbf B}\right) = f\left(\mathbf A, \mathbf B\right). \end{align*}\] Since \(\varepsilon\) can also be made arbitrarily small and \(\overline{\mathbf A} \to \mathbf A\) as \(\varepsilon \to 0\), \(\left(\mathbf A, \mathbf B\right)\) is a saddle point.

When \(r = p\), we have \[\begin{align*} \mathbf A = \mathbf U_{\mathcal J} \mathbf C, \quad \mathbf B = \mathbf C^{-1} \mathbf U_{\mathcal J}^\top \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} \end{align*}\] for some index set \(\left|\mathcal J\right| = p\) and invertible \(\mathbf C \in \mathbb{R}^{p \times p}\). Hence there are \(\binom{n}{p}\) possible choice for \(\mathcal J\) in this case and so, up to equivalence, \(\binom{n}{p}\) critical points with full rank. Assuming the natural ordering induced by the eigenvalues of \(\mathbf \Sigma\), we’ll show that whenever \(\mathcal J \neq \left\{1, \dots, p\right\}\), the corresponding critical point is a saddle point. First notice that when \(\mathbf A\) is full rank, \(\mathbf B\) is uniquely defined by \(\eqref{eq:crit_ch1}\), so \[\begin{align*} f\left(\mathbf A, \mathbf B\left(\mathbf A\right)\right) & = \left\| \mathbf Y - \mathbf A \mathbf B\left(\mathbf A\right) \mathbf X \right\|_{F}^2 /2 \newline & = \left\| \mathbf Y - \mathcal P_{\mathbf A} \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} \mathbf X \right\|_{F}^2 /2 \newline & = \left\| \mathbf Y \right\|_{F}^2/2 - \left\langle \mathbf \Sigma_{\mathbf Y \mathbf X}, \mathcal P_{\mathbf A} \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} \right \rangle + \left \langle \mathbf \Sigma_{\mathbf Y \mathbf X}^\top \mathcal P_{\mathbf A}, \mathbf \Sigma_{\mathbf X \mathbf X}^{-1} \mathbf \Sigma_{\mathbf X \mathbf Y} \mathcal P_{\mathbf A} \right \rangle/2 \newline & = \left\| \mathbf Y \right\|_{F}^2/2 - \left\langle \mathcal P_{\mathbf A}, \mathbf \Sigma \right\rangle + \left\langle \mathcal P_{\mathbf A}, \mathbf \Sigma \right\rangle/2 \newline & = \left\| \mathbf Y \right\|_{F}^2/2 - \left\langle \mathcal P_{\mathbf A}, \mathbf \Sigma \right\rangle/2 = \left\| \mathbf Y \right\|_{F}^2/2 - \left\langle \mathcal P_{\mathbf U^\top \mathbf A}, \mathbf \Lambda \right\rangle/2 \end{align*}\] where in the simplification we have used \(\eqref{eq:crit_ch3}\) and \(\eqref{eq:crit_ch4}\). Now by \(\eqref{eq:diagonal_constraint}\) and results following that we have \[\begin{align} f\left(\mathbf A, \mathbf B\left(\mathbf A\right)\right) = \left\| \mathbf Y \right\|_{F}^2/2 - \frac{1}{2}\sum_{j \in \mathcal J} \lambda_j. \end{align}\] Now if \(\mathcal J \neq \left\{1, \cdots, p\right\}\), there exists some index \(\ell \in [p]\) but \(\ell \notin \mathcal J\). For some index \(k \in \mathcal J\) such that \(k > \ell\), consider an \(\varepsilon\)-perturbation to \(\mathbf u_k\): \(\widehat{\mathbf u}_k = \left(\mathbf u_k + \varepsilon \mathbf u_{\ell}\right) / \sqrt{1+\varepsilon^2}\) and replace \(\mathbf u_k\) in \(\mathbf U_{\mathcal J}\) by \(\widehat{\mathbf u}_k\) to form \(\widehat{\mathbf U}_{\mathcal J}\). It is obvious still we have \(\widehat{\mathbf U}_{\mathcal J}^\top \widehat{\mathbf U}_{\mathcal J} = \mathbf I_{\mathcal J}\). Now let \(\widehat{\mathbf A} = \widehat{\mathbf U}_{\mathcal J} \mathbf C\) and \(\widehat{\mathbf B} = \mathbf C^{-1} \widehat{\mathbf U}_{\mathcal J}^\top \mathbf \Sigma_{\mathbf Y \mathbf X} \mathbf \Sigma_{\mathbf X \mathbf X}^{-1}\). Now the diagonal elements of \(\mathbf M \doteq \mathcal P_{\mathbf U_{\mathcal J}^\top \widehat{\mathbf A}}\) are: \[\begin{align*} M_{ii} = \begin{cases} 0 & i \notin \mathcal J \cup \left\{\ell\right\} \newline 1 & i \in \mathcal J \; \text{and}\; i \neq k \newline 1/\left(1+\varepsilon^2\right) & i = k \newline \varepsilon^2 /\left(1+\varepsilon^2\right) & i = \ell \end{cases}. \end{align*}\] So the perturbed objective \[\begin{align*} f\left(\widehat{\mathbf A}, \widehat{\mathbf B}\right) & = \left\| \mathbf Y \right\|_{F}^2/2 - \left\langle \mathcal P_{\mathbf U^\top \widehat{\mathbf A}}, \mathbf \Lambda \right\rangle/2 \newline & = \left\| \mathbf Y \right\|_{F}^2/2 - \frac{1}{2}\sum_{j \in \mathcal J} \lambda_j - \varepsilon^2\left(\lambda_{\ell} - \lambda_{k}\right)/\left(1+\varepsilon^2\right). \end{align*}\] Since by our construction \(\lambda_{\ell} - \lambda_k > 0\), for all \(\varepsilon > 0\) \[\begin{align} f\left(\widehat{\mathbf A}, \widehat{\mathbf B}\right) < f \left(\mathbf A, \mathbf B\right), \end{align}\] and also \(\left(\widehat{\mathbf A}, \widehat{\mathbf B}\right)\) can be arbitrarily close to \(\left(\mathbf A, \mathbf B\right)\) as \(\varepsilon \to 0\). So \(\left(\mathbf A, \mathbf B\right)\) is a saddle point. \(\Box\)

One may naturally think of using second-order geometry directly, i.e., deriving the Hessian matrix and looking at its definiteness, to investigate the behaviors of the critical points. It is likely to be a principled but cumbersome approach, as compared to the analysis presented here.

Baldi, Pierre, and Kurt Hornik. “Neural networks and principal component analysis: Learning from examples without local minima." Neural networks 2, no. 1 (1989): 53-58. ftp://canuck.seos.uvic.ca/CFD-NN/1-s2.0-0893608089900142-main.pdf↩

From simulation, it seems one needs at most \(m \leq O\left(n \log n\right)\) to ensure the desired coverage with high probability. Here I present a heuristic argument showing \(m\) only needs to be at most a low-order polynomial in \(n\). Suppose we require something stronger: all edges get colored. By the celebrated Balinski’s theorem, this implies vertex coverage we desire.

Theorem 1 (Balinski’s Theorem)The undirected graph from the vertices and edges of a \(d\)-dimensional convex polytope is \(d\)-connected, i.e., the graph remains connected if one removes any \(d-1\) or fewer vertices and the associated edges from the graph.

Now from the generation model, we expect \(\mathcal K_n\) to be regular both in combinatorial and geometric properties. So on average, the spherical simplex cut by each facet from the sphere has normalized spherical measure \(1/\#\mathrm{facets}\) and each edge is touched by \(\#\mathrm{facets}/\#\mathrm{edges}\) facets. So \[\begin{align*} & \mathbb P\left[\exists\; \text{an edge not colored with $m$ samples}\right] \newline \leq\; & \#\mathrm\; \times \; \mathbb P\left[\text{a fixed edge not colored with $m$ samples}\right] \newline \approx \; & \#\mathrm\; \times \; \left(1-\frac{1}{\#\mathrm{facets}} \frac{\#\mathrm{facets}}{\#\mathrm{edges}}\right)^m \newline \leq \; & \#\mathrm\times \exp\left(-\frac{m}{\#\mathrm{edges}}\right) \newline =\; & \exp\left(-\frac{m}{\#\mathrm{edges}} + \log \#\mathrm{edges}\right). \end{align*}\] Note that \(\#\mathrm{edges} \leq n^2/2\), and it is enough to take \[\begin{align} m = O\left(n^2 \log n\right) \end{align}\] to ensure all edges get colored with some nontrivial constant probability or \(m = O\left(n^3\right)\) to ensure high probability. The order is likely to be loose, as in the random model \(\#\mathrm{edges}\) could be order-of-magnitude smaller than \(O\left(n^2\right)\), and also one may only need to cover a fraction of edges to ensure vertex coverage. Of course to make the argument rigorous, one needs in addition to also work on extreme quantities such as 1) minimum normalized spherical measure one spherical simplex subtends; and 2) minimum number of facets one edge will touch in \(\mathcal K_n\).

]]>There are many questions of probabilistic and combinatorial nature that can be posed for this setting. Recent research on this culminates in proving some concentration results to the \(\mathrm{volume}\left(\mathcal K\right)\) in this paper^{1}. In fact, the result applies to points sampled i.i.d. uniform from the boundary of any convex body with smooth boundaries:

Theorem 1For any compact convex body \(\mathbf K\) with nonempty interior and \(\mathcal C^2\) boundaries and everywhere positive Gauss-Kronecker curvature, there are positive constants \(\alpha\) and \(c\) such that the following holds: consider the convex polytope \(\mathcal K_n = \operatorname{\mathbf{conv}}\left(\mathbf x_1, \cdots, \mathbf x_n\right)\), where \(x_i\)’s are sampled i.i.d. uniform from the boundary of \(\mathbf K\). For any constant \(0 < \eta < \frac{d-1}{3d+1}\) and \(0 < \lambda \leq \frac{\alpha}{4} n^{\left(d-1\right)/\left(3d+1\right) + 2\left(d+1\right)\eta/\left(d-1\right)}\), we have \[\begin{align} \mathbb P\left[\left|\mathrm{Vol}\left(\mathcal K_n\right)- \mathbb E\mathrm{Vol}\left(\mathcal K_n\right) \right| > \sqrt{\lambda V_0}\right] \leq 2\exp\left(-\lambda/4\right) + \exp\left(-cn^{\left(d-1\right)/\left(3d+1\right) - \eta}\right) \end{align}\] where \(V_0 = \alpha n^{-\left(d+3\right)/\left(d-1\right)}\).

This development is in line with the more developed results that achieves central limit theorems (CLT) of various quantities of interest for the model in which random points sampled directly from the mother convex body \(\mathbf K\), instead of its boundary. See this paper^{2}.

One problem from my research project recently is more leaned toward the combinatorial side. A simple version is like this: consider \(\mathcal K\) as in the first paragraph, how many facets one needs to randomly color (say in red), such that ultimately 1) the colored regions are connected; and 2) any vertex lie in at least colored facet. Here by “randomly” we mean uniform as measured by the “volume of the spherical simplex” subtended by facets, normalized by the surface area of \(\mathbb S^{d-1}\) of course. It is very easy to see the lower bound is \[\begin{align*} \Omega\left(n/d\right) \end{align*}\] as this is the least required if one were to cover all vertices. It is not immediately clear what the upper bound is. Intuitively, since the random polytopes tend to behave like very “regular objects”, and our sampling is “uniformly” random, one expects substantially fewer such colored facets to cover all vertices, rather than all facets, the number of which is often exponential in \(d\). Some simulations (of course, computation quickly becomes prohibitive even one’s interested in computing the convex hull of many points in large dimensions) suggest the upper bound is at most \[\begin{align} O\left(n\log n\right), \end{align}\] which is very mild as compared to the crude estimate of the number of facets one’d have \[\begin{align} \binom{n}{d} \leq \frac{n^d}{d!}. \end{align}\]

An Inscribing Model for Random Polytopes, by Richardson, Vu, and Wu, http://link.springer.com/article/10.1007%2Fs00454-007-9012-3↩

Sharp Concentration of Random Polytopes, by Vu, http://link.springer.com/article/10.1007%2Fs00039-005-0541-8↩

This (a) has explained things pretty clearly, except for that you won’t find **ntfsprogs** in recent versions of Linux distributions, as the ntfsprogs project has been merged with the **ntfs-3g** project under Tuxera Inc. according to Wiki. The real attack works as described in Answer in this (b), except that one needs to check out the mount point by “**sudo fdisk -l**” and “**sudo ntfsfix -b XXX**” to get the program to start the fix, where **XXX** is to be replaced by the real mount point.

内容我就不总结了，以防制造笑料，毕竟还不知后续剧情(我就只等3D了)。主要总结各位痴人之梦语如下，仅供娱己。

一、科学家很危险，特别是狂热型的。当然尤其是搞生物和物理的一一此影片在验证前者，金三世的小朝廷正在验证后者。科学加政治是毁灭。

一、码工不可靠，尤其胖子码工。人家需要花钱畜油，你当老板的让人家总差钱是要把人逼成反派的。

一、戴墨镜的爱耍流氓。那个试验为名搞亲密接触的家伙，当着人家老公求交往。不过这家伙相对于那个倔强的老头还是有点科学良知的。

一、律师招人厌，也最没节操。坐在马桶上 死，是不是剧组为迎合观众的深度发泄和调侃。就像骂人“生孩子没屁眼”，这个是“死在马桶上”。

一、科学家很现世，特别是看到子funding 。对于在沙地里挖恐龙的主，钱比啥的风力都大。

一、话多的熊孩子是美剧的黄金导线；你觉的可有可无又控场的角色，如挖恐龙的夫妇，铁定的英雄好汉。主人英雄主义横行乃王道。

一、恐龙也有素食主义者的，它们可爱温驯。

]]>It seems to me "

**sudo apt-get install libopencv-dev**" for dependencies is not necessary - this will be superseded by the subsequent installation of opencv from source. At the end of the scripting installation, the package seems to be in some cache place and be not in effect:It seems "

**sudo apt-get install libavcodec-dev libavformat-dev libswscale-dev**" is also unnecessary. These libraries are to be provided by the next fresh installation of ffmpeg follows. Checking their location seems to also confirm my guess:About version of ffmpeg: ffmepg is under active development as always. The choice of its version is sometimes tricky, especially in running together with OpenCV. I remember the 1.x series was not compatible with OpenCV 2.3 in my previous trial (compilation errors were thrown due to some problem with ffmpeg libraries). The current 2.4.3 version seems to be at least fine with ffmpeg 0.11.2 in compilation.

[Update] From test by Mr. **Alok Singh Mahor ** (comments in the original post), installing Ubuntu package “libopencv-dev” would already get things work, though the linking order with g++ has to be taken care of as discussed (also comments in the original post and here) . Of course, it’s not bad idea to install from source that always guarantees to bring in the newest feature of actively developed OpenCV. [Dec 28 2012 / Oct 30 2013]

Florence Gallery ( Google photos or 百度相册 )

]]>Bear Mountain Gallery (Oct 27 2012) ( Google photos or 百度相册 )

Several that are of interest to most (after a very random walk of recent entries):

Randomness in Matlab (discussing why using the new random generator controller

**rng()**is important, here also)Pandas toolbox (Python data analysis package frequently mentioned in the blog)

The Julia language (A new high-level language targeted for high-performance with the slightest coding pain)

Disagreement on operator precedence for 2

^{3}4 (Don’t be afraid to use brackets!)Nonlinear eigenvalue problems and a toolbox (Solving problem of the form \(F(\lambda) \mathbf{ x} = 0\) where \(F\) maps complex scalars to complex matrices (e.g., matrix-coefficient polynomials). Applications can be found in many minimal problems in computer vision, e.g., the relative pose problem.

Will 2012 be the year when OpenCL adoption really takes off? (Describing trends in CPU-contained GPU units and relative lack in OpenCL’s stand in scientific computing arena)

弄到手头这辆自行车之后，心里就一直痒着要牵出去溜溜。网上查到这个Greenway 之后，忍不住激动，拉着车就出发了！（哥也是有车的人啦！哈哈哈）

按官方的说法，这条Greenway 全长32 Miles （约51.2 公里）。西边沿Hudson River 一直蜿蜒至Battery Park, 远眺自由女神后，折向东北，继而继续蜿蜒前行，经Manhattan Bridge, Willamburg Bridge, 联合国， 至Harlem River 上游。

我从West 103 街的入口，取道Riverside Park。时值下午3点左右，众多的New Yorkers在公园里玩弄时光：跑步的人最多好像，然后慢性散步的，脱了上衣晒太阳的，轮滑的，当然也有谈请说爱的。哥关注玩Soccer的小朋友们

从一座天桥下穿过后，Hudsen River已尽在眼前。来去匆匆的跑步的和骑行者，点缀着这座都市的一个典型周末。河风，在单车旁拍打，渗透，直至心灵。好清爽的下午！河上的泛舟者，伴着偶尔的帆船，让人们诉说风中点点惺味的来由。

当然，水中河畔也有这座曾经的工业城市留下的蛛丝马迹。

桩桩短木，是为时光留下的墓志，没有碑文当然。一路向南的骑行，基本上就是在现代和生锈的历史间穿梭，Hudsen River 作为纽带。然后西南角的地表，应该就是那时隐时现的自由女神了（不好意思，这个太远了，你得仔细的找，哈哈）

还有愁云翻滚下的911遗址上新建中的大厦。

对了，还忘了说空中时常掠过的直升机，这应该是纽约上层人士的交通工具吧（手机上的相机，也就只能这么浪费有效空间了。）

W42 街的Circle Line 码头（应该是去看自由女神像的水上线路）， 以及对面的Chinese Consult in New York (五星红旗很亲切的，你看多了星条旗之后）

离自由女神越来越近啦！西南转角处众多的游艇，和远方渺茫矗立的女神！

曼哈顿大桥，和隔岸相望的新泽西州。

后来，后来，Greenway 开始频繁穿梭于市区之中，而且经常消失了去处，相当的路段还在维修。暮色逐渐深了，我就没再拍了。最过瘾的事情是骑行穿梭了两次Queensboro bridge, 这是全程坡度最大的地方。不过后来验证我确实是走错了路，哈哈。

就这样，下次好好计划，多拍点。

]]>Version 2 of CVX has featured extended support for commercial solvers, MOSEK and Gurobi. This is really good news for academic users, since 1) both solvers provide free licenses to academic users; and 2) both solvers are mostly more optimized than the free solvers currently bundles with CVX, namely, SDPT3 and SeDuMi. Related part of the announcement as follows:

Academic usersmay utilize the CVX Professional capabilityat no charge.Users with a valid academic license for Gurobi 5.0 or later may use it with CVX

without obtaining a CVX Professional license. This is because CVX is able to detect the presence of an academic Gurobi license.In order to use MOSEK, a CVX Professional license is required. We intend to provide such licenses at no charge, but we have not yet completed the licensing architecture. We will make an announcement once this has been completed.

In addition, mailing list support has changed into a Q&A forum (StackExchange style …) and the documentation has changed to online html version with better cross references.

]]>In fact, Nikon has added in rechargeable support not long after this camera was released. You can follow these steps to get your L100 live well with the rechargeable:

**Firmware Version**. Press the MENU botton, and go to Set Up –> Firmware version. If the firmware is COOLPIX L100 Ver.1.0, you will need to update your firmware to V1.1. following these instructions (*Coolpix**L100*1.1 firmware update from Nikon website).**Battery Type**. After the updating, one has to select Set up –> Battery type –> COOLPIX (Ni-MH) to enable this lovely feature.

Now you should be able to power your L100 with your rechargeable batteries (and also still be able to use the disposable)！

]]>Stay tuned!

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PS： 今天presentation 把without loss of generality 说成 without loss of generosity 。。。被John 掐住好，很肿的补了句 Anyway we cannot lose it …

]]>One possibility is observe the non-negativity and continuity of the function \(\arctan x\) over \(j \in {\mathbb Z}_+\) and apply an integral lower bound: \[ \begin{align*} & \sum_{j=1}^k \arctan \frac{1}{j} \newline \geq \; & \int_{1}^{k+1} \arctan 1/x \; dx = \left. x\arctan 1/x + \frac{1}{2} \ln \left(1+x^2\right) \right|_{x = 1}^{k+1} \newline = \; & \frac{1}{2}\ln \left[\left(k+1\right)^2 +1\right] + \left(k+1\right) \arctan \frac{1}{k+1} - \frac{\pi}{4} - \frac{1}{2} \ln 2 \newline \geq \; & \frac{1}{2}\ln \left[\left(k+1\right)^2 +1\right] + \left(k+1\right) \left(\frac{1}{k+1} - \frac{1}{3} \frac{1}{\left(k+1\right)^3}\right) - 1.132 \newline = \; & \frac{1}{2}\ln \left[\left(k+1\right)^2 +1\right] - 0.132 - \frac{1}{3} \frac{1}{\left(k+1\right)^2}, \end{align*} \] where in the third line we have retained the first two terms of series expansion for \(\arctan x\) for \(\left|x\right| \leq 1\). The integral approximation gives very messy terms and a somewhat loose lower bound. We can obtain a much neater one: \[ \begin{align} \sum_{j=1}^k \arctan \frac{1}{j} \geq \log \left(k+1\right). \end{align} \]

Before the proof, we may look at a plot to see how tight the lower bound is. Here it goes!

The trick of proof lies with series expansion.

**Proof**: It is true for \(k=1\) as \(\pi/4 > \log 2\). Now suppose the claim holds for \(k-1\), i.e., \(\sum_{j=1}^{k-1} \arctan\left(1/j\right) \geq \log\left(k\right)\), we need to show it holds for \(k\). It suffices to show \(\arctan(1/k) \geq \log\left(1+1/k\right)\). Now we consider the series expansions of \(\arctan\left(x\right)\) and \(\log\left(1+x\right)\): \[
\begin{align*}
\arctan\left(x\right)
& = x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \frac{1}{7}x^7 + \frac{1}{9}x^9 + \cdots, \forall \left|x\right|\leq 1, \newline
\log\left(x+1\right) & = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 + \cdots, \forall -1 < x \leq 1.
\end{align*}
\] So we have \[
\begin{align*}
&\arctan\left(x\right) - \log\left(1+x\right) \newline
= & \left(\frac{1}{2}x^2 + \frac{1}{4}x^4 + \frac{1}{6}x^6 + \cdots\right) - \left(\frac{2}{3}x^3 + \frac{2}{7}x^7 + \frac{2}{11}x^{11} + \cdots \right) \newline
= &\frac{2x^2}{3}\left(\frac{3}{4}- x\right) + \frac{2x^4}{7}\left(\frac{7}{8}-x^3\right) + \frac{2x^6}{11}\left(\frac{11}{12} - x^5\right) + \cdots \geq 0
\end{align*}
\] if $ 0 < x $, and \(1/k, \forall k > 1\) satisfies the condition. \(\Box\)

**Updated to the Proof**: There is a simpler way to see \(\arctan \left(1/k\right) \geq \log\left(1+1/k\right)\) (Thanks to Dai Liang! ). Basically it follows from the fact that \(\arctan x \geq \log \left(1+x\right), \forall x\leq 1\), where the latter can be observed as follows: \(\arctan 0 = \log \left(1+0\right)\) and \(\left(\arctan x\right)' \geq \left[\log \left(1+x\right)\right]'\) for \(x \leq 1\). (03/07/2012)

Recall that \(\arctan x \leq x\). Thus we have \[ \boxed{\sum_{j=1}^{k} \frac{1}{j} \geq \sum_{j=1}^{k} \arctan \frac{1}{j} \geq \log \left(k+1\right)} \] Acknowledgement: Dr. Tewodros Amdeberhan, at Math Department of Tulane University, has kindly provided the original proof and permitted me to share this on my blog. You may want to read their interesting paper on techniques of evaluating sum of arctangent: Sum of Arctangents and Some Formula of Ramanujan.

]]>Qualification Exam is ahead within two plus weeks. It’s the time to get refreshed my mind with the concepts of circuits, Fourier transformation, registers, and bit error probability, pretty everything a typical electrical engineering graduate needs to know. Yuh… another fighting! Of course the words your professor would often spend on this issue is > make sure you pass, but not spend too much time! It’s an interesting balance a graduate student most likely always tries to strike…

Visits to this blog keep roll up, even after my absence in the past few months. I’m pleased to be noticed of the 20,000 hits by the system. I’ve decided to put reasonable amount of time in future to update this blog with more technical expositions and discussions. This would considerably improve my ability to understand technical materials, which I would need to sharpen as soon as possible.

I was just having a Christmas dinner with John and one master student. It’s pretty fun! We consumed some 1.5 bottles of red wines. And John was asking one theorem for each bottle we had for next semester, jokingly …

]]>From the cyber-discussion being fired there, some interesting disagreements amongst these top vision researchers are already significant: conservatism versus radicalism, and pragmatism versus idealism. For a research community as ambitious and diverse and young as vision, nevertheless, consensus rarely occurs and ideological debates prevail – it is not surprising; in fact, this signals an active research field in my opinion. But my humble mind is really seeking some fuels the workshop could potentially generate for these topics:

- Object recognition
- Culture of scholarship in vision

The former has been central on the spot for the past 10 years with least success, while the latter has partially contributed to the dismal stagnation.

]]>PS：翻译纯属搞笑，不喜勿拍！

This paper is desperate. Please reject it completely and then block the author’s email ID so they can’t use the online system in future.

这文章太不给力了…请万勿发表。此外，建议锁定该作者的电子邮件ID，避免此人日后继续投稿。。。

The writing and data presentation are so bad that I had to leave work and go home early and then spend time to wonder what life is about.

写作水平和展示的数据无敌了，哥不得不提前下班，匆匆回家，然后花时间思考下人活着是为了虾米。。。

Reject – More holes than my grandad’s string vest!

拒发，必须的。本文的漏洞比我爷爷网眼背心上的洞还多！

I would suggest that EM set up a fund that pays for the red wine reviewers may need to digest manuscripts like this one.

哥建议贵刊（环境微生物学杂志）成立基金，以买单审稿人审阅时可能需要的红酒，哥上火呀。( Great! – Ju Sun)

The biggest problem with this manuscript, which has nearly sucked the will to live out of me, is the terrible writing style.

这篇文章问题多多，写作格式尤其可怕，简直摧残了哥身体里求生的意志。

Hopeless – Seems like they have been asleep and are not up on recent work on metagenomics.

哥真的不能淡定了。显然作者要么睡着了，要么完全没跟上宏基因组学的前沿发展。

A weak paper, poor experimental design, comparison of sequences using different primers, no statistical analysis possible, carelessly written, poorly thought through.

文章很弱。实验设计很锉，使用不同的引物序列比较；统计分析不置可否；写作粗犷。彻头彻尾的悲剧亚。。。

I agreed to review this Ms whilst answering e-mails in the golden glow of a balmy evening on the terrace of our holiday hotel on Lake Como. Back in the harsh light of reality in Belfast I realize that it’s just on the limit of my comfort zone and that it would probably have been better not to have volunteered.

迷人的傍晚，金色的夕照，如果此时哥在科莫湖的假日酒店那铺洒着余辉的露台上，我会欣然同意审阅这篇文章。然而，在贝尔法斯特残酷的日光里，我想刚才我那么想是被酒店啥的爽到了，文章么，咱能不提这事行吗。。。

The presentation is of a standard that I would reject from an undergraduate student.

哥从本科开始就不看这种操性的文章了。

The lack of negative controls… . results in the authors being lost in the funhouse. Unfortunately, I do not think they even realize this.

阴性对照实验的缺乏导致作者在游乐场里完全迷失了，不幸的是，我看他们现在还没意识到。

References

**The truth about scientific peer review**

http://www.sciscoop.com/the-truth-about-scientific-peer-review.html

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