十年磨一剑

该方法不用求梯度, 面向于每个未知量有边界核心算法对于一维来说就是找中心,分三段, 选最小的那一段继续迭代其中找最小那一段就是step2的选择过程关于多维参考:An Implementation of the DIRECT algorithm同样找最小点也就是相当于找最右下角的区域, 这样才能找条直线位于所有点之下....

note: this passage serves for the analysis of Alec Jacobson’s thesis1.what’s stiffness matrixAccording to (2.16), stiffness matrix is given by:Lij=∫Ω−∇ϕi∇ϕjdAL_{ij}=\int\limits_{\Omega}-\nabla\phi_i\na

Here, I would like to derive the formula (10) For triangle vjvivmv_jv_iv_m, we have the following three equation which is related to viv_iwij∥(v′i−v′j)−Ri(vi−vj)∥2w_{ij}\|(v'_i-v'_j)-R_i(v_i-v_j)\|^

先看高数曲面积分的定义 推导下如何将曲面上的面元投影到xOy平面上, 曲面z−z(x,y)=0z-z(x,y)=0 的法向量为(−zx,−zy,1)(-z_x, -z_y, 1), 单位化后与xOy平面的法向量(0,0,1)(0,0,1)相乘得到cosθ=1/1+z2x+z2y−−−−−−−−−√cos \theta = 1 / \sqrt{1+z_x^2 + z_y^2}dxdy=cosθdSd

Hadamard product Main article: Hadamard product (matrices) For two matrices of the same dimensions, there is the Hadamard product, also known as the element-wise product, pointwise product, entrywise

0-norm 非零元素的个数应用,比如要想找出不重叠的数据, 那么表示数据的向量应该尽量正交, 而正交呢就是向量相乘应该为0, 也就是让非零元素的个数尽量少, 所以可以min 0-norm去设计优化函数, 0-norm一般不能解,通常会用1-norm去解1-norm是元素绝对值相加, 以二维来讲所以优化的时候会优化到棱形的四个点上, 所以1-norm的优化结果就是大的有值,  小的没值, 应用: ...

假设XX为未知矩阵,那么对于矩阵XX的积分 ∫XdX=12∥X∥2F\int X dX = \frac{1}{2}\|X\|_F^2另外一种方式理解对于XX中的每个元素xijx_{ij}, 对其积分∫xijdxij=12x2ij\int x_{ij}dx_{ij} = \frac{1}{2}x_{ij}^2 所以 ∫XdX=∑i∑j∫xijdxij=12∑i∑jx2ij=12∑i∑jxij∗

先来复习下高数里面的第二类积分变换 以前对conformal mapping的理解都是盲人摸象, 片面局部的. 因为看的资料都是从不同的角度对解释它的. 随着阅历的丰富和理解的深入, 现已能将原来零星的知识点串连起来并找出其中的关系, 于是开篇博客讲讲自己对conformal mapping的学习历程以及深化后的理解, 好了, 开始正题.以前看的资料呢对conformal mapping有

注意购造incident matrix时, 边上点索引小的为-1,大的为1这里的L没有标准化处理,要标准化处理的话不需要对称的话,则可以表示为laplacian coordinates

Min12xTAx+xTb+cMin \frac{1}{2}x^TAx+x^Tb+c Equals to solve: 12(A+AT)x=−b\frac{1}{2}(A+A^T)x = -bMin12xTAx+xTb+cMin\frac{1}{2}x^TAx+x^Tb+c subject to: A′x=b′A'x=b' Equals to solve: ⎡⎣12(A+AT)A′A′

{ 推导：7式 (3) vec(X)TKvec(X)vec(X)^TKvec(X) (6) K=diag(vec(Kp))+(G2⊗G1)diag(vec(Kq))(H2⊗H1)TK = diag(vec(K_p))+(G_2\otimes G_1)diag(vec(K_q))(H_2\otimes H_1)^T 将(6)式代入(3)式得到: \begin{align} &vec(X)^

average diffusion distance(ADD) 定义见: [1]说明见: [2]主要代码:% test for farthest point sampling on 3D meshes%%%%%%%%%%%%%% sphere %%%%%%%%%%%%%%%%% name = 'sphere40.o

公式为:cotC=a2+b2−c24A\cot C = \frac{a^2 + b^2 - c^2}{4A} 证明 根据余弦定理cosC=a2+b2−c22ab\cos C = \frac{a^2 + b^2 - c^2}{2ab} 根据面积公式 A=12absinCA = \frac{1}{2}ab \sin C 所以 sinC=2Aab\sin C = \frac{2A}{ab}所以co

laplace-beltrami operator 是 laplace operator在riemann manifolds上的拓展. 至于他们两到底有什么区别和联系,请看[2]中的一段话: 可以看到, laplace-beltrami operator与普通laplace operator的区别就在于一个metric G,那么这个metric G究竟是个什么东西?注意这里的局部坐标系底面的两个

meshlab只基于vcglib写的, 最近需要检测mesh的自相交, 于是来看看它的源代码 static bool SelfIntersections(MeshType &m, std::vector<FaceType*> &ret) { RequirePerFaceMark(m); ret.clear(); int r

参照文献https://pdfs.semanticscholar.org/0141/b1416bb749bb5ba94210a30d70f0824760a4.pdf这里不同的是, 这里的法向量不是三角形的法向量, 而是给定点上的法向量.

assume the size of block is 2values = [1 0 2 1 6 7 8 2 1 4 5 1 4 3 0 0 7 2 0 0]columns = [0 1 1 1 2]rowIndex = [0 2 3 5]rowIndex[i] records the offset (in block) for each block-...

先从alec jacobson代码入手, 这里求得的G是一个#Face*dim by #V, 所以他求的结果是以按面来算的,然后dim分别是x,y,z上的分量.关于梯度求法的理论请看 http://blog.youkuaiyun.com/seamanj/article/details/52070611 只是说这里把e1h1e1h1\frac{e_1}{h_1}根据高线性质(具体看http://blog...

Some time ago, I quoted a passage from a website which is able to give a general idea about the Laplacian interpolation. The address is http://blog.youkuaiyun.com/seamanj/article/details/50701112Here, I wan

Relate this question to one of our blogs（http://blog.youkuaiyun.com/seamanj/article/details/50727219）here:

对于二项式的计算[u1u2u3]⎡⎣⎢a11a21a31a12a22a32a13a23a33⎤⎦⎥⎡⎣⎢v1v2v3⎤⎦⎥\begin{bmatrix}u_1&u_2&u_3\end{bmatrix}\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\begin{bm

Tutorial comes from https://www.youtube.com/watch?v=jFe6dMnpQho 0-forms : f:R3→Rf : \mathbb R^3 \rightarrow \mathbb R 1-forms : α=adx+bdy+cdz=w<a,b,c> \alpha =adx + bdy + cdz = w_{<a,b,c>} 2-forms

写这篇博客主要来源ARAP paper中的一段话 那么为什么第一基本范式和第二基本范式可以描述surface的stretching和bending. 设在平面uov中, 向量(1,0)和(0,1)经过f和f’的映射后分别变成

与fast marching on 3D meshes with euclidian distance 不同(http://blog.youkuaiyun.com/seamanj/article/details/52077197), 基于欧氏距离是在欧氏空间算geodesic的距离, 而基于diffusion distance,在是diffusion space里面采用类似geodesic的算法, 只不过这里

与基于geodesic distance的farthest points samping不同(http://blog.youkuaiyun.com/seamanj/article/details/52099358)下一次是采样点的选取gedesic distance最远的点最终结果为:注意跟以geodesic距离为基础的最远采样方法的区别matlab源代码

3D mesh的fast marching 跟2D图片的fast marching类似. 2D图片是规则的平面网格,点的ux,uyu_x, u_y是通过上或下,左或右(具体哪个,是通过距离小的点去确定),具体请参考http://blog.youkuaiyun.com/seamanj/article/details/51991067 而3D mesh上点的ux,uyu_x,u_y是通过与它位于同一个三角形的其

这个问题要追溯到,我看normal curvature的时候, 里面涉及到了一个叫做Shaper Operator, 在wolfram mathworld里面的定义为:而normal vurvature在里面的定义为:按道理讲,curvature应该是个标量, 所以这个的Shape operator的结果应该为一向量,然而按照它的定义, 最开始我理解成向量N在v方向上的方向

prove det(I+uvt)=1+utvdet(I+uv^t) = 1 + u^tv First, let us list several basic formulations: Assume Aa=λaAa = \lambda a, then we have A−1a=1λaA^{-1}a = \frac{1}{\lambda}a, meanwhile, we can also get (

矩阵分解是指根据一定的原理用某种算法将一个矩阵分解成若干个矩阵的乘积。常见的矩阵分解有LU分解（三角分解）、QR分解（正交变换）、Cholesky分解，以及Schur分解、Hessenberg分解、奇异分解等。 (1) LU分解（三角分解）矩阵的LU分解就是将一个矩阵表示为一个交换下三角矩阵和一个上三角矩阵的乘积形式。线性代数中已经证明，只要方阵A是非奇异（即行列式不等

More details please refer to https://www.youtube.com/watch?v=08vJyA-XD3Q

哈密顿算符：▽≡i*d/dx+j*d/dy+k*d/dz 运算规则: 一、▽A=(i*d/dx+j*d/dy+k*d/dz)A=i*dA/dx+j*dA/dy+k*dA/dz 这样标量场A通过▽的这个运算就形成了一个矢量场，该矢量场反应了标量场A的分布。二、 ▽·A=(i*d/dx+j*d/dy+k*d/dz)·(Ax*i+Ay*j+Az*k)=dAx/dx+dAy/dy+dA

Consider a (two dimensional) data matrix with some values missing and you want to fill the holes by interpolated values. One particularly simple but at the same time powerful method is called Laplace

let's look at a question first.how can we derive (2.10) from (2.9)?Give the definition of integration by part in high dimension from wiki first.{[from https://en.wikipedia.org/wiki/Integra

引用于 https://en.wikipedia.org/wiki/Second_partial_derivative_test#Functions_of_many_variablesIn mathematics, the second partial derivative test is a method in multivariable calculus used to determine i

∂xTa∂x=∂aTx∂x=a\frac{\partial x^Ta}{\partial x} = \frac{\partial a^Tx}{\partial x} = a (引用于matrix cookbook 2.4.1) (应用于 Lesson 4 Part 2 Softmax Regression)∇fTg=fT∇g+gT∇f\nabla f^Tg = f^T\nabla g+g^T\n

引用于Matrix Differential Calculus with Applications in Statistics and Econometrics, 99

There are six kinds of derivatives that can be expressed as matrices:The partials with respect to the numerator are laid out according to the shape ofY while the partials with respect

the original tutor is like:there are some points I wanna mention:1. the discrete Gaussian curvature formula it listed is wrong. How? listen to me explain.According to the paper "Disc

this material is from http://math.arizona.edu/~cais/223Page/hout/236w06fields.pdf

Fast Marching method 跟 dijkstra 方法类似,只不过dijkstra方法的路径只能沿网格,而Fast Marching method的方法可以沿斜线. [Level Set Methods and Fast Marching Methods p94-95 ]这里uu理解为到达点的时间, FijkF_{ijk}理解为在点ijkijk的流速. 然后就可以跟Bound