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13

A more structured way of providing a basis or quadrature (which may replace MC in many cases) in multiple dimensions is that of sparse grids, which combines some family of one dimensional rules of varying order in such a way as to have merely exponential growth in dimension, $2^d$, rather than having it be that dimension is an exponent of the resolution $N^d$...

11

First of all, you should specify whether you want all components or the most significant ones? Denote your matrix $A \in \mathbb{R}^{N \times M}$ with $N$ being number of samples and $M$ dimensionality. In case you want all components the classical way to go is to compute covariance matrix $C \in \mathbb{R}^{M\times M}$ (which has time complexity of $O(NM^... 8 First, recognize than the functions you are working with must have some decay for any "efficient" method to be possible. If your functions have rich structure in all modes of the full tensor product, then you can't do better. But many functions of interest have decay in their mixed derivatives, in which case efficient methods are available. The term "... 6 As a general rule, it's easy to understand why regular grids can't go much beyond 3 or 4-dimensional problems: in d dimensions, if you want to have a minimum of N points per coordinate direction, you'll get N^d points overall. Even for relatively nice functions in 1d, you need at least N=10 grid points to resolve them at all, so the overall number of points ... 6 The solutions for the equation are in $$\psi \in \mathbb{C}^{3M}\times\mathbb{R}^+ \enspace .$$ If the number of electrons is small enough you can just use any traditional method. Like a domain discretization method (Finite Difference, Finite Element, Boundary Element), or a pseudospectral method. Since solving this equation is not more difficult than ... 5 As you suggest, the binary representation works because the binary numbers with$d$digits can be thought of as vectors representing the coordinates of the vertices of the$d$-dimensional hypercube. Now notice that the coordinates of an edge midpoint can be obtained by averaging the coordinates of two (adjacent) vertices and the coordinates of a face ... 5 If you know the Gram matrix$G_{ik}=(x_i-x_0,x_k-x_0)$of inner products with respect to the origin$x_0$, you can get the shifted Gram matrix$G_{ik}'=(x_i-x,x_k-x)$with respect to any affine combination$x=x_0+\sum a_j (x_j-x_0)$by expanding the inner products using$x_i-x=x_i-x_0-\sum a_j (x_j-x_0)$and expressing the result in terms of the$G_{ik}$... 5 This feature seems to be available in CGAL 5 Take a look at active subspaces, e.g., Active Subspace Methods in Theory and Practice: http://epubs.siam.org/doi/abs/10.1137/130916138 And a PDF here: http://inside.mines.edu/~pconstan/docs/constantine-asm.pdf I have a SIAM book (Active Subspaces: Emerging Ideas for Dimension Reduction in Parameter Studies) coming out in March. Suppose$f$maps$\mathbb{...

4

This depends on how many dimensions you have an how many points. Also on how the points are structured. If the number of dimensions are high and the points are randomly chosen, you essentially have a Monte Carlo integration procedure. If the points are randomly chosen from a distribution, then you have a kind of importance sampling. If the number of points ...

4

I think you can do this using convex hull software (e.g. QHull) via the lifting algorithm. At least, the documentation of matlab's "delaunayn" command seems to indicate as much.

4

I don't have time to get all the details down, but maybe this answer can give some helpful intuition. Basically, something that will work is taking the first $N$ binary binary of your number in $[0,1]$ and assigning them as the most significant digits for each of the $N$ dimensions (so, the first decimal digit is the most significant digit in the first ...

4

Suppose that you know the orthogonal polynomial basis in a single dimension $(x)$ of each degree $i$ up to some desired order $K$. That is, we know $$p_0(x),p_1(x),...p_i(x),...,p_K(x)$$ To extend this into a two dimensions $(x,y)$, we need only consider the product between 1D polynomials in (x) and (y) and collect only the products whose total degree is ...

3

That all depends. Not having read the psychology literature at all, the minimum you probably need is 2-3 semesters of calculus and 1 semester of differential equations, and that's to understand something on the level of Nonlinear Dynamics And Chaos: With Applications To Physics, Biology, Chemistry, And Engineering (Studies in Nonlinearity) by Steven ...

3

Your proposal will give meaningless results. This can be seen already with simple examples consisting of two points only. Indeed, the matrix that you want to decompose is not even positive semidefinite as it is nonzero but has zero diagonal entries. Note that you can't cancel $J$ since it is singular. In general, $Ju=Jv$ implies $u=v$ if and only if $J$ ...

3

If the dimension is not too high, you can use a triangulation library (e.g. CGAL) to decompose the polytope into n-dimensional simplices; calculating the volume of a simplex has a simple formula.

3

In general, the principal components of $X$ and $X'$ with rows shifted by a fixed vector will not have any simple relationship. In applications it may therefore be meaningful (or even important) to shift all rows of $X$ by subtraction their mean before doing the prinicpal component analysis. This makes the result independent of the placement of the ...

3

In order to answer your question, I need to clarify a few things, especially because I've noticed that in at least three of your previous questions, the word "projection" tends to be used very loosely. Projection is a technical term, and as such, it's vitally important that it have a precise meaning, so here it is: Definition: A projection matrix $P$ is a ...

3

From suitable linear combinations of the entries of the distance matrix (the matrix of squared distances) one can construct the Gram matrix of the direction vectors from one point (taken as zero) to the others. To be embeddable into a Euclidean space of dimension $r$, this matrix must be positive semidefinite of rank $r$; then the Cholesky factor provides an ...

3

EDIT: Rewritten after clearing up some notational confusion in the comments. Let $X = \left[x_1, x_2, \ldots x_r\right]$ and $B = \left[b_1, b_2, \ldots b_r\right]$ where $r$ is the numbers of vectors. Your problem can then be written as $$\text{minimize}_{A,X\geq 0} ||AX-B||$$ This looks very much like a version of the non-negative matrix ...

3

You can transform this into a standard form quadratic program by expanding the square term, then distributing the sum (assuming your set B is finite). You can then use one of the standard solvers (see the links at the bottom of the wikipedia page for several options). If you are looking for more specific advice, I'd suggest adding more detail about your ...

3

PCA is called this way since it picks the principal components. If you happen to have several components with the same or almost the same eigenvalue and you pick one but not the other, then you can't claim that you picked the principal components. You picked a subset of the principal components. In other words, if your second eigenvalue is doubled, then it's ...

3

In addition to labeling all edges and faces there are also other questions I have found important in the same context: Numbering them in some particular way. deal.II uses a numbering where we first consider all edges that are parallel to the x-axis and sort them by their y,z-values (either 0 or 1) in lexicographic order, then all edges parallel to the y-...

3

I'd calculate the mean $\overline x$ and the covariance matrix $C$ of the joint data set, and then do a K/S test on the univariate quantity $V(x):=(x-\overline x)^TC^{-1}(x-\overline x)$ evaluated on the parts. If the K/S test give a significant difference between the parts, there is one. If it gives no significant difference, the test is to be regarded as ...

3

ROOT supports Kolmogorov tests on higher dimensional histograms, and the notes (for the 2D version) suggest that there is a ambiguity--which they deal with by punting: calculate it both ways. I don't know if the code contains anymore details, but the comments sometimes have references to papers and the like. There are some additional interesting comments in ...

3

You want to solve for 3 to 10 particle systems (3D per particle)? As far as I am aware, mean field theories do not work especially well for so few particles, but it seems there has been DFT work on diatomic molecules. Is this a system where Born-Oppenheimer is valid? If so, I might be inclined to expand the electronic wavefunction using a linear ...

2

Monte Carlo has a very attractive property: it has the same convergence rate independent of the dimension of the problem. So, even for problems of dimension $d=4,...,100$, they will converge at the same rate as problems with dimension $d=100,101,...\infty$. So, even as your dimension increases from 4 to 100, monte carlo will probably be faster anyways.

2

If the largest eigenvalue has multiplicity, the PCA solution really depends on your choice of the eigenvectors as 'the first two dominant'. Different choice will give you different representation of the original high-dimensional data, but they all will have the same maximal variance.

2

I guess you only need a few (or a few hundred) dominant singular value/vector pairs. Then it is best to use an iterative method, which will be much faster and consume far less memory. In Matlab, see help svds

2

For $d>1$, you need to determine all distances of a simplex consisting of $d+1$ points, and then the distances from each other point to these points. ($d$ points and distances already reduce the possibilities to a finite number, but then one needs additional disambiguation information). This makes a total of $d(d+1)/2+(n-d-1)(d+1) = n(d+1)-(d+2)(d+1)/2$ ...

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