I have a function f(x) = 1/(x + a+ b*I*sign(x)) and I want to calculate the
integral of
dx dy dz f(x) f(y) f(z) f(x+y+z) f(x-y - z)
over the entire R^3 (b>0 and a,- b are of order unity). This is just a representative example -- in practice I have n<7 variables and 2n-1 instances of f(), n of them involving the n integration variables and n-1 of them involving some linear combintation of the integration variables. At this stage I'm only interested in a rough estimate with relative error of 1e-3 or so.
I have tried the following libraries :
Steven Johnson's cubature code: the hcubature algorithm works but is abysmally slow, taking hundreds of millions of integrand evaluations for even n=2.
HintLib: I tried adaptive integration with a Genz-Malik rule, the cubature routines, VEGAS and MISER with the Mersenne twister RNG. For n=3 only the first seems to be somewhat viable option but it again takes hundreds of millions of integrand evaluations for n=3 and relerr = 1e-2, which is not encouraging.
For the region of integration I have tried both approaches: Integrating over [-200, 200]^n (i.e. a region so large that it essentially captures most of the integral) and the substitution x = sinh(t) which seems to be a standard trick.
I do not have much experience with numerical analysis but presumably the difficulty lies in the discontinuities from the sign() term. For n=2 and f(x)f(y)f(x-y) there are discontinuities along x=0, y=0, x=y. These create a very sharp peak around the origin (with a different sign in the various quadrants) and sort of 'ridges' at x=0,y=0,x=y along which the integrand is large in absolute value and changes sign as you cross them. So at least I know which regions are important. I was thinking that maybe I could do Monte Carlo but somehow "tell" the algorithm in advance where to focus. But I'm not quite sure how to do that.
I would be very grateful if you had any advice on how to evaluate the integral with a reasonable amount of computing power or how to make my Monte Carlo "idea" work. I've been stuck on this for a while so any input would be welcome. Thanks in advance.
One thing you can do is to use a guiding function for your Monte Carlo integration: given an integral (am writing it in 1D for simplicity) of ∫ f(x) dx, write it as ∫ f(x)/g(x) g(x) dx, and use g(x) as a distribution from which you sample x.
Since g(x) is arbitrary, construct it such that (1) it has peaks where you expect them to be in f(x), and (2) such that you can sample x from g(x) (e.g., a gaussian, or 1/(1+x^2)).
Alternatively, you can use a Metropolis-type Markov chain MC. It will find the relevant regions of the integrand (almost) by itself.
Here are a couple of trivial examples.
Related
I am trying to implement nn.MultiheadAttention in my network. According to the docs,
embed_dim – total dimension of the model.
However, according to the source file,
embed_dim must be divisible by num_heads
and
self.q_proj_weight = Parameter(torch.Tensor(embed_dim, embed_dim))
If I understand properly, this means each head takes only a part of features of each query, as the matrix is quadratic. Is it a bug of realization or is my understanding wrong?
Each head uses a different part of the projected query vector. You can imagine it as if the query gets split into num_heads vectors that are independently used to compute the scaled dot-product attention. So, each head operates on a different linear combination of the features in queries (and keys and values, too). This linear projection is done using the self.q_proj_weight matrix and the projected queries are passed to F.multi_head_attention_forward function.
In F.multi_head_attention_forward, it is implemented by reshaping and transposing the query vector, so that the independent attentions for individual heads can be computed efficiently by matrix multiplication.
The attention head sizes are a design decision of PyTorch. In theory, you could have a different head size, so the projection matrix would have a shape of embedding_dim × num_heads * head_dims. Some implementations of transformers (such as C++-based Marian for machine translation, or Huggingface's Transformers) allow that.
So, I have a vector that corresponds to a given feature (same dimensionality). Is there a package in Julia that would provide a mathematical function that fits these data points, in relation to the original feature? In other words, I have x and y (both vectors) and need to find a decent mapping between the two, even if it's a highly complex one. The output of this process should be a symbolic formula that connects x and y, e.g. (:x)^3 + log(:x) - 4.2454. It's fine if it's just a polynomial approximation.
I imagine this is a walk in the park if you employ Genetic Programming, but I'd rather opt for a simpler (and faster) approach, if it's available. Thanks
Turns out the Polynomials.jl package includes the function polyfit which does Lagrange interpolation. A usage example would go:
using Polynomials # install with Pkg.add("Polynomials")
x = [1,2,3] # demo x
y = [10,12,4] # demo y
polyfit(x,y)
The last line returns:
Poly(-2.0 + 17.0x - 5.0x^2)`
which evaluates to the correct values.
The polyfit function accepts a maximal degree for the output polynomial, but defaults to using the length of the input vectors x and y minus 1. This is the same degree as the polynomial from the Lagrange formula, and since polynomials of such degree agree on the inputs only if they are identical (this is a basic theorem) - it can be certain this is the same Lagrange polynomial and in fact the only one of such a degree to have this property.
Thanks to the developers of Polynomial.jl for leaving me just to google my way to an Answer.
Take a look to MARS regression. Multi adaptive regression splines.
I am attempting to fit a circle to some data. This requires numerically solving a set of three non-linear simultaneous equations (see the Full Least Squares Method of this document).
To me it seems that the NEWTON function provided by IDL is fit for solving this problem. NEWTON requires the name of a function that will compute the values of the equation system for particular values of the independent variables:
FUNCTION newtfunction,X
RETURN, [Some function of X, Some other function of X]
END
While this works fine, it requires that all parameters of the equation system (in this case the set of data points) is hard coded in the newtfunction. This is fine if there is only one data set to solve for, however I have many thousands of data sets, and defining a new function for each by hand is not an option.
Is there a way around this? Is it possible to define functions programmatically in IDL, or even just pass in the data set in some other manner?
I am not an expert on this matter, but if I were to solve this problem I would do the following. Instead of solving a system of 3 non-linear equations to find the three unknowns (i.e. xc, yc and r), I would use an optimization routine to converge to a solution by starting with an initial guess. For this steepest descent, conjugate gradient, or any other multivariate optimization method can be used.
I just quickly derived the least square equation for your problem as (please check before use):
F = (sum_{i=1}^{N} (xc^2 - 2 xi xc + xi^2 + yc^2 - 2 yi yc + yi^2 - r^2)^2)
Calculating the gradient for this function is fairly easy, since it is just a summation, and therefore writing a steepest descent code would be trivial, to calculate xc, yc and r.
I hope it helps.
It's usual to use a COMMON block in these types of functions to pass in other parameters, cached values, etc. that are not part of the calling signature of the numeric routine.
I recently wanted to use a simple CUDA matrix-vector multiplication. I found a proper function in cublas library: cublas<<>>gbmv. Here is the official documentation
But it is actually very poor, so I didn't manage to understand what the kl and ku parameters mean. Moreover, I have no idea what stride is (it must also be provided).
There is a brief explanation of these parameters (Page 37), but it looks like I need to know something else.
A search on the internet doesn't provide tons of useful information on this question, mostly references to different version of documentation.
So I have several questions to GPU/CUDA/cublas gurus:
How do I find more understandable docs or guides about using cublas?
If you know how to use this very function, couldn't you explain me how do I use it?
Maybe cublas library is somewhat extraordinary and everyone uses something more popular, better documented and so on?
Thanks a lot.
So BLAS (Basic Linear Algebra Subprograms) generally is an API to, as the name says, basic linear algebra routines. It includes vector-vector operations (level 1 blas routines), matrix-vector operations (level 2) and matrix-matrix operations (level 3). There is a "reference" BLAS available that implements everything correctly, but most of the time you'd use an optimized implementation for your architecture. cuBLAS is an implementation for CUDA.
The BLAS API was so successful as an API that describes the basic operations that it's become very widely adopted. However, (a) the names are incredibly cryptic because of architectural limitations of the day (this was 1979, and the API was defined using names of 8 characters or less to ensure it could widely compile), and (b) it is successful because it's quite general, and so even the simplest function calls require a lot of extraneous arguments.
Because it's so widespread, it's often assumed that if you're doing numerical linear algebra, you already know the general gist of the API, so implementation manuals often leave out important details, and I think that's what you're running into.
The Level 2 and 3 routines generally have function names of the form TMMOO.. where T is the numerical type of the matrix/vector (S/D for single/double precision real, C/Z for single/double precision complex), MM is the matrix type (GE for general - eg, just a dense matrix you can't say anything else about; GB for a general banded matrix, SY for symmetric matrices, etc), and OO is the operation.
This all seems slightly ridiculous now, but it worked and works relatively well -- you quickly learn to scan these for familiar operations so that SGEMV is a single-precision general-matrix times vector multiplication (which is probably what you want, not SGBMV), DGEMM is double-precision matrix-matrix multiply, etc. But it does take some practice.
So if you look at the cublas sgemv instructions, or in the documentation of the original, you can step through the argument list. First, the basic operation is
This function performs the matrix-vector multiplication
y = a op(A)x + b y
where A is a m x n matrix stored in column-major format, x and y
are vectors, and and are scalars.
where op(A) can be A, AT, or AH. So if you just want y = Ax, as is the common case, then a = 1, b = 0. and transa == CUBLAS_OP_N.
incx is the stride between different elements in x; there's lots of situations where this would come in handy, but if x is just a simple 1d array containing the vector, then the stride would be 1.
And that's about all you need for SGEMV.
I want to invert a 4x4 matrix. My numbers are stored in fixed-point format (1.15.16 to be exact).
With floating-point arithmetic I usually just build the adjoint matrix and divide by the determinant (e.g. brute force the solution). That worked for me so far, but when dealing with fixed point numbers I get an unacceptable precision loss due to all of the multiplications used.
Note: In fixed point arithmetic I always throw away some of the least significant bits of immediate results.
So - What's the most numerical stable way to invert a matrix? I don't mind much about the performance, but simply going to floating-point would be to slow on my target architecture.
Meta-answer: Is it really a general 4x4 matrix? If your matrix has a special form, then there are direct formulas for inverting that would be fast and keep your operation count down.
For example, if it's a standard homogenous coordinate transform from graphics, like:
[ux vx wx tx]
[uy vy wy ty]
[uz vz wz tz]
[ 0 0 0 1]
(assuming a composition of rotation, scale, translation matrices)
then there's an easily-derivable direct formula, which is
[ux uy uz -dot(u,t)]
[vx vy vz -dot(v,t)]
[wx wy wz -dot(w,t)]
[ 0 0 0 1 ]
(ASCII matrices stolen from the linked page.)
You probably can't beat that for loss of precision in fixed point.
If your matrix comes from some domain where you know it has more structure, then there's likely to be an easy answer.
I think the answer to this depends on the exact form of the matrix. A standard decomposition method (LU, QR, Cholesky etc.) with pivoting (an essential) is fairly good on fixed point, especially for a small 4x4 matrix. See the book 'Numerical Recipes' by Press et al. for a description of these methods.
This paper gives some useful algorithms, but is behind a paywall unfortunately. They recommend a (pivoted) Cholesky decomposition with some additional features too complicated to list here.
I'd like to second the question Jason S raised: are you certain that you need to invert your matrix? This is almost never necessary. Not only that, it is often a bad idea. If you need to solve Ax = b, it is more numerically stable to solve the system directly than to multiply b by A inverse.
Even if you have to solve Ax = b over and over for many values of b, it's still not a good idea to invert A. You can factor A (say LU factorization or Cholesky factorization) and save the factors so you're not redoing that work every time, but you'd still solve the system each time using the factorization.
You might consider doubling to 1.31 before doing your normal algorithm. It'll double the number of multiplications, but you're doing a matrix invert and anything you do is going to be pretty tied to the multiplier in your processor.
For anyone interested in finding the equations for a 4x4 invert, you can use a symbolic math package to resolve them for you. The TI-89 will do it even, although it'll take several minutes.
If you give us an idea of what the matrix invert does for you, and how it fits in with the rest of your processing we might be able to suggest alternatives.
-Adam
Let me ask a different question: do you definitely need to invert the matrix (call it M), or do you need to use the matrix inverse to solve other equations? (e.g. Mx = b for known M, b) Often there are other ways to do this w/o explicitly needing to calculate the inverse. Or if the matrix M is a function of time & it changes slowly then you could calculate the full inverse once, & there are iterative ways to update it.
If the matrix represents an affine transformation (many times this is the case with 4x4 matrices so long as you don't introduce a scaling component) the inverse is simply the transpose of the upper 3x3 rotation part with the last column negated. Obviously if you require a generalized solution then looking into Gaussian elimination is probably the easiest.