I'm writing a program to do the Newton Raphson Method for n variable (System of equation) using Datagridview. My problem is to determine the inverse for Jacobian Matrix. I've search in internet to find a solution but a real couldn't get it until now so if someone can help me I will real appreciate. Thanks in advance.
If you are asking for a recommendation of a library, that is explicitly off topic in Stack Overflow. However below I mention some algorithms that are commonly used; this may help you to find, or write, what you need. I would, though, not recommend writing something, unless you really want to, as it can be tricky to get these algorithms right. If you do decide to write something I'd recommend the QR method, as the easiest to write, though the theory is a little subtle.
First off do you really need to compute the inverse? If, for example, what you need to do is to compute
x = inv(J)*y
then it's faster and more accurate to treat this problem as
solve J*x = y for x
The methods below all factor J into other matrices, for which this solution can be done. A good package that implements the factorisation will also have the code to perform the solution.
If you do really really need the inverse often the best way is to solve, one column at a time
J*K = I for K, where I is the identity matrix
LU decomposition
This may well be the fastest of the algorithms described here but is also the least accurate. An important point is that the algorithm must include (partial) pivoting, or it will not work on all invertible matrices, for example it will fail on a rotation through 90 degrees.
What you get is a factorisation of J into:
J = P*L*U
where P is a permutation matrix,
L lower triangular,
U upper triangular
So having factorised, to solve for x we do three steps, each straightforward, and each can be done in place (ie all the x's can be the same variable)
Solve P*x1 = y for x1
Solve L*x2 = x1 for x2
Solve U*x = x2 for x
QR decomposition
This may be somewhat slower than LU but is more accurate. Conceptually this factorises J into
J = Q*R
Where Q is orthogonal and R upper triangular. However as it is usually implemented you in fact pass y as well as J to the routine, and it returns R (in J) and Q'*y (in the passed y), so to solve for x you just need to solve
R*x = y
which, given that R is upper triangular, is easy.
SVD (Singular value decomposition)
This is the most accurate, but also the slowest. Moreover unlike the others you can make progress even if J is singular (you can compute the 'generalised inverse' applied to y).
I recommend reading up on this, but advise against implementing it yourself.
Briefly you factorise J as
J = U*S*V'
where U and V are orthogonal and S diagonal.
There are, of course, many other ways of solving this problem. For example if your matrices are very large (dimension in the thousands) an it may, particularly if they are sparse (lots of zeroes), be faster to use an iterative method.
Related
When we need to optimize a function on the positive real half-line, and we only have non-constraints optimization routines, we use y = exp(x), or y = x^2 to map to the real line and still optimize on the log or the (signed) square root of the variable.
Can we do something similar for linear constraints, of the form Ax = b where, for x a d-dimensional vector, A is a (N,n)-shaped matrix and b is a vector of length N, defining the constraints ?
While, as Ervin Kalvelaglan says this is not always a good idea, here is one way to do it.
Suppose we take the SVD of A, getting
A = U*S*V'
where if A is n x m
U is nxn orthogonal,
S is nxm, zero off the main diagonal,
V is mxm orthogonal
Computing the SVD is not a trivial computation.
We first zero out the elements of S which we think are non-zero just due to noise -- which can be a slightly delicate thing to do.
Then we can find one solution x~ to
A*x = b
as
x~ = V*pinv(S)*U'*b
(where pinv(S) is the pseudo inverse of S, ie replace the non zero elements of the diagonal by their multiplicative inverses)
Note that x~ is a least squares solution to the constraints, so we need to check that it is close enough to being a real solution, ie that Ax~ is close enough to b -- another somewhat delicate thing. If x~ doesn't satisfy the constraints closely enough you should give up: if the constraints have no solution neither does the optimisation.
Any other solution to the constraints can be written
x = x~ + sum c[i]*V[i]
where the V[i] are the columns of V corresponding to entries of S that are (now) zero. Here the c[i] are arbitrary constants. So we can change variables to using the c[] in the optimisation, and the constraints will be automatically satisfied. However this change of variables could be somewhat irksome!
When I reading this paper http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1976ApJ...209..214B&data_type=PDF_HIGH&whole_paper=YES&type=PRINTER&filetype=.pdf
I try to solve eq(49) numerically, it seems a fokker-planck equation, I find finite difference method doesn't work, it's unstable.
Does any one know how to solve it?
computational science stack exchange is where you can ask and hope for an answer. Or you could try its physics cousin. The equation, you quote, is integro-differential equation, fairly non-linear... Fokker-Plank looking equation. Definitely not the typical Fokker-Plank.
What you can try is to discretize the space part of the function g(x,t) using finite differences or finite-elements. After all, 0 < x < x_max and you have boundary conditions. You also have to discretize the corresponding integration. So maybe finite elements might be more appropriate? Finite elements means you can write g(x, t) as a series of a well chosen basis of compactly supported simple enough functions Bj(x) : j = 1...N in the interval [0, x_max]
g(x,t) = sum_j=1:N gj(t)*Bj(x)
That will turn your function into a (large) vector gj(t) = g(x_j, t), for j = 1, 1, ...., N. As a result, you will obtain a non-linear system of ODEs
dgj(t)/dt = Qj(g1(t), g2(t), ..., gN(t))
j = 1 ... N
After that use something like Runge-Kutta to integrate numerically the ODE system.
I have implemented an algorithm that uses two other algorithms for calculating the shortest path in a graph: Dijkstra and Bellman-Ford. Based on the time complexity of the these algorithms, I can calculate the running time of my implementation, which is easy giving the code.
Now, I want to experimentally verify my calculation. Specifically, I want to plot the running time as a function of the size of the input (I am following the method described here). The problem is that I have two parameters - number of edges and number of vertices.
I have tried to fix one parameter and change the other, but this approach results in two plots - one for varying number of edges and the other for varying number of vertices.
This leads me to my question - how can I determine the order of growth based on two plots? In general, how can one experimentally determine the running time complexity of an algorithm that has more than one parameter?
It's very difficult in general.
The usual way you would experimentally gauge the running time in the single variable case is, insert a counter that increments when your data structure does a fundamental (putatively O(1)) operation, then take data for many different input sizes, and plot it on a log-log plot. That is, log T vs. log N. If the running time is of the form n^k you should see a straight line of slope k, or something approaching this. If the running time is like T(n) = n^{k log n} or something, then you should see a parabola. And if T is exponential in n you should still see exponential growth.
You can only hope to get information about the highest order term when you do this -- the low order terms get filtered out, in the sense of having less and less impact as n gets larger.
In the two variable case, you could try to do a similar approach -- essentially, take 3 dimensional data, do a log-log-log plot, and try to fit a plane to that.
However this will only really work if there's really only one leading term that dominates in most regimes.
Suppose my actual function is T(n, m) = n^4 + n^3 * m^3 + m^4.
When m = O(1), then T(n) = O(n^4).
When n = O(1), then T(n) = O(m^4).
When n = m, then T(n) = O(n^6).
In each of these regimes, "slices" along the plane of possible n,m values, a different one of the terms is the dominant term.
So there's no way to determine the function just from taking some points with fixed m, and some points with fixed n. If you did that, you wouldn't get the right answer for n = m -- you wouldn't be able to discover "middle" leading terms like that.
I would recommend that the best way to predict asymptotic growth when you have lots of variables / complicated data structures, is with a pencil and piece of paper, and do traditional algorithmic analysis. Or possibly, a hybrid approach. Try to break the question of efficiency into different parts -- if you can split the question up into a sum or product of a few different functions, maybe some of them you can determine in the abstract, and some you can estimate experimentally.
Luckily two input parameters is still easy to visualize in a 3D scatter plot (3rd dimension is the measured running time), and you can check if it looks like a plane (in log-log-log scale) or if it is curved. Naturally random variations in measurements plays a role here as well.
In Matlab I typically calculate a least-squares solution to two-variable function like this (just concatenates different powers and combinations of x and y horizontally, .* is an element-wise product):
x = log(parameter_x);
y = log(parameter_y);
% Find a least-squares fit
p = [x.^2, x.*y, y.^2, x, y, ones(length(x),1)] \ log(time)
Then this can be used to estimate running times for larger problem instances, ideally those would be confirmed experimentally to know that the fitted model works.
This approach works also for higher dimensions but gets tedious to generate, maybe there is a more general way to achieve that and this is just a work-around for my lack of knowledge.
I was going to write my own explanation but it wouldn't be any better than this.
I am computing a similarity matrix based on Euclidean distance in MATLAB. My code is as follows:
for i=1:N % M,N is the size of the matrix x for whose elements I am computing similarity matrix
for j=1:N
D(i,j) = sqrt(sum(x(:,i)-x(:,j)).^2)); % D is the similarity matrix
end
end
Can any help with optimizing this = reducing the for loops as my matrix x is of dimension 256x30000.
Thanks a lot!
--Aditya
The function to do so in matlab is called pdist. Unfortunately it is painfully slow and doesnt take Matlabs vectorization abilities into account.
The following is code I wrote for a project. Let me know what kind of speed up you get.
Qx=repmat(dot(x,x,2),1,size(x,1));
D=sqrt(Qx+Qx'-2*x*x');
Note though that this will only work if your data points are in the rows and your dimensions the columns. So for example lets say I have 256 data points and 100000 dimensions then on my mac using x=rand(256,100000) and the above code produces a 256x256 matrix in about half a second.
There's probably a better way to do it, but the first thing I noticed was that you could cut the runtime in half by exploiting the symmetry D(i,j)==D(i,j)
You can also use the function norm(x(:,i)-x(:,j),2)
I think this is what you're looking for.
D=zeros(N);
jIndx=repmat(1:N,N,1);iIndx=jIndx'; %'# fix SO's syntax highlighting
D(:)=sqrt(sum((x(iIndx(:),:)-x(jIndx(:),:)).^2,2));
Here, I have assumed that the distance vector, x is initalized as an NxM array, where M is the number of dimensions of the system and N is the number of points. So if your ordering is different, you'll have to make changes accordingly.
To start with, you are computing twice as much as you need to here, because D will be symmetric. You don't need to calculate the (i,j) entry and the (j,i) entry separately. Change your inner loop to for j=1:i, and add in the body of that loop D(j,i)=D(i,j);
After that, there's really not much redundancy left in what that code does, so your only room left for improvement is to parallelize it: if you have the Parallel Computing Toolbox, convert your outer loop to a parfor and before you run it, say matlabpool(n), where n is the number of threads to use.
Greetings. I'm trying to approximate the function
Log10[x^k0 + k1], where .21 < k0 < 21, 0 < k1 < ~2000, and x is integer < 2^14.
k0 & k1 are constant. For practical purposes, you can assume k0 = 2.12, k1 = 2660. The desired accuracy is 5*10^-4 relative error.
This function is virtually identical to Log[x], except near 0, where it differs a lot.
I already have came up with a SIMD implementation that is ~1.15x faster than a simple lookup table, but would like to improve it if possible, which I think is very hard due to lack of efficient instructions.
My SIMD implementation uses 16bit fixed point arithmetic to evaluate a 3rd degree polynomial (I use least squares fit). The polynomial uses different coefficients for different input ranges. There are 8 ranges, and range i spans (64)2^i to (64)2^(i + 1).
The rational behind this is the derivatives of Log[x] drop rapidly with x, meaning a polynomial will fit it more accurately since polynomials are an exact fit for functions that have a derivative of 0 beyond a certain order.
SIMD table lookups are done very efficiently with a single _mm_shuffle_epi8(). I use SSE's float to int conversion to get the exponent and significand used for the fixed point approximation. I also software pipelined the loop to get ~1.25x speedup, so further code optimizations are probably unlikely.
What I'm asking is if there's a more efficient approximation at a higher level?
For example:
Can this function be decomposed into functions with a limited domain like
log2((2^x) * significand) = x + log2(significand)
hence eliminating the need to deal with different ranges (table lookups). The main problem I think is adding the k1 term kills all those nice log properties that we know and love, making it not possible. Or is it?
Iterative method? don't think so because the Newton method for log[x] is already a complicated expression
Exploiting locality of neighboring pixels? - if the range of the 8 inputs fall in the same approximation range, then I can look up a single coefficient, instead of looking up separate coefficients for each element. Thus, I can use this as a fast common case, and use a slower, general code path when it isn't. But for my data, the range needs to be ~2000 before this property hold 70% of the time, which doesn't seem to make this method competitive.
Please, give me some opinion, especially if you're an applied mathematician, even if you say it can't be done. Thanks.
You should be able to improve on least-squares fitting by using Chebyshev approximation. (The idea is, you're looking for the approximation whose worst-case deviation in a range is least; least-squares instead looks for the one whose summed squared difference is least.) I would guess this doesn't make a huge difference for your problem, but I'm not sure -- hopefully it could reduce the number of ranges you need to split into, somewhat.
If there's already a fast implementation of log(x), maybe compute P(x) * log(x) where P(x) is a polynomial chosen by Chebyshev approximation. (Instead of trying to do the whole function as a polynomial approx -- to need less range-reduction.)
I'm an amateur here -- just dipping my toe in as there aren't a lot of answers already.
One observation:
You can find an expression for how large x needs to be as a function of k0 and k1, such that the term x^k0 dominates k1 enough for the approximation:
x^k0 +k1 ~= x^k0, allowing you to approximately evaluate the function as
k0*Log(x).
This would take care of all x's above some value.
I recently read how the sRGB model compresses physical tri stimulus values into stored RGB values.
It basically is very similar to the function I try to approximate, except that it's defined piece wise:
k0 x, x < 0.0031308
k1 x^0.417 - k2 otherwise
I was told the constant addition in Log[x^k0 + k1] was to make the beginning of the function more linear. But that can easily be achieved with a piece wise approximation. That would make the approximation a lot more "uniform" - with only 2 approximation ranges. This should be cheaper to compute due to no longer needing to compute an approximation range index (integer log) and doing SIMD coefficient lookup.
For now, I conclude this will be the best approach, even though it doesn't approximate the function precisely. The hard part will be proposing this change and convincing people to use it.