I have some simulation code that draws from various distributions. To facilitate some sanity checks, is there a way to make a Distribution that returns only a single floating-point value? That way I can test without changing code that calls rand on the distribution. Right now I'm doing something like, supposing I want to always get the value 2.2
mydist = Normal(2.2, 0.000001)
But this seems kind of silly. Of course, if I change the variance to 0 I get an error.
The Distibutions.jl docs have an extends section, so you can see what needs to be defined. An incomplete implementation of a new Distribution starts
using Distributions
struct OneFloatDistribution <: Distribution{Univariate,Continuous}
v::Float64
end
Base.rand(x::OneFloatDistribution) = x.v
You can get down to two possible floating point numbers with Uniform(1.0,nextfloat(1.0))
Related
In traditional Simplex Algorithm notation, we have x at the current basis selection B as so:
xB = AB-1b - AB-1ANxN. How can I compute the AB-1AN term inside a separator in SCIP, or at least iterate over its columns?
I see three helpful methods: getLPColsData, getLPRowsData, getLPBasisInd. I'm just not sure exactly what data those methods represent, particularly the last one, with its negative row indexes. How do I use those to get the value I want?
Do those methods return the same data no matter what LP algorithm is used? Or do I need to account for dual vs primal? How does the use of the "revised" algorithm play into my calculation?
Update: I discovered the getLPBInvARow and getLPBInvRow. That seems to be much closer to what I'm after. I don't yet understand their results; they seem to include more/less dimensions than expected. I'm still looking for understanding at how to use them to get the rays away from the corner.
you are correct that getLPBInvRow or getLPBInvARow are the methods you want. getLPBInvARow directly returns you a of the simplex tableau, but it is not more efficient to use than getLPBInvRow and doing the multiplication yourself since the LP solver needs to also compute the actual tableau first.
I suggest you look into either sepa_gomory.c or sepa_gmi.c for examples of how to use these methods. How do they include less dimensions than expected? They both return sparse vectors.
This will be a strange question: I know what to do, and I am actually doing it, and it works, but I don't know how to write about it. Looking for solutions to a homogeneous matrix equation, say AX=0, I use the kernel of the parameter matrix A. But, the world being imperfect as it is, the matrix does not have a "perfect" kernel; it does have an "imperfect" one if you set a nonzero "tolerance" parameter. FWIW I'm using Scilab, the function is kernel(A,tol).
Now what are the correct terms for "imperfect kernel", or "tolerance" (of what?), how should this whole process be described in correct English and maths terminology? Should I say something like a "least-squares kernel"? "Approximate kernel"? Is tol the "tolerance of kernel-determination algorithm"? Sounds lame to me...
Depending on the method used (QR or SVD, third flag allows to choose this in Scilab implementation) the tolerance is used to determine when pivots (QR case) or singular values (SVD case) are consider to be zero. The kernel is then considered to be the associated subspace.
I have a function that I need to maximize that contains 3 parameters, one of which is an integer.
How do I let the optim function know to maximize (instead of minimize which is the default).
And how do I let it know that one of the parameters in an integer?
Will it work if one of the parameters is a binary or categorical?
Max vs min is easy (set fnscale=-1 in the control parameter).
Integer parameters are not easy. I don't know of a simple out-of-the-box solution for this, hopefully someone else does.
Most of the methods implemented in optim assume continuous parameter spaces. (method="SANN" will work since you can give it explicit rules for updating - see the examples - but it's tricky to get it to work efficiently.) Most of the optimizers listed in the Optimization Task View are for continuous optimization - the section on global/stochastic gives the most options for mixed discrete/continuous problems.
If the range of plausible integers is reasonably small you can use brute force (i.e., optimize over the two continuous parameters for each of a range of fixed integer values); you could also use bisection search over the integers.
I have a function f(x,y) whose outcome is random (I take mean from 20 random numbers depending on x and y). I see no way to modify this function to make it symbolic.
And when I run
x,y = var('x,y')
d = plot_vector_field((f(x),x), (x,0,1), (y,0,1))
it says it can't cast symbolic expression to real or rationa number. In fact it stops when I write:
a=matrix(RR,1,N)
a[0]=x
What is the way to change this variable to real numbers in the beginning, compute f(x) and draw a vector field? Or just draw a lot of arrows with slope (f(x),x)?
I can create something sort of like yours, though with no errors. At least it doesn't do what you want.
def f(m,n):
return m*randint(100,200)-n*randint(100,200)
var('x,y')
plot_vector_field((f(x,y),f(y,x)),(x,0,1),(y,0,1))
The reason is because Python functions immediately evaluate - in this case, f(x,y) was 161*x - 114*y, though that will change with each invocation.
My suspicion is that your problem is similar, the immediate evaluation of the Python function once and for all. Instead, try lambda functions. They are annoying but very useful in this case.
var('x,y')
plot_vector_field((lambda x,y: f(x,y), lambda x,y: f(y,x)),(x,0,1),(y,0,1))
Wow, I now I have to find an excuse to show off this picture, cool stuff. I hope your error ends up being very similar.
Someone somewhere has had to solve this problem. I can find many a great website explaining this problem and how to solve it. While I'm sure they are well written and make sense to math whizzes, that isn't me. And while I might understand in a vague sort of way, I do not understand how to turn that math into a function that I can use.
So I beg of you, if you have a function that can do this, in any language, (sure even fortran or heck 6502 assembler) - please help me out.
prefer an analytical to iterative solution
EDIT: Meant to specify that its a cubic bezier I'm trying to work with.
What you're asking for is the inverse of the arc length function. So, given a curve B, you want a function Linv(len) that returns a t between 0 and 1 such that the arc length of the curve between 0 and t is len.
If you had this function your problem is really easy to solve. Let B(0) be the first point. To find the next point, you'd simply compute B(Linv(w)) where w is the "equal arclength" that you refer to. To get the next point, just evaluate B(Linv(2*w)) and so on, until Linv(n*w) becomes greater than 1.
I've had to deal with this problem recently. I've come up with, or come across a few solutions, none of which are satisfactory to me (but maybe they will be for you).
Now, this is a bit complicated, so let me just give you the link to the source code first:
http://icedtea.classpath.org/~dlila/webrevs/perfWebrev/webrev/raw_files/new/src/share/classes/sun/java2d/pisces/Dasher.java. What you want is in the LengthIterator class. You shouldn't have to look at any other parts of the file. There are a bunch of methods that are defined in another file. To get to them just cut out everything from /raw_files/ to the end of the URL. This is how you use it. Initialize the object on a curve. Then to get the parameter of a point with arc length L from the beginning of the curve just call next(L) (to get the actual point just evaluate your curve at this parameter, using deCasteljau's algorithm, or zneak's suggestion). Every subsequent call of next(x) moves you a distance of x along the curve compared to your last position. next returns a negative number when you run out of curve.
Explanation of code: so, I needed a t value such that B(0) to B(t) would have length LEN (where LEN is known). I simply flattened the curve. So, just subdivide the curve recursively until each curve is close enough to a line (you can test for this by comparing the length of the control polygon to the length of the line joining the end points). You can compute the length of this sub-curve as (controlPolyLength + endPointsSegmentLen)/2. Add all these lengths to an accumulator, and stop the recursion when the accumulator value is >= LEN. Now, call the last subcurve C and let [t0, t1] be its domain. You know that the t you want is t0 <= t < t1, and you know the length from B(0) to B(t0) - call this value L0t0. So, now you need to find a t such that C(0) to C(t) has length LEN-L0t0. This is exactly the problem we started with, but on a smaller scale. We could use recursion, but that would be horribly slow, so instead we just use the fact that C is a very flat curve. We pretend C is a line, and compute the point at t using P=C(0)+((LEN-L0t0)/length(C))*(C(1)-C(0)). This point doesn't actually lie on the curve because it is on the line C(0)->C(1), but it's very close to the point we want. So, we just solve Bx(t)=Px and By(t)=Py. This is just finding cubic roots, which has a closed source solution, but I just used Newton's method. Now we have the t we want, and we can just compute C(t), which is the actual point.
I should mention that a few months ago I skimmed through a paper that had another solution to this that found an approximation to the natural parameterization of the curve. The author has posted a link to it here: Equidistant points across Bezier curves