I have two - likely simple - questions that are bothering me, both related to quadratic programming:
1). There are two "standard" forms of the objective function I have found, differing by multiplication of negative 1.
In the R package quadprog, the objective function to be minimized is given as −dTb+12bTDb and in Matlab the objective is given as dTb+12bTDb. How can these be the same? It seems that one has been multiplied through by a negative 1 (which as I understand it would change from a min problem to a max problem.
2). Related to the first question, in the case of using quadprog for minimizing least squares, in order to get the objective function to match the standard form, it is necessary to multiply the objective by a positive 2. Does multiplication by a positive number not change the solution?
EDIT: I had the wrong sign for the Matlab objective function.
Function f(b)=dTb is a linear function thus it is both convex and concave. From optimization standpoint it means you can maximize or minimize it. Nevertheless minimizer of −dTb+12bTDb will be different from dTb+12bTDb, because there is additional quadratic term. Matlab implementation will find the one with plus sign. So if you are using different optimization software you will need to change d→−d to get the same result.
The function −dTb+12bTDb where D is symmetric and convex and thus has unique minimum. In general that is called standard quadratic programming form, but that doesn't really matter. The other function dTb−12bTDb is concave function which has unique maximum. It is easy to show that for, say, bounded function f(x) from above the following holds:
argmaxxf=argminx−f
Using the identity above value b∗1 where −dTb+12bTDb achieves minimum is the same as the value b∗2 which achieves maximum at dTb−12bTDb, that is b∗1=b∗2.
Programmatically it doesn't matter if you are minimizing −dTb+12bTDb or maximizing the other one. These are implementation-dependent details.
No it does not. ∀α>0 if x∗=argmaxxf(x), then x∗=argmaxxαf(x). This can be showed by contradiction.
Related
My problem is described in this picture(It's like a Pyramid structure):
The objective function is below:
In this problem, D is known, A is the object that I want to get. It is a layered structure, each block in the upper layer is divided into four sub-blocks in the layer below. And the value of the upper layer node is equal to the sum of the four child nodes of the lower layer. In above example, I used only 2 layers.
What I want to do is simulate the distribution of D with A, so in the objective function is the ratio of two adjacent squares in each row in A compared to the value in D. I do this comparison on each layer and sum them. Then it is all of my objective function. But in the finest layer, the value in A has a constrain A<=1, the value in A can be a number between 0 and 1. I have tried to solve it using Quadratic programming in python library CVXPY. However, it seems the speed is slow.
So I want to solve it in another way, because this is a convex optimization problem, which can guarantee the global optimal solution. What I think is whether it is possible to use the method of derivation. There are two unknown variables in each item, that is, the two items with A in the formula. Partial derivatives are obtained for them, and the restriction of A<=1 is added, then solve using gradient descent method. Is this mathematically feasible, because I don't know much about optimization, and if it is possible, how should I do it? If not possible, what other methods can I use?
how are you?.
I'm trying to create a function that calculates the z-transform of a transfer function using the residues method but for that, I need the factors of the characteristic equation and the powers of the factors, so, in order to do that I tried to factorize polynomials with non-integer coefficients but after trying everything that I read I couldn't factorize make maxima to factorize those polynomials the way I need it.
For giving an example, I have this characteristic equation: "s·(s^2+0.1·s)", the factors should be "s^2" and "s + 0.1" but maxima allways gives me "(s^2·(10·s + 1))/10".
Why I'm signalling this?, well, as I learned that maxima treates the outputs equation as list so I can have its dimension and separate the factos by its positions in the list to measure the powers of the factors and do what I need, but like maxima gives me the result that is shown above then the dimension of the list is different and it will make my function to work differently and possibly have errors.
The result that is shown is given by maxima no matter if I use factor, gfactor, or expand or whatever other function that I found and I know that result is because maxima are rationalizing the polynomial before working with it but I don't need that behavior, I only need the pure factors, so, how can I have the result that I want?.
Thanks in advance for the help.
I use the scipy.optimize.minimize ( https://docs.scipy.org/doc/scipy/reference/tutorial/optimize.html ) function with method='L-BFGS-B.
An example of what it returns is here above:
fun: 32.372210618549758
hess_inv: <6x6 LbfgsInvHessProduct with dtype=float64>
jac: array([ -2.14583906e-04, 4.09272616e-04, -2.55795385e-05,
3.76587650e-05, 1.49213975e-04, -8.38440428e-05])
message: 'CONVERGENCE: REL_REDUCTION_OF_F_<=_FACTR*EPSMCH'
nfev: 420
nit: 51
status: 0
success: True
x: array([ 0.75739412, -0.0927572 , 0.11986434, 1.19911266, 0.27866406,
-0.03825225])
The x value correctly contains the fitted parameters. How do I compute the errors associated to those parameters?
TL;DR: You can actually place an upper bound on how precisely the minimization routine has found the optimal values of your parameters. See the snippet at the end of this answer that shows how to do it directly, without resorting to calling additional minimization routines.
The documentation for this method says
The iteration stops when (f^k - f^{k+1})/max{|f^k|,|f^{k+1}|,1} <= ftol.
Roughly speaking, the minimization stops when the value of the function f that you're minimizing is minimized to within ftol of the optimum. (This is a relative error if f is greater than 1, and absolute otherwise; for simplicity I'll assume it's an absolute error.) In more standard language, you'll probably think of your function f as a chi-squared value. So this roughly suggests that you would expect
Of course, just the fact that you're applying a minimization routine like this assumes that your function is well behaved, in the sense that it's reasonably smooth and the optimum being found is well approximated near the optimum by a quadratic function of the parameters xi:
where Δxi is the difference between the found value of parameter xi and its optimal value, and Hij is the Hessian matrix. A little (surprisingly nontrivial) linear algebra gets you to a pretty standard result for an estimate of the uncertainty in any quantity X that's a function of your parameters xi:
which lets us write
That's the most useful formula in general, but for the specific question here, we just have X = xi, so this simplifies to
Finally, to be totally explicit, let's say you've stored the optimization result in a variable called res. The inverse Hessian is available as res.hess_inv, which is a function that takes a vector and returns the product of the inverse Hessian with that vector. So, for example, we can display the optimized parameters along with the uncertainty estimates with a snippet like this:
ftol = 2.220446049250313e-09
tmp_i = np.zeros(len(res.x))
for i in range(len(res.x)):
tmp_i[i] = 1.0
hess_inv_i = res.hess_inv(tmp_i)[i]
uncertainty_i = np.sqrt(max(1, abs(res.fun)) * ftol * hess_inv_i)
tmp_i[i] = 0.0
print('x^{0} = {1:12.4e} ± {2:.1e}'.format(i, res.x[i], uncertainty_i))
Note that I've incorporated the max behavior from the documentation, assuming that f^k and f^{k+1} are basically just the same as the final output value, res.fun, which really ought to be a good approximation. Also, for small problems, you can just use np.diag(res.hess_inv.todense()) to get the full inverse and extract the diagonal all at once. But for large numbers of variables, I've found that to be a much slower option. Finally, I've added the default value of ftol, but if you change it in an argument to minimize, you would obviously need to change it here.
One approach to this common problem is to use scipy.optimize.leastsq after using minimize with 'L-BFGS-B' starting from the solution found with 'L-BFGS-B'. That is, leastsq will (normally) include and estimate of the 1-sigma errors as well as the solution.
Of course, that approach makes several assumption, including that leastsq can be used and may be appropriate for solving the problem. From a practical view, this requires the objective function return an array of residual values with at least as many elements as variables, not a cost function.
You may find lmfit (https://lmfit.github.io/lmfit-py/) useful here: It supports both 'L-BFGS-B' and 'leastsq' and gives a uniform wrapper around these and other minimization methods, so that you can use the same objective function for both methods (and specify how to convert the residual array into the cost function). In addition, parameter bounds can be used for both methods. This makes it very easy to first do a fit with 'L-BFGS-B' and then with 'leastsq', using the values from 'L-BFGS-B' as starting values.
Lmfit also provides methods to more explicitly explore confidence limits on parameter values in more detail, in case you suspect the simple but fast approach used by leastsq might be insufficient.
It really depends what you mean by "errors". There is no general answer to your question, because it depends on what you're fitting and what assumptions you're making.
The easiest case is one of the most common: when the function you are minimizing is a negative log-likelihood. In that case the inverse of the hessian matrix returned by the fit (hess_inv) is the covariance matrix describing the Gaussian approximation to the maximum likelihood.The parameter errors are the square root of the diagonal elements of the covariance matrix.
Beware that if you are fitting a different kind of function or are making different assumptions, then that doesn't apply.
I'm using
R=QQ['x'];
to declare x as variable such that I can calculate with polynomials in x. What I need is another variable, for example t, to represent an integer which can also be used as exponent. For example I want to consider the polynomial (t+1)x^t. Is that somehow possible in SageMath?
EDIT: I want to explain a little bit the reason why I'm looking for such a feature. I've got a couple of really complex rational functions in a few variables and want SageMath to help me to show that they are identical. They are written down in a different way and if you would do it with pen and paper you would need hours and have a huge change of making mistakes. Actually it are not only a few rational functions but infinitely many. But using variables in the coefficients you can cover infinitely many with only one rational function. A simple example is the polynomial 1+x+x^2+...+x^t. For every non-negative integer you get a different polynomial. But you can write (x^(t+1)-1)/(x-1) as rational function instead. With taking t as a variable you cover infinitely many cases with just one rational function.
Is there a way to do such stuff in SageMath?
To create the polynomial x^t as an element of a polynomial ring, Sage needs to know what integer t is equal to. If polynomials of undetermined degree were introduced, most of the Sage methods for polynomials would not work for them: no way to get the list of coefficients with coefficients(), or to find the factors, or the GCD of two polynomials, etc.
However, you can manipulate and simplify polynomials and rational functions just like any other symbolic expressions. For example:
x,t,k = var('x,t,k')
sum(x^k, k, 1, t)
returns (x^(t + 1) - x)/(x - 1).
The relevant articles are Symbolic Computation and Symbolic Expressions.
I searched the site but did not find exactly what I was looking for... I wanted to generate a discrete random number from normal distribution.
For example, if I have a range from a minimum of 4 and a maximum of 10 and an average of 7. What code or function call ( Objective C preferred ) would I need to return a number in that range. Naturally, due to normal distribution more numbers returned would center round the average of 7.
As a second example, can the bell curve/distribution be skewed toward one end of the other? Lets say I need to generate a random number with a range of minimum of 4 and maximum of 10, and I want the majority of the numbers returned to center around the number 8 with a natural fall of based on a skewed bell curve.
Any help is greatly appreciated....
Anthony
What do you need this for? Can you do it the craps player's way?
Generate two random integers in the range of 2 to 5 (inclusive, of course) and add them together. Or flip a coin (0,1) six times and add 4 to the result.
Summing multiple dice produces a normal distribution (a "bell curve"), while eliminating high or low throws can be used to skew the distribution in various ways.
The key is you are going for discrete numbers (and I hope you mean integers by that). Multiple dice throws famously generate a normal distribution. In fact, I think that's how we were first introduced to the Gaussian curve in school.
Of course the more throws, the more closely you approximate the bell curve. Rolling a single die gives a flat line. Rolling two dice just creates a ramp up and down that isn't terribly close to a bell. Six coin flips gets you closer.
So consider this...
If I understand your question correctly, you only have seven possible outcomes--the integers (4,5,6,7,8,9,10). You can set up an array of seven probabilities to approximate any distribution you like.
Many frameworks and libraries have this built-in.
Also, just like TokenMacGuy said a normal distribution isn't characterized by the interval it's defined on, but rather by two parameters: Mean μ and standard deviation σ. With both these parameters you can confine a certain quantile of the distribution to a certain interval, so that 95 % of all points fall in that interval. But resticting it completely to any interval other than (−∞, ∞) is impossible.
There are several methods to generate normal-distributed values from uniform random values (which is what most random or pseudorandom number generators are generating:
The Box-Muller transform is probably the easiest although not exactly fast to compute. Depending on the number of numbers you need, it should be sufficient, though and definitely very easy to write.
Another option is Marsaglia's Polar method which is usually faster1.
A third method is the Ziggurat algorithm which is considerably faster to compute but much more complex to program. In applications that really use a lot of random numbers it may be the best choice, though.
As a general advice, though: Don't write it yourself if you have access to a library that generates normal-distributed random numbers for you already.
For skewing your distribution I'd just use a regular normal distribution, choosing μ and σ appropriately for one side of your curve and then determine on which side of your wanted mean a point fell, stretching it appropriately to fit your desired distribution.
For generating only integers I'd suggest you just round towards the nearest integer when the random number happens to fall within your desired interval and reject it if it doesn't (drawing a new random number then). This way you won't artificially skew the distribution (such as you would if you were clamping the values at 4 or 10, respectively).
1 In testing with deliberately bad random number generators (yes, worse than RANDU) I've noticed that the polar method results in an endless loop, rejecting every sample. Won't happen with random numbers that fulfill the usual statistic expectations to them, though.
Yes, there are sophisticated mathematical solutions, but for "simple but practical" I'd go with Nosredna's comment. For a simple Java solution:
Random random=new Random();
public int bell7()
{
int n=4;
for (int x=0;x<6;++x)
n+=random.nextInt(2);
return n;
}
If you're not a Java person, Random.nextInt(n) returns a random integer between 0 and n-1. I think the rest should be similar to what you'd see in any programming language.
If the range was large, then instead of nextInt(2)'s I'd use a bigger number in there so there would be fewer iterations through the loop, depending on frequency of call and performance requirements.
Dan Dyer and Jay are exactly right. What you really want is a binomial distribution, not a normal distribution. The shape of a binomial distribution looks a lot like a normal distribution, but it is discrete and bounded whereas a normal distribution is continuous and unbounded.
Jay's code generates a binomial distribution with 6 trials and a 50% probability of success on each trial. If you want to "skew" your distribution, simply change the line that decides whether to add 1 to n so that the probability is something other than 50%.
The normal distribution is not described by its endpoints. Normally it's described by it's mean (which you have given to be 7) and its standard deviation. An important feature of this is that it is possible to get a value far outside the expected range from this distribution, although that will be vanishingly rare, the further you get from the mean.
The usual means for getting a value from a distribution is to generate a random value from a uniform distribution, which is quite easily done with, for example, rand(), and then use that as an argument to a cumulative distribution function, which maps probabilities to upper bounds. For the standard distribution, this function is
F(x) = 0.5 - 0.5*erf( (x-μ)/(σ * sqrt(2.0)))
where erf() is the error function which may be described by a taylor series:
erf(z) = 2.0/sqrt(2.0) * Σ∞n=0 ((-1)nz2n + 1)/(n!(2n + 1))
I'll leave it as an excercise to translate this into C.
If you prefer not to engage in the exercise, you might consider using the Gnu Scientific Library, which among many other features, has a technique to generate random numbers in one of many common distributions, of which the Gaussian Distribution (hint) is one.
Obviously, all of these functions return floating point values. You will have to use some rounding strategy to convert to a discrete value. A useful (but naive) approach is to simply downcast to integer.