I am solving general black-box optimization problems like:
x*: f(x) -> min, where x are permutations of length N (N = 50 for example, so brute force search is not possible). Objective function f(x) is represented by stand-alone computer code and x represents configuration of complex system with the response simulated by f(x).
I learned, that in this case I can use many heuristic methods. But, most of these methods use always some kind of local search, which require suitable distance metric at search space (space of permutations x in my case). Under suitable distance metric I mean the metric which fulfill the "locality" property, e.g. small change of permutation x produce small change of objective function f(x). In my case is not known any suitable distance metric with this property, so any kind of local search is nearly the random search.
I have a few questions:
Are there available any heuristic black-box combinatorial optimization methods, which does not use local search and/or any distance metric at search space? I need to overcome the low "locality" of the problem or simply the fact, that any suitable distance metric at search space is unknown.
Is the "locality" property really so restricted at combinatorial optimization in general? May be I miss something..., but the most of real-world black-box combinatorial problem has low or very low "locality" due to the fact, that the common permutation distance metrics (Hamming, Kendal, etc.) are not suitable metrics in general.
Is there any general method how to find suitable distance metric at search space to satisfy at least approximately the "locality"?
Additional remarks:
In real, the black-box function f(x) is realized by stand-alone deterministic simulation code, where x plays a role of discrete configuration of the simulated physical system. So, function f(x) has definitely well defined properties, but this properties are so difficult, that is not possible to simple exploit it.
Because of above mentioned complicated internal properties of function f(x) is not possible to find proper distance metric d(x,x') in search space which fulfill "locality" (similar x and x' in a sense of any distance metric produce similar responses f(x) and f(x'))
So, finally, I am looking for any optimization heuristics, which are able to find any suitable sub-optimal solutions only by informations available by properties of f(x) at fitness space. Like EDA's (Estimation of Distribution Algorithms) for example.
The main reason of this question is, what types of optimization heuristics are suitable to solve this kind of problems.
Related
I am learning statistics, and have some basic yet core questions on SD:
s = sample size
n = total number of observations
xi = ith observation
μ = arithmetic mean of all observations
σ = the usual definition of SD, i.e. ((1/(n-1))*sum([(xi-μ)**2 for xi in s])**(1/2) in Python lingo
f = frequency of an observation value
I do understand that (1/n)*sum([xi-μ for xi in s]) would be useless (= 0), but would not (1/n)*sum([abs(xi-μ) for xi in s]) have been a measure of variation?
Why stop at power of 1 or 2? Would ((1/(n-1))*sum([abs((xi-μ)**3) for xi in s])**(1/3) or ((1/(n-1))*sum([(xi-μ)**4 for xi in s])**(1/4) and so on have made any sense?
My notion of squaring is that it 'amplifies' the measure of variation from the arithmetic mean while the simple absolute difference is somewhat a linear scale notionally. Would it not amplify it even more if I cubed it (and made absolute value of course) or quad it?
I do agree computationally cubes and quads would have been more expensive. But with the same argument, the absolute values would have been less expensive... So why squares?
Why is the Normal Distribution like it is, i.e. f = (1/(σ*math.sqrt(2*pi)))*e**((-1/2)*((xi-μ)/σ))?
What impact would it have on the normal distribution formula above if I calculated SD as described in (1) and (2) above?
Is it only a matter of our 'getting used to the squares', it could well have been linear, cubed or quad, and we would have trained our minds likewise?
(I may not have been 100% accurate in my number of opening and closing brackets above, but you will get the idea.)
So, if you are looking for an index of dispersion, you actually don't have to use the standard deviation. You can indeed report mean absolute deviation, the summary statistic you suggested. You merely need to be aware of how each summary statistic behaves, for example the SD assigns more weight to outlying variables. You should also consider how each one can be interpreted. For example, with a normal distribution, we know how much of the distribution lies between ±2SD from the mean. For some discussion of mean absolute deviation (and other measures of average absolute deviation, such as the median average deviation) and their uses see here.
Beyond its use as a measure of spread though, SD is related to variance and this is related to some of the other reasons it's popular, because the variance has some nice mathematical properties. A mathematician or statistician would be able to provide a more informed answer here, but squared difference is a smooth function and is differentiable everywhere, allowing one to analytically identify a minimum, which helps when fitting functions to data using least squares estimation. For more detail and for a comparison with least absolute deviations see here. Another major area where variance shines is that it can be easily decomposed and summed, which is useful for example in ANOVA and regression models generally. See here for a discussion.
As to your questions about raising to higher powers, they actually do have uses in statistics! In general, the mean (which is related to average absolute mean), the variance (related to standard deviation), skewness (related to the third power) and kurtosis (related to the fourth power) are all related to the moments of a distribution. Taking differences raised to those powers and standardizing them provides useful information about the shape of a distribution. The video I linked provides some easy intuition.
For some other answers and a larger discussion of why SD is so popular, See here.
Regarding the relationship of sigma and the normal distribution, sigma is simply a parameter that stretches the standard normal distribution, just like the mean changes its location. This is simply a result of the way the standard normal distribution (a normal distribution with mean=0 and SD=variance=1) is mathematically defined, and note that all normal distributions can be derived from the standard normal distribution. This answer illustrates this. Now, you can parameterize a normal distribution in other ways as well, but I believe you do need to provide sigma, whether using the SD or precisions. I don't think you can even parametrize a normal distribution using just the mean and the mean absolute difference. Now, a deeper question is why normal distributions are so incredibly useful in representing widely different phenomena and crop up everywhere. I think this is related to the Central Limit Theorem, but I do not understand the proofs of the theorem well enough to comment further.
I was wondering if there was a direct way of computing the iteration matrix for nth Linear Block Gauss Seidel iteration within OpenMDAO?
thank you
If I understand you correctly, you are referring to the matrix-form of the Gauss Seidel algorithm where you take Ax=b, and break A up into the Diagonal (D), Lower (L) and Upper (U) parts, then use those parts to compute the next iterate.
Specifically you compute [D-L]^-1. This, I believe is what you are referring to as the "iteration matrix" (I am not familiar with this terminology, but based on the algorithm I'm comfortable making an educated guess).
This formulation of the algorithm is useful to think about and a simple way to implement it, but OpenMDAO takes a different approach. The LBGS algorithm implemented in OpenMDAO is set up to work in a matrix-free manner. That means it only interacts with the linear operator methods solve_linear and apply_linear and never explicitly assembles the A matrix at all. Hence there isn't an opportunity to split A up into D, L, U.
Depending on the way you constructed the model, the A matrix you would need might or might not be there at all because OpenMDAO is capable of working in a completely matrix free context. However, if all of your components use the compute_partials or linearize methods to provide partial derivatives then the data you would need for the A matrix does exist in memory.
You'll have to dig for it a bit, and ironically the best place to see how to do that is in the direct solver which does actually require the matrix be formed to compute a factorization.
Also, in that code you'll see a function can iteratively call the linear operator to construct a dense matrix even if the underlying components don't provide their partials directly. Please note that this approach for assembling the matrix is extremely slow and is not recommended for normal operations.
Despite going through lots of similar question related to this I still could not understand why some algorithm is susceptible to it while others are not.
Till now I found that SVM and K-means are susceptible to feature scaling while Linear Regression and Decision Tree are not.Can somebody please elaborate me why? in general or relating to this 4 algorithm.
As I am a beginner, please explain this in layman terms.
One reason I can think of off-hand is that SVM and K-means, at least with a basic configuration, uses an L2 distance metric. An L1 or L2 distance metric between two points will give different results if you double delta-x or delta-y, for example.
With Linear Regression, you fit a linear transform to best describe the data by effectively transforming the coordinate system before taking a measurement. Since the optimal model is the same no matter the coordinate system of the data, pretty much by definition, your result will be invariant to any linear transform including feature scaling.
With Decision Trees, you typically look for rules of the form x < N, where the only detail that matters is how many items pass or fail the given threshold test - you pass this into your entropy function. Because this rule format does not depend on dimension scale, since there is no continuous distance metric, we again have in-variance.
Somewhat different reasons for each, but I hope that helps.
A friend is using Factor Graphs to do text mining (identifying references to people in text), and it got me interested in this tool, but I'm having a hard time finding an intuitive explanation of what Factor Graphs are and how to use them.
Can anyone provide an explanation of Factor Graphs that isn't math heavy, and which focusses on practical applications rather than abstract theory?
They are used extensively for breaking down a problem into pieces. One very interesting application of factor graphs (and message passing on them) is the XBox Live TrueSkill algorithm. I wrote extensively about it on my blog where I tried to go for an introductory explanation rather than an overly academic one.
A factor graph is the graphical representation of the dependencies between variables and factors (parts of a formula) that are present in a particular kind of formula.
Suppose you have a function f(x_1,x_2,...,x_n) and you want to compute the marginalization of this function for some argument x_i, thus summing over all assignments to the remaining formula. Further f can be broken into factors, e.g.
f(x_1,x_2,...,x_n)=f_1(x_1,x_2)f_2(x_5,x_8,x_9)...f_k(x_1,x_10,x_11)
Then in order to compute the marginalization of f for some of the variables you can use a special algorithm called sum product (or message passing), that breaks the problem into smaller computations. For this algortithm, it is very important which variables appear as arguments to which factor. This information is captured by the factor graph.
A factor graph is a bipartite graph with both factor nodes and variable nodes. And there is an edge between a factor and a variable node if the variable appears as an argument of the factor. In our example there would be an edge between the factor f_2 and the variable x_5 but not between f_2 and x_1.
There is a great article: Factor graphs and the sum-product algorithm.
Factor graph is math model, and can be explained only with math equations. In nutshell it is way to explain complex relations between interest variables in your model. Example: A is temperature, B is pressure, components C,D,E are depends on B,A in some way, and component K is depends on B,A. And you want to predict value K based on A and B. So you know only visible states. Basic ML libraries don't allow to model such structure. Neural network do it better. And Factor Graph is exactly solve that problem.
Factor graph is an example of deep learning. When it is impossible to present model with features and output, Factor models allow to build hidden states, layers and complex structure of variables to fit real world behavior. Examples are Machine translation alignment, fingerprint recognition, co-reference etc.
I am using Octave and I would like to use the anderson_darling_test from the Octave forge Statistics package to test if two vectors of data are drawn from the same statistical distribution. Furthermore, the reference distribution is unlikely to be "normal". This reference distribution will be the known distribution and taken from the help for the above function " 'If you are selecting from a known distribution, convert your values into CDF values for the distribution and use "uniform'. "
My question therefore is: how would I convert my data values into CDF values for the reference distribution?
Some background information for the problem: I have a vector of raw data values from which I extract the cyclic component (this will be the reference distribution); I then wish to compare this cyclic component with the raw data itself to see if the raw data is essentially cyclic in nature. If the the null hypothesis that the two are the same can be rejected I will then know that most of the movement in the raw data is not due to cyclic influences but is due to either trend or just noise.
If your data has a specific distribution, for instance beta(3,3) then
p = betacdf(x, 3, 3)
will be uniform by the definition of a CDF. If you want to transform it to a normal, you can just call the inverse CDF function
x=norminv(p,0,1)
on the uniform p. Once transformed, use your favorite test. I'm not sure I understand your data, but you might consider using a Kolmogorov-Smirnov test instead, which is a nonparametric test of distributional equality.
Your approach is misguided in multiple ways. Several points:
The Anderson-Darling test implemented in Octave forge is a one-sample test: it requires one vector of data and a reference distribution. The distribution should be known - not come from data. While you quote the help-file correctly about using a CDF and the "uniform" option for a distribution that is not built in, you are ignoring the next sentence of the same help file:
Do not use "uniform" if the distribution parameters are estimated from the data itself, as this sharply biases the A^2 statistic toward smaller values.
So, don't do it.
Even if you found or wrote a function implementing a proper two-sample Anderson-Darling or Kolmogorov-Smirnov test, you would still be left with a couple of problems:
Your samples (the data and the cyclic part estimated from the data) are not independent, and these tests assume independence.
Given your description, I assume there is some sort of time predictor involved. So even if the distributions would coincide, that does not mean they coincide at the same time-points, because comparing distributions collapses over the time.
The distribution of cyclic trend + error would not expected to be the same as the distribution of the cyclic trend alone. Suppose the trend is sin(t). Then it never will go above 1. Now add a normally distributed random error term with standard deviation 0.1 (small, so that the trend is dominant). Obviously you could get values well above 1.
We do not have enough information to figure out the proper thing to do, and it is not really a programming question anyway. Look up time series theory - separating cyclic components is a major topic there. But many reasonable analyses will probably be based on the residuals: (observed value - predicted from cyclic component). You will still have to be careful about auto-correlation and other complexities, but at least it will be a move in the right direction.