I need to write a function that returns on of the numbers (-2,-1,0,1,2) randomly, but I need the average of the output to be a specific number (say, 1.2).
I saw similar questions, but all the answers seem to rely on the target range being wide enough.
Is there a way to do this (without saving state) with this small selection of possible outputs?
UPDATE: I want to use this function for (randomized) testing, as a stub for an expensive function which I don't want to run. The consumer of this function runs it a couple of hundred times and takes an average. I've been using a simple randint function, but the average is always very close to 0, which is not realistic.
Point is, I just need something simple that won't always average to 0. I don't really care what the actual average is. I may have asked the question wrong.
Do you really mean to require that specific value to be the average, or rather the expected value? In other words, if the generated sequence were to contain an extraordinary number of small values in its initial part, should the rest of the sequence atempt to compensate for that in an attempt to get the overall average right? I assume not, I assume you want all your samples to be computed independently (after all, you said you don't want any state), in which case you can only control the expected value.
If you assign a probability pi for each of your possible choices, then the expected value will be the sum of these values, weighted by their probabilities:
EV = − 2p−2 − p−1 + p1 + 2p2 = 1.2
As additional constraints you have to require that each of these probabilities is non-negative, and that the above four add up to a value less than 1, with the remainder taken by the fifth probability p0.
there are many possible assignments which satisfy these requirements, and any one will do what you asked for. Which of them are reasonable for your application depends on what that application does.
You can use a PRNG which generates variables uniformly distributed in the range [0,1), and then map these to the cases you described by taking the cumulative sums of the probabilities as cut points.
Related
RSME calculates how close the predicted value is compared to the actual value, but in a point cloud, there are 2 things that I am confused about:
How do we know which point corresponds to which point, to be subtracted from?
Point clouds are 3-dimensional since it has xyz values, but how do people turn those 3 values to one RSME value?
First of all, it's RMSE, not RSME. It stands for Root Mean Square Error:
https://en.wikipedia.org/wiki/Root-mean-square_deviation
With 3D coordinates you can compare component wise, or however else you choose to define a distance measure. Then you plug this into the RMSE formula. Essentially this means comparing an expected value to your observed value.
As for the point correspondence - this depends on the algorithm of choice. Probably one of the most famous examples is ICP:
https://de.wikipedia.org/wiki/Iterative_Closest_Point_Algorithm
In a nutshell for every point of one cloud, the closest point of the other cloud is determined. Then an error measure is calculated and lastly points are transformed. This is done an arbitrary number of times, depending on the desired precision.
Since I strongly suspect that you are indeed looking for ICP, here is the description as to how they are put together:
https://en.wikipedia.org/wiki/Iterative_closest_point
Other than that you will have to do some reading yourself.
Can anybody please look at the image and tell me how the time is calculated for exponential algorithms i.e., 2^n and 3^n.
From the top row, we can see that when n = 10, it takes 10μs to perform the work. That means that each operation takes one microsecond.
The rows with 2n and 3n are computed by listing 2nμs and 3nμs in more convenient units. For example, 210μs = 1024μs is about 0.001s.
(It would have been nice for the table designer to explicitly indicate that each operation is one microsecond, since that would let you interpret the data more clearly or adjust it for cases where, say, each operation took one nanosecond.)
Hope this helps!
I am using the finite difference scheme to find gradients.
Lets say i have 2 outputs (y1,y2) and 1 input (x) in a single component. And in advance I know that the sensitivity of y1 with respect to x is not same as the sensitivity of y2 to x. And thus i could potentially have two different steps for those as in ;
self.declare_partials(of=y1, wrt=x, method='fd',step=0.01, form='central')
self.declare_partials(of=y2, wrt=x, method='fd',step=0.05, form='central')
There is nothing that stops me (algorithmically) but it is not clear what would openmdao gradient calculation exactly do in this case?
does it exchange information from the case where the steps are different by looking at the steps ratios or simply treating them independently and therefore doubling computational time ?
I just tested this, and it does the finite difference twice with the two different step sizes, and only saves the requested outputs for each step. I don't think we could do anything with the ratios as you suggested, as the reason for using different stepsizes to resolve individual outputs is because you don't trust the accuracy of the outputs at the smaller (or large) stepsize.
This is a fair question about the effect of the API. In typical FD applications you would get only 1 function call per design variable for forward and backward difference and 2 function calls for central difference.
However in this case, you have asked for two different step sizes for two different outputs, both with central difference. So here, you'll end up with 4 function calls to compute all the derivatives. dy1_dx will be computed using the step size of .01 and dy2_dx will be computed with a step size of .05.
There is no crosstalk between the two different FD calls, and you do end up with more function calls than you would have if you just specified a single step size via:
self.declare_partials(of='*', wrt=x, method='fd',step=0.05, form='central')
If the cost is something you can bear, and you get improved accuracy, then you could use this method to get different step sizes for different outputs.
Problem
I want to find
The first root
The first local minimum/maximum
of a black-box function in a given range.
The function has following properties:
It's continuous and differentiable.
It's combination of constant and periodic functions. All periods are known.
(It's better if it can be done with weaker assumptions)
What is the fastest way to get the root and the extremum?
Do I need more assumptions or bounds of the function?
What I've tried
I know I can use root-finding algorithm. What I don't know is how to find the first root efficiently.
It needs to be fast enough so that it can run within a few miliseconds with precision of 1.0 and range of 1.0e+8, which is the problem.
Since the range could be quite large and it should be precise enough, I can't brute-force it by checking all the possible subranges.
I considered bisection method, but it's too slow to find the first root if the function has only one big root in the range, as every subrange should be checked.
It's preferable if the solution is in java, but any similar language is fine.
Background
I want to calculate when arbitrary celestial object reaches certain height.
It's a configuration-defined virtual object, so I can't assume anything about the object.
It's not easy to get either analytical solution or simple approximation because various coordinates are involved.
I decided to find a numerical solution for this.
For a general black box function, this can't really be done. Any root finding algorithm on a black box function can't guarantee that it has found all the roots or any particular root, even if the function is continuous and differentiable.
The property of being periodic gives a bit more hope, but you can still have periodic functions with infinitely many roots in a bounded domain. Given that your function relates to celestial objects, this isn't likely to happen. Assuming your periodic functions are sinusoidal, I believe you can get away with checking subranges on the order of one-quarter of the shortest period (out of all the periodic components).
Maybe try Brent's Method on the shortest quarter period subranges?
Another approach would be to apply your root finding algorithm iteratively. If your range is (a, b), then apply your algorithm to that range to find a root at say c < b. Then apply your algorithm to the range (a, c) to find a root in that range. Continue until no more roots are found. The last root you found is a good candidate for your minimum root.
Black box function for any range? You cannot even be sure it has the continuous domain over that range. What kind of solutions are you looking for? Natural numbers, integers, real numbers, complex? These are all the question that greatly impact the answer.
So 1st thing should be determining what kind of number you accept as the result.
Second is having some kind of protection against limes of function that will try to explode your calculations as it goes for plus or minus infinity.
Since we are touching the limes topics you could have your solution edge towards zero and look like a solution but never touch 0 and become a solution. This depends on your margin of error, how close something has to be to be considered ok, it's good enough.
I think for this your SIMPLEST TO IMPLEMENT bet for real number solutions (I assume those) is to take an interval and this divide and conquer algorithm:
Take lower and upper border and middle value (or approx middle value for infinity decimals border/borders)
Try to calculate solution with all 3 and have some kind of protection against infinities
remember all 3 values in an array with results from them (3 pair of values)
remember the current best value (one its closest to solution) in seperate variable (a pair of value and result for that value)
STEP FORWARD - repeat above with 1st -2nd value range and 2nd -3rd value range
have a new pair of value and result to be closest to solution.
clear the old value-result pairs, replace them with new ones gotten from this iteration while remembering the best value solution pair (total)
Repeat above for how precise you wish to get and look at that memory explode with each iteration, keep in mind you are gonna to have exponential growth of values there. It can be further improved if you lets say take one interval and go as deep as you wanna, remember best value-result pair and then delete all other memory and go for next interval and dig deep.
Here's some pseudocode:
count = 0
for every item in a list
1/20 chance to add one to count
This is more or less my current code, but there could be hundreds of thousands of items in that list; therefore, it gets inefficient fast. (isn't this called like, 0(n) or something?)
Is there a way to compress this into one equation?
Let's look at the properties of the random variable you've described. Quoting Wikipedia:
The binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent yes/no experiments, each of which yields success with probability p.
Let N be the number of items in the list, and C be a random variable that represents the count you're obtaining from your pseudocode. C will follow a binomial probability distribution (as shown in the image below), with p = 1/20:
The remaining problem is how to efficently poll a random variable with said probability distribution. There are a number of libraries that allow you to draw samples from random variables with a specified PDF. I've never had to implement it myself, so I don't exactly know the details, but many are open source and you can refer to the implementation for yourself.
Here's how you would calculate count with the numpy library in Python:
n, p = 10, 0.05 # 10 trials, probability of success is 0.05
count = np.random.binomial(n, p) # draw a single sample
Apparently the OP was asking for a more efficient way to generate random numbers with the same distribution this will give. I though the question was how to do the exact same operation as the loop, but as a one liner (and preferably with no temporary list that exists just to be iterated over).
If you sample a random number generator n times, it's going to have at best O(n) run time, regardless of how the code looks.
In some interpreted languages, using more compact syntax might make a noticeable difference in the constant factors of run time. Other things can affect the run time, like whether you store all the random values and then process them, or process them on the fly with no temporary storage.
None of this will allow you to avoid having your run time scale up linearly with n.