In my class, for a final project, we are working on improving an algorithm that matches a prefix with a power of 2
(I.E. input="25", output="2^8=256", input="99", output="2^99=9903520314283042199192993792...")
Anyways, we are relying on logarithms to identify matching prefixes. Logarithm precision actually does matter and we are looking for better log functions. The standard log function and the calc-function both have the same precision. Are there any options if I wanted even better precision?
Based off several google searches, I've come to the conclusion that one currently does not exist. If someone where to inform me otherwise, I would happily change my best answer however.
Related
I am coding a spell-casting system where you draw a symbol with your wand (mouse), and it can recognize said symbol.
There are two methods I believe might work; neural networking and an "invisible grid system"
The problem with the neural networking system is that It would be (likely) suboptimal in Roblox Luau, and not be able to match the performance nor speed I wish for. (Although, I may just be lacking in neural networking knowledge. Please let me know whether I should continue to try implementing it this way)
For the invisible grid system, I thought of converting the drawing into 1s and 0s (1 = drawn, 0 = blank), then seeing if it is similar to one of the symbols. I create the symbols by making a dictionary like:
local Symbol = { -- "Answer Key" shape, looks like a tilted square
00100,
01010,
10001,
01010,
00100,
}
The problem is that user error will likely cause it to be inaccurate, like this "spell"'s blue boxes, showing user error/inaccuracy. I'm also sure that if I have multiple Symbols, comparing every value in every symbol will surely not be quick.
Do you know an algorithm that could help me do this? Or just some alternative way of doing this I am missing? Thank you for reading my post.
I'm sorry if the format on this is incorrect, this is my first stack-overflow post. I will gladly delete this post if it doesn't abide to one of the rules. ( Let me know if there are any tags I should add )
One possible approach to solving this problem is to use a template matching algorithm. In this approach, you would create a "template" for each symbol that you want to recognize, which would be a grid of 1s and 0s similar to what you described in your question. Then, when the user draws a symbol, you would convert their drawing into a grid of 1s and 0s in the same way.
Next, you would compare the user's drawing to each of the templates using a similarity metric, such as the sum of absolute differences (SAD) or normalized cross-correlation (NCC). The template with the lowest SAD or highest NCC value would be considered the "best match" for the user's drawing, and therefore the recognized symbol.
There are a few advantages to using this approach:
It is relatively simple to implement, compared to a neural network.
It is fast, since you only need to compare the user's drawing to a small number of templates.
It can tolerate some user error, since the templates can be designed to be tolerant of slight variations in the user's drawing.
There are also some potential disadvantages to consider:
It may not be as accurate as a neural network, especially for complex or highly variable symbols.
The templates must be carefully designed to be representative of the expected variations in the user's drawings, which can be time-consuming.
Overall, whether this approach is suitable for your use case will depend on the specific requirements of your spell-casting system, including the number and complexity of the symbols you want to recognize, the accuracy and speed you need, and the resources (e.g. time, compute power) that are available to you.
Matlab offers multiple algorithms for solving Linear Programs.
For example Matlab R2012b offers: 'active-set', 'trust-region-reflective', 'interior-point', 'interior-point-convex', 'levenberg-marquardt', 'trust-region-dogleg', 'lm-line-search', or 'sqp'.
But other versions of Matlab support different algorithms.
I would like to run a loop over all algorithms that are supported by the users Matlab-Version. And I would like them to be ordered like the recommendation order of Matlab.
I would like to implement something like this:
i=1;
x=[];
while (isempty(x))
options=optimset(options,'Algorithm',Here_I_need_a_list_of_Algorithms(i))
x = linprog(f,A,b,Aeq,beq,lb,ub,x0,options);
end
In 99% this code should be equivalent to
x = linprog(f,A,b,Aeq,beq,lb,ub,x0,options);
but sometimes the algorithm gives back an empty array because of numerical problems (exitflag -4). If there is a chance that one of the other algorithms can find a solution I would like to try them too.
So my question is:
Is there a possibility to automatically get a list of all linprog-algorithms that are supported by the installed Matlab-version ordered like Matlab recommends them.
I think looping through all algorithms can make sense in other scenarios too. For example when you need very precise data and have a lot of time, you could run them all and than evaluate which gives the best results.
Or one would like to loop through all algorithms, if one wants to find which algorithms is the best for LPs with a certain structure.
There's no automatic way to do this as far as I know. If you really want to do it, the easiest thing to do would be to go to the online documentation, and check through previous versions (online documentation is available for old versions, not just the most recent release), and construct some variables like this:
r2012balgos = {'active-set', 'trust-region-reflective', 'interior-point', 'interior-point-convex', 'levenberg-marquardt', 'trust-region-dogleg', 'lm-line-search', 'sqp'};
...
r2017aalgos = {...};
v = ver('matlab');
switch v.Release
case '(R2012b)'
algos = r2012balgos;
....
case '(R2017a)'
algos = r2017aalgos;
end
% loop through each of the algorithms
Seems boring, but it should only take you about 30 minutes.
There's a reason MathWorks aren't making this as easy as you might hope, though, because what you're asking for isn't a great idea.
It is possible to construct artificial problems where one algorithm finds a solution and the others don't. But in practice, typically if the recommended algorithm doesn't find a solution this doesn't indicate that you should switch algorithms, it indicates that your problem wasn't well-formulated, and you should consider modifying it, perhaps by modifying some constraints, or reformulating the objective function.
And after all, why stop with just looping through the alternative algorithms? Why not also loop through lots of values for other options such as constraint tolerances, optimality tolerances, maximum number of function evaluations, etc.? These may have just as much likelihood of affecting things as a choice of algorithm. And soon you're running an optimisation algorithm to search through the space of meta-parameters for your original optimisation.
That's not a great plan - probably better to just choose one of the recommended algorithms, stick to that, and if things don't work out then focus on improving your formulation of the problems rather than over-tweaking the optimisation itself.
I'm trying to write a basic digit counter (an integer is inputted and the number of digits of that integer is outputted) for positive integers. This is my general formula:
dig(x) := Math.floor(Math.log(x,10))
I tried implementing the equivalent of dig(x) in Ruby, and found that when I was computing dig(1000) I was getting 2 instead of 3 because Math.log was returning 2.9999999999999996 which would then be truncated down to 2. What is the proper way to handle this problem? (I'm assuming this problem can occur regardless of the language used to implement this approach, but if that's not the case then please explain that in your answer).
To get an exact count of the number of digits in an integer, you can do the usual thing: (in C/C++, assuming n is non-negative)
int digits = 0;
while (n > 0) {
n = n / 10; // integer division, just drops the ones digit and shifts right
digits = digits + 1;
}
I'm not certain but I suspect running a built-in logarithm function won't be faster than this, and this will give you an exact answer.
I thought about it for a minute and couldn't come up with a way to make the logarithm-based approach work with any guarantees, and almost convinced myself that it is probably a doomed pursuit in the first place because of floating point rounding errors, etc.
From The Art of Computer Programming volume 2, we will eliminate one bit of error before the floor function is applied by adding that one bit back in.
Let x be the result of log and then do x += x / 0x10000000 for a single precision floating point number (C's float). Then pass the value into floor.
This is guaranteed to be the fastest (assuming you have the answer in numerical form) because it uses only a few floating point instructions.
Floating point is always subject to roundoff error; that's one of the hazards you need to be aware of, and actively manage, when working with it. The proper way to handle it, if you must use floats is to figure out what the expected amount of accumulated error is and allow for that in comparisons and printouts -- round off appropriately, compare for whether the difference is within that range rather than comparing for equality, etcetera.
There is no exact binary-floating-point representation of simple things like 1/10th, for example.
(As others have noted, you could rewrite the problem to avoid using the floating-point-based solution entirely, but since you asked specifically about working log() I wanted to address that question; apologies if I'm off target. Some of the other answers provide specific suggestions for how you might round off the result. That would "solve" this particular case, but as your floating operations get more complicated you'll have to continue to allow for roundoff accumulating at each step and either deal with the error at each step or deal with the cumulative error -- the latter being the more complicated but more accurate solution.)
If this is a serious problem for an application, folks sometimes use scaled fixed point instead (running financial computations in terms of pennies rather than dollars, for example). Or they use one of the "big number" packages which computes in decimal rather than in binary; those have their own round-off problems, but they round off more the way humans expect them to.
so I have the following Integral that i need to do numerically:
Int[Exp(0.5*(aCosx + bSinx + cCos2x + dSin2x))] x=0..2Pi
The problem is that the output at any given value of x can be extremely large, e^2000, so larger than I can deal with in double precision.
I havn't had much luck googling for the following, how do you deal with large numbers in fortran, not high precision, i dont care if i know it to beyond double precision, and at the end i'll just be taking the log, but i just need to be able to handle the large numbers untill i can take the log..
Are there integration packes that have the ability to handle arbitrarily large numbers? Mathematica clearly can.. so there must be something like this out there.
Cheers
This is probably an extended comment rather than an answer but here goes anyway ...
As you've already observed Fortran isn't equipped, out of the box, with the facility for handling such large numbers as e^2000. I think you have 3 options.
Use mathematics to reduce your problem to one which does (or a number of related ones which do) fall within the numerical range that your Fortran compiler can compute.
Use Mathematica or one of the other computer algebra systems (eg Maple, SAGE, Maxima). All (I think) of these can be integrated into a Fortran program (with varying degrees of difficulty and integration).
Use a library for high-precision (often called either arbitray-precision or multiple-precision too) arithmetic. Your favourite search engine will turn up a number of these for you, some written in Fortran (and therefore easy to integrate), some written in C/C++ or other languages (and therefore slightly harder to integrate). You might start your search at Lawrence Berkeley or the GNU bignum library.
(Yes I know that I wrote that you have 3 options, but your question suggests that you aren't ready to consider this yet) You could write your own high-/arbitrary-/multiple-precision functions. Fortran provides everything you need to construct such a library, there is a lot of work already done in the field to learn from, and it might be something of interest to you.
In practice it generally makes sense to apply as much mathematics as possible to a problem before resorting to a computer, that process can not only assist in solving the problem but guide your selection or construction of a program to solve what's left of the problem.
I agree with High Peformance Mark that the best option here numerically is to use analytics to scale or simplify the result first.
I will mention that if you do want to brute force it, gfortran (as of 4.6, with the libquadmath library) has support for quadruple precision reals, which you can use by selecting the appropriate kind. As long as your answers (and the intermediate results!) don't get too much bigger than what you're describing, that may work, but it will generally be much slower than double precision.
This requires looking deeper at the problem you are trying to solve and the behavior of the underlying mathematics. To add to the good advice already provided by Mark and Jonathan, consider expanding the exponential and trig functions into Taylor series and truncating to the desired level of precision.
Also, take a step back and ask why you are trying to accomplish by calculating this value. As an example, I recently had to debug why I was getting outlandish results from a property correlation which was calculating vapor pressure of a fluid to see if condensation was occurring. I spent a long time trying to understand what was wrong with the temperature being fed into the correlation until I realized the case causing the error was a simulation of vapor detonation. The problem was not in the numerics but in the logic of checking for condensation during a literal explosion; physically, a condensation check made no sense. The real problem was the code was asking an unnecessary question; it already had the answer.
I highly recommend Forman Acton's Numerical Methods That (Usually) Work and Real Computing Made Real. Both focus on problems like this and suggest techniques to tame ill-mannered computations.
I'm currently writing an optimization algorithm in MATLAB, at which I completely suck, therefore I could really use your help. I'm really struggling to find a good way of representing a graph (or well more like a tree with several roots) which would look more or less like this:
alt text http://img100.imageshack.us/img100/3232/graphe.png
Basically 11/12/13 are our roots (stage 0), 2x is stage1, 3x stage2 and 4x stage3. As you can see nodes from stageX are only connected to several nodes from stage(X+1) (so they don't have to be connected to all of them).
Important: each node has to hold several values (at least 3-4), one will be it's number and at least two other variables (which will be used to optimize the decisions).
I do have a simple representation using matrices but it's really hard to maintain, so I was wondering is there a good way to do it?
Second question: when I'm done with that representation I need to calculate how good each route (from roots to the end) is (like let's say I need to compare is 11-21-31-41 the best or is 11-21-31-42 better) to do that I will be using the variables that each node holds. But the values will have to be calculated recursively, let's say we start at 11 but to calcultate how good 11-21-31-41 is we first need to go to 41, do some calculations, go to 31, do some calculations, go to 21 do some calculations and then we can calculate 11 using all the previous calculations. Same with 11-21-31-42 (we start with 42 then 31->21->11). I need to check all the possible routes that way. And here's the question, how to do it? Maybe a BFS/DFS? But I'm not quite sure how to store all the results.
Those are some lengthy questions, but I hope I'm not asking you for doing my homework (as I got all the algorithms, it's just that I'm not really good at matlab and my teacher wouldn't let me to do it in java).
Granted, it may not be the most efficient solution, but if you have access to Matlab 2008+, you can define a node class to represent your graph.
The Matlab documentation has a nice example on linked lists, which you can use as a template.
Basically, a node would have a property 'linksTo', which points to the index of the node it links to, and a method to calculate the cost of each of the links (possibly with some additional property that describe each link). Then, all you need is a function that moves down each link, and brings the cost(s) with it when it moves back up.