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f = #(w) test(w, phis, t, r_coeff);
function test(w, phis, t, r_coeff)
M = size(phis, 2);
expt = exp(-t .* (phis * w'));
coeff = expt .* t .^ 2 ./ (1 + expt) .^ 2;
averaging_coef = 1.0 / M; % mean replace
G = bsxfun(#times, phis', coeff' * averaging_coef) * phis + 2 * r_coeff * eye(M);
end
w - size is (1xM)
phis - size is (NxM)
t - size is (Nx1)
r_coeff is const
Help me please to optimize this piece of code, because function f runs a thousand times and N is around 300-800, M have mostly the same size. When I'm doing this * phis performance goes down.
As you can see it depends only on w - which I don't know.
It already is fairly well "optimized". Just because you want it to run faster does not make that possible.
You can buy/find/borrow/lease a faster/bigger computer. You can choose to solve smaller problems. You can just let it run overnight. Or you can change your problem to be something simpler and approximate, that runs more quickly.
This is what happens in research problems. They expand to the limits of your capability and just a bit more, because solving something easy is not worth a paper or a thesis. Computers make it easy to gather much data too, so problems get surprisingly large very quickly.
A special, rare skill in mathematics and modeling is knowing how to simplify your problem, deleting terms that are not truly important, while retaining the same basic behavior that you wish to study. This often involves linear approximations to nonlinear terms.
Related
I've looked and tried but i cant find anything really helpful so thank you in advance.
My problem is i have a changing variable, "balance" for the moment i have it represented as 200. I need to use this equation to find how much money i should withdraw in a game, but I don't know how to write a LUA script that solves algebra
The equation is: 200/(x+x^2+x^3+x^4+x^5)=0.00001001 how would i set about solving for x?
I have tried adding .0000001 if 200/(x+x^2+x^3+x^4+x^5) doesn't equal 0.00001001 but it is very impractical and I haven't gotten it to work. This is The only way I can come up with at the moment. Any help would be appreciated.
This solution finds zero of any continuous function (not only algebraical and not only differentiable) and requires knowing the diapazone of the root to be found.
local function find_zero(f, x_left, x_right, eps)
eps = eps or 0.0000000001 -- precision
local f_left, f_right = f(x_left), f(x_right)
assert(x_left <= x_right and f_left * f_right <= 0, "Wrong diapazone")
while x_right - x_left > eps do
local x_middle = (x_left + x_right) / 2
local f_middle = f(x_middle)
if f_middle * f_left > 0 then
x_left, f_left = x_middle, f_middle
else
x_right, f_right = x_middle, f_middle
end
end
return (x_left + x_right) / 2
end
local function my_func(x)
return 200/(x+x^2+x^3+x^4+x^5) - 0.00001001
end
-- Assuming that the root is between 1 and 1000
local x = find_zero(my_func, 1.0, 1000.0)
print(x) --> 28.643931367544
200/(x+x^2+x^3+x^4+x^5)=0.00001001 is equivalent to 200 = 0.00001001 * (x+x^2+x^3+x^4+x^5), so you have a polynomial equation to solve, and traditionally it is this form of the equation that people like to deal with.
If you want to stay in Lua, then if the form of the equation is predictable enough that you can find a place where the right side is always less than the left (e.g. x = 0) and a place where the right sight is always greater than the left (e.g. very large values of x) then you can use binary search - not terribly efficient, but certain and easy to code.
For general polynomial equations, one well known method is https://en.wikipedia.org/wiki/Newton's_method. Given f(x) = 0 and a guess for x, a better guess might be x - f(x) / f'(x), where f'(x) is the derivative of f(x). There are a few pathological cases where this fails for various reasons, though, so again you probably want to know that your equations is reliably tractable.
Since you have Lua, you may be able to bring in C code that calls out to a maths library such as http://commons.apache.org/proper/commons-math/. They have a routine called LaguerreSolver() which will reasonably reliably solve polynomial equations for you, defending itself against all of the pathological cases. Most math libraries contain a lot more work than any single person is likely to put in for an individual problem, and are of correspondingly higher quality than do it yourself approach such as I describe above.
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I have a math problem that I can't solve: I don't know how to find the value of n so that
365! / ((365-n)! * 365^n) = 50%.
I am using the Casio 500ms scientific calculator but I don't know how.
Sorry because my question is too easy, I am changing my career so I have to review and upgrade my math, the subject that I have neglected for years.
One COULD in theory use a root-finding scheme like Newton's method, IF you could take derivatives. But this function is defined only on the integers, since it uses factorials.
One way out is to recognize the identity
n! = gamma(n+1)
which will effectively allow you to extend the function onto the real line. The gamma function is defined on the positive real line, though it does have singularities at the negative integers. And of course, you still need the derivative of this expression, which can be done since gamma is differentiable.
By the way, a danger with methods like Newton's method on problems like this is it may still diverge into the negative real line. Choose poor starting values, and you may get garbage out. (I've not looked carefully at the shape of this function, so I won't claim for what set of starting values it will diverge on you.)
Is it worth jumping through the above set of hoops? Of course not. A better choice than Newton's method might be something like Brent's algorithm, or a secant method, which here will not require you to compute the derivative. But even that is a waste of effort.
Recognizing that this is indeed a problem on the integers, one could use a tool like bisection to resolve the solution extremely efficiently. It never requires derivatives, and it will work nicely enough on the integers. Once you have resolved the interval to be as short as possible, the algorithm will terminate, and take vary few function evaluations in the process.
Finally, be careful with this function, as it does involve some rather large factorials, which could easily overflow many tools to evaluate the factorial. For example, in MATLAB, if I did try to evaluate factorial(365):
factorial(365)
ans =
Inf
I get an overflow. I would need to move into a tool like the symbolic toolbox, or my own suite of variable precision integer tools. Alternatively, one could recognize that many of the terms in these factorials will cancel out, so that
365! / (365 - n)! = 365*(365-1)*(365-2)*...*(365-n+1)
The point is, we get an overflow for such a large value if we are not careful. If you have a tool that will not overflow, then use it, and use bisection as I suggested. Here, using the symbolic toolbox in MATLAB, I get a solution using only 7 function evaluations.
f = #(n) vpa(factorial(sym(365))/(factorial(sym(365 - n))*365^sym(n)));
f(0)
ans =
1.0
f(365)
ans =
1.4549552156187034033714015903853e-157
f(182)
ans =
0.00000000000000000000000095339164972764493041114884521295
f(91)
ans =
0.000004634800180846641815683109605743
f(45)
ans =
0.059024100534225072005461014516788
f(22)
ans =
0.52430469233744993108665513602619
f(23)
ans =
0.49270276567601459277458277166297
Or, if you can't take an option like that, but do have a tool that can evaluate the log of the gamma function, AND you have a rootfinder available as MATLAB does...
f = #(n) exp(gammaln(365+1) - gammaln(365-n + 1) - n*log(365));
fzero(#(n) f(n) - .5,10)
ans =
22.7677
As you can see here, I used the identity relating gamma and the factorial function, then used the log of the gamma function, in MATLAB, gammaln. Once all the dirty work was done, then I exponentiated the entire mess, which will be a reasonable number. Fzero tells us that the cross-over occurs between 22 and 23.
If a numerical approximation is ok, ask Wolfram Alpha:
n ~= -22.2298272...
n ~= 22.7676903...
I'm going to assume you have some special reason for wanting an actual algorithm, even though you only have one specific problem to solve.
You're looking for a value n where...
365! / ((365-n)! * 365^n) = 0.5
And therefore...
(365! / ((365-n)! * 365^n)) - 0.5 = 0.0
The general form of the problem is to find a value x such that f(x)=0. One classic algorithm for this kind of thing is the Newton-Raphson method.
[EDIT - as woodchips points out in the comment, the factorial is an integer-only function. My defence - for some problems (the birthday problem among them) it's common to generalise using approximation functions. I remember the Stirling approximation of factorials being used for the birthday problem - according to this, Knuth uses it. The Wikipedia page for the Birthday problem mentions several approximations that generalise to non-integer values.
It's certainly bad that I didn't think to mention this when I first wrote this answer.]
One problem with that is that you need the derivative of that function. That's more a mathematics issue, though you can estimate the derivative at any point by taking values a short distance either side.
You can also look at this as an optimisation problem. The general form of optimisation problems is to find a value x such that f(x) is maximised/minimised. In your case, you could define your function as...
f(x)=((365! / ((365-n)! * 365^n)) - 0.5)^2
Because of the squaring, the result can never be negative, so try to minimise. Whatever value of x gets you the smallest f(x) will also give you the result you want.
There isn't so much an algorithm for optimisation problems as a whole field - the method you use depends on the complexity of your function. However, this case should be simple so long as your language can cope with big numbers. Probably the simplest optimisation algorithm is called hill-climbing, though in this case it should probably be called rolling-down-the-hill. And as luck would have it, Newton-Raphson is a hill-climbing method (or very close to being one - there may be some small technicality that I don't remember).
[EDIT as mentioned above, this won't work if you need an integer solution for the problem as actually stated (rather than a real-valued approximation). Optimisation in the integer domain is one of those awkward issues that helps make optimisation a field in itself. The branch and bound is common for complex functions. However, in this case hill-climbing still works. In principle, you can even still use a tweaked version of Newton-Raphson - you just have to do some rounding and check that you don't keep rounding back to the same place you started if your moves are small.]
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Closed 9 years ago.
My problem is that I am trying to decrypt following cipher text:
AWGNEYDTBSREAHGIDEIEEIHKIWABHPRCROIWSSBHTE
OCSNASLAIGTTOTYIOBNANNOOCHENEHSADUAIEOSNTH
TEOOAWUTOVTISNTVIFIHSNADVOGOREAPTBETSIT
DCSNASAAHOOULITUREICBNOERSNETTNOITTNTSDEOO
RERNINTOVTINTEFKGITPRCROEEIGREIGREHKFN
HOCWHPDTOSREGTOUREINDIUATHCHIEOSN
I know that it is English language. I know that it was encrypted only once by substitution, monoalphabetic or transposition.
I checked the Index of Coincidence and I received: 0.06833754056978002
I also checked frequency of letter and I received:
T 11.02, E 10.59, O 10.16, I 9.74, N 8.47, S 6.77, A 6.35, R 5.93, H 5.93, G 3.38, D 3.38, C 3.38, U 2.54, B 2.54, W 2.11, V 1.69, P 1.69, K 1.27, F 1.27, Y 0.84, L 0.84.
So I claim that it is transposition cipher. Am I right?
I checked all combinations of Rail Fence Transposition. I did not receive something sensible.
I checked all combinations of Columnar Transposition without changing order. I did not receive something sensible.
I checked a few combinations of Columnar Transposition with changing order. I did not receive something sensible.
Do you have idea how I can resolve this interesting task?
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Using continued fractions, I'm generating integer ratios between frequencies to a certain precision, along with the error (difference from integer ratio to the real ratio). So I end up with things like:
101 Hz with 200 Hz = 1:2 + 0.0005
61 Hz with 92 Hz = 2:3 - 0.0036
However, I've run into a snag on actually deciding which of these will be more dissonant than others. At first I thought low numbers = better, but something like 1:51 would likely be not very dissonant since it's a frequency up 51 octaves from the other. It might be a screaming high, ear bleeding pitch, but I don't think it would have dissonance.
It seems to me that it must be related to the product of the two sides of the ratio compared to the constituents somehow. 1 * 51 = 51, which doesn't "go up much" from one side. 2 * 3 = 6, which I would think would indicate higher dissonance than 1:51. But I need to turn this feeling into an actual number, so I can compare 5:7 vs 3:8, or any other combinations.
And how could I work error into this? Certainly 1:2 + 0 would be less dissonant than 1:2 + 1. Would it be easier to apply an algorithm that works for the above integer ratios directly to the frequencies themselves? Or does having the integer ratio with an error allow for a simpler calculation?
edit: Thinking on it, an algorithm that could extend to any set of N frequencies in a chord would be awesome, but I get the feeling that would be much more difficult...
edit 2: Clarification:
Let's consider that I am dealing with pure sine waves, and either ignoring the specific thresholds of the human ear or abstracting them into variables. If there are severe complications, then they are ignored. My question is how it could be represented in an algorithm, in that case.
Have a look at Chapter 4 of http://homepages.abdn.ac.uk/mth192/pages/html/maths-music.html. From memory:
1) If two sine waves are just close enough for the human ear to be confused, but not so close that the human ear cannot tell they are different, there will be dissonance.
2) Pure sine waves are extremely rare - most tones have all sorts of harmonics. Dissonance is very likely to occur from colliding harmonics, rather than colliding main tones - to sort of follow your example, two tones many octaves apart are unlikely to be dissonant because their harmonics may not meet, whereas with just a couple of octaves different and loads of harmonics a flute could sound out of tune with a double bass. Therefore dissonance or not depends not only on the frequencies of the main tones, but on the harmonics present, and this has been experimentally demonstrated by constructing sounds with peculiar pseudo-harmonics.
The answer is in Chapter 4 of Music: a Mathematical Offering. In particular, see the following two figures:
consonance / dissonance plotted against the x critical bandwidth in 4.3. History of consonance and dissonance
dissonance vs. frequency in 4.5. Complex tones
Of course you still have to find a nice way to turn these data into a formula / program that gives you a measure of dissonance but I believe this gives you a good start. Good luck!
This will help:
http://www.acs.psu.edu/drussell/demos/superposition/superposition.html
You want to look at superposition.
Discrete or Fast Fourier Transform is the most generic means to get what you're asking for.
[Environment: MATLAB 64 bit, Windows 7, Intel I5-2320]
I would like to RMS-fit a function to experimental data y, so I am minimizing the following function (by using fminsearch):
minfunc = rms(y - fitfunc)
From the general point of view, does it make sense to minimize:
minfunc = sum((y - fitfunc) .^ 2)
instead and then (after minimization) just do minfunc = sqrt(minfunc / N) to get the fit RMS error?
To reformulate the question, how much time (roughly, in percent) would fminsearch save by not doing sqrt and 1/(N - 1) each time? I wouldn't like to decrease readability of my code if my CPU / MATLAB are so fast that it wouldn't improve performance by at least a percent.
Update: I've tried simple tests, but the results are not clear: depending on the actual value of the minfunc, fminsearch takes more or less time.
The general answer for performance questions:
If you just want to figure out what is faster, design a benchmark and run it a few times.
By just providing general information it is not likely that you will determine which method is 1 percent faster.