So while playing with the game framework Love2D lua engine I noticed the following formula inside HUMPS vector code (https://github.com/vrld/hump/blob/master/vector.lua):
return atan2(self.y, self.x) - atan2(other.y, other.x)
Note: atan2 is described here https://en.wikipedia.org/wiki/Atan2
:basically it is a safe atan function that protects against divide by zero and some other atan pitfalls - to my understanding.
I've been drawing some graphs of two separate vectors and trying to figure out how someone would come up with a formula like this. The context for this formula, for me at least, is rotating an image (image_pos) to where the mouse has clicked (mouse_pos).
Can anyone please ELI5 how this formula was derived/how it works?
As you can see in the picture it is the difference of angles between two vectors.
atan2(self.y, self.x) == \beta
atan2(other.y, other.x) == \alpha
Hence:
atan2(self.y, self.x) - atan2(other.y, other.x) == \theta
Related
For a game that I am getting started making I have had to learn vector math to calculate forces. To convert a vector from x and y to magnitude and angle I have read that I for the angle have to use the function tan^-1(y/x). Is this correct and if so how do I implement it into godots GDscript?
For a game that I am getting started making I have had to learn vector math to calculate forces.
First, this isn't necessarily true. Depending on your use case, it may be most effective to use Rigid Bodies and let Godot compute the effects of forces for you.
To convert a vector from x and y to magnitude and angle I have read that I for the angle have to use the function tan^-1(y/x).
To convert a vector from x and y to magnitude and angle, use Vector2.length() and Vector2.angle().
That being said, if you want to learn vector math, do it! Godot has its own vector math docs, and I'm sure you can find plenty of other similar lessons online.
However, it is good to be aware that the engine provides much of this functionality for you.
Writing a bunch of vector math instead of just calling a bultin function will just make your code more complicated.
I have a function that takes a floating point number and returns a floating point number. It can be assumed that if you were to graph the output of this function it would be 'n' shaped, ie. there would be a single maximum point, and no other points on the function with a zero slope. We also know that input value that yields this maximum output will lie between two known points, perhaps 0.0 and 1.0.
I need to efficiently find the input value that yields the maximum output value to some degree of approximation, without doing an exhaustive search.
I'm looking for something similar to Newton's Method which finds the roots of a function, but since my function is opaque I can't get its derivative.
I would like to down-thumb all the other answers so far, for various reasons, but I won't.
An excellent and efficient method for minimizing (or maximizing) smooth functions when derivatives are not available is parabolic interpolation. It is common to write the algorithm so it temporarily switches to the golden-section search (Brent's minimizer) when parabolic interpolation does not progress as fast as golden-section would.
I wrote such an algorithm in C++. Any offers?
UPDATE: There is a C version of the Brent minimizer in GSL. The archives are here: ftp://ftp.club.cc.cmu.edu/gnu/gsl/ Note that it will be covered by some flavor of GNU "copyleft."
As I write this, the latest-and-greatest appears to be gsl-1.14.tar.gz. The minimizer is located in the file gsl-1.14/min/brent.c. It appears to have termination criteria similar to what I implemented. I have not studied how it decides to switch to golden section, but for the OP, that is probably moot.
UPDATE 2: I googled up a public domain java version, translated from FORTRAN. I cannot vouch for its quality. http://www1.fpl.fs.fed.us/Fmin.java I notice that the hard-coded machine efficiency ("machine precision" in the comments) is 1/2 the value for a typical PC today. Change the value of eps to 2.22045e-16.
Edit 2: The method described in Jive Dadson is a better way to go about this. I'm leaving my answer up since it's easier to implement, if speed isn't too much of an issue.
Use a form of binary search, combined with numeric derivative approximations.
Given the interval [a, b], let x = (a + b) /2
Let epsilon be something very small.
Is (f(x + epsilon) - f(x)) positive? If yes, the function is still growing at x, so you recursively search the interval [x, b]
Otherwise, search the interval [a, x].
There might be a problem if the max lies between x and x + epsilon, but you might give this a try.
Edit: The advantage to this approach is that it exploits the known properties of the function in question. That is, I assumed by "n"-shaped, you meant, increasing-max-decreasing. Here's some Python code I wrote to test the algorithm:
def f(x):
return -x * (x - 1.0)
def findMax(function, a, b, maxSlope):
x = (a + b) / 2.0
e = 0.0001
slope = (function(x + e) - function(x)) / e
if abs(slope) < maxSlope:
return x
if slope > 0:
return findMax(function, x, b, maxSlope)
else:
return findMax(function, a, x, maxSlope)
Typing findMax(f, 0, 3, 0.01) should return 0.504, as desired.
For optimizing a concave function, which is the type of function you are talking about, without evaluating the derivative I would use the secant method.
Given the two initial values x[0]=0.0 and x[1]=1.0 I would proceed to compute the next approximations as:
def next_x(x, xprev):
return x - f(x) * (x - xprev) / (f(x) - f(xprev))
and thus compute x[2], x[3], ... until the change in x becomes small enough.
Edit: As Jive explains, this solution is for root finding which is not the question posed. For optimization the proper solution is the Brent minimizer as explained in his answer.
The Levenberg-Marquardt algorithm is a Newton's method like optimizer. It has a C/C++ implementation levmar that doesn't require you to define the derivative function. Instead it will evaluate the objective function in the current neighborhood to move to the maximum.
BTW: this website appears to be updated since I last visited it, hope it's even the same one I remembered. Apparently it now also support other languages.
Given that it's only a function of a single variable and has one extremum in the interval, you don't really need Newton's method. Some sort of line search algorithm should suffice. This wikipedia article is actually not a bad starting point, if short on details. Note in particular that you could just use the method described under "direct search", starting with the end points of your interval as your two points.
I'm not sure if you'd consider that an "exhaustive search", but it should actually be pretty fast I think for this sort of function (that is, a continuous, smooth function with only one local extremum in the given interval).
You could reduce it to a simple linear fit on the delta's, finding the place where it crosses the x axis. Linear fit can be done very quickly.
Or just take 3 points (left/top/right) and fix the parabola.
It depends mostly on the nature of the underlying relation between x and y, I think.
edit this is in case you have an array of values like the question's title states. When you have a function take Newton-Raphson.
Someone somewhere has had to solve this problem. I can find many a great website explaining this problem and how to solve it. While I'm sure they are well written and make sense to math whizzes, that isn't me. And while I might understand in a vague sort of way, I do not understand how to turn that math into a function that I can use.
So I beg of you, if you have a function that can do this, in any language, (sure even fortran or heck 6502 assembler) - please help me out.
prefer an analytical to iterative solution
EDIT: Meant to specify that its a cubic bezier I'm trying to work with.
What you're asking for is the inverse of the arc length function. So, given a curve B, you want a function Linv(len) that returns a t between 0 and 1 such that the arc length of the curve between 0 and t is len.
If you had this function your problem is really easy to solve. Let B(0) be the first point. To find the next point, you'd simply compute B(Linv(w)) where w is the "equal arclength" that you refer to. To get the next point, just evaluate B(Linv(2*w)) and so on, until Linv(n*w) becomes greater than 1.
I've had to deal with this problem recently. I've come up with, or come across a few solutions, none of which are satisfactory to me (but maybe they will be for you).
Now, this is a bit complicated, so let me just give you the link to the source code first:
http://icedtea.classpath.org/~dlila/webrevs/perfWebrev/webrev/raw_files/new/src/share/classes/sun/java2d/pisces/Dasher.java. What you want is in the LengthIterator class. You shouldn't have to look at any other parts of the file. There are a bunch of methods that are defined in another file. To get to them just cut out everything from /raw_files/ to the end of the URL. This is how you use it. Initialize the object on a curve. Then to get the parameter of a point with arc length L from the beginning of the curve just call next(L) (to get the actual point just evaluate your curve at this parameter, using deCasteljau's algorithm, or zneak's suggestion). Every subsequent call of next(x) moves you a distance of x along the curve compared to your last position. next returns a negative number when you run out of curve.
Explanation of code: so, I needed a t value such that B(0) to B(t) would have length LEN (where LEN is known). I simply flattened the curve. So, just subdivide the curve recursively until each curve is close enough to a line (you can test for this by comparing the length of the control polygon to the length of the line joining the end points). You can compute the length of this sub-curve as (controlPolyLength + endPointsSegmentLen)/2. Add all these lengths to an accumulator, and stop the recursion when the accumulator value is >= LEN. Now, call the last subcurve C and let [t0, t1] be its domain. You know that the t you want is t0 <= t < t1, and you know the length from B(0) to B(t0) - call this value L0t0. So, now you need to find a t such that C(0) to C(t) has length LEN-L0t0. This is exactly the problem we started with, but on a smaller scale. We could use recursion, but that would be horribly slow, so instead we just use the fact that C is a very flat curve. We pretend C is a line, and compute the point at t using P=C(0)+((LEN-L0t0)/length(C))*(C(1)-C(0)). This point doesn't actually lie on the curve because it is on the line C(0)->C(1), but it's very close to the point we want. So, we just solve Bx(t)=Px and By(t)=Py. This is just finding cubic roots, which has a closed source solution, but I just used Newton's method. Now we have the t we want, and we can just compute C(t), which is the actual point.
I should mention that a few months ago I skimmed through a paper that had another solution to this that found an approximation to the natural parameterization of the curve. The author has posted a link to it here: Equidistant points across Bezier curves
In C the atan2 function has the following signature:
double atan2( double y, double x );
Other languages do this as well. This is the only function I know of that takes its arguments in Y,X order rather than X,Y order, and it screws me up regularly because when I think coordinates, I think (X,Y).
Does anyone know why atan2's argument order convention is this way?
Because I believe it is related to arctan(y/x), so y appears on top.
Here's a nice link talking about it a bit: Angles and Directions
My assumption has always been that this is because of the trig definition, ie that
tan(theta) = opposite / adjacent
When working with the canonical angle from the origin, opposite is always Y and adjacent is always X, so:
atan2(opposite, adjacent) = theta
Ie, it was done that way so there's no ordering confusion with respect to the mathematical definition.
Suppose a rectangle triangle with its opposite side called y, adjacent side called x:
tan(angle) = y/x
arctan(tan(angle)) = arctan(y/x)
It's because in school, the mnemonic for calculating the gradient
is rise over run, or in other words dy/dx, or more briefly y/x.
And this order has snuck into the arguments of arctangent functions.
So it's a historical artefact. For me it depends on what I'm thinking
about when I use atan2. If I'm thinking about differentials, I get it right
and if I'm thinking about coordinate pairs, I get it wrong.
The order is atan2(X,Y) in excel so I think the reverse order is a programming thing. atan(Y/X) can easily be changed to atan2(Y,X) by putting a '2' between the 'n' and the '(', and replacing the '/' with a ',', only 2 operations. The opposite order would take 4 operations and some of the operations would be more complex (cut and paste).
I often work out my math in Excel then port it to .NET, so will get hung up on atan2 sometimes. It would be best if atan2 could be standardized one way or the other.
It would be more convenient if atan2 had its arguments reversed. Then you wouldn't need to worry about flipping the arguments when computing polar angles. The Mathematica equivalent does just that: https://reference.wolfram.com/language/ref/ArcTan.html
Way back in the dawn of time, FORTRAN had an ATAN2 function with the less convenient argument order that, in this reference manual, is (somewhat inaccurately) described as arctan(arg1 / arg2).
It is plausible that the initial creator was fixated on atan2(arg1, arg2) being (more or less) arctan(arg1 / arg2), and that the decision was blindly copied from FORTRAN to C to C++ and Python and Java and JavaScript.
What is the usual method or algorithm used to plot implicit equations of 2 variables?
I am talking about equations such as,
sin(x*y)*y = 20
x*x - y*y = 1
Etc.
Does anyone know how Maple or Matlab do this? My target language is C#.
Many thanks!
One way to do this is to sample the function on a regular, 2D grid. Then you can run an algorithm like marching squares on the resulting 2D grid to draw iso-contours.
In a related question, someone also linked to the gnuplot source code. It's fairly complex, but might be worth going through. You can find it here: http://www.gnuplot.info/
Iterate the value of x across the range you want to plot. For each fixed value of x, solve the equation numerically using a method such as interval bisection or the Newton-Raphson method (for which you can calculate the derivative using implicit differentiation, or perhaps differentiate numerically). This will give you the corresponding value of y for a given x. In most cases, you won't need too many iterations to get a very precise result, and it's very efficient anyway.
Note that you will need to transform the equation into the form f(x) = 0, though this is always trivial. The nice thing about this method is that it works just as well the other way round (i.e. taking a fixed range of y and computing x per value).
There're multiple methods. The easiest algorithm I could find is descripted here:
https://homepages.warwick.ac.uk/staff/David.Tall/pdfs/dot1986b-implicit-fns.pdf and describes what Noldorin has described you.
The most complex one, and seems to be the one that can actually solve a lot of special cases is described here:
https://academic.oup.com/comjnl/article/33/5/402/480353
i think,
in matlab you give array as input for x.
then for every x, it calculates y.
then draws line from x0,y0 to x1, y1
then draws line from x1,y1 to x2, y2
...
...