I am trying to solve an integral with this form (where a, b, c, d, e, f, and g are constants):
integral from f to g( (ax^4 + bx^3 + cx^2 + dx + e)^0.5dx )
I've come across this when trying to find the length of a Cubic Bezier curve (without using recursive subdivision). My goal is to have a symbolic equation that can find the length of a Bezier between any range of t (i.e f and g in the equation above).
My first guess was to complete the square, but Wikipedia was clear that that only works with quadratic equations.
Integration by parts... doesn't seem like it'd work for this.
The next question (though, not the subject of this topic), is using that really faster than recursively subdividing the Bezier?
Just in case anyone else finds this question, there is no closed form solution, but there are very efficient ways to find the length without doing recursive subdivision, such as described on http://pomax.github.com/bezierinfo/#arclength
Related
There is a theory that says six degrees of seperations is the highest
degree for people to be connected through a chain of acquaintances.
(You know the Baker - Degree of seperation 1, the Baker knows someone
you don't know - Degree of separation 2)
We have a list of People P, list A of corresponding acquaintances
among these people, and a person x
We are trying to implement an algorithm to check if person x respects
the six degrees of separations. It returns true if the distance from x
to all other people in P is at most six, false otherwise.
We are tying to accomplish O(|P| + |A|) in the worst-case.
To implement this algorithm, I thought about implementing an adjacency list over an adjacency matrix to represent the Graph G with vertices P and edges A, because an Adjacency Matrix would take O(n^2) to traverse.
Now I thought about using either BFS or DFS, but I can't seem to find a reason as to why the other is more optimal for this case.
I want to use BFS or DFS to store the distances from x in an array d, and then loop over the array d to look if any Degree is larger than 6.
DFS and BFS have the same Time Complexity, but Depth is better(faster?) in most cases at finding the first Degree larger than 6, whereas Breadth is better at excluding all Degrees > 6 simultaneously.
After DFS or BFS I would then loop over the array containing the distances from person x, and return true if there was no entry >6 and false when one is found.
With BFS, the degrees of separations would always be at the end of the Array, which would maybe lead to a higher time complexity?
With DFS, the degrees of separations would be randomly scattered in the Array, but the chance to have a degree of separation higher than 6 early in the search is higher.
I don't know if it makes any difference to the Time Complexity if using DFS or BFS here.
Time complexity of BFS and DFS is exactly the same. Both methods visit all connected vertices of the graph, so in both cases you have O(V + E), where V is the number of vertices and E is the number of edges.
That being said, sometimes one algorithm can be preferred over the other precisely because the order of vertex visitation is different. For instance, if you were to evaluate a mathematical expression, DFS would be much more convenient.
In your case, BFS could be used to optimize graph traversal, because you can simply cut-off BFS at the required degree of separation level. All the people who have the required (or bigger) degree of separation would be left unmarked as visited.
The same trick would be much more convoluted to implement with DFS, because as you've astutely noticed, DFS first gets "to the bottom" of the graph, and then it goes back recursively (or via stack) up level by level.
I believe that you can use the the Dijkstra algorithm.
Is a BFS approach that updates your path, is the path have a smaller value. Think as distance have always a cost of 1 and, if you have two friends (A and B) for a person N.
Those friends have a common friend C but, in a first time your algorithm checks a distance for friend A with cost 4 and mark as visited, they can't check the friend B that maybe have a distance of 3. The Dijkstra will help you doing checking this.
The Dijkstra solve this in O(|V|+|E|log|V)
See more at https://en.wikipedia.org/wiki/Dijkstra%27s_algorithm
Given a set points P in the plane, and given a threshold t, I'd like to compute a connected graph G to minimize the sum of the lengths of its edges, subject to the following constraints:
The vertices of G contain all the points in P.
For every pair of points u and v in P, their distance in G is no greater than t times their Euclidean distance.
When t=1, this problem is solved by constructing a complete graph on P. When t is infinite (or simply large enough), this problem is the Euclidean Steiner Tree Problem.
If there already a name for this problem, I'm curious what it is. More than that, does anyone have any suggestions for how to make an algorithm for this? Since it contains the Euclidean Steiner Tree Problem as a special case, it can't be simpler, so I'm not looking for anything particularly time efficient. Thanks!
I would like to ask how I can define / declare a variable in Scilab. In some PDFs that I read, it says that I can just type it in and Scilab will take care of the declaration. Not so. I want to set up a matrix equation of something like:
Ax + By + Cz = D
Mx + Ny + Pz = E
Rx + Sy + Tz = F
And then I want to get the general value of x, y, x in terms of A, B, C, D, E, F, M, N, P, R, S, T. I remember this is possible with Matlab. And later on, I want to plug in these values to get actual numbers. Please help.
Scilab is much more oriented at numerical computation than algebra solving, but you can still do it.
In your case you first should define the system in the form M1*x=M2, being M1 upper triangular.
I suggest you look at help for solve() and trianfml(), there are nice examples.
After that you can evaluate the expressions giving any value you want for A, B, C, ..., using evstr()
For symbolic algebra, I recommend Wolfram mathematica, Maple, or Maxima (this last one is open-source like Scilab)
OK, this is what I found. SciLab requires "symbolic math toolbox" in order to do symbolic math. the scimax/overload toolbox (by Calixte Denizet) can do that by integrating Maxima with SciLab. however, it is only available on Linux/Unix OS. another way to do it is the OVLD/SYM toolbox (by the deceased Jean-François Magni) which works with Windows (even Win 7). however, support for this toolbox has ceased due to its author's demise. the installation guide on spoken-tutorial.org no longer exists. thus, I am left with using Maxima by itself to solve symbolic equations and calculus problems.
I'm trying to figure out the most efficient/fast way to add a large number of convex quads (four given x,y points) into an array/list and then to check against those quads if a point is within or on the border of those quads.
I originally tried using ray casting but thought that it was a little overkill since I know that all my polygons will be quads and that they are also all convex.
currently, I am splitting each quad into two triangles that share an edge and then checking if the point is on or in each of those two triangles using their areas.
for example
Triangle ABC and test point P.
if (areaPAB + areaPAC + areaPBC == areaABC) { return true; }
This seems like it may run a little slow since I need to calculate the area of 4 different triangles to run the check and if the first triangle of the quad returns false, I have to get 4 more areas. (I include a bit of an epsilon in the check to make up for floating point errors)
I'm hoping that there is an even faster way that might involve a single check of a point against a quad rather than splitting it into two triangles.
I've attempted to reduce the number of checks by putting the polygon's into an array[,]. When adding a polygon, it checks the minimum and maximum x and y values and then using those, places the same poly into the proper array positions. When checking a point against the available polygons, it retrieves the proper list from the array of lists.
I've been searching through similar questions and I think what I'm using now may be the fastest way to figure out if a point is in a triangle, but I'm hoping that there's a better method to test against a quad that is always convex. Every polygon test I've looked up seems to be testing against a polygon that has many sides or is an irregular shape.
Thanks for taking the time to read my long winded question to what's prolly a simple problem.
I believe that fastest methods are:
1: Find mutual orientation of all vector pairs (DirectedEdge-CheckedPoint) through cross product signs. If all four signs are the same, then point is inside
Addition: for every edge
EV[i] = V[i+1] - V[i], where V[] - vertices in order
PV[i] = P - V[i]
Cross[i] = CrossProduct(EV[i], PV[i]) = EV[i].X * PV[i].Y - EV[i].Y * PV[i].X
Cross[i] value is positive, if point P lies in left semi-plane relatively to i-th edge (V[i] - V[i+1]), and negative otherwise. If all the Cross[] values are positive, then point p is inside the quad, vertices are in counter-clockwise order. f all the Cross[] values are negative, then point p is inside the quad, vertices are in clockwise order. If values have different signs, then point is outside the quad.
If quad set is the same for many point queries, then dmuir suggests to precalculate uniform line equation for every edge. Uniform line equation is a * x + b * y + c = 0. (a, b) is normal vector to edge. This equation has important property: sign of expression
(a * P.x + b * Y + c) determines semi-plane, where point P lies (as for crossproducts)
2: Split quad to 2 triangles and use vector method for each: express CheckedPoint vector in terms of basis vectors.
P = a*V1+b*V2
point is inside when a,b>=0 and their sum <=1
Both methods require about 10-15 additions, 6-10 multiplications and 2-7 comparisons (I don't consider floating point error compensation)
If you could afford to store, with each quad, the equation of each of its edges then you could save a little time over MBo's answer.
For example if you have an inward pointing normal vector N for each edge of the quad, and a constant d (which is N.p for one of the vertcies p on the edge) then a point x is in the quad if and only if N.x >= d for each edge. So thats 2 multiplications, one addition and one comparison per edge, and you'll need to perform up to 4 tests per point.This technique works for any convex polygon.
edit
So based on the answers so far (thanks for taking your time) I'm getting the sense that I'm probably NOT looking for a Normal Distribution function. Perhaps I'll try to re-describe what I'm looking to do.
Lets say I have an object that returns a number of 0 to 10. And that number controls "speed". However instead of 10 being the top speed, I need 5 to be the top speed, and anything lower or higher would slow down accordingly. (with easing, thus the bell curve)
I hope that's clearer ;/
-original question
These are the times I wish I remembered something from math class.
I'm trying to figure out how to write a function in obj-C where I define the boundries, ex (0 - 10) and then if x = foo y = ? .... where x runs something like 0,1,2,3,4,5,6,7,8,9,10 and y runs 0,1,2,3,4,5,4,3,2,1,0 but only on a curve
Something like the attached image.
I tried googling for Normal Distribution but its way over my head. I was hoping to find some site that lists some useful algorithms like these but wasn't very successful.
So can anyone help me out here ? And if there is some good sites which shows useful mathematical functions, I'd love to check them out.
TIA!!!
-added
I'm not looking for a random number, I'm looking for.. ex: if x=0 y should be 0, if x=5 y should be 5, if x=10 y should be 0.... and all those other not so obvious in between numbers
alt text http://dizy.cc/slider.gif
Okay, your edit really clarifies things. You're not looking for anything to do with the normal distribution, just a nice smooth little ramp function. The one Paul provides will do nicely, but is tricky to modify for other values. It can be made a little more flexible (my code examples are in Python, which should be very easy to translate to any other language):
def quarticRamp(x, b=10, peak=5):
if not 0 <= x <= b:
raise ValueError #or return 0
return peak*x*x*(x-b)*(x-b)*16/(b*b*b*b)
Parameter b is the upper bound for the region you want to have a slope on (10, in your example), and peak is how high you want it to go (5, in the example).
Personally I like a quadratic spline approach, which is marginally cheaper computationally and has a different curve to it (this curve is really nice to use in a couple of special applications that don't happen to matter at all for you):
def quadraticSplineRamp(x, a=0, b=10, peak=5):
if not a <= x <= b:
raise ValueError #or return 0
if x > (b+a)/2:
x = a + b - x
z = 2*(x-a)/b
if z > 0.5:
return peak * (1 - 2*(z-1)*(z-1))
else:
return peak * (2*z*z)
This is similar to the other function, but takes a lower bound a (0 in your example). The logic is a little more complex because it's a somewhat-optimized implementation of a piecewise function.
The two curves have slightly different shapes; you probably don't care what the exact shape is, and so could pick either. There are an infinite number of ramp functions meeting your criteria; these are two simple ones, but they can get as baroque as you want.
The thing you want to plot is the probability density function (pdf) of the normal distribution. You can find it on the mighty Wikipedia.
Luckily, the pdf for a normal distribution is not difficult to implement - some of the other related functions are considerably worse because they require the error function.
To get a plot like you showed, you want a mean of 5 and a standard deviation of about 1.5. The median is obviously the centre, and figuring out an appropriate standard deviation given the left & right boundaries isn't particularly difficult.
A function to calculate the y value of the pdf given the x coordinate, standard deviation and mean might look something like:
double normal_pdf(double x, double mean, double std_dev) {
return( 1.0/(sqrt(2*PI)*std_dev) *
exp(-(x-mean)*(x-mean)/(2*std_dev*std_dev)) );
}
A normal distribution is never equal to 0.
Please make sure that what you want to plot is indeed a
normal distribution.
If you're only looking for this bell shape (with the tangent and everything)
you can use the following formula:
x^2*(x-10)^2 for x between 0 and 10
0 elsewhere
(Divide by 125 if you need to have your peek on 5.)
double bell(double x) {
if ((x < 10) && (x>0))
return x*x*(x-10.)*(x-10.)/125.;
else
return 0.;
}
Well, there's good old Wikipedia, of course. And Mathworld.
What you want is a random number generator for "generating normally distributed random deviates". Since Objective C can call regular C libraries, you either need a C-callable library like the GNU Scientific Library, or for this, you can write it yourself following the description here.
Try simulating rolls of dice by generating random numbers between 1 and 6. If you add up the rolls from 5 independent dice rolls, you'll get a surprisingly good approximation to the normal distribution. You can roll more dice if you'd like and you'll get a better approximation.
Here's an article that explains why this works. It's probably more mathematical detail than you want, but you could show it to someone to justify your approach.
If what you want is the value of the probability density function, p(x), of a normal (Gaussian) distribution of mean mu and standard deviation sigma at x, the formula is
p(x) = exp( ((x-mu)^2)/(2*sigma^2) ) / (sigma * 2 * sqrt(pi))
where pi is the area of a circle divided by the square of its radius (approximately 3.14159...). Using the C standard library math.h, this is:
#include <math>
double normal_pdf(double x, double mu, double sigma) {
double n = sigma * 2 * sqrt(M_PI); //normalization factor
p = exp( -pow(x-mu, 2) / (2 * pow(sigma, 2)) ); // unnormalized pdf
return p / n;
}
Of course, you can do the same in Objective-C.
For reference, see the Wikipedia or MathWorld articles.
It sounds like you want to write a function that yields a curve of a specific shape. Something like y = f(x), for x in [0:10]. You have a constraint on the max value of y, and a general idea of what you want the curve to look like (somewhat bell-shaped, y=0 at the edges of the x range, y=5 when x=5). So roughly, you would call your function iteratively with the x range, with a step that gives you enough points to make your curve look nice.
So you really don't need random numbers, and this has nothing to do with probability unless you want it to (as in, you want your curve to look like a the outline of a normal distribution or something along those lines).
If you have a clear idea of what function will yield your desired curve, the code is trivial - a function to compute f(x) and a for loop to call it the desired number of times for the desired values of x. Plot the x,y pairs and you're done. So that's your algorithm - call a function in a for loop.
The contents of the routine implementing the function will depend on the specifics of what you want the curve to look like. If you need help on functions that might return a curve resembling your sample, I would direct you to the reading material in the other answers. :) However, I suspect that this is actually an assignment of some sort, and that you have been given a function already. If you are actually doing this on your own to learn, then I again echo the other reading suggestions.
y=-1*abs(x-5)+5