i have implemented a simple image resampler in OpenCL which uses the Lanczos function.
Lanczos is defined by:
Written in C:
inline
float lanczos(float x, float a) {
if( x > fabs(a) ) return 0.0f;
if( x == 0.0f ) return 1.0f;
float pix = pi * x;
return sinc(pix)*sinc(pix/a);
}
Why is there a special case for 0? When i pass 0 to the formular it returns 1. But if i don't include the check for x == 0 it doesn't work.
Could someone shed some light for me?
Florian
Paul already answered, but in case OP wants to know why 0 is special case =>
1) x->0, sin(x)/x = 0/0 and this is indeterminate form.
2) One way to solve this problem is to expand sin(x)/x into Taylor series about zero point, by doing this we get:
x2 x4 x6 x8
1 - ----- + ----- - ------ + ----------- + ...
6 120 5040 362880
3) By substituting 0 into x we see that series converges to 1.
Oh man ... i have been looking at the lanczos function for hours ... and haven't noticed that sinc actually is:
sinc -> sin(x)/x
so the special case for 0 is to prevent a division by zero ... plain and simple ...
Related
Is there a standard sigmoidal function used to increase the contrast in a gray level bitmap?
Currently I am using the following. This would be applied to gray levels represented at values between 0 and 1 inclusive.
static double ContrastCurve(double val, double k = 1)
{
Func<double,double> logistic_func = (double x) => 1.0 / (1.0 + Math.Exp(-k * (x - 0.5)));
var low = logistic_func(0);
var high = logistic_func(1);
var range = high - low;
var value = logistic_func(val);
return (value - low) / range;
}
This is the logistic function applied to a value between 0 and 1 with the output normalized so that the output is also in [0...1]. This function works but it is completely arbitrary, something I just made up, so the k param has no official name or meaning in image processing literature and so forth.
If there is a function that is standard I would prefer that but haven't found anything that seems definitive. Code such as this link seems as ad hoc to me.
As Mark Setchell's comment notes, ImageMagick uses the following function citing "Fundamentals of Image Processing", Hany Farid:
g(u) = 1 / [1 + exp(-α*u + β)]
scaled such that for domain [0..1] its range is [0..1].
This is essentially a two parameter version of the function defined in the code in the question above i.e. the code in the question implements the same function but makes the substitution α = k and β = -k/2 which yields a one parameter function f where f(0.5) = 0.5 when scaled such that f(0) = 0 and f(1) = 1.
I have the following problem:
Solve the following recurrence relation, simplifying your final answer
using 'O' notation.
f(0)=3
f(1)=12
f(n)=6f(n-1)-9f(n-2)
We know this is a homogeneous 2nd order relation so we write the characteristic equation: a^2-6a+9=0 and the solutions are a1,2=3.
The problem is when I replace these values I get:
f(n)=c1*3^n+c2*3^n
and using the 2 initial relations I have:
f(0)=c1+c2=3
f(1)=3(c1+c2)=12
which gives me that there no values such that c1 and c2 such that these 2 relation are true.
Am I doing something wrong? Is the way it should be solved different when it comes to identical roots for the characteristic equation?
You can't solve it this way, because your matrix A is not diagonalizable.
However, here is what you get if you use Jordan's normal form instead:
f(n) = 3^{n-1}(3n + 9)
The Jordan matrix and the basis (with notation from wikipedia + Octave) is:
J := [3,1;0,3]
P := [3,4;1,1]
such that PJP^{-1} = A, where
A := [6,-9;1,0]
is your recurrence matrix. Furthermore, the Jordan matrix is almost as good as a diagonal matrix for computing powers:
J^n = 3^(n-1) * [3,n;0,3].
The recurrence is then:
[f(n+1); f(n)] = A^n [12,3] = PJ^nP^-1[12,3] = (<whatever>, 3^(n-1)*(3n+9)).
Here a quick numerical check (Scala, but you can take whatever you want, Octave or I whatever you like):
scala> def f(n: Int): Int = { if (n == 0) 3 else if (n == 1) 12 else (6 * f(n-1) - 9 * f(n-2)) }
f: (n: Int)Int
scala> for (i <- 0 until 20) println(f(i))
3
12
45
162
567
1944
6561
21870
72171
236196
767637
2480058
7971615
25509168
81310473
258280326
817887699
^
scala> def explicit(n: Int): Int = (Math.pow(3, n -1) * (3 * n + 9)).toInt
explicit: (n: Int)Int
scala> for (i <- 0 until 20) println(explicit(i))
3
12
45
162
567
1944
6561
21870
72171
236196
767637
2480058
7971615
25509168
81310473
258280326
817887699
I've been working on a project that renders a Mandelbrot fractal. For those of you who know, it is generated by iterating through the following function where c is the point on a complex plane:
function f(c, z) return z^2 + c end
Iterating through that function produces the following fractal (ignore the color):
When you change the function to this, (z raised to the third power)
function f(c, z) return z^3 + c end
the fractal should render like so (again, the color doesn't matter):
(source: uoguelph.ca)
However, when I raised z to the power of 3, I got an image extremely similar as to when you raise z to the power of 2. How can I make the fractal render correctly? This is the code where the iterations are done: (the variables real and imaginary simply scale the screen from -2 to 2)
--loop through each pixel, col = column, row = row
local real = (col - zoomCol) * 4 / width
local imaginary = (row - zoomRow) * 4 / width
local z, c, iter = 0, 0, 0
while math.sqrt(z^2 + c^2) <= 2 and iter < maxIter do
local zNew = z^2 - c^2 + real
c = 2*z*c + imaginary
z = zNew
iter = iter + 1
end
So I recently decided to remake a Mandelbrot fractal generator, and it was MUCH more successful than my attempt last time, as my programming skills have increased with practice.
I decided to generalize the mandelbrot function using recursion for anyone who wants it. So, for example, you can do f(z, c) z^2 + c or f(z, c) z^3 + c
Here it is for anyone that may need it:
function raise(r, i, cr, ci, pow)
if pow == 1 then
return r + cr, i + ci
end
return raise(r*r-i*i, 2*r*i, cr, ci, pow - 1)
end
and it's used like this:
r, i = raise(r, i, CONSTANT_REAL_PART, CONSTANT_IMAG_PART, POWER)
I have to create a path between two given points in a grid in Prolog. The code I have so far is:
createPath(GridSize, BeginPosition, EndPosition, VisitedPoints, Path):-
nextStep(BeginPosition, NextStep, GridSize),
(
NextStep \== EndPosition,
->
nonmember(NextStep, VisitedPoints),
add(NextStep, VisitedPoints, NewVisitedPoints),
add(NextStep, Path, NewPath),
createPath(GridSize, NextStep, EndPosition, NewVisitedPoints, NewPath)
;
???
).
A little bit of explanation of my code:
GridSize is just an integer. If it is 2, the grid is a 2x2 grid. So all the grids are square.
The BeginPosition and EndPosition are shown like this: pos(X,Y).
The function nextStep looks for a valid neigbor of a given position. The values of X and Y have to be between 1 and the grid size. I've declared 4 different predicates of nextStep: X + 1, X - 1, Y + 1 and Y - 1.
This is the code:
nextStep(pos(X,Y),pos(X1,Y),GridSize):-
X1 is X + 1,
X1 =< GridSize.
nextStep(pos(X,Y),pos(X1,Y),_):-
X1 is X - 1,
X1 >= 1.
nextStep(pos(X,Y),pos(X,Y1),GridSize):-
Y1 is Y + 1,
Y1 =< GridSize.
nextStep(pos(X,Y),pos(X,Y1),_):-
Y1 is Y - 1,
Y1 >= 1.
nonmember returns true if a given element doesn't occur in a given list.
add adds an element to a given list, and returns the list with that element in it.
Another thing to know about VisitedPoints: Initially the BeginPosition and EndPosition are stored in that list. For example, if I want to find a path in a 2x2 grid, and I have to avoid point pos(2,1), then I will call the function like this:
createPath(2, pos(1,1), pos(2,2), [pos(1,1),pos(2,2),pos(2,1)], X).
The result I should get of it, should be:
X = [pos(1,2)]
Because that is the point needed to connect pos(1,1) and pos(2,2).
My question is, how can I stop the code from running when NextStep == EndPosition. In other words, what do I have to type at the location of the '???' ? Or am I handling this problem the wrong way?
I'm pretty new to Prolog, and making the step from object oriented languages to this is pretty hard.
I hope somebody can answer my question.
Kind regards,
Walle
I think you just placed the 'assignment' to path at the wrong place
createPath(GridSize, BeginPosition, EndPosition, VisitedPoints, Path):-
nextStep(BeginPosition, NextStep, GridSize),
(
NextStep \== EndPosition,
->
nonmember(NextStep, VisitedPoints),
add(NextStep, VisitedPoints, NewVisitedPoints),
% add(NextStep, Path, NewPath),
% createPath(GridSize, NextStep, EndPosition, NewVisitedPoints, NewPath)
createPath(GridSize, NextStep, EndPosition, NewVisitedPoints, Path)
;
% ???
% bind on success the output variable, maybe add EndPosition
Path = VisitedPoints
).
Maybe this is not entirely worth an answer, but a comment would be a bit 'blurry'
If I am given a floating point number but do not know beforehand what range the number will be in, is it possible to scale that number in some meaningful way to be in another range? I am thinking of checking to see if the number is in the range 0<=x<=1 and if not scale it to that range and then scale it to my final range. This previous post provides some good information, but it assumes the range of the original number is known beforehand.
You can't scale a number in a range if you don't know the range.
Maybe what you're looking for is the modulo operator. Modulo is basically the remainder of division, the operator in most languages is is %.
0 % 5 == 0
1 % 5 == 1
2 % 5 == 2
3 % 5 == 3
4 % 5 == 4
5 % 5 == 0
6 % 5 == 1
7 % 5 == 2
...
Sure it is not possible. You can define range and ignore all extrinsic values. Or, you can collect statistics to find range in run time (i.e. via histogram analysis).
Is it really about image processing? There are lots of related problems in image segmentation field.
You want to scale a single random floating point number to be between 0 and 1, but you don't know the range of the number?
What should 99.001 be scaled to? If the range of the random number was [99, 100], then our scaled-number should be pretty close to 0. If the range of the random number was [0, 100], then our scaled-number should be pretty close to 1.
In the real world, you always have some sort of information about the range (either the range itself, or how wide it is). Without further info, the answer is "No, it can't be done."
I think the best you can do is something like this:
int scale(x) {
if (x < -1) return 1 / x - 2;
if (x > 1) return 2 - 1 / x;
return x;
}
This function is monotonic, and has a range of -2 to 2, but it's not strictly a scaling.
I am assuming that you have the result of some 2-dimensional measurements and want to display them in color or grayscale. For that, I would first want to find the maximum and minimum and then scale between these two values.
static double[][] scale(double[][] in, double outMin, double outMax) {
double inMin = Double.POSITIVE_INFINITY;
double inMax = Double.NEGATIVE_INFINITY;
for (double[] inRow : in) {
for (double d : inRow) {
if (d < inMin)
inMin = d;
if (d > inMax)
inMax = d;
}
}
double inRange = inMax - inMin;
double outRange = outMax - outMin;
double[][] out = new double[in.length][in[0].length];
for (double[] inRow : in) {
double[] outRow = new double[inRow.length];
for (int j = 0; j < inRow.length; j++) {
double normalized = (inRow[j] - inMin) / inRange; // 0 .. 1
outRow[j] = outMin + normalized * outRange;
}
}
return out;
}
This code is untested and just shows the general idea. It further assumes that all your input data is in a "reasonable" range, away from infinity and NaN.