Recently I posted another thread and the answers were really helpful. From this reference I tried to write the code for two intersecting circle.
I think I did something wrong with my code and the result it's giving me the mid point of the connecting line of the intersection points. I have no clue what's wrong with my code.
I'm finding it difficult to handle the loop. I need the actual length of the vectors but if I wrote it to the loop, for i+1 it is showing me the error (which is very natural as I don't find any '<' operator :(()
Here's my code -
Xs = [150,130];% x coordinates of center of circles
Ys = [100,250];% y coordinates of center of circles
Rs = [100,70];% radius of circles
C=Imread('spinks_map.png');
figure, Imshow(C)
hold on;
simple_draw_circle( Xs, Ys, Rs);
for i=1:length(Xs)-1
dsq(i) = ((Xs(i+1)- Xs(i))^2) + ((Ys(i+1) - Ys(i))^2);
K(i)= (sqrt((((Rs(i)+Rs(i+1))^2)-dsq(i))-(dsq(i)-(Rs(i)-Rs(i+1))^2)))/4;
x(i) = ((Xs(i+1)+Xs(i))/2) + ((Xs(i+1)-Xs(i))*(Rs(i)^2-Rs(i+1)^2)/dsq(i))/2 - 2*(Ys(i+1)-Ys(i))*(K(i)/dsq(i));
y(i) = ((Ys(i+1)+Ys(i))/2) + ((Ys(i+1)-Ys(i))*(Rs(i)^2-Rs(i+1)^2)/dsq(i))/2 +(-2)*(Xs(i+1)-Xs(i))*(K(i)/dsq(i));
end
scatter(x(1),y(1),'filled');
You have an error in your use of Heron's formula. Look at the link you posted that gives the calculation for K. There should be a product between the two terms inside the square root, not a difference:
K(i)= (sqrt((((Rs(i)+Rs(i+1))^2)-dsq(i))*(dsq(i)-(Rs(i)-Rs(i+1))^2)))/4;
Also, with regards to your second question. I've found it to be a best practice for me if I explicitly declare order of operations when using an expression with ":".
For your loop, using the appropriate choice of either
for i=(1:length(Xs))-1
or
for i=1:(length(Xs)-1)
can cut down on unexpected behaviors during indexing. This was the issue as I understood it from your post. If it was something else, then post the error text that Matlab spits up and I'll take a look at it.
There is a "<" operator in Matlab. If you're trying to use it in the style of the C equivalent for loop:
for(iter=0;iter++,iter<2){
something();
}
then look into the syntax for the Matlab while loop.
Best of luck,
Andrew
Related
I'd like to declare first of all, that I'm a mathematician. This might be a stupid stupid question; but I've gone through all the matlab tutorials--they've gotten me nowhere. I imagine I could code this in C (it'd be exhausting); but I need matlab for this particular function. And I don't get exactly how to do it.
Here is the pasted Matlab code of where I'm running into trouble:
function y = TAU(z,n)
y=0;
for i =[1,n]
y(z) = log(beta(z+1,i) + y(z+1)) - beta(z,i);
end
end
(beta is an arbitrary "float" to "float" function with an index i.)
I'm having trouble declaring y as a function, in which we call the function at a different argument. I want to define y_n(z) with something something y_{n-1}(z+1). This is all done in a recursive process to create the function. I really feel like I'm missing something stupid.
As a default function it assigns y to be an array (or whatever you call the default index assignment). But I don't want an array. I want y to be assigned as a "function" class (i.e. takes "float" to "float"). And then I'm defining a sequence of y_n : "float" to "float". So that z to z+1 is a map on "float" to "float".
I don't know if I'm asking too much of matlab...
Help a poor mathematician who hasn't coded since the glory days of X-box mods.
...Please don't tell me I have to go back to Pari-GP/C drawing boards over something so stupid.
Please help!
EDIT: At rahnema1 & mimocha's request, I'll describe the math, and of what I am trying to do with my program. I can't see how to implement latex in here. So I'll write the latex code in a generator and upload a picture. I'm not so sure if there even is a work around to what I want to do.
As to the expected output. We'd want,
beta(z+1,i) + TAU(z+1,i) = exp(beta(z,i) + TAU(z,i+1))
And we want to grow i to a fixed value n. Again, I haven't programmed in forever, so I apologize if I'm speaking a little nonsensically.
EDIT2:
So, as #rahnema1 suggests; I should produce a reproducible example. In order to do this, I'll write the code for my beta function. It's surprisingly simple. This is for the case where the "multiplier" variable is set to log(2); but you don't need to worry about any of that.
function f = beta(z,n)
f=0;
for i = 0:n-1
f = exp(f)/(1+exp(log(2)*(n-i-z)));
end
end
This will work fine for z a float no greater than 4. Once you make z larger it'll start to overflow. So for example, if you put in,
beta(2,100)
1.4242
beta(3,100)
3.3235
beta(3,100) - exp(beta(2,100))/(1/4+1)
0
The significance of the 100, is simply how many iterations we perform; it converges fast so even setting this to 15 or so will still produce the same numerical accuracy. Now, the expected output I want for TAU is pretty straight forward,
TAU(z,1) = log(beta(z+1,1)) - beta(z,1)
TAU(z,2) = log(beta(z+1,2) + TAU(z+1,1)) - beta(z,2)
TAU(z,3) = log(beta(z+1,3) + TAU(z+1,2)) - beta(z,3)
...
TAU(z,n) = log(beta(z+1,n) + TAU(z+1,n-1)) -beta(z,n)
I hope this helps. I feel like there should be an easy way to program this sequence, and I must be missing something obvious; but maybe it's just not possible in Matlab.
At mimocha's suggestion, I'll look into tail-end recursion. I hope to god I don't have to go back to Pari-gp; but it looks like I may have to. Not looking forward to doing a deep dive on that language, lol.
Thanks, again!
Is this what you are looking for?
function out = tau(z,n)
% Ends recursion when n == 1
if n == 1
out = log(beta(z+1,1)) - beta(z,1);
return
end
out = log(beta(z+1,n) + tau(z+1,n-1)) - beta(z,n);
end
function f = beta(z,n)
f = 0;
for i = 0:n-1
f = exp(f) / (1 + exp(log(2)*(n-i-z)));
end
end
This is basically your code from the most recent edit, but I've added a simple catch in the tau function. I tried running your code and noticed that n gets decremented infinitely (no exit condition).
With the modification, the code runs successfully on my laptop for smaller integer values of n, where 1e5 > n >= 1; and for floating values of z, real and complex. So the code will unfortunately break for floating values of n, since I don't know what values to return for, say, tau(1,0) or tau(1,0.9). This should easily be fixable if you know the math though.
However, many of the values I get are NaNs or Infs. So I'm not sure if your original problem was Out of memory error (infinite recursion), or values blowing up to infinity / NaN (numerical stability issue).
Here is a quick 100x100 grid calculation I made with this code.
Then I tested on negative values of z, and found the imaginary part of the output to looks kinda cool.
Not to mention I'm slightly geeking out over the fact that pi is showing up in the imaginary part as well :)
tau(-0.3,2) == -1.45179335740446147085 +3.14159265358979311600i
Incredibly simple question, but I think I'm unable to come up with the correct terminology to google search it.
If I have a snippet of code that relies on three independent variables:
code(x,y,z)
That produces two values, i.e.:
output1, output2
How do I go about iterating like so (pseudocode):
for x
for y
for z
code(x,y,z)
end
end
end
And have data I can parse to generate 3D graphs such as
surf(x,y,output1)
A naive solution I came up with was just to create a bin of n length and then iterating one variable n times to come up with a 2D graph, i.e:
x_axis = zeros(1,25)
for m = 1:25
xm = x + 1
x_axis(m) = xm
code(x,y,z)
Even a referral to some documentation would be extremely helpful.
Thanks!
Brute force approach:
for x=[1:50]
for y=[1:50]
for z=[1:50]
result(y,x,z)=code(x,y,z);
end
end
end
More paradigmatic approach (in MATLAB) is to meshgrid it, and pump those in.
[XX,YY,ZZ]=meshgrid([1:50],[1:50],[1:50]);
result=code(XX,YY,ZZ);
I have a Matlab project in which I need to make a GUI that receives two mathematical functions from the user. I then need to find their intersection point, and then plot the two functions.
So, I have several questions:
Do you know of any algorithm I can use to find the intersection point? Of course I prefer one to which I can already find a Matlab code for in the internet. Also, I prefer it wouldn't be the Newton-Raphson method.
I should point out I'm not allowed to use built in Matlab functions.
I'm having trouble plotting the functions. What I basically did is this:
fun_f = get(handles.Function_f,'string');
fun_g = get(handles.Function_g,'string');
cla % To clear axes when plotting new functions
ezplot(fun_f);
hold on
ezplot(fun_g);
axis ([-20 20 -10 10]);
The problem is that sometimes, the axes limits do not allow me to see the other function. This will happen, if, for example, I will have one function as log10(x) and the other as y=1, the y=1 will not be shown.
I have already tried using all the axis commands but to no avail. If I set the limits myself, the functions only exist in certain limits. I have no idea why.
3 . How do I display numbers in a static text? Or better yet, string with numbers?
I want to display something like x0 = [root1]; x1 = [root2]. The only solution I found was turning the roots I found into strings but I prefer not to.
As for the equation solver, this is the code I have so far. I know it is very amateurish but it seemed like the most "intuitive" way. Also keep in mind it is very very not finished (for example, it will show me only two solutions, I'm not so sure how to display multiple roots in one static text as they are strings, hence question #3).
function [Sol] = SolveEquation(handles)
fun_f = get(handles.Function_f,'string');
fun_g = get(handles.Function_g,'string');
f = inline(fun_f);
g = inline(fun_g);
i = 1;
Sol = 0;
for x = -10:0.1:10;
if (g(x) - f(x)) >= 0 && (g(x) - f(x)) < 0.01
Sol(i) = x;
i = i + 1;
end
end
solution1 = num2str(Sol(1));
solution2 = num2str(Sol(2));
set(handles.roots1,'string',solution1);
set(handles.roots2,'string',solution2);
The if condition is because the subtraction will never give me an absolute zero, and this seems to somewhat solve it, though it's really not perfect, sometimes it will give me more than two very similar solutions (e.g 1.9 and 2).
The range of x is arbitrary, chosen by me.
I know this is a long question, so I really appreciate your patience.
Thank you very much in advance!
Question 1
I think this is a more robust method for finding the roots given data at discrete points. Looking for when the difference between the functions changes sign, which corresponds to them crossing over.
S=sign(g(x)-f(x));
h=find(diff(S)~=0)
Sol=x(h);
If you can evaluate the function wherever you want there are more methods you can use, but it depends on the size of the domain and the accuracy you want as to what is best. For example, if you don't need a great deal of accurac, your f and g functions are simple to calculate, and you can't or don't want to use derivatives, you can get a more accurate root using the same idea as the first code snippet, but do it iteratively:
G=inline('sin(x)');
F=inline('1');
g=vectorize(G);
f=vectorize(F);
tol=1e-9;
tic()
x = -2*pi:.001:pi;
S=sign(g(x)-f(x));
h=find(diff(S)~=0); % Find where two lines cross over
Sol=zeros(size(h));
Err=zeros(size(h));
if ~isempty(h) % There are some cross-over points
for i=1:length(h) % For each point, improve the approximation
xN=x(h(i):h(i)+1);
err=1;
while(abs(err)>tol) % Iteratively improve aproximation
S=sign(g(xN)-f(xN));
hF=find(diff(S)~=0);
xN=xN(hF:hF+1);
[~,I]=min(abs(f(xN)-g(xN)));
xG=xN(I);
err=f(xG)-g(xG);
xN=linspace(xN(1),xN(2),15);
end
Sol(i)=xG;
Err(i)=f(xG)-g(xG);
end
else % No crossover points - lines could meet at tangents
[h,I]=findpeaks(-abs(g(x)-f(x)));
Sol=x(I(abs(f(x(I))-g(x(I)))<1e-5));
Err=f(Sol)-g(Sol)
end
% We also have to check each endpoint
if abs(f(x(end))-g(x(end)))<tol && abs(Sol(end)-x(end))>1e-12
Sol=[Sol x(end)];
Err=[Err g(x(end))-f(x(end))];
end
if abs(f(x(1))-g(x(1)))<tol && abs(Sol(1)-x(1))>1e-12
Sol=[x(1) Sol];
Err=[g(x(1))-f(x(1)) Err];
end
toc()
Sol
Err
This will "zoom" in to the region around each suspected root, and iteratively improve the accuracy. You can tweak the parameters to see whether they give better behaviour (the tolerance tol, the 15, number of new points to generate, could be higher probably).
Question 2
You would probably be best off avoiding ezplot, and using plot, which gives you greater control. You can vectorise inline functions so that you can evaluate them like anonymous functions, as I did in the previous code snippet, using
f=inline('x^2')
F=vectorize(f)
F(1:5)
and this should make plotting much easier:
plot(x,f(x),'b',Sol,f(Sol),'ro',x,g(x),'k',Sol,G(Sol),'ro')
Question 3
I'm not sure why you don't want to display your roots as strings, what's wrong with this:
text(xPos,yPos,['x0=' num2str(Sol(1))]);
In Matlab, I've got several points in the 3D space. Those points represent a rope and I would like to draw a line linking all those points. Here is my problem: How the oraganization of those points should be done to have a "simple" and "more or less straigth line". In other words I would like to draw a line linking all the points from the first to the last one but not "going back". Maybe with a simple image i can explain better my problem:
This is what the code should do:
This is what the code shouldn't do:
Some idea of how can I achieve the intended result? How can I organize the points? I'm working with Matlab but if you know any paper where I can read how to do this it will be fine. Thank you.
If you just don't want to go back in the upper direction, the solution that #Dan suggested should be fine. Sort them in that direction before connecting.
However, if that is not a requirement but you are actually looking for a solution without ugly crossings that is as short as possible, you may want to look into solutions for the travelling salesman problem.
If you define the distance between 1 and 9 to be zero, the solution you find to the travelling salesman problem should give you a nice and short rope.
If you want to go for the TSP approach but get a little lost, just try a 'furthest insertion' start position or '2 opt' improvements as that is likely going to give a big improvement over random path selection already with minimal effort.
OK. I've found the right answer (by Gunther Struyf): Sort Coordinates Points in Matlab
In order to visualize the final result just add to Gunther's code the following lines:
row=data(:,1);
col=data(:,2);
figure
scatter(row,col,'r');
hold on
line(row(result),col(result));
The following code shows the same algorithm but for a 3D implementation:
close all;
data = [2,2, 2; 2,3,2 ; 1,2,3 ; 1,3,3 ; 2,1,3 ; 1,1,3 ; 3,2,4 ; 3,3,4 ; 3,1,5; 5,4,6; 7,3,8; 8,9,7; 9,4,7; 6,2,5; 5,8,6; 9,3,8;6,9,2];
row=data(:,1);
col=data(:,2);
frame=data(:,3);
dist = pdist2(data,data);
N = size(data,1);
result = NaN(1,N);
result(1) = 1; % first point is first row in data matrix
for ii=2:N
dist(:,result(ii-1)) = Inf;
[~, closest_idx] = min(dist(result(ii-1),:));
result(ii) = closest_idx;
end
figure
%Ploting the points directly (without sorting)
subplot(2,1,1);
scatter3(row,col,frame,'r');
hold on;
line(row,col,frame);
%Ploting after sorting
subplot(2,1,2);
scatter3(row,col,frame,'r');
hold on
line(row(result),col(result),frame(result));
I have recently been tasked with using a first derivative filter on an image of myself. The instructor said that I should first fix the value of y and preform f(x+1) - f(x) on the rows and then fix the new "X" values and preform f(y+1)-f(y) on the columns.
Note: I have been asked to do this task manually, not using filter2() or any other programmed function, so please do not suggest that I use filter2() or similar. Thanks!
I tried calling up all the pixels and subtracting each successive one by doing
fid = fopen('image.raw')
myimage = fread(fid,[512 683], '*int8')
fclose(fid)
imsz = size(myimage)
x = imsz(1)
for I = 1:512
for J = 1:683
X(I) - X(I-1) = XX
But it doesnt seem to work, and I dont quite understand why. If you could help me, or point me in the right direction, I would be very appreciative.
First of all, your code is syntatically incorrect:
There is no end statement to any of your loops, and besides, you don't even need loops here.
You seem to read your image into the variable myimage, but you're using an undefined variable X when attempting to calculate the derivative.
The order of your assignment statements is reversed. The variable you wish to assign to should be written in the left hand part of the assignement.
I strongly suggest that you read online tutorials and get yourself familiar with MATLAB basics before taking on more complicated tasks.
As to your specific problem:
MATLAB encourages vectorized operations, i.e operations on entire arrays (vectors or matrices) at once. To subtract adjacent values in an array, what you're basically doing is subtracting two arrays, shifted by one element with respect to each other. For one dimensional arrays, that would translate in MATLAB to:
a(2:end) - a(1:end-1)
where a is your array. The end keyword specifies the last index in the array.
To compute the derivative of an image (a 2-D matrix), you need to decide along which axis you want to perform that operation. To approximate the derivate along the y-axis, do this:
X(2:end, :) - X(1:end-1, :)
You can verify that this gives you the same result as diff(X, 1) (or simply diff(X)). To compute the approximate derivative along the x-axis, which is equivalent to diff(X, 2), do this:
X(:, 2:end) - X(:, 1:end-1)
The colon (:) is the same as writing 1:end as the array subscript for the corresponding dimension.
If your filtered image is div then
for Y = 1:682
for X = 1:511
div(X, Y) = myimage(X + 1, Y + 1) - myimage(X,Y);
end
end
Remember the last row and the last column are not filtered!