i am having problem with writing equations.
r = 25, k= 2, R = 50:25:600, DR = 0.5:0.5:4.0
h= r*[1-cos(asin((sqrt(2*R*DR+DR^2))+r*sin(acos(r-k)/r)/r))]-k
but as a resault i get this: h = 1.9118e+001 +1.7545e+002i.
I just start with Matlab. Thanks
What I get from what you've written is actually
??? Error using ==> mtimes
Inner matrix dimensions must agree.
which is correct because you're trying to multiply two row vectors by one another. Could you please show us the actual code you used?
Anyway, supposing that's dealt with somehow, it looks to me as if you're feeding something to asin that's much bigger than 1. That'll give you complex results. Is the thing you're passing to asin perhaps meant to be divided by R^2 or DR^2 or something of the kind? You have a similar issue a bit later with the argument to acos.
I also suspect that some of your * and ^ and / operators should actually be elementwise ones .*, .^, ./.
If you're trying to do as you said:
so in first equation i used R= 50, DR
= 0.5, r= 25, k=2 and i need to get h. In second equation i used R=75,
DR=1.0, r=25, k=2...for a last
equation i used
R=600,DR=4.0,r=25,k=2.
DR and R need to be the same length... so if R goes between 50 and 600 in increments of 25, DR should go from 0.5 to 12.5 in increments of 0.5, or 0.5 to 4.0 in increments of 0.1522...
once you figure that out, be sure the add a period before every matrix multiplication operation (e.g. * or ^)
EDIT: formula adjusted slightly (bracketing) to reflect success in comment.
When you say you want a table, I guess it is to be an R by DR table (since you have to vectors of different length). To do that you need to use R as a column vector (R' below) and multiply with * (not .*). When R doesn't appear in a term multiply by ones(size(R)) (or use repmat) to get DR into the right shape. To square DR by element, you need DR.^2. There seems to be a misplaced bracket for the acos, surely you divide by r before taking the acos. There must be a division by something like r in the asin (not r^2 because you've taken the sqrt). Finally, the last division by r is redundant as written, since you multiply by r at the same level just before. Anyway, if I do the following:
h= r*(1-cos(asin((sqrt(2*R'*DR+ones(size(R))'*DR.^2)/r)+sin(acos((r-k)/r)))))-k
I get an R by DR table. Results for small R,DR are real; higher R,DR are complex due to the argument of the first asin being >1. The first entry in the table is 4.56, as you require.
Related
I noticed that SciPy has an implementation of the Discrete Sine Transform, and I was comparing it to the one that's in MATLAB. The MATLAB documentation notes that for best performance, the size of the inputs should be 2^p -1, presumably for a divide and conquer strategy. Is this also true for the SciPy implementation?
Although this question is old, I happen to have just ran some tests and then stumbled upon this question.
The answer is yes. Internally, scipy seems to converts the array to size M = 2*(N+1).
Ideally, M = 2^i, for some integer i. Therefore, N should follow N = 2^i - 1. The following picture shows how timings scale with fft-size. Note that the orange line is much smoother, indicating no unexpected memory overhead.
Green line: N = 2^i
Blue line: N = 2^i + 1
Orange line: N = 2^i - 1
UPDATE
After digging some more into the documentation of scipy.fftpack, I found that the above answer is only partly true. According to the documentation, "SciPy’s FFTPACK has efficient functions for radix {2, 3, 4, 5}". This means that instead of efficiently doing arrays of size M = 2^i, it can handle any M = 2^i * 3^j * 5^k (4 is not a prime). The optimum for scipy.fftpack.dst (or dct) is then M - 1. Finding those numbers can be a little awkward, but luckily there's a function for that, too!
Please note that the above graph is log-log scale, so speedups of 40 or so are not uncommon. Thus, choosing a fast size can make you calculations orders of magnitudes faster! (I found this out the hard way).
I'm just learning matlab and I have a snippet of code which I don't understand the syntax of. The x is an n x 1 vector.
Code is below
p = (min(x):(max(x)/300):max(x))';
The p vector is used a few lines later to plot the function
plot(p,pp*model,'r');
It generates an arithmetic progression.
An arithmetic progression is a sequence of numbers where the next number is equal to the previous number plus a constant. In an arithmetic progression, this constant must stay the same value.
In your code,
min(x) is the initial value of the sequence
max(x) / 300 is the increment amount
max(x) is the stopping criteria. When the result of incrementation exceeds this stopping criteria, no more items are generated for the sequence.
I cannot comment on this particular choice of initial value and increment amount, without seeing the surrounding code where it was used.
However, from a naive perspective, MATLAB has a linspace command which does something similar, but not exactly the same.
Certainly looks to me like an odd thing to be doing. Basically, it's creating a vector of values p that range from the smallest to the largest values of x, which is fine, but it's using steps between successive values of max(x)/300.
If min(x)=300 and max(x)=300.5 then this would only give 1 point for p.
On the other hand, if min(x)=-1000 and max(x)=0.3 then p would have thousands of elements.
In fact, it's even worse. If max(x) is negative, then you would get an error as p would start from min(x), some negative number below max(x), and then each element would be smaller than the last.
I think p must be used to create pp or model somehow as well so that the plot works, and without knowing how I can't suggest how to fix this, but I can't think of a good reason why it would be done like this. using linspace(min(x),max(x),300) or setting the step to (max(x)-min(x))/299 would make more sense to me.
This code examines an array named x, and finds its minimum value min(x) and its maximum value max(x). It takes the maximum value and divides it by the constant 300.
It doesn't explicitly name any variable, setting it equal to max(x)/300, but for the sake of explanation, I'm naming it "incr", short for increment.
And, it creates a vector named p. p looks something like this:
p = [min(x), min(x) + incr, min(x) + 2*incr, ..., min(x) + 299*incr, max(x)];
My goal is to solve for a matrix [A] that satisfies [A]*[B]=[C] where [C] is known and [B] is randomly generated. Below is an example:
C=[1/3 1/3 1/3]'*[1/3 1/6 1/6 1/6 1/6];
B=rand(5,5);
A=C*pinv(B);
A*B=C_test;
norm(C-C_test);
ans =
4.6671e-16
Here the elements of [C_test] are within 1e-15 to the original [C], but when [B] has less rows than columns, the error dramatically increases (not sure is norm() is the best way to show the error, but I think it illustrates the problem). For example:
B=rand(4,5);
A=C*pinv(B);
A*B=C_test;
norm(C-C_test);
ans =
0.0173
Additional methods:
QR-Factorization
[Q,R,P]=qr(B);
A=((C*P)/R))*Q';
norm(C-A*B);
ans =
0.0173
/ Operator
A=C/B;
norm(C-A*B);
ans =
0.0173
Why does this happen? In both cases [B]*pinv([B])=[I] so it seems like the process should work.
If this is a numerical or algebraic fact of life associated with pinv() or the other methods, is there another way I can generate [A] to satisfy the equation? Thank you!
Since C is 3×5, the number of elements in C and hence the number of equations is equal to 15. If B is 5×5, the number of unknowns (the elements in A) equals 3×5 = 15 as well, and the solution will be accurate.
If on the other hand B is for instance 3×5, the number of elements in A is equal to 3×3 = 9 and hence the system is overdetermined, which means that the resulting A will be the least-squares solution.
See for general information wikipedia: System of linear equations, and Matlabs Overdetermined system.
The resulting matrix A is the best fit and there is no way to improve (in a least square sense).
In response to your second question: you are measuring the quality of A*B as an approximation of C by applying the 2-norm to A*B-C: which is equivalent to least-squares fitting. In this measure, all the approaches that you use provide the optimal answer.
If you however would prefer some other measure, such as the 1-norm, the Infinity-norm or any other measure (for instance by picking different weights for column, row or element), the obtained answers from the original approach will of course not be necessarily optimal with respect to this new measure.
The most general approach would be to use some optimization routine, like this:
x = fminunc(f, zeros(3*size(B,1),1));
A = reshape(x,3,size(B,1));
where f is some (any) measure. The least-square measure should result in the same A. So if you try this one:
f = #(x) norm(reshape(x,3,size(B,1))*B - C);
A should match the results in your approaches.
But you could use any f here. For instance, try the 1-norm:
f = #(x) norm(reshape(x,3,size(B,1))*B - C, 1);
Or something crazy like:
f = #(x) sum(abs(reshape(x,3,size(B,1))*B - C)*[1 10 100 1000 10000]');
This will give different results, which are according to the new measure f optimal. That being said, I would stick to the least squares ;)
I have a Problem. I have a Matrix A with integer values between 0 and 5.
for example like:
x=randi(5,10,10)
Now I want to call a filter, size 3x3, which gives me the the most common value
I have tried 2 solutions:
fun = #(z) mode(z(:));
y1 = nlfilter(x,[3 3],fun);
which takes very long...
and
y2 = colfilt(x,[3 3],'sliding',#mode);
which also takes long.
I have some really big matrices and both solutions take a long time.
Is there any faster way?
+1 to #Floris for the excellent suggestion to use hist. It's very fast. You can do a bit better though. hist is based on histc, which can be used instead. histc is a compiled function, i.e., not written in Matlab, which is why the solution is much faster.
Here's a small function that attempts to generalize what #Floris did (also that solution returns a vector rather than the desired matrix) and achieve what you're doing with nlfilter and colfilt. It doesn't require that the input have particular dimensions and uses im2col to efficiently rearrange the data. In fact, the the first three lines and the call to im2col are virtually identical to what colfit does in your case.
function a=intmodefilt(a,nhood)
[ma,na] = size(a);
aa(ma+nhood(1)-1,na+nhood(2)-1) = 0;
aa(floor((nhood(1)-1)/2)+(1:ma),floor((nhood(2)-1)/2)+(1:na)) = a;
[~,a(:)] = max(histc(im2col(aa,nhood,'sliding'),min(a(:))-1:max(a(:))));
a = a-1;
Usage:
x = randi(5,10,10);
y3 = intmodefilt(x,[3 3]);
For large arrays, this is over 75 times faster than colfilt on my machine. Replacing hist with histc is responsible for a factor of two speedup. There is of course no input checking so the function assumes that a is all integers, etc.
Lastly, note that randi(IMAX,N,N) returns values in the range 1:IMAX, not 0:IMAX as you seem to state.
One suggestion would be to reshape your array so each 3x3 block becomes a column vector. If your initial array dimensions are divisible by 3, this is simple. If they don't, you need to work a little bit harder. And you need to repeat this nine times, starting at different offsets into the matrix - I will leave that as an exercise.
Here is some code that shows the basic idea (using only functions available in FreeMat - I don't have Matlab on my machine at home...):
N = 100;
A = randi(0,5*ones(3*N,3*N));
B = reshape(permute(reshape(A,[3 N 3 N]),[1 3 2 4]), [ 9 N*N]);
hh = hist(B, 0:5); % histogram of each 3x3 block: bin with largest value is the mode
[mm mi] = max(hh); % mi will contain bin with largest value
figure; hist(B(:),0:5); title 'histogram of B'; % flat, as expected
figure; hist(mi-1, 0:5); title 'histogram of mi' % not flat?...
Here are the plots:
The strange thing, when you run this code, is that the distribution of mi is not flat, but skewed towards smaller values. When you inspect the histograms, you will see that is because you will frequently have more than one bin with the "max" value in it. In that case, you get the first bin with the max number. This is obviously going to skew your results badly; something to think about. A much better filter might be a median filter - the one that has equal numbers of neighboring pixels above and below. That has a unique solution (while mode can have up to four values, for nine pixels - namely, four bins with two values each).
Something to think about.
Can't show you a mex example today (wrong computer); but there are ample good examples on the Mathworks website (and all over the web) that are quite easy to follow. See for example http://www.shawnlankton.com/2008/03/getting-started-with-mex-a-short-tutorial/
I'm having trouble creating a random vector V in Matlab subject to the following set of constraints: (given parameters N,D, L, and theta)
The vector V must be N units long
The elements must have an average of theta
No 2 successive elements may differ by more than +/-10
D == sum(L*cosd(V-theta))
I'm having the most problems with the last one. Any ideas?
Edit
Solutions in other languages or equation form are equally acceptable. Matlab is just a convenient prototyping tool for me, but the final algorithm will be in java.
Edit
From the comments and initial answers I want to add some clarifications and initial thoughts.
I am not seeking a 'truly random' solution from any standard distribution. I want a pseudo randomly generated sequence of values that satisfy the constraints given a parameter set.
The system I'm trying to approximate is a chain of N links of link length L where the end of the chain is D away from the other end in the direction of theta.
My initial insight here is that theta can be removed from consideration until the end, since (2) in essence adds theta to every element of a 0 mean vector V (shifting the mean to theta) and (4) simply removes that mean again. So, if you can find a solution for theta=0, the problem is solved for all theta.
As requested, here is a reasonable range of parameters (not hard constraints, but typical values):
5<N<200
3<D<150
L==1
0 < theta < 360
I would start by creating a "valid" vector. That should be possible - say calculate it for every entry to have the same value.
Once you got that vector I would apply some transformations to "shuffle" it. "Rejection sampling" is the keyword - if the shuffle would violate one of your rules you just don't do it.
As transformations I come up with:
switch two entries
modify the value of one entry and modify a second one to keep the 4th condition (Theoretically you could just shuffle two till the condition is fulfilled - but the chance that happens is quite low)
But maybe you can find some more.
Do this reasonable often and you get a "valid" random vector. Theoretically you should be able to get all valid vectors - practically you could try to construct several "start" vectors so it won't take that long.
Here's a way of doing it. It is clear that not all combinations of theta, N, L and D are valid. It is also clear that you're trying to simulate random objects that are quite complex. You will probably have a hard time showing anything useful with respect to these vectors.
The series you're trying to simulate seems similar to the Wiener process. So I started with that, you can start with anything that is random yet reasonable. I then use that as a starting point for an optimization that tries to satisfy 2,3 and 4. The closer your initial value to a valid vector (satisfying all your conditions) the better the convergence.
function series = generate_series(D, L, N,theta)
s(1) = theta;
for i=2:N,
s(i) = s(i-1) + randn(1,1);
end
f = #(x)objective(x,D,L,N,theta)
q = optimset('Display','iter','TolFun',1e-10,'MaxFunEvals',Inf,'MaxIter',Inf)
[sf,val] = fminunc(f,s,q);
val
series = sf;
function value= objective(s,D,L,N,theta)
a = abs(mean(s)-theta);
b = abs(D-sum(L*cos(s-theta)));
c = 0;
for i=2:N,
u =abs(s(i)-s(i-1)) ;
if u>10,
c = c + u;
end
end
value = a^2 + b^2+ c^2;
It seems like you're trying to simulate something very complex/strange (a path of a given curvature?), see questions by other commenters. Still you will have to use your domain knowledge to connect D and L with a reasonable mu and sigma for the Wiener to act as initialization.
So based on your new requirements, it seems like what you're actually looking for is an ordered list of random angles, with a maximum change in angle of 10 degrees (which I first convert to radians), such that the distance and direction from start to end and link length and number of links are specified?
Simulate an initial guess. It will not hold with the D and theta constraints (i.e. specified D and specified theta)
angles = zeros(N, 1)
for link = 2:N
angles (link) = theta(link - 1) + (rand() - 0.5)*(10*pi/180)
end
Use genetic algorithm (or another optimization) to adjust the angles based on the following cost function:
dx = sum(L*cos(angle));
dy = sum(L*sin(angle));
D = sqrt(dx^2 + dy^2);
theta = atan2(dy/dx);
the cost is now just the difference between the vector given by my D and theta above and the vector given by the specified D and theta (i.e. the inputs).
You will still have to enforce the max change of 10 degrees rule, perhaps that should just make the cost function enormous if it is violated? Perhaps there is a cleaner way to specify sequence constraints in optimization algorithms (I don't know how).
I feel like if you can find the right optimization with the right parameters this should be able to simulate your problem.
You don't give us a lot of detail to work with, so I'll assume the following:
random numbers are to be drawn from [-127+theta +127-theta]
all random numbers will be drawn from a uniform distribution
all random numbers will be of type int8
Then, for the first 3 requirements, you can use this:
N = 1e4;
theta = 40;
diffVal = 10;
g = #() randi([intmin('int8')+theta intmax('int8')-theta], 'int8') + theta;
V = [g(); zeros(N-1,1, 'int8')];
for ii = 2:N
V(ii) = g();
while abs(V(ii)-V(ii-1)) >= diffVal
V(ii) = g();
end
end
inline the anonymous function for more speed.
Now, the last requirement,
D == sum(L*cos(V-theta))
is a bit of a strange one...cos(V-theta) is a specific way to re-scale the data to the [-1 +1] interval, which the multiplication with L will then scale to [-L +L]. On first sight, you'd expect the sum to average out to 0.
However, the expected value of cos(x) when x is a random variable from a uniform distribution in [0 2*pi] is 2/pi (see here for example). Ignoring for the moment the fact that our limits are different from [0 2*pi], the expected value of sum(L*cos(V-theta)) would simply reduce to the constant value of 2*N*L/pi.
How you can force this to equal some other constant D is beyond me...can you perhaps elaborate on that a bit more?