There are 2 matrices:
A: (6 x 78) max=22.2953324329113, min=0
B: (6 x 6 ) max=2187.9013214004 , min=-377.886378385521
B is symmetric and as a result, C = A' * B * A must be a symmetric matrix (theoretically), but this is not the case when I calculate them in Matlab. In fact:
max(max(abs(C - C'))) = 2.3283064365386963e-010
How can I multiply them and get an accurate result?
or
What is a safe way to round the elements of C?
I read this question : efficient-multiplication-of-very-large-matrices-in-matlab, but my problem is not speed or memory. I need an accurate result
Thanks.
You can consider cholesky decomposition of B since it is symmetric
B = R'R
R = chol(A) % // in matlab
then C = A'R'R A =D'D, where D = RA.
With C=D'D you should have machine epsilon precision, although you introduce a possible error due to the accuracy of the decomposition.
You need to read "What Every Computer Scientist Should Know About Floating-Point Arithmetic":
http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html
Realize that computers will never be able to give perfect floating point results, and that leaves you with a few options:
Do as few operations as possible, that is, pick the order of operations for the purpose of having the fewest rounding errors
Fixed decimal point arithmetic - or integer arithmetic - this isn't always practical for all applications, but in some applications, you can get away with this. Financial applications are the commonly cited example (multiply by 100 to make pennies go away! divide by 100 when you are done!).
There are other tricks I can't think of this late.
I'm going to have to give your operations a spin - on my machine, eps gives me 2.2204e-16, which is six orders of magnitude lower than what you are getting. See what eps is on your machine - it should be similar - if it is something like 1e-12 or so, I'd say your result is exactly what you'd expect from those operations.
When I do this with random numbers, I get
a = rand(6, 78);
b = rand(6, 6);
b = b + b'; % To make b symmetric
c = a' * b * a;
max(max(abs(c - c')))
ans =
7.1054e-15
Which is a bit closer to what I'd expect with rounding errors after that many operations, but I am not sure of your input, your machine, and I have no idea what else might be affecting things.
Cheers,--
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).
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'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?
MATLAB does not satisfy matrix arithmetic for inverse, that is;
(ABC)-1 = C-1 * B-1 * A-1
in MATLAB,
if inv(A*B*C) == inv(C)*inv(B)*inv(A)
disp('satisfied')
end
It does not qualify. When I made it format long, I realized that there is difference in points, but it even does not satisfy when I make it format rat.
Why is that so?
Very likely a floating point error. Note that the format function affects only how numbers display, not how MATLAB computes or saves them. So setting it to rat won't help the inaccuracy.
I haven't tested, but you may try the Fractions Toolbox for exact rational number arithmetics, which should give an equality to above.
Consider this (MATLAB R2011a):
a = 1e10;
>> b = inv(a)*inv(a)
b =
1.0000e-020
>> c = inv(a*a)
c =
1.0000e-020
>> b==c
ans =
0
>> format hex
>> b
b =
3bc79ca10c924224
>> c
c =
3bc79ca10c924223
When MATLAB calculates the intermediate quantities inv(a), or a*a (whether a is a scalar or a matrix), it by default stores them as the closest double precision floating point number - which is not exact. So when these slightly inaccurate intermediate results are used in subsequent calculations, there will be round off error.
Instead of comparing floating point numbers for direct equality, such as inv(A*B*C) == inv(C)*inv(B)*inv(A), it's often better to compare the absolute difference to a threshold, such as abs(inv(A*B*C) - inv(C)*inv(B)*inv(A)) < thresh. Here thresh can be an arbitrary small number, or some expression involving eps, which gives you the smallest difference between two numbers at the precision at which you're working.
The format command only controls the display of results at the command line, not the way in which results are internally stored. In particular, format rat does not make MATLAB do calculations symbolically. For this, you might take a look at the Symbolic Math Toolbox. format hex is often even more useful than format long for diagnosing floating point precision issues such as the one you've come across.
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.