I fail to understand why, in the below example, only x1 turns into a 1000 column array while y is a single number.
x = [0:1:999];
y = (7.5*(x))/(18000+(x));
x1 = exp(-((x)*8)/333);
Any clarification would be highly appreciated!
Why is x1 1x1000?
As given in the documentation,
exp(X) returns the exponential eˣ for each element in array X.
Since x is 1x1000, so -(x*8)/333 is 1x1000 and when exp() is applied on it, exponentials of all 1000 elements are computed and hence x1 is also 1x1000. As an example, exp([1 2 3]) is same as [exp(1) exp(2) exp(3)].
Why is y a single number?
As given in the documentation,
If A is a rectangular m-by-n matrix with m~= n, and B is a matrix
with n columns, then x = B/A returns a least-squares solution of the
system of equations x*A = B.
In your case,
A is 18000+x and size(18000+x) is 1x1000 i.e. m=1 and n=1000, and m~=n
and B is 7.5*x which has n=1000 columns.
⇒(7.5*x)/(18000+x) is returning you least-squares solution of equations x*(18000+x) = 7.5*x.
Final Remarks:
x = [0:1:999];
Brackets are unnecessary here and it should better be use like this: x=0:1:999 ;
It seems that you want to do element-wise division for computing x1 for which you should use ./ operator like this:
y=(7.5*x)./(18000+x); %Also removed unnecessary brackets
Also note that addition is always element-wise. .+ is not a valid MATLAB syntax (It works in Octave though). See valid arithmetic array and matrix operators in MATLAB here.
3. x1 also has some unnecessary brackets.
The question has already been answered by other people. I just want to point out a small thing. You do not need to write x = 0:1:999. It is better written as x = 0:999 as the default increment value used by MATLAB or Octave is 1.
Try explicitly specifying that you want to do element-wise operations rather than matrix operations:
y = (7.5.*(x))./(18000+(x));
In general, .* does elementwise multiplication, ./ does element-wise division, etc. So [1 2] .* [3 4] yields [3 8]. Omitting the dots will cause Matlab to use matrix operations whenever it can find a reasonable interpretation of your inputs as matrices.
I am having some trouble understanding Multi-Valued Dependencies. The definition being: A multivalued dependency exists when there are at least 3 attributes (like X,Y and Z) in a relation and for value of X there is a well defined set of values of Y and a well defined set of values of Z. However, the set of values of Y is independent of set Z and vice versa.
Suppose we have a relation R(A,B,C,D,E) that satisfies the MVD's
A →→ B and B →→ D
How does MVD play into A->B and B->D here? Honestly I'm not sure I really understand the definition after looking at example problems.
If R contains the tuples (0,1,2,3,4) and (0,5,6,7,8), what other tuples must
necessarily be in R? Identify one such tuple in the list below.
a) (0,5,2,3,8)
b) (0,5,6,3,8)
c) (0,5,6,7,4)
d) (0,1,6,3,4)
I would have thought AB is 0,1 and 0,5 and BD is 1,3 and 5,7. None of the answers have 0,1,3,5,7.
MVDs (multi-valued dependencies) have nothing to do with "at least 3 attributes". (You seem to be quoting the Wikipedia article but that informal definition is partly wrong and partly unintelligible.) You need to read and think through a clear, precise and correct definition. (Which you were probably given.)
MVD X ↠ Y mentions two subsets of the set S of all attributes, X & Y. There are lots of ways to define when a MVD holds in a relation but the simplest to state & envisage is probably that the two projections XY and X(S-Y) join to the original relation. Which also mentions a third subset, S-Y. Which is what the (binary) JD (join dependency) {XY, X(S-Y)} says.
Wikipedia (although that article is a mess):
A decomposition of R into (X, Y) and (X, R−Y) is a lossless-join decomposition if and only if X ↠ Y holds in R.
From this answer:
MVDs always come in pairs. Suppose MVD X ↠ Y holds in a relation with attributes S, normalized to components XY & X(S-Y). Notice that S-XY is the set of non-X non-Y attributes, and X(S-Y) = X(S-XY). Then there is also an MVD X ↠ S-XY, normalized to components X(S-XY) & X(S-(S-XY)), ie X(S-XY) & XY, ie X(S-Y) & XY. Why? Notice that both MVDs give the same component pair. Ie both MVDs describe the same condition, that S = XY JOIN X(S-XY). So when an MVD holds, that partner holds too. We can write the condition expressed by each of the MVDs using the special explicit & symmetrical notation X ↠ Y | S-XY.
For R =
A,B,C,D,E
0,1,2,3,4
0,5,6,7,8
...
A ↠ B tells us that the following join to R:
A,B A,C,D,E
0,1 0,2,3,4
0,5 0,6,7,8
... ...
so R has at least
A,B,C,D,E
0,1,2,3,4
0,1,6,7,8
0,5,2,3,4
0,5,6,7,8
of which two are given and two are new but not choices.
B ↠ D tells us that the following join to R:
B,D A,B,C,E
1,3 0,1,2,4
5,7 0,5,6,8
... ...
so R has at least
A,B,C,D,E
0,1,2,3,4
0,5,6,7,8
which we already know.
So we don't yet know whether any of the choices are in R. But we do now know R is
A,B,C,D,E
0,1,2,3,4
0,1,6,7,8
0,5,2,3,4
0,5,6,7,8
...
Repeating, A ↠ B adds no new tuples but B ↠ D now gives this join:
B,D A,B,C,E
1,3 0,1,2,4
1,7 0,1,6,8
5,3 0,5,2,4
5,7 0,5,6,8
... ...
And one of the tuples in that join is choice b) (0,5,6,3,8).
The way the question is phrased, they are probably expecting you to use a definition that they will have given you that is like another two in Wikipedia. One says that α ↠ β holds in R when
[...] if we denote by (x, y, z) the tuple having values for α, β, R − α − β collectively equal to x, y, z, then whenever the tuples (a, b, c) and (a, d, e) exist in r, the tuples (a, b, e) and (a, d, c) should also exist in r.
(The only sense in which this gives the "formal" definition "in more simple words" is that this is also a definition. Because this isn't actually paraphrasing it, because this uses R − α − β whereas it uses R − β.)
By applying this rule to repeatedly generate further tuples for R starting from the given ones, we end up generating b) (0,5,6,3,8) much as we did above.
PS I would normally suggest that you review the (unsound) reasoning that led you to expect "AB is 0,1 and 0,5 and BD is 1,3 and 5,7" (whatever that means) or "0,1,3,5,7". But the "definition" you give (from Wikipedia) doesn't make any sense. So I suggest that you consider what you were doing with it.
I have to write a program that solves linear equations with the Cramer method, and ask specifically to find the determinant with the Laplace expansion.
det A = sum on i=1...N:(-1)**(i+1) a_i1 det ||A||_i1
where ||A||_i1 is the cofactor matrix of A, a n-1 X n-1 matrix, created by eliminating the i row and 1 column.
and that's where I'm stuck.
This is what I wrote so far
integer, parameter :: rk= selected_real_kind(6)
end module prec
module lap
use prec
implicit none
contains
recursive function det(a,n) result (d)
real(kind=rk), intent(in), dimension(n,n) :: a
real(kind=rk), dimension(n-1,n-1) :: b
real(kind=rk) :: d
integer ::i
integer, intent(in)::n
if (size(a) > 4) then
do i=1,n
b(1:(i-1),:) = a(1:(i-1),:)
b(i:n,:) = a((i+1):n,:)
b(:,:) = a(:,2:n)
d= ((-1)**(i+1))*a(i,1)*det(b,n)
end do
else
d = a(1,1)*a(2,2)-a(1,2)*a(2,1)
end if
end function det
end module lap
program sistema
it keeps telling me that I have non-conformant arrays, even though I'm using the subsets (and my professor says it's quite easy to obtain |A| with subsets).
It would be nice to see the portions of your code that are missing. However, I think I see the problem:
With dimensions a(n,n) and b(n-1,n-1), you cannot do b(1:(i-1),:) = a(1:(i-1),:) because the sizes of second dimension do not match. You should check the definition of a cofactor matrix - you should be removing a column as well as a row. Instead of the the three lines setting b, you should try:
b(1:(i-1),:) = a(1:(i-1),2:n)
b(i:n,:) = a((i+1):n,2:n)
Edit: damn, I always mix up row and column. Perhaps I mean remove a row instead of a column?
Suppose that f(x,y) is a bivariate function as follows:
function [ f ] = f(x,y)
UN=(g)1.6*(1-acos(g)/pi)-0.8;
f= 1+UN(cos(0.5*pi*x+y));
end
How to improve execution time for function F(N) with the following code:
function [VAL] = F(N)
x=0:4/N:4;
y=0:2*pi/1000:2*pi;
VAL=zeros(N+1,3);
for i = 1:N+1
val = zeros(1,N+1);
for j = 1:N+1
val(j) = trapz(y,f(0,y).*f(x(i),y).*f(x(j),y))/2/pi;
end
val = fftshift(fft(val))/N;
l = (length(val)+1)/2;
VAL(i,:)= val(l-1:l+1);
end
VAL = fftshift(fft(VAL,[],1),1)/N;
L = (size(VAL,1)+1)/2;
VAL = VAL(L-1:L+1,:);
end
Note that N=2^p where p>10, so please consider the memory limitations while optimizing the code using ndgrid, arrayfun, etc.
FYI: The code intends to find the central 3-by-3 submatrix of the fftn of
fun=#(a,b) trapz(y,f(0,y).*f(a,y).*f(b,y))/2/pi;
where a,b are in [0,4]. The key idea is that we can save memory using the code above specially when N is very large. But the execution time is still an issue because of nested loops. See the figure below for N=2^2:
This is not a full answer, but some possibly helpful hints:
0) The trivial: Are you sure you need numerics? Can't you do the computation analytically?
1) Do not use function handles:
function [ f ] = f(x,y)
f= 1+1.6*(1-acos(cos(0.5*pi*x+y))/pi)-0.8
end
2) Simplify analytically: acos(cos(x)) is the same as abs(mod(x + pi, 2 * pi) - pi), which should compute slightly faster. Or, instead of sampling and then numerically integrating, first integrate analytically and sample the result.
3) The FFT is a very efficient algorithm to compute the full DFT, but you don't need the full DFT. Since you only want the central 3 x 3 coefficients, it might be more efficient to directly apply the DFT definition and evaluate the formula only for those coefficients that you want. That should be both fast and memory-efficient.
4) If you repeatedly do this computation, it might be helpful to precompute DFT coefficients. Here, dftmtx from the Signal Processing toolbox can assist.
5) To get rid of the loops, think about the problem not in the form of computation instructions, but a single matrix operation. If you consider your input N x N matrix as a vector with N² elements, and your output 3 x 3 matrix as a 9-element vector, then the whole operation you apply (numerical integration via trapz and DFT via fft) appears to be a simple linear transform, which it should be possible to express as an N² x 9 matrix.
I have a rather unusual challenge for lapack, and I have spent hours searching for a solution to it.
I have a generalized eigenvalue problem of the traditional form (A - x B = 0). Normally I would use for instance ?hegvx or ?hegvd to calculate the eigenvalues and the eigenvectors.
However the challenge I am facing now, is now that I already know the eigenvalues from the construction of the problem, and therefore I need an efficient lapack routine for calculating the eigenvectors only?
Anyone got a hack for this?
Given the generalised eigenvalue problem
(A - y B) x = 0
And an eigenvalue yn:
(A - yn B) xn = 0
We know A, B and yn, so we can form a new matrix Cn
Cn = A - yn B
Cn xn = 0
You can solve this with any linear algebra solver individually for each eigenvalue. According to the LAPACK docs on linear equations, for a general matrix, double precision, you might use DGETRS
Edit for degenerate eigenvalues:
The null space of the matrix Cn is what we are solving for here (as MvG commented). If
Cn j = 0 and
Cn k = 0
(i.e. degenerate e-vals) then given jTk = 0 (both are still eigenvectors of the AB system) we can say
Call a row of Cn r:
r.k = r.k - jT.k = (r-jT).k
Thus forming a matrix J whose rows are each jT (there must be a name for this, but I don't know it):
(Cn - J) k = 0
Define
Dnj = Cn - J
And now solve for the new matrix Dnj. By construction this will be a new, orthogonal eigenvector of the original matrix with the same degenerate eigenvalue.