As per the title, is the best way to calculate the n-dimensional cross product just using the determinant definition and using the LU Decomposition method of doing as such or could you guys suggest a better one?
Thanks
Edit: for clarity I mean http://en.wikipedia.org/wiki/Cross_product and not the Cartesian Product
Edit: It also seems that using the Leibniz Formula might help - though I don't know how that compares to LU Decomp. at the moment.
From your comment, it seems like you are looking for an operation which takes n −1 vectors as input and computes a single vector as its result, which will be orthogonal to all the input vectors and perhaps have a well-defined length as well.
With defined length
You can characterize the 3-dimensional cross product v =a ×b using the identity v ∙w =det(a,b,w). In other words, taking the cross product of the input vectors and then computing the dot product with any other vector w is the same as plugging the input vectors and that other vector into a matrix and computing its determinant.
This definition can be generalized to arbitrary dimensions. Due to the way a determinant can be computed using Laplace expansion along the last column, the resulting coordinates of that cross product will be the values of all (n −1)×(n −1) sub-determinants you can form from the input vectors, with alternating signs. So yes, Leibniz might be useful in theory, although it is hardly suitable for real-world computations. In practice, you'll soon have to figure out ways to avoid repeating computationswhile computing these n determinants. But wait for the last section of this answer…
Just the direction
Most applications however can do with a weaker requirement. They don't care about the length of the resulting vector, but only about its direction. In that case, what you are asking for is the kernel of the (n −1)×n matrix you can form by taking the input vectors as rows. Any element of that kernel will be orthogonal to the input vectors, and since computing kernels is a common task, you can build on a lot of existing implementations, e.g. Lapack. Details might depend on the language you are using.
Combining these
You can even combine the two approaches above: compute one element of the kernel, and for a non-zero entry of that vector, also compute the corresponding (n −1)×(n −1) determinant which would give you that single coordinate using the first approach. You can then simply scale the vector so that the selected coordinate reaches the computed value, and all the other coordinates will match that one.
Related
In traditional Simplex Algorithm notation, we have x at the current basis selection B as so:
xB = AB-1b - AB-1ANxN. How can I compute the AB-1AN term inside a separator in SCIP, or at least iterate over its columns?
I see three helpful methods: getLPColsData, getLPRowsData, getLPBasisInd. I'm just not sure exactly what data those methods represent, particularly the last one, with its negative row indexes. How do I use those to get the value I want?
Do those methods return the same data no matter what LP algorithm is used? Or do I need to account for dual vs primal? How does the use of the "revised" algorithm play into my calculation?
Update: I discovered the getLPBInvARow and getLPBInvRow. That seems to be much closer to what I'm after. I don't yet understand their results; they seem to include more/less dimensions than expected. I'm still looking for understanding at how to use them to get the rays away from the corner.
you are correct that getLPBInvRow or getLPBInvARow are the methods you want. getLPBInvARow directly returns you a of the simplex tableau, but it is not more efficient to use than getLPBInvRow and doing the multiplication yourself since the LP solver needs to also compute the actual tableau first.
I suggest you look into either sepa_gomory.c or sepa_gmi.c for examples of how to use these methods. How do they include less dimensions than expected? They both return sparse vectors.
I have a set of vectors. I'm working on ways to reduce a n-dimensional vector to a unary value (1-d), say
(x1,x2,....,xn) ------> y
This single value needs to be the characteristic value of the vector. Each unique vector produces a unique output value. Which of the following methods is appropriate:
1- norm of the vector - square root of sum of squares that measures euclidian distance from origin
2- compute hash of F, using some hashing techniques avoiding collision
3- use linear regression to compute, y = w1*x1 + w2*x2 + ... + wn*xn - unlikely to be good if there is no good dependence of input values on output
4- feature extraction technique like PCA that assigns weights to each of x1,x2,..xn based on
the set of input vectors
It's unclear from the method what properties you need this transform to have, so I'm making a guess that you don't need the transformation to preserve any properties other than uniqueness, and possibly invertibility.
None of the techniques you suggest can in general avoid collisions:
Norm - two vectors pointing in opposite directions have the same norm.
Hash - if the input isn't known apriori - what is generally meant by hash function has a finite image, and you have an infinite number of possible vectors - no good.
It's easy to find to vectors which give the same result for any linear regression result (think about it).
PCA is a specific kind of linear transformation - hence the same problem as with linear regression.
So - if you're just looking for uniqueness, you could "stringify" your vectors. One way to do it is to write them down as text strings, with the different coordinates separated by a special character (underscore, for example). Then take the binary value of this string as your representation.
If space is important and you need a more efficient representation, you could think of a more efficient bit encoding: each character in the set 0,1,...,9,'.','' can be represented by 4 bits - a hexadecimal digit (map '.' to A and '' to B). Now encode this string as a hexadecimal number, saving half the space.
Maybe this question is better suited in the math section of the site but I guess stackoverflow is suited too. In mathematics, a vector has a position and a direction, but in programming, a vector is usually defined as:
Vector v (3, 1, 5);
Where is the direction and magnitude? For me, this is a point, not a vector... So what gives? Probably I am not getting something so if anybody can explain this to me it would be very appreciated.
If we are working in cartesian coordinates, and assume (0,0,0) to be the origin, then a point p=(3,1,5) can be written as
where i, j and k are the unit vectors in the x, y and z directions. For convenience sake, the unit vectors are dropped from programming constructs.
The magnitude of the vector is
and its direction cosines are
respectively, both of which can be done programmatically. You can also take dot products and cross-products, which I'm sure you know about. So the usage is consistent between programming and mathematics. The difference in notations is mostly because of convenience.
However as Tomas pointed out, in programming, it is also common to define a vector of strings or objects, which really have no mathematical meaning. You can consider such vectors to be a one dimensional array or a list of items that can be accessed or manipulated easily by indexing.
In mathematics, it is easy to represent a vector by a point - just say that the "base" of the vector is implied to be the origin. Thus, a mathematical point for all practical purposes is also a representation of a mathematical vector, and the vector in your example has the magnitude sqrt(3^2 + 1^2 + 5^2) = 6 and the direction (1/2, 1/6, 5/6) (a normalized vector from the origin).
However, a vector in programming usually has no geometrical use, which means you really aren't interested in things like magnitude or direction. A vector in programming is rather just an ordered list of items. Important here is that the items need not be numbers - it can be anything handled by the language in question! Thus, ("Hello", "little", "world") is also a vector in programming, although it (obviously) has no vector interpretation in the mathematical sense.
Practically speaking (!):
A vector in mathematics is only a direction without a position (actually something more general, but to stay in your terminology). In programming you often use vectors for points. You can think of your vector as the vector pointing from the origin (0,0,0) to the point (3,1,5), called the location vector of the point. Consult texts on analytical and affine geometry for more insight.
A Vector in computer science is an "one dimensional" data structure (array) (can be thought as direction) with an usually dynamic size (length/magnitude). For that reason it is called as vector. But it's an array at least.
A vector also means a set of coordinates. This is how it is used in programming. Just as a set of numbers. You might want to represent position vectors, velocity vectors, momentum vectors, force vectors with a vector object, or you may wish to represent it any way that suits you.
Many times vector quantities may be represented by 4 coordinates instead of 3 (see homogeneous coordinates in computer graphics) so a physical vector is represented by a computer vector with 4 elements. Alternatively you can store direction and magnitude separately, or encode them with 3, 4 or more coordinates.
I guess what I am getting to, is that computer languages are designed to represent physical models, but abstract data containers that the programmer use as tools for his/hers modeling.
Vector in math is an element of n-dimensional space over some field(e.g. real/complex number, functions, string). It may have infinite dimension, e.g. functional space L^2. I don't remember infite-dimensional vectors were used in programming (infinite vectors are not vectors with non-limited length, but vector with infite number of elements)
The most rigorous statement is that a mathematical vector is a first-order tensor that transforms from one coordinate system to another according to tensor transformation rules. The physical idea to keep in mind is that vectors have both magnitude and direction.
Programming vectors are data structures that need not transform according to any rules and may or may not have a notion of a coordinate system as reference. If you happen to use a vector data structure to hold numbers, they may conform to the mathematical definition. But if you have a vector of objects, it's unlikely that they have anything to do with coordinate transformations.
I have a sparse matrix that represents a 3D rectangular space. Along some of the boundaries, I know what the value is going to be (it's a constant). The other boundaries may be reflective, differential, etc.
Should I just set the problem up as if all the boundaries were say, differential, and then go back and set the nodes in the solution vector b to be the constants?
Thanks!
In the finite element method you treat Dirchelet (value constraints) and Neumann (derivative constraints) differently. Usually you assemble the matrix without consideration for boundary conditions first, then apply boundary conditions, then do LU decomposition to solve.
You apply boundary conditions by modifying both the assembled matrix and the RHS vector. I'd have to know more details to tell you exactly what you need to do.
Disclaimer
This is not strictly a programming question, but most programmers soon or later have to deal with math (especially algebra), so I think that the answer could turn out to be useful to someone else in the future.
Now the problem
I'm trying to check if m vectors of dimension n are linearly independent. If m == n you can just build a matrix using the vectors and check if the determinant is != 0. But what if m < n?
Any hints?
See also this video lecture.
Construct a matrix of the vectors (one row per vector), and perform a Gaussian elimination on this matrix. If any of the matrix rows cancels out, they are not linearly independent.
The trivial case is when m > n, in this case, they cannot be linearly independent.
Construct a matrix M whose rows are the vectors and determine the rank of M. If the rank of M is less than m (the number of vectors) then there is a linear dependence. In the algorithm to determine the rank of M you can stop the procedure as soon as you obtain one row of zeros, but running the algorithm to completion has the added bonanza of providing the dimension of the spanning set of the vectors. Oh, and the algorithm to determine the rank of M is merely Gaussian elimination.
Take care for numerical instability. See the warning at the beginning of chapter two in Numerical Recipes.
If m<n, you will have to do some operation on them (there are multiple possibilities: Gaussian elimination, orthogonalization, etc., almost any transformation which can be used for solving equations will do) and check the result (eg. Gaussian elimination => zero row or column, orthogonalization => zero vector, SVD => zero singular number)
However, note that this question is a bad question for a programmer to ask, and this problem is a bad problem for a program to solve. That's because every linearly dependent set of n<m vectors has a different set of linearly independent vectors nearby (eg. the problem is numerically unstable)
I have been working on this problem these days.
Previously, I have found some algorithms regarding Gaussian or Gaussian-Jordan elimination, but most of those algorithms only apply to square matrix, not general matrix.
To apply for general matrix, one of the best answers might be this:
http://rosettacode.org/wiki/Reduced_row_echelon_form#MATLAB
You can find both pseudo-code and source code in various languages.
As for me, I transformed the Python source code to C++, causes the C++ code provided in the above link is somehow complex and inappropriate to implement in my simulation.
Hope this will help you, and good luck ^^
If computing power is not a problem, probably the best way is to find singular values of the matrix. Basically you need to find eigenvalues of M'*M and look at the ratio of the largest to the smallest. If the ratio is not very big, the vectors are independent.
Another way to check that m row vectors are linearly independent, when put in a matrix M of size mxn, is to compute
det(M * M^T)
i.e. the determinant of a mxm square matrix. It will be zero if and only if M has some dependent rows. However Gaussian elimination should be in general faster.
Sorry man, my mistake...
The source code provided in the above link turns out to be incorrect, at least the python code I have tested and the C++ code I have transformed does not generates the right answer all the time. (while for the exmample in the above link, the result is correct :) -- )
To test the python code, simply replace the mtx with
[30,10,20,0],[60,20,40,0]
and the returned result would be like:
[1,0,0,0],[0,1,2,0]
Nevertheless, I have got a way out of this. It's just this time I transformed the matalb source code of rref function to C++. You can run matlab and use the type rref command to get the source code of rref.
Just notice that if you are working with some really large value or really small value, make sure use the long double datatype in c++. Otherwise, the result will be truncated and inconsistent with the matlab result.
I have been conducting large simulations in ns2, and all the observed results are sound.
hope this will help you and any other who have encontered the problem...
A very simple way, that is not the most computationally efficient, is to simply remove random rows until m=n and then apply the determinant trick.
m < n: remove rows (make the vectors shorter) until the matrix is square, and then
m = n: check if the determinant is 0 (as you said)
m < n (the number of vectors is greater than their length): they are linearly dependent (always).
The reason, in short, is that any solution to the system of m x n equations is also a solution to the n x n system of equations (you're trying to solve Av=0). For a better explanation, see Wikipedia, which explains it better than I can.