For affine 4*4 transformation, I saw two representation in different text
one is
L T
0 1
Another is
L 0
T 1
L is the linear part, T is the translation part; I am wondering which is correct?
Both forms are correct. The first is used in left-multiply matrix by column vector
ResultVector = Matrix*V (for example, in OpenGL), and the second - in right-multiply convention with row vector V*Matrix (for example, in DirectX)
Related
As part of my journey to learn j I implemented a technique for computing the area of a polygon I came across in Futility Closet. I came up with a solution, but it's quite inelegant, so I'm interested in better methods:
polyarea =: -:#((+/#((1&{&|:)*(0{&|:1&|.)))-(+/#((0&{&|:)*(1{&|:1&|.))))
y =: 2 7 9 5 6,.5 7 1 0 4
polyarea y
20
This technique rotates one column and takes the dot product of the columns, then does the same after rotating the other column. The area is half the difference of these two results.
Interested in suggestions!
I think that their technique boils down to using the determinant to find the area of the polygon http://mathworld.wolfram.com/PolygonArea.html
But using the Futility Closet technique I would first close the polygon by adding the first point to the end.
y =: 2 7 9 5 6,.5 7 1 0 4
close=: (, {.)
close is a hook that takes the first pair and appends it to the end
Then take the determinants two points at a time, which is essentially what they are doing with their columns and rotations
dets=: 2 (-/ . *)\ close
dets takes the determinant of each pair of points - result is negative if the points are in clockwise order
Then take those values and process for the answer.
clean=: |#:-:#:(+/)
clean sums up the determinants, divides by 2 and returns the absolute value of the result.
clean #: dets y
20
To see the result in complete tacit form we can lean on the f. adverb (Fix) to flatten our definitions.
clean #: dets f.
|#:-:#:(+/)#:(2 -/ .*\ (, {.))
It is just a different way of looking at what they are doing, but it allows J to use the . conjunction (Dot Product) and \ adverb (Infix) to handle all of those rotations with determinants.
Hope this helps.
Suppose you have Matrix A.
Suppose also that we have Matrix C
If we have A = B x C and we want to find out the B matrix values which I believe should be 3x3 (Correct me if I am wrong)
Do we need to use matrix inversion here? I did not use algebra since many years.
I do not have a code yet but if someone can provide a snippet that will be great.
This is a problem that I have in image processing where the A , C hold RGB values.
The submitted matrices are just for illustration.
I am trying to solve this problem using Python numpy
I hope that someone can help with it.
Your matrix should be 5x5. As we are dealing with non-square matrices, you could use the generalized inverse of C to obtain B:
import numpy as np
np.random.seed(10)
A = np.random.randint(0,9,(5,3))
C = np.random.randint(0,9,(5,3))
B = np.matmul(A,np.linalg.pinv(C))
print B
Building on percusse's comment, you can do this with numpy.linalg.lstsq. However, this assumes that we are performing matrix left division but the situation is your question is for right division.
Using the fact that you are solving for B with B = A / C, lstsq solves problems of the type A \ C. To convert this into a form for lstsq, we can convert it into the latter problem by:
B = A / C = (C' \ A')'
The ' operator is the transpose. The above is found by linear algebra rules. Specifically, perform two transposes: ((A / C)')' where transposing a matrix twice is simply the result of itself. Also, knowing that (AC)' is equal to C'A' and for a matrix, the inverse of the transpose is equal to the transpose of the inverse you should get the above relationship.
Therefore:
B = numpy.linalg.lstsq(C.T, A.T)[0].T
The output of lstsq is a tuple where the first element is the actual solution.
Take note that for your particular example, C is a rank-deficient matrix so you won't be able to reconstruct A properly from B and C.
I have an S1 (21x21) matrix and a W (21x21) matrix given. I define a matrix results with each element as a matrix as results = {W};
and then, I have two regular for loops such that it runs all the values in index1 and then goes to the second index; but each time it should take a specific value of k for example.
There are also two given vectors cos and ens each having dimension 21x1. Here is the code:
rowsP=21;
M=0;
beta=0.9;
p=0.5;
q=0.5;
k= [1:rowsP-1];
for j=1:rowsP-k
for i=1:rowsP-k
R(i,j) = ( S1(i,end-k) - cos(j+k) ) *ens(j)-0.001*M +
beta*(p*results{k}(i,end-j)+q*results{k}(i+1,end-j));
results{k+1}=fliplr(R);
end
end
I am getting the error
Matrix Dimensions must agree.
So I am trying to calculate a matrix results each time using two for loops given results{1}=W (a given matrix) given k=1.
Then flipping the matrix left to right, I get results{2} which then helps to calculate R again but for k=2. And this is then repeated until k=21.
As you see, I keep dropping the last column of each successive R, the matrix results should be appended each time giving a row of 21 elements each cell having 21x21 matrix (the given matrix W) and then a matrix of 20x20 and then 19x19 and so on... until a matrix of 1x1. I am unable to solve the problem as Matlab only does 1 iteration and then does not compute the correct answer. I keep getting two cells in results with a 21x21 matrix (the one given) and the next 20x20 matrix.
I tried with another for loop for k, but in that case, for a given k, starting from k=1, it runs the whole code for j and then i, but it does not solve my problem.
This is a question in MatLab...
I have two matrices, one being a (5 x 1 double) :
1
2
3
1
3
And the second matrix being a (5 x 3 string), with spaces where no character appears :
a
bc
def
g
hij
I am trying to get an output such that a (5 x 1 string) is created and outputs the nth value from each line of matrix two, where n is the value in matrix one. I am unsure how to do this using a mask which would be able to handle much larger matrces. My target matrix would have the following :
a
c
f
g
j
Thank you very much for the help!!!
There are so many ways you can accomplish this task. I'll give you two.
Method #1 - Generate linear indices and access elements
Use sub2ind to generate a set of linear indices that correspond to the row and column locations you want to access in your matrix. You'll note that the column locations are the ones changing, but the row locations are always increasing by 1 as you want to access each row. As such, given your string matrix A, and your columns you want to access stored in ind, just do this:
A = ['a '; 'bc '; 'def'; 'g ';'hij'];
ind = [1 2 3 1 3];
out = A(sub2ind(size(A), (1:numel(ind)).', ind(:)))
out =
a
c
f
g
j
Method #2 - Create a sparse matrix, convert to logical and access
Alternatively, you can create a sparse matrix through sparse where the non-zero entries are rows vary from 1 up to as many elements as you have in ind and the columns vary like what you have given us.
S = sparse((1:numel(ind)).',ind(:),true,size(A,1),size(A,2));
A = A.'; out = A(S.');
Be mindful that you are trying to access each element in a row-major fashion, yet MATLAB will do this in a column-major format. As such, we would need to transpose our data matrix, and also take our sparse matrix and transpose that too. The end result should give you the same order as Method #1.
for back face culling algorithm I need to find the normal vector for each polygon.
given 3 points, I want to find the normal of a plane.
so I know how to do it :
find 2 vectors on the plane
find their cross vector - which will give me the normal vector (a,b,c)
my question is, does it matter what is the order of the points when I find 2 vectors?
for ex: given 3 points: p1(0,0,0), p2(5,0,0), p3(10,10,10)
does it matter if I choose vector
V1=(p2-p1)=(5, 0, 0)-(0, 0, 0)=(5, 0, 0)
V2=(p3-p1)=(10,10,10)-(0, 0, 0)=(10, 10, 10)
or
v1=(p1-p2)
v2=(p1-p3)
your polygon has vertexes a, b, c.
you calculate the vectors:
v1 = a-c
v2 = b-c
this refers a and b to c. It would be the same if you decided to refer, say, b and c to a.
calculate the cross product v1*v2 (this gives a vector perpendicular to v1 and v2) and normalize it.
If, you did calculate (a-b)(a-c) instead of (b-a)(c-a), the resulting vector will be mirrored (ie, pointing to the wrong direction).
OT: normalize with/see http://en.wikipedia.org/wiki/Fast_inverse_square_root that was developed exactly to calculate face normals