i have a n-dimensional matrix from which i want the n-2 -dimensional submatrix
For example:
if i have a matrix A that is 6x5x4x3x2, i want to get for example the matrix
B = A(1,:,:,:,2);
This is easy if the amount of dimensions is fixed, but how can i do this for variable dimensions without having to handle a specific case for each number of dimensions?
Bad:
n = length(size(A));
if (n == 2)
B = A(1,2)
else if (n == 3)
B = A(1,:,2);
else if (n == 4)
B = A(1,:,:,2);
else if (n == 5)
B = A (1,:,:,:,2);
...
Good:
B=A(1,<some cool operator/expression>,2);
Thanks to the link provided in the comment, i was able to solve it like this:
n = ndim(A);
if (n <= 2)
B = A(1,2);
else
colons = repmat({':'},[1 n-2]);
B = A(1,colons{:},2);
Related
so I'm new to Matlab and had to draw the Impulsefunction with y(n) is only 1 if n == 3, else 0. The following code works:
n = -5:5;
f = n; % allocate f
for i = 1 : length(n)
f(i) = dd1(n(i)-3);
end
stem(n, f);
function y = dd1(n)
y = 0;
if n == 0
y = 1;
end
end
But I feel like it's to complicated, so I tried the following:
n = -5:5
stem(n, fo)
function y = fo(n)
y = 0
if n == 3
y=1
end
end
This returns
Not enough input arguments.
Error in alternative>fo (line 5)
if n == 3
Error in alternative (line 2)
stem(n, fo)
I feel like I'm missing something trivial here.
if is no vector-wise operation but expects a single boolean (or at least a scalar that it can cast to a boolean).
But you can do this vector-wise:
lg = n == 3;
This produces a logical (MATLAB's name for boolean) array (because n is an array and not a vector), which is true where n is equal (==) to three. So you don't need a function, because you can make use of MATLAB's ability to work with vectors and arrays implicitly.
(for your code it would be f = (n-3) == 3)
A last hint: if you have a state-space system (ss-object), you can use the function step to get the step-response as a plot.
I am trying to code the Cramer's Rule on MATLAB for a square matrix. I understand the Rule perfectly and also believe the logic behind my code is alright. However, could you please check where I might be going wrong that my result is not being displayed correctly? Any help would be appreciated! Thanks :)
function x = cramer(A, b)
r = size(A, 1);
c = size(A, 2);
n = size(b, 1);
if r ~= c
disp('Oops! Please enter a square matrix');
end
if r == c == n
D = det(A);
if D == 0
disp('Oops! Either, there are a family of soultions or no unique solution')
end
if D ~= 0
result = zeros(n, 1);
for (i = 1:n)
A_x = A;
A_x(:, i) = b;
Dx = det(A_x);
result(i,1) = Dx/D;
end
x = result;
end
end
end
You have an error in your if r == c == n check.
The expression r == c == n can be expanded as (r == c) == n
So it compares the value of logical(r == c) with n.
As an example:
>> 2 == 2 == 1
ans =
logical
1
Rewrite your check as r == c && c == n or isequal(r,c,n) and you should be good.
EDIT: You can simplify your logic quite some, say for example by:
function x = cramers(A, b)
if diff(size(A)) && size(A,1) == n
D = det(A);
if D == 0
disp('Oops! Either, there are a family of solutions or no unique solution')
return
else
result = zeros(n, 1);
for i = 1:n
A_x = A;
A_x(:, i) = b;
Dx = det(A_x);
result(i,1) = Dx/D;
end
x = result;
end
else
disp('Dimension error')
return
end
end
I want to put a small matrix (p-by-q) called B into a bigger matrix (m-by-n) called A. How can I do it? Matrix B should be put on the right corner of matrix A:
You can do this with basic array indexing, for example:
m = 3;
n = 4;
A = rand(m,n)
p = 2;
q = 3;
B = rand(p,q)
A(end-p+1:end,end-q+1:end) = B
... assuming that p <= m and q <= n of course.
I need to create arbitrary perpendicular vector n with components (a, b, c) to another known vector k with components (x,y,z).
The following code creates arbitrary vector n, but I need random numbers for components in the range [-inf, inf] how can I acheive that? (because otherwise vector components created may not exceed some value in given case 10^11 ) Or maybe concept "arbitrary vector" does not require that?
function [a,b,c] = randomOrghogonalVector(x,y,z)
a = 0;
b = 0;
c = 0;
randomDistr = rand * 10^11 * 2 - 10^11; % issue 1
% excluding trivial solution
if x == 0 && y == 0 && z ==0
a = NaN; b = a; c = a;
else
if z ~=0
a = randomDistr;
b = randomDistr;
c = - (x * a + b * y ) / z;
else
if z == 0 && x ~= 0
c = randomDistr;
b = randomDistr;
a = - (z * c + b * y ) / x;
else
if z == 0 && x == 0 && y ~= 0
c = randomDistr;
a = randomDistr;
b = - (z * c + a * x ) / y;
end
end
end
end
The easiest solution I see is to first find a random vector that is orthogonal to your original vector, and then give it a random length. In Matlab, this can be done by defining the following function
function [a, b, c] = orthoVector(x, y, z)
xin = [x;y;z];
e = xin;
while ((e'*xin)==xin'*xin)
e = 2.*rand(3,1)-1;
end
xout = cross(xin, e);
xout = 1.0/(rand()) * xout;
a = xout(1);
b = xout(2);
c = xout(3);
end
Line-by-line, here's what I'm doing:
you asked for this format [a,b,c] = f(x,y,z). I would recommend using function xout = orthoVector(xin), which would make this code even shorter.
Since Matlab handles vectors best as vectors, I'm creating vector xin.
e will be one random vector, different from xin used to compute the orthogonal vector. Since we're dealing with random vectors, we initialize it to be equal to xin.
For this algorithm to work, we need to make sure that e and xin are pointing in different directions. Until this is the case...
...create a new random vector e. Note that rand will give values between 0 and 1. Thus, each component of e will be between -1 and 1.
Ok, if we end, e and xin are pointing in different directions
Our vector xout will be orthogonal to xin and e.
Let's multiply vector xout by a random number between 1 and "very large"
a is first component of xout
b is second component of xout
c is third component of xout
all done.
Optional: if you want to have very large vectors, you could replace line 8 by
xout = exp(1./rand())/(rand()) * xout;
This will give you a very large spread of values.
Hope this helps, cheers!
I would like to generate the adjacency matrix of an undirected graph with N nodes.
In particular, this graph should have a fixed degree (each node is connected to a fixed number of node d).
If a set d = N-1, the solution is trivial:
A = ones(N) - eye(N);
How can I generalize it for any d?
ADD:
Here is a solution (thanks to Oli Charlesworth):
function A = fixedDegreeGraph(N, d)
A = zeros(N);
for i=1:N
b = i;
f = i;
for k=1:floor(d/2)
f = f + 1;
if (f == N + 1)
f = 1;
end
A(i, f) = 1;
A(f, i) = 1;
b = b - 1;
if (b == 0)
b = N;
end
A(i, b) = 1;
A(b, i) = 1;
end
end
For even d, here's a way to visualise the approach.
Draw the vertices out arranged in a circle.
Each vertex is connected to its immediate (d/2) left-hand neighbours, and its immediate (d/2) right-hand neighbours.
It should be fairly obvious how to turn this into an adjacency matrix (hint: it will be a circulant matrix, so you may find the toeplitz function useful).
Extending this to odd d is not much harder... (although note there is no solution if both N and d are odd)