How can I determine what matrix is needed to flip an image upside down?

10 Ansichten (letzte 30 Tage)
I need to write a program that will flip an image upside down. This is for a linear algebra class so I am not allowed to use flipud(). I must find a transformation matrix (T) which can be multiplied by an image (X) to give the flipped image.
T*X_original = X_flipped
I'm not sure where to start with this. Thanks!

Akzeptierte Antwort

Image Analyst
Image Analyst am 27 Okt. 2013
Bearbeitet: Image Analyst am 27 Okt. 2013
Try this:
flippedMatrix = originalMatrix(end:-1:1, :);
% Alternatively (less efficient) but what you want.
T = zeros(size(originalMatrix, 1)); % INitialize.
for k = 1 : size(originalMatrix, 1)
T(k, end-k+1) = 1;
end
T % Print to command window.
flippedMatrix2 = T * originalMatrix

Weitere Antworten (1)

Youssef  Khmou
Youssef Khmou am 27 Okt. 2013
Bearbeitet: Youssef Khmou am 27 Okt. 2013
Generally that type of matrices are for euclidean space where a vector or a set of vector can be rotated with a cos and sin rotation matrix, however there might a real matrix that provide what you described , in the mean while you try this way :
X=im2double(imread('circuit.tif')); % Original image
[m n]=size(X);
Y=X(m:-1:1,n:-1:1); % flipped image upside down
% gievn T*X=Y , is T bounded operator ??????? can you answer this equation
T=Y*pinv(X); % Pseudo inverse because Y is not square .

Kategorien

Mehr zu Images finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by