MATLAB exercise program (image rotation, bilinear interpolation)

Many algorithms have long been implemented, but they are not implemented for various reasons. This bilinear interpolation rotating image is one of them.

Previously I wrote about the nearest interpolation rotating image and the portal. It works well when combined.

Clear all; close all; clc; jiaodu = 45; degrees of rotation to be rotated, the direction of rotation is instant needle img1_imread('lena.jpg '); % here V is the height of the original image, U is the width of the original image imshow (IMG ); % here Y is the height of the transformed image, X is the width of the transformed image [H w] = size (IMG); Theta = jiaodu/180 * PI; rot = [cos (theta)-sin (theta) 0; sin (theta) Cos (theta) 0; 0 0 1]; pix1 = [1 1 1] * rot; % coordinate pix2 = [1 W 1] * rot of the upper left point of the transformed image; % coordinate pix3 = [H 1 1] * rot of the upper right point of the transformed image; % coordinate pix4 = [h w 1] * rot; % coordinate Height = round (max ([ABS (pix1 (1) -pix4 (1) + 0.5 ABS (pix2 (1)-pix3 (1) + 0.5]); % width of the transformed image = round (max ([ABS (pix1 (2)-pix4 (2) + 0.5 ABS (pix2 (2)-pix3 (2 )) + 0.5]); % converted Image Width imgn = zeros (height, width); delta_y = ABS (min ([pix1 (1) pix2 (1) pix3 (1) pix4 (1)]); % get the offset of the negative axis exceeding the y direction delta_x = ABS (min ([pix1 (2) pix2 (2) pix3 (2) pix4 (2)]); % get the offset of the negative axis in the X direction beyond for I = 1-delta_y: height-delta_y for J = 1-delta_x: width-delta_x pix = [I J 1]/rot; % use the coordinates of the transformed image points to find the coordinates of the original image points. % otherwise, the pixels of some transformed images cannot be completely filled with float_y = pix (1) -Floor (pix (1); float_x = pix (2)-floor (pix (2); If pix (1)> = 1 & pix (2)> = 1 & pix (1) <= H & pix (2) <= W pix_up_left = [floor (pix (1) floor (pix (2)]; % four adjacent points pix_up_right = [floor (pix (1) Ceil (pix (2)]; pix_down_left = [Ceil (pix (1 )) floor (pix (2)]; pix_down_right = [Ceil (pix (1) Ceil (pix (2)]; value_up_left = (1-float_x) * (1-float_y ); % calculate the weights of neighboring four points value_up_right = float_x * (bytes); value_down_left = (1-float_x) * float_y; value_down_right = float_x * float_y; imgn (I + delta_y, J + delta_x) = value_up_left * IMG (pix_up_left (1), pix_up_left (2) +... value_up_right * IMG (pix_up_right (1), pix_up_right (2) +... value_down_left * IMG (pix_down_left (1), pix_down_left (2) +... value_down_right * IMG (pix_down_right (1), pix_down_right (2); End endfigure, imshow (uint8 (imgn ))

Source image

Nearest interpolation Rotation

Bilinear interpolation Rotation

Postscript:

The above cannot pass through the limit. If the rotation is 90 degrees or 180 degrees, the boundary will have black pixels. Modify as follows:

Main. m

Clear all; close all; clc; jiaodu = 90; degrees of rotation to be rotated, the direction of rotation is instant needle img1_imread('lena.jpg '); % here V is the height of the original image, U is the width of the original image imshow (IMG ); % here Y is the height of the transformed image, X is the width of the transformed image [H w] = size (IMG); Theta = jiaodu/180 * PI; rot = [cos (theta)-sin (theta) 0; sin (theta) Cos (theta) 0; 0 0 1]; pix1 = [1 1 1] * rot; % coordinate pix2 = [1 W 1] * rot of the upper left point of the transformed image; % coordinate pix3 = [H 1 1] * rot of the upper right point of the transformed image; % coordinate pix4 = [h w 1] * rot; % coordinate Height = round (max ([ABS (pix1 (1) -pix4 (1) + 0.5 ABS (pix2 (1)-pix3 (1) + 0.5]); % width of the transformed image = round (max ([ABS (pix1 (2)-pix4 (2) + 0.5 ABS (pix2 (2)-pix3 (2 )) + 0.5]); % converted Image Width imgn = zeros (height, width); delta_y = ABS (min ([pix1 (1) pix2 (1) pix3 (1) pix4 (1)]); % get the offset of the negative axis exceeding the y direction delta_x = ABS (min ([pix1 (2) pix2 (2) pix3 (2) pix4 (2)]); % get the offset imgm = img_extend (IMG, 1) of the negative axis in the X direction; % extend the boundary to get the image for I = 1-delta_y: height-delta_y for J = 1-delta_x: width-delta_x pix = [I J 1]/rot; % use the coordinates of the transformed image points to find the coordinates of the original image points, % otherwise, the pixels of some transformed images cannot be fully filled with float_y = pix (1)-floor (pix (1); float_x = pix (2) -Floor (pix (2); If pix (1)> =-1 & pix (2)> =-1 & pix (1) <= H + 1 & pix (2) <= W + 1 pix_up_left = [floor (pix (1) floor (pix (2)]; % four adjacent points pix_up_right = [floor (pix (1) Ceil (pix (2)]; pix_down_left = [Ceil (pix (1 )) floor (pix (2)]; pix_down_right = [Ceil (pix (1) Ceil (pix (2)]; value_up_left = (1-float_x) * (1-float_y ); % calculate the weights of neighboring four points value_up_right = float_x * (bytes); value_down_left = (1-float_x) * float_y; value_down_right = float_x * float_y; imgn (I + delta_y, J + delta_x) = value_up_left * imgm (pix_up_left (1) + 2, pix_up_left (2) + 2) +... value_up_right * imgm (pix_up_right (1) + 2, pix_up_right (2) + 2) +... value_down_left * imgm (pix_down_left (1) + 2, pix_down_left (2) + 2) +... value_down_right * imgm (pix_down_right (1) + 2, pix_down_right (2) + 2); End endfigure, imshow (uint8 (imgn ))

Img_extend.m

function imgm=img_extend(img,r)    [m n]=size(img);    imgm=zeros(m+2*r+1,n+2*r+1);    imgm(r+1:m+r,r+1:n+r)=img;    imgm(1:r,r+1:n+r)=img(1:r,1:n);     imgm(1:m+r,n+r+1:n+2*r+1)=imgm(1:m+r,n:n+r);    imgm(m+r+1:m+2*r+1,r+1:n+2*r+1)=imgm(m:m+r,r+1:n+2*r+1);    imgm(1:m+2*r+1,1:r)=imgm(1:m+2*r+1,r+1:2*r);end