MATLAB exercise program (generate a Hilbert curve)

This line can be used to traverse all pixels in the image, but this is not done here, but only generates such a curve.

In the program, H and W are the height and width of the final image, and N is the order of the Hilbert curve.

Here, if n is equal to log2 (h) or log2 (W), the image will be completely white, and it will just traverse all pixels.

Of course, if n is large, the image is completely white. However, that is not just a traversal.

The core functions for generating curves in the code can be found here.

The generated image is as follows:

The Matlab code is as follows:

Main. m

clear all;close all;clc;h=256;w=256;n=5;   imgn=zeros(h,w);[x,y]=hilbert(n);       x=floor((x+0.5)*w)+1;y=floor((y+0.5)*h)+1;l=length(x);for i=1:l-1    imgn=drawline(imgn,x(i),y(i),x(i+1),y(i+1));  endimshow(imgn)

Helbert. m

function [x,y] = hilbert(n)%HILBERT Hilbert curve.%% [x,y]=hilbert(n) gives the vector coordinates of points%   in n-th order Hilbert curve of area 1.%% Example: plot of 5-th order curve%% [x,y]=hilbert(5);line(x,y)%%   Copyright (c) by Federico Forte%   Date: 2000/10/06 if n<=0  x=0;  y=0;else  [xo,yo]=hilbert(n-1);  x=.5*[-.5+yo -.5+xo .5+xo  .5-yo];  y=.5*[-.5+xo  .5+yo .5+yo -.5-xo];end

Drawline. m

Function IMG = drawline (IMG, X1, Y1, X2, Y2) % because the orientation of the image coordinate is opposite to that of the Y axis of the mathematical function, the inverse Number of all Y variables is Y1 =-Y1; y2 =-Y2; If X1 ~ = X2 k = (y2-y1)/(x2-x1); B = y1-k * x1; MI = min (x1, x2); MA = max (x1, x2 ); for I = mi: Ma IMG (-round (I * k + B), I) = 1; end if Y1 ~ = Y2 k = (x2-x1)/(y2-y1); B = x1-k * Y1; MI = min (Y1, Y2); MA = max (Y1, Y2 ); for I = mi: Ma IMG (-I, round (I * k + B) = 1; end

 

MATLAB exercise program (generate a Hilbert curve)