我想找到代数重建方法的权重矩阵.为此,我必须找到与网格的线交点.我可以找到直线与线的交点,但我必须存储相交的线段网格数.所以假设如果在网格第一个方格中不与网格相交,则将零置于权重矩阵的第一个元素上.这里为线路交叉尝试的代码:
ak = 3:6
aka = 3:6
x = zeros(size(aka))
y = zeros(size(ak))
for k = 1:length(ak)
line([ak(1) ak(end)], [aka(k) aka(k)],'color','r')
end
% Vertical grid
for k = 1:length(aka)
line([ak(k) ak(k)], [aka(1) aka(end)],'color','r')
end
hold on;
X =[0 15.5]
Y = [2.5 8.5]
m = (Y(2)-Y(1))/(X(2)-X(1)) ;
c = 2.5 ;
plot(X,Y)
axis([0 10 0 10])
axis square
% plotting y intercept
for i = 1:4
y(i) = m * ak(i) + c
if y(i)<2 || y(i)>6
y(i) = 0 …Run Code Online (Sandbox Code Playgroud)