如何获得线矩形交叉段？

Question

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。因此，假设在网格第一平方不与网格相交，则将零置于权重矩阵的第一个元素上。

这是我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
    end
end
% plotting x intercept
for i = 1:4
   x(i) = (y(i) - c)/m 
    if x(i)<2 || x(i)>6
        x(i) = 0
    end
end  
z = [x' y']

我有一条线，由参数m, h定义，其中y = m*x + h此线穿过网格（即像素）。

对于网格的每个方格(a, b)（即方格[a, a+1]x[b, b+1]），我想确定是否给定的线条穿过此方格，如果是，什么是正方形中<段>的长度，这样我就可以构造权重矩阵，这对于代数重建方法至关重要。

Answer 1

这里有一个很好的方法来使一条线与矩形网格相交并获得每个交叉段的长度：我使用了来自此link的第三个答案中的伪代码的线条交点。

% create some line form the equation y=mx+h
m = 0.5; h = 0.2;
x = -2:0.01:2;
y = m*x+h;
% create a grid on the range [-1,1]
[X,Y] = meshgrid(linspace(-1,1,10),linspace(-1,1,10));
% create a quad mesh on this range
fvc = surf2patch(X,Y,zeros(size(X)));
% extract topology
v = fvc.vertices(:,[1,2]);
f = fvc.faces;
% plot the grid and the line
patch(fvc,'EdgeColor','g','FaceColor','w'); hold on;
plot(x,y);
% use line line intersection from the link
DC = [f(:,[1,2]);f(:,[2,3]);f(:,[3,4]);f(:,[4,1])];
D = v(DC(:,1),:);
C = v(DC(:,2),:);
A = repmat([x(1),y(1)],size(DC,1),1);
B = repmat([x(end),y(end)],size(DC,1),1);
E = A-B;
F = D-C;
P = [-E(:,2),E(:,1)];
h = dot(A-C,P,2)./dot(F,P,2);
% calc intersections
idx = (0<=h & h<=1);
intersections = C(idx,:)+F(idx,:).*repmat(h(idx),1,2);
intersections = uniquetol(intersections,1e-8,'ByRows',true);
% sort by x axis values
[~,ii] = sort(intersections(:,1));
intersections = intersections(ii,:);
scatter(intersections(:,1),intersections(:,2));
% get segments lengths
directions = diff(intersections);
lengths = sqrt(sum(directions.^2,2));
directions = directions./repmat(sqrt(sum(directions.^2,2)),1,2);
directions = directions.*repmat(lengths,1,2);
quiver(intersections(1:end-1,1),intersections(1:end-1,2),directions(:,1),directions(:,2),'AutoScale','off','Color','k');

这是结果（图像中箭头的长度是段长度）