Connect 2 points in matrix

I have matrix zeros(5,5) and i know coordinates of 2 points example: [1,1],[5,5]. And now i want to connect this 2 points to get shortest road from one point to another and get something like this in matrix:

matrix =[1 0 0 0 0
         0 1 0 0 0
         0 0 1 0 0
         0 0 0 1 0
         0 0 0 0 1]

I completed code for this but some points are in strange coordinates and dont fit in my algorithm. I need simple solution. I want only to write path in this matrix with value 1. In my project i have matrix 800x800. I dont know if I explained well ;p

Answer 1

[Image Analyst]

You can use imline() if you have the Image Processing Toolbox. See attached demo. The key lines are

hLine = imline(gca);
singleLineBinaryImage = hLine.createMask();
burnedImage(singleLineBinaryImage) = 255; % Burn line into existing image.

Answer 2

[Andrew Crawford]

%%Recreate Your Matrix: but this will work with any logical matrix with only two % 1's.

   I=zeros(5);
   I(1)=1;
   I(end)=1;

%%Operate on binary matrix with only two true's

   pts=nonzeros((1:numel(I)).*I(:)');

   [x,y]=ind2sub(size(I),pts);

   px = polyfit(x,y,1);
   py = polyfit(y,x,1);

   if x(1)<(end)
        xi=x(1):x(end);
   else
        xi=x(1):-1:x(end);
   end
   if y(1)<y(end)
        yi=y(1):y(end);
   else
        yi=y(1):-1:y(end);
   end

   yo=round(polyval(px,xi));
   xo=round(polyval(py,yi));

   xy=[[xi(:),yo(:)];[xo(:),yi(:)]];
   xy=unique(xy,'rows');

   idx=sub2ind(size(I),xy(:,2),xy(:,1));
   I(idx)=1;

%%Display output:

   figure;
   imagesc(I);

[Jan]

I assume if x(1)<(end) should be if x(1)<x(end) .

Answer 3

[Walter Roberson]

If there are no obstacles, then one way to get a shortest path is:

• if the x coordinate and the y coordinate of the target are both different than the current location, then move along the diagonal to reduce the difference in both x and y coordinates by 1
• if only one of the x coordinate or y coordinate are different than the current location, then move along the coordinate that is different

This is not the only possible shortest path if abs(x difference) is not the same as abs(y difference), but it is easy to program.