shortest path in 2D matrix between two coordinate points

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As in the attached image, i have a 2D matrix of 50 by 50. The matrix contains zeros (blue) and ones(yellow). Imagine it as a floor, The yellow or ones mean the walls. In this whole matrix a transmitter(T) and receiver (R) can be placed anywhere and i need to know how many walls the signal penetrates when it reaches from T to R. Lets suppose the T is at (x,y) and R is at (i,j). I take the shortest path between T and R. If i get the shortest path i can calculate the number of 1s in the path and that will be the number of walls which i need.
Can anyone help in this regard.. Thank you

Accepted Answer

KSSV on 5 Jul 2018
YOu have the following options to check:
1. Do interpolation of the points along the path, if any point has 1, it is crossing the wall. Note that, for this you need to generate lot many points along the path so that a point lies on the yellow path.
2. Generate the points along the path, get the nearest neigboours to those points and see, any point is from yellow path.
3. Find the intersection between shortest paths and yellow paths. This would be better a solution.
Sohaib  Bin Altaf
Sohaib Bin Altaf on 9 Jul 2018
you can see in the attached image i have entered as [15,10] but it is plotted as [10,15]. Please check the figure attached to this comment

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More Answers (2)

Image Analyst
Image Analyst on 5 Jul 2018
Use bwdistgeodesic().
See Steve's 5 part blog on shortest paths.

Sohaib  Bin Altaf
Sohaib Bin Altaf on 10 Jul 2018
This thing worked for me:
clear all
close all
M = zeros(50);
% draw lines
for y = [20,30]
M(:,y) = 1;
for x = [13,25,37]
for y = [1:20, 30:50]
M(x,y) = 1;
p1 = [12,48];
p2 = [39,5];
% get direction
d = p1-p2;
d = d/max(abs(d)); % limit to 1 px step size
steps = 0:1:max(abs(p2-p1));
for i=length(p1):-1:1
p_line(:,i) = round(p2(i) + steps.*d(i));
idx = sub2ind(size(M), p_line(:,1), p_line(:,2));
walls = sum(M(idx));
M(idx) = 2;
M = flipud(M)
M = rot90(M,-1)
fprintf('You passed %i walls', walls)

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