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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
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.
  8 Comments
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)


Sohaib  Bin Altaf
Sohaib Bin Altaf on 10 Jul 2018
This thing worked for me:
clc
clear all
close all
M = zeros(50);
% draw lines
for y = [20,30]
M(:,y) = 1;
end
for x = [13,25,37]
for y = [1:20, 30:50]
M(x,y) = 1;
end
end
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));
end
idx = sub2ind(size(M), p_line(:,1), p_line(:,2));
walls = sum(M(idx));
M(idx) = 2;
M = flipud(M)
M = rot90(M,-1)
imagesc(M)
set(gca,'YDir','normal')
set(gca,'XDir','normal')
fprintf('You passed %i walls', walls)

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