Line of sight between points in a logical matrix

Hi all,

I am trying to work out the 'line of sight' between points in a logical matrix, i.e. if zeros represent floor space and ones represent walls, I would like to know all the points in the matrix than could be observed from a specific row,column location directly, without the line of sight crossing a wall.

I have tried using bwdist and bwdistgeodesic to solve this problem, my first thought was to take the difference between the maps produced using these which would give me a map of all pixels where the euclidean distance and distance traversing around walls are the same (directly visible). Alas this doesn't seem to be the case (see code below).

There must be an easy way to do this but I just can't seem to work it out...

Thanks for any help,

Rod.

emap = true(30,30);
emap = padarray(emap,[1 1],false,'both');
emap(1:15,16) = false;

spoint = [9,6];

figure
subplot(2,2,1)
imshow(emap)
daspect([1 1 1])
axis xy
title('Shape of interest')

subplot(2,2,2)
D1 = bwdistgeodesic(emap,spoint(2),spoint(1),'quasi-euclidean');
imagesc(D1)
daspect([1 1 1])
axis xy
c = caxis;
title('Geodesic distance')

subplot(2,2,3)
bw = false(size(emap));
bw(spoint(1),spoint(2)) = true;
D2 = bwdist(bw,'quasi-euclidean');
D2(~emap) = 0;
imagesc(D2)
daspect([1 1 1])
axis xy
caxis(c)
title('Euclidean distance')

subplot(2,2,4)
imagesc(D1-D2)
daspect([1 1 1])
axis xy
title('Difference')

Answer 1

[John BG]

Hi Roddy

this is John BG <mailto:[e-mail] [e-mail]>

1.- the map

    clear all;close all;clc

    S1=30;S2=30;
    emap = true(S1,S2);
    emap = padarray(emap,[1 1],false,'both');
    emap(1:15,16) = false;

2.-

obstacle definition

    K=[16*ones(15,1) [1:15]'];

3.-

the spotter

    nx_spoint=9;
    ny_spoint=6;
    spoint=[nx_spoint,ny_spoint];

    figure(1);imshow(emap);ax1=gca

4.- map points free of obstacles

    [nx,ny]=find(emap==true);

5.- checking the start area free of obstacles has the correct points

    A=zeros(size(emap));                  % 
    hf2=figure(2);imshow(A);ax2=hf2.CurrentAxes
    plot(ax2,ny,nx,'r*');
    axis(ax2,'equal');

6.- spotter on the map

    hold(ax1,'all');plot(ax1,spoint(1),spoint(2),'g.');        % 'LineSpec' Style-Marker-Color 'LineStyle'
    hold(ax2,'all');plot(ax2,spoint(1),spoint(2),'g*');       % 'LineSpec' Style-Marker-Color 'LineStyle'

    P=[nx ny];                      % free space points, excluding obstacles

7.-

remove spotter point from area to sweep

    L1=[];   
    for k=1:1:size(P,1)
        if P(k,1)==nx_spoint && P(k,2)==ny_spoint
            L1=[L1 k];
        end
    end
    P(L1,:)=[];

8.-

defining outer perimeter: the fence along which the spotter is going to sweep along

    left_side=[ones(1,S1)' [1:1:S1]']                          
    top_side=[[2:1:S2-1]' S1*ones(1,S2-2)']
    right_side=[S2*ones(1,S1)' [S1:-1:1]']
    bottom_side=[[S2-1:-1:2]' ones(1,S2-2)']
    Fence=[top_side;right_side;bottom_side;left_side];

9.-

check obstacle points correctly on map

    plot(ax2,K(:,1),K(:,2),'c*')              
    for k2=1:1:size(Fence,1)
        P0=Fence(k2,:);

        nxP0=P0(1);nyP0=P0(2);
        plot(ax2,nxP0,nyP0,'b*');

10.-

define one ray points, spotter - fence

        Lx=floor(linspace(nx_spoint,nxP0,max(abs(nxP0-nx_spoint),abs(nyP0-ny_spoint))));  
        Ly=floor(linspace(ny_spoint,nyP0,max(abs(nxP0-nx_spoint),abs(nyP0-ny_spoint))));
        L1=[Lx' Ly'];

11.-

intersect finds, if any, intersect points for 2D vectors, intersect.m attached along with this script.

        [U,n1,n2]=intersectN(L1,K)

12.-

have to decide whether truncate ray or not

        if isempty(U)
              plot(ax2,Lx,Ly,'y.');drawnow;       %   ray without obstacle
    %          pause(.25)
        else
            if size(U,2)>1   % solve intersectN may return more than 1 point
                  U=U(1,:);    
            end

13.-

Following bank of ifs to make sure rays cover from all 4 quadrants

            if spoint(1)<U(1)    
                Lx2obst_=[spoint(1):1:U(1)]';
            end
            if spoint(1)>U(1)
                Lx2obst_=[spoint(1):-1:U(1)]';
            end
            if spoint(1)==U(1)
                Lx2obst=spoint(1);
            end
              if spoint(2)<U(2) 
                Ly2obst_=[spoint(2):1:U(2)]';
            end
            if spoint(2)>U(2)
                Ly2obst_=[spoint(2):-1:U(2)]';
            end
            if spoint(2)==U(2)
                Ly2obst=spoint(2);
            end

            Lx2obst=floor(linspace(spoint(1),U(1),max(numel(Lx2obst_),numel(Ly2obst_))))
            Ly2obst=floor(linspace(spoint(2),U(2),max(numel(Lx2obst_),numel(Ly2obst_))))

            plot(ax2,Lx2obst',Ly2obst','y.')    % ray up to obstacle
    %         pause(.25)
        end

    end

.

.

.

Note that the plot function shows the image reversed compared to the imshow of logical 2D maps that you start with, but the coordinates are all the same.

Roddy

f you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance for time and attention

John BG

<mailto:[e-mail] [e-mail]>

[Neuropragmatist]

Hi John,

Thanks a lot for your help, it is actually quite similar to something I used previously (with help from the Geom2D toolbox), but for speed I'm trying to use matrix operations rather than geometric ones.

I have, I think, discovered an easier and hopefully faster approach which I have added as an answer here if you are interested.

Thanks again,

Roddy.

[John BG]

Grieves

this is John BG [e-mail]

your code doesn't work for 2 out of 4 quadrants.

Have a look at what happens when you place the spotter in a point belonging a to each quadrant as shown below

[Neuropragmatist]

Hi John,

I have added some new code as a comment to my own answer, mainly I changed the interpolation to nearest and added a tolerance limit for detecting wall crossings (although I don't think the tolerance is necessary if the interpolation is 'nearest'). On testing this seems to work for any point I try now.

I didn't know there was a rule about accepting your own answer - very often my questions go unanswered, but if I find a solution I like to come back and add the info for further people... I have unaccepted my answer if you would like to accept it instead.

Rod.

[Walter Roberson]

You cannot accept your own answer to someone else's question. You can accept your own answer to your own question.

Answer 2

[Neuropragmatist]

I think I have found a relatively simple solution that should also be quite fast. I use the inbuilt function improfile which was mentioned by image analyst somewhere (I have lost which post it was!) to interpolate out the values between my starting position and any pixel in the map.

Next I say that if this interpolated vector contains any zeros (the value assigned to walls) the pixel is given a value representing 'not visible'.

This approach results in some missing pixels, I assume this is because the interpolation passes between wall pixels perfectly and the zeros are missed out. To rectify this I use bwareaopen to fill in these 'holes'.

Here is the code and a plot of the result:

% create environment
S1=30;
S2=30;
emap = true(S1,S2);
emap = padarray(emap,[1 1],false,'both');
emap(1:15,16) = false;

% define current position
nx_spoint=9;
ny_spoint=6;
spoint=[nx_spoint,ny_spoint];

% show matrix
figure(1);
subplot(1,3,1)
imshow(emap);
daspect([1 1 1])
axis xy
title('environment')

% create 'viewmap'
vmap = zeros(size(emap));
indx = find(emap);
for ii = 1:length(indx)
    [y,x] = ind2sub(size(emap),indx(ii));

      xi = [spoint(2),x];
      yi = [spoint(1),y];
      C = improfile(emap,xi,yi,'bilinear');

      if sum(~C)>0
          vmap(indx(ii)) = 5;
      else
          vmap(indx(ii)) = 10;        
      end
  end

% show resulting viewmap
subplot(1,3,2)
imagesc(vmap);
daspect([1 1 1])
axis xy
title('viewmap')

% close single pixel 'holes' in viewmap caused by lines passing between pixels without hitting them
vmap2 = vmap;
vmap2(vmap<10) = 0;
vmap2 = logical(vmap2);
vmap2 = bwareaopen(vmap2,2);

% show resulting viewmap
subplot(1,3,3)
imshow(vmap2);
daspect([1 1 1])
axis xy
title('final viewmap')

[Neuropragmatist]

As per John's testing, the correct code should actually be:

% create environment
S1=30;
S2=30;
emap = true(S1,S2);
emap = padarray(emap,[1 1],false,'both');
emap(1:15,16) = false;

% define current position
nx_spoint=25;
ny_spoint=5;
spoint=[nx_spoint,ny_spoint];

% show matrix
figure(1);
subplot(1,3,1)
imshow(emap);
hold on
plot(ny_spoint,nx_spoint,'kx')
daspect([1 1 1])
axis xy
title('environment')

% create 'viewmap'
vmap = zeros(size(emap));
indx = find(emap);
tolerance = 0.3;
for ii = 1:length(indx)
    [y,x] = ind2sub(size(emap),indx(ii));

      xi = [spoint(2),x];
      yi = [spoint(1),y];
      C = improfile(emap,xi,yi,'nearest');

      if sum(C<tolerance)>0
        vmap(indx(ii)) = 5;
      else
        vmap(indx(ii)) = 10;        
      end
  end

% show resulting viewmap
subplot(1,3,2)
imagesc(vmap);
hold on
plot(ny_spoint,nx_spoint,'kx')
daspect([1 1 1])
axis xy
title('viewmap')

% close single pixel 'holes' in viewmap caused by lines passing between pixels without hitting them
vmap2 = vmap;
vmap2(vmap<10) = 0;
vmap2 = logical(vmap2);
vmap2 = bwareaopen(vmap2,2);

% show resulting viewmap
subplot(1,3,3)
imshow(vmap2);
hold on
plot(ny_spoint,nx_spoint,'kx')
daspect([1 1 1])
axis xy
title('final viewmap')

[John BG]

Hi Roddy

this is John BG [e-mail]

You have fixed the quadrants that looked as if spotter out of square, following is with spotter [52 25] which is ok because the observer is not supposed to be out of square

Yet,

.

.

the few pixels on the left of the ray for spotter on [25 25] are pixels that shouldn't see but apparently your answer claims to see beyond the cut ray. Is this ok to you?

.

.

With these simple answers we are supply it's understandable a +-1 pixels error away from the correct pixels, but +-2 pixels away, why is it?

2 pixels too many may be the difference between missing or hitting a pedestrian, don't you agree?

regards

John BG

[e-mail]