What we want is to draw several solid circles at random locations, with random gray scale colors, on a dark gray background. How can we do this? Also, if the circles overlap, we need them to change color in the overlapping part.
Since this is an assignment for school, we are not looking for ready-made answers, but for a guide which tools to use in MATLAB!
Here's a checklist of things I would investigate if you want to do this properly:
Figure out how to draw circles in MATLAB. Because you don't have the Image Processing Toolbox (see comments), you will probably have to make a function yourself. I'll give you some starter code:
function [xout, yout] = circle(x,y,r,rows,cols)
[X,Y] = meshgrid(x-r:x+r, y-r:y+r);
ind = find(X.^2 + Y.^2 <= r^2 & X >= 1 & X <= cols & Y >= 1 & Y <= rows);
xout = X(ind);
yout = Y(ind);
end
What the above function does is that it takes in an (x,y) co-ordinate as well as the radius of
the circle. You also will need to specify how many rows and how many columns you want in your image. The reason why is because this function will prevent giving you co-ordinates that are out of bounds in the image that you can't draw. The final output of this will give you co-ordinates of all values inside and along the boundary of the circle. These co-ordinates will already be in integer so there's no need for any rounding and such things. In addition, these will perfectly fit when you're assigning these co-ordinates to locations in your image. One caveat to note is that the co-ordinates assume an inverted Cartesian. This means that the top left corner is the origin (0,0). x values increase from left to right, and y values increase from top to bottom. You'll need to keep this convention in mind when drawing circles in your image.
Take a look at the rand class of functions. rand will generate random values for you and so you can use these to generate a random set of co-ordinates - each of these co-ordinates can thus serve as your centre. In addition, you can use this class of functions to help you figure out how big you want your circles and also what shade of gray you want your circles to be.
Take a look at set operations (logical AND, logical OR) etc. You can use a logical AND to find any circles that are intersecting with each other. When you find these areas, you can fill each of these areas with a different shade of gray. Again, the rand functions will also be of use here.
As such, here is a (possible) algorithm to help you do this:
Take a matrix of whatever size you want, and initialize all of the elements to dark gray. Perhaps an intensity of 32 may work.
Generate a random set of (x,y) co-ordinates, a random set of radii and a random set of intensity values for each circle.
For each pair of circles, check to see if there are any co-ordinates that intersect with each other. If there are such co-ordinates, generate a random shade of gray and fill in these co-ordinates with this new shade of gray. A possible way to do this would be to take each set of co-ordinates of the two circles and draw them on separate temporary images. You would then use the logical AND operator to find where the circles intersect.
Now that you have your circles, you can plot them all. Take a look at how plot works with plotting matrices. That way you don't have to loop through all of the circles as it'll be inefficient.
Good luck!
Let's get you home, shall we? Now this stays away from the Image Processing Toolbox functions, so hopefully these must work for you too.
Code
%%// Paramters
numc = 5;
graph_size = [300 300];
max_r = 100;
r_arr = randperm(max_r/2,numc)+max_r/2
cpts = [randperm(graph_size(1)-max_r,numc)' randperm(graph_size(2)-max_r,numc)']
color1 = randperm(155,numc)+100
prev = zeros(graph_size(1),graph_size(2));
for k = 1:numc
r = r_arr(k);
curr = zeros(graph_size(1),graph_size(2));
curr(cpts(k,1):cpts(k,1)+r-1,cpts(k,2):cpts(k,2)+r-1)= color1(k)*imcircle(r);
common_blob = prev & curr;
curr = prev + curr;
curr(common_blob) = min(color1(1),color1(2))-50;
prev = curr;
end
figure,imagesc(curr), colormap gray
%// Please note that the code uses a MATLAB file-exchange tool called
%// imcircle, which is available at -
%// http://www.mathworks.com/matlabcentral/fileexchange/128-imcircle
Screenshot of a sample run
As you said that your problem is an assignment for school I will therefore not tell you exactly how to do it but what you should look at.
you should be familiar how 2d arrays (matrices) work and how to plot them using image/imagesc/imshow ;
you should look at the strel function ;
you should look at the rand/randn function;
such concepts should be enough for the assignment.
Related
I have written a MATLAB code to create the figure attached using the voronoi. My region of interest is the red circle. Hence the seeds for voronoi were kept within the region.
Idea: One approach would be to use a homothetic transformation of the Voronoi cell C{k} about the corresponding point X(k,:), with a ratio R such as 0 < R < 1. The shape of the cells—the number of corners and their associated angles—will be preserved, and the areas will be reduced proportionally (i.e. by a factor R2, and not by a constant value).
Please note that this will "destroy" your cells, because the reduced Voronoi cells will not share anymore vertices/edges, thus the [V,C] representation doesn't work anymore as it is. Also, the distances between what were once common edges will depend on the areas of the original cells (bigger cells, bigger distances between adjacent edges).
Transformation example for 2 2D points:
A = [1,2]; %'Center'
B = [10,1]; %'To be transformed'
R = 0.8; %'Transformation ratio'
trB = A + R*(B-A); %'Transformed'
couldn't follow your implementation of CST-link's idea, but here is one that works (i tested it in matlab, but not yet in abaqus, the code it spits out looks like abaqus should be happy with it)
rng(0);
x=rand(40,2);
plot(x(:,1),x(:,2),'x')
[v,c]=voronoin(x);
fact=0.9;
for i=1:length(c)
cur_cell=c{i};
coords=v(cur_cell,:);
if isfinite(coords)
%fact=somefunctionofarea?;
centre=x(i,:); % i used the voronoi seeds as my centres
coords=bsxfun(#minus,coords,centre); %move everything to a local coord sys centred on the seed point
[theta,rho] = cart2pol(coords(:,1),coords(:,2));
[xnew, ynew]= pol2cart(theta,rho*fact);
xnew=xnew+centre(1); % put back in real coords.
ynew=ynew+centre(2);
xnew2=circshift(xnew,1);
ynew2=circshift(ynew,1);
fprintf('s1.Line(point1=(%f,%f),point2=(%f,%f))\n',...
[xnew, ynew, xnew2,ynew2]');
line(xnew,ynew); %testing purposes - doesn't plot last side in matlab
end
end
Having seen the results of this one, i think you will need a different way factor to shrink your sides. either to subtract a fixed area or some other formula.
I want to detect edges (with sub-pixel accuracy) in images like the one displayed:
The resolution would be around 600 X 1000.
I came across a comment by Mark Ransom here, which mentions about edge detection algorithms for vertical edges. I haven't come across any yet. Will it be useful in my case (since the edge isn't strictly a straight line)? It will always be a vertical edge though. I want it to be accurate till 1/100th of a pixel at least. I also want to have access to these sub-pixel co-ordinate values.
I have tried "Accurate subpixel edge location" by Agustin Trujillo-Pino. But this does not give me a continuous edge.
Are there any other algorithms available? I will be using MATLAB for this.
I have attached another similar image which the algorithm has to work on:
Any inputs will be appreciated.
Thank you.
Edit:
I was wondering if I could do this:
Apply Canny / Sobel in MATLAB and get the edges of this image (note that it won't be a continuous line). Then, somehow interpolate this Sobel edges and get the co-ordinates in subpixel. Is it possible?
A simple approach would be to project your image vertically and fit the projected profile with an appropriate function.
Here is a try, with an atan shape:
% Load image
Img = double(imread('bQsu5.png'));
% Project
x = 1:size(Img,2);
y = mean(Img,1);
% Fit
f = fit(x', y', 'a+b*atan((x0-x)/w)', 'Startpoint', [150 50 10 150])
% Display
figure
hold on
plot(x, y);
plot(f);
legend('Projected profile', 'atan fit');
And the result:
I get x_0 = 149.6 pix for your first image.
However, I doubt you will be able to achieve a subpixel accuracy of 1/100th of pixel with those images, for several reasons:
As you can see on the profile, your whites are saturated (grey levels at 255). As you cut the real atan profile, the fit is biased. If you have control over the experiments, I suggest you do it again again with a smaller exposure time for instance.
There are not so many points on the transition, so there is not so many information on where the transition is. Typically, your resolution will be the square root of the width of the atan (or whatever shape you prefer). In you case this limits the subpixel resolution at 1/5th of a pixel, at best.
Finally, your edges are not stricly vertical, they are slightly titled. If you choose to use this projection method, to increase the accuracy you should look for a way to correct this tilt before projecting. This won't increase your accuracy by several orders of magnitude, though.
Best,
There is a problem with your image. At pixel level, it seems like there are four interlaced subimages (odd and even rows and columns). Look at this zoomed area close to the edge.
In order to avoid this artifact, I just have taken the even rows and columns of your image, and compute subpixel edges. And finally, I look for the best fitting straight line, using the function clsq whose code is in this page:
%load image
url='http://i.stack.imgur.com/bQsu5.png';
image = imread(url);
imageEvenEven = image(1:2:end,1:2:end);
imshow(imageEvenEven, 'InitialMagnification', 'fit');
% subpixel detection
threshold = 25;
edges = subpixelEdges(imageEvenEven, threshold);
visEdges(edges);
% compute fit line
A = [ones(size(edges.x)) edges.x edges.y];
[c n] = clsq(A,2);
y = [1,200];
x = -(n(2)*y+c) / n(1);
hold on;
plot(x,y,'g');
When executing this code, you can see the green line that best aproximate all the edge points. The line is given by the equation c + n(1)*x + n(2)*y = 0
Take into account that this image has been scaled by 1/2 when taking only even rows and columns, so the right coordinates must be scaled.
Besides, you can try with the other tree subimages (imageEvenOdd, imageOddEven and imageOddOdd) and combine the four straigh lines to obtain the best solution.
I was working on my image processing problem with detecting coins.
I have some images like this one here:
and wanted to separate the falsely connected coins.
We already tried the watershed method as stated on the MATLAB-Homepage:
the-watershed-transform-strategies-for-image-segmentation.html
especially since the first example is exactly our problem.
But instead we get a somehow very messed up separation as you can see here:
We already extracted the area of the coin using the regionprops Extrema parameter and casting the watershed only on the needed area.
I'd appreciate any help with the problem or even another method of getting it separated.
If you have the Image Processing Toolbox, I can also suggest the Circular Hough Transform through imfindcircles. However, this requires at least version R2012a, so if you don't have it, this won't work.
For the sake of completeness, I'll assume you have it. This is a good method if you want to leave the image untouched. If you don't know what the Hough Transform is, it is a method for finding straight lines in an image. The circular Hough Transform is a special case that aims to find circles in the image.
The added advantage of the circular Hough Transform is that it is able to detect partial circles in an image. This means that those regions in your image that are connected, we can detect them as separate circles. How you'd call imfindcircles is in the following fashion:
[centers,radii] = imfindcircles(A, radiusRange);
A would be your binary image of objects, and radiusRange is a two-element array that specifies the minimum and maximum radii of the circles you want to detect in your image. The outputs are:
centers: A N x 2 array that tells you the (x,y) co-ordinates of each centre of a circle that is detected in the image - x being the column and y being the row.
radii: For each corresponding centre detected, this also gives the radius of each circle detected. This is a N x 1 array.
There are additional parameters to imfindcircles that you may find useful, such as the Sensitivity. A higher sensitivity means that it is able to detect circular shapes that are more non-uniform, such as what you are showing in your image. They aren't perfect circles, but they are round shapes. The default sensitivity is 0.85. I set it to 0.9 to get good results. Also, playing around with your image, I found that the radii ranged from 50 pixels to 150 pixels. Therefore, I did this:
im = im2bw(imread('http://dennlinger.bplaced.net/t06-4.jpg'));
[centers,radii] = imfindcircles(im, [50 150], 'Sensitivity', 0.9);
The first line of code reads in your image directly from StackOverflow. I also convert this to logical or true black and white as the image you uploaded is of type uint8. This image is stored in im. Next, we call imfindcircles in the method that we described.
Now, if we want to visualize the detected circles, simply use imshow to show your image, then use the viscircles to draw the circles in the image.
imshow(im);
viscircles(centers, radii, 'DrawBackgroundCircle', false);
viscircles by default draws the circles with a white background over the contour. I want to disable this because your image has white circles and I don't want to show false contouring. This is what I get with the above code:
Therefore, what you can take away from this is the centers and radii variables. centers will give you the centre of each detected circle while radii will tell you what the radii is for each circle.
Now, if you want to simulate what regionprops is doing, we can iterate through all of the detected circles and physically draw them onto a 2D map where each circle would be labeled by an ID number. As such, we can do something like this:
[X,Y] = meshgrid(1:size(im,2), 1:size(im,1));
IDs = zeros(size(im));
for idx = 1 : numel(radii)
r = radii(idx);
cen = centers(idx,:);
loc = (X - cen(1)).^2 + (Y - cen(2)).^2 <= r^2;
IDs(loc) = idx;
end
We first define a rectangular grid of points using meshgrid and initialize an IDs array of all zeroes that is the same size as the image. Next, for each pair of radii and centres for each circle, we define a circle that is centered at this point that extends out for the given radius. We then use these as locations into the IDs array and set it to a unique ID for that particular circle. The result of IDs will be that which resembles the output of bwlabel. As such, if you want to extract the locations of where the idx circle is, you would do:
cir = IDs == idx;
For demonstration purposes, this is what the IDs array looks like once we scale the IDs such that it fits within a [0-255] range for visibility:
imshow(IDs, []);
Therefore, each shaded circle of a different shade of gray denotes a unique circle that was detected with imfindcircles.
However, the shades of gray are probably a bit ambiguous for certain coins as this blends into the background. Another way that we could visualize this is to apply a different colour map to the IDs array. We can try using the cool colour map, with the total number of colours to be the number of unique circles + 1 for the background. Therefore, we can do something like this:
cmap = cool(numel(radii) + 1);
RGB = ind2rgb(IDs, cmap);
imshow(RGB);
The above code will create a colour map such that each circle gets mapped to a unique colour in the cool colour map. The next line applies a mapping where each ID gets associated with a colour with ind2rgb and we finally show the image.
This is what we get:
Edit: the following solution is more adequate to scenarios where one does not require fitting the exact circumferences, although simple heuristics could be used to approximate the radii of the coins in the original image based on the centers found in the eroded one.
Assuming you have access to the Image Processing toolbox, try imerode on your original black and white image. It will apply an erosion morphological operator to your image. In fact, the Matlab webpage with the documentation of that function has an example strikingly similar to your problem/image and they use a disk structure.
Run the following code (based on the example linked above) assuming the image you submitted is called ima.jpg and is local to the code:
ima=imread('ima.jpg');
se = strel('disk',50);
eroded = imerode(ima,se);
imshow(eroded)
and you will see the image that follows as output. After you do this, you can use bwlabel to label the connected components and compute whatever properties you may want, for example, count the number of coins or detect their centers.
How do I separate the two connected circles in the image below, using MATLAB? I have tried using imerode, but this does not give good results. Eroding does not work, because in order to erode enough to separate the circles, the lines disappear or become mangled. In other starting pictures, a circle and a line overlap, so isolating the overlapping objects won't work either.
The image shows objects identified by bwboundaries, each object painted a different color. As you can see, the two light blue circles are joined, and I want to disjoin them, producing two separate circles. Thanks
I would recommend you use the Circular Hough Transform through imfindcircles. However, you need version 8 of the Image Processing Toolbox, which was available from version R2012a and onwards. If you don't have this, then unfortunately this won't work :(... but let's go with the assumption that you do have it. However, if you are using something older than R2012a, Dev-iL in his/her comment above linked to some code on MATLAB's File Exchange on an implementation of this, most likely created before the Circular Hough Transform was available: http://www.mathworks.com/matlabcentral/fileexchange/9168-detect-circles-with-various-radii-in-grayscale-image-via-hough-transform/
This is a special case of the Hough Transform where you are trying to find circles in your image rather than lines. The beauty with this is that you are able to find circles even when the circle is partially completed or overlapping.
I'm going to take the image that you provided above and do some post-processing on it. I'm going to convert the image to binary, and remove the border, which is white and contains the title. I'm also going to fill in any holes that result so that all of the objects are filled in with solid white. There is also some residual quantization noise after I do this step, so I'm going to a small opening with a 3 x 3 square element. After, I'm going to close the shapes with a 3 x 3 square element, as I see that there are noticeable gaps in the shapes. Therefore:
Therefore, directly reading in your image from where you've posted it:
im = imread('http://s29.postimg.org/spkab8oef/image.jpg'); %// Read in the image
im_gray = im2double(rgb2gray(im)); %// Convert to grayscale, then [0,1]
out = imclearborder(im_gray > 0.6); %// Threshold using 0.6, then clear the border
out = imfill(out, 'holes'); %// Fill in the holes
out = imopen(out, strel('square', 3));
out = imclose(out, strel('square', 3));
This is the image I get:
Now, apply the Circular Hough Transform. The general syntax for this is:
[centres, radii, metric] = imfindcircles(img, [start_radius, end_radius]);
img would be the binary image that contains your shapes, start_radius and end_radius would be the smallest and largest radius of the circles you want to find. The Circular Hough Transform is performed such that it will find any circles that are within this range (in pixels). The outputs are:
centres: Which returns the (x,y) positions of the centres of each circle detected
radii: The radius of each circle
metric: A measure of purity of the circle. Higher values mean that the shape is more probable to be a circle and vice-versa.
I searched for circles having a radius between 30 and 60 pixels. Therefore:
[centres, radii, metric] = imfindcircles(out, [30, 60]);
We can then demonstrate the detected circles, as well as the radii by a combination of plot and viscircles. Therefore:
imshow(out);
hold on;
plot(centres(:,1), centres(:,2), 'r*'); %// Plot centres
viscircles(centres, radii, 'EdgeColor', 'b'); %// Plot circles - Make edge blue
Here's the result:
As you can see, even with the overlapping circles towards the top, the Circular Hough Transform was able to detect two distinct circles in that shape.
Edit - November 16th, 2014
You wish to ensure that the objects are separated before you do bwboundaries. This is a bit tricky to do. The only way I can see you do this is if you don't even use bwboundaries at all and do this yourself. I'm assuming you'll want to analyze each shape's properties by themselves after all of this, so what I suggest you do is iterate through every circle you have, then place each circle on a new blank image, do a regionprops call on that shape, then append it to a separate array. You can also keep track of all of the circles by having a separate array that adds the circles one at a time to this array.
Once you've finished with all of the circles, you'll have a structure array that contains all of the measured properties for all of the measured circles you have found. You would use the array that contains only the circles from above, then use these and remove them from the original image so you get just the lines. You'd then call one more regionprops on this image to get the information for the lines and append this to your final structure array.
Here's the first part of the procedure I outlined above:
num_circles = numel(radii); %// Get number of circles
struct_reg = []; %// Save the shape analysis per circle / line here
%// For creating our circle in the temporary image
[X,Y] = meshgrid(1:size(out,2), 1:size(out,1));
%// Storing all of our circles in this image
circles_img = false(size(out));
for idx = 1 : num_circles %// For each circle we have...
%// Place our circle inside a temporary image
r = radii(idx);
cx = centres(idx,1); cy = centres(idx,2);
tmp = (X - cx).^2 + (Y - cy).^2 <= r^2;
% // Save in master circle image
circles_img(tmp) = true;
%// Do regionprops on this image and save
struct_reg = [struct_reg; regionprops(tmp)];
end
The above code may be a bit hard to swallow, but let's go through it slowly. I first figure out how many circles we have, which is simply looking at how many radii we have detected. I keep a separate array called struct_reg that will append a regionprops struct for each circle and line we have in our image. I use meshgrid to determine the (X,Y) co-ordinates with respect to the image containing our shapes so that I can draw one circle onto a blank image at each iteration. To do this, you simply need to find the Euclidean distance with respect to the centre of each circle, and set the pixels to true only if that location has its distance less than r. After doing this operation, you will have created only one circle and filtered all of them out. You would then use regionprops on this circle, add it to our circles_img array, which will only contain the circles, then continue with the rest of the circles.
At this point, we will have saved all of our circles. This is what circles_img looks like so far:
You'll notice that the circles drawn are clean, but the actual circles in the original image are a bit jagged. If we tried to remove the circles with this clean image, you will get some residual pixels along the border and you won't completely remove the circles themselves. To illustrate what I mean, this is what your image looks like if I tried to remove the circles with circles_img by itself:
... not good, right?
If you want to completely remove the circles, then do a morphological reconstruction through imreconstruct where you can use this image as the seed image, and specify the original image to be what we're working on. The job of morphological reconstruction is essentially a flood fill. You specify seed pixels, and an image you want to work on, and the job of imreconstruct is from these seeds, flood fill with white until we reach the boundaries of the objects that the seed pixels resided in. Therefore:
out_circles = imreconstruct(circles_img, out);
Therefore, we get this for our final reconstructed circles image:
Great! Now, use this and remove the circles from the original image. Once you do this, run regionprops again on this final image and append to your struct_reg variable. Obviously, save a copy of the original image before doing this:
out_copy = out;
out_copy(out_circles) = false;
struct_reg = [struct_reg; regionprops(out_copy)];
Just for sake of argument, this is what the image looks like with the circles removed:
Now, we have analyzed all of our shapes. Bear in mind I did the full regionprops call because I don't know exactly what you want in your analysis... so I just decided to give you everything.
Hope this helps!
erosion is the way to go. You should probably use a larger structuring element.
How about
1 erode
2 detect your objects
3 dilate each object for itself using the same structuring element
I have a set of points which I want to propagate on to the edge of shape boundary defined by a binary image. The shape boundary is defined by a 1px wide white edge.
I have the coordinates of these points stored in a 2 row by n column matrix. The shape forms a concave boundary with no holes within itself made of around 2500 points. I have approximately 80 to 150 points that I wish to propagate on the shape boundary.
I want to cast a ray from each point from the set of points in an orthogonal direction and detect at which point it intersects the shape boundary at. The orthogonal direction has already been determined. For the required purposes it is calculated taking the normal of the contour calculated for point, using point-1 and point+1.
What would be the best method to do this?
Are there some sort of ray tracing algorithms that could be used?
Thank you very much in advance for any help!
EDIT: I have tried to make the question much clearer and added a image describing the problem. In the image the grey line represents the shape contour, the red dots the points
I want to propagate and the green line an imaginary orthongally cast ray.
alt text http://img504.imageshack.us/img504/3107/orth.png
ANOTHER EDIT: For clarification I have posted the code used to calculate the normals for each point. Where the xt and yt are vectors storing the coordinates for each point. After calculating the normal value it can be propagated by using the linspace function and the requested length of the orthogonal line.
%#derivaties of contour
dx=[xt(2)-xt(1) (xt(3:end)-xt(1:end-2))/2 xt(end)-xt(end-1)];
dy=[yt(2)-yt(1) (yt(3:end)-yt(1:end-2))/2 yt(end)-yt(end-1)];
%#normals of contourpoints
l=sqrt(dx.^2+dy.^2);
nx = -dy./l;
ny = dx./l;
normals = [nx,ny];
It depends on how many unit vectors you want to test against one shape. If you have one shape and many tests, the easiest thing to do is probably to convert your shape coordinates to polar coordinates which implicitly represent your solution already. This may not be a very effective solution however if you have different shapes and only a few tests for every shape.
Update based on the edited question:
If the rays can start from arbitrary points, not only from the origin, you have to test against all the points. This can be done easily by transforming your shape boundary such that your ray to test starts in the origin in either coordinate direction (positive x in my example code)
% vector of shape boundary points (assumed to be image coordinates, i.e. integers)
shapeBoundary = [xs, ys];
% define the start point and direction you want to test
startPoint = [xsp, ysp];
testVector = unit([xv, yv]);
% now transform the shape boundary
shapeBoundaryTrans(:,1) = shapeBoundary(:,1)-startPoint(1);
shapeBoundaryTrans(:,2) = shapeBoundary(:,2)-startPoint(2);
rotMatrix = [testVector(2), testVector(1); ...
testVector(-1), testVector(2)];
% somewhat strange transformation to keep it vectorized
shapeBoundaryTrans = shapeBoundaryTrans * rotMatrix';
% now the test is easy: find the points close to the positive x-axis
selector = (abs(shapeBoundaryTrans(:,2)) < 0.5) & (shapeBoundaryTrans(:,1) > 0);
shapeBoundaryTrans(:,2) = 1:size(shapeBoundaryTrans, 1)';
shapeBoundaryReduced = shapeBoundaryTrans(selector, :);
if (isempty(shapeBoundaryReduced))
[dummy, idx] = min(shapeBoundaryReduced(:, 1));
collIdx = shapeBoundaryReduced(idx, 2);
% you have a collision with point collIdx of your shapeBoundary
else
% no collision
end
This could be done in a nicer way probably, but you get the idea...
If I understand your problem correctly (project each point onto the closest point of the shape boundary), you can
use sub2ind to convert the "2 row by n column matrix" description to a BW image with white pixels, something like
myimage=zeros(imagesize);
myimage(imagesize, x_coords, y_coords) = 1
use imfill to fill the outside of the boundary
run [D,L] = bwdist(BW) on the resulting image, and just read the answers from L.
Should be fairly straightforward.