Plot image on dome - image

I am struggling to plot a round image on the surface of the dome in Matlab. Here is what I have. This is the png image:
And this is the dome:
Now, I need to project this image on the dome. I've written a code to place the image on the surface:
r = 10;
r2 = 9;
cdata = imread('circle_image.png');
props.EdgeColor = 'none';
figure();
n = 50;
[X,Y,Z] = sphere(n) ;
X1 = X * r;
Y1 = Y * r;
Z1 = Z * r;
for i = 1:n+1
for j = 1:n+1
if Z1(i,j) < r2
X1(i,j) = NaN;
Y1(i,j) = NaN;
Z1(i,j) = NaN;
end
end
end
my_dome = surf(X1,Y1,Z1,props) ;
alpha = 1;
set(kopula, 'FaceColor', 'texturemap', 'CData', cdata, 'FaceAlpha', alpha, 'EdgeColor', 'none');
axis equal
What I am getting looks like this:
It seems like the image is centred in the wrong place, or even in the wrong axis. How can I fix that?

Yes, the image is centered in the wrong place.
The texture mapping is applied on the whole surface, not just the "active" points (the one you didn't NaN). Basically, what you are getting is the image being spread onto the full sphere, and then when you crop the top of the sphere to get your dome, the image is cropped as well:
What you need to do, is actually remove all those points you converted to NaN, so they are not part of the surface at all and the texture mapping is aplied only on the top dome surface.
So replace your nested for loop with the following code:
idx_Raws2crop = Z1(:,1) < r2 ;
X1(idx_Raws2crop,:) = [] ;
Y1(idx_Raws2crop,:) = [] ;
Z1(idx_Raws2crop,:) = [] ;
Then continue with your code, you'll get:
Actually I also added the instruction:
cdata = flipud(cdata) ;
in order to have the THE CIRCLEwriting in the proper orientation (otherwise it appears upside down).
Edit:
To render the picture on the dome in the way you'd like according to your comment, I can see 2 options:
Options 1: Build upon what we have already
This will consist of:
Extend the [X,Y] domain on a square grid (so the picture is not distorted when texture mapped onto the surface).
Extend the picture itself (add some transparent margin), so the margin will cover the domain extension we introduced and the actual visible part of the picture will be nicely centered on the dome.
With the same X1, Y1 and Z1 than we obtained with the code above:
% Create a square grid large enough to cover the dome and a bit more:
[X2,Y2] = meshgrid(linspace(-5,5,100),linspace(-5,5,100)) ;
% reinterpolate Z1 over this new grid
Z2 = griddata(X1,Y1,Z1,X2,Y2) ;
Now your surface looks like this:
As I warned you, the image is now properly applied but the edge of the dome looks rather ugly. To ease that you could replace the NaN with the base value for your dome, this will transition a lot better between dome and flat domain:
Z2(isnan(Z2)) = 9 ;
Will now yield:
The next problem, as you can see, is that (as in your first question), the image is stretched onto the whole surface. So part of the writing is now on the flat part of the surface. To simply alleviate that, you can modify the picture (add a bit of transparent margin on every side), until the visible part of the image matches the dome size.
Options 2: Build your surface differently
This is actually easier and more straighforward. We will build a dome out of a square surface in the first place (no trimming and NaNing).
This code does not require any part of your previous code, it is self contained:
r = 10 ;
% Build a square grid
[X2,Y2] = meshgrid(linspace(-5,5,100),linspace(-5,5,100)) ;
% build Z2 according to the sphere equation.
Z2 = sqrt( r^2 - X2.^2 - Y2.^2) ;
figure();
my_dome = surf(X2,Y2,Z2,'EdgeColor', 'none', 'FaceColor', 'texturemap', 'CData', cdata, 'FaceAlpha', 1) ;
axis equal
Which yields a nicely centered texture map:
Note: The texture mapping works so well because your actual surface is still a square, only bent a bit to conform to a sphere. The texture mapping does not display the part of the surface where the picture is transparent, so it is invisible, but if you look at your surface without the texture mapping you'll understand what went on in the background:
hs=surf(X2,Y2,Z2) ; shading interp ; axis equal

Related

Rotating an image matrix around its center in MATLAB

Assume I have a 2x2 matrix filled with values which will represent a plane. Now I want to rotate the plane around itself in a 3-D way, in the "z-Direction". For a better understanding, see the following image:
I wondered if this is possible by a simple affine matrix, thus I created the following simple script:
%Create a random value matrix
A = rand*ones(200,200);
%Make a box in the image
A(50:200-50,50:200-50) = 1;
Now I can apply transformations in the 2-D room simply by a rotation matrix like this:
R = affine2d([1 0 0; .5 1 0; 0 0 1])
tform = affine3d(R);
transformed = imwarp(A,tform);
However, this will not produce the desired output above, and I am not quite sure how to create the 2-D affine matrix to create such behavior.
I guess that a 3-D affine matrix can do the trick. However, if I define a 3-D affine matrix I cannot work with the 2-D representation of the matrix anymore, since MATLAB will throw the error:
The number of dimensions of the input image A must be 3 when the
specified geometric transformation is 3-D.
So how can I code the desired output with an affine matrix?
The answer from m3tho correctly addresses how you would apply the transformation you want: using fitgeotrans with a 'projective' transform, thus requiring that you specify 4 control points (i.e. 4 pairs of corresponding points in the input and output image). You can then apply this transform using imwarp.
The issue, then, is how you select these pairs of points to create your desired transformation, which in this case is to create a perspective projection. As shown below, a perspective projection takes into account that a viewing position (i.e. "camera") will have a given view angle defining a conic field of view. The scene is rendered by taking all 3-D points within this cone and projecting them onto the viewing plane, which is the plane located at the camera target which is perpendicular to the line joining the camera and its target.
Let's first assume that your image is lying in the viewing plane and that the corners are described by a normalized reference frame such that they span [-1 1] in each direction. We need to first select the degree of perspective we want by choosing a view angle and then computing the distance between the camera and the viewing plane. A view angle of around 45 degrees can mimic the sense of perspective of normal human sight, so using the corners of the viewing plane to define the edge of the conic field of view, we can compute the camera distance as follows:
camDist = sqrt(2)./tand(viewAngle./2);
Now we can use this to generate a set of control points for the transformation. We first apply a 3-D rotation to the corner points of the viewing plane, rotating around the y axis by an amount theta. This rotates them out of plane, so we now project the corner points back onto the viewing plane by defining a line from the camera through each rotated corner point and finding the point where it intersects the plane. I'm going to spare you the mathematical derivations (you can implement them yourself from the formulas in the above links), but in this case everything simplifies down to the following set of calculations:
term1 = camDist.*cosd(theta);
term2 = camDist-sind(theta);
term3 = camDist+sind(theta);
outP = [-term1./term2 camDist./term2; ...
term1./term3 camDist./term3; ...
term1./term3 -camDist./term3; ...
-term1./term2 -camDist./term2];
And outP now contains your normalized set of control points in the output image. Given an image of size s, we can create a set of input and output control points as follows:
scaledInP = [1 s(1); s(2) s(1); s(2) 1; 1 1];
scaledOutP = bsxfun(#times, outP+1, s([2 1])-1)./2+1;
And you can apply the transformation like so:
tform = fitgeotrans(scaledInP, scaledOutP, 'projective');
outputView = imref2d(s);
newImage = imwarp(oldImage, tform, 'OutputView', outputView);
The only issue you may come across is that a rotation of 90 degrees (i.e. looking end-on at the image plane) would create a set of collinear points that would cause fitgeotrans to error out. In such a case, you would technically just want a blank image, because you can't see a 2-D object when looking at it edge-on.
Here's some code illustrating the above transformations by animating a spinning image:
img = imread('peppers.png');
s = size(img);
outputView = imref2d(s);
scaledInP = [1 s(1); s(2) s(1); s(2) 1; 1 1];
viewAngle = 45;
camDist = sqrt(2)./tand(viewAngle./2);
for theta = linspace(0, 360, 360)
term1 = camDist.*cosd(theta);
term2 = camDist-sind(theta);
term3 = camDist+sind(theta);
outP = [-term1./term2 camDist./term2; ...
term1./term3 camDist./term3; ...
term1./term3 -camDist./term3; ...
-term1./term2 -camDist./term2];
scaledOutP = bsxfun(#times, outP+1, s([2 1])-1)./2+1;
tform = fitgeotrans(scaledInP, scaledOutP, 'projective');
spinImage = imwarp(img, tform, 'OutputView', outputView);
if (theta == 0)
hImage = image(spinImage);
set(gca, 'Visible', 'off');
else
set(hImage, 'CData', spinImage);
end
drawnow;
end
And here's the animation:
You can perform a projective transformation that can be estimated using the position of the corners in the first and second image.
originalP='peppers.png';
original = imread(originalP);
imshow(original);
s = size(original);
matchedPoints1 = [1 1;1 s(1);s(2) s(1);s(2) 1];
matchedPoints2 = [1 1;1 s(1);s(2) s(1)-100;s(2) 100];
transformType = 'projective';
tform = fitgeotrans(matchedPoints1,matchedPoints2,'projective');
outputView = imref2d(size(original));
Ir = imwarp(original,tform,'OutputView',outputView);
figure; imshow(Ir);
This is the result of the code above:
Original image:
Transformed image:

Crop and transform image in Matlab

I'm trying to crop an image but not with a rectangle (like in imcrop()) but with a polygon that has four corners. I've searched a lot and discovered that I need to perform a homography to reajust the cropped polygon into a rectangle.
So I've used imcrop() to select a polygon in an image :
img = imread('pout.tif');
imshow(img);
h = impoly;
position = wait(h);
x1 = min(position(:, 1));
x2 = max(position(:, 1));
y1 = min(position(:, 2));
y2 = max(position(:, 2));
BW = createMask(h);
How could I use these two things to crop out an area in the shape of a polygon with four corners ?
First of all, it is a bad idea to transform the image for cropping. It will results in changing the content of the ROI with artifacts due to interpolation when applying the homography. In addition, if one day you want to turn into a ROI defined by more than 4 points, this approach doesn't apply anylonger.
Second, I make some minor changes to your script, like this:
img = imread('circuit.tif');
imshow(img);
h = impoly;
position = wait(h);
boundbox = [min(position(:,1)), ....
min(position(:,2)), ....
max(position(:,1))-min(position(:,1)), ....
max(position(:,2))-min(position(:,2))];
BW = createMask(h);
img = imcrop(uint8(BW).*img, boundbox);
imshow(img)
You were almost there ... just mask the ROI of the image you want and crop with the bounding box of the ROI. Here it puts 0 outside the mask; you can adapt differently if you want.
Try "impoly" function in MATLAB
refer http://www.mathworks.in/help/images/ref/impoly.html

Love2d Rotating an image

I would like to rotate an image in Love2D.
I have found a documentation on love2d.org: https://love2d.org/wiki/love.graphics.rotate
But I can't seem to get it to work when I try to load an image.
Heres my code:
local angle = 0
function love.load()
g1 = love.graphics.newImage("1.png")
end
function love.draw()
width = 100
height = 100
love.graphics.translate(width/2, height/2)
love.graphics.rotate(angle)
love.graphics.translate(-width/2, -height/2)
love.graphics.draw(g1, width, height)
end
function love.update(dt)
love.timer.sleep(10)
angle = angle + dt * math.pi/2
angle = angle % (2*math.pi)
end
Could anyone show me an simple example of rotating an image in love2d?
https://love2d.org/wiki/love.graphics.draw
You may be better off using the fourth argument, shown as 'r' to rotate images, such as:
love.graphics.draw(image, x, y, math.pi/4)
It's not worth the trouble of using the translate functions for a single draw, and keeping those for when you're batching many draws at once.
Your code worked perfectly for me, aside from a small unrelated issue (love.timer.sleep uses seconds in LÖVE 0.8.0).
We will be able to help you better, and perhaps reproduce your error, if you provide us with more information.
When you say
I can't seem to get it to work when I try to load an image
..what is the result?
Is the image a white box? Does the application crash? Is there nothing on the screen?
All of these imply a image loading issue, rather than a rotation issue. Although, it could be the case that the image is rotating off of the screen.
If you continue to use translate, rotate, and scale (which is usually a good idea), I recommend you take a look at the push and pop functions.
They allow you to 'stack' transformations so you can render sub elements.
Example uses are rendering a GUI (each child pushes its translation and then renders the children) and drawing sprites on a scrolling map (the camera translates the entire map and then does for entity in entities do push() entity:draw() pop() end. Each entity can translate and rotate in local coordinates (0,0 = centre of sprite)).
love.graphics.draw( drawable, x, y, r, sx, sy, ox, oy, kx, ky )
the R is the rotation.. why don't you just set it to a variable and change it as you please? ... I'm new to programming so I may be wrong but this is how I would do it.
Example of rotating at center of image using LOVE 11.3 (Mysterious Mysteries):
function love.draw()
love.graphics.draw(img, 400,300, wheel.r, wheel.sx, wheel.sy, wheel.w / 2, wheel.h / 2)
end
function love.update(dt)
wheel.r = wheel.r + dt
end
function love.load()
wheel = {x = 0, y = 0, w = 0, h = 0, sx = 0.5, sy = 0.5, r = 0, image = "wheel.png"}
img = love.graphics.newImage(wheel.image)
wheel.w = img:getWidth()
wheel.h = img:getHeight()
end
Normaly the axis for rotating is the upper left corner. To center the axis to the middle of an image you have to use the parameters after the r parameter to half of width and half of height of the image.

How to position an axes in a figure relative to another axes?

When laying out a figure in MATLAB, typing axis equal ensures that no matter what the figure dimensions, the axes will always be square:
My current problem is that I want to add a second axes to this plot. Usually, that's no problem; I would just type axes([x1 y1 x2 y2]), and a new square figure would be added with corners at (x1, y1), (x2, y2), which is a fixed location relative to the figure. The problem is, I want this new axes to be located at a fixed location relative to the first axes.
So, my questions are:
Does anyone know how I can position an axes in a figure by specifying the location relative to another axes?
Assuming I can do 1, how can I have this new axes remain in the same place even if I resize the figure?
An axis position property is relative to its parent container. Therefore, one possibility is to create a transparent panel with the same size as the first axis, then inside it create the second axis, and set its location and size as needed. The position specified would be as if it were relative to the first axis.
Now we need to always maintain the panel to be the same size/location as the first axis. Usually this can be done using LINKPROP which links a property of multiple graphic objects (panel and axis) to be the same, namely the 'Position' property.
However, this would fail in your case: when calling axis image, it fixes the data units to be the same in every direction by setting aspect ratio properties like 'PlotBoxAspectRatio' and 'DataAspectRatio'. The sad news is that the 'Position' property will not reflect the change in size, thus breaking the above solution. Here is an example to illustrate the problem: if you query the position property before/after issuing the axis image call, it will be the same:
figure, plot(1:10,1:10)
get(gca,'Position')
pause(1)
axis image
get(gca,'Position')
Fortunately for us, there is a submission on FEX (plotboxpos) that solves this exact issue, and returns the actual position of the plotting region of the axis. Once we have that, it's a matter of syncing the panel position to the axis position. One trick is to create a event listener for when the axis changes size (it appears that the 'TightInset' property changes unlike the 'Position' property, so that could be the trigger in our case).
I wrapped the above in a function AXESRELATIVE for convenience: you call it as you would the builtin AXES function. The only difference is you give it as first argument the handle to the axis you want to relatively-position the newly created axis against. It returns handles to both the new axis and its containing panel.
Here is an example usage:
%# automatic resize only works for normalized units
figure
hParentAx = axes('Units','normalized');
axis(hParentAx, 'image')
%# create a new axis positioned at normalized units with w.r.t the previous axis
%# the axis should maintain its relative position on resizing the figure
[hAx hPan] = axesRelative(hParentAx, ...
'Units','normalized', 'Position',[0.7 0.1 0.1 0.1]);
set(hAx, 'Color','r')
And the function implementation:
function [hAx hPan] = axesRelative(hParentAx, varargin)
%# create panel exactly on top of parent axis
s = warning('off', 'MATLAB:hg:ColorSpec_None');
hPan = uipanel('Parent',get(hParentAx, 'Parent'), ...
'BorderType','none', 'BackgroundColor','none', ...
'Units',get(hParentAx,'Units'), 'Position',plotboxpos(hParentAx));
warning(s)
%# sync panel to always match parent axis position
addlistener(handle(hParentAx), ...
{'TightInset' 'Position' 'PlotBoxAspectRatio' 'DataAspectRatio'}, ...
'PostSet',#(src,ev) set(hPan, 'Position',plotboxpos(hParentAx)) );
%# create new axis under the newly created panel
hAx = axes('Parent',hPan, varargin{:});
end
On a completely different note: before you recent edit, I got the impression that you were trying to produce a scatter plot of images (i.e like a usual scatter plot, but with full images instead of points).
What you suggested (from what I understand) is creating one axis for each image, and setting its position corresponding to the x/y coordinates of the point.
My solution is to use the IMAGE/IMAGESC functions and draw the small images by explicitly setting the 'XData' and 'YData' properties to shift and scale the images appropriately. The beauty of this is it require a single axis, and doesn't suffer from having to deal with resizing issues..
Here is a sample implementation for that:
%# create fan-shaped coordinates
[R,PHI] = meshgrid(linspace(1,2,5), linspace(0,pi/2,10));
X = R.*cos(PHI); Y = R.*sin(PHI);
X = X(:); Y = Y(:);
num = numel(X);
%# images at each point (they don't have to be the same)
img = imread('coins.png');
img = repmat({img}, [num 1]);
%# plot scatter of images
SCALE = 0.2; %# image size along the biggest dimension
figure
for i=1:num
%# compute XData/YData vectors of each image
[h w] = size(img{i});
if h>w
scaleY = SCALE;
scaleX = SCALE * w/h;
else
scaleX = SCALE;
scaleY = SCALE * h/w;
end
xx = linspace(-scaleX/2, scaleX/2, h) + X(i);
yy = linspace(-scaleY/2, scaleY/2, w) + Y(i);
%# note: we are using the low-level syntax of the function
image('XData',xx, 'YData',yy, 'CData',img{i}, 'CDataMapping','scaled')
end
axis image, axis ij
colormap gray, colorbar
set(gca, 'CLimMode','auto')
This is usually the sort of thing you can take care of with a custom 'ResizeFcn' for your figure which will adjust the position and size of the smaller axes with respect the the larger. Here's an example of a resize function that maintains the size of a subaxes so that it is always 15% the size of the larger square axes and located in the bottom right corner:
function resizeFcn(src,event,hAxes,hSubAxes)
figurePosition = get(get(hAxes,'Parent'),'Position');
axesPosition = get(hAxes,'Position').*figurePosition([3 4 3 4]);
width = axesPosition(3);
height = axesPosition(4);
minExtent = min(width,height);
newPosition = [axesPosition(1)+(width-minExtent)/2+0.8*minExtent ...
axesPosition(2)+(height-minExtent)/2+0.05*minExtent ...
0.15*minExtent ...
0.15*minExtent];
set(hSubAxes,'Units','pixels','Position',newPosition);
end
And here's an example of its use:
hFigure = figure('Units','pixels'); %# Use pixel units for figure
hAxes = axes('Units','normalized'); %# Normalized axes units so it auto-resizes
axis(hAxes,'image'); %# Make the axes square
hSubAxes = axes('Units','pixels'); %# Use pixel units for subaxes
set(hFigure,'ResizeFcn',{#resizeFcn,hAxes,hSubAxes}); %# Set resize function

How can I draw a circle on an image in MATLAB?

I have an image in MATLAB:
im = rgb2gray(imread('some_image.jpg');
% normalize the image to be between 0 and 1
im = im/max(max(im));
And I've done some processing that resulted in a number of points that I want to highlight:
points = some_processing(im);
Where points is a matrix the same size as im with ones in the interesting points.
Now I want to draw a circle on the image in all the places where points is 1.
Is there any function in MATLAB that does this? The best I can come up with is:
[x_p, y_p] = find (points);
[x, y] = meshgrid(1:size(im,1), 1:size(im,2))
r = 5;
circles = zeros(size(im));
for k = 1:length(x_p)
circles = circles + (floor((x - x_p(k)).^2 + (y - y_p(k)).^2) == r);
end
% normalize circles
circles = circles/max(max(circles));
output = im + circles;
imshow(output)
This seems more than somewhat inelegant. Is there a way to draw circles similar to the line function?
You could use the normal PLOT command with a circular marker point:
[x_p,y_p] = find(points);
imshow(im); %# Display your image
hold on; %# Add subsequent plots to the image
plot(y_p,x_p,'o'); %# NOTE: x_p and y_p are switched (see note below)!
hold off; %# Any subsequent plotting will overwrite the image!
You can also adjust these other properties of the plot marker: MarkerEdgeColor, MarkerFaceColor, MarkerSize.
If you then want to save the new image with the markers plotted on it, you can look at this answer I gave to a question about maintaining image dimensions when saving images from figures.
NOTE: When plotting image data with IMSHOW (or IMAGE, etc.), the normal interpretation of rows and columns essentially becomes flipped. Normally the first dimension of data (i.e. rows) is thought of as the data that would lie on the x-axis, and is probably why you use x_p as the first set of values returned by the FIND function. However, IMSHOW displays the first dimension of the image data along the y-axis, so the first value returned by FIND ends up being the y-coordinate value in this case.
This file by Zhenhai Wang from Matlab Central's File Exchange does the trick.
%----------------------------------------------------------------
% H=CIRCLE(CENTER,RADIUS,NOP,STYLE)
% This routine draws a circle with center defined as
% a vector CENTER, radius as a scaler RADIS. NOP is
% the number of points on the circle. As to STYLE,
% use it the same way as you use the rountine PLOT.
% Since the handle of the object is returned, you
% use routine SET to get the best result.
%
% Usage Examples,
%
% circle([1,3],3,1000,':');
% circle([2,4],2,1000,'--');
%
% Zhenhai Wang <zhenhai#ieee.org>
% Version 1.00
% December, 2002
%----------------------------------------------------------------
Funny! There are 6 answers here, none give the obvious solution: the rectangle function.
From the documentation:
Draw a circle by setting the Curvature property to [1 1]. Draw the circle so that it fills the rectangular area between the points (2,4) and (4,6). The Position property defines the smallest rectangle that contains the circle.
pos = [2 4 2 2];
rectangle('Position',pos,'Curvature',[1 1])
axis equal
So in your case:
imshow(im)
hold on
[y, x] = find(points);
for ii=1:length(x)
pos = [x(ii),y(ii)];
pos = [pos-0.5,1,1];
rectangle('position',pos,'curvature',[1 1])
end
As opposed to the accepted answer, these circles will scale with the image, you can zoom in an they will always mark the whole pixel.
Hmm I had to re-switch them in this call:
k = convhull(x,y);
figure;
imshow(image); %# Display your image
hold on; %# Add subsequent plots to the image
plot(x,y,'o'); %# NOTE: x_p and y_p are switched (see note below)!
hold off; %# Any subsequent plotting will overwrite the image!
In reply to the comments:
x and y are created using the following code:
temp_hull = stats_single_object(k).ConvexHull;
for k2 = 1:length(temp_hull)
i = i+1;
[x(i,1)] = temp_hull(k2,1);
[y(i,1)] = temp_hull(k2,2);
end;
it might be that the ConvexHull is the other way around and therefore the plot is different. Or that I made a mistake and it should be
[x(i,1)] = temp_hull(k2,2);
[y(i,1)] = temp_hull(k2,1);
However the documentation is not clear about which colum = x OR y:
Quote: "Each row of the matrix contains the x- and y-coordinates of one vertex of the polygon. "
I read this as x is the first column and y is the second colum.
In newer versions of MATLAB (I have 2013b) the Computer Vision System Toolbox contains the vision.ShapeInserter System object which can be used to draw shapes on images. Here is an example of drawing yellow circles from the documentation:
yellow = uint8([255 255 0]); %// [R G B]; class of yellow must match class of I
shapeInserter = vision.ShapeInserter('Shape','Circles','BorderColor','Custom','CustomBorderColor',yellow);
I = imread('cameraman.tif');
circles = int32([30 30 20; 80 80 25]); %// [x1 y1 radius1;x2 y2 radius2]
RGB = repmat(I,[1,1,3]); %// convert I to an RGB image
J = step(shapeInserter, RGB, circles);
imshow(J);
With MATLAB and Image Processing Toolbox R2012a or newer, you can use the viscircles function to easily overlay circles over an image. Here is an example:
% Plot 5 circles at random locations
X = rand(5,1);
Y = rand(5,1);
% Keep the radius 0.1 for all of them
R = 0.1*ones(5,1);
% Make them blue
viscircles([X,Y],R,'EdgeColor','b');
Also, check out the imfindcircles function which implements the Hough circular transform. The online documentation for both functions (links above) have examples that show how to find circles in an image and how to display the detected circles over the image.
For example:
% Read the image into the workspace and display it.
A = imread('coins.png');
imshow(A)
% Find all the circles with radius r such that 15 ≤ r ≤ 30.
[centers, radii, metric] = imfindcircles(A,[15 30]);
% Retain the five strongest circles according to the metric values.
centersStrong5 = centers(1:5,:);
radiiStrong5 = radii(1:5);
metricStrong5 = metric(1:5);
% Draw the five strongest circle perimeters.
viscircles(centersStrong5, radiiStrong5,'EdgeColor','b');
Here's the method I think you need:
[x_p, y_p] = find (points);
% convert the subscripts to indicies, but transposed into a row vector
a = sub2ind(size(im), x_p, y_p)';
% assign all the values in the image that correspond to the points to a value of zero
im([a]) = 0;
% show the new image
imshow(im)

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