I have a 5000x5000 grid, and I'm trying to implement a simple model of cancer division in MATLAB. Initially, it picks a random point (x,y) and makes that cell a cancer cell. On the first iteration, it divides - the parent cell stays in it's place, the daughter cell is randomly assigned to any neighbouring cell.
Easy so far.
My problem is this: on successive iterations, a daughter cell will often be assigned to a cell that already has a cancer cell. In this case, I want the daughter cell to take its place and "bump" the cell already there to an adjacent cell. If that adjacent cell is empty, it is filled and the process stops. If not, the cell already in that place is bumped and so on until the last cell finds an empty space and the process stops.
This should be simple, but I have no idea how to code it up and what kind of loops to use.
I'm a physical scientists rather than a programmer, so please treat me like a simpleton!
Here is a function I hacked together that roughly meets the specs you provided.
I does slow down as the number of cancerous cells gets large.
Basically I have a few variables, the NxN matrix that represents the grid of cell locations (i call this a plate as grid is the name of an existing matlab function)
A vector of points that I can iterate through quickly. I pick a seed location and then run a while loop until the grid is full.
On each loop iteration I perform the following for each cell:
Generate a random number to determine if that cell should divide
Generate a random direction to divide
Find the first open plate position in that direction
Populate that position
I haven't tested it extensively but it appears to work.
function simulateCancer(plateSize, pDivide)
plate = zeros(plateSize, plateSize);
nCells = 1;
cellLocations = zeros(plateSize*plateSize,2);
initX = randi(plateSize);
initY = randi(plateSize);
cellLocations(nCells,:) = [initX, initY];
plate(initX, initY) = 1;
f = figure;
a = axes('Parent', f);
im = imagesc(plate, 'Parent', a);
while(nCells < (plateSize * plateSize))
currentGeneration = currentGeneration+1;
for i = 1:nCells
divide = rand();
if divide <= pDivide
divideLocation = cellLocations(i,:);
divideDir = randi(4);
[x, y, v] = findNewLocation(divideLocation(1), divideLocation(2), plate, divideDir);
if (v==1)
nCells = nCells+1;
plate(x,y) = 1;
cellLocations(nCells,:) = [x,y];
end
end
end
set(im,'CData', plate);
pause(.1);
end
end
function [x,y, valid] = findNewLocation(xin, yin, plate, direction)
x = xin;
y = yin;
valid = 1;
% keep looking for new spot if current spot is occupied
while( plate(x, y) == 1)
switch direction
case 1 % divide up
y = y-1;
case 2 % divide down
y = y+1;
case 3 % divide left
x = x-1;
case 4 % divide down
x = x+1;
otherwise
warning('Invalid direction')
x = xin;
y = yin;
return;
end
%if there has been a collision with a wall then just quit
if y==0 || y==size(plate,2)+1 || x==0 || x==size(plate,1)+1 % hit the top
x = xin; %return original values to say no division happend
y = yin;
valid = 0;
return;
end
end
end
Note: Instead of thinking of pushing cells, I coded this in a way that leaves cells where they currently are and creates the new cell at the end of the row/column. Semantically its different but logically it has the same end result, as long as you don't care about the generations.
Inspired by an another question, I though of using image processing techniques to implement this simulation. Specifically we can use morphological dilation to spread the cancerous cells.
The idea is to dilate each pixel using a structuring element that looks like:
1 0 0
0 1 0
0 0 0
where the center is fixed, and the other 1 is placed at random at one of the other eight remaining positions. This would effectively extend the pixel in that direction.
The way the dilation is performed is by created a blank image, with only one pixel set, then accumulating all the results using a simple OR operation.
To speed things up, we don't need to consider every pixel, only those on the perimeter of the current blocks formed by the clusters of cancerous cells. The pixels on the inside are already surrounded by cancer cells, and would have no effect if dilated.
To speed even further, we perform the dilation on all pixels that are chosen to be extended in the same direction in one call. Thus every iteration, we perform at most 8 dilation operations.
This made the code relatively fast (I tested up to 1000x1000 grid). Also it maintains the same timing across all iterations (will not slow down as the grid starts to fill up).
Here is my implementation:
%# initial grid
img = false(500,500);
%# pick 10 random cells, and set them as cancerous
img(randi(numel(img),[10 1])) = true;
%# show initial image
hImg = imshow(img, 'Border','tight', 'InitialMag',100);
%# build all possible structing elements
%# each one dilates in one of the 8 possible directions
SE = repmat([0 0 0; 0 1 0; 0 0 0],[1 1 8]);
SE([1:4 6:9] + 9*(0:7)) = 1;
%# run simulation until all cells have cancer
BW = false(size(img));
while ~all(img(:)) && ishandle(hImg)
%# find pixels on the perimeter of all "blocks"
on = find(bwperim(img,8));
%# percentage chance of division
on = on( rand(size(on)) > 0.5 ); %# 50% probability of cell division
if isempty(on), continue; end
%# decide on a direction for each pixel
d = randi(size(SE,3),[numel(on) 1]);
%# group pixels according to direction chosen
dd = accumarray(d, on, [8 1], #(x){x});
%# dilate each group of pixels in the chosen directions
%# to speed up, we perform one dilation for all pixels with same direction
for i=1:8
%# start with an image with only those pixels set
BW(:) = false;
BW(dd{i}) = true;
%# dilate in the specified direction
BW = imdilate(BW, SE(:,:,i));
%# add results to final image
img = img | BW;
end
%# show new image
set(hImg, 'CData',img)
drawnow
end
I also created an animation of the simulation on a 500x500 grid, with 10 random initial cancer cells (warning: the .gif image is approximately 1MB in size, so may take some time to load depending on your connection)
Related
I am trying to create a dynamic random dots stereogram by updating the position of the dots, so that I can get an inward motion of the dots (optic flow).
I first created a disk filled with dots whose position is randomised:
se = strel('disk',4);
x = linspace(-1,1,768); %generates 768 points in a vector with a [-1:1] interval
[X,Y] = meshgrid(x,x); %generates a rectangular grid in replicating the x vector.
[T,R] = cart2pol(X,Y);
threshold = 1 - (dot_nb / 768^2); %dot_nb=5;
Rd = double(rand(768,768)>threshold);
Rd(R>0.95 | R<0.06)=0; %remove some of the dots in order to get a circle and no dots in the middle of it
I am now trying to update the location of the dots at a specific refresh rate (20Hz) but I am having troubles with it.
I thought the following lines of code (which are in a bigger loop controlling the frame rate) could work but obviously something is missing and I do not know how else I could do it:
update = 0.1;
for i = 1:size(R,1) %R is the matrix with the radius values of the dots
for j = 1:size(R,2)
if R(i,j)<0.95 && R(i,j)>0.06
R(i,j) = R(i,j)-update;
else
R(i,j)=0;
end
end
end
Has anyone any idea how I could do to get what I'm looking for?
Thank you in advance!
I'm making an image processing project and I have stuck in one the project's steps. Here is the situation;
This is my mask:
and I want to detect the maximum-sized rectangle that can fit into this mask like this.
I'm using MATLAB for my project. Do you know any fast way to accomplish this aim. Any code sample, approach or technique would be great.
EDIT 1 : The two algorithms below are works with lot's of the cases. But both of them give wrong results in some difficult cases. I'am using both of them in my project.
This approach starts with the entire image and shrinks each border in turn pixel-by-pixel until it finds an acceptable rectangle.
It takes ~0.02 seconds to run on the example image, so it's reasonably fast.
EDIT: I should clarify that this isn't meant to be a universal solution. This algorithm relies on the rectangle being centered and having roughly the same aspect ratio as the image itself. However, in the cases where it is appropriate, it is fast. #DanielHsH offered a solution which they claim works in all cases.
The code:
clear; clc;
tic;
%% // read image
imrgb= imread('box.png');
im = im2bw(rgb2gray(imrgb)); %// binarize image
im = 1-im; %// convert "empty" regions to 0 intensity
[rows,cols] = size(im);
%% // set up initial parameters
ULrow = 1; %// upper-left row (param #1)
ULcol = 1; %// upper-left column (param #2)
BRrow = rows; %// bottom-right row (param #3)
BRcol = cols; %// bottom-right column (param #4)
parameters = 1:4; %// parameters left to be updated
pidx = 0; %// index of parameter currently being updated
%% // shrink region until acceptable
while ~isempty(parameters); %// update until all parameters reach bounds
%// 1. update parameter number
pidx = pidx+1;
pidx = mod( pidx-1, length(parameters) ) + 1;
p = parameters(pidx); %// current parameter number
%// 2. update current parameter
if p==1; ULrow = ULrow+1; end;
if p==2; ULcol = ULcol+1; end;
if p==3; BRrow = BRrow-1; end;
if p==4; BRcol = BRcol-1; end;
%// 3. grab newest part of region (row or column)
if p==1; region = im(ULrow,ULcol:BRcol); end;
if p==2; region = im(ULrow:BRrow,ULcol); end;
if p==3; region = im(BRrow,ULcol:BRcol); end;
if p==4; region = im(ULrow:BRrow,BRcol); end;
%// 4. if the new region has only zeros, stop shrinking the current parameter
if isempty(find(region,1))
parameters(pidx) = [];
end
end
toc;
params = [ULrow ULcol BRrow BRcol]
area = (BRrow-ULrow)*(BRcol-ULcol)
The results for this image:
Elapsed time is 0.027032 seconds.
params =
10 25 457 471
area =
199362
Code to visualize results:
imrgb(params(1):params(3),params(2):params(4),1) = 0;
imrgb(params(1):params(3),params(2):params(4),2) = 255;
imrgb(params(1):params(3),params(2):params(4),3) = 255;
imshow(imrgb);
Another example image:
Here is a correct answer.
You must use dynamic programming! Other methods of direct calculation (like cutting 1 pixel from each edge) might produce sub-optimal results. My method guarantees that it selects the largest possible rectangle that fits in the mask. I assume that the mask has 1 big convex white blob of any shape with black background around it.
I wrote 2 methods. findRect() which finds the best possible square (starting on x,y with length l). The second method LargestInscribedImage() is an example of how to find any rectangle (of any aspect ratio). The trick is to resize the mask image, find a square and resize it back.
In my example the method finds the larges rectangle that can be fit in the mask having the same aspect ration as the mask image. For example if the mask image is of size 100x200 pixels than the algorithm will find the largest rectangle having aspect ratio 1:2.
% ----------------------------------------------------------
function LargestInscribedImage()
% ----------------------------------------------------------
close all
im = double(imread('aa.bmp')); % Balck and white image of your mask
im = im(:,:,1); % If it is colored RGB take only one of the channels
b = imresize(im,[size(im,1) size(im,1)]); Make the mask square by resizing it by its aspect ratio.
SC = 1; % Put 2..4 to scale down the image an speed up the algorithm
[x1,y1,l1] = findRect(b,SC); % Lunch the dyn prog algorithm
[x2,y2,l2] = findRect(rot90(b),SC); % rotate the image by 90deg and solve
% Rotate back: x2,y2 according to rot90
tmp = x2;
x2 = size(im,1)/SC-y2-l2;
y2 = tmp;
% Select the best solution of the above (for the original image and for the rotated by 90degrees
if (l1>=l2)
corn = sqCorn(x1,y1,l1);
else
corn = sqCorn(x2,y2,l2);
end
b = imresize(b,1/SC);
figure;imshow(b>0); hold on;
plot(corn(1,:),corn(2,:),'O')
corn = corn*SC;
corn(1,:) = corn(1,:)*size(im,2)/size(im,1);
figure;imshow(im); hold on;
plot(corn(1,:),corn(2,:),'O')
end
function corn = sqCorn(x,y,l)
corn = [x,y;x,y+l;x+l,y;x+l,y+l]';
end
% ----------------------------------------------------------
function [x,y,l] = findRect(b,SC)
b = imresize(b,1/SC);
res = zeros(size(b,1),size(b,2),3);
% initialize first col
for i = 1:1:size(b,1)
if (b(i,1) > 0)
res(i,1,:) = [i,1,0];
end
end
% initialize first row
for i = 1:1:size(b,2)
if (b(1,i) > 0)
res(1,i,:) = [1,i,0];
end
end
% DynProg
for i = 2:1:size(b,1)
for j = 2:1:size(b,2)
isWhite = b(i,j) > 0;
if (~isWhite)
res(i,j,:)=res(i-1,j-1,:); % copy
else
if (b(i-1,j-1)>0) % continuous line
lineBeg = [res(i-1,j-1,1),res(i-1,j-1,2)];
lineLenght = res(i-1,j-1,3);
if ((b(lineBeg(1),j)>0)&&(b(i,lineBeg(2))>0)) % if second diag is good
res(i,j,:) = [lineBeg,lineLenght+1];
else
res(i,j,:)=res(i-1,j-1,:); % copy since line has ended
end
else
res(i,j,:) = [i,j,0]; % Line start
end
end
end
end
% check last col
[maxValCol,WhereCol] = max(res(:,end,3));
% check last row
[maxValRow,WhereRow] = max(res(end,:,3));
% Find max
x= 0; y = 0; l = 0;
if (maxValCol>maxValRow)
y = res(WhereCol,end,1);
x = res(WhereCol,end,2);
l = maxValCol;
else
y = res(end,WhereRow,1);
x = res(end,WhereRow,2);
l = maxValRow;
end
corn = [x,y;x,y+l;x+l,y;x+l,y+l]';
% figure;imshow(b>0); hold on;
% plot(corn(1,:),corn(2,:),'O')
return;
end
The black boundaries in your image are curved and not closed. For example, in the top right corner, the black boundaries won't meet and form a closed contour. Therefore, a simple strategy in one of my comments will not work.
I am now providing you with a skeleton of a code which you can play with and add conditions as per your need. My idea is as follows:
To find left-side x-coordinate of the rectangle, first count the white pixels each column of the image contains:
%I assume that the image has already been converted to binary.
whitePixels=sum(img,1);
Then find the rate of change:
diffWhitePixels=diff(whitePixels);
If you see the bar plot of diffWhitePixels then you will observe various large entries (which indicate that the white region is still not in a straight line, and it is not a proper place to put the rectangles left vertical edge). Small entries (in your image, less than 5) indicate you can put the rectangle edge there.
You can do similar things to determine right, top and bottom edge positions of the rectangles.
Discussion:
First of all, the problem is ill-posed in my opinion. What do you mean by maximum-sized rectangle? Is it maximum area or length of side? In all possible cases, I don't think above method can get the correct answer. I can think of two or three cases right now where above method would fail, but it will at least give you the right answer on images similar to the given image, provided you adjust the values.
You can put some constraints once you know how your images are going to look. For example, if the black boundary curves inside, you can say that you don't want a column such as [0;0;...0;1;1;...0;0;...;0;1;1;...;1] i.e. zeros surrounded by ones. Another constraint could be how many black pixels do you want to allow? You can also crop the image till to remove extra black pixels. In your image, you can crop the image (programmatically) from the left and the bottom edge. Cropping an image is probably necessary, and definitely the better thing to do.
I am implementing a simple algorithm to do in-painting on a "damaged" image. I have a predefined mask that specifies the area which needs to be fixed. My strategy is to start at the border of the masked area and in-paint each pixel with the central mean of its neighboring non-zero pixels, repeating until there's no unknown pixels left.
function R = inPainting(I, mask)
H = [1 2 1; 2 0 2; 1 2 1];
R = I;
n = 1;
[row,col,~] = find(~mask); %Find zeros in mask (area to be inpainted)
unknown = horzcat(row, col)';
while size(unknown,2) > 0
new_unknown = [];
new_R = R;
for u = unknown
r = u(1);
c = u(2);
nb = R(max((r-n), 1):min((r+n), end), max((c-n),1):min((c+n),end));
nz = nb~=0;
nzs = sum(nz(:));
if nzs ~= 0 %We have non-zero neighbouring pixels. In-paint with average.
new_R(r,c) = sum(nb(:)) / nzs;
else
new_unknown = horzcat(new_unknown, u);
end
end
unknown = new_unknown;
R = new_R;
end
This works well, but it's not very efficient. Is it possible to vectorize such an approach, using mostly matrix operations? Does someone know of a more efficient way to implement this algorithm?
If I understand your problem statement, you are given a mask and you wish to fill in these pixels in this mask with the mean of the neighbourhood pixels that surround each pixel in the mask. Another constraint is that the image is defined such that any pixels that belong to the mask in the same spatial locations are zero in this mask. You are starting from the border of the mask and are propagating information towards the innards of the mask. Given this algorithm, there is unfortunately no way you can do this with standard filtering techniques as the current time step is dependent on the previous time step.
Image filtering mechanisms, like imfilter or conv2 can't work here because of this dependency.
As such, what I can do is help you speed up what is going on inside your loop and hopefully this will give you some speed up overall. I'm going to introduce you to a function called im2col. This is from the image processing toolbox, and given that you can use imfilter, we can use this function.
im2col creates a 2D matrix such that each column is a pixel neighbourhood unrolled into a single vector. How it works is that each pixel neighbourhood in column major order is grabbed, so we get a pixel neighbourhood at the top left corner of the image, then move down one row, and another row and we keep going until we reach the last row. We then move one column over and repeat the same process. For each pixel neighbourhood that we have, it gets unrolled into a single vector, and the output would be a MN x K matrix where you have a neighbourhood size of M x N for each pixel neighbourhood and there are K neighbourhoods.
Therefore, at each iteration of your loop, we can unroll the current inpainted image's pixel neighbourhoods into single vectors, determine which pixel neighborhoods are non-zero and from there, determine how many zero values there are for each of these selected pixel neighbourhood. After, we compute the mean for these non-zero columns disregarding the zero elements. Once we're done, we update the image and move to the next iteration.
What we're going to need to do first is pad the image with a 1 pixel border so that we're able to grab neighbourhoods that extend beyond the borders of the image. You can use padarray, also from the image processing toolbox.
Therefore, we can simply do this:
function R = inPainting(I, mask)
R = double(I); %// For precision
n = 1;
%// Change - column major indices
unknown = find(~mask); %Find zeros in mask (area to be inpainted)
%// Until we have searched all unknown pixels
while numel(unknown) ~= 0
new_R = R;
%// Change - take image at current iteration and
%// create columns of pixel neighbourhoods
padR = padarray(new_R, [n n], 'replicate');
cols = im2col(padR, [2*n+1 2*n+1], 'sliding');
%// Change - Access the right pixel neighbourhoods
%// denoted by unknown
nb = cols(:,unknown);
%// Get total sum of each neighbourhood
nbSum = sum(nb, 1);
%// Get total number of non-zero elements per pixel neighbourhood
nzs = sum(nb ~= 0, 1);
%// Replace the right pixels in the image with the mean
new_R(unknown(nzs ~= 0)) = nbSum(nzs ~= 0) ./ nzs(nzs ~= 0);
%// Find new unknown pixels to look at
unknown = unknown(nzs == 0);
%// Update image for next iteration
R = new_R;
end
%// Cast back to the right type
R = cast(R, class(I));
I'm making an image processing project and I have stuck in one the project's steps. Here is the situation;
This is my mask:
and I want to detect the maximum-sized rectangle that can fit into this mask like this.
I'm using MATLAB for my project. Do you know any fast way to accomplish this aim. Any code sample, approach or technique would be great.
EDIT 1 : The two algorithms below are works with lot's of the cases. But both of them give wrong results in some difficult cases. I'am using both of them in my project.
This approach starts with the entire image and shrinks each border in turn pixel-by-pixel until it finds an acceptable rectangle.
It takes ~0.02 seconds to run on the example image, so it's reasonably fast.
EDIT: I should clarify that this isn't meant to be a universal solution. This algorithm relies on the rectangle being centered and having roughly the same aspect ratio as the image itself. However, in the cases where it is appropriate, it is fast. #DanielHsH offered a solution which they claim works in all cases.
The code:
clear; clc;
tic;
%% // read image
imrgb= imread('box.png');
im = im2bw(rgb2gray(imrgb)); %// binarize image
im = 1-im; %// convert "empty" regions to 0 intensity
[rows,cols] = size(im);
%% // set up initial parameters
ULrow = 1; %// upper-left row (param #1)
ULcol = 1; %// upper-left column (param #2)
BRrow = rows; %// bottom-right row (param #3)
BRcol = cols; %// bottom-right column (param #4)
parameters = 1:4; %// parameters left to be updated
pidx = 0; %// index of parameter currently being updated
%% // shrink region until acceptable
while ~isempty(parameters); %// update until all parameters reach bounds
%// 1. update parameter number
pidx = pidx+1;
pidx = mod( pidx-1, length(parameters) ) + 1;
p = parameters(pidx); %// current parameter number
%// 2. update current parameter
if p==1; ULrow = ULrow+1; end;
if p==2; ULcol = ULcol+1; end;
if p==3; BRrow = BRrow-1; end;
if p==4; BRcol = BRcol-1; end;
%// 3. grab newest part of region (row or column)
if p==1; region = im(ULrow,ULcol:BRcol); end;
if p==2; region = im(ULrow:BRrow,ULcol); end;
if p==3; region = im(BRrow,ULcol:BRcol); end;
if p==4; region = im(ULrow:BRrow,BRcol); end;
%// 4. if the new region has only zeros, stop shrinking the current parameter
if isempty(find(region,1))
parameters(pidx) = [];
end
end
toc;
params = [ULrow ULcol BRrow BRcol]
area = (BRrow-ULrow)*(BRcol-ULcol)
The results for this image:
Elapsed time is 0.027032 seconds.
params =
10 25 457 471
area =
199362
Code to visualize results:
imrgb(params(1):params(3),params(2):params(4),1) = 0;
imrgb(params(1):params(3),params(2):params(4),2) = 255;
imrgb(params(1):params(3),params(2):params(4),3) = 255;
imshow(imrgb);
Another example image:
Here is a correct answer.
You must use dynamic programming! Other methods of direct calculation (like cutting 1 pixel from each edge) might produce sub-optimal results. My method guarantees that it selects the largest possible rectangle that fits in the mask. I assume that the mask has 1 big convex white blob of any shape with black background around it.
I wrote 2 methods. findRect() which finds the best possible square (starting on x,y with length l). The second method LargestInscribedImage() is an example of how to find any rectangle (of any aspect ratio). The trick is to resize the mask image, find a square and resize it back.
In my example the method finds the larges rectangle that can be fit in the mask having the same aspect ration as the mask image. For example if the mask image is of size 100x200 pixels than the algorithm will find the largest rectangle having aspect ratio 1:2.
% ----------------------------------------------------------
function LargestInscribedImage()
% ----------------------------------------------------------
close all
im = double(imread('aa.bmp')); % Balck and white image of your mask
im = im(:,:,1); % If it is colored RGB take only one of the channels
b = imresize(im,[size(im,1) size(im,1)]); Make the mask square by resizing it by its aspect ratio.
SC = 1; % Put 2..4 to scale down the image an speed up the algorithm
[x1,y1,l1] = findRect(b,SC); % Lunch the dyn prog algorithm
[x2,y2,l2] = findRect(rot90(b),SC); % rotate the image by 90deg and solve
% Rotate back: x2,y2 according to rot90
tmp = x2;
x2 = size(im,1)/SC-y2-l2;
y2 = tmp;
% Select the best solution of the above (for the original image and for the rotated by 90degrees
if (l1>=l2)
corn = sqCorn(x1,y1,l1);
else
corn = sqCorn(x2,y2,l2);
end
b = imresize(b,1/SC);
figure;imshow(b>0); hold on;
plot(corn(1,:),corn(2,:),'O')
corn = corn*SC;
corn(1,:) = corn(1,:)*size(im,2)/size(im,1);
figure;imshow(im); hold on;
plot(corn(1,:),corn(2,:),'O')
end
function corn = sqCorn(x,y,l)
corn = [x,y;x,y+l;x+l,y;x+l,y+l]';
end
% ----------------------------------------------------------
function [x,y,l] = findRect(b,SC)
b = imresize(b,1/SC);
res = zeros(size(b,1),size(b,2),3);
% initialize first col
for i = 1:1:size(b,1)
if (b(i,1) > 0)
res(i,1,:) = [i,1,0];
end
end
% initialize first row
for i = 1:1:size(b,2)
if (b(1,i) > 0)
res(1,i,:) = [1,i,0];
end
end
% DynProg
for i = 2:1:size(b,1)
for j = 2:1:size(b,2)
isWhite = b(i,j) > 0;
if (~isWhite)
res(i,j,:)=res(i-1,j-1,:); % copy
else
if (b(i-1,j-1)>0) % continuous line
lineBeg = [res(i-1,j-1,1),res(i-1,j-1,2)];
lineLenght = res(i-1,j-1,3);
if ((b(lineBeg(1),j)>0)&&(b(i,lineBeg(2))>0)) % if second diag is good
res(i,j,:) = [lineBeg,lineLenght+1];
else
res(i,j,:)=res(i-1,j-1,:); % copy since line has ended
end
else
res(i,j,:) = [i,j,0]; % Line start
end
end
end
end
% check last col
[maxValCol,WhereCol] = max(res(:,end,3));
% check last row
[maxValRow,WhereRow] = max(res(end,:,3));
% Find max
x= 0; y = 0; l = 0;
if (maxValCol>maxValRow)
y = res(WhereCol,end,1);
x = res(WhereCol,end,2);
l = maxValCol;
else
y = res(end,WhereRow,1);
x = res(end,WhereRow,2);
l = maxValRow;
end
corn = [x,y;x,y+l;x+l,y;x+l,y+l]';
% figure;imshow(b>0); hold on;
% plot(corn(1,:),corn(2,:),'O')
return;
end
The black boundaries in your image are curved and not closed. For example, in the top right corner, the black boundaries won't meet and form a closed contour. Therefore, a simple strategy in one of my comments will not work.
I am now providing you with a skeleton of a code which you can play with and add conditions as per your need. My idea is as follows:
To find left-side x-coordinate of the rectangle, first count the white pixels each column of the image contains:
%I assume that the image has already been converted to binary.
whitePixels=sum(img,1);
Then find the rate of change:
diffWhitePixels=diff(whitePixels);
If you see the bar plot of diffWhitePixels then you will observe various large entries (which indicate that the white region is still not in a straight line, and it is not a proper place to put the rectangles left vertical edge). Small entries (in your image, less than 5) indicate you can put the rectangle edge there.
You can do similar things to determine right, top and bottom edge positions of the rectangles.
Discussion:
First of all, the problem is ill-posed in my opinion. What do you mean by maximum-sized rectangle? Is it maximum area or length of side? In all possible cases, I don't think above method can get the correct answer. I can think of two or three cases right now where above method would fail, but it will at least give you the right answer on images similar to the given image, provided you adjust the values.
You can put some constraints once you know how your images are going to look. For example, if the black boundary curves inside, you can say that you don't want a column such as [0;0;...0;1;1;...0;0;...;0;1;1;...;1] i.e. zeros surrounded by ones. Another constraint could be how many black pixels do you want to allow? You can also crop the image till to remove extra black pixels. In your image, you can crop the image (programmatically) from the left and the bottom edge. Cropping an image is probably necessary, and definitely the better thing to do.
I wonder whether it would be possible to extract only hands from a video with matlab. In the video hands perform some gesture. Because first frames are only background I tried in this way:
readerObj = VideoReader('VideoWithHands.mp4');
nFrames = readerObj.NumberOfFrames;
fr = get(readerObj, 'FrameRate');
writerObj = VideoWriter('Hands.mp4', 'MPEG-4');
set(writerObj, 'FrameRate', fr);
open(writerObj);
bg = read(readerObj, 1); %background
for k = 1 : nFrames
frame = read(readerObj, k);
hands = imabsdiff(frame,bg);
writeVideo(writerObj,hands);
end
close(writerObj);
But I realized that colors of the hands are not "real" and they are transparent. Is there a better way to extract them from video keeping colors and opacity level exploiting the first frames (background)?
EDIT: Well, I have found a good setting for vision.ForegroundDetector object, now hands are white logical regions but when I try to visualize them with:
videoSource = vision.VideoFileReader('VideoWithHands.mp4', 'VideoOutputDataType', 'uint8');
detector = vision.ForegroundDetector('NumTrainingFrames', 46, 'InitialVariance', 4000, 'MinimumBackgroundRatio', 0.2);
videoplayer = vision.VideoPlayer();
hands = uint8(zeros(720,1280,3));
while ~isDone(videoSource)
frame = step(videoSource);
fgMask = step(detector, frame);
[m,n] = find(fgMask);
a = [m n];
if isempty(a)==true
hands(:,:,:) = uint8(zeros(720,1280,3));
else
hands(m,n,1) = frame(m,n,1);
hands(m,n,2) = frame(m,n,2);
hands(m,n,3) = frame(m,n,3);
end
step(videoplayer, hands)
end
release(videoplayer)
release(videoSource)
or put them into a videofile with:
eaderObj = VideoReader('Video 9.mp4');
nFrames = readerObj.NumberOfFrames;
fr = get(readerObj, 'FrameRate');
writerObj = VideoWriter('hands.mp4', 'MPEG-4');
set(writerObj, 'FrameRate', fr);
detector = vision.ForegroundDetector('NumTrainingFrames', 46, 'InitialVariance', 4000, 'MinimumBackgroundRatio', 0.2);
open(writerObj);
bg = read(readerObj, 1);
frame = uint8(zeros(size(bg)));
for k = 1 : nFrames
frame = read(readerObj, k);
fgMask = step(detector, frame);
[m,n] = find(fgMask);
hands = uint8(zeros(720,1280));
if isempty([m n]) == true
hands(:,:) = uint8(zeros(720,1280));
else
hands(m,n) = frame(m,n);
end
writeVideo(writerObj,mani);
end
close(writerObj);
...my PC crashes. Some suggestion?
So you're trying to cancel out the background, making it black, right?
The easiest way to do this should be to filter it, you can do that by comparing your difference data to a threshold value and then using the result as indices to set a custom background.
filtered = imabsdiff(frame,bg);
bgindex = find( filtered < 10 );
frame(bgindex) = custombackground(bgindex);
where custombackground is whatever image file you want to put into the background. If you want it to be just black or white, use 0 or 255 instead of custombackground(bgindex). Note that the numbers depend on your video data's format and could be inaccurate (except 0, this one should always be right). If too much gets filtered out, lower the 10 above, if too much remains unfiltered, increase the 10.
At the end, you write your altered frame back into the video, so it just replaces the hands variable in your code.
Also, depending on your format, you might have to do the comparison across RGB values. This is slightly more complicated as it involves checking 3 values at the same time and doing some magic with the indices. This is the RGB version (works with anything containing 3 color bands):
filtered = imabsdiff(frame,bg); % differences at each pixel in each color band
totalfiltered = sum(filtered,3); % sums up the differences
% in each color band (RGB)
bgindex = find( totalfiltered < 10 ); % extracts indices of pixels
% with color close to bg
allind = sub2ind( [numel(totalfiltered),3] , repmat(bgindex,1,3) , ...
repmat(1:3,numel(bgindex),1) ); % index magic
frame(allind) = custombackground(allind); % copy custom background into frame
EDIT :
Here's a detailed explanation of the index magic.
Let's assume a 50x50 image. Say the pixel at row 2, column 5 is found to be background, then bgindex will contain the number 202 (linear index corresponding to [2,5] = (5-1)*50+2 ). What we need is a set of 3 indices corresponding to the matrix coordinates [2,5,1], [2,5,2] and [2,5,3]. That way, we can change all 3 color bands corresponding to that pixel. To make calculations easier, this approach actually assumes linear indexing for the image and thus converts it to a 2500x1 image. Then it expands the 3 color bands, creating a 2500x3 matrix. We now construct the indices [202,1], [202,2] and [202,3] instead.
To do that, we first construct a matrix of indices by repeating our values. repmat does this for us, it creates the matrices [202 202 202] and [1 2 3]. If there were more pixels in bgindex, the first matrix would contain more rows, each repeating the linear pixel coordinates 3 times. The second matrix would contain additional [1 2 3] rows. The first argument to sub2ind is the size of the matrix, in this case, 2500x3, so we calculate the number of pixels with numel applied to the sum vector (which collapses the image's 3 bands into 1 value and thus has 1 value per pixel) and add a static 3 in the second dimension.
sub2ind now takes each element from the first matrix as a row index, each corresponding element from the second matrix as a column index and converts them to linear indices into a matrix of the size we determined earlier. In our example, this results in the indices [202 2702 5202]. sub2ind preserves the shape of the inputs, so if we had 10 background pixels, this result would have the size 10x3. But since linear indexing doesn't care about the shape of the index matrix, it just takes all of those values.
To confirm this is correct, let's revert the values in the example. The original image data would have the size 50x50x3. For an NxMxP matrix, a linear index to the subscript [n m p] can be calculated as ind = (p-1)*M*N + (m-1)*N + n. Using our values, we get the following:
[2 5 1] => 202
[2 5 2] => 2702
[2 5 3] => 5202
ind2sub confirms this.
Yes, there is a better way. The computer vision system toolbox includes a vision.ForegroundDetector object that does what you need. It implements the Gaussian Mixture Model algorithm for background subtraction.