I know its not appropriate to ask questions related to hacker rank here. But this Easy section question is giving me a very hard time and I started to doubt my image processing concepts.
They have given an image
0 0 0 0
0 1 1 0
0 0 0 0
and the structuring element as
1 0
1 1
with the origin as bottom left.
And I got the dilated output as
0 0 0 0
1 1 1 0
0 1 1 0
(Since according to the structuring element, I have to see for bright pixels to origin's top and to right side). I have even verified with Matlab.
But why hacker rank is not accepting my output ? Or am I missing a
core concept of binary image dilation ?
I got a slightly different result
bw = [ 0 0 0 0; 0 1 1 0; 0 0 0 0];
imdilate(bw,[1 0;1 1;0 0])
ans =
0 1 1 0
0 1 1 1
0 0 0 0
Note that I used a slightly different strel: I added a row of zeros at the bottom. I did that in order to adhere to the requirement "origin as bottom left".
I hope this figure explains a bit more:
As you can see the requirement is that the "origin" of the strel would be the bottom left corner, according to matlab's strel doc:
The center (or origin) of nhood is its center element, given by floor((size(nhood) + 1)/2).
In order to have the vertical center in the second row (instead of the first), I added an empty line to the strel.
Related
I want to visualize all of the paths on 2D square in matlab.
This code gives me the following figure which consists of a 2D square and randomly distributed of 1 and 0.
https://i.hizliresim.com/Ey4G4D.png
Each 1's must connected with lines from top to bottom.
If there is 1, then there is a way and I have to plot line. Otherwise therre is no way and stop.
Without the boundary elements, there are 3 way for each elements. Each element can go left side, right side or down side.
The top left hand corner's element can go right and down direction.
The top right hand corner's element can go left and down direction.
This is the algorithm of the modelling.
https://i.hizliresim.com/Dy0z0y.jpg
How can I write this code ?
I am waiting your advise :)
Problem analysis
To get an information on the possible paths in your matrix/image you can use the diff function. It calculates the difference between two neighbouring matrix elements along the specified dimension.
The conditions for the existence of a path are:
The difference between the element and its neighbour must be 0
The element itself must be 1
Solution
The following matlab program will create 3 matrices containing the value 1 or true for each element with a path existing to its neighbour.
matrix = logical([1 1 1 1 0; ...
1 1 0 1 1; ...
0 0 0 1 0; ...
0 0 1 1 0])
hasPathtoRight = false(size(matrix));
hasPathtoRight(:,1:end-1) = (diff(matrix,1,2)==0) & (matrix(:,1:end-1)==1)
hasPathtoLeft = false(size(matrix));
hasPathtoLeft(:,2:end) = (diff(matrix,1,2)==0) & (matrix(:,2:end)==1)
hasPathDown = false(size(matrix));
hasPathDown(1:end-1,:) = (diff(matrix,1,1)==0) & (matrix(1:end-1,:)==1)
Result
The result for the example matrix is shown here:
matrix =
1 1 1 1 0
1 1 0 1 1
0 0 0 1 0
0 0 1 1 0
hasPathtoRight =
1 1 1 0 0
1 0 0 1 0
0 0 0 0 0
0 0 1 0 0
hasPathtoLeft =
0 1 1 1 0
0 1 0 0 1
0 0 0 0 0
0 0 0 1 0
hasPathDown =
1 1 0 1 0
0 0 0 1 0
0 0 0 1 0
0 0 0 0 0
You can use these matrices to draw the paths in a graphical display.
I have spent all day reading up on the above MATLAB functions. I can't seem to find any good explanations online, even on the MathWorks website!
I would be very grateful if anyone could explain bwlabel, regionprops and centroid. How do they work if applied to a grayscale image?
Specifically, they are being used in this code below. How do the above functions apply to the code below?
fun=#minutie; L = nlfilter(K,[3 3],fun);
%% Termination LTerm=(L==1);
figure; imshow(LTerm)
LTermLab=bwlabel(LTerm);
propTerm=regionprops(LTermLab,'Centroid');
CentroidTerm=round(cat(1,LTerm(:).Centroid));
figure; imshow(~K)
set(gcf,'position',[1 1 600 600]); hold on
plot(CentroidTerm(:,1),CentroidTerm(:,2),'ro')
That's quite a mouthful to explain!... nevertheless, I'd love to explain it to you. However, I'm a bit surprised that you couldn't understand the documentation from MathWorks. It's actually quite good at explaining a lot (if not all...) of their functions.
BTW, bwlabel and regionprops are not defined for grayscale images. You can only apply these to binary images.
Update: bwlabel still has the restriction of accepting a binary image but regionprops no longer has this restriction. It can also take in a label matrix that is usually output from bwlabel as well as binary images.
Assuming binary images is what you want, my explanations for each function is as follows.
bwlabel
bwlabel takes in a binary image. This binary image should contain a bunch of objects that are separated from each other. Pixels that belong to an object are denoted with 1 / true while those pixels that are the background are 0 / false. For example, suppose we have a binary image that looks like this:
0 0 0 0 0 1 1 1 0 0
0 1 0 1 0 0 1 1 0 0
0 1 1 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 1 1
0 0 1 1 1 1 0 0 1 1
You can see in this image that there are four objects in this image. The definition of an object are those pixels that are 1 that are connected in a chain by looking at local neighbourhoods. We usually look at 8-pixel neighbourhoods where you look at the North, Northeast, East, Southeast, South, Southwest, West, Northwest directions. Another way of saying this is that the objects are 8-connected. For simplicity, sometimes people look at 4-pixel neighbourhoods, where you just look at the North, East, South and West directions. This woudl mean that the objects are 4-connected.
The output of bwlabel will give you an integer map where each object is assigned a unique ID. As such, the output of bwlabel would look something like this:
0 0 0 0 0 3 3 3 0 0
0 1 0 1 0 0 3 3 0 0
0 1 1 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 4
0 0 0 0 0 0 0 0 4 4
0 0 2 2 2 2 0 0 4 4
Because MATLAB processes things in column major, that's why the labelling is how you see above. As such, bwlabel gives you the membership of each pixel. This tells you where each pixel belongs to if it falls on an object. 0 in this map corresponds to the background. To call bwlabel, you can do:
L = bwlabel(img);
img would be the binary image that you supply to the function and L is the integer map I just talked about. Additionally, you can provide 2 outputs to bwlabel, and the second parameter tells you how many objects exist in the image. As such:
[L, num] = bwlabel(img);
With our above example, num would be 4. As another method of invocation, you can specify the connected pixel neighbourhoods you would examine, and so you can do this:
[L, num] = bwlabel(img, N);
N would be the pixel neighbourhood you want to examine (i.e. 4 or 8).
regionprops
regionprops is a very useful function that I use daily. regionprops measures a variety of image quantities and features in a black and white image. Specifically, given a black and white image it automatically determines the properties of each contiguous white region that is 8-connected. One of these particular properties is the centroid. This is also the centre of mass. You can think of this as the "middle" of the object. This would be the (x,y) locations of where the middle of each object is located. As such, the Centroid for regionprops works such that for each object that is seen in your image, this would calculate the centre of mass for the object and the output of regionprops would return a structure where each element of this structure would tell you what the centroid is for each of the objects in your black and white image. Centroid is just one of the properties. There are other useful features as well, but I'm assuming you don't want to do this. To call regionprops, you would do this:
s = regionprops(img, 'Centroid');
The above code will calculate the centroids of each of your objects in the image. You can specify additional flags to regionprops to specify each feature that you want. I do highly encourage that you take a look at all of the possible features that regionprops can calculate, as there are many that are useful in a variety of different applications and situations.
Also, by omitting any flags as input into the function, you would calculate all of the features in your image by default. Therefore, if we were to declare the image that we have seen above in MATLAB, this is what would happen after I run regionprops. After, let's calculate what the centroids are:
img = logical(...
[0 0 0 0 0 1 1 1 0 0;
0 1 0 1 0 0 1 1 0 0;
0 1 1 1 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 1;
0 0 0 0 0 0 0 0 1 1;
0 0 1 1 1 1 0 0 1 1]);
s = regionprops(img, 'Centroid');
... and finally when we display the centroids:
>> disp(cat(1,s.Centroid))
3.0000 2.6000
4.5000 6.0000
7.2000 1.4000
9.6000 5.2000
As such, the first centroid is located at (x,y) = (3, 2.6), the next centroid is located at (x,y) = (4.5, 6) and so on. Take special note that the x co-ordinate is the column while the y co-ordinate is the row.
Hope this is clear!
I have got a 2D matrix. There is some region in the matrix where the elements are non-zero, in particular everywhere around the edge they are zero.
I plot the matrix using image as a colorplot and would like to add the curve that shows the boundary between non-zero values to zero values in the matrix. Is there any neat way to do this without loops?
This looks like a job for convhull :
To illustrate this code i'll take a dummy example :
A=zeros(10);
B=binornd(1,0.5,8,8);
A(2:end-1,2:end-1)=B
A =
0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 1 1 1 1 1 1 0 0
0 0 1 1 0 0 0 0 1 0
0 0 0 1 0 0 0 1 0 0
0 1 0 0 0 0 0 1 0 0
0 0 0 1 1 1 1 1 1 0
0 0 1 0 1 1 1 1 0 0
0 1 0 1 1 1 1 0 1 0
0 0 0 0 0 0 0 0 0 0
1/ Find the locations of all non zero entries :
[row,col]=find(A);
2/ Take the convex hull of these locations
k=convhull(row,col);
3/ Plot the convex hull (I plot the non zero points aswell but in your problem it will be your image points)
plot(row(k),col(k),'r-',row,col,'b*')
Result :
Another option is using the image processing toolbox and the bwperim function. This will work if you know that your area is completely closed (i.e. has no holes in the boundary)
This is an example using a black and white image, and you have 2 options: fill the inner gaps before, or not. You can see in the result the differences.
A = imread('circles.png');
Afill=imfill(A,'holes'); % optional
Abound1=bwperim(Afill);
Abound2=bwperim(A);
imshow([A,Abound, Abound2])
You can plot one on top of the other with:
[x,y]= find(Abound2);
hold on
image(A*255) %// If A is logical, else use just A (not *255)
colormap('gray')
plot(y,x,'r.')
hold off
axis tight
If you have a gray-scale image (or a matrix with a single value in each position (2D matrix), then you can binarize it first by either:
If you know everything outside your object is EXACTLY zero
A=yourA>0;
If you want to separate your object from the background, and the background is not exactly zero by A=im2bw(yourA,level), by choosing your own level, or letting Otsu do it for you with level=graythresh(yourA)
I am very interested in fingerprint verification and studying minutia extraction at present. I have found the following code online and wonder if someone would be kind enough to explain it? I have looked up centroid, regionprops etc, I understand these a little but the code below has me puzzled!
fun=#minutie;
L = nlfilter(K,[3 3],fun);
%% Termination
LTerm=(L==1);
imshow(LTerm)
LTermLab=bwlabel(LTerm);
propTerm=regionprops(LTermLab,'Centroid');
CentroidTerm=round(cat(1,propTerm(:).Centroid));
imshow(~K)
set(gcf,'position',[1 1 600 600]);
hold on
plot(CentroidTerm(:,1),CentroidTerm(:,2),'ro')
%% Bifurcation
LBif=(L==3);
LBifLab=bwlabel(LBif);
propBif=regionprops(LBifLab,'Centroid','Image');
CentroidBif=round(cat(1,propBif(:).Centroid));
plot(CentroidBif(:,1),CentroidBif(:,2),'go')
The code first filters the binary image with a neighborhood of 3x3 pixels. nfilter is a moving filter function. It will go through all the pixels in the image given as argument and apply an operation based on the values of the neighboring pixels.
I don't know the exact content of the minutie filter, but judging by the rest of the code, it probably counts the pixels with a value of 1 in the neighborhood of all 1s. In other words it will be equal to one at the end of a segment, and equal to 3 when there are 3 branches (a bifurcation).
Example:
Let a filter sum up the ones in the neighborhood, like this:
sum(block(1,1:3), block(3,1:3), block(2,1), block(2,3))*block(2, 2);
where block denotes a neighborhood around each pixel of the binary image.
In the left matrix below (if you ignore the boundary exceptions) there is one position with a one that has exactly one 1 in its 3x3 neighborhood, in the right matrix, there is one position with a one that has exactly three 1s in its 3x3 neighborhood.
[0 0 0 0 0 [0 0 1 0 0
0 0 0 0 0 0 0 1 0 0
0 0 1 0 0 1 1 1 0 0
0 0 1 0 0 0 0 1 0 0
0 0 1 0 0] 0 0 1 0 0]
The filtered output would be:
[0 0 0 0 0 [0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 3 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0] 0 0 0 0 0]
It found a termination in the left matrix, and a bifurcation in the right matrix.
The filtered image are then thresholded at the value 1 and 3, then the use of bwlabel and regionprops is somewhat mysterious to me† since bifurcations and terminations are single points, their position is simply their index. I think you could simply achieve the detection of the coordinates of the terminations and bifurcation using something like:
[It Jt]= find(L==1);
[Ib Jb]= find(L==3);
† one reason I can think of is that coordinates in images and arrays are different in matlab, and these two function output coordinates in the image format, which is easier to plot on top of the original image.
Consider the following operation:
a =
0 0 0 0
0 1 0 0
0 0 1 0
0 0 0 0
b=imdilate(a,[1,1;1,1])
b =
0 0 0 0
0 1 1 0
0 1 1 1
0 0 1 1
The above result indicates the origin of the structual element is [2,2]. If the origin is floor((size([1,1;1,1])+1)/2)=[1,1] by the definition (http://www-rohan.sdsu.edu/doc/matlab/toolbox/images/morph4.html)
b=
1 1 0 0
1 1 1 0
0 1 1 0
0 0 0 0
What's wrong with my understanding? Thank you!
You are correct that the origin is [1,1], but the origin is in the structuring matrix (2nd input), not the image matrix a (1st input).
Think about placing element [1,1] of your 2x2 structuring matrix on the non-zero values of a and you will see how the imdilate result is correct.
The Mathworks online help has a really nice explanation of dilation.
OK,Ｉｔｈｉｎｋ Ｉ ｈａｖｅ ｆｏｕｎｄ ｔｈｅ ａｎｓｗｅｒ．
Similar to convolution, in dilation the structuring element is reflected through its center (rotated 180 degrees) before the max operation is applied.