Perspective correction of UIImage from Points - iphone

I'm working on a app where I'll let the user take a picture e.g of a business card or photograph.
The user will then mark the four corners of the object (which they took a picture off) - Like it is seen in a lot of document/image/business card scanning apps:
My question is how do i crop and fix the perspective according to these four points? I've been searching for days and looked at several image proccessing libraries without any luck.
Any one who can point me in the right direction?

From iOS8+ there is Filter for Core Image called CIPerspectiveCorrection. All you need to do is pass the image and four points.
Also there is one more filter supporting iOS6+ called CIPerspectiveTransform which can be used in similar way (skewing image).

If this image were loaded in as a texture, it'd be extremely simple to skew it using OpenGL. You'd literally just draw a full-screen quad and use the yellow correction points as the UV coordinate at each point.

I'm not sure if you've tried the Opencv library yet, but it has a very nice way to deskew an image. I've got here a small snippet that takes an array of corners, your four corners for example, and a final size to map it into.
You can read the man page for warpPerspective on the OpenCV site.
cv::Mat deskew(cv::Mat& capturedFrame, cv::Point2f source_points[], cv::Size finalSize)
{
cv::Point2f dest_points[4];
// Output of deskew operation has same color space as source frame, but
// is proportional to the area the document occupied; this is to reduce
// blur effects from a scaling component.
cv::Mat deskewedMat = cv::Mat(finalSize, capturedFrame.type());
cv::Size s = capturedFrame.size();
// Deskew to full output image corners
dest_points[0] = cv::Point2f(0,s.height); // lower left
dest_points[1] = cv::Point2f(0,0); // upper left
dest_points[2] = cv::Point2f(s.width,0); // upper right
dest_points[3] = cv::Point2f(s.width,s.height); // lower right
// Build quandrangle "de-skew" transform matrix values
cv::Mat transform = cv::getPerspectiveTransform( source_points, dest_points );
// Apply the deskew transform
cv::warpPerspective( capturedFrame, deskewedMat, transform, s, cv::INTER_CUBIC );
return deskewedMat;
}

I don't know exact solution of your case, but there is approach for trapezoid: http://www.comp.nus.edu.sg/~tants/tsm/TSM_recipe.html - the idea is to continuously build transformation matrix. Theoretically you can add transformation that converts your shape into trapecy.
And there are many questions like this: https://math.stackexchange.com/questions/13404/mapping-irregular-quadrilateral-to-a-rectangle , but I didn't check solutions.

Related

Subtract background in image with Matlab

I have a list of images that i expect to have uneven illumination across the x axis only. The background estimation/model will be calculated in advance but only its shape will be given (so i will not have an empty background image that i can just subtract), I will just have the form/shape of it (for example linear).
Is there a way to subtract the background and get an even illumination of the image with just knowing its shape (and without having an actual background image)? I have attached an image from the Matlab library that was created using a linear background (background=-3*x+0.5). Can someone show me how to go from this example to the original image with just using the background shape?
I am also including the original image.
If you have the shape, and the conditions you state are correct, then you have the full background information. It is, however, not possible to undo the clipping that happened (where the image has values 255, you don't know what the original value was).
This is how you would create a background image given the shape:
img = imread('https://i.stack.imgur.com/0j2Gi.jpg');
img = img(27:284,91:370); % OP posted a screen shot? What is this margin?
x = 0:size(img,2)-1;
background = 0.3*x+0.5;
background = repmat(background,size(img,1),1);
imshow(img-background)
The results looks not exactly like the original image, but most of that can be explained by the clipping.
I replaced your -3*x+0.5 with 0.3*x+0.5 because the -3 makes no sense, and 0.3 makes it so that the background values remain in a meaningful range.
If, on the other hand, you are asking about fitting the linear model to the image data and estimating the exact background to subtract, then the problem is a bit more difficult, but not impossible. You can, for example, make an assumption about intensity uniformity across the x-axis if the illumination had been uniform across the image. You can then take the mean intensity along the x-axis and fit your model to it:
meanintensity = mean(img,1)';
plot(m)
X = [ones(length(meanintensity),1),(0:length(meanintensity)-1)']; % meanintensity = X*params
params = X\meanintensity;
hold on, plot(X*params)
background = X*params;
background = repmat(background',size(img,1),1);
imshow(img-background)

How to correlate properly a moving sample in 2 images of different size?

I am currently recording on a single camera the images, one aside of the other one, of the same sample out of a microscope.
I have 2 issues with that, and I figured out that in post procesing with Matlab I could arrange these questions.
-First, the 2 images on the camera are supposed to have the same pixel size, or one is just a litle bigger than the other one, probably because of optical pathways. What is the adapted Matlab function or way to correlate the two images so they will have exactly the same pixel size in X and Y ?
Two images on same camera , one bigger or smaller compared to the other one
-Secondly, my sample is moving a litle during the recording ( while still staying in my field of view of course ). To make my analysis easier, it would be suitable that I could correct the images so the sample remain at the same place as in the first image, to perform calculations on it easier. What would be the adapted Matlab function or way to correct this movement in the image ?
Sample moving in the image on the camera
Sorry for the poor quality of my drawings !
Thank you very much for your advices and help.
First zero-pad the images to a sufficient degree, to get them both to double the size of the bigger one.
size_padding = max(size(fig1),size(fig2));
fig1_pad = padarray(fig1,size_padding-size(fig1),'post');
fig2_pad = padarray(fig2,size_padding-size(fig2),'post');
Assuming the sample is the only feature present in the images, the best way to proceed would be to use the xcorr2() function and find the lag corresponding to the maximum correlation, to get the space shift between the two images:
xc = xcorr2(fig1_pad,fig2_pad);
[max_cc, imax] = max(abs(xc(:)));
[ypeak, xpeak] = ind2sub(size(xc),imax(1));
corr_offset = [ (ypeak-size(fig2_pad,1)) (xpeak-size(fig2_pad,2)) ];
You then use circshift() to shift one of the images using the lag you obtained in the last step.
fig2_shift = circshift(fig2_pad,corr_offset);
You now have two images of the same size, where hopefully the sample is in the same position. If you want to remove the padding zeroes, crop the images to your liking with respect to the center using imcrop().

How to auto-crop a barrel-distorted image using ImageMagick?

Using ImageMagick's convert to barrel-distort a photo to correct a strongly visible pincushion distortion, I provide positive a, b or c values (from a database for my lens + focal length). This results in an image that is corrected, has the original width and height, but includes a non-rectangular, bent/distorted border, as the image is corrected towards its center. Simplified example:
convert rose: -virtual-pixel black -distort Barrel '+0.0 +0.1 +0.0' out.png
How can I automatically crop the black, bent border to the largest possible rectangle in the original aspect ratio within the rose?
The ImageMagick website says, that a parameter "d" is automatically calculated, that could do this (resulting in linear distortion effectively zooming into the image and pushing the bent border right outside the image bounds), but the imagemagick-calculated value seems to aim for something different (v6.6.9 on ubuntu 12.04). If I guess and manually specify a "d", I can get the intended result:
convert rose: -virtual-pixel black -distort Barrel '+0.0 +0.1 +0.0 +0.6' out.png
The given formular a+b+c+d=1 does not seem to be a proper d for my cropping case. Also, d seems to depend on the aspect ratio of the image and not only on a/b/c. How do I make ImageMagick crop the image, or, how to I calculate a proper d?
Update
I found Fred's ImageMagick script innercrop (http://www.fmwconcepts.com/imagemagick/innercrop/index.php) that does a bit what I need, but has drawbacks and is no solution for me. It asumes arbitrary outer areas, so it takes long to find the cropping rectangle. It does not work within Unix pipes, and it does not keep the original aspect ratio.
Update 2
Contemplating on the problem makes me think that calculating a "d" is not the solution, as changing d introduces more or less bending and seems to do more than just zoom. The d=1-(a+b+c) that is calculated by imagemagick results in the bent image touching the upper/lower bounds (for landscape images) or the left/right bounds (for portrait images). So I think the proper solution would be to calculate where one of the new 4 corners will be given a/b/c/d, and then crop to those new corners.
The way I understand the docs, you do not use commas to separate the parameters for the barrel-distort operator.
Here is an example image, alongside the output of the two commands you gave:
convert o.png -virtual-pixel black -distort Barrel '+0.0 +0.1 +0.0' out0.png
convert o.png -virtual-pixel black -distort Barrel '+0.0 +0.1 +0.0 +0.6' out1.png
I created the example image in order to better visualize what you possibly want to achieve.
However, I do not see the point you stated about the automatically calculated parameter 'd', and I do not see the effect you stated about using 'd=+0.6'...
I'm not sure I understand your wanted result correctly, so I'm assuming you want the area marked by the yellow rectangle cropped.
The image on the left is out0.png as created by the first command above.
In order to guess the required coordinates, we have to determine the image dimensions first:
identify out0.png
out0.png PNG 700x700 700x700+0+0 8-bit sRGB 36KB 0.000u 0:00.000
The image in the center is marked up with the white rectangle. The rectangle is there so you can look at it and tell me if that is the region you want cropped. The image on the right is the cropped image (without scaling it back to the original size).
Is this what you want? If yes, I can possibly update the answer in order to automatically determine the required coordinates of the cropping. (For now I've done it based on guessing.)
Update
I think you may have mis-understood the purpose of the barrel-distortion operation. It is meant for correcting a barrel (slight) distortion, as is produced by camera lenses. The 3 parameters a, b and c to be used for any specific combination of camera, lens and current zoom could possibly be stated in your photo's EXIF data. The formula were a+b+c+d = 1 is meant to be used when the new, distortion-corrected image should have the same dimensions as the original (distorted) image.
So to imitate the barrel-correction, we should probably use the second image from the last row above as our input:
convert out3.png -virtual-pixel gray -distort barrel '0 -0.2 0' corrected.png
Result:

remove some top, down rows and right, and left some columns of jpg image border using matlab

I have RGB museum JPG Images. most of them have image footnotes on one or more sides, and I'd like to remove them. I do that manually using paint software. now I applied the following matlab code to remove the image footnotes automatically. I get a good result for some images but for others it not remove any border. Please, can any one help me by update this code to apply it for all images?
'rgbIm = im2double(imread('A3.JPG'));
hsv=rgb2hsv(rgbIm);
m = hsv(:,:,2);
foreground = m > 0.06; % value of background
foreground = bwareaopen(foreground, 1000); % or whatever.
labeledImage = bwlabel(foreground);
measurements = regionprops(labeledImage, 'BoundingBox');
ww = measurements.BoundingBox;
croppedImage = imcrop(rgbImage, ww);'
In order to remove the boundaries you could use "imclearborder", where it checks for labelled components at boundaries and clears them. Caution! if the ROI touches the boundary, it may remove. To avoid such instance you can use "imerode" with desired "strel" -( a line or disc) before clearing the borders. The accuracy or generalizing the method to work for all images depends entirely on "threshold" which separates the foreground and background.
More generic method could be - try to extract the properties of footnotes. For instance, If they are just some texts, you can easily remove them by using a edge detection and morphology opening with line structuring element along the cols. (basic property for text detection)
Hope it helps.
I could give you a clear idea or method if you upload the image.

Extract Rectangular Image from Scanned Image

I have scanned copies of currency notes from which I need to extract only the rectangular notes.
Although the scanned copies have a very blank background, the note itself can be rotated or aligned correctly. I'm using matlab.
Example input:
Example output:
I have tried using thresholding and canny/sobel edge detection to no avail.
I also tried the solution given here but it detects the entire image for cropping and it would not work for rotated images.
PS: My primary objective is to determine the denomination of the currency. There are a couple of methods I thought I could use:
Color based, since all currency notes have varying primary colors.
The advantage of this method is that it's independent of the
rotation or scale of the input image.
Detect the small black triangle on the lower left corner of the note. This shape is unique
for each denomination.
Calculating the difference between 2 images. Since this is a small project, all input images will be of the same dpi and resolution and hence, once aligned, the difference between the input and the true images can give a rough estimate.
Which method do you think is the most viable?
It seems you are further advanced than you looked (seeing you comments) which is good! Im going to show you more or less the way you can go to solve you problem, however im not posting the whole code, just the important parts.
You have an image quite cropped and segmented. First you need to ensure that your image is without holes. So fill them!
Iinv=I==0; % you want 1 in money, 0 in not-money;
Ifill=imfill(Iinv,8,'holes'); % Fill holes
After that, you want to get only the boundary of the image:
Iedge=edge(Ifill);
And in the end you want to get the corners of that square:
C=corner(Iedge);
Now that you have 4 corners, you should be able to know the angle of this rotated "square". Once you get it do:
Irotate=imrotate(Icroped,angle);
Once here you may want to crop it again to end up just with the money! (aaah money always as an objective!)
Hope this helps!