ttOM123
-
Posts
15 -
Joined
-
Last visited
Posts posted by ttOM123
-
-
Your algorithm is backwards. When looping though the y (rows) and x (columns) you should be calculating the destination pixel for that x,y address.
You might want to review: http://boltbait.com/pdn/CodeLab/help/overview.asp
Yeah I found that out, code should be this which produces the image I posted above, the transformation maps from the selected quadrilateral to the region bounded by the canvas. So this function is nearly there, just need to change to the correct aspect ratio. To run it, copy the picture in first post into paint.net, and copy into Codelab script.
Another question, is it possible to enlarge or shrink the canvas size from this script? Or even better, produce the output onto a new image where the script defines the canvas size?
Still cannot get the "if" statement to work so to only grab pixels inside the canvas.
//if ((XX > rect.Left) && (XX < rect.Right)) { //if ((YY > rect.Top) && (YY < rect.Bottom)) { //dst[x,y] = src[XX,YY]; //} //}To me, it looks like the boundaries are correct, took me a while to work out the origin in the top left of the canvas, not bottom left, and that the y axis is positive downwards. You don't need to check with this particular transformation as I'm always grabbing from inside the canvas, but it would be nice to have.
#region UICode int Amount1=0; //[0,100]Slider 1 Description int Amount2=0; //[0,100]Slider 2 Description int Amount3=0; //[0,100]Slider 3 Description #endregion void Render(Surface dst, Surface src, Rectangle rect) { // Delete any of these lines you don't need Rectangle selection = EnvironmentParameters.GetSelection(src.Bounds).GetBoundsInt(); int CenterX = ((selection.Right - selection.Left) / 2)+selection.Left; int CenterY = ((selection.Bottom - selection.Top) / 2)+selection.Top; ColorBgra PrimaryColor = (ColorBgra)EnvironmentParameters.PrimaryColor; ColorBgra SecondaryColor = (ColorBgra)EnvironmentParameters.SecondaryColor; int BrushWidth = (int)EnvironmentParameters.BrushWidth; ColorBgra CurrentPixel; double a = 0.300191; double b = -0.190286; double c = 509.000000; double d = 0.013433; double e = 0.408391; double f = 172.000000; double g = 0.000026; double h = -0.000291; double X, Y; int XX, YY; for (int y = rect.Top; y < rect.Bottom; y++) { if (IsCancelRequested) return; for (int x = rect.Left; x < rect.Right; x++) { X = (a*x + b*y + c) / (g*x + h*y + 1); Y = (d*x + e*y + f) / (g*x + h*y + 1); XX = Convert.ToInt32(X); YY = Convert.ToInt32(Y); dst[x,y] = src[XX,YY]; //if ((XX > rect.Left) && (XX < rect.Right)) { //if ((YY > rect.Top) && (YY < rect.Bottom)) { //dst[x,y] = src[XX,YY]; //} //} } } } -
I didn't use diagrams; I just solved the equations. First, I knew that it would be a homogeneous matrix, since that's a well-known fact, so I just had to find the matrix.
I assumed a solution of the form:
/ Mx * x1 - x0 My * x3 - x0 x0 \ | | | Mx * y1 - y0 My * y3 - y0 y0 | | | \ Mx - 1 My - 1 1 /
That may not be an obvious choice at first glance, but I chose it because you can quickly verify that it gives the correct results for (0,0)->(x0,y0), (1,0)->(x1,y1), and (0,1)->(x3,y3). So I just need to find Mx, My so that (1,1)->(x2, y2).
I need:
(Mx * x1 + My * x3 - x0) / (Mx + My - 1) = x2
(Mx * y1 + My * y3 - y0) / (Mx + My - 1) = y2
Which is:
Mx * (x1 - x2) + My * (x3 - x2) = x0 - x2
Mx * (y1 - y2) + My * (y3 - y2) = y0 - y2
Those are just simultaneous equations that can be solved by Cramer's Rule.
EDIT: Note I didn't solve the same problem (sort of) solved in the PDF file. That was a quadrilateral to another quadrilateral. I just solved a unit square to a quadrilateral. The quadrilateral to quadrilateral could be solved by finding the matrix that maps a quadrilateral to a unit square; then the solution is simply to map the first quadrilateral to a unit square, then map the unit square to the second quadrilateral. That second mapping can be found by inverting the matrix for the first method. Maybe I'll add a comment later explaining how that can be done fairly easily.
Thanks for your explanation, I haven't got time to go though it in depth at the moment but I appreciate it.
-
-
YES got it working!! Sorry for all the spamming posts. All my if checks are not working for some reason, not sure why?
Here's the first result, I copied the photo into paint.net and run the script:
// Name: // Submenu: // Author: // Title: // Desc: // Keywords: // URL: // Help: #region UICode int Amount1=0; //[0,100]Slider 1 Description int Amount2=0; //[0,100]Slider 2 Description int Amount3=0; //[0,100]Slider 3 Description #endregion void Render(Surface dst, Surface src, Rectangle rect) { // Delete any of these lines you don't need Rectangle selection = EnvironmentParameters.GetSelection(src.Bounds).GetBoundsInt(); int CenterX = ((selection.Right - selection.Left) / 2)+selection.Left; int CenterY = ((selection.Bottom - selection.Top) / 2)+selection.Top; ColorBgra PrimaryColor = (ColorBgra)EnvironmentParameters.PrimaryColor; ColorBgra SecondaryColor = (ColorBgra)EnvironmentParameters.SecondaryColor; int BrushWidth = (int)EnvironmentParameters.BrushWidth; ColorBgra CurrentPixel; double a = 0.375007; double b = 0.214097; double c = 471.000000; double d = 0.014862; double e = -0.531855; double f = 640.000000; double g = 0.000022; double h = 0.000337; double X, Y; int XX, YY; for (int y = rect.Top; y < rect.Bottom; y++) { if (IsCancelRequested) return; for (int x = rect.Left; x < rect.Right; x++) { CurrentPixel = src[x,y]; X = (a*x + b*y + c) / (g*x + h*y + 1); Y = (d*x + e*y + f) / (g*x + h*y + 1); XX = Convert.ToInt32(X); YY = Convert.ToInt32(Y); dst[XX,YY] = CurrentPixel; //if (XX > rect.Right) { //if (XX < rect.Left) { //if (YY > rect.Top) { //if (YY <rect.Bottom) { //dst[XX,YY] = CurrentPixel; //} //} //} //} } } } -
My math was wrong; these numbers for a to h should now work, but still script isn't doing anything:
// Name: // Submenu: // Author: // Title: // Desc: // Keywords: // URL: // Help: #region UICode int Amount1=0; //[0,100]Slider 1 Description int Amount2=0; //[0,100]Slider 2 Description int Amount3=0; //[0,100]Slider 3 Description #endregion void Render(Surface dst, Surface src, Rectangle rect) { // Delete any of these lines you don't need Rectangle selection = EnvironmentParameters.GetSelection(src.Bounds).GetBoundsInt(); int CenterX = ((selection.Right - selection.Left) / 2)+selection.Left; int CenterY = ((selection.Bottom - selection.Top) / 2)+selection.Top; ColorBgra PrimaryColor = (ColorBgra)EnvironmentParameters.PrimaryColor; ColorBgra SecondaryColor = (ColorBgra)EnvironmentParameters.SecondaryColor; int BrushWidth = (int)EnvironmentParameters.BrushWidth; ColorBgra CurrentPixel; double a = 0.375007; double b = 0.214097; double c = 471.000000; double d = 0.014862; double e = -0.531855; double f = 640.000000; double g = 0.000022; double h = 0.000337; double X, Y; int XX, YY; for (int y = rect.Top; y < rect.Bottom; y++) { if (IsCancelRequested) return; for (int x = rect.Left; x < rect.Right; x++) { CurrentPixel = src[x,y]; X = (a*x + b*y + c) / (g*x + h*y + 1); Y = (d*x + e*y + f) / (g*x + h*y + 1); XX = Convert.ToInt32(X); YY = Convert.ToInt32(Y); if (XX < rect.Right) { if (XX > rect.Left) { if (YY < rect.Top) { if (YY > rect.Bottom) { dst[XX,YY] = CurrentPixel; } } } } } } } -
Installed Codelab, made a rough script but nothing changes? Not sure why? Can you print out values to check where the pixels get mapped to? I'm trying to fix the photo to the booklet in the first post. EDIT: yes looks like its mapping to out of bounds regions
// Name: // Submenu: // Author: // Title: // Desc: // Keywords: // URL: // Help: #region UICode int Amount1=0; //[0,100]Slider 1 Description int Amount2=0; //[0,100]Slider 2 Description int Amount3=0; //[0,100]Slider 3 Description #endregion void Render(Surface dst, Surface src, Rectangle rect) { // Delete any of these lines you don't need Rectangle selection = EnvironmentParameters.GetSelection(src.Bounds).GetBoundsInt(); int CenterX = ((selection.Right - selection.Left) / 2)+selection.Left; int CenterY = ((selection.Bottom - selection.Top) / 2)+selection.Top; ColorBgra PrimaryColor = (ColorBgra)EnvironmentParameters.PrimaryColor; ColorBgra SecondaryColor = (ColorBgra)EnvironmentParameters.SecondaryColor; int BrushWidth = (int)EnvironmentParameters.BrushWidth; ColorBgra CurrentPixel; double a = 3.7; double b = 0.3; double c = -1912.5; double d = 0; double e = -1.8; double f = 1149.9; double g = 0; double h = 0; double X, Y; int XX, YY; for (int y = rect.Top; y < rect.Bottom; y++) { if (IsCancelRequested) return; for (int x = rect.Left; x < rect.Right; x++) { CurrentPixel = src[x,y]; X = (a*x + b*y + c) / (g*x + h*y + 1); Y = (d*x + e*y + f) / (g*x + h*y + 1); XX = Convert.ToInt32(X); YY = Convert.ToInt32(Y); if (XX < rect.Right) { if (XX > rect.Left) { if (YY < rect.Top) { if (YY > rect.Bottom) { dst[XX,YY] = CurrentPixel; } } } } } } } -
I still can't see how they get the transformation matrix...
This is how they do the maths: http://www.visualopticslab.com/OPTI600C/Lecture/Lectures6_7.pdf
-
The perspective matrix that maps {(0,0), (1,0), (1,1), (0,1)} to {(x0,y0), (x1,y1), (x2,y2), (x3,y3)} is:
/ x' \ / T230 * x1 - T123 * x0 T012 * x3 - T123 * x0 T123 * x0 \ / x \ | y' | = | T230 * y1 - T123 * y0 T012 * y3 - T123 * y0 T123 * y0 | | y | \ w' / \ T230 - T123 T012 - T123 T123 / \ 1 /
I'm curious as how you got this transformation matrix. Do you have any diagrams or something?
Looks like OpenCV uses this method:
http://answers.opencv.org/question/5018/getperspectivetransform-mathematical-explanation/
This explantion here doesn't look too bad: http://stackoverflow.com/a/3190725/2750236
so my take on it: we know for n=4 (x1,y1), (x2,y2), (x3,y3), (x4,y4) which are our rectangle edges we are transforming to and (X1, Y1) ... (X4, Y4) which are the defined Quadrilateral edges.
Solve for the vector a ... h
then transforming from a rectangle to a Quadrilateral is
in fact this method looks similar to your method haha. With this method, I still can't see how they get the transformation matrix.
Interesting tutorial: http://opencv-code.com/tutorials/automatic-perspective-correction-for-quadrilateral-objects/
-
Are you saying you would want to manually type in values for the 4 points? If so, I think that would be possible. Otherwise, please elaborate.
Yes, I was thinking noting the 4 edges of the Quadrilateral then enter them somehow.
-
Yeah time, the most limiting resource. I would also look into it if I wasn't also busy.
Searching quadrilateral to rectangle transformation brings up some interesting conversations: https://math.stackexchange.com/questions/13404/mapping-irregular-quadrilateral-to-a-rectangle
Is it possible to feed the Quadrilateral Reshape plugin a set of points, or do you need the source code? It seems to do the transform and interpolation already.
-
... I have never looked into image processing but if one could list the coordinates of a quadrilateral shape, then apply some translation of the pixels to some normalized rectangle, it might work. Probably not that simple I guess.
Something like Quadrilateral crop (I haven't thought out all cases):
1) Define edges of Quadrilateral
2) Find the 2 diagonal points of Quadrilateral with the smallest distance. Call them p1 = (x1, y1), p2 = (x2, y2). The other point closest horizontally to p1 is called p3 = (x3, y3), remaining point is p4 = (x4, y4)
3) these points p1 p2 remain the same after some translation. These new points once translated become P1 = (x1, y1), P2 = (x2, y2)
4) The other diagonal pair p3 = (x3, y3), p4 = (x4, y4) get mapped to P3 = (x1, y2), P4 = (x2, y1).
5) Apply crop to rectangle P1 P2 P3 P4
-
Well there is no easy way to do the job. I usually use the select tool and then press delete. Unfortunaly, you will be left with a skewed shape.
....
Yeah I got an similar effect by using Quadrilateral Reshape to get rectangular then cropping. I've been looking for the last few hours on how to do this nicely in Windows in any program but haven't found that much.
I have never looked into image processing but if one could list the coordinates of a quadrilateral shape, then apply some translation of the pixels to some normalized rectangle, it might work. Probably not that simple I guess.
-
You can use the Quadrilateral Reshape plugin to adjust the perspective. Then, just select with rectangular marquee and hit ctrl+shift+x to crop to the selection.
Tried Quadrilateral Reshape to make rectangular then a rectangular crop, works ok I guess but is quite fiddly. I was hoping for a "Quadrilateral" select tool, which then from edges makes rectangular and crops.
-
Hi there,
I was wondering if there was a way in paint.net to crop a photo of a document to the documents edges, like the many mobile apps for example:
https://play.google.com/store/apps/details?id=com.liuxqsmile.imagecorrect
This is actually a pretty hard thing to search for, no idea of what terms to search for.
How to crop to document edges in photo?
in Paint.NET Discussion and Questions
Posted
Haha I can provide a dll which only does one photo
.
I've been flat out. Haven't touched it for a week apart from MJW's advice which works nicely. The next hurdle is some method of entering the 4 coordinates of the quadrangle, which I haven't looked into yet. I was thinking of somehow clicking on the canvas 4 times so one can zoom in and be precise then run the plugin, so no menus hopefully.