Get Object distributed with a specific pixels distance in photoshop.! - object

There are 5 objects in a row which i want to distribute in a line
but i want 15 pixels distance between edges of objects in photoshop
what I am getting is Distribution Object by center, but not same distance between objects, How can i get it?
Here's what i want (case 2) & what i get (case 1).
Thanks in Advance!!!!

What you want to do is zoom in to the pictures to the point of where you can literally count each pixel. And start adjusting each picture with the arrow keys.
Or draw a square that measures 15 pixels wide and duplicate it. Once you are done with adjustment you can delete the square.

Related

How can I split image as pizza in cv2?

I wanna split an image in cv2 as shown in the picture and get the dominant color for each of the parts of the picture. Is there any function so I can give it parameters (a, b, c, d) for the count of lines to up, right, bottom and down of the picture and it returns color in rgb space?
Usually this is done not by the count of lines (up, down, left, right), but by the angle of one pizza slice. The whole pizza is 360 degrees, so you may choose a number like 10-20-30 degrees, to make slices even. Then take the x, y coordinates of the image pixel and after subtracting the center pixel coordinates calculate the slice number as atan((y - y_center)/(x - x_center)).
To avoid calculation complications, you may want to split the calculation into the two similar, but separate cases when x > x_center and x < x_center to avoid ZeroDivide and make life easier.
Regarding the dominant colors, there are a bunch of articles on this site, just use the search, I don't want to copy-paste someone else's work.

Is there a simple algorithm that can find the envelope of several circles?

Given a number of points on a 2d surface and radiuses for these points I can easily paint circles for them. What I need is an algorithm that only paints the envelope (right word for what I am looking for?) or outer bound of these combined circles. Additionally a second set of circles can 'encroach' on these circles, resulting in a kind of 'border'.
Image of what I am looking for
A quick way to draw the outline of the union of the disks is to
fill all disks in yellow, then
fill all disks in white with a smaller radius.
This can be adapted to the "encroached" circles, provided you only fill the remaining portions of the disks. Unfortunately, in a general setting finding the remaining portions can be an uneasy geometric problem.
There is an alternative approach which can work in all cases:
fill an image with zeroes, then for all disks fill every pixel with the value of the distance to the circumference (maximum at the center), but only keep the highest value so far.
while you do this, fill a second image with the color of the disk that achieved that highest value. (Initialize the image with the background color.)
At the end of this process, the first image will represent a "terrain" made of intersecting cones; and for every point of the terrain, you will know the color.
finally, color the pixels that have a height smaller than the desired stroke width, using the color map.
You can do the drawing in two steps.
1) Draw the outline using the following method: For each point, draw a circle using your favorite circle-drawing method, but before drawing a pixel, ensure that it is not contained inside any other circle. Do this for every point and you will get your outline.
2) Draw the borders between different sets using the following method: For each pair of points from different sets, calculate the two intersection points of the circles. If there is an intersection, the border can be drawn as a segment joining these two points. However, you have to make two lines, one for circle A, and another for circle B. To draw the line for circle A, slightly offset the segment towards point A. Then, use your favorite line-drawing method, but before drawing a pixel, ensure that it is closer to point A that any other point of the opposite set. After drawing the line, repeat the process for circle B. Note that both segment are not guaranteed to be the same length since the asymmetry of the points of the different sets. It will, however, always form a closed shape when all outlines and borders are drawn.

Find contour of 2D unorganized pointcloud

I have a set of 2D points, unorganized, and I want to find the "contour" of this set (not the convex hull). I can't use alpha shapes because I have a speed objective (less than 10ms on an average computer).
My first approach was to compute a grid and find the outline squares (squares which have an empty square as a neighbor). So I think I downsized efficiently my numbers of points (from 22000 to 3000 roughly). But I still need to refine this new set.
My question is : how do I find the real outlines points among my green points ?
After a weekend full of reflexions, I may have found a convenient solution.
So we need a grid, we need to fill it with our points, no difficulty here.
We have to decide which squares are considered as "Contour". Our criteria is : at least one empty neighbor and at least 3 non empty neighbors.
We lack connectivity information. So we choose a "Contour" square which as 2 "Contour" neighbors or less. We then pick one of the neighbor. From that, we can start the expansion. We just circle around the current square to find the next "Contour" square, knowing the previous "Contour" squares. Our contour criteria prevent us from a dead end.
We now have vectors of connected squares, and normally if our shape doesn't have a hole, only one vector of connected squares !
Now for each square, we need to find the best point for the contour. We select the one which is farther from the barycenter of our plane. It works for most of the shapes. Another technique is to compute the barycenter of the empty neighbors of the selected square and choose the nearest point.
The red points are the contour of the green one. The technique used is the plane barycenter one.
For a set of 28000 points, this techniques take 8 ms. CGAL's Alpha shapes would take an average 125 ms for 28000 points.
PS : I hope I made myself clear, English is not my mothertongue :s
You really should use the alpha shapes. Maybe use only green points as inputs of the alpha alpha algorithm.

Finding a point clicked in a grid

Given this grid ( http://i.stack.imgur.com/Nz39I.jpg is a trapezium/trapezoid, not a square), how do you find the point clicked by the user? I.e. When the user clicks a point in the grid, it should return the coordinates like A1 or D5.
I am trying to write pseudo code for this and I am stuck. Can anyone help me? Thanks!
EDIT: I am still stuck... Does anyone know of any way to find the height of the grid?
If it is a true perspective projection, you can run the click-point through the inverse projection to find it's X,Z coordinates in the 3D world. That grid has regular spacing and you can use simple math to get the A1,D5,etc.
If it's just something you drew, then you'll have to compare the Y coordinates to the positions of the horizontal lines to figure out which row. Then you'll need to check its position (left/right) relative to the angled lines to get the column - for that, you'll need either coordinates of the end-points, or equations for the lines.
Yet another option is to store an identical image where each "square" is flood-filled with a different color. You then check the color of the pixel where the user clicked but in this alternate image. This method assumes that it's a fixed image and is the least flexible.
If you have the coordinates of end points of the grid lines then
Try using the inside-outside test for each grid line and find the position
Since this grid is just a 3D view of a 2D grid plane, there is a projective transform that transforms the coordinates on the grid into coordinates on the 2D plane. To find this transform, it is sufficient to mark 4 different points on the plane (say, the edges), assign them coordinates on the 2D plane and solve the resulting linear equation system.

How to calculate mid point vertices?

I have a set of vertices to draw a circle, but I want to draw a high-res circle by drawing twice the number of vertices, I cant just increase the number of vertices what I need is to calculate the mid points from the supplied vertices, if that makes sense
So from that image how can I calculate the points A, B, C, ... given the points V0, V1, V2, ... and the center point of the circle ?
Please note that I cant just calculate the mid-points by rotating the vertices they need to be calculated using their position
Thanks
The center of the circle can be determined by making a perpendicular line to two neighboring "sides", and intersecting them.
If there are an even number of vertices, just pick two which are opposite to each other, and "avarage them" - calculate the midpoint.
Then, you can just rotate all the vertices to either way by 180°/No.vertices around this center, so you get the ones you are looking for. Of course, you should keep the existing ones too.

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