I am trying various visualizations for an Igraph in R (version.3.3.1).
Currently my visualizing is as shown as below, 2 nodes (blue and green) in circular layout.
Circular Layout
visNetwork(data$nodes,data$edges) %>% visIgraphLayout(layout="layout_in_circle")
Now I want to have a semicircle structure instead of a full circle as in the pic. All blue nodes form a semicircle, green nodes another semicircle. Each semicircle separated by a small distance as well. How can i achieve this. I found grid package has an option for semicircle, but i couldnt make it work with igraph. Please provide some pointers.
The layout argument accepts an arbitrary matrix with two columns and N rows if your graph has N vertices; all you need to do is to create a list of coordinates that correspond to a semicircle. You can make use of the fact that a vertex at angle alpha around a circle with radius r centered at (0, 0) is to be found at (r * cos(alpha), r * sin(alpha)). Since you are using R, alpha should be specified in radians, spaced evenly between 0 and pi (which corresponds to 180 degrees).
Related
I'm looking for an algorithm (or pseudo code) that can calculate the maximum number of (smaller) circles with diameter "s" that can be squeezed into the circumference of another (larger) circle with radius "r" ...
Image: http://teasy.space/images/terracolony-squeezingcircles2.jpg
You can alternate between radius/diameter etc if you wish -- as these are the only 2 parameters (other than the center (large circle) coordinate) that i have, i.e. that are known ...
The outer circles may not overlap but can fit "snug" together ...
After various upgrades to my routine through the years, I'm currently using an algorithm that is not perfect (and it needs to be accurate or the galaxy breaks down lol)
which does a broad interpolation between small outside circle diameter and large inside circle circumference, to somewhat accurately plot the circle count in a polygon style fitting pattern, which causes problems (i.e. overlaps) when using larger outside circles ...
; try to fit a random number of circles
num_Circles = Rand( min,max )
; check if the number of circles exceed the maximum that can fit
If num_Circles * SmallCircle_Diameter > LargeCircle_Circumference
; adjust the amount accordingly
num_Circles = LargeCircle_Circumference / SmallCircle_Diameter
End If
Another assumption is that the size of the smaller outer circles never exceeds that of the larger inner circle ...
something less to worry about ;)
I'm using this algorithm for one of my projects called Terra Colony, based on Gravity Well, a 2D space/gravity realtime colonization simulation game with moons, planets, stars, black/white holes, etc
Image: http://teasy.space/images/terracolony-squeezingcircles1.jpg
This is an issue that has plagued this project for over a decade!
Hopefully you can point me in the right direction :D
I have previously done many experiments and wrote different programs to find a solution, and have traveled the internet looking for formulas and solutions which in the end are very close, but not close enough! :P
Thank you! <3
Teasy
P.S. I tried to add the tag "circumference" but it apparently requires "1500 reputation" (points i guess, maybe to prevent spam)
There is formula that establishes relation between radius of big circle R, radius of small circle r and number of (touching) small circles N
R = r / Sin(Pi/N)
So maximum number of small circles might be found as
Sin(Pi/N) = r / R
Pi / N = arcsin(r / R)
and finally
N = Pi / arcsin(r / R)
Example:
R=5
r=2.5
so
N = Pi / arcsin(1/2) =
Pi / (Pi/6) =
6
Given the diam. of the small circle 'd' and the number of them 'c'
then the dia. of the large circle 'D' is
D=d/sin(180/c)
I need to find 4 points in Latitude/Longitude format surrounding a given center point and a resulting algorithm (if possible).
Known information:
Equal distances for each "bin" from center of point (Radar) outward.
Example = .54 nautical miles.
1 Degree beam width.
Center point of the "bin"
This image is in Polar coordinates (I think this is similar to Radial coordinates???):
I need to convert from Polar/Radial to Cartesian and I should be able to do that with this formula.
x = r × cos( θ )
y = r × sin( θ )
So now all I need to do is find the "bin" outline coordinates (4 corners) so I can draw a polygon in a Cartesian coordinate space.
I'm using Delphi/Pascal for coding, but I might be able to convert other languages if you have a sample algorithm.
Thanks for any suggestions or sample algorithms.
Regards,
Bryan
You need to convert everything to the same coordinate system and then impose the distance criteria as follows:
Convert your center point from geographic coordinates to polar coordinates to yield (rC, θC)
Convert your center point from polar to Cartesian coordinates using your equations yielding (xC, yC)
The corner points on the right side of the center points (xR, yR) satisfy the equation
(xR - xC)2 + (yR - yC)2 = D2
[rRcos(θC+0.5o) - xC]2 + [rRsin(θC+0.5o) - yC]2 = D2
where D=distance between the center point and corner points
Everything is known in the above equation except rR. This should yield a quadratic equation with two solutions which you can easily solve. Those are your two corner points on the right side.
Repeat step 3 with angle θC-0.5o to get the corner points on the left side.
please , see following image, here you can see blue rectangle is custom shape bounds and custom shape is shoe , i want to find area of a portion written in image and i want that area in form of rectangle
do is there any path iterator concept ?
Note
custom shape i derived from image of the same size.
I would do it like this:
1.create table for all bounding box-rect perimeter lines
each value in it will represent the empty space length form border line to shape
something like this:
the values are found by simple image scanning until first non space color found
2.now bruteforce find the biggest rectangle area
x,y = top left corner
for xs = 1 to bounding box width
now scan the max valid height of rectangle from x to x + xs (x grows to the right)
// it should be the min y0[x..x+xs]
remember the biggest valid area/size combination
do this for all 4 combinations (star from the other corners)
I now Brute-force is slow but
you can divide perimeter lines not by pixels but with some step instead
also I am sure this can be optimized somehow
for example by derivation of perimeter find the extremes and check from them backwards
when the size will start shrinking then stop ...
of course take in mind that on complicated shapes this optimization will not work ...
EDIT - Thanks for all the answers everyone. I think I accidentally led you slightly wrong as the square in the picture below should be a rectangle (I see most of you are referencing squares which seems like it would make my life a lot easier). Also, the x/y lines could go in any direction, so the red dot won't always be at the top y boundary. I was originally going for a y = mx + b solution, but then I got stuck trying to figure out how I know whether to plug in the x or the y (one of them has to be known, obviously).
I have a very simple question (I think) that I'm currently struggling with for some reason. I'm trying to have a type of minimap in my game which shows symbols around the perimeter of the view, pointing towards objectives off-screen.
Anyway, I'm trying to find the value of the red point (while the black borders and everything in green is known):
It seems like simple trigonometry, but for some reason I can't wrap my head around it. I just need to find the "new" x value from the green point to the red point, then I can utilize basic math to get the red point, but how I go about finding that new x is puzzling me.
Thanks in advance!
scale = max(abs(x), abs(y))
x = x / scale
y = y / scale
This is the simple case, for a square from (-1, -1) to (1, 1). If you want a different sized square, multiply the coordinates by sidelen / 2.
If you want a rectangle instead of a square, use the following formula. (This is another solution to the arbitrarily-sized square version)
scale = max(abs(x) / (width / 2), abs(y) / (height / 2))
x = x / scale
y = y / scale
Let's call the length of one side of the square l. The slope of the line is -y/x. That means, if you move along the line and rise a distance y toward the top of the square, then you'll move a distance x to the left. But since the green point is at the center of the square, you can rise only l/2. You can express this as a ratio:
-y -l/2
——— = ———
x d
Where d is the distance you'll move to the left. Solving for d, we have
d = xl/2y
So if the green dot is at (0, 0), the red dot is at (-l/2, xl/2y).
All you need is the angle and the width of the square w.
If the green dot is at (0,0), then the angle is a = atan(y/x), the y-coordinate of the dot is w/2, and therefore the x-coordinate of the dot is tan(1/a) * (w/2). Note that tan(1/a) == pi/2 - tan(a), or in other words the angle you really want to plug into tan is the one outside the box.
Edit: yes, this can be done without trig, too. All you need is to interpolate the x-coordinate of the dot on the line. So you know the y-coordinate is w/2, then the x-coordinate is (w/2) * x/y. But, be careful which quadrant of the square you're working with. That formula is only valid for -y<x<y, otherwise you want to reverse x and y.
My dataset consists of three vectors (x,y and z). I plot these values as dots in a 3d-plot with plot3(x,y,z), which is fine. I also want to show a plane in the same plot. To get the data of this plot I use linear regression on x and y to get a new z.
This is how it looks:
(source: bildr.no)
I want the surf to be filled with only one color (say light blue or gray) and set the opacity, to make it see-through. How can I do this?
The easiest way to create a surface that has just 1 color and a given transparency value is to set the 'FaceColor' and 'FaceAlpha' properties of the surface object:
hSurface = surf(...your arguments to create the surface object...);
set(hSurface,'FaceColor',[1 0 0],'FaceAlpha',0.5);
This example sets the surface color to be red and the transparency to 0.5. You can also set the edge properties too (with 'EdgeColor' and 'EdgeAlpha').
It is not clear to me what you want to do. When you say one color for the surf, do you mean exactly one color, or do you mean you want shades of gray?
Here is some code that will do a variety of things, you can choose which lines to use:
x = rand(1,20);
y = rand(1,20);
z = rand(1,20);
[X,Y] = meshgrid(linspace(0,1,10),linspace(0,1,10));
Z = rand(10)*0.1;
clf
plot3(x,y,z,'.');
hold on
h = surf(X,Y,Z)
hold off
%% This will change the color
colormap(copper)
%% This will remove colordata
set(h, 'cdata',zeros(10))
%% This will make transparent
alpha(0.5)
Completing the answer from gnovice, an extra ingredient in set(hsurface...) may be required (Matlab R2010b 64):
hSurface = surf(...your arguments to create the surface object...);
set(hSurface, 'FaceColor',[1 0 0], 'FaceAlpha',0.5, 'EdgeAlpha', 0);
to make invisible the point-to-point edges of the plotted surface
#matlabDoug has what you need, I think. The property cdata holds color data that gets a color map applied to it. Setting it to an array the same size as your surface data, with each element in that array having the same value, will make your surface one color. With the default color map, setting everything in cdata to zero will make your surface blue, and setting everything to 1 will make the surface red. Then you can play with the alpha to make it transparent.