I need to move some objects lets say 50 in a space (i.e a grid of [-5,5]) and making sure that if the grid is divided into 100 portions most of the portions (90% or more) are once visited by any object
constraints :
object should move in random directions in the grid changing their velocities frequently (change speed and direction in each iteration)
I was thinking of bouncing balls ( BUT moving in random directions even if not hit by anything in space, not they way a real ball moves) , if we could leave them into space in different positions with different forces and each time they hit each other (or getting closer to a specific distance ) they move to different directions with different speed and could give us a result near to 90% hit of portions in the grid .
I also need to make sure objects are not getting out of grid ( could make lb and ub limits and get them back in case they try to leave the grid)
My code is different from the idea I have written above ...
ux = 1;
uy = 15;
g = 9.81;
t = 0; x(1) = 0;
y(1) = 0;
tf = 2.0 * uy / g; % time of flight back to the ground
dt = tf / 20; % time increment - taking 20 steps
while t < tf
t = t + dt;
if((uy - 0.5 * g * t) * t >= 0)
x(end + 1) = ux * t;
y(end + 1) = (uy - 0.5 * g * t) * t;
end
end
plot(x,y)
this code makes the ball to go with Newton's law which is not the case
Bottom line i just need to be able to visit many portions of grid in a short time so this is why i want the objects to moves in a chaotic way in the space in a random manner (each time running the code i need different result so it needs to be random path) and to get a better result i could make the objects bounce to different directions if they hit or visit each other in the same portions , this probably give me a better result .
Related
Check the following gif: https://i.gyazo.com/72998b8e2e3174193a6a2956de2ed008.gif
I want the cylinder to instantly change location to the nearest empty space on the plane as soon as I put a cube on the cylinder. The cubes and the cylinder have box colliders attached.
At the moment the cylinder just gets stuck when I put a cube on it, and I have to click in some direction to make it start "swimming" through the cubes.
Is there any easy solution or do I have to create some sort of grid with empty gameobjects that have a tag which tells me if there's an object on them or not?
This is a common problem in RTS-like video games, and I am solving it myself. This requires a breadth-first search algorithm, which means that you're checking the closest neighbors first. You're fortunate to only have to solve this problem in a gridded-environment.
Usually what programmers will do is create a queue and add each node (space) in the entire game to that queue until an empty space is found. It will start with e.g. the above, below, and adjacent spaces to the starting space, and then recursively move out, calling the same function inside of itself and using the queue to keep track of which spaces still need to be checked. It will also need to have a way to know whether a space has already been checked and avoid those spaces.
Another solution I'm conceiving of would be to generate a (conceptual) Archimedean spiral from the starting point and somehow check each space along that spiral. The tricky part would be generating the right spiral and checking it at just the right points in order to hit each space once.
Here's my quick-and-dirty solution for the Archimedean spiral approach in c++:
float x, z, max = 150.0f;
vector<pair<float, float>> spiral;
//Generate the spiral vector (run this code once and store the spiral).
for (float n = 0.0f; n < max; n += (max + 1.0f - n) * 0.0001f)
{
x = cos(n) * n * 0.05f;
z = sin(n) * n * 0.05f;
//Change 1.0f to 0.5f for half-sized spaces.
//fmod is float modulus (remainder).
x = x - fmod(x, 1.0f);
z = z - fmod(z, 1.0f);
pair<float, float> currentPoint = make_pair(x, z);
//Make sure this pair isn't at (0.0f, 0.0f) and that it's not already in the spiral.
if ((x != 0.0f || z != 0.0f) && find(spiral.begin(), spiral.end(), currentPoint) == spiral.end())
{
spiral.push_back(currentPoint);
}
}
//Loop through the results (run this code per usage of the spiral).
for (unsigned int n = 0U; n < spiral.size(); ++n)
{
//Draw or test the spiral.
}
It generates a vector of unique points (float pairs) that can be iterated through in order, which will allow you to draw or test every space around the starting space in a nice, outward (breadth-first), gridded spiral. With 1.0f-sized spaces, it generates a circle of 174 test points, and with 0.5f-sized spaces, it generates a circle of 676 test points. You only have to generate this spiral once and then store it for usage numerous times throughout the rest of the program.
Note:
This spiral samples differently as it grows further and further out from the center (in the for loop: n += (max + 1.0f - n) * 0.0001f).
If you use the wrong numbers, you could very easily break this code or cause an infinite loop! Use at your own risk.
Though more memory intensive, it is probably much more time-efficient than the traditional queue-based solutions due to iterating through each space exactly once.
It is not a 100% accurate solution to the problem, however, because it is a gridded spiral; in some cases it may favor the diagonal over the lateral. This is probably negligible in most cases though.
I used this solution for a game I'm working on. More on that here. Here are some pictures (the orange lines in the first are drawn by me in Paint for illustration, and the second picture is just to demonstrate what the spiral looks like if expanded):
I have a voxel based game in development right now and I generate my world by using Simplex Noise so far. Now I want to generate some other structures like rivers, cities and other stuff, which can't be easily generated because I split my world (which is practically infinite) into chunks of 64x128x64. I already generated trees (the leaves can grow into neighbouring chunks), by generating the trees for a chunk, plus the trees for the 8 chunks surrounding it, so leaves wouldn't be missing. But if I go into higher dimensions that can get difficult, when I have to calculate one chunk, considering chunks in an radius of 16 other chunks.
Is there a way to do this a better way?
Depending on the desired complexity of the generated structure, you may find it useful to first generate it in a separate array, perhaps even a map (a location-to-contents dictionary, useful in case of high sparseness), and then transfer the structure to the world?
As for natural land features, you may want to google how fractals are used in landscape generation.
I know this thread is old and I suck at explaining, but I'll share my approach.
So for example 5x5x5 trees. What you want is for your noise function to return the same value for an area of 5x5 blocks, so that even outside of the chunk, you can still check if you should generate a tree or not.
// Here the returned value is different for every block
float value = simplexNoise(x * frequency, z * frequency) * amplitude;
// Here it will return the same value for an area of blocks (you should use floorDiv instead of dividing, or you it will get negative coordinates wrong (-3 / 5 should be -1, not 0 like in normal division))
float value = simplexNoise(Math.floorDiv(x, 5) * frequency, Math.floorDiv(z, 5) * frequency) * amplitude;
And now we'll plant a tree. For this we need to check what x y z position this current block is relative to the tree's starting position, so we can know what part of the tree this block is.
if(value > 0.8) { // A certain threshold (checking if tree should be generated at this area)
int startX = Math.floorDiv(x, 5) * 5; // flooring the x value to every 5 units to get the start position
int startZ = Math.floorDiv(z, 5) * 5; // flooring the z value to every 5 units to get the start position
// Getting the starting height of the trunk (middle of the tree , that's why I'm adding 2 to the starting x and starting z), which is 1 block over the grass surface
int startY = height(startX + 2, startZ + 2) + 1;
int relx = x - startX; // block pos relative to starting position
int relz = z - startZ;
for(int j = startY; j < startY + 5; j++) {
int rely = j - startY;
byte tile = tree[relx][rely][relz]; // Get the needing block at this part of the tree
tiles[i][j][k] = tile;
}
}
The tree 3d array here is almost like a "prefab" of the tree, which you can use to know what block to set at the position relative to the starting point. (God I don't know how to explain this, and having english as my fifth language doesn't help me either ;-; feel free to improve my answer or create a new one). I've implemented this in my engine, and it's totally working. The structures can be as big as you want, with no chunk pre loading needed. The one problem with this method is that the trees or structures will we spawned almost within a grid, but this can easily be solved with multiple octaves with different offsets.
So recap
for (int i = 0; i < 64; i++) {
for (int k = 0; k < 64; k++) {
int x = chunkPosToWorldPosX(i); // Get world position
int z = chunkPosToWorldPosZ(k);
// Here the returned value is different for every block
// float value = simplexNoise(x * frequency, z * frequency) * amplitude;
// Here it will return the same value for an area of blocks (you should use floorDiv instead of dividing, or you it will get negative coordinates wrong (-3 / 5 should be -1, not 0 like in normal division))
float value = simplexNoise(Math.floorDiv(x, 5) * frequency, Math.floorDiv(z, 5) * frequency) * amplitude;
if(value > 0.8) { // A certain threshold (checking if tree should be generated at this area)
int startX = Math.floorDiv(x, 5) * 5; // flooring the x value to every 5 units to get the start position
int startZ = Math.floorDiv(z, 5) * 5; // flooring the z value to every 5 units to get the start position
// Getting the starting height of the trunk (middle of the tree , that's why I'm adding 2 to the starting x and starting z), which is 1 block over the grass surface
int startY = height(startX + 2, startZ + 2) + 1;
int relx = x - startX; // block pos relative to starting position
int relz = z - startZ;
for(int j = startY; j < startY + 5; j++) {
int rely = j - startY;
byte tile = tree[relx][rely][relz]; // Get the needing block at this part of the tree
tiles[i][j][k] = tile;
}
}
}
}
So 'i' and 'k' are looping withing the chunk, and 'j' is looping inside the structure. This is pretty much how it should work.
And about the rivers, I personally haven't done it yet, and I'm not sure why you need to set the blocks around the chunk when generating them ( you could just use perlin worms and it would solve problem), but it's pretty much the same idea, and for your cities too.
I read something about this on a book and what they did in these cases was to make a finer division of chunks depending on the application, i.e.: if you are going to grow very big objects, it may be useful to have another separated logic division of, for example, 128x128x128, just for this specific application.
In essence, the data resides is in the same place, you just use different logical divisions.
To be honest, never did any voxel, so don't take my answer too serious, just throwing ideas. By the way, the book is game engine gems 1, they have a gem on voxel engines there.
About rivers, can't you just set a level for water and let rivers autogenerate in mountain-side-mountain ladders? To avoid placing water inside mountain caveats, you could perform a raycast up to check if it's free N blocks up.
I would like to have a function where I can input a radius value and have said function spit out the area for that size circle. The catch is I want it to do so for integer based coordinates only.
I was told elsewhere to look at Gauss's circle problem, which looks to be exactly what I'm interested in, but I don't really understand the math behind it (assuming it is actually accurate in calculating what I'm wanting).
As a side note, I currently use a modified circle drawing algorithm which does indeed produce the results I desire, but it just seems so incredibly inefficient (both the algorithm and the way in which I'm using it to get the area).
So, possible answers for this to me would be actual code or pseudocode for such a function if such a thing exists or something like a thorough explanation of Gauss's circle problem and why it is/isn't what I'm looking for.
The results I would hope the function would produce:
Input: Output
0: 1
1: 5
2: 13
3: 29
4: 49
5: 81
6: 113
7: 149
8: 197
9: 253
I too had to solve this problem recently and my initial approach was that of Numeron's - iterate on x axis from the center outwards and count the points within the upper right quarter, then quadruple them.
I then improved the algorithm around 3.4 times.
What I do now is just calculating how many points there are within an inscribed square inside that circle, and what's between that square and the edge of the circle (actually in the opposite order).
This way I actually count one-eighth of the points between the edge of the circle, the x axis and the right edge of the square.
Here's the code:
public static int gaussCircleProblem(int radius) {
int allPoints=0; //holds the sum of points
double y=0; //will hold the precise y coordinate of a point on the circle edge for a given x coordinate.
long inscribedSquare=(long) Math.sqrt(radius*radius/2); //the length of the side of an inscribed square in the upper right quarter of the circle
int x=(int)inscribedSquare; //will hold x coordinate - starts on the edge of the inscribed square
while(x<=radius){
allPoints+=(long) y; //returns floor of y, which is initially 0
x++; //because we need to start behind the inscribed square and move outwards from there
y=Math.sqrt(radius*radius-x*x); // Pythagorean equation - returns how many points there are vertically between the X axis and the edge of the circle for given x
}
allPoints*=8; //because we were counting points in the right half of the upper right corner of that circle, so we had just one-eightth
allPoints+=(4*inscribedSquare*inscribedSquare); //how many points there are in the inscribed square
allPoints+=(4*radius+1); //the loop and the inscribed square calculations did not touch the points on the axis and in the center
return allPoints;
}
Here's a picture to illustrate that:
Round down the length of the side of an inscribed square (pink) in the upper right quarter of the circle.
Go to next x coordinate behind the inscribed square and start counting orange points until you reach the edge.
Multiply the orange points by eight. This will give you the yellow
ones.
Square the pink points. This will give you the dark-blue ones. Then
multiply by four, this will get you the green ones.
Add the points on the axis and the one in the center. This gives you
the light-blue ones and the red one.
This is an old question but I was recently working on the same thing. What you are trying to do is as you said, Gauss's circle problem, which is sort of described here
While I too have difficulty understaning the serious maths behind it all, what it more or less pans out to when not using wierd alien symbols is this:
1 + 4 * sum(i=0, r^2/4, r^2/(4*i+1) - r^2/(4*i+3))
which in java at least is:
int sum = 0;
for(int i = 0; i <= (radius*radius)/4; i++)
sum += (radius*radius)/(4*i+1) - (radius*radius)/(4*i+3);
sum = sum * 4 + 1;
I have no idea why or how this works and to be honest Im a bit bummed I have to use a loop to get this out rather than a single line, as it means the performance is O(r^2/4) rather than O(1).
Since the math wizards can't seem to do better than a loop, I decided to see whether I could get it down to O(r + 1) performance, which I did. So don't use the above, use the below. O(r^2/4) is terrible and will be slower even despite mine using square roots.
int sum = 0;
for(int x = 0; x <= radius; x++)
sum += Math.sqrt(radius * radius - x * x);
sum = sum * 4 + 1;
What this code does is loop from centre out to the edge along an orthogonal line, and at each point adding the distance from line to edge in a perpendicualr direction. At the end it will have the number of points in a quater, so it quadruples the result and adds one because there is also central point. I feel like the wolfram equation does something similar, since it also multiplies by 4 and adds one, but IDK why it loops r^2/4.
Honestly these aren't great solution, but it seems to be the best there is. If you are calling a function which does this regularly then as new radii come up save the results in a look-up table rather than doing a full calc each time.
Its not a part of your question, but it may be relevant to someone maybe so I'll add it in anyway. I was personally working on finding all the points within a circle with cells defined by:
(centreX - cellX)^2 + (centreY - cellY)^2 <= radius^2 + radius
Which puts the whole thing out of whack because the extra +radius makes this not exactly the pythagorean theorem. That extra bit makes the circles look a whole lot more visually appealing on a grid though, as they don't have those little pimples on the orthogonal edges. It turns out that, yes my shape is still a circle, but its using sqrt(r^2+r) as radius instead of r, which apparently works but dont ask me how. Anyway that means that for me, my code is slightly different and looks more like this:
int sum = 0;
int compactR = ((radius * radius) + radius) //Small performance boost I suppose
for(int j = 0; j <= compactR / 4; j++)
sum += compactR / (4 * j + 1) - compactR / (4 * j + 3);
sum = sum * 4 + 1;
I am trying to rotate the image manually using the following code.
clc;
m1 = imread('owl','pgm'); % a simple gray scale image of order 260 X 200
newImg = zeros(500,500);
newImg = int16(newImg);
rotationMatrix45 = [cos((pi/4)) -sin((pi/4)); sin((pi/4)) cos((pi/4))];
for x = 1:size(m1,1)
for y = 1:size(m1,2)
point =[x;y] ;
product = rotationMatrix45 * point;
product = int16(product);
newx =product(1,1);
newy=product(2,1);
newImg(newx,newy) = m1(x,y);
end
end
imshow(newImg);
Simply I am iterating through every pixel of image m1, multiplying m1(x,y) with rotation matrix, I get x',y', and storing the value of m1(x,y) in to `newImg(x',y')' BUT it is giving the following error
??? Attempted to access newImg(0,1); index must be a positive integer or logical.
Error in ==> at 18
newImg(newx,newy) = m1(x,y);
I don't know what I am doing wrong.
Part of the rotated image will get negative (or zero) newx and newy values since the corners will rotate out of the original image coordinates. You can't assign a value to newImg if newx or newy is nonpositive; those aren't valid matrix indices. One solution would be to check for this situation and skip such pixels (with continue)
Another solution would be to enlarge the newImg sufficiently, but that will require a slightly more complicated transformation.
This is assuming that you can't just use imrotate because this is homework?
The problem is simple, the answer maybe not : Matlab arrays are indexed from one to N (whereas in many programming langages it's from 0 to (N-1) ).
Try newImg( max( min(1,newX), m1.size() ) , max( min(1,newY), m1.size() ) ) maybe (I don't have Matlab at work so I can tell if it's gonna work), but the resulting image will be croped.
this is an old post so I guess it wont help the OP but as I was helped by his attempt I post here my corrected code.
basically some freedom in the implementation regarding to how you deal with unassigned pixels as well as wether you wish to keep the original size of the pic - which will force you to crop areas falling "outside" of it.
the following function rotates the image around its center, leaves unassigned pixels as "burned" and crops the edges.
function [h] = rot(A,ang)
rotMat = [cos((pi.*ang/180)) sin((pi.*ang/180)); -sin((pi.*ang/180)) cos((pi.*ang/180))];
centerW = round(size(A,1)/2);
centerH = round(size(A,2)/2);
h=255.* uint8(ones(size(A)));
for x = 1:size(A,1)
for y = 1:size(A,2)
point =[x-centerW;y-centerH] ;
product = rotMat * point;
product = int16(product);
newx =product(1,1);
newy=product(2,1);
if newx+centerW<=size(A,1)&& newx+centerW > 0 && newy+centerH<=size(A,2)&& newy+centerH > 0
h(newx+centerW,newy+centerH) = A(x,y);
end
end
end
I'm solving the following problem: I have an object and I know its position now and its position 300ms ago. I assume the object is moving. I have a point to which I want the object to get.
What I need is to get the angle from my current object to the destination point in such a format that I know whether to turn left or right.
The idea is to assume the current angle from the last known position and the current position.
I'm trying to solve this in MATLAB. I've tried using several variations with atan2 but either I get the wrong angle in some situations (like when my object is going in circles) or I get the wrong angle in all situations.
Examples of code that screws up:
a = new - old;
b = dest - new;
alpha = atan2(a(2) - b(2), a(1) - b(1);
where new is the current position (eg. x = 40; y = 60; new = [x y];), old is the 300ms old position and dest is the destination point.
Edit
Here's a picture to demonstrate the problem with a few examples:
In the above image there are a few points plotted and annotated. The black line indicates our estimated current facing of the object.
If the destination point is dest1 I would expect an angle of about 88°.
If the destination point is dest2 I would expect an angle of about 110°.
If the destination point is dest3 I would expect an angle of about -80°.
Firstly, you need to note the scale on the sample graph you show above. The x-axis ticks move in steps of 1, and the y-axis ticks move in steps of 20. The picture with the two axes appropriately scaled (like with the command axis equal) would be a lot narrower than you have, so the angles you expect to get are not right. The expected angles will be close to right angles, just a few degrees off from 90 degrees.
The equation Nathan derives is valid for column vector inputs a and b:
theta = acos(a'*b/(sqrt(a'*a) * sqrt(b'*b)));
If you want to change this equation to work with row vectors, you would have to switch the transpose operator in both the calculation of the dot product as well as the norms, like so:
theta = acos(a*b'/(sqrt(a*a') * sqrt(b*b')));
As an alternative, you could just use the functions DOT and NORM:
theta = acos(dot(a,b)/(norm(a)*norm(b)));
Finally, you have to account for the direction, i.e. whether the angle should be positive (turn clockwise) or negative (turn counter-clockwise). You can do this by computing the sign of the z component for the cross product of b and a. If it's positive, the angle should be positive. If it's negative, the angle should be negative. Using the function SIGN, our new equation becomes:
theta = sign(b(1)*a(2)-b(2)*a(1)) * acos(dot(a,b)/(norm(a)*norm(b)));
For your examples, the above equation gives an angle of 88.85, 92.15, and -88.57 for your three points dest1, dest2, and dest3.
NOTE: One special case you will need to be aware of is if your object is moving directly away from the destination point, i.e. if the angle between a and b is 180 degrees. In such a case you will have to pick an arbitrary turn direction (left or right) and a number of degrees to turn (180 would be ideal ;) ). Here's one way you could account for this condition using the function EPS:
theta = acos(dot(a,b)/(norm(a)*norm(b))); %# Compute theta
if abs(theta-pi) < eps %# Check if theta is within some tolerance of pi
%# Pick your own turn direction and amount here
else
theta = sign(b(1)*a(2)-b(2)*a(1))*theta; %# Find turn direction
end
You can try using the dot-product of the vectors.
Define the vectors 'a' and 'b' as:
a = new - old;
b = dest - new;
and use the fact that the dot product is:
a dot b = norm2(a) * norm2(b) * cos(theta)
where theta is the angle between two vectors, and you get:
cos(theta) = (a dot b)/ (norm2(a) * norm2(b))
The best way to calculate a dot b, assuming they are column vectors, is like this:
a_dot_b = a'*b;
and:
norm2(a) = sqrt(a'*a);
so you get:
cos(theta) = a'*b/(sqrt((a'*a)) * sqrt((b'*b)))
Depending on the sign of the cosine you either go left or right
Essentially you have a line defined by the points old and new and wish to determine if dest is on right or the left of that line? In which case have a look at this previous question.