I made this small game similar to flappy birds, where one can use to fly up and down using the mouse and keyboard.
I won't be posting the full code because it's a university project and I don't want to get caught for any possible plagiarism.
What I did was, I used two objects:
A rectangle as the bird:
r= rectangle('Position', pos, 'curvature', [0.99,0.1], 'FaceColor', 'r', 'EraseMode','xor');
Thick lines to represent the walls:
line([ 100 100], [10 400], 'color', 'm', 'LineWidth', 10, 'EraseMode', 'background')
My problem:
The problem is that the bird moves through the walls, as if the walls were transparent. As you can imagine, I want to break the game and show something like "game over", when the bird hits the wall (not go through them). How do I make it such that my game breaks when the bird (object 1) collides with the walls (other objects)?
Thank You Very Much for reading my question!
If you have the Mapping Toolbox installed, it provides a function called polyxpoly that allows you to intersect polylines, so you can find the intersection of the whole rectangle against every wall, and you will not have to worry about splitting every edge of the rectangle.
Here you have a full working example of a collision with one wall:
% Bird (rectangle).
% Position.
xr = 30;
yr = 100;
% Length of rectangle edges.
deltaxr = 10;
deltayr = 10;
% Vector for rectangle object.
vr = [xr, yr, deltaxr, deltayr];
% Bird polyline.
a = [xr, yr];
b = [xr + deltaxr, yr];
c = [xr + deltaxr, yr + deltayr];
d = [xr, yr + deltayr];
r = [a; b; c; d];
% Wall (line).
% Wall polyline.
l = [40 0; 40 105];
% Draw objects.
r1 = rectangle('Position',vr,'LineWidth',2);
line(l(:,1), l(:,2),'LineWidth',2,'Color','r');
axis equal;
% Find intersections.
[xi,yi] = polyxpoly(r(:,1),r(:,2),l(:,1),l(:,2));
% Are there any intersections? If so, GAME OVER.
if ~isempty(xi)
% Stop the game and display GAME OVER.
text(xr-20,yr-20,'GAME OVER','Color','b','FontSize', 20);
end
You can try with different positions of the bird to test the collision detection:
You'd want to use an intersection algorithm to check whether there is an intersection of one of the wall-lines (check it four times) with the rectangle. The rectangle basically also consists of 4 lines, that means you should check the 4 rectangle-lines against the four wall-lines (if you use a line-line-intersection algorithm).
E.g. check this topic:
How to find the intersection point between a line and a rectangle?
It's basically a simple mathematical equation to solve, see https://en.wikipedia.org/wiki/Line%E2%80%93line_intersection.
Related
I want to write a code where an arc stops rotating as soon as it comes in contact with a semi-circle.
I have written a code to do so, my arc does not just touch the circle but it slightly intersects it.
I have put rotation of arc inside a while loop by using linspace() to change theta. I used polyxpoly() for finding intersection. Condition for the while loop is that as long as I have empty array the loop continues, but as soon as I get a value from polyxpoly() my loop stops.
However, at the place of touch the theta value exceeds what I needed, so as a result I get an intersection.
How do I modify the code so that the arc will touch the semi-circle and not intersect it?
Here is the output. Click the link below
Image of intersection but I need touch and not intersection
clc,clear
R = 5; % radius of a circle
r = 10; % radius of arc
aa = 60*pi/180; % arc angle
ap = 0*pi/180; % arc position angle
% defining the semi-circle about the origin
t = linspace(0,pi);
[x,y] = pol2cart(t,R); % circle data
% Shifting circle centre to (3.5,0)
x=x+3.5;
y=y+0;
% defining the arc about the origin
t1 = linspace(0,aa)-aa/2+ap;
[x1,y1] = pol2cart(t1,r); % arc data
% shifting arc-lower-end to (14,0)
delx=14-x1(1); % Finding the x difference between arc-lower-end x-coordinate & 14
dely=0-y1(1); % Finding the y difference between arc-lower-end y-coordinate & 0
x1=x1+delx;
y1=y1+dely;
theta =linspace(0,pi,1000);
i=1;
xc=[];
yc=[];
while isempty(xc)&& isempty(yc)
% create a matrix of these points, which will be useful in future calculations
v = [x1;y1];
% choose a point which will be the center of rotation
x_center = 14;
y_center = 0;
% create a matrix which will be used later in calculations
center = repmat([x_center; y_center], 1, length(x1));
% define a 60 degree counter-clockwise rotation matrix
R = [cos(theta(i)) -sin(theta(i)); sin(theta(i)) cos(theta(i))];
% do the rotation...
s = v - center; % shift points in the plane so that the center of rotation is at the origin
so = R*s; % apply the rotation about the origin
vo = so + center; % shift again so the origin goes back to the desired center of rotation
% this can be done in one line as:
% vo = R*(v - center) + center
% pick out the vectors of rotated x- and y-data
x_rotated = vo(1,:);
y_rotated = vo(2,:);
[xc,yc] = polyxpoly(x_rotated,y_rotated,x,y)
[xc1,yc1] = polyxpoly(x1,y1,x,y)
i=i+1;
end
% make a plot
plot(x,y)
hold on
plot(x1, y1, 'k-', x_rotated, y_rotated, 'r-', x_center, y_center, 'bo');
axis equal
I need to find way it contacts a circle and does not intersect it.
The code is in matlab.
Any suggestions are welcome.
This is the problem of collision detection. Most, if not all, methods I know of in collision detection requires the computer to check for intersections of some sort. It's very difficult (if not impossible) to have two objects "just touch" in simulations, because you'll need the precise (analytically solved) location of the boundaries of those objects.
polyxpoly() is a function to return the intersection of two polygons. So unfortunately, if you insist that the arc cannot touch, then you cannot use polyxpoly(). In that extreme case, you'll have to solve some mathematical equation for when the tip of the arc coincides with a point on the circle perfectly, then simulate up until that point in time.
But realistically, what you need is a finer simulation (although I personally think what you have alone is good enough). So in every simulation step, you calculate a smaller movement, so that when the arc eventually intersects, only a very small amount of the arc intersects.
I currently have to create a sphere from a stack of 2D slices (2D matrices in MATLAB, which represent 2D gray scale images). I have created a bunch of slices using the code below. To create the sphere I have repeatedly created slices of circles of increasing size till 100, and then of decreasing sizes. All of these slices are added to a 3D matrix.
Circle = ones(200,400,400);
for i = 1:100
[rr cc] = meshgrid(1:400);
C = sqrt((rr-200).^2+(cc-200).^2)<=i;
for j = 1:400
for k = 1:400
Circle(i,j,k) = C(j,k);
end
end
end
index = 100;
for i = 1:100
[rr cc] = meshgrid(1:400);
C = sqrt((rr-200).^2+(cc-200).^2)<=index;
for j = 1:400
for k = 1:400
Circle(i+100,j,k) = C(j,k);
end
end
index = index - 1;
end
viewer3d(Circle);
viewer3d is 3rd part library that helps you view your 3D image stacks of slices as 3d objects. Once I visualized this 'supposed' sphere, I realized that it is a diamond shape top and not a sphere.
Therefore I do not understand how to vary the size of circles till the center point of the sphere in the y plane and then decrease it with the same algorithm.
Thank you for your answers and please do not hesitate to ask me to clarify anything within this question.
Alternatively, create a sphere directly, without using loops:
Circle = zeros(200,400,400);
[x,y,z]=meshgrid(1:size(Circle,1),1:size(Circle,2),1:size(Circle,3));
radius=??; %// e.g. radius=100;
%//this sphere is centered in the middle of the image
Circle(sqrt((x-size(Circle,1)/2).^2+(y-size(Circle,2)/2).^2..
+(z-size(Circle,2)/2).^2)<radius)=1;
Yes, the radius along the Z axis is not linear but varies with cos/sin function. Using this representation:
your radius is "Radius = r sin(Theta)", with "Theta = arccos(r / z)". So "r" is the radius of your sphere, and "z" the level/slice you want to draw in. Don't forget to that "z" goes from -"r" to "r". I've tested the formulae and it works fine for an image stack (slices).
l have generated a topology of a network with 200 nodes.than l want to draw a circle with black color around a choosen node(satisfying a certain condition IF (condition) ).
to simplify let the user introduce the the index of the node to circle with the black color.
here is my code of the generated topology.l need to add wich instructions to draw the circle around the choosen node ?
X=100;
Y=100;
N=200; %number of nodes
nodesX(1)=rand*X;
nodesY(1)=rand*Y;
for i=2:N
nodesX(i)=rand*X;
nodesY(i)=rand*Y;
d(i-1) =((nodesX(i)-nodesX(i-1)).^2+(nodesY(i)-nodesY(i-1)).^2).^0.5;
while (d(i-1)>200)
nodesX(i)=rand*X;
nodesY(i)=rand*Y;
d(i-1) =((nodesX(i)-nodesX(i-1)).^2+(nodesY(i)-nodesY(i-1)).^2).^0.5;
end
end
h_old=plot(nodesX,nodesY,'m*');
labels=[1:N]';
labels=num2str(labels);
text(nodesX,nodesY,labels);
xlabel('X (Meters)');
ylabel('Y (Meters)');
title(['Network Topology with',num2str(N),'nodes']);
hold on
for k=1:N;
for j=1:N;
if (k~=j)
d=((nodesX(k)-nodesX(j))^2+(nodesY(k)-nodesY(j))^2)^0.5;
end
if (k~=j);
if(d < 50);
line([nodesX(k),nodesX(j)],[nodesY(k),nodesY(j)]);
end
end
end;
end;
Split it into two tasks:
Task 1:
Define a simple function to draw a circle. A cheap and dirty way is to use polar expressions.
function [] = circle(center_x,center_y,r)
theta = 0:0.01:2*pi;
x = center_x + r*cos(theta);
y = center_y + r*sin(theta);
plot(x,y,'k','LineWidth',2)
Task 2:
Pass this function the NodeX and NodeY values of the indexed node as the point at which the circle is centered. You can set the radius of the circle as per your choice. Using r=1 and picking an arbitrary index, I got:
Just one caveat: Make sure your axis are square. Otherwise, a circle might look like an ellipse. If you want to draw a circle around non-square axis, then you can modify the code to generate an ellipse instead.
Below is an arbitrary hand-drawn Intensity profile of a line in an image:
The task is to draw the line. The profile can be approximated to an arc of a circle or ellipse.
This I am doing for camera calibration. Since I do not have the actual industrial camera, I am trying to simulate the correction needed for calibration.
The question can be rephrased as I want pixel values which will follow a plot similar to the above. I want to do this using program (Preferably using opencv) and not manually enter these values because I have thousands of pixels in the line.
An algorithm/pseudo code will suffice. Also please note that I do not have any actual Intensity profile, otherwise I would have read those values.
When will you encounter such situation ?
Suppose you take a picture (assuming complete white) from a Camera, your object being placed on table, and camera just above it in vertical direction. The light coming on the center of the picture vertically downward from the camera will be stronger in intensity as compared to the light reflecting at the edges. You measure pixel values across any line in the Image, you will find intensity curve like shown above. Since I dont have camera for the time being, I want to emulate this situation. How to achieve this?
This is not exactly image processing, rather image generation... but anyways.
Since you want an arc, we still need three points on that arc, lets take the first, middle and last point (key characteristics in my opinion):
N = 100; % number of pixels
x1 = 1;
x2 = floor(N/2);
x3 = N;
y1 = 242;
y2 = 255;
y3 = 242;
and now draw a circle arc that contains these points.
This problem is already discussed here for matlab: http://www.mathworks.nl/matlabcentral/newsreader/view_thread/297070
x21 = x2-x1; y21 = y2-y1;
x31 = x3-x1; y31 = y3-y1;
h21 = x21^2+y21^2; h31 = x31^2+y31^2;
d = 2*(x21*y31-x31*y21);
a = x1+(h21*y31-h31*y21)/d; % circle center x
b = y1-(h21*x31-h31*x21)/d; % circle center y
r = sqrt(h21*h31*((x3-x2)^2+(y3-y2)^2))/abs(d); % circle radius
If you assume the middle value is always larger (and thus it's the upper part of the circle you'll have to plot), you can draw this with:
x = x1:x3;
y = b+sqrt(r^2-(x-a).^ 2);
plot(x,y);
you can adjust the visible window with
xlim([1 N]);
ylim([200 260]);
which gives me the following result:
I have a figure of size 14 x 14 square drawn inside an axis of 20 x 20, in matlab.
I am trying to draw circles of radius 0.7 inside the square and need to arrange them evenly. I need to draw 233 circles. Please let me know how can I do it?
Currently I can draw them randomly but couldn't get 233 circle. Please see my below code.
Your reply is appreciated.
% Urban, sub urban, Rural areas
x_area =[3, 12, 6];
y_area = [6, 8, 16];
r_area = [1, 7, 2];
f = figure;
hAxs = axes('Parent',f);
hold on, box on, axis equal
xlabel('x')
ylabel('y','Rotation',0)
title('Compute the area of circles a vectorized way for several cicles')
axis([0 20 0 20])
rectangle('Position',[5,1,14,14])
rectangle('Position',[3,1,2,2])
rectangle('Position',[1,3,4,4])
hold on, box on, axis equal
a = 233;
x_base_urban = randi([6 18], 1, a);
b = rand([10 8], 1);
y_base_urban = randi([2 14],1, a);
r_base_urban = 0.9;
size_x = size(x_base_urban);
size_x = size_x(2);
size_y = size(y_base_urban);
size_y = size_y(2);
colour = rand(size_x,3);
for t = 1: size_x
plot(x_base_urban(t)+ r_base_urban.*cos(0:2*pi/100:2*pi),...
y_base_urban(t)+ r_base_urban.*sin(0:2*pi/100:2*pi),'parent',hAxs)
plot(x_base_urban(t),y_base_urban(t),'+','parent',hAxs)
end
Thanks
Randomly plotting everything won't work. Actually, if your circles can't overlap, nothing will work. To show that, just compare the results of following calculations:
lSquare = 14;
rCircle = 0.7;
nCircles = 233;
areaCircles = nCircles * pi * rCircle^2
areaSquare = lSquare^2
You will see that areaCircles > areaSquare, so it is impossible to fit them all in. On the other hand, if areaSquare >= areaCircles does not guarantee you that a solution exists!
Try your setup with a smaller example to come up with a solution. E.g. take a square box and a bunch of spherical objects (balls, marbles, oranges, apples, ... if need be) and try to fit as much of those in your box. If that works, you might even want to draw their positions on a sheet of paper before trying to implement it.
If you do this correctly, you will get an idea of how to stack round objects in a square container. That is also exactly what you need to do in your exercise. Then try to make a model/algorithm of what you did manually and implement that in MATLAB. It won't be hard, but you will need some small calculations: Pythagoras and intersection of circles.
I also suggest you use a function to draw a circle as #Andrey shows, so something of the form function drawCircle(center, radius). That allows you to keep complexity down.
If your circles can overlap, then the solution is quite easy: look at a circle as an object with a center point and distribute these center points evenly over the square. Don't use rand to do this, but calculate their positions yourself.
If you can't find a solution, I might expand my answer in a few days.
Without diving too deep into your code, I think that you need to add a hold function, after the first plot
for t = 1: size_x
plot(x_base_urban(t)+ r_base_urban.*cos(0:2*pi/100:2*pi),...
y_base_urban(t)+ r_base_urban.*sin(0:2*pi/100:2*pi),'parent',hAxs)
plot(x_base_urban(t),y_base_urban(t),'+','parent',hAxs)
hold(hAxs,'on');
end
By the way, the best way to draw circle is by using rectangle command.
rectangle('Curvature',[1 1],'Position',[1 3 4 5])
So you can create a PlotCircle function (like #EgonGeerardyn suggests) like this:
function plotCircle(x,y,r)
rectangle('Position',[x-r y-r 2*r 2*r],'Curvature',[1 1]);
end