In the image shown, number of possible routes should be calculated (for example: between (0,0) and (4,3)). Condition: no diagonal direction
I have tried this using adjacency matrix and digraph in MATALB, but I require output in Octave where digraph is not supported. Please suggest.
I actually have 8 possible ways to move between any two points:
Down, left
Down, right
Up, left
Up, right
Right, up
Right, down
Left, up
Left, down
%%%%%%%%%%%%%% Make grid points.
x = -1:6;
y = -1:5;
[xx, yy] = meshgrid(x, y, indexing = 'ij');
G = plot(xx(:),yy(:), 'b.');
grid on;
drawnow;
%%%%%%%%%% Set up figure properties: Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0, 0, 1, 1]);
%%%%%%%%%%%%%%% Get rid of tool bar and pulldown menus that are along top of figure.
set(gcf, 'Toolbar', 'none', 'Menu', 'none');
%%%%%%%%%%% Give a name to the title bar.
set(gcf, 'Name', 'Mega Constellation:Geographical separation', 'NumberTitle', 'Off');
%%%%%%%%%%%%%%% Print (x,y) values at each point.
for k = 1 : numel(xx)
str = sprintf('(%i, %i)', xx(k), yy(k));
text(xx(k), yy(k), str);
end
Related
I want to crate an pygame.Rect object from a center point (xc, yc) and a size (w, h).
pygame.Rect just provides a constructor with the top left point and the size.
Of course I can calculate the top left point:
rect = pygame.Rect(xc - w // 2, yc - h // 2, w, h)
Or I can set the location via the virtual attribute center:
rect = pygame.Rect(0, 0, w, h)
rect.center = xc, yc
If I want to completely confuse someone, I use inflate:
rect = pygame.Rect(xc, yc, 0, 0).inflate(w, h)
Or even clamp:
rect = pygame.Rect(0, 0, w, h).clamp((xc, yc, 0, 0))
Not any of this methods satisfies me. Either I have to calculate something, I have to write several lines of code, or I have to use a function that completely hides what is happening.
I also don't want to write a function (or lambda) as I think this is completely over the top for creating a simple rectangle.
So my question is:
How do you usually create such a rectangle with a self-explanatory line of code so everyone can see what is happening at a glance?
Is there a much easier method? Do I miss something?
Interesting question Rabbid76.
I personally try to write code such that a person with only a general understanding of programming concepts can read the code. This includes absolute beginners (95% of people making PyGame questions), and converts from other languages.
This is why I mostly shy-away from using Python's if x < y < z:, and blah = [ x for x in some-complex-iff-loop ], et.al. syntax. (And that's also why I always put my if conditions in brackets.) Sure if you know python well it doesn't matter, but for an example of why it's important, go try to read a Perl script from the mid 2010's and you see stuff like:
print #$_, "\n" foreach ( #tgs );
It didn't have to be written like that, they could have used a loop-block with some instructive variable names, and not $_, etc.
So bearing the above in mind, the question comes down to - Which is the easiest to read and understand.
So for my 2-cents worth, it has to be option #2:
rect = pygame.Rect(0, 0, w, h)
rect.center = xc, yc
It's absolutely clear to a syntax-ignorant code reader that a rectangle is being created, and some kind of centre-point is being set.
But to make the code more "self documenting", it could be wrapped in a function call:
def getRectAround( centre_point, width, height ):
""" Return a pygame.Rect of size width by height,
centred around the given centre_point """
rectangle = pygame.Rect( 0, 0, w, h ) # make new rectangle
rectangle.center = centre_point # centre rectangle
return rectangle
# ...
rect = getRectAround( ( x, y ), w, h )
Sometimes more code is better.
I have two sets of coordinates (both positive and negative values, not necessarily in increasing order, and in many cases there are different values of y for the same value of x) which I can load into two row vectors of equal size.
I want to calculate the area enclosed by the curve.
How to do it with octave?
I tried this answer but it does not work because it seems that the area printed (204.64) is too high (see picture).
I tried the code:
function showdata(fName)
M = dlmread(fName);
H = M(2:end, 1); % starting row number is 2
B = M(2:end, 2);
aux = figure();
plot(H, B,'linewidth',2);
xlabel ("Auxilary field H (A/m)");
ylabel ("Magnetic Field B (Tesla)");
area = polyarea(H,B)
axis([min(H), max(H), min(B), max(B)]);
grid on;
grid minor on;
title (area,"fontsize",20);
Then I am calling showdata('data.txt') in Octave.
Picture of Data points:
This is the data file I am using.
There is a function for computing convex hull called "convhull" in Octave. It returns the indices of the points formming convex hull data.
M = dlmread("data.txt"); #I removed the header in data.txt
x = M(:,1);
y = M(:,2);
k = convhull(x,y);
plot (x(k), y(k), "r-", x, y, "b+");
n = rows(k);
x_prime = vertcat(x(k(n)), x(k(1:n-1)));
y_prime = vertcat(y(k(n)), y(k(1:n-1)));
A = .5*abs(x_prime'*y(k)-y_prime'*x(k)); #80.248
polyarea(x(k), y(k)) == A and true
Maybe convex hull is not good estimate of area because the top left and the down-right lines are a little far away from the points. There are other ways to form a polygon from data
, one of which could be alpha shape. However, alpha shape are more complicated and there is no corresponding pre-built function in Octave.
Update:
Each x corresponds to at least one y cordinate. I marked the highest point and lowest point laying on the same x and estimate the area again.
There is the code:
[uni, ~] = sort(unique(x));
n = rows(uni);
outline = [];
for i = 1:n
y_list = y(x==uni(i));
[y_max, ~] = max(y_list);
outline(i, :)= [uni(i), y_max];
[y_min, ~] = min(y_list);
outline(2*n-i+1,:)= [uni(i), y_min];
endfor
figure;
plot (x(k), y(k), "r-", x, y, "b+", outline(:,1), outline(:,2), "g-", "linewidth", 3);
polyarea(outline(:,1), outline(:,2)) #74.856
By the way, if the arguments of function polyarea do not form a close curve function polyarea would return wrong area.
Four point on a unit square:
[(0,0), (1,0), (1,1), (0,1)], [(0,0), (1,1), (1,0), (0,1)]
polyarea([0,1,1,0],[0,0,1,1])!==polyarea([0,1,1,0],[0,1,0,1]).
I'm working with the Ipython widgets to modify dynamically plotted data, so i get the result as seen in the image; however i dont like the arrangment of the widgets (sliders and float boxes) so i have tried to align them horizontally with Box and here is the trouble because i don't know how to pass this arrangment to the function, this is how i obtain what you see in the image
#I'm just going to put two sliders
x= FloatSlider(0,1000)
y= FloatSlider(0,1000)
interactive(Function,Xo=x,Yo=y)
if i try
x= FloatSlider(0,1000,10)
y= FloatSlider(-1,1,0.1)
#this gives me the desired arrangment
co = HBox(children=[x,y])
interactive(Function,Xo=Co.children[0],Yo=Co.children[1])
this just gives me the same unordered result, and here is where i got stuck, i don't know how to input the container to make it work with my plot. Where is my error? what can i do?
The following might be a solution for your problem:
from IPython.html.widgets import *
x = FloatSlider(value = 500.0, min=0.0, max=1000.0, step=10)
y = FloatSlider(value = 0.0, min=-1.0, max=1.0, step=0.1)
plt.xlim(0,1)
plt.ylim(-1,1)
def Function(x, y):
plt.plot([0.0, 1.0], [-1.0, 1.0], 'b.')
plt.plot(x/1000.0, y, 'o')
#this gives me the desired arrangment
f = interactive(Function, x=x, y=y)
Co = HBox(f.children)
display(Co)
I hope someone can help me here, I have been asked to write some code for an Lua script for a game. Firstly i am not an Lua Scripter and I am defiantly no mathematician.
What i need to do is generate random points within a parallelogram, so over time the entire parallelogram becomes filled. I have played with the scripting and had some success with the parallelogram (rectangle) positioned on a straight up and down or at 90 degrees. My problem comes when the parallelogram is rotated.
As you can see in the image, things are made even worse by the coordinates originating at the centre of the map area, and the parallelogram can be positioned anywhere within the map area. The parallelogram itself is defined by 3 pairs of coordinates, start_X and Start_Y, Height_X and Height_Y and finally Width_X and Width_Y. The random points generated need to be within the bounds of these coordinates regardless of position or orientation.
Map coordinates and example parallelogram
An example of coordinates are...
Start_X = 122.226
Start_Y = -523.541
Height_X = 144.113
Height_Y = -536.169
Width_X = 128.089
Width_Y = -513.825
In my script testing i have eliminated the decimals down to .5 as any smaller seems to have no effect on the final outcome. Also in real terms the start width and height could be in any orientation when in final use.
Is there anyone out there with the patients to explain what i need to do to get this working, my maths is pretty basic, so please be gentle.
Thanks for reading and in anticipation of a reply.
Ian
In Pseudocode
a= random number with 0<=a<=1
b= random number with 0<=b<=1
x= Start_X + a*(Width_X-Start_X) + b*(Height_X-Start_X)
y= Start_Y + a*(Width_Y-Start_Y) + b*(Height_Y-Start_Y)
this should make a random point at coordinates x,y within the parallelogram
The idea is that each point inside the parallelogram can be specified by saying how far you go from Start in the direction of the first edge (a) and how far you go in the direction of the second edge (b).
For example, if you have a=0, and b=0, then you do not move at all and are still at Start.
If you have a=1, and b=0, then you move to Width.
If you have a=1, and b=1, then you move to the opposite corner.
You can use something like "texture coordinates", which are in the range [0,1], to generate X,Y for a point inside your parallelogram. Then, you could generate random numbers (u,v) from range [0,1] and get a random point you want.
To explain this better, here is a picture:
The base is formed by vectors v1 and v2. The four points A,B,C,D represent the corners of the parallelogram. You can see the "texture coordinates" (which I will call u,v) of the points in parentheses, for example A is (0,0), D is (1,1). Every point inside the parallelogram will have coordinates within (0,0) and (1,1), for example the center of the parallelogram has coordinates (0.5,0.5).
To get the vectors v1,v2, you need to do vector subtraction: v1 = B - A, v2 = C - A. When you generate random coordinates u,v for a random point r, you can get back the X,Y using this vector formula: r = A + u*v1 + v*v2.
In Lua, you can do this as follows:
-- let's say that you have A,B,C,D defined as the four corners as {x=...,y=...}
-- (actually, you do not need D, as it is D=v1+v2)
-- returns the vector a+b
function add(a,b)
return {x = a.x + b.x, y = a.y + b.y} end
end
-- returns the vector a-b
function sub(a,b)
return {x = a.x - b.x, y = a.y - b.y} end
end
-- returns the vector v1*u + v2*v
function combine(v1,u,v2,v)
return {x = v1.x*u + v2.x*v, y = v1.y*u + v2.y*v}
end
-- returns a random point in parallelogram defined by 2 vectors and start
function randomPoint(s,v1,v2)
local u,v = math.random(), math.random() -- these are in range [0,1]
return add(s, combine(v1,u,v2,v))
end
v1 = sub(B,A) -- your basis vectors v1, v2
v2 = sub(C,A)
r = randomPoint(A,v1,v2) -- this will be in your parallelogram defined by A,B,C
Note that this will not work with your current layout - start, width, height. How do you want to handle rotation with these parameters?
I have one character which can face several directions: t, b, l, r, tl, tr, bl, br
The type of camera is top-down 2d
T stands for Top, which is when he is looking to the top of the screen
B = Bottom
L = Left
R = Right
The character can look to other directions, like the Top-Left of the screen, which is tl
How do I calculate which direction he should look to if I only have one variable, rotation?
Assuming your rotation property is from -180 to 180 (this is the way DisplayObject.rotation behaves)
// this gives us an index of 0-7 for rotation.
var direction:int = (character.rotation + 180) / 45;
// should you need a string representation, use an array like this:
var direction_labels:Array = ['b', 'bl', 'l', 'tl', 't', 'tr', 'r', 'br'];
trace(direction_labels[direction]);