Outdated code from a tutorial. Can't figure it out - pygame

I was following a tutorial about how to 3D render objects in pygame. However it seems that the code is outdated and I don't know how to fix it.
I've spent quite a bit of time looking for the solution, but to no avail, as I don't know what I'm looking for.
def rotateX(self, (cx,cy,cz), radians):
for node in self.nodes:
y = node.y - cy
z = node.z - cz
d = math.hypot(y, z)
theta = math.atan2(y, z) + radians
node.z = cz + d * math.cos(theta)
node.y = cy + d * math.sin(theta)
It is supposed to show me a game window with a cube that I can move around and such, but instead it throws me an error.
Traceback (most recent call last):
File "index.py", line 1, in <module>
import wireframe
File "C:\Users\elee4018\Desktop\Python\pyGame01\wireframe.py", line 34
def scale(self, (centre_x, centre_y), scale):
^
SyntaxError: invalid syntax

Aside from the fact that your error message isn't pointing to the code you've pasted (def scale vs def rotateX). They share the same issue.
It's an older Python2 syntax. Simply just replace it with rotateX(self, cx, cy, cz, radians) or rotateX(self, axies, radians) and add a line that does cx, cy, cz = axies at the start of the function. Either way, Python3 simply just doesn't take a tuple expansion by default. At least not in this way.
So either do:
def rotateX(self, cx, cy, cz, radians):
for node in self.nodes:
y = node.y - cy
z = node.z - cz
d = math.hypot(y, z)
theta = math.atan2(y, z) + radians
node.z = cz + d * math.cos(theta)
node.y = cy + d * math.sin(theta)
And make sure you update the call to the rotateX function, or you could do:
def rotateX(self, axies, radians):
cx, cy, cz = axies
for node in self.nodes:
y = node.y - cy
z = node.z - cz
d = math.hypot(y, z)
theta = math.atan2(y, z) + radians
node.z = cz + d * math.cos(theta)
node.y = cy + d * math.sin(theta)
And you shouldn't need to update the call to the function in any way. There's probably a neater way to do the same thing as Python2 did. But I have never needed or used it. But this should solve your problem at least.
Oh, and submit a issue report to Pygame. They should be aware or instructed if the instructions are off/old.

Related

How to fix: "anonymous function bodies must be single expressions" error on Octave

I am trying to make a function in Octave where you give octave a function f(x,y) as a string, a change in X, a change in Y, a starting point, and the size of a matrix, the function will create a matrix populated with the values of f(x,y) at each point in the matrix.
This is for an application that displays a 3d graph, using the matrix to map each value to a block
# funcStr: The function whose Z values are being calculated
# dx: the change in x that each block in the x direction represents
# dy: the change in y that each block in the y direction represents
# startPt: the point (in an array of x, y) that center block represents
# res: the side length (in blocks) of the plane
pkg load symbolic
syms x y
function[zValues] = calculateZValues(funcStr, dx, dy, startPt, res)
zValues = zeros(res);
eqn = #(x, y) inline(funcStr);
startX = startPt{1};
startY = startPt{2};
for yOffset = 1:res
for xOffset = 1:res
xCoord = startX + dx * xOffset;
yCoord = startY + dy * yOffset;
zValues(res * yOffset + xOffset) = double(subs(eqn, #(x, y), {xCoord, yCoord}));
endfor
endfor
endfunction
The error I am getting is:
>> calculateZValues("x*y", 1, 1, {0,0}, 10)
parse error near line 20 of file /home/rahul/Documents/3dGraph/graph/calculateZValues.m
anonymous function bodies must be single expressions
>>> zValues(res * yOffset + xOffset) = double(subs(eqn, #(x, y), {xCoord, yCoord}));
I have no idea what the issue is. I have replaced the #(x,y) part with {x,y} in the line referenced by the error but it says nothing or it raises an error about the function subs not being declared. I have also tried moving the pkg and syms lines above the function header

find point of intersection between two arc

I'm creating a game for kids. It's creating a triangle using 3 lines. How I approached this is I create two arcs(semi circle) from two end points of a base line. But I couldn't figure how to find the point of intersection of those two arc. I've search about it but only found point of intersection between two straight lines. Is there any method to find this point of intersection? Below is the figure of two arcs drawn from each end of the baseline.
Assume centers of the circle are (x1, y1) and (x2, y2), radii are R1 and R2. Let the ends of the base be A and B and the target point be T. We know that AT = R1 and BT = R2. IMHO the simplest trick to find T is to notice that difference of the squares of the distances is a known constant (R1^2 - R2^2). And it is easy to see that the line the contains points meeting this condition is actually a straight line perpendicular to the base. Circles equations:
(x - x1)^2 + (y-y1)^2 = R1^2
(x - x2)^2 + (y-y2)^2 = R2^2
If we subtract one from another we'll get:
(x2 - x1)(2*x - x1 - x2) + (y2 - y1)(2*y - y1 - y2) = R1^2 - R2^2
Let's x0 = (x1 + x2)/2 and y0 = (y1 + y2)/2 - the coordinates of the center. Let also the length of the base be L and its projections dx = x2 - x1 and dy = y2 - y1 (i.e. L^2 = dx^2 + dy^2). And let's Q = R1^2 - R2^2 So we can see that
2 * (dx * (x-x0) + dy*(y-y0)) = Q
So the line for all (x,y) pairs with R1^2 - R2^2 = Q = const is a straight line orthogonal to the base (because coefficients are exactly dx and dy).
Let's find the point C on the base that is the intersection with that line. It is easy - it splits the base so that difference of the squares of the lengths is Q. It is easy to find out that it is the point on a distance L/2 + Q/(2*L) from A and L/2 - Q/(2*L) from B. So now we can find that
TC^2 = R1^2 - (L/2 + Q/(2*L))^2
Substituting back Q and simplifying a bit we can find that
TC^2 = (2*L^2*R1^2 + 2*L^2*R2^2 + 2*R1^2*R2^2 - L^4 - R1^4 - R2^4) / (4*L^2)
So let's
a = (R1^2 - R2^2)/(2*L)
b = sqrt(2*L^2*R1^2 + 2*L^2*R2^2 + 2*R1^2*R2^2 - L^4 - R1^4 - R2^4) / (2*L)
Note that formula for b can also be written in a different form:
b = sqrt[(R1+R2+L)*(-R1+R2+L)*(R1-R2+L)*(R1+R2-L)] / (2*L)
which looks quite similar to the Heron's formula. And this is not a surprise because b is effectively the length of the height to the base AB from T in the triangle ABT so its length is 2*S/L where S is the area of the triangle. And the triangle ABT obviously has sides of lengths L, R1 and R2 respectively.
To find the target T we need to move a along the base and b in a perpendicular direction. So coordinates of T calculated from the middle of the segment are:
Xt = x0 + a * dx/L ± b * dy / L
Yt = y0 + a * dy/L ± b * dx / L
Here ± means that there are two solutions: one on either side of the base line.
Partial case: if R1 = R2 = R, then a = 0 and b = sqrt(R^2 - (L/2)^2) which makes obvious sense: T lies on the segment bisector on a length of sqrt(R^2 - (L/2)^2) from the middle of the segment.
Hope this helps.
While you have not stated clearly, I assume that you have points with coordinates (A.X, A.Y) and (B.X, B.Y) and lengths of two sides LenA and LenB and need to find coordinates of point C.
So you can make equation system exploiting circle equation:
(C.X - A.X)^2 + (C.Y - A.Y)^2 = LenA^2
(C.X - B.X)^2 + (C.Y - B.Y)^2 = LenB^2
and solve it for unknowns C.X, C.Y.
Not that it is worth to subtract A coordinates from all others, make and solve simpler system (the first equation becomes C'.X^2 + C'.Y^2 = LenA^2), then add A coordinates again
So I actually needed this to design a hopper to lift grapes during the wine harvest. Tried to work it out myself but the algebra is horrible, so I had a look on the web -in the end I did it myself but introduced some intermediate variables (that I calculate in Excel - this should also work for the OP since the goal was a calculated solution). In fairness this is really much the same as previous solutions but hopefully a little clearer.
Problem:
What are the coordinates of a point P(Xp,Yp) distance Lq from point Q(Xq,Yq) and distance Lr from point R(Xr,Yr)?
Let us first map the problem onto to new coordinate system where Lq is the origin, thus Q’ = (0,0), let (x,y) = P’(Xp-Xq,Yp-Yq) and let (a,b) = R’(Xr-Xq,Yr-Yq).
We may now write:
x^2 + y^2 = Lq^2 -(1)
(x-a)^2 + (y-b)^2 = Lr^2 -(2)
Expanding 2:
x^2 – 2ax + a^2 + y^2 -2ay + b^2 =Lr^2
Subtracting 1 and rearranging
2by = -2ax + a2 + b2 - Lr^2+ Lq^2
For convenience, let c = a^2 + b^2 + Lq^2 + Lr^2 (these are all known constants so c may be easily computed), thus we obtain:
y = -ax/b + c/2b
Substituting into 1 we obtain:
x^2 + (-a/b x + c/2b)^2 = Lq^2
Multiply the entire equation by b^2 and gather terms:
(a^2 + b^2) x2 -ac x + c/4 + Lq^2 b^2 = 0
Let A = (a2 + b2), B= -ac ,and C= c/4 + Lq^2 b^2
Use the general solution for a quadratic
x = (-B +-SQRT(B^2-4AC))/2A
Substitute back into 1 to get:
y= SQRT(Lq^2 - x^2 )
(This avoids computational difficulties where b = 0)
Map back to original coordinate system
P = (x+Xq, y + Yq)
Hope this helps, sorry about the formatting, I had this all pretty in Word, but lost it

Scilab plotting functions with if

I have a problen in scilab
How can I plot functions containing if and < like
function y = alpha(t)
if (t < 227.8) then
y = 0.75;
elseif (t < 300) then
y = 2.8 - 0.009 .* t;
else
y = 0.1;
end
endfunction
and
function [r]=minus_alpha(t)
r = 1 - alpha(t)
endfunction
When I use
x = linspace(0,300)
plot(x, alpha(x))
I got the error message
WARNING: Transposing row vector X to get compatible dimensions
plot2d: falsche Größe für Eingangsargument: inkompatible Größen.
Error 999 : in plot2d called by plot
Sorry for german mix. Thank you.
You can avoid explicit loop and be more efficient using the followin code
function y = alpha(t)
y=0.1*ones(t);
y(t<227.8)=0.75;
i=t>=227.8&t<300;
y(i)=2.8 - 0.009 .* t(i);
endfunction
It is really sad to see a great majority of Scilab community is not aware of vectorized operations. You can change your function to:
function y = alpha(t)
y = 0.1;
if t < 227.8 then
y = 0.75;
elseif t < 300 then
y = 2.8 - 0.009 * t;
end
y = 1 - y;
endfunction
and then use feval to broadcast the function over the sequence:
x = linspace(0, 300);
plot2d(x, feval(x, alpha));
which results:
rule of thumb if you are using for loop you need to revise your code and if someone is offering you a code where there is unnecessary for loop you shouldn't probably use it.
All the proposed answers are overcomplicated considering that the function alpha in the original demand is piecewise-affine. In Scilab in can be coded that way:
x = linspace(0,400,1000);
plot(x,linear_interpn(x,[227.8 300],[0.75 0.1]))
i.e. you just have to know the nodes coordinates (here abscissae) and value of the function at nodes. The function linear_interpn does also multilinear interpolation, it is worth knowing it guys...
If you check the output of your alpha(x), you will see that it is just a scalar (not a vector). I guess you wanted something like this, so it's necessary to iterate through t to compute each value of y based on the value of t:
clc;
clear;
function y = alpha(t)
for i=1:size(t,"*")
if t(i) < 227.8 then
y(i) = 0.75;
elseif t(i) < 300 then
y(i) = 2.8 - 0.009 * t(i);
else
y(i) = 0.1;
end
end
endfunction
x = linspace(0,300);
plot2d(x,alpha(x));
If you find the answer useful, please do not forget to accept it, so others will see that your problem is solved.
Before your answers (thank you) my workaround was a combination of indicator functions composed with floor and exp( -t^2):
function y = alpha(t)
y = floor(exp(-(t .* (t-T1)) / (T1*T1))) * 0.75
+ floor(exp(-((t-T2) .* (t- T1) / (2000)))) .* (2.8-0.009 .* t)
+ floor(exp(-((t-T2) .* (t-1000) / (200000))))*0.1
endfunction

Matlab second order derivative

I have followed the tutorial on http://www.mit.edu/people/abbe/matlab/ode.html and prepared a function as follows:
function dxy = diffxy(xy)
%
%split xy into variables in our equations
%
x = xy(1);
xdot = xy(2);
y = xy(3);
%
% define the derivatives of these variables from equations
%
xdot = xdot;
ydot = 3*x + 2*y + 5;
xdoubledot = 3 - ydot + 2*xdot;
%
%return the derivatives in dxy in the right order
%
dxy = [xdot; xdoubledot; ydot]
end
When I call it using
[T, XY] = ode45('diffxy',0,10,[0 1 0])
I get an error
??? Error using ==> diffxy
Too many input arguments.
I also tried
XY= ode45(#diffxy,[0 10],[0;1;0])
Anybody have any idea?
haven't read the whole tutorial but aren't you supposed to define your function as
function dxy = diffxy(t, xy)
where t is time vector

Solving two coupled non-linear second order differentially equations numerically

I have encountered the following system of differential equations in lagrangian mechanics. Can you suggest a numerical method, with relevant links and references on how can I solve it. Also, is there a shorter implementation on Matlab or Mathematica?
mx (y dot)^2 + mgcosy - Mg - (M=m)(x double dot) =0
gsiny + 2(x dot)(y dot + x (y double dot)=0
where (x dot) or (y dot)= dx/dt or dy/dt, and the double dot indicated a double derivative wrt time.
You can create a vector Y = (x y u v)' so that
dx/dt = u
dy/dt = v
du/dt = d²x/dt²
dv/dt = d²y/dt²
It is possible to isolate the second derivatives from the equations, so you get
d²x/dt² = (m*g*cos(y) + m*x*v² - M*g)/(M-m)
d²y/dt² = -(g*sin(y) - 2*u*v)/x
Now, you can try to solve it using standard ODE solvers, such as Runge-Kutta methods. Matlab has a set of solvers, such as ode23. I didn't test he following, but it would be something like it:
function f = F(Y)
x = Y(1); y = Y(2); u = Y(3); v = Y(4);
f = [0,0,0,0];
f(1) = u;
f(2) = v;
f(3) = (m*g*cos(y) + m*x*v*v - M*g)/(M-m);
f(4) = -(g*sin(y) - 2*u*v)/x;
[T,Y] = ode23(F, time_period, Y0);