I was recently tracing a line of code:
x -= 353 - 350
However, the answer came up to be -3.
To my surprise, I figure the -= operator would follow as:
x = x - 353 - 350
which then would equal to -703
Why is the actual answer -3 and not -703?
I was searching for reference on this website:
http://www.adobe.com/devnet/actionscript/learning/as3-fundamentals/operators.html
The example it gave led me to believe that the operator -= should produce -703.
var x:uint = 5; x -= 5; // x is now 0
Wouldn't the above example represent how x = x - 5 is 0? Or is there an alternative code/logic that I'm missing?
Operator precendece:
https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Operators/Operator_Precedence
-= has precedence of 3, while + and - have precedence 13 (higher),
so the code executes as the equivalent of
x -= (353 - 350)
x -= (3)
x -= 3;
x = x - 3;
And as per basic mathematics:
x -= (353 - 350)
x += -(353 - 350)
x += (-353 + 350)
x += (-3);
I've tested it in Python. It appears to be just basic arithmetics..
Assuming x is currently zero
x -= 353 - 350 which is equal to saying x = x - (353 - 350).
Get the result of brackets first then equation is simply x = 0 - 3 which gives result : -3
Daniel and Marc are right.
You use the wrong operator :
To get -703 you should use -= like this
var x:int = 0;
x -= 353 + 350;
// x -= (353 + 350);
// So => x -= 703
trace(x); //output -703
Because
x -= 353 - 350;
Means x -= 3;
Same logic.
Related
I am reading "Numerical Methods for Engineers" by Chapra and Canale. In it, they've provided pseudocode for the implementation of Euler's method (for solving ordinary differential equations). Here is the pseucode:
Pseucode for implementing Euler's method
I tried implementing this code in GNU Octave, but depending on the input values, I am getting one of two errors:
The program doesn't give any output at all. I have to press 'Ctrl + C' in order to break execution.
The program gives this message:
error: 'ynew' undefined near line 5 column 21
error: called from
Integrator at line 5 column 9
main at line 18 column 7
I would be very grateful if you could get this program to work for me. I am actually an amateur in GNU Octave. Thank you.
Edit 1: Here is my code. For main.m:
%prompt user
y = input('Initial value of y:');
xi = input('Initial value of x:');
xf = input('Final value of x:');
dx = input('Step size:');
xout = input('Output interval:');
x = xi;
m = 0;
xpm = x;
ypm = y;
while(1)
xend = x + xout;
if xend > xf
xend = xf;
h = dx;
Integrator(x,y,h,xend);
m = m + 1;
xpm = x;
ypm = y;
if x >= xf
break;
endif
endif
end
For Integrator.m:
function Integrator(x,y,h,xend)
while(1)
if xend - x < h
h = xend - x;
Euler(x,y,h,ynew);
y = ynew;
if x >= xend
break;
endif
endif
end
endfunction
For Euler.m:
function Euler(x,y,h,ynew)
Derivs(x,y,dydx);
ynew = y + dydx * h;
x = x + h;
endfunction
For Derivs.m:
function Derivs(x,y,dydx)
dydx = -2 * x^3 + 12 * x^2 - 20 * x + 8.5;
endfunction
Edit 2: I shoud mention that the differential equation which Chapra and Canale have given as an example is:
y'(x) = -2 * x^3 + 12 * x^2 - 20 * x + 8.5
That is why the 'Derivs.m' script shows dydx to be this particular polynomial.
Here is my final code. It has four different M-files:
main.m
%prompt the user
y = input('Initial value of y:');
x = input('Initial value of x:');
xf = input('Final value of x:');
dx = input('Step size:');
xout = dx;
%boring calculations
m = 1;
xp = [x];
yp = [y];
while x < xf
[x,y] = Integrator(x,y,dx,min(xf, x+xout));
m = m+1;
xp(m) = x;
yp(m) = y;
end
%plot the final result
plot(xp,yp);
title('Solution using Euler Method');
ylabel('Dependent variable (y)');
xlabel('Independent variable (x)');
grid on;
Integrator.m
%This function takes in 4 inputs (x,y,h,xend) and returns 2 outputs [x,y]
function [x,y] = Integrator(x,y,h,xend)
while x < xend
h = min(h, xend-x);
[x,y] = Euler(x,y,h);
end
endfunction
Euler.m
%This function takes in 3 inputs (x,y,h) and returns 2 outputs [x,ynew]
function [x,ynew] = Euler(x,y,h)
dydx = Derivs(x,y);
ynew = y + dydx * h;
x = x + h;
endfunction
Derivs.m
%This function takes in 2 inputs (x,y) and returns 1 output [dydx]
function [dydx] = Derivs(x,y)
dydx = -2 * x^3 + 12 * x^2 - 20 * x + 8.5;
endfunction
Your functions should look like
function [x, y] = Integrator(x,y,h,xend)
while x < xend
h = min(h, xend-x)
[x,y] = Euler(x,y,h);
end%while
end%function
as an example. Depending on what you want to do with the result, your main loop might need to collect all the results from the single steps. One variant for that is
m = 1;
xp = [x];
yp = [y];
while x < xf
[x,y] = Integrator(x,y,dx,min(xf, x+xout));
m = m+1;
xp(m) = x;
yp(m) = y;
end%while
I wrote following code for gradientDescent in Octave in .m file as follows:
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
% Test values:
X = [1 5; 1 2; 1 4; 1 5];
y = [1 6 4 2]';
theta = [0 0]';
alpha = 0.01;
num_iters = 1000;
% Initialize some useful values:
m = length(y); % number of training examples
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
x = X(:,2);
h = theta(1) + (theta(2)*x);
theta_zero = theta(1) - alpha * (1/m) * sum(h-y);
theta_one = theta(2) - alpha * (1/m) * sum((h - y) .* x);
theta = [theta_zero; theta_one];
% ============================================================
% Save the cost J in every iteration
J_history(iter) = computeCost(X, y, theta); % History of J
end
disp(min(J_history));
end
% Code for computeCost function is as follows:
function J = computeCost(X, y, theta)
data =
6.1101 17.5920
5.5277 9.1302
8.5186 13.6620
7.0032 11.8540
5.8598 6.8233
8.3829 11.8860
7.4764 4.3483
8.5781 12.0000
6.4862 6.5987
m = length(y);
J = 0;
X = data(:, 1);
y = data(:, 2);
predictions = X*theta'; % predictions of hypothesis on examples
sqrErrors = (predictions - y).^2; % squared errors
J = 1/(2*m) * sum(sqrErrors);
end
When I run this from octave workspace I get the following error:
Error: A(I) = X: X must have the same size as I
error: called from
gradientDescent at line 55 column 21
I tried many things but unsuccessfully and mentors never replied properly.
Can you please tell me where I may be making a mistake.
Thanks in advance.
Bharat.
I got an object (called tempEnemy) which is flying around and shooting.
The problem is that I can't keep the value tempEnemy.rotateTo positive, i.e.
it shall be between 0 and 359 degrees. Currently rotateTo ranges from:
rotateTo < 0 (bug) && rotateTo > 0 && rotateTo > 359 (bug).
tempEnemy.dX = tempEnemy.destX - tempEnemy.x;
tempEnemy.dY = tempEnemy.destY - tempEnemy.y;
//I added 180 because my tempEnemy object was looking and shooting to the wrong direction
tempEnemy.rotateTo = (toDegrees(getRadians(tempEnemy.dX, tempEnemy.dY))) + 180;
if (tempEnemy.rotateTo > tempEnemy.frame + 180) tempEnemy.rotateTo -= 360;
if (tempEnemy.rotateTo < tempEnemy.frame - 180) tempEnemy.rotateTo += 360;
tempEnemy.incFrame = int((tempEnemy.rotateTo - tempEnemy.frame) / tempEnemy.rotateSpeed);
You can always use the modulo operator (%) to keep a value positive. The module calculates the rest of a division.
E.g. (example works with integers there for a division always has a left over.)
19 % 5 = 4
Because in the number 19 5 only fits 3 times (3 * 5 = 15,, 4 * 5 = 20,, 20 is too high) the left over is 4 (19 - 15). That is the modulo.
Extra examples:
7 % 3 = 1
15 % 4 = 3
21 % 9 = 3
The output of a modulo operation is never higher then the right hand operator - 1 There for it is perfect for your problem.
If your object is rotated 1234 degrees,, then operate it with a modulo 360 to get the respective number between 0 and 360 for it.
1234 % 360 = 154
Other more easier examples:
720 % 360 = 0
360 % 360 = 0
540 % 360 = 180
-180 % 360 = 180
-720 % 360 = 0
-540 % 360 = 180
Sounds like a classic angle averaging issue. Here's a formula that works for averaging angles
private function averageNums($a:Number, $b:Number):Number {
return = (Math.atan2( Math.sin($a) + Math.sin($b) , Math.cos($a) + Math.cos($b) ));
}
I have an equation for a parabolic curve intersecting a specified point, in my case where the user clicked on a graph.
// this would typically be mouse coords on the graph
var _target:Point = new Point(100, 50);
public static function plot(x:Number, target:Point):Number{
return (x * x) / target.x * (target.y / target.x);
}
This gives a graph such as this:
I also have a series of line segments defined by start and end coordinates:
startX:Number, startY:Number, endX:Number, endY:Number
I need to find if and where this curve intersects these segments (A):
If it's any help, startX is always < endX
I get the feeling there's a fairly straight forward way to do this, but I don't really know what to search for, nor am I very well versed in "proper" math, so actual code examples would be very much appreciated.
UPDATE:
I've got the intersection working, but my solution gives me the coordinate for the wrong side of the y-axis.
Replacing my target coords with A and B respectively, gives this equation for the plot:
(x * x) / A * (B/A)
// this simplifies down to:
(B * x * x) / (A * A)
// which i am the equating to the line's equation
(B * x * x) / (A * A) = m * x + b
// i run this through wolfram alpha (because i have no idea what i'm doing) and get:
(A * A * m - A * Math.sqrt(A * A * m * m + 4 * b * B)) / (2 * B)
This is a correct answer, but I want the second possible variation.
I've managed to correct this by multiplying m with -1 before the calculation and doing the same with the x value the last calculation returns, but that feels like a hack.
SOLUTION:
public static function intersectsSegment(targetX:Number, targetY:Number, startX:Number, startY:Number, endX:Number, endY:Number):Point {
// slope of the line
var m:Number = (endY - startY) / (endX - startX);
// where the line intersects the y-axis
var b:Number = startY - startX * m;
// solve the two variatons of the equation, we may need both
var ix1:Number = solve(targetX, targetY, m, b);
var ix2:Number = solveInverse(targetX, targetY, m, b);
var intersection1:Point;
var intersection2:Point;
// if the intersection is outside the line segment startX/endX it's discarded
if (ix1 > startX && ix1 < endX) intersection1 = new Point(ix1, plot(ix1, targetX, targetY));
if (ix2 > startX && ix2 < endX) intersection2 = new Point(ix2, plot(ix2, targetX, targetY));
// somewhat fiddly code to return the smallest set intersection
if (intersection1 && intersection2) {
// return the intersection with the smaller x value
return intersection1.x < intersection2.x ? intersection1 : intersection2;
} else if (intersection1) {
return intersection1;
}
// this effectively means that we return intersection2 or if that's unset, null
return intersection2;
}
private static function solve(A:Number, B:Number, m:Number, b:Number):Number {
return (m + Math.sqrt(4 * (B / (A * A)) * b + m * m)) / (2 * (B / (A * A)));
}
private static function solveInverse(A:Number, B:Number, m:Number, b:Number):Number {
return (m - Math.sqrt(4 * (B / (A * A)) * b + m * m)) / (2 * (B / (A * A)));
}
public static function plot(x:Number, targetX:Number, targetY:Number):Number{
return (targetY * x * x) / (targetX * targetX);
}
Or, more explicit yet.
If your parabolic curve is
y(x)= A x2+ B x + C (Eq 1)
and your line is
y(x) = m x + b (Eq 2)
The two possible solutions (+ and -) for x are
x = ((-B + m +- Sqrt[4 A b + B^2 - 4 A C - 2 B m + m^2])/(2 A)) (Eq 3)
You should check if your segment endpoints (in x) contains any of these two points. If they do, just replace the corresponding x in the y=m x + b equation to get the y coordinate for the intersection
Edit>
To get the last equation you just say that the "y" in eq 1 is equal to the "y" in eq 2 (because you are looking for an intersection!).
That gives you:
A x2+ B x + C = m x + b
and regrouping
A x2+ (B-m) x + (C-b) = 0
Which is a quadratic equation.
Equation 3 are just the two possible solutions for this quadratic.
Edit 2>
re-reading your code, it seems that your parabola is defined by
y(x) = A x2
where
A = (target.y / (target.x)2)
So in your case Eq 3 becomes simply
x = ((m +- Sqrt[4 A b + m^2])/(2 A)) (Eq 3b)
HTH!
Take the equation for the curve and put your line into y = mx +b form. Solve for x and then determine if X is between your your start and end points for you line segment.
Check out: http://mathcentral.uregina.ca/QQ/database/QQ.09.03/senthil1.html
Are you doing this often enough to desire a separate test to see if an intersection exists before actually computing the intersection point? If so, consider the fact that your parabola is a level set for the function f(x, y) = y - (B * x * x) / (A * A) -- specifically, the one for which f(x, y) = 0. Plug your two endpoints into f(x,y) -- if they have the same sign, they're on the same side of the parabola, while if they have different signs, they're on different sides of the parabola.
Now, you still might have a segment that intersects the parabola twice, and this test doesn't catch that. But something about the way you're defining the problem makes me feel that maybe that's OK for your application.
In other words, you need to calulate the equation for each line segment y = Ax + B compare it to curve equation y = Cx^2 + Dx + E so Ax + B - Cx^2 - Dx - E = 0 and see if there is a solution between startX and endX values.
Out of curiosity, is there a (language independent*) way to make these two generic statements into one statement?
if (a < b) a += x;
else if (a > b) a -= x;
May also be used as
if (abs(b - a) < x) a = b;
else if (a < b) a += x;
else if (a > b) a -= x;
Oh, now that I'm writing this I'm thinking of something like this:
if (a != b) a += x * (a < b ? 1 : -1);
But that doesn't really benefit the readability. Other ways perhaps?
= Normal languages, not Brainfuck and the likes or really ancient ones
a += x * sgn(b-a)
Where sgn is the sign function:
sgn x | x > 0 = 1
| x < 0 = -1
| otherwise = 0
You should really wonder why you want this, the compiler will do fine optimizing and the given statement is very readable.
Language independent is a bit tricky, but if you have cmp or similar, you can use that:
a += x * cmp(b, a)
cmp(b, a) returns:
0 if b==a
1 if b > a
-1 if a < b
Also, there is a bug in your suggested implementation:
a = 9;
b = 10;
x = 2;
if (a < b) a += x;
if (a > b) a -= x;
print a;
Output: 9 (expected 11)
You need an else to fix this. Alternatively, if you want to prevent a going past b, do this:
if (a < b)
{
a = min(b, a + x);
}
else if (a > b)
{
a = max(b, a - x);
}
If you want to do this in one expression you can:
a = (a < b) ? min(b, a + x) : max(b, a - x);
I think the first way is clearer though.
a += (b - a) / abs(b - a) * x
But this is not language independent and does not really help readiness if you ask me....