newTheta Vector is of size 4
-6.3702
0.5712
0.2067
0.3877
It can plot:
ezplot("-6.3702 + 0.5712*x + 0.2067*y + 0.3877*x*y", [-10 10]);
But the problem occurs when I pass values of newTheta vector into the equation.
Octave can plot the following command:
ezplot(#(x,y) newTheta(1) + newTheta(2)*x + newTheta(3)*y, [-10 10]);
But does not plot this command:
ezplot(#(x,y) newTheta(1) + newTheta(2)*x + newTheta(3)*y + newTheta(4)*x *y, [-10 10])
Consider using a dot, .*, when multiplying for vectorization because x and y are vectors. This should work:
newTheta = [-6.3702
0.5712
0.2067
0.3877];
ezplot(#(x,y) newTheta(1) + newTheta(2)*x + newTheta(3)*y + newTheta(4)*x .*y, [-10 10])
Related
I implemented a custom loss function, which looks like this:
However, the gradient of this function is always zero and I don't understand why.
The code for the objective function:
def objective(p, output):
x,y = p
a = minA
b = minB
r = 0.1
XA = 1/2 -1/2 * torch.tanh(100*((x - a[0])**2 + (y - a[1])**2 - (r + 0.02)**2))
XB = 1/2 -1/2 * torch.tanh(100*((x - b[0])**2 + (y - b[1])**2 - (r + 0.02)**2))
q = (1-XA)*((1-XB)* output + (XB))
output_grad, _ = torch.autograd.grad(q, (x,y))
output_grad.requires_grad_()
q = output_grad**2
return q
And the code for training the model (which is a simple, fully connected NN):
model = NN(input_size)
optimizer = optim.SGD(model.parameters(), lr=learning_rate)
for e in range(epochs) :
for configuration in total:
print("Train for configuration", configuration)
# Training pass
optimizer.zero_grad()
#output is q~
output = model(configuration)
#loss is the objective function we defined
loss = objective(configuration, output.item())
loss.backward()
optimizer.step()
I really think the problem is in the output_grad, _ = torch.autograd.grad(q, (x,y)).
(During he training, "configuration" is a point sampled from a distribution identified by the coordinates x and y).
Thanks!!
Here I provide the code on a google colab session:
Google colab
Tanh is a bounded function and converges quite quickly to 1. Your XA and XB points are defined as
XA = 1/2 - 1/2 * torch.tanh(100*(z1 + z2 - z0))
XB = 1/2 - 1/2 * torch.tanh(100*(z3 + z4 - z0))
Since z1 + z2 - z0 and z3 + z4 - z0 are rather close to 1, you will end up with an input close to 100. This means the tanh will output 1, resulting in XA and XB begin zeros. You might not want to have this 100 coefficient if you want to have non zero outputs.
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
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
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);
I would like to plot the shifted logistic function as shown from Wolfram Alpha.
In particular, I would like the function to be of the form
y = exp(x - t) / (1 + exp(x - t))
where t > 0. In the link, for example, t is 6. I had originally tried the following:
x = 0:.1:12;
y = exp(x - 6) ./ (1 + exp(x - 6));
plot(x, y);
axis([0 6 0 1])
However, this is not the same as the result from Wolfram Alpha. Here is an export of my plot.
I do not understand what the difference is between what I am trying to do here vs. plotting shifted sin and cosine functions (which works using the same technique).
I am not completely new to Matlab but I do not usually use it in this way.
Edit: My values for x in the code should have been from 0 to 12.
fplot takes as inputs a function handle and a range to plot for:
>> fplot(#(x) exp(x-6) / (1 + exp(x-6)), [0 12])
The beauty of fplot in this case is you don't need to spend time calculating y-values beforehand; you could also extract values from the graph after the fact if you want (by getting the line handle's XData and YData properties).
Your input to Wolfram Alpha is incorrect. It is interpreted as e*(x-6)/(1-e*(x-6)). Use plot y = exp(x - 6) / (1 + exp(x - 6)) for x from 0 to 12 in Wolfram Alpha (see here) for the same results as in MATLAB. Also use axis([0 12 0 1]) (or no axis statement at all on a new plot) to see the full results in MATLAB.
In reply to your comment: use y = exp(1)*(x - 6) ./ (1 + exp(1)*(x - 6)); to do in MATLAB what you were doing in Wolfram Alpha.