3dsMAX Wire Param. expression: How to specify moving in positive direction? - syntax

I have a wire parameter expression that I am trying to make do one thing when the value is moving in the Negative X direction and another when moving in the Positive X. How do I specify this?
if <moving in positive direction>
then A
else B
I assume it would be something like if the new difference from zero is greater or less than the current one. That would work, but only if you stay in either negative or positive x-space. I just do not know how to write this.
I apologize if this is simple math. I am a bit new to this.

You will need a more advanced approach than wire parameters. Using a script controller is the easiest way. It has to save current position in a variable, let's say prevPos (and the same for frame number, if you want it to work correctly when scrubbing timeslider backwards) - which will be the last step, as you want to first subtract the prevPos from current pos and if the result is positive, do A, else do B.

Related

Maple: How to define new elements of a matrix?

In Maple, I have a matrix N and its elements N[i,j], If I modify the elements of this matrix as follows for example
>for j from 1 to 4 do
>print(F[i,j]=(diff(N[i,j],x)));
>od;od;
where the matrix elements are functions of x.
I've wanted to define new matrix elements
>BA[i,j]:=(diff(N[i,j],x)));
but I can't do this with Maple, through the above command. Can someone help me ?
Better than using a loop is simply BA:= diff~(N,x). The ~ can be appended to any operator to mean "apply the operator to each member of the container and return a new container containing the modified members."
Also, be careful about using print. Its only purpose is to print stuff on the screen from the middle (not the end) of a computation. It can't be used to change any stored values. Good programs use print very sparingly, if at all. The end result of a computation is displayed automatically, without needing a print command.

How to check if an Eigen::Matrix4f is close to identity matrix?

Is there a good practice to check if my result Eigen::Matrix4f is almost identity? Since due to floating point errors I don't get some times exactly only zeros and ones.
One brute force method would be, to check each value in the matrix if it is between certain EPSILON and if just one of them fails, then it is not an identity matrix. Is there a better solution?
First, you have to define in what sense they shall be "close". There can be many different definitions of closeness, depending on your specific task. One of the most used is:
norm( A - I ) < eps
where norm is some matrix norm. Most common are 2-norm, 1-norm, inf-norm and Frobenius norm.
Your method is also possible. It is equivalent to the method above with max-norm (where norm(A) = max abs Aij). It can be implemented in Eigen using:
(A - Matrix4f::Identity()).cwiseAbs().max() < eps;
Update:
Actually, in Eigen there is a special method to check that: isIdentity. You give it the threshold value:
A.isIdentity(eps)

Trouble implementing Perceptron in Scala

I'm taking the CalTech online course Learning From Data, and I'm stumped with creating a Perceptron in Scala. I chose Scala because I'm learning it and wanted to challenge myself. I understand the theory, and I also understand others' solutions in Python and Ruby. But I can't figure out why my own Scala code doesn't work.
For a background in the Perceptron code: Learning_algorithm
I'm running Scala 2.11 on OSX 10.10.
Per the algorithm, I start off with weights (0.0, 0.0, 0.0), where weight[2] is a learned bias component. I've already generated a test set in the space [-1, 1],[-1,1] on the X-Y plane. I do this by a) picking two random points and drawing a line through them, then b) generating some other random points and calculating if they are on one side of the line or the other. As far as I can tell by plotting it in Python, this generates linearly separable data.
My next step is to take my initialized weights and check against every point to find miss-classified points, i.e. points that don't generate the right +1 or -1 result. Here is the code that simply calculates dot-product of the weight and the vector x:
def h(weight:List[Double], p:Point ): Double = if ( (weight(0)*p.x + weight(1)*p.y + weight(2)) > 0) 1 else -1
It's the initial weights, so they are all miss-classified. I then update the weights, like so:
def newH(weight:List[Double], p:Point, y:Double): List[Double] = {
val newWt = scala.collection.mutable.ArrayBuffer[Double](0.0, 0.0, 0.0)
newWt(0) = weight(0) + p.x*y
newWt(1) = weight(1) + p.y*y
newWt(2) = weight(2) + 1*y
return newWt.toList
}
Then I identify miss-classified points again by checking the test set against the value output by h() above, and continue iterating.
This follows the algorithm (or is supposed to, at least) that Prof Yaser shows here: Library
The problem is that the algorithm never converges. My weights -- the third component of which is the bias -- keep getting more negative or more positive. My weight vector after every adjustment resembles this:
Weights: List(16.43341624736786, 11627.122008800507, -34130.0)
Weights: List(15.533397436141968, 11626.464265227318, -34131.0)
Weights: List(14.726969361305237, 11626.837346673012, -34132.0)
Weights: List(14.224745154380798, 11627.646470665932, -34133.0)
Weights: List(14.075232982635498, 11628.026384592056, -34134.0)
I'm a Scala newbie so my code is probably atrocious. But am I missing something in Scala, e.g. reassignment, that could be causing my weight to be messed up? Or have I completely misunderstood how the Perceptron even operates? Is my weight update just wrong?
Thanks for any help you can give me on this!
Thanks Till. I've discovered the two problems with my code and I'll share them, but to address your point: Someone else asked about this on the class's forum and it looks like what the Wiki formula does is simply to change the learning rate. Alpha can be picked randomly, and y-h(weight, p) would give you weights like
-1-1 = 2
In the case that y=-1 and h()=1, or
1-(-1) = 2
In the case that y=1 and h()=-1
My/the class formula takes 1*p.x instead of alpha*2, which seems to be a matter of different learning rates. Hope that makes sense.
My two problems were as follows:
The y value passed into the recalculation formula newH needs to be the target value of y, that is, the "correct y" that was discovered while generating the test points. I was passing in the y that was generated through h(), which is the guessed-at function. This makes sense obviously since we are looking to correc the weight by using the target y, not the incorrect y.
I was doing a comparison of target y and h()=yin Scala, but was comparison an element obtained from a map through .get(). My Scala map looks like Map[Point,Double] where the Double value refers to the y value generated during the test set creation. But doing a .get() gives you Option[Double] and not a Double value at all. This is explained in Scala Map#get and the return of Some() and makes a lot of sense now. I did map.get(<some Point>).get() for now, since I was focusing on debugging and not code perfection, and then I was accurately able to compare two Double values.

How to find the maximum value of a function in Sympy?

These days I am trying to redo shock spectrum of single degree of freedom system using Sympy. The problem can reduce to find maximum value of a function. Following are two cases I cannot figure out how to do.
The first one is
tau,t,t_r,omega,p0=symbols('tau,t,t_r,omega,p0',positive=True)
h=expand(sin(omega*(t-tau)))
f=simplify(integrate(p0*tau/t_r*h,(tau,0,t_r))+integrate(p0*h,(tau,t_r,t)))
The final goal is to obtain maximum absolute value of f (The variable is t). The direct way is
df=diff(f,t)
sln=solve(simplify(df),t)
simplify(f.subs(t,sln[1]))
Here is the result, I tried many ways, but I can not simplify any further.
Therefore, I tried another way. Because I need the maximum absolute value and the location where abs(f) is maximum happens at the same location of square of f, we can calculate square of f first.
df=expand_trig(diff(expand(f)**2,t))
sln=solve(df,t)
simplify(f.subs(t,sln[2]))
It seems the answer is almost the same, just in another form.
The expected answer is a sinc function plus a constant as following:
Therefore, the question is how to get the final presentation.
The second one may be a little harder. The question can be reduced to find the maximum value of f=sin(pi*t/t_r)-T/2/t_r*sin(2*pi/T*t), in which t_r and T are two parameters. The maximum located at different peak when the ratio of t_r and T changes. And I do not find a way to solve it in Sympy. Any suggestion? The answer can be represented in following figure.
The problem is the log(exp(I*omega*t_r/2)) term. SymPy is not reducing this to I*omega*t_r/2. SymPy doesn't simplify this because in general, log(exp(x)) != x, but rather log(exp(x)) = x + 2*pi*I*n for some integer n. But in this case, if you replace log(exp(I*omega*t_r/2)) with omega*t_r/2 or omega*t_r/2 + 2*pi*I*n, it will be the same, because it will just add a 2*pi*I*n inside the sin.
I couldn't figure out any functions that force this simplification, but the easiest way is to just do a substitution:
In [18]: print(simplify(f.subs(t,sln[1]).subs(log(exp(I*omega*t_r/2)), I*omega*t_r/2)))
p0*(omega*t_r - 2*sin(omega*t_r/2))/(omega**2*t_r)
That looks like the answer you are looking for, except for the absolute value (I'm not sure where they should come from).

AI: selecting immediate acceleration/rotation to get to a final point

I'm working on a game where on each update of the game loop, the AI is run. During this update, I have the chance to turn the AI-controlled entity and/or make it accelerate in the direction it is facing. I want it to reach a final location (within reasonable range) and at that location have a specific velocity and direction (again it doesn't need to be exact) That is, given a current:
P0(x, y) = Current position vector
V0(x, y) = Current velocity vector (units/second)
θ0 = Current direction (radians)
τmax = Max turn speed (radians/second)
αmax = Max acceleration (units/second^2)
|V|max = Absolute max speed (units/second)
Pf(x, y) = Target position vector
Vf(x, y) = Target velocity vector (units/second)
θf = Target rotation (radians)
Select an immediate:
τ = A turn speed within [-τmax, τmax]
α = An acceleration scalar within [0, αmax] (must accelerate in direction it's currently facing)
Such that these are minimized:
t = Total time to move to the destination
|Pt-Pf| = Distance from target position at end
|Vt-Vf| = Deviation from target velocity at end
|θt-θf| = Deviation from target rotation at end (wrapped to (-π,π))
The parameters can be re-computed during each iteration of the game loop. A picture says 1000 words so for example given the current state as the blue dude, reach approximately the state of the red dude within as short a time as possible (arrows are velocity):
Pic http://public.blu.livefilestore.com/y1p6zWlGWeATDQCM80G6gaDaX43BUik0DbFukbwE9I4rMk8axYpKwVS5-43rbwG9aZQmttJXd68NDAtYpYL6ugQXg/words.gif
Assuming a constant α and τ for Δt (Δt → 0 for an ideal solution) and splitting position/velocity into components, this gives (I think, my math is probably off):
Equations http://public.blu.livefilestore.com/y1p6zWlGWeATDTF9DZsTdHiio4dAKGrvSzg904W9cOeaeLpAE3MJzGZFokcZ-ZY21d0RGQ7VTxHIS88uC8-iDAV7g/equations.gif
(EDIT: that last one should be θ = θ0 + τΔt)
So, how do I select an immediate α and τ (remember these will be recomputed every iteration of the game loop, usually > 100 fps)? The simplest, most naieve way I can think of is:
Select a Δt equal to the average of the last few Δts between updates of the game loop (i.e. very small)
Compute the above 5 equations of the next step for all combinations of (α, τ) = {0, αmax} x {-τmax, 0, τmax} (only 6 combonations and 5 equations for each, so shouldn't take too long, and since they are run so often, the rather restrictive ranges will be amortized in the end)
Assign weights to position, velocity and rotation. Perhaps these weights could be dynamic (i.e. the further from position the entity is, the more position is weighted).
Greedily choose the one that minimizes these for the location Δt from now.
Its potentially fast & simple, however, there are a few glaring problems with this:
Arbitrary selection of weights
It's a greedy algorithm that (by its very nature) can't backtrack
It doesn't really take into account the problem space
If it frequently changes acceleration or turns, the animation could look "jerky".
Note that while the algorithm can (and probably should) save state between iterations, but Pf, Vf and θf can change every iteration (i.e. if the entity is trying to follow/position itself near another), so the algorithm needs to be able to adapt to changing conditions.
Any ideas? Is there a simple solution for this I'm missing?
Thanks,
Robert
sounds like you want a PD controller. Basically draw a line from the current position to the target. Then take the line direction in radians, that's your target radians. The current error in radians is current heading - line heading. Call it Eh. (heading error) then you say the current turn rate is going to be KpEh+d/dt EhKd. do this every step with a new line.
thats for heading
acceleration is "Accelerate until I've reached max speed or I wont be able to stop in time". you threw up a bunch of integrals so I'm sure you'll be fine with that calculation.
I case you're wondering, yes I've solved this problem before, PD controller works. don't bother with PID, don't need it in this case. Prototype in matlab. There is one thing I've left out, you need to have a trigger, like "i'm getting really close now" so I should start turning to get into the target. I just read your clarification about "only accelerating in the direction we're heading". that changes things a bit but not too much. that means to need to approach the target "from behind" meaning that the line target will have to be behind the real target, when you get near the behind target, follow a new line that will guide you to the real target. You'll also want to follow the lines, rather than just pick a heading and try to stick with it. So don't update the line each frame, just say the error is equal to the SIGNED DISTANCE FROM THE CURRENT TARGET LINE. The PD will give you a turn rate, acceleration is trivial, so you're set. you'll need to tweak Kd and Kp by head, that's why i said matlab first. (Octave is good too)
good luck, hope this points you in the right direction ;)
pun intended.
EDIT: I just read that...lots of stuff, wrote real quick. this is a line following solution to your problem, just google line following to accompany this answer if you want to take this solution as a basis to solving the problem.
I would like to suggest that yout consider http://en.wikipedia.org/wiki/Bang%E2%80%93bang_control (Bang-bang control) as well as a PID or PD. The things you are trying to minimise don't seem to produce any penalty for pushing the accelerator down as far as it will go, until it comes time to push the brake down as far as it will go, except for your point about how jerky this will look. At the very least, this provides some sort of justification for your initial guess.

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