I'm implementing the system detailed in this paper.
On page 3, section 4 it shows the form that tensors take within the system:
R [ cos(2t), sin(2t); sin(2t), -cos(2t) ]
In my system, I only store R and t, since everything can be calculated from them.
However, I've got to the point where I need to sum two of these tensors (page 4, section 5.2). How can I find values for R and t after summing two tensors of this form?
I guess that's what you are looking for:
x = R_1*cos(2*t_1) + R_2*cos(2*t_2)
y = R_1*sin(2*t_1) + R_2*sin(2*t_2)
R_result = sqrt(x*x+y*y)
t_result = atan2(y,x)/2
Each term reduces to
R_1 trg(2 t_1) + R_2 trg(2 t_2) = R_1 trg_1 + R_2 trg_2
where trg represents either sin or cos and the indexed version takes the obvious meaning. So this is a just an ordinary problem in trigonometric identities repeated a couple of times.
Let
Q = (R_1 + R_2)/2
S = (R_1 - R_2)/2
then
R_1 trg(2 t_1) + R_2 trg(2 t_2) = (Q+S)(trg_1 + trg_2) + (Q-S)(trg_1 - trg_2)
which involves identities you can look up.
Sorry, adding two tensors is nothing more than algebra. The two matricies have to be the same size, and you add them term by term.
You can't just add the radii and angles and plug them back into the tensor. Do the addition properly and it'll work. Here's the first term:
R1*cost(2t1) + R2*cos(2t2) = ?
Here's the answer from Wolfram Alpha. As you can see, it doesn't simplify into a nice, neat expression with an R and a T for you.
In case you haven't thought of it, put the tensor sum into Wolfram Alpha and see what it gives you. They're better at algebra than anyone at this site. Why not get an independent check of your work?
Related
In the beginning of the function I calculate the total weight of a protein sequence and define it as seq_weight.
After that I calculate the weight of several fragments and make combinations of those weights that sum to the total weight of the first proteins sequence.
The first print statement prints the total weight correctly, but near the end of the function it seems to forget that value when I want to define it as the result of the sum.
When I type the value manually I get the result I want:
def fragmentcombinations(sequence, fragments):
for seq_rec in sequence:
seq_weight = 0.0
for i in seq_rec.seq:
seq_weight += SeqUtils.molecular_weight(i, "protein")
print("The protein sequence: " + seq_rec.seq)
print("The molecular weight: " + str(round(seq_weight, 2)) + " Da.")
nums = []
for a in fragments:
fragment_weights = 0.0
for aa in a.seq:
fragment_weights += SeqUtils.molecular_weight(aa, 'protein')
nums.append(round(fragment_weights, 2))
print(nums)
weights_array = []
combs = []
if len(nums) > 0:
for r in range(0,len(nums)+1):
weights_array += list(combinations(nums, r))
for item in weights_array:
if sum(item) == 4364.85: #Sequence weight needs to inserted manually -> not ideal
combs.append(item)
print(" ")
print("The possible combinations of fragment weights that may cover the protein sequence without overlap are: ")
for row in combs:
print(*row, sep = ", ")
fragmentcombinations(seq_list3, seq_list4)
This is the result:
The protein sequence: IEEATHMTPCYELHGLRWVQIQDYAINVMQCL
The molecular weight: 4364.85 Da.
[3611.86, 2269.63, 469.53, 556.56, 1198.41, 2609.88, 547.69, 1976.23, 2306.48, 938.01, 1613.87, 789.87, 737.75, 2498.71, 2064.25, 1184.39, 1671.87]
The possible combinations of fragment weights that may cover the protein sequence without overlap are:
556.56, 1198.41, 2609.88
469.53, 2609.88, 547.69, 737.75
556.56, 1198.41, 938.01, 1671.87
469.53, 547.69, 938.01, 737.75, 1671.87
If I write
if sum(item) == seq_weight:
the result doesn't print the combination of weights like I intended.
Sorry if the code is kind of messy, I'm still a beginner.
Thanks in advance!
The problem is not that your variable is not remembered anymore. The problem is that you perform an exact comparison between floating point numbers. In programming floating point numbers are the "decimal" numbers, but they are not the exact presentation of your numbers. They only are up to an arbitrary precision.
Let's do some basic maths Python.
>>> a = 0.2 + 0.1
>>> a
0.30000000000000004
>>> a == 0.3
False
As you can see, there is clearly happening something weird here. But this is just how floating point arithmetic works.
Now we have explained that. What should you do to make your program work? There are multiple solutions.
One way to deal with it, is to compare your numbers to some fixed difference. ie
if abs(sum(item) - seq_weight) < 0.00001
Another way to deal with this is using fixed precision decimal objects, but that can be more difficult than you think it is. https://docs.python.org/3/library/decimal.html
If rewriting
x = 2
y = 2
z = x + y
to
z = 2 + 2
can be described by 'hardcoding' ('hard-coding'?) values of x and y,
what's do you call its opposite, e.g.
z = 2 + 2
to
x = 2
y = 2
z = x + y
(I've never heard of 'softcoding', so I assume it's inappropriate)?
#melpomene's comment is correct, both of your examples are hard-coding.
Anyway, in my team we call this process parameterization, e.g. "we need to parameterize the threshold value used in function X", though I'm not sure if this is a common term.
For what it's worth, there's actually a Wikipedia entry on Softcoding, which means what you asked for, but in a negative connotation:
The term is generally used where softcoding becomes an anti-pattern. Abstracting too many values and features can introduce more complexity and maintenance issues than would be experienced with changing the code when required.
Pytorch provides a good example of using actor-critic to play Cartpole in the OpenAI gym environment.
I'm confused about several of their equations in the code snippet found at https://github.com/pytorch/examples/blob/master/reinforcement_learning/actor_critic.py#L67-L79:
saved_actions = model.saved_actions
value_loss = 0
rewards = []
for r in model.rewards[::-1]:
R = r + args.gamma * R
rewards.insert(0, R)
rewards = torch.Tensor(rewards)
rewards = (rewards - rewards.mean()) / (rewards.std() + np.finfo(np.float32).eps)
for (action, value), r in zip(saved_actions, rewards):
action.reinforce(r - value.data.squeeze())
value_loss += F.smooth_l1_loss(value, Variable(torch.Tensor([r])))
optimizer.zero_grad()
final_nodes = [value_loss] + list(map(lambda p: p.action, saved_actions))
gradients = [torch.ones(1)] + [None] * len(saved_actions)
autograd.backward(final_nodes, gradients)
optimizer.step()
What do r and value mean in this case? Why do they run REINFORCE on the action space with the reward equal to r - value? And why do they try to set the value so that it matches r?
Thanks for your help!
First the rewards a collected for a time, along with the state:action that resulted in the reward
Then r - value is the difference between the expected reward and actual
That difference is used to adjust the expected value of that action from that state
So if in state "middle", the expected reward for action "jump" was 10 and the actual reward was only 2, then the AI was off by -8 (2-10). Reinforce means "adjust expectations". So if we adjust them by half, we'll new expected reward is 10-(8 *.5), or 6. meaning the AI really thought it would get 10 for that, but now it's less confident and thinks 6 is a better guess. So if the AI is not off by much, 10 - ( 2 *.5) = 9, it will adjust by a smaller amount.
Within my daily work, I have got to maximize a particular function making use of fminsearch; the code is:
clc
clear all
close all
f = #(x,c,k) -(x(2)/c)^3*(((exp(-(x(1)/c)^k)-exp(-(x(2)/c)^k))/((x(2)/c)^k-(x(1)/c)^k))-exp(-(x(3)/c)^k))^2;
c = 10.1;
k = 2.3;
X = fminsearch(#(x) f(x,c,k),[4,10,20]);
It works fine, as I expect, but not the issue is coming up: I need to bound x within certain limits, as:
4 < x(1) < 5
10 < x(2) < 15
20 < x(3) < 30
To achieve the proper results, I should use the optimization toolbox, that I unfortunately cannot hand.
Is there any way to get the same analysis by making use of only fminsearch?
Well, not using fminsearch directly, but if you are willing to download fminsearchbnd from the file exchange, then yes. fminsearchbnd does a bound constrained minimization of a general objective function, as an overlay on fminsearch. It calls fminsearch for you, applying bounds to the problem.
Essentially the idea is to transform your problem for you, in a way that your objective function sees as if it is solving a constrained problem. It is totally transparent. You call fminsearchbnd with a function, a starting point in the parameter space, and a set of lower and upper bounds.
For example, minimizing the rosenbrock function returns a minimum at [1,1] by fminsearch. But if we apply purely lower bounds on the problem of 2 for each variable, then fminsearchbnd finds the bound constrained solution at [2,4].
rosen = #(x) (1-x(1)).^2 + 105*(x(2)-x(1).^2).^2;
fminsearch(rosen,[3 3]) % unconstrained
ans =
1.0000 1.0000
fminsearchbnd(rosen,[3 3],[2 2],[]) % constrained
ans =
2.0000 4.0000
If you have no constraints on a variable, then supply -inf or inf as the corresponding bound.
fminsearchbnd(rosen,[3 3],[-inf 2],[])
ans =
1.4137 2
Andrey has the right idea, and the smoother way of providing a penalty isn't hard: just add the distance to the equation.
To keep using the anonymous function:
f = #(x,c,k, Xmin, Xmax) -(x(2)/c)^3*(((exp(-(x(1)/c)^k)-exp(-(x(2)/c)^k))/((x(2)/c)^k-(x(1)/c)^k))-exp(-(x(3)/c)^k))^2 ...
+ (x< Xmin)*(Xmin' - x' + 10000) + (x>Xmax)*(x' - Xmax' + 10000) ;
The most naive way to bound x, would be giving a huge penalty for any x that is not in the range.
For example:
function res = f(x,c,k)
if x(1)>5 || x(1)<4
penalty = 1000000000000;
else
penalty = 0;
end
res = penalty - (x(2)/c)^3*(((exp(-(x(1)/c)^k)-exp(-(x(2)/c)^k))/((x(2)/c)^k-(x(1)/c)^k))-exp(-(x(3)/c)^k))^2;
end
You can improve this approach, by giving the penalty in a smoother way.
I'm trying to solve Kepler's Equation as a step towards finding the true anomaly of an orbiting body given time. It turns out though, that Kepler's equation is difficult to solve, and the wikipedia page describes the process using calculus. Well, I don't know calculus, but I understand that solving the equation involves an infinite number of sets which produce closer and closer approximations to the correct answer.
I can't see from looking at the math how to do this computationally, so I was hoping someone with a better maths background could help me out. How can I solve this beast computationally?
FWIW, I'm using F# -- and I can calculate the other elements necessary for this equation, it's just this part I'm having trouble with.
I'm also open to methods which approximate the true anomaly given time, periapsis distance, and eccentricity
This paper:
A Practical Method for Solving the Kepler Equation http://murison.alpheratz.net/dynamics/twobody/KeplerIterations_summary.pdf
shows how to solve Kepler's equation using an iterative computing method. It should be fairly straightforward to translate it to the language of your choice.
You might also find this interesting. It's an ocaml program, part of which claims to contain a Kepler Equation solver. Since F# is in the ML family of languages (as is ocaml), this might provide a good starting point.
wanted to drop a reply in here in case this page gets found by anyone else looking for similar materials.
The following was written as an "expression" in Adobe's After Effects software, so it's javascriptish, although I have a Python version for a different app (cinema 4d). The idea is the same: execute Newton's method iteratively until some arbitrary precision is reached.
Please note that I'm not posting this code as exemplary or meaningfully efficient in any way, just posting code we produced on a deadline to accomplish a specific task (namely, move a planet around a focus according to Kepler's laws, and do so accurately). We don't write code for a living, and so we're also not posting this for critique. Quick & dirty is what meets deadlines.
In After Effects, any "expression" code is executed once - for every single frame in the animation. This restricts what one can do when implementing many algorithms, due to the inability to address global data easily (other algorithms for Keplerian motion use interatively updated velocity vectors, an approach we couldn't use). The result the code leaves behind is the [x,y] position of the object at that instant in time (internally, this is the frame number), and the code is intended to be attached to the position element of an object layer on the timeline.
This code evolved out of material found at http://www.jgiesen.de/kepler/kepler.html, and is offered here for the next guy.
pi = Math.PI;
function EccAnom(ec,am,dp,_maxiter) {
// ec=eccentricity, am=mean anomaly,
// dp=number of decimal places
pi=Math.PI;
i=0;
delta=Math.pow(10,-dp);
var E, F;
// some attempt to optimize prediction
if (ec<0.8) {
E=am;
} else {
E= am + Math.sin(am);
}
F = E - ec*Math.sin(E) - am;
while ((Math.abs(F)>delta) && (i<_maxiter)) {
E = E - F/(1.0-(ec* Math.cos(E) ));
F = E - ec * Math.sin(E) - am;
i = i + 1;
}
return Math.round(E*Math.pow(10,dp))/Math.pow(10,dp);
}
function TrueAnom(ec,E,dp) {
S=Math.sin(E);
C=Math.cos(E);
fak=Math.sqrt(1.0-ec^2);
phi = 2.0 * Math.atan(Math.sqrt((1.0+ec)/(1.0-ec))*Math.tan(E/2.0));
return Math.round(phi*Math.pow(10,dp))/Math.pow(10,dp);
}
function MeanAnom(time,_period) {
curr_frame = timeToFrames(time);
if (curr_frame <= _period) {
frames_done = curr_frame;
if (frames_done < 1) frames_done = 1;
} else {
frames_done = curr_frame % _period;
}
_fractime = (frames_done * 1.0 ) / _period;
mean_temp = (2.0*Math.PI) * (-1.0 * _fractime);
return mean_temp;
}
//==============================
// a=semimajor axis, ec=eccentricity, E=eccentric anomaly
// delta = delta digits to exit, period = per., in frames
//----------------------------------------------------------
_eccen = 0.9;
_delta = 14;
_maxiter = 1000;
_period = 300;
_semi_a = 70.0;
_semi_b = _semi_a * Math.sqrt(1.0-_eccen^2);
_meananom = MeanAnom(time,_period);
_eccentricanomaly = EccAnom(_eccen,_meananom,_delta,_maxiter);
_trueanomaly = TrueAnom(_eccen,_eccentricanomaly,_delta);
r = _semi_a * (1.0 - _eccen^2) / (1.0 + (_eccen*Math.cos(_trueanomaly)));
x = r * Math.cos(_trueanomaly);
y = r * Math.sin(_trueanomaly);
_foc=_semi_a*_eccen;
[1460+x+_foc,540+y];
You could check this out, implemented in C# by Carl Johansen
Represents a body in elliptical orbit about a massive central body
Here is a comment from the code
True Anomaly in this context is the
angle between the body and the sun.
For elliptical orbits, it's a bit
tricky. The percentage of the period
completed is still a key input, but we
also need to apply Kepler's
equation (based on the eccentricity)
to ensure that we sweep out equal
areas in equal times. This
equation is transcendental (ie can't
be solved algebraically) so we
either have to use an approximating
equation or solve by a numeric method.
My implementation uses
Newton-Raphson iteration to get an
excellent approximate answer (usually
in 2 or 3 iterations).