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I need to code a function with two inputs (x,Tau). Let's suppose that the vector is a simple x=sin(t). The function must detect the local maximum. If a local max is spotted, it should deteriorate from there with the exp(-t/Tau) until the sin(x) is greater again. From there, the x should be followed until the next local max. It is a kind of a low pass filter, but I cannot code it properly in Octave.
Here is a sample picture drawn by hand:
This question is more mathematics than computing. The following solution will work when your first curve is a smooth function, but I'm not sure how well it would work if your input is a set of experimentally measured values with a high degree of scatter. In that case you might need something more sophisticated.
I assume you want to make the decay curve deviate from the underlying curve not at the maximum but at the point slightly past the maximum where the downward slope of the underlying curve first exceeds that of exponential decay with time constant tau. This seems like a much more physical situation. Switching to exponential decay at the literal maximum could/would result in the decay curve actually crossing the underlying curve.
If your 'sine' function is x(t), and if the decay constant is tau, then you need a new symbol for the red curve in your graph. Call this y(t). Representing the time axis by a vector of numbers t = [t(1), t(2), ..., t(n)] and the corresponding values of your 'sine' function by a vector x = [x(1), x(2), ..., x(n)] and setting y(1) = x(1), then the (i > 1)th member of the vector y is given by max(x(i), y(i-1) exp(-(t(i) - t(i-1))/tau)).
Implementing this in Octave is straightforward:
clear
t = linspace(0, 6*pi, 1000); %Or whatever
x = sin(t); %Could be any function
function y = decay(x, t, tau)
y = zeros(1, length(t));
y(1) = x(1);
for i = 2:length(t)
y(i) = max(x(i), y(i-1)*exp(-(t(i)-t(i-1))/tau));
endfor
endfunction
tau = 10;
y = decay(x, t, tau);
clf
hold on
plot(t, x)
plot(t, y)
hold off
If you want to use the above function regularly it would probably be a good idea to add checks to make sure that the correct number of arguments are passed, that x and t are the same length and that tau is a scalar.
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I'm working with SVR, and using this resource. Erverything is super clear, with epsilon intensive loss function (from figure). Prediction comes with tube, to cover most training sample, and generalize bounds, using support vectors.
Then we have this explanation. This can be described by introducing (non-negative) slack variables , to measure the deviation of training samples outside -insensitive zone. I understand this error, outside tube, but don't know, how we can use this in optimization. Could somebody explain this?
In local source. I'm trying to achieve very simple optimization solution, without libraries. This what I have for loss function.
import numpy as np
# Kernel func, linear by default
def hypothesis(x, weight, k=None):
k = k if k else lambda z : z
k_x = np.vectorize(k)(x)
return np.dot(k_x, np.transpose(weight))
.......
import math
def boundary_loss(x, y, weight, epsilon):
prediction = hypothesis(x, weight)
scatter = np.absolute(
np.transpose(y) - prediction)
bound = lambda z: z \
if z >= epsilon else 0
return np.sum(np.vectorize(bound)(scatter))
First, let's look at the objective function. The first term, 1/2 * w^2 (wish this site had LaTeX support but this will suffice) correlates with the margin of the SVM. The article you linked doesn't, in my opinion, explain this very well and calls this term describing "the model's complexity", but perhaps this is not the best way of explaining it. Minimizing this term maximizes the margin (while still representing the data well), which is the predominant goal of using SVM's doing regression.
Warning, Math Heavy Explanation: The reason this is the case is that when maximizing the margin, you want to find the "farthest" non-outlier points right on the margin and minimize its distance. Let this farthest point be x_n. We want to find its Euclidean distance d from the plane f(w, x) = 0, which I will rewrite as w^T * x + b = 0 (where w^T is just the transpose of the weights matrix so that we can multiply the two). To find the distance, let us first normalize the plane such that |w^T * x_n + b| = epsilon, which we can do WLOG as w is still able to form all possible planes of the form w^T * x + b= 0. Then, let's note that w is perpendicular to the plane. This is obvious if you have dealt a lot with planes (particularly in vector calculus), but can be proven by choosing two points on the plane x_1 and x_2, then noticing that w^T * x_1 + b = 0, and w^T * x_2 + b = 0. Subtracting the two equations we get w^T(x_1 - x_2) = 0. Since x_1 - x_2 is just any vector strictly on the plane, and its dot product with w is 0, then we know that w is perpendicular to the plane. Finally, to actually calculate the distance between x_n and the plane, we take the vector formed by x_n' and some point on the plane x' (The vectors would then be x_n - x', and projecting it onto the vector w. Doing this, we get d = |w * (x_n - x') / |w||, which we can rewrite as d = (1 / |w|) * | w^T * x_n - w^T x'|, and then add and subtract b to the inside to get d = (1 / |w|) * | w^T * x_n + b - w^T * x' - b|. Notice that w^T * x_n + b is epsilon (from our normalization above), and that w^T * x' + b is 0, as this is just a point on our plane. Thus, d = epsilon / |w|. Notice that maximizing this distance subject to our constraint of finding the x_n and having |w^T * x_n + b| = epsilon is a difficult optimization problem. What we can do is restructure this optimization problem as minimizing 1/2 * w^T * w subject to the first two constraints in the picture you attached, that is, |y_i - f(x_i, w)| <= epsilon. You may think that I have forgotten the slack variables, and this is true, but when just focusing on this term and ignoring the second term, we ignore the slack variables for now, I will bring them back later. The reason these two optimizations are equivalent is not obvious, but the underlying reason lies in discrimination boundaries, which you are free to read more about (it's a lot more math that frankly I don't think this answer needs more of). Then, note that minimizing 1/2 * w^T * w is the same as minimizing 1/2 * |w|^2, which is the desired result we were hoping for. End of the Heavy Math
Now, notice that we want to make the margin big, but not so big that includes noisy outliers like the one in the picture you provided.
Thus, we introduce a second term. To motivate the margin down to a reasonable size the slack variables are introduced, (I will call them p and p* because I don't want to type out "psi" every time). These slack variables will ignore everything in the margin, i.e. those are the points that do not harm the objective and the ones that are "correct" in terms of their regression status. However, the points outside the margin are outliers, they do not reflect well on the regression, so we penalize them simply for existing. The slack error function that is given there is relatively easy to understand, it just adds up the slack error of every point (p_i + p*_i) for i = 1,...,N, and then multiplies by a modulating constant C which determines the relative importance of the two terms. A low value of C means that we are okay with having outliers, so the margin will be thinned and more outliers will be produced. A high value of C indicates that we care a lot about not having slack, so the margin will be made bigger to accommodate these outliers at the expense of representing the overall data less well.
A few things to note about p and p*. First, note that they are both always >= 0. The constraint in your picture shows this, but it also intuitively makes sense as slack should always add to the error, so it is positive. Second, notice that if p > 0, then p* = 0 and vice versa as an outlier can only be on one side of the margin. Last, all points inside the margin will have p and p* be 0, since they are fine where they are and thus do not contribute to the loss.
Notice that with the introduction of the slack variables, if you have any outliers then you won't want the condition from the first term, that is, |w^T * x_n + b| = epsilon as the x_n would be this outlier, and your whole model would be screwed up. What we allow for, then, is to change the constraint to be |w^T * x_n + b| = epsilon + (p + p*). When translated to the new optimization's constraint, we get the full constraint from the picture you attached, that is, |y_i - f(x_i, w)| <= epsilon + p + p*. (I combined the two equations into one here, but you could rewrite them as the picture is and that would be the same thing).
Hopefully after covering all this up, the motivation for the objective function and the corresponding slack variables makes sense to you.
If I understand the question correctly, you also want code to calculate this objective/loss function, which I think isn't too bad. I have not tested this (yet), but I think this should be what you want.
# Function for calculating the error/loss for a SVM. I assume that:
# - 'x' is 2d array representing the vectors of the data points
# - 'y' is an array representing the values each vector actually gives
# - 'weights' is an array of weights that we tune for the regression
# - 'epsilon' is a scalar representing the breadth of our margin.
def optimization_objective(x, y, weights, epsilon):
# Calculates first term of objective (note that norm^2 = dot product)
margin_term = np.dot(weight, weight) / 2
# Now calculate second term of objective. First get the sum of slacks.
slack_sum = 0
for i in range(len(x)): # For each observation
# First find the absolute distance between expected and observed.
diff = abs(hypothesis(x[i]) - y[i])
# Now subtract epsilon
diff -= epsilon
# If diff is still more than 0, then it is an 'outlier' and will have slack.
slack = max(0, diff)
# Add it to the slack sum
slack_sum += slack
# Now we have the slack_sum, so then multiply by C (I picked this as 1 aribtrarily)
C = 1
slack_term = C * slack_sum
# Now, simply return the sum of the two terms, and we are done.
return margin_term + slack_term
I got this function working on my computer with small data, and you may have to change it a little to work with your data if, for example, the arrays are structured differently, but the idea is there. Also, I am not the most proficient with python, so this may not be the most efficient implementation, but my intent was to make it understandable.
Now, note that this just calculates the error/loss (whatever you want to call it). To actually minimize it requires going into Lagrangians and intense quadratic programming which is a much more daunting task. There are libraries available for doing this but if you want to do this library free as you are doing with this, I wish you good luck because doing that is not a walk in the park.
Finally, I would like to note that most of this information I got from notes I took in my ML class I took last year, and the professor (Dr. Abu-Mostafa) was a great help to have me learn the material. The lectures for this class are online (by the same prof), and the pertinent ones for this topic are here and here (although in my very biased opinion you should watch all the lectures, they were a great help). Leave a comment/question if you need anything cleared up or if you think I made a mistake somewhere. If you still don't understand, I can try to edit my answer to make more sense. Hope this helps!
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the question is simple one object is moving from east-west with a velocity of v1 and another from south-north with velocity v2.
I just need the algorithm(formula) to calculate the shortest distance between them so I can write a program for it.
I do have distance between them and the meting point of their paths they are d1 and d2.
Assuming you are asking for 2-d space, at t=0, let the starting points be (d1,0) and (0,d2) on the coordinate axes. We can assume this because one object is always moving horizontally (E-W direction, along X-axis) and other vertically (S-N direction, along Y-axis). Now, after some time t, their positions will be,(d1-t*v1) and (0,d2-t*v2). (Speed-Distance-Time relation).
Now, distance between them at this time t will be,
D = d^2 = (d1-t*v1)^2 + (d2-t*v2)^2
So, differentiating both sides wrt t,
dD/dt = 2(-v1)(d1-t*v1) + 2(-v2)(d2-t*v2) ....(1)
For D to be minimum, dD/dt = 0 and second differential must be positive. Now, second differential :
d2D/dt2 = 2*v1^2 + 2*v2^2 which is positive for all real v1/v2. So, if `dD/dt = 0`, distance will be minimum.
So, equating (1) = 0, we get
t = (d1v1 + d2v2)/(v1^2 + v2^2)
So, get sqrt(D) at t = --above value-- and that shall be your answer.
PS: ask these type of questions on mathematics stackexchange.
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I am trying to understand this pattern, how the rows and columns equation could be so that I could make a loop to give me the same picture.
I don't necessarily need the code, just the way the pattern go, I can't seem to understand it have tried coding it still my results are bad...
When my input is 1 this is the result
When my input is 3 this is the result
And when my input is 15 this is the result
It appears to be a rough pixelated circle. So the equation is x^2 + y^2 <= r^2. Where the coordinate values are taken relative to the center, the radius given as input.
Edit: Upon further inspection, these are very wide circles, presumably to account for the rectangular shape of the character cels. So, it should be more like (x/scale)^2 + y^2 <= r^2
As mentioned in the comments, the keyword to learn more is "Bresenham's circle-drawing algorithm", which is a fast way to do the calculations for one quadrant of the circle (and the rest you get by mirroring). Using Bresenham, you'll probably also need a flood fill. You might also search for "scan-line rasterization of circle", which can skip the flood-fill step.
This is an ASCII-art circle with radius (r) equal to the input number.
Double X and double dot (XX and ..) are used as pixel contents.
We can just impose a coordinate system [-r..r, -r..r] on the pixel matrix and test each one for membership inside the circle of given radius as we print it. The test is x^2 + y^2 <= r^2.
The following C function does all this. It produces correct output for all of your inputs (1,3,15).
void print_filled_circle(int r)
{
for (int x = -r; x <= r; x++)
{
for (int y = -r; y <= r; y++)
{
if (x*x + y*y <= r*r)
printf("XX");
else
printf("..");
}
printf("\n");
}
}
I am trying to create an android smartphone application which uses Apples iBeacon technology to determine the current indoor location of itself. I already managed to get all available beacons and calculate the distance to them via the rssi signal.
Currently I face the problem, that I am not able to find any library or implementation of an algorithm, which calculates the estimated location in 2D by using 3 (or more) distances of fixed points with the condition, that these distances are not accurate (which means, that the three "trilateration-circles" do not intersect in one point).
I would be deeply grateful if anybody can post me a link or an implementation of that in any common programming language (Java, C++, Python, PHP, Javascript or whatever). I already read a lot on stackoverflow about that topic, but could not find any answer I were able to convert in code (only some mathematical approaches with matrices and inverting them, calculating with vectors or stuff like that).
EDIT
I thought about an own approach, which works quite well for me, but is not that efficient and scientific. I iterate over every meter (or like in my example 0.1 meter) of the location grid and calculate the possibility of that location to be the actual position of the handset by comparing the distance of that location to all beacons and the distance I calculate with the received rssi signal.
Code example:
public Location trilaterate(ArrayList<Beacon> beacons, double maxX, double maxY)
{
for (double x = 0; x <= maxX; x += .1)
{
for (double y = 0; y <= maxY; y += .1)
{
double currentLocationProbability = 0;
for (Beacon beacon : beacons)
{
// distance difference between calculated distance to beacon transmitter
// (rssi-calculated distance) and current location:
// |sqrt(dX^2 + dY^2) - distanceToTransmitter|
double distanceDifference = Math
.abs(Math.sqrt(Math.pow(beacon.getLocation().x - x, 2)
+ Math.pow(beacon.getLocation().y - y, 2))
- beacon.getCurrentDistanceToTransmitter());
// weight the distance difference with the beacon calculated rssi-distance. The
// smaller the calculated rssi-distance is, the more the distance difference
// will be weighted (it is assumed, that nearer beacons measure the distance
// more accurate)
distanceDifference /= Math.pow(beacon.getCurrentDistanceToTransmitter(), 0.9);
// sum up all weighted distance differences for every beacon in
// "currentLocationProbability"
currentLocationProbability += distanceDifference;
}
addToLocationMap(currentLocationProbability, x, y);
// the previous line is my approach, I create a Set of Locations with the 5 most probable locations in it to estimate the accuracy of the measurement afterwards. If that is not necessary, a simple variable assignment for the most probable location would do the job also
}
}
Location bestLocation = getLocationSet().first().location;
bestLocation.accuracy = calculateLocationAccuracy();
Log.w("TRILATERATION", "Location " + bestLocation + " best with accuracy "
+ bestLocation.accuracy);
return bestLocation;
}
Of course, the downside of that is, that I have on a 300m² floor 30.000 locations I had to iterate over and measure the distance to every single beacon I got a signal from (if that would be 5, I do 150.000 calculations only for determine a single location). That's a lot - so I will let the question open and hope for some further solutions or a good improvement of this existing solution in order to make it more efficient.
Of course it has not to be a Trilateration approach, like the original title of this question was, it is also good to have an algorithm which includes more than three beacons for the location determination (Multilateration).
If the current approach is fine except for being too slow, then you could speed it up by recursively subdividing the plane. This works sort of like finding nearest neighbors in a kd-tree. Suppose that we are given an axis-aligned box and wish to find the approximate best solution in the box. If the box is small enough, then return the center.
Otherwise, divide the box in half, either by x or by y depending on which side is longer. For both halves, compute a bound on the solution quality as follows. Since the objective function is additive, sum lower bounds for each beacon. The lower bound for a beacon is the distance of the circle to the box, times the scaling factor. Recursively find the best solution in the child with the lower lower bound. Examine the other child only if the best solution in the first child is worse than the other child's lower bound.
Most of the implementation work here is the box-to-circle distance computation. Since the box is axis-aligned, we can use interval arithmetic to determine the precise range of distances from box points to the circle center.
P.S.: Math.hypot is a nice function for computing 2D Euclidean distances.
Instead of taking confidence levels of individual beacons into account, I would instead try to assign an overall confidence level for your result after you make the best guess you can with the available data. I don't think the only available metric (perceived power) is a good indication of accuracy. With poor geometry or a misbehaving beacon, you could be trusting poor data highly. It might make better sense to come up with an overall confidence level based on how well the perceived distance to the beacons line up with the calculated point assuming you trust all beacons equally.
I wrote some Python below that comes up with a best guess based on the provided data in the 3-beacon case by calculating the two points of intersection of circles for the first two beacons and then choosing the point that best matches the third. It's meant to get started on the problem and is not a final solution. If beacons don't intersect, it slightly increases the radius of each up until they do meet or a threshold is met. Likewise, it makes sure the third beacon agrees within a settable threshold. For n-beacons, I would pick 3 or 4 of the strongest signals and use those. There are tons of optimizations that could be done and I think this is a trial-by-fire problem due to the unwieldy nature of beaconing.
import math
beacons = [[0.0,0.0,7.0],[0.0,10.0,7.0],[10.0,5.0,16.0]] # x, y, radius
def point_dist(x1,y1,x2,y2):
x = x2-x1
y = y2-y1
return math.sqrt((x*x)+(y*y))
# determines two points of intersection for two circles [x,y,radius]
# returns None if the circles do not intersect
def circle_intersection(beacon1,beacon2):
r1 = beacon1[2]
r2 = beacon2[2]
dist = point_dist(beacon1[0],beacon1[1],beacon2[0],beacon2[1])
heron_root = (dist+r1+r2)*(-dist+r1+r2)*(dist-r1+r2)*(dist+r1-r2)
if ( heron_root > 0 ):
heron = 0.25*math.sqrt(heron_root)
xbase = (0.5)*(beacon1[0]+beacon2[0]) + (0.5)*(beacon2[0]-beacon1[0])*(r1*r1-r2*r2)/(dist*dist)
xdiff = 2*(beacon2[1]-beacon1[1])*heron/(dist*dist)
ybase = (0.5)*(beacon1[1]+beacon2[1]) + (0.5)*(beacon2[1]-beacon1[1])*(r1*r1-r2*r2)/(dist*dist)
ydiff = 2*(beacon2[0]-beacon1[0])*heron/(dist*dist)
return (xbase+xdiff,ybase-ydiff),(xbase-xdiff,ybase+ydiff)
else:
# no intersection, need to pseudo-increase beacon power and try again
return None
# find the two points of intersection between beacon0 and beacon1
# will use beacon2 to determine the better of the two points
failing = True
power_increases = 0
while failing and power_increases < 10:
res = circle_intersection(beacons[0],beacons[1])
if ( res ):
intersection = res
else:
beacons[0][2] *= 1.001
beacons[1][2] *= 1.001
power_increases += 1
continue
failing = False
# make sure the best fit is within x% (10% of the total distance from the 3rd beacon in this case)
# otherwise the results are too far off
THRESHOLD = 0.1
if failing:
print 'Bad Beacon Data (Beacon0 & Beacon1 don\'t intersection after many "power increases")'
else:
# finding best point between beacon1 and beacon2
dist1 = point_dist(beacons[2][0],beacons[2][1],intersection[0][0],intersection[0][1])
dist2 = point_dist(beacons[2][0],beacons[2][1],intersection[1][0],intersection[1][1])
if ( math.fabs(dist1-beacons[2][2]) < math.fabs(dist2-beacons[2][2]) ):
best_point = intersection[0]
best_dist = dist1
else:
best_point = intersection[1]
best_dist = dist2
best_dist_diff = math.fabs(best_dist-beacons[2][2])
if best_dist_diff < THRESHOLD*best_dist:
print best_point
else:
print 'Bad Beacon Data (Beacon2 distance to best point not within threshold)'
If you want to trust closer beacons more, you may want to calculate the intersection points between the two closest beacons and then use the farther beacon to tie-break. Keep in mind that almost anything you do with "confidence levels" for the individual measurements will be a hack at best. Since you will always be working with very bad data, you will defintiely need to loosen up the power_increases limit and threshold percentage.
You have 3 points : A(xA,yA,zA), B(xB,yB,zB) and C(xC,yC,zC), which respectively are approximately at dA, dB and dC from you goal point G(xG,yG,zG).
Let's say cA, cB and cC are the confidence rate ( 0 < cX <= 1 ) of each point.
Basically, you might take something really close to 1, like {0.95,0.97,0.99}.
If you don't know, try different coefficient depending of distance avg. If distance is really big, you're likely to be not very confident about it.
Here is the way i'll do it :
var sum = (cA*dA) + (cB*dB) + (cC*dC);
dA = cA*dA/sum;
dB = cB*dB/sum;
dC = cC*dC/sum;
xG = (xA*dA) + (xB*dB) + (xC*dC);
yG = (yA*dA) + (yB*dB) + (yC*dC);
xG = (zA*dA) + (zB*dB) + (zC*dC);
Basic, and not really smart but will do the job for some simple tasks.
EDIT
You can take any confidence coef you want in [0,inf[, but IMHO, restraining at [0,1] is a good idea to keep a realistic result.
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(This question could be better off on math, but im not sure)
http://i.imgur.com/TVINP.png
This is probably very simple but the way I'm thinking of doing it doesn't seem very easy and there must be a simpler method.I've got an image and I want to find some points that fall on a line so in this example image below the starting point of my line is (39,75) and the ending point is (75,142) from there I want to find 5 points (or any number really 5 is just an example) that are all on that line.
Is there some equation I can use that will get me a certain amount of points given any start and end coordinates?
yes.
suppose (x0,y0) and (x1,y1) are the starting and ending points on the line.
t*(x0,y0) + (t-1)*(x1,y1) are also going to be points on that line where t ranges from 0 to 1.
note:
if t = 0, you get (x0,y0)
if t = 1, you get (x1,y1)
if t is any value inside (0,1) you get that "percentage" of the way from (x0,y0) to (x1,y1)
(if t = 0.5, you are halfway between the points)
this is what is often called "tweening" in computer graphics
Yes. Your line segment can be described by this equation:
x = 39 + t * (75 - 39)
y = 75 + t * (142 - 75)
where, t can take on any value between 0 and 1.
So, to get random points on the line, just choose a random value for t (between 0 and 1), and calculate what x, y are.
The idea is, x is traveling from 39 to 75, while y is traveling from 75 to 142, and t represents the fraction of travel that has been completed.
A line can be defined by the function y = mx + b where x and y are coordinates on the cartesian plane, m is the slope of the line defined by (y2 - y1)/(x2 - x1), and b is the point where the line intersects the y-axis.
Given that information and two points on the line, you can fill in the blanks with some basic algebra to determine a function that defines the line. Note that if your coordinate plane for the image places (0, 0) in the top left corner, you may have to flip the sign of the y-coordinate.