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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");
}
}
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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!
I'm currently developing an application that will alert users of incoming rain. To do this I want to check certain area around user location for rainfall (different pixel colours for intensity on rainfall radar image). I would like the checked area to be a circle but I don't know how to do this efficiently.
Let's say I want to check radius of 50km. My current idea is to take subset of image with size 100kmx100km (user+50km west, user+50km east, user+50km north, user+50km south) and then check for each pixel in this subset if it's closer to user than 50km.
My question here is, is there a better solution that is used for this type of problems?
If the occurrence of the event you are searching for (rain or anything) is relatively rare, then there's nothing wrong with scanning a square or pixels and then, only after detecting rain in that square, checking whether that rain is within the desired 50km circle. Note that the key point here is that you don't need to check each pixel of the square for being inside the circle (that would be very inefficient), you have to search for your event (rain) first and only when you found it, check whether it falls into the 50km circle. To implement this efficiently you also have to develop some smart strategy for handling multi-pixel "stains" of rain on your image.
However, since you are scanning a raster image, you can easily implement the well-known Bresenham circle algorithm to find the starting and the ending point of the circle for each scan line. That way you can easily limit your scan to the desired 50km radius.
On the second thought, you don't even need the Bresenham algorithm for that. For each row of pixels in your square, calculate the points of intersection of that row with the 50km circle (using the usual schoolbook formula with square root), and then check all pixels that fall between these intersection points. Process all rows in the same fashion and you are done.
P.S. Unfortunately, the Wikipedia page I linked does not present Bresenham algorithm at all. It has code for Michener circle algorithm instead. Michener algorithm will also work for circle rasterization purposes, but it is less precise than Bresenham algorithm. If you care for precision, find a true Bresenham on somewhere. It is actually surprisingly diffcult to find on the net: most search hits erroneously present Michener as Bresenham.
There is, you can modify the midpoint circle algorithm to give you an array of for each y, the x coordinate where the circle starts (and ends, that's the same thing because of symmetry). This array is easy to compute, pseudocode below.
Then you can just iterate over exactly the right part, without checking anything.
Pseudo code:
data = new int[radius];
int f = 1 - radius, ddF_x = 1;
int ddF_y = -2 * radius;
int x = 0, y = radius;
while (x < y)
{
if (f >= 0)
{
y--;
ddF_y += 2; f += ddF_y;
}
x++;
ddF_x += 2; f += ddF_x;
data[radius - y] = x; data[radius - x] = y;
}
Maybe you can try something that will speed up your algorithm.
In brute force algorithm you will probably use equation:
(x-p)^2 + (y-q)^2 < r^2
(p,q) - center of the circle, user position
r - radius (50km)
If you want to find all pixels (x,y) that satisfy above condition and check them, your algorithm goes to O(n^2)
Instead of scanning all pixels in this circle I will check only only pixels that are on border of the circle.
In that case, you can use some more clever way to define circle.
x = p+r*cos(a)
y = q*r*sin(a)
a - angle measured in radians [0-2pi]
Now you can sample some angles, for example twenty of them, iterate and find all pairs (x,y) that are border for radius 50km. Now check are they on the rain zone and alert user.
For more safety I recommend you to use multiple radians (smaller than 50km), because your whole rain cloud can be inside circle, and your app will not recognize him. For example use 3 incircles (r = 5km, 15km, 30km) and do same thing. Efficiency of this algorithm only depends on number of angles and number of incircles.
Pseudocode will be:
checkRainDanger()
p,q <- position
radius[] <- array of radii
for c = 1 to length(radius)
a=0
while(a<2*pi)
x = p + radius[c]*cos(a)
y = q + radius[c]*sin(a)
if rainZone(x,y)
return true
else
a+=pi/10
end_while
end_for
return false //no danger
r2=r*r
for x in range(-r, +r):
max_y=sqrt(r2-x*x)
for y in range(-max_y, +max_y):
# x,y is in range - check for rain
Currently I'm doing benchmarks on time series indexing algorithms. Since most of the time no reference implementations are available, I have to write my own implementations (all in Java). At the moment I am stuck a little at section 6.2 of a paper called Indexing multi-dimensional time-series with support for multiple distance measures available here in PDF : http://hadjieleftheriou.com/papers/vldbj04-2.pdf
A MBR (minimum bounding rectangle) is basically a rectanglular cubiod with some coordinates and directions. As an example P and Q are two MBRs with P.coord={0,0,0} and P.dir={1,1,3} and Q.coords={0.5,0.5,1} and Q.dir={1,1,1} where the first entries represent the time dimension.
Now I would like to calculate the MINDIST(Q,P) between Q and P :
However I am not sure how to implement the "intersection of two MBRs in the time dimension" (Dim 1) since I am not sure what the intersection in the time dimension actually means. It is also not clear what h_Q, l_Q, l_P, h_P mean, since this notation is not explained (my guess is they mean something like highest or lowest value of a dimension in the intersection).
I would highly appreciate it, if someone could explain to me how to calculate the intersection of two MBRs in the first dimension and maybe enlighten me with an interpretation of the notation. Thanks!
Well, Figure 14 in your paper explains the time intersection. And the rectangles are axis-aligned, thus it makes sense to use high and low on each coordinate.
The multiplication sign you see is not a cross product, just a normal multiplication, because on both sides of it you have a scalar, and not vectors.
However I must agree that the discussions on page 14 are rather fuzzy, but they seem to tell us that both types of intersections (complete and partial), when they are have a t subscript, mean the norm of the intersection along the t coordinate.
Thus it seems you could factorize the time intersection to get a formula that would be :
It is worth noting that, maybe counter-intuitively, when your objects don't intersect on the time plane, their MINDIST is defined to be 0.
Hence the following pseudo-code ;
mindist(P, Q)
{
if( Q.coord[0] + Q.dir[0] < P.coord[0] ||
Q.coord[0] > P.coord[0] + P.dir[0] )
return 0;
time = min(Q.coord[0] + Q.dir[0], P.coord[0] + P.dir[0]) - max(Q.coord[0], P.coord[0]);
sum = 0;
for(d=1; d<D; ++d)
{
if( Q.coord[d] + Q.dir[d] < P.coord[d] )
x = Q.coord[d] + Q.dir[d] - P.coord[d];
else if( P.coord[d] + P.dir[d] < Q.coord[d] )
x = P.coord[d] + P.dir[d] - Q.coord[d];
else
x = 0;
sum += x*x;
}
return sqrt(time * sum);
}
Note the absolute values in the paper are unnecessary since we just checked which values where bigger, and we thus know we only add positive numbers.
Probably an easy question, but I could not find an easy solution so far. I'm working on a simple image recognition software for a very specific use case.
Given is a bunch of points that are supposedly on a straight line. However, some of the points are mistakenly placed and away from the line. Especially near the ends of the line, points may happen to be more or less inaccurate.
Example:
X // this guy is off
X // this one even more
X // looks fine
X
X
X // a mistake in the middle
X
X // another mistake, not as bad as the previous
X
X
X
X
X // we're off the line again
The general direction of the line is known, in this case, it's vertical. The actual line in the example is in fact vertical with slight diagonal slope.
I'm only interested in the infinite line (that is, it's slope and offset), the position of the endpoints is not important.
As additional information (not sure if it is important), it is impossible for 2 points to lie next to each other horizontally. Example:
X
X
X
X X // cannot happen
X
X
Performance is not important. I'm working in C#, but I'm fine with any language or just a generic idea, too.
I think you're looking for Least squares fit via Linear Regression
Linear regression (as mentioned by others) is good if you know you do not have outliers.
If you do have outliers, then one of my favorite methods is the median median line method:
http://education.uncc.edu/droyster/courses/spring00/maed3103/Median-Median_Line.htm
Basically you sort the points by the X values and then split the points up into three equal sized groups (smallest values, medium values, and largest values). The final slope is the slope of the line going through the median of the small group and through the median of the large group. The median of the middle group is used with the other medians to calculate the final offset/intercept.
This is a simple algorithm that can be found on several graphing calculators.
By taking the three medians, you are completely ignoring any outliers (either on the far left, far right, far up, or far down).
The image below shows the linear regression and median-median lines for a set of data with a couple of large outliers.
Mike is spot on! Use the following:
double[] xVals = {...};
double[] yVals = {...};
double xMean = 0;
double yMean = 0;
double Sxy = 0;
double Sxx = 0;
double beta0, beta1;
int i;
for (i = 0; i < xVals.Length; i++)
{
xMean += xVals[i]/xVals.Length;
yMean += yVals[i]/yVals.Length;
}
for (i = 0; i < xVals.Length; i++)
{
Sxy += (xVals[i]-xMean)*(yVals[i]-yMean);
Sxx += (xVals[i]-xMean)*(xVals[i]-xMean);
}
beta1 = Sxy/Sxx;
beta0 = yMean-beta1*xMean;
Use beta1 as the slope and beta0 as the y-intercept!