I am working on a genetic algorithm. Here is how it works :
Input : a list of 2D points
Input : the degree of the curve
Output : the equation of the curve that passes through points the best way (try to minimize the sum of vertical distances from point's Ys to the curve)
The algorithm finds good equations for simple straight lines and for 2-degree equations.
But for 4 points and 3 degree equations and more, it gets more complicated. I cannot find the right combination of parameters : sometimes I have to wait 5 minutes and the curve found is still very bad. I tried modifying many parameters, from population size to number of parents selected...
Do famous combinations/theorems in GA programming can help me ?
Thank you ! :)
Based on what is given, you would need a polynomial interpolation in which, the degree of the equation is number of points minus 1.
n = (Number of points) - 1
Now having said that, let's assume you have 5 points that need to be fitted and I am going to define them in a variable:
var points = [[0,0], [2,3], [4,-1], [5,7], [6,9]]
Please be noted the array of the points have been ordered by the x values which you need to do.
Then the equation would be:
f(x) = a1*x^4 + a2*x^3 + a3*x^2 + a4*x + a5
Now based on definition (https://en.wikipedia.org/wiki/Polynomial_interpolation#Constructing_the_interpolation_polynomial), the coefficients are computed like this:
Now you need to used the referenced page to come up with the coefficient.
It is not that complicated, for the polynomial interpolation of degree n you get the following equation:
p(x) = c0 + c1 * x + c2 * x^2 + ... + cn * x^n = y
This means we need n + 1 genes for the coefficients c0 to cn.
The fitness function is the sum of all squared distances from the points to the curve, below is the formula for the squared distance. Like this a smaller value is obviously better, if you don't want that you can take the inverse (1 / sum of squared distances):
d_squared(xi, yi) = (yi - p(xi))^2
I think for faster conversion you could limit the mutation, e.g. when mutating choose a new value with 20% probability between min and max (e.g. -1000 and 1000) and with 80% probabilty a random factor between 0.8 and 1.2 with which you multiply the old value.
Related
I'm now trying to use Kendall's distance to improve sets of rankings based on Borda counts method.
I'm asked to follow a specific document's instructions. In the document it states that :
"The Kendall's distance counts the pairwise disagreements between items from two rankings as :
where
The Kendall's distance is normalized by its maximum value C2n. The less the Kendall’s distance is, the greater the similarity degree between the rankings is.
The Kendall's tau is another method for measuring the similarity degree between rankings, which is easy to be confused with the Kendall's distance.
The Kendall's tau is defined as:
The Kendall's tau is defined based on the normalized Kendall's distance. Note that the greater the Kendall's tau is, the greater the similarity degree between the compared rankings is. In this paper, we use the Kendall's distance rather than the Kendall's tau."
My goal is to improve the following ranking by using Kendall's distance :
x1 x2 x3 x4
A1 4 1 3 2
A2 4 1 3 2
A3 4 3 2 1
A4 1 4 3 2
A5 1 2 4 3
In this ranking, the ith row represents the ranking obtained based on Ai, and each column represents the ranking position of the corresponding item in each ranking. (i.e. xn represents the items to be ranked, Ai represents the ones who rank the items.)
I don't understand what's the difference between the two distances despite the explanation of the doc. And what what does the "(j,s), j != s" beneath the sigma symbol stand for? And finally how to implement Kendall's distance in the ranking provided above?
Distance and similarity are two related concepts, but for distance, exact identity means distance 0, and as things get more different, the distance between them gets greater, with no very obvious fixed limit. A well-behaved distance will obey the rules for a metric - see https://en.wikipedia.org/wiki/Metric_(mathematics). For a similarity, exact identity means similarity 1, and similarity decreases as things get greater, but usually never decreases below 0. Kendall's tau seems to be a way of turning Kendall's distance into a similarity.
"(j,s), j != s" means consider all possibilities for j and s except those for which j = s.
You can compute Kendall's distance by simply summing over all possibilities for j not equal to s - but the time taken for this goes up with the square of the number of items. There are ways for which the time taken only goes up as n * log(n) where n is the number of items - for this and much other stuff on Kendall see https://en.wikipedia.org/wiki/Kendall_rank_correlation_coefficient
From what I understood, the classical KNN algorithm works like this (for discrete data):
Let x be the point you want to classify
Let dist(a,b) be the Euclidean distance between points a and b
Iterate through the training set points pᵢ, taking the distances dist(pᵢ,x)
Classify x as the most frequent class between the K points closest (according to dist) to x.
How would I introduce weights on this classic KNN? I read that more importance should be given to nearer points, and I read this, but couldn't understand how this would apply to discrete data.
For me, first of all, using argmax doesn't make any sense, and if the weight acts increasing the distance, than it would make the distance worse. Sorry if I'm talking nonsense.
Consider a simple example with three classifications (red green blue) and the six nearest neighbors denoted by R, G, B. I'll make this linear to simplify visualization and arithmetic
R B G x G R R
The points listed with distance are
class dist
R 3
B 2
G 1
G 1
R 2
R 3
Thus, if we're using unweighted nearest neighbours, the simple "voting" algorithm is 3-2-1 in favor of Red. However, with the weighted influences, we have ...
red_total = 1/3^2 + 1/2^2 + 1/3^2 = 1/4 + 2/9 ~= .47
blue_total = 1/2^2 ..............................= .25
green_total = 1/1^2 + 1/1^2 ......................= 2.00
... and x winds up as Green due to proximity.
That lower-delta function is merely the classification function; in this simple example, it returns red | green | blue. In a more complex example, ... well, I'll leave that to later tutorials.
Okay, off the bat let me say I am not the fan of the link you provided, it has image equations and follows a different notation in the images and the text.
So leaving that off let's look at the regular k-NN algorithm. regular k-NN is actually just a special case of weighted k-NN. You assign a weight of 1 to k neighbors and 0 to the rest.
Let Wqj denote the weight associated with a point j relative to a point q
Let yj be the class label associated with the data point j. For simplicity let us assume we are classifying birds as either crows, hens or turkeys => discrete classes. So for all j, yj <- {crow, turkey, hen}
A good weight metric is the inverse of the distance , whatever distance be it Euclidean, Mahalanobis etc.
Given all this, the class label yq you would associate with the point q you are trying to predict would be the the sum of the wqj . yj terms diviided by the sum of all weights. You do not have to the division if you normalize the weights first.
You would end up with an equation as follows somevalue1 . crow + somevalue2 . hen + somevalue3 . turkey
One of these classes will have a higher somevalue. The class witht he highest value is what you will predict for point q
For the purpose of training you can factor in the error anyway you want. Since the classes are discrete there are a limited number of simple ways you can adjust the weight to improve accuracy
Given n points in 2-D plane, like (0,0),(1,1), ... We can select any three points from them to construct angle. For example, we choose A(0, 0), B(1, 1), C(1, 0), then we get angle ABC = 45 degree, ACB = 90 degree and CAB = 45 degree.
My question is how to calculate max or min angle determined by three points selected from n points.
Obviously, we can use brute-force algorithm - calculate all angels and find maximal and minimal value, using Law Of Cosines to calculate angles and Pythagorean theorem to calculate distances. But does efficient algorithm exist?
If I'm correct, brute-force runs in O(n^3): you basically take every triplet, compute the 3 angles, and store the overall max.
You can improve slightly to O(n^2 * log(n)) but it's trickier:
best_angle = 0
for each point p1:
for each point p2:
compute vector (p1, p2), and the signed angle it makes with the X-axis
store it in an array A
sort array A # O(n * log(n))
# Traverse A to find the best choice:
for each alpha in A:
look for the element beta in A closest to alpha+Pi
# Takes O(log n) because it's sorted. Don't forget to take into account the fact that A represents a circle: both ends touch...
best_angle = max(best_angle, abs(beta - alpha))
The complexity is O(n * (n + nlog(n) + n * log(n))) = O(n^2 * log(n))
Of course you can also retrieve the pt1, pt2 that obtained the best angle during the loops.
There is probably still better, this feels like doing too much work overall, even if you re-use your previous computations of pt1 for pt2, ..., ptn ...
The problem:
N points are given on a 2-dimensional plane. What is the maximum number of points on the same straight line?
The problem has O(N2) solution: go through each point and find the number of points which have the same dx / dy with relation to the current point. Store dx / dy relations in a hash map for efficiency.
Is there a better solution to this problem than O(N2)?
There is likely no solution to this problem that is significantly better than O(n^2) in a standard model of computation.
The problem of finding three collinear points reduces to the problem of finding the line that goes through the most points, and finding three collinear points is 3SUM-hard, meaning that solving it in less than O(n^2) time would be a major theoretical result.
See the previous question on finding three collinear points.
For your reference (using the known proof), suppose we want to answer a 3SUM problem such as finding x, y, z in list X such that x + y + z = 0. If we had a fast algorithm for the collinear point problem, we could use that algorithm to solve the 3SUM problem as follows.
For each x in X, create the point (x, x^3) (for now we assume the elements of X are distinct). Next, check whether there exists three collinear points from among the created points.
To see that this works, note that if x + y + z = 0 then the slope of the line from x to y is
(y^3 - x^3) / (y - x) = y^2 + yx + x^2
and the slope of the line from x to z is
(z^3 - x^3) / (z - x) = z^2 + zx + x^2 = (-(x + y))^2 - (x + y)x + x^2
= x^2 + 2xy + y^2 - x^2 - xy + x^2 = y^2 + yx + x^2
Conversely, if the slope from x to y equals the slope from x to z then
y^2 + yx + x^2 = z^2 + zx + x^2,
which implies that
(y - z) (x + y + z) = 0,
so either y = z or z = -x - y as suffices to prove that the reduction is valid.
If there are duplicates in X, you first check whether x + 2y = 0 for any x and duplicate element y (in linear time using hashing or O(n lg n) time using sorting), and then remove the duplicates before reducing to the collinear point-finding problem.
If you limit the problem to lines passing through the origin, you can convert the points to polar coordinates (angle, distance from origin) and sort them by angle. All points with the same angle lie on the same line. O(n logn)
I don't think there is a faster solution in the general case.
The Hough Transform can give you an approximate solution. It is approximate because the binning technique has a limited resolution in parameter space, so the maximum bin will give you some limited range of possible lines.
Again an O(n^2) solution with pseudo code. Idea is create a hash table with line itself as the key. Line is defined by slope between the two points, point where line cuts x-axis and point where line cuts y-axis.
Solution assumes languages like Java, C# where equals method and hashcode methods of the object are used for hashing function.
Create an Object (call SlopeObject) with 3 fields
Slope // Can be Infinity
Point of intercept with x-axis -- poix // Will be (Infinity, some y value) or (x value, 0)
Count
poix will be a point (x, y) pair. If line crosses x-axis the poix will (some number, 0). If line is parallel to x axis then poix = (Infinity, some number) where y value is where line crosses y axis.
Override equals method where 2 objects are equal if Slope and poix are equal.
Hashcode is overridden with a function which provides hashcode based on combination of values of Slope and poix. Some pseudo code below
Hashmap map;
foreach(point in the array a) {
foeach(every other point b) {
slope = calculateSlope(a, b);
poix = calculateXInterception(a, b);
SlopeObject so = new SlopeObject(slope, poix, 1); // Slope, poix and intial count 1.
SlopeObject inMapSlopeObj = map.get(so);
if(inMapSlopeObj == null) {
inMapSlopeObj.put(so);
} else {
inMapSlopeObj.setCount(inMapSlopeObj.getCount() + 1);
}
}
}
SlopeObject maxCounted = getObjectWithMaxCount(map);
print("line is through " + maxCounted.poix + " with slope " + maxCounted.slope);
Move to the dual plane using the point-line duality transform for p=(a,b) p*:y=a*x + b.
Now using a line sweep algorithm find all intersection points in NlogN time.
(If you have points which are one above the other just rotate the points to some small angle).
The intersection points corresponds in the dual plane to lines in the primer plane.
Whoever said that since 3SUM have a reduction to this problem and thus the complexity is O(n^2). Please note that the complexity of 3SUM is less than that.
Please check https://en.wikipedia.org/wiki/3SUM and also read
https://tmc.web.engr.illinois.edu/reduce3sum_sosa.pdf
As already mentioned, there probably isn't a way to solve the general case of this problem better than O(n^2). However, if you assume a large number of points lie on the same line (say the probability that a random point in the set of points lie on the line with the maximum number of points is p) and don't need an exact algorithm, a randomized algorithm is more efficient.
maxPoints = 0
Repeat for k iterations:
1. Pick 2 random, distinct points uniformly at random
2. maxPoints = max(maxPoints, number of points that lies on the
line defined by the 2 points chosen in step 1)
Note that in the first step, if you picked 2 points which lies on the line with the maximum number of points, you'll get the optimal solution. Assuming n is very large (i.e. we can treat the probability of finding 2 desirable points as sampling with replacement), the probability of this happening is p^2. Therefore the probability of finding a suboptimal solution after k iterations is (1 - p^2)^k.
Suppose you can tolerate a false negative rate rate = err. Then this algorithm runs in O(nk) = O(n * log(err) / log(1 - p^2)). If both n and p are large enough, this is significantly more efficient than O(n^2). (i.e. Supposed n = 1,000,000 and you know there are at least 10,000 points that lie on the same line. Then n^2 would required on the magnitude of 10^12 operations, while randomized algorithm would require on the magnitude of 10^9 operations to get a error rate of less than 5*10^-5.)
It is unlikely for a $o(n^2)$ algorithm to exist, since the problem (of even checking if 3 points in R^2 are collinear) is 3Sum-hard (http://en.wikipedia.org/wiki/3SUM)
This is not a solution better than O(n^2), but you can do the following,
For each point convert first convert it as if it where in the (0,0) coordinate, and then do the equivalent translation for all the other points by moving them the same x,y distance you needed to move the original choosen point.
2.Translate this new set of translated points to the angle with respect to the new (0,0).
3.Keep stored the maximum number (MSN) of points that are in each angle.
4.Choose the maximum stored number (MSN), and that will be the solution
Using assorted matrix math, I've solved a system of equations resulting in coefficients for a polynomial of degree 'n'
Ax^(n-1) + Bx^(n-2) + ... + Z
I then evaulate the polynomial over a given x range, essentially I'm rendering the polynomial curve. Now here's the catch. I've done this work in one coordinate system we'll call "data space". Now I need to present the same curve in another coordinate space. It is easy to transform input/output to and from the coordinate spaces, but the end user is only interested in the coefficients [A,B,....,Z] since they can reconstruct the polynomial on their own. How can I present a second set of coefficients [A',B',....,Z'] which represent the same shaped curve in a different coordinate system.
If it helps, I'm working in 2D space. Plain old x's and y's. I also feel like this may involve multiplying the coefficients by a transformation matrix? Would it some incorporate the scale/translation factor between the coordinate systems? Would it be the inverse of this matrix? I feel like I'm headed in the right direction...
Update: Coordinate systems are linearly related. Would have been useful info eh?
The problem statement is slightly unclear, so first I will clarify my own interpretation of it:
You have a polynomial function
f(x) = Cnxn + Cn-1xn-1 + ... + C0
[I changed A, B, ... Z into Cn, Cn-1, ..., C0 to more easily work with linear algebra below.]
Then you also have a transformation such as: z = ax + b that you want to use to find coefficients for the same polynomial, but in terms of z:
f(z) = Dnzn + Dn-1zn-1 + ... + D0
This can be done pretty easily with some linear algebra. In particular, you can define an (n+1)×(n+1) matrix T which allows us to do the matrix multiplication
d = T * c ,
where d is a column vector with top entry D0, to last entry Dn, column vector c is similar for the Ci coefficients, and matrix T has (i,j)-th [ith row, jth column] entry tij given by
tij = (j choose i) ai bj-i.
Where (j choose i) is the binomial coefficient, and = 0 when i > j. Also, unlike standard matrices, I'm thinking that i,j each range from 0 to n (usually you start at 1).
This is basically a nice way to write out the expansion and re-compression of the polynomial when you plug in z=ax+b by hand and use the binomial theorem.
If I understand your question correctly, there is no guarantee that the function will remain polynomial after you change coordinates. For example, let y=x^2, and the new coordinate system x'=y, y'=x. Now the equation becomes y' = sqrt(x'), which isn't polynomial.
Tyler's answer is the right answer if you have to compute this change of variable z = ax+b many times (I mean for many different polynomials). On the other hand, if you have to do it just once, it is much faster to combine the computation of the coefficients of the matrix with the final evaluation. The best way to do it is to symbolically evaluate your polynomial at point (ax+b) by Hörner's method:
you store the polynomial coefficients in a vector V (at the beginning, all coefficients are zero), and for i = n to 0, you multiply it by (ax+b) and add Ci.
adding Ci means adding it to the constant term
multiplying by (ax+b) means multiplying all coefficients by b into a vector K1, multiplying all coefficients by a and shifting them away from the constant term into a vector K2, and putting K1+K2 back into V.
This will be easier to program, and faster to compute.
Note that changing y into w = cy+d is really easy. Finally, as mattiast points out, a general change of coordinates will not give you a polynomial.
Technical note: if you still want to compute matrix T (as defined by Tyler), you should compute it by using a weighted version of Pascal's rule (this is what the Hörner computation does implicitely):
ti,j = b ti,j-1 + a ti-1,j-1
This way, you compute it simply, column after column, from left to right.
You have the equation:
y = Ax^(n-1) + Bx^(n-2) + ... + Z
In xy space, and you want it in some x'y' space. What you need is transformation functions f(x) = x' and g(y) = y' (or h(x') = x and j(y') = y). In the first case you need to solve for x and solve for y. Once you have x and y, you can substituted those results into your original equation and solve for y'.
Whether or not this is trivial depends on the complexity of the functions used to transform from one space to another. For example, equations such as:
5x = x' and 10y = y'
are extremely easy to solve for the result
y' = 2Ax'^(n-1) + 2Bx'^(n-2) + ... + 10Z
If the input spaces are linearly related, then yes, a matrix should be able to transform one set of coefficients to another. For example, if you had your polynomial in your "original" x-space:
ax^3 + bx^2 + cx + d
and you wanted to transform into a different w-space where w = px+q
then you want to find a', b', c', and d' such that
ax^3 + bx^2 + cx + d = a'w^3 + b'w^2 + c'w + d'
and with some algebra,
a'w^3 + b'w^2 + c'w + d' = a'p^3x^3 + 3a'p^2qx^2 + 3a'pq^2x + a'q^3 + b'p^2x^2 + 2b'pqx + b'q^2 + c'px + c'q + d'
therefore
a = a'p^3
b = 3a'p^2q + b'p^2
c = 3a'pq^2 + 2b'pq + c'p
d = a'q^3 + b'q^2 + c'q + d'
which can be rewritten as a matrix problem and solved.