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I want to analyse some data in order to program a pricing algorithm.
Following dates are available:
I need a function/correlationfactor of the three variables/dimension which show the change of the Median (price) while the three dimensions (pers_capacity, amount of bedrooms, amount of bathrooms) grow.
e.g. Y(#pers_capacity,bedroom,bathroom) = ..
note:
- in the screenshot below are not all the data available (just a part of it)
- median => price per night
- yellow => #bathroom
e.g. For 2 persons, 2 bedrooms and 1 bathroom is the median price 187$ per night
Do you have some ideas how I can calculate the correlation/equation (f(..)=...) in order to get a reliable factor?
Kind regards
One typical approach would be formulating this as a linear model. Given three variables x, y and z which explain your observed values v, you assume v ≈ ax + by + cz + d and try to find a, b, c and d which match this as closely as possible, minimizing the squared error. This is called a linear least squares approximation. You can also refer to this Math SE post for one example of a specific linear least squares approximation.
If your your dataset is sufficiently large, you may consider more complicated formulas. Things like
v ≈
a1x2 +
a2y2 +
a3z2 +
a4xy +
a5xz +
a6yz +
a7x +
a8y +
a9z +
a10
The above is non-linear in the variables but still linear in the coefficients ai so it's still a linear least squares problem.
Or you could apply transformations to your variables, e.g.
v ≈
a1x +
a2y +
a3z +
a4exp(x) +
a5exp(y) +
a6exp(z) +
a7
Looking at the residual errors (i.e. difference between predicted and observed values) in any of these may indicate terms worth adding.
Personally I'd try all this in R, since computing linear models is just one line in that language, and visualizing data is fairly easy as well.
I am running a speech enhancement algorithm based on Gaussian Mixture Model. The problem is that the estimation algorithm underflows during the training processing.
I am trying to calculate the PDF of a log spectrum frame X given a Gaussian cluster which is a product of the PDF of each frequnecy component X_k (fft is done for k=1..256)
what i get is a product of 256 exp(-v(k)) such that v(k)>=0
Here is a snippet of the MATLAB calculation:
N - number of frames; M- number of mixtures; c_i weight for each mixture;
gamma(n,i) = c_i*f(X_n|I = i)
for i=1 : N
rep_DataMat(:,:,i) = repmat(DataMat(:,i),1,M);
gamma_exp(:,:) = (1./sqrt((2*pi*sigmaSqr_curr))).*exp(((-1)*((rep_DataMat(:,:,i) - mue_curr).^2)./(2*sigmaSqr_curr)));
gamma_curr(i,:) = c_curr.*(prod(10*gamma_exp(:,:),1));
alpha_curr(i,:) = gamma_curr(i,:)./sum(gamma_curr(i,:));
end
The product goes quickly to zero due to K = 256 since the numbers being smaller then one. Is there a way I can calculate this with causing an underflow (like logsum or similar)?
You can perform the computations in the log domain.
The conversion of products into sums is straightforward.
Sums on the other hand can be converted with something such as logsumexp.
This works using the formula:
log(a + b) = log(exp(log(a)) + exp(log(b)))
= log(exp(loga) + exp(logb))
Where loga and logb are the respective representation of a and b in the log domain.
The basic idea is then to factorize the exponent with the largest argument (eg. loga for sake of illustration):
log(exp(loga)+exp(logb)) = log(exp(loga)*(1+exp(logb-loga)))
= loga + log(1+exp(logb-loga))
Note that the same idea applies if you have more than 2 terms to add.
I want to implement an iterative algorithm, which calculates weighted average. The specific weight law does not matter, but it should be close to 1 for the newest values and close to 0 to the oldest.
The algorithm should be iterative. i.e. it should not remember all previous values. It should know only one newest value and any aggregative information about past, like previous values of the average, sums, counts etc.
Is it possible?
For example, the following algorithm can be:
void iterate(double value) {
sum *= 0.99;
sum += value;
count++;
avg = sum / count;
}
It will give exponential decreasing weight, which may be not good. Is it possible to have step decreasing weight or something?
EDIT 1
The the requirements for weighing law is follows:
1) The weight decreases into past
2) I has some mean or characteristic duration so that values older this duration matters much lesser than newer ones
3) I should be able to set this duration
EDIT 2
I need the following. Suppose v_i are values, where v_1 is the first. Also suppose w_i are weights. But w_0 is THE LAST.
So, after first value came I have first average
a_1 = v_1 * w_0
After the second value v_2 came, I should have average
a_2 = v_1 * w_1 + v_2 * w_0
With next value I should have
a_3 = v_1 * w_2 + v_2 * w_1 + v_3 * w_0
Note, that weight profile is moving with me, while I am moving along value sequence.
I.e. each value does not have it's own weight all the time. My goal is to have this weight lower while going to past.
First a bit of background. If we were keeping a normal average, it would go like this:
average(a) = 11
average(a,b) = (average(a)+b)/2
average(a,b,c) = (average(a,b)*2 + c)/3
average(a,b,c,d) = (average(a,b,c)*3 + d)/4
As you can see here, this is an "online" algorithm and we only need to keep track of pieces of data: 1) the total numbers in the average, and 2) the average itself. Then we can undivide the average by the total, add in the new number, and divide it by the new total.
Weighted averages are a bit different. It depends on what kind of weighted average. For example if you defined:
weightedAverage(a,wa, b,wb, c,wc, ..., z,wz) = a*wa + b*wb + c*wc + ... + w*wz
or
weightedAverage(elements, weights) = elements·weights
...then you don't need to do anything besides add the new element*weight! If however you defined the weighted average akin to an expected-value from probability:
weightedAverage(elements,weights) = elements·weights / sum(weights)
...then you'd need to keep track of the total weights. Instead of undividing by the total number of elements, you undivide by the total weight, add in the new element*weight, then divide by the new total weight.
Alternatively you don't need to undivide, as demonstrated below: you can merely keep track of the temporary dot product and weight total in a closure or an object, and divide it as you yield (this can help a lot with avoiding numerical inaccuracy from compounded rounding errors).
In python this would be:
def makeAverager():
dotProduct = 0
totalWeight = 0
def averager(newValue, weight):
nonlocal dotProduct,totalWeight
dotProduct += newValue*weight
totalWeight += weight
return dotProduct/totalWeight
return averager
Demo:
>>> averager = makeAverager()
>>> [averager(value,w) for value,w in [(100,0.2), (50,0.5), (100,0.1)]]
[100.0, 64.28571428571429, 68.75]
>>> averager(10,1.1)
34.73684210526316
>>> averager(10,1.1)
25.666666666666668
>>> averager(30,2.0)
27.4
> But my task is to have average recalculated each time new value arrives having old values reweighted. –OP
Your task is almost always impossible, even with exceptionally simple weighting schemes.
You are asking to, with O(1) memory, yield averages with a changing weighting scheme. For example, {values·weights1, (values+[newValue2])·weights2, (values+[newValue2,newValue3])·weights3, ...} as new values are being passed in, for some nearly arbitrarily changing weights sequence. This is impossible due to injectivity. Once you merge the numbers in together, you lose a massive amount of information. For example, even if you had the weight vector, you could not recover the original value vector, or vice versa. There are only two cases I can think of where you could get away with this:
Constant weights such as [2,2,2,...2]: this is equivalent to an on-line averaging algorithm, which you don't want because the old values are not being "reweighted".
The relative weights of previous answers do not change. For example you could do weights of [8,4,2,1], and add in a new element with arbitrary weight like ...+[1], but you must increase all the previous by the same multiplicative factor, like [16,8,4,2]+[1]. Thus at each step, you are adding a new arbitrary weight, and a new arbitrary rescaling of the past, so you have 2 degrees of freedom (only 1 if you need to keep your dot-product normalized). The weight-vectors you'd get would look like:
[w0]
[w0*(s1), w1]
[w0*(s1*s2), w1*(s2), w2]
[w0*(s1*s2*s3), w1*(s2*s3), w2*(s3), w3]
...
Thus any weighting scheme you can make look like that will work (unless you need to keep the thing normalized by the sum of weights, in which case you must then divide the new average by the new sum, which you can calculate by keeping only O(1) memory). Merely multiply the previous average by the new s (which will implicitly distribute over the dot-product into the weights), and tack on the new +w*newValue.
I think you are looking for something like this:
void iterate(double value) {
count++;
weight = max(0, 1 - (count / 1000));
avg = ( avg * total_weight * (count - 1) + weight * value) / (total_weight * (count - 1) + weight)
total_weight += weight;
}
Here I'm assuming you want the weights to sum to 1. As long as you can generate a relative weight without it changing in the future, you can end up with a solution which mimics this behavior.
That is, suppose you defined your weights as a sequence {s_0, s_1, s_2, ..., s_n, ...} and defined the input as sequence {i_0, i_1, i_2, ..., i_n}.
Consider the form: sum(s_0*i_0 + s_1*i_1 + s_2*i_2 + ... + s_n*i_n) / sum(s_0 + s_1 + s_2 + ... + s_n). Note that it is trivially possible to compute this incrementally with a couple of aggregation counters:
int counter = 0;
double numerator = 0;
double denominator = 0;
void addValue(double val)
{
double weight = calculateWeightFromCounter(counter);
numerator += weight * val;
denominator += weight;
}
double getAverage()
{
if (denominator == 0.0) return 0.0;
return numerator / denominator;
}
Of course, calculateWeightFromCounter() in this case shouldn't generate weights that sum to one -- the trick here is that we average by dividing by the sum of the weights so that in the end, the weights virtually seem to sum to one.
The real trick is how you do calculateWeightFromCounter(). You could simply return the counter itself, for example, however note that the last weighted number would not be near the sum of the counters necessarily, so you may not end up with the exact properties you want. (It's hard to say since, as mentioned, you've left a fairly open problem.)
This is too long to post in a comment, but it may be useful to know.
Suppose you have:
w_0*v_n + ... w_n*v_0 (we'll call this w[0..n]*v[n..0] for short)
Then the next step is:
w_0*v_n1 + ... w_n1*v_0 (and this is w[0..n1]*v[n1..0] for short)
This means we need a way to calculate w[1..n1]*v[n..0] from w[0..n]*v[n..0].
It's certainly possible that v[n..0] is 0, ..., 0, z, 0, ..., 0 where z is at some location x.
If we don't have any 'extra' storage, then f(z*w(x))=z*w(x + 1) where w(x) is the weight for location x.
Rearranging the equation, w(x + 1) = f(z*w(x))/z. Well, w(x + 1) better be constant for a constant x, so f(z*w(x))/z better be constant. Hence, f must let z propagate -- that is, f(z*w(x)) = z*f(w(x)).
But here again we have an issue. Note that if z (which could be any number) can propagate through f, then w(x) certainly can. So f(z*w(x)) = w(x)*f(z). Thus f(w(x)) = w(x)/f(z).
But for a constant x, w(x) is constant, and thus f(w(x)) better be constant, too. w(x) is constant, so f(z) better be constant so that w(x)/f(z) is constant. Thus f(w(x)) = w(x)/c where c is a constant.
So, f(x)=c*x where c is a constant when x is a weight value.
So w(x+1) = c*w(x).
That is, each weight is a multiple of the previous. Thus, the weights take the form w(x)=m*b^x.
Note that this assumes the only information f has is the last aggregated value. Note that at some point you will be reduced to this case unless you're willing to store a non-constant amount of data representing your input. You cannot represent an infinite length vector of real numbers with a real number, but you can approximate them somehow in a constant, finite amount of storage. But this would merely be an approximation.
Although I haven't rigorously proven it, it is my conclusion that what you want is impossible to do with a high degree of precision, but you may be able to use log(n) space (which may as well be O(1) for many practical applications) to generate a quality approximation. You may be able to use even less.
I tried to practically code something (in Java). As has been said, your goal is not achievable. You can only count average from some number of last remembered values. If you don't need to be exact, you can approximate the older values. I tried to do it by remembering last 5 values exactly and older values only SUMmed by 5 values, remembering the last 5 SUMs. Then, the complexity is O(2n) for remembering last n+n*n values. This is a very rough approximation.
You can modify the "lastValues" and "lasAggregatedSums" array sizes as you want. See this ascii-art picture trying to display a graph of last values, showing that the first columns (older data) are remembered as aggregated value (not individually), and only the earliest 5 values are remembered individually.
values:
#####
##### ##### #
##### ##### ##### # #
##### ##### ##### ##### ## ##
##### ##### ##### ##### ##### #####
time: --->
Challenge 1: My example doesn't count weights, but I think it shouldn't be problem for you to add weights for the "lastAggregatedSums" appropriately - the only problem is, that if you want lower weights for older values, it would be harder, because the array is rotating, so it is not straightforward to know which weight for which array member. Maybe you can modify the algorithm to always "shift" values in the array instead of rotating? Then adding weights shouldn't be a problem.
Challenge 2: The arrays are initialized with 0 values, and those values are counting to the average from the beginning, even when we haven't receive enough values. If you are running the algorithm for long time, you probably don't bother that it is learning for some time at the beginning. If you do, you can post a modification ;-)
public class AverageCounter {
private float[] lastValues = new float[5];
private float[] lastAggregatedSums = new float[5];
private int valIdx = 0;
private int aggValIdx = 0;
private float avg;
public void add(float value) {
lastValues[valIdx++] = value;
if(valIdx == lastValues.length) {
// count average of last values and save into the aggregated array.
float sum = 0;
for(float v: lastValues) {sum += v;}
lastAggregatedSums[aggValIdx++] = sum;
if(aggValIdx >= lastAggregatedSums.length) {
// rotate aggregated values index
aggValIdx = 0;
}
valIdx = 0;
}
float sum = 0;
for(float v: lastValues) {sum += v;}
for(float v: lastAggregatedSums) {sum += v;}
avg = sum / (lastValues.length + lastAggregatedSums.length * lastValues.length);
}
public float getAvg() {
return avg;
}
}
you can combine (weighted sum) exponential means with different effective window sizes (N) in order to get the desired weights.
Use more exponential means to define your weight profile more detailed.
(more exponential means also means to store and calculate more values, so here is the trade off)
A memoryless solution is to calculate the new average from a weighted combination of the previous average and the new value:
average = (1 - P) * average + P * value
where P is an empirical constant, 0 <= P <= 1
expanding gives:
average = sum i (weight[i] * value[i])
where value[0] is the newest value, and
weight[i] = P * (1 - P) ^ i
When P is low, historical values are given higher weighting.
The closer P gets to 1, the more quickly it converges to newer values.
When P = 1, it's a regular assignment and ignores previous values.
If you want to maximise the contribution of value[N], maximize
weight[N] = P * (1 - P) ^ N
where 0 <= P <= 1
I discovered weight[N] is maximized when
P = 1 / (N + 1)
I am reading about Dynamic programming in Cormen etc book on algorithms. following is text from book
Suppose we have motor car factory with two assesmly lines called as line 1 and line 2. We have to determine fastest time to get chassis all the way.
Ultimate goal is to determine the fastest time to get a chassis all the way through the factory, which we denote by Fn. The chasssis has to get all the way through station "n" on either line 1 or line 2 and then to factory exit. Since the faster of these ways is the fastest way through the entire factory, we have
Fn = min(f1[n] + x1, f2[n]+x2) ---------------- Eq1
Above x1 and x2 final additional time for comming out from line 1 and line 2
I have following recurrence equations. Consider following are Eq2.
f1[j] = e1 + a1,1 if j = 1
min(f1[j-1] + a1,j, f2[j-1] + t2,j-1 + a1,j if j >= 2
f2[j] = e2 + a2,1 if j = 1
min(f2[j-1] + a2,j, f1[j-1] + t1,j-1 + a2,j if j >= 2
Let Ri(j) be the number of references made to fi[j] in a recursive algorithm.
From equation R1(n) = R2(n) = 1
From equation 2 above we have
R1(j) = R2(j) = R1(j+1) + R2(j+1) for j = 1, 2, ...n-1
My question is how author came with R(n) =1 because usally we have base case as 0 rather than n, here then how we will write recursive functions in code
for example C code?
Another question is how author came up with R1(j) and R2(j)?
Thanks for all the help.
If you solve the problem in a recursive way, what would you do?
You'd start calculating F(n). F(n) would recursively call f1(n-1) and f2(n-1) until getting to the leaves (f1(0), f2(0)), right?
So, that's the reason the number of references to F(n) in the recursive solution is 1, because you'd need to compute f1(n) and f2(n) only once. This is not true to f1(n-1), which is referenced when you compute f1(n) and when you compute f2(n).
Now, how did he come up with R1(j) = R2(j) = R1(j+1) + R2(j+1)?
well, computing it in a recursive way, every time you need f1(i), you have to compute f1(j), f2(j), for every j in the interval [0, i) -- AKA for every j smaller than i.
In other words, the value of f1,2(i) depends on the value of f1,2(0..i-1), so every time you compute a f_(i), you're computing EVERY f1,2(1..i-1) - (because it depends on their value).
For this reason, the number of times you compute f_(i) depends on how many f1,2 there are "above him".
Hope that's clear.
Suppose you have a list of floating point numbers that are approximately multiples of a common quantity, for example
2.468, 3.700, 6.1699
which are approximately all multiples of 1.234. How would you characterize this "approximate gcd", and how would you proceed to compute or estimate it?
Strictly related to my answer to this question.
You can run Euclid's gcd algorithm with anything smaller then 0.01 (or a small number of your choice) being a pseudo 0. With your numbers:
3.700 = 1 * 2.468 + 1.232,
2.468 = 2 * 1.232 + 0.004.
So the pseudo gcd of the first two numbers is 1.232. Now you take the gcd of this with your last number:
6.1699 = 5 * 1.232 + 0.0099.
So 1.232 is the pseudo gcd, and the mutiples are 2,3,5. To improve this result, you may take the linear regression on the data points:
(2,2.468), (3,3.7), (5,6.1699).
The slope is the improved pseudo gcd.
Caveat: the first part of this is algorithm is numerically unstable - if you start with very dirty data, you are in trouble.
Express your measurements as multiples of the lowest one. Thus your list becomes 1.00000, 1.49919, 2.49996. The fractional parts of these values will be very close to 1/Nths, for some value of N dictated by how close your lowest value is to the fundamental frequency. I would suggest looping through increasing N until you find a sufficiently refined match. In this case, for N=1 (that is, assuming X=2.468 is your fundamental frequency) you would find a standard deviation of 0.3333 (two of the three values are .5 off of X * 1), which is unacceptably high. For N=2 (that is, assuming 2.468/2 is your fundamental frequency) you would find a standard deviation of virtually zero (all three values are within .001 of a multiple of X/2), thus 2.468/2 is your approximate GCD.
The major flaw in my plan is that it works best when the lowest measurement is the most accurate, which is likely not the case. This could be mitigated by performing the entire operation multiple times, discarding the lowest value on the list of measurements each time, then use the list of results of each pass to determine a more precise result. Another way to refine the results would be adjust the GCD to minimize the standard deviation between integer multiples of the GCD and the measured values.
This reminds me of the problem of finding good rational-number approximations of real numbers. The standard technique is a continued-fraction expansion:
def rationalizations(x):
assert 0 <= x
ix = int(x)
yield ix, 1
if x == ix: return
for numer, denom in rationalizations(1.0/(x-ix)):
yield denom + ix * numer, numer
We could apply this directly to Jonathan Leffler's and Sparr's approach:
>>> a, b, c = 2.468, 3.700, 6.1699
>>> b/a, c/a
(1.4991896272285252, 2.4999594813614263)
>>> list(itertools.islice(rationalizations(b/a), 3))
[(1, 1), (3, 2), (925, 617)]
>>> list(itertools.islice(rationalizations(c/a), 3))
[(2, 1), (5, 2), (30847, 12339)]
picking off the first good-enough approximation from each sequence. (3/2 and 5/2 here.) Or instead of directly comparing 3.0/2.0 to 1.499189..., you could notice than 925/617 uses much larger integers than 3/2, making 3/2 an excellent place to stop.
It shouldn't much matter which of the numbers you divide by. (Using a/b and c/b you get 2/3 and 5/3, for instance.) Once you have integer ratios, you could refine the implied estimate of the fundamental using shsmurfy's linear regression. Everybody wins!
I'm assuming all of your numbers are multiples of integer values. For the rest of my explanation, A will denote the "root" frequency you are trying to find and B will be an array of the numbers you have to start with.
What you are trying to do is superficially similar to linear regression. You are trying to find a linear model y=mx+b that minimizes the average distance between a linear model and a set of data. In your case, b=0, m is the root frequency, and y represents the given values. The biggest problem is that the independent variables X are not explicitly given. The only thing we know about X is that all of its members must be integers.
Your first task is trying to determine these independent variables. The best method I can think of at the moment assumes that the given frequencies have nearly consecutive indexes (x_1=x_0+n). So B_0/B_1=(x_0)/(x_0+n) given a (hopefully) small integer n. You can then take advantage of the fact that x_0 = n/(B_1-B_0), start with n=1, and keep ratcheting it up until k-rnd(k) is within a certain threshold. After you have x_0 (the initial index), you can approximate the root frequency (A = B_0/x_0). Then you can approximate the other indexes by finding x_n = rnd(B_n/A). This method is not very robust and will probably fail if the error in the data is large.
If you want a better approximation of the root frequency A, you can use linear regression to minimize the error of the linear model now that you have the corresponding dependent variables. The easiest method to do so uses least squares fitting. Wolfram's Mathworld has a in-depth mathematical treatment of the issue, but a fairly simple explanation can be found with some googling.
Interesting question...not easy.
I suppose I would look at the ratios of the sample values:
3.700 / 2.468 = 1.499...
6.1699 / 2.468 = 2.4999...
6.1699 / 3.700 = 1.6675...
And I'd then be looking for a simple ratio of integers in those results.
1.499 ~= 3/2
2.4999 ~= 5/2
1.6675 ~= 5/3
I haven't chased it through, but somewhere along the line, you decide that an error of 1:1000 or something is good enough, and you back-track to find the base approximate GCD.
The solution which I've seen and used myself is to choose some constant, say 1000, multiply all numbers by this constant, round them to integers, find the GCD of these integers using the standard algorithm and then divide the result by the said constant (1000). The larger the constant, the higher the precision.
This is a reformulaiton of shsmurfy's solution when you a priori choose 3 positive tolerances (e1,e2,e3)
The problem is then to search smallest positive integers (n1,n2,n3) and thus largest root frequency f such that:
f1 = n1*f +/- e1
f2 = n2*f +/- e2
f3 = n3*f +/- e3
We assume 0 <= f1 <= f2 <= f3
If we fix n1, then we get these relations:
f is in interval I1=[(f1-e1)/n1 , (f1+e1)/n1]
n2 is in interval I2=[n1*(f2-e2)/(f1+e1) , n1*(f2+e2)/(f1-e1)]
n3 is in interval I3=[n1*(f3-e3)/(f1+e1) , n1*(f3+e3)/(f1-e1)]
We start with n1 = 1, then increment n1 until the interval I2 and I3 contain an integer - that is floor(I2min) different from floor(I2max) same with I3
We then choose smallest integer n2 in interval I2, and smallest integer n3 in interval I3.
Assuming normal distribution of floating point errors, the most probable estimate of root frequency f is the one minimizing
J = (f1/n1 - f)^2 + (f2/n2 - f)^2 + (f3/n3 - f)^2
That is
f = (f1/n1 + f2/n2 + f3/n3)/3
If there are several integers n2,n3 in intervals I2,I3 we could also choose the pair that minimize the residue
min(J)*3/2=(f1/n1)^2+(f2/n2)^2+(f3/n3)^2-(f1/n1)*(f2/n2)-(f1/n1)*(f3/n3)-(f2/n2)*(f3/n3)
Another variant could be to continue iteration and try to minimize another criterium like min(J(n1))*n1, until f falls below a certain frequency (n1 reaches an upper limit)...
I found this question looking for answers for mine in MathStackExchange (here and here).
I've only managed (yet) to measure the appeal of a fundamental frequency given a list of harmonic frequencies (following the sound/music nomenclature), which can be useful if you have a reduced number of options and is feasible to compute the appeal of each one and then choose the best fit.
C&P from my question in MSE (there the formatting is prettier):
being v the list {v_1, v_2, ..., v_n}, ordered from lower to higher
mean_sin(v, x) = sum(sin(2*pi*v_i/x), for i in {1, ...,n})/n
mean_cos(v, x) = sum(cos(2*pi*v_i/x), for i in {1, ...,n})/n
gcd_appeal(v, x) = 1 - sqrt(mean_sin(v, x)^2 + (mean_cos(v, x) - 1)^2)/2, which yields a number in the interval [0,1].
The goal is to find the x that maximizes the appeal. Here is the (gcd_appeal) graph for your example [2.468, 3.700, 6.1699], where you find that the optimum GCD is at x = 1.2337899957639993
Edit:
You may find handy this JAVA code to calculate the (fuzzy) divisibility (aka gcd_appeal) of a divisor relative to a list of dividends; you can use it to test which of your candidates makes the best divisor. The code looks ugly because I tried to optimize it for performance.
//returns the mean divisibility of dividend/divisor as a value in the range [0 and 1]
// 0 means no divisibility at all
// 1 means full divisibility
public double divisibility(double divisor, double... dividends) {
double n = dividends.length;
double factor = 2.0 / divisor;
double sum_x = -n;
double sum_y = 0.0;
double[] coord = new double[2];
for (double v : dividends) {
coordinates(v * factor, coord);
sum_x += coord[0];
sum_y += coord[1];
}
double err = 1.0 - Math.sqrt(sum_x * sum_x + sum_y * sum_y) / (2.0 * n);
//Might happen due to approximation error
return err >= 0.0 ? err : 0.0;
}
private void coordinates(double x, double[] out) {
//Bhaskara performant approximation to
//out[0] = Math.cos(Math.PI*x);
//out[1] = Math.sin(Math.PI*x);
long cos_int_part = (long) (x + 0.5);
long sin_int_part = (long) x;
double rem = x - cos_int_part;
if (cos_int_part != sin_int_part) {
double common_s = 4.0 * rem;
double cos_rem_s = common_s * rem - 1.0;
double sin_rem_s = cos_rem_s + common_s + 1.0;
out[0] = (((cos_int_part & 1L) * 8L - 4L) * cos_rem_s) / (cos_rem_s + 5.0);
out[1] = (((sin_int_part & 1L) * 8L - 4L) * sin_rem_s) / (sin_rem_s + 5.0);
} else {
double common_s = 4.0 * rem - 4.0;
double sin_rem_s = common_s * rem;
double cos_rem_s = sin_rem_s + common_s + 3.0;
double common_2 = ((cos_int_part & 1L) * 8L - 4L);
out[0] = (common_2 * cos_rem_s) / (cos_rem_s + 5.0);
out[1] = (common_2 * sin_rem_s) / (sin_rem_s + 5.0);
}
}