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I'm trying to get a single element of an adjugate A_adj of a matrix A, both of which need to be symbolic expressions, where the symbols x_i are binary and the matrix A is symmetric and sparse. Python's sympy works great for small problems:
from sympy import zeros, symbols
size = 4
A = zeros(size,size)
x_i = [x for x in symbols(f'x0:{size}')]
for i in range(size-1):
A[i,i] += 0.5*x_i[i]
A[i+1,i+1] += 0.5*x_i[i]
A[i,i+1] = A[i+1,i] = -0.3*(i+1)*x_i[i]
A_adj_0 = A[1:,1:].det()
A_adj_0
This calculates the first element A_adj_0 of the cofactor matrix (which is the corresponding minor) and correctly gives me 0.125x_0x_1x_2 - 0.28x_2x_2^2 - 0.055x_1^2x_2 - 0.28x_1x_2^2, which is the expression I need, but there are two issues:
This is completely unfeasible for larger matrices (I need this for sizes of ~100).
The x_i are binary variables (i.e. either 0 or 1) and there seems to be no way for sympy to simplify expressions of binary variables, i.e. simplifying polynomials x_i^n = x_i.
The first issue can be partly addressed by instead solving a linear equation system Ay = b, where b is set to the first basis vector [1, 0, 0, 0], such that y is the first column of the inverse of A. The first entry of y is the first element of the inverse of A:
b = zeros(size,1)
b[0] = 1
y = A.LUsolve(b)
s = {x_i[i]: 1 for i in range(size)}
print(y[0].subs(s) * A.subs(s).det())
print(A_adj_0.subs(s))
The problem here is that the expression for the first element of y is extremely complicated, even after using simplify() and so on. It would be a very simple expression with simplification of binary expressions as mentioned in point 2 above. It's a faster method, but still unfeasible for larger matrices.
This boils down to my actual question:
Is there an efficient way to compute a single element of the adjugate of a sparse and symmetric symbolic matrix, where the symbols are binary values?
I'm open to using other software as well.
Addendum 1:
It seems simplifying binary expressions in sympy is possible with a simple custom substitution which I wasn't aware of:
A_subs = A_adj_0
for i in range(size):
A_subs = A_subs.subs(x_i[i]*x_i[i], x_i[i])
A_subs
You should make sure to use Rational rather than floats in sympy so S(1)/2 or Rational(1, 2) rather than 0.5.
There is a new (undocumented and for the moment internal) implementation of matrices in sympy called DomainMatrix. It is likely to be a lot faster for a problem like this and always produces polynomial results in a fully expanded form. I expect that it will be much faster for this kind of problem but it still seems to be fairly slow for this because is is not sparse internally (yet - that will probably change in the next release) and it does not take advantage of the simplification from the symbols being binary-valued. It can be made to work over GF(2) but not with symbols that are assumed to be in GF(2) which is something different.
In case it is helpful though this is how you would use it in sympy 1.7.1:
from sympy import zeros, symbols, Rational
from sympy.polys.domainmatrix import DomainMatrix
size = 10
A = zeros(size,size)
x_i = [x for x in symbols(f'x0:{size}')]
for i in range(size-1):
A[i,i] += Rational(1, 2)*x_i[i]
A[i+1,i+1] += Rational(1, 2)*x_i[i]
A[i,i+1] = A[i+1,i] = -Rational(3, 10)*(i+1)*x_i[i]
# Convert to DomainMatrix:
dM = DomainMatrix.from_list_sympy(size-1, size-1, A[1:, 1:].tolist())
# Compute determinant and convert back to normal sympy expression:
# Could also use dM.det().as_expr() although it might be slower
A_adj_0 = dM.charpoly()[-1].as_expr()
# Reduce powers:
A_adj_0 = A_adj_0.replace(lambda e: e.is_Pow, lambda e: e.args[0])
print(A_adj_0)
I'm looking for a efficient, uniformly distributed PRNG, that generates one random integer for any whole number point in the plain with coordinates x and y as input to the function.
int rand(int x, int y)
It has to deliver the same random number each time you input the same coordinate.
Do you know of algorithms, that can be used for this kind of problem and also in higher dimensions?
I already tried to use normal PRNGs like a LFSR and merged the x,y coordinates together to use it as a seed value. Something like this.
int seed = x << 16 | (y & 0xFFFF)
The obvious problem with this method is that the seed is not iterated over multiple times but is initialized again for every x,y-point. This results in very ugly non random patterns if you visualize the results.
I already know of the method which uses shuffled permutation tables of some size like 256 and you get a random integer out of it like this.
int r = P[x + P[y & 255] & 255];
But I don't want to use this method because of the very limited range, restricted period length and high memory consumption.
Thanks for any helpful suggestions!
I found a very simple, fast and sufficient hash function based on the xxhash algorithm.
// cash stands for chaos hash :D
int cash(int x, int y){
int h = seed + x*374761393 + y*668265263; //all constants are prime
h = (h^(h >> 13))*1274126177;
return h^(h >> 16);
}
It is now much faster than the lookup table method I described above and it looks equally random. I don't know if the random properties are good compared to xxhash but as long as it looks random to the eye it's a fair solution for my purpose.
This is what it looks like with the pixel coordinates as input:
My approach
In general i think you want some hash-function (mostly all of these are designed to output randomness; avalanche-effect for RNGs, explicitly needed randomness for CryptoPRNGs). Compare with this thread.
The following code uses this approach:
1) build something hashable from your input
2) hash -> random-bytes (non-cryptographically)
3) somehow convert these random-bytes to your integer range (hard to do correctly/uniformly!)
The last step is done by this approach, which seems to be not that fast, but has strong theoretical guarantees (selected answer was used).
The hash-function i used supports seeds, which will be used in step 3!
import xxhash
import math
import numpy as np
import matplotlib.pyplot as plt
import time
def rng(a, b, maxExclN=100):
# preprocessing
bytes_needed = int(math.ceil(maxExclN / 256.0))
smallest_power_larger = 2
while smallest_power_larger < maxExclN:
smallest_power_larger *= 2
counter = 0
while True:
random_hash = xxhash.xxh32(str((a, b)).encode('utf-8'), seed=counter).digest()
random_integer = int.from_bytes(random_hash[:bytes_needed], byteorder='little')
if random_integer < 0:
counter += 1
continue # inefficient but safe; could be improved
random_integer = random_integer % smallest_power_larger
if random_integer < maxExclN:
return random_integer
else:
counter += 1
test_a = rng(3, 6)
test_b = rng(3, 9)
test_c = rng(3, 6)
print(test_a, test_b, test_c) # OUTPUT: 90 22 90
random_as = np.random.randint(100, size=1000000)
random_bs = np.random.randint(100, size=1000000)
start = time.time()
rands = [rng(*x) for x in zip(random_as, random_bs)]
end = time.time()
plt.hist(rands, bins=100)
plt.show()
print('needed secs: ', end-start)
# OUTPUT: needed secs: 15.056888341903687 -> 0,015056 per sample
# -> possibly heavy-dependence on range of output
Possible improvements
Add additional entropy from some source (urandom; could be put into str)
Make a class and initialize to memorize preprocessing (costly if done for each sampling)
Handle negative integers; maybe just use abs(x)
Assumptions:
the ouput-range is [0, N) -> just shift for others!
the output-range is smaller (bits) than the hash-output (may use xxh64)
Evaluation:
Check randomness/uniformity
Check if deterministic regarding input
You can use various randomness extractors to achieve your goals. There are at least two sources you can look for a solution.
Dodis et al, "Randomness Extraction and Key Derivation
Using the CBC, Cascade and HMAC Modes"
NIST SP800-90 "Recommendation for the Entropy Sources Used for
Random Bit Generation"
All in all, you can preferably use:
AES-CBC-MAC using a random key (may be fixed and reused)
HMAC, preferably with SHA2-512
SHA-family hash functions (SHA1, SHA256 etc); using a random final block (eg use a big random salt at the end)
Thus, you can concatenate your coordinates, get their bytes, add a random key (for AES and HMAC) or a salt for SHA and your output has an adequate entropy.
According to NIST, the output entropy relies on the input entropy:
Assuming you use SHA1; thus n = 160bits. Let's suppose that m = input_entropy (your coordinates' entropy)
if m >= 2n then output_entropy=n=160 bits
if 2n < m <= n then maximum output_entropy=m (but full entropy is not guaranteed).
if m < n then maximum output_entropy=m (this is your case)
see NIST sp800-90c (page 11)
I am very new in Apache Spark Scala. Can you help me with some operations?
I have two distributed matrix H and Y in Spark Scala.
I want to compute the pseudo-inverse of H and then multiply H and Y.
How can I do this?
Here is an implementation for the inverse.
import org.apache.spark.mllib.linalg.{Vectors,Vector,Matrix,SingularValueDecomposition,DenseMatrix,DenseVector}
import org.apache.spark.mllib.linalg.distributed.RowMatrix
def computeInverse(X: RowMatrix): DenseMatrix = {
val nCoef = X.numCols.toInt
val svd = X.computeSVD(nCoef, computeU = true)
if (svd.s.size < nCoef) {
sys.error(s"RowMatrix.computeInverse called on singular matrix.")
}
// Create the inv diagonal matrix from S
val invS = DenseMatrix.diag(new DenseVector(svd.s.toArray.map(x => math.pow(x,-1))))
// U cannot be a RowMatrix
val U = new DenseMatrix(svd.U.numRows().toInt,svd.U.numCols().toInt,svd.U.rows.collect.flatMap(x => x.toArray))
// If you could make V distributed, then this may be better. However its alreadly local...so maybe this is fine.
val V = svd.V
// inv(X) = V*inv(S)*transpose(U) --- the U is already transposed.
(V.multiply(invS)).multiply(U)
}
To calculate the pseudo-inverse of non-square matrices you need to be able to calculate the transpose (easy) and the matrix inverse (others have supplied that functionality). There are two different calculations, depending on whether M has full column rank or full row rank.
Full column rank means that the columns of the matrix are linearly independent which requires that the the number of columns is less than or equal to the number of rows. (In pathological cases, an mxn matrix with m>=n might still not have full column rank, but we'll ignore that statistical impossibility. If it is a possibility in your case, the matrix inversion step below will fail.) For full column rank, the pseudo-inverse is
M^+ = (M^T M)^{-1} M^T
where M^T is the transpose of M. Matrix multiply M^T by M, then take the inverse, then matrix multiply by M^T again. (I'm assuming M has real number entries; if the entries are complex numbers, you also have to take complex conjugates.)
A quick check to make sure you have calculated the psuedo-inverse correctly is to check M^+ M. It should be the identity matrix (up to floating point error).
On the other hand, if M has full row rank, in other words M is mxn with m<=n, the pseudo-inverse is
M^+ = M^T (M M^T)^{-1}
To check whether you have the correct pseudo-inverse in this case, right multiply with the original matrix: M M^+. That should equal the identity matrix, up to floating point error.
Matrix multiplication is the easier one: there are several Matrix implementations with a multiply method in packages org.apache.spark.mllib.linalg and org.apache.spark.mllib.linalg.distributed. Pick whatever fits your needs most.
I have not seen (pseudo-)inverse anywhere in the Spark API. But RowMatrix is able to compute the singular value decomposition which can be used to calculate the inverse of a matrix. Here is a very naive implementation, inspired by How can we compute Pseudoinverse for any Matrix (warning: dimensions of the 2x2 matrix are hard-coded):
val m = new RowMatrix(sc.parallelize(Seq(Vectors.dense(4, 3), Vectors.dense(3, 2))))
val svd = m.computeSVD(2, true)
val v = svd.V
val sInvArray = svd.s.toArray.toList.map(x => 1.0 / x).toArray
val sInverse = new DenseMatrix(2, 2, Matrices.diag(Vectors.dense(sInvArray)).toArray)
val uArray = svd.U.rows.collect.toList.map(_.toArray.toList).flatten.toArray
val uTranspose = new DenseMatrix(2, 2, uArray) // already transposed because DenseMatrix is column-major
val inverse = v.multiply(sInverse).multiply(uTranspose)
// -1.9999999999998297 2.999999999999767
// 2.9999999999997637 -3.9999999999996767
Unfortunately, a lot of conversion from Matrix to Array and so forth is necessary. If you need a fully distributed implementation, try using DistributedMatrix instead of DenseMatrix. If not, maybe using Breeze is preferable here.
Let's say that I know the probability of a "success" is P. I run the test N times, and I see S successes. The test is akin to tossing an unevenly weighted coin (perhaps heads is a success, tails is a failure).
I want to know the approximate probability of seeing either S successes, or a number of successes less likely than S successes.
So for example, if P is 0.3, N is 100, and I get 20 successes, I'm looking for the probability of getting 20 or fewer successes.
If, on the other hadn, P is 0.3, N is 100, and I get 40 successes, I'm looking for the probability of getting 40 our more successes.
I'm aware that this problem relates to finding the area under a binomial curve, however:
My math-fu is not up to the task of translating this knowledge into efficient code
While I understand a binomial curve would give an exact result, I get the impression that it would be inherently inefficient. A fast method to calculate an approximate result would suffice.
I should stress that this computation has to be fast, and should ideally be determinable with standard 64 or 128 bit floating point computation.
I'm looking for a function that takes P, S, and N - and returns a probability. As I'm more familiar with code than mathematical notation, I'd prefer that any answers employ pseudo-code or code.
Exact Binomial Distribution
def factorial(n):
if n < 2: return 1
return reduce(lambda x, y: x*y, xrange(2, int(n)+1))
def prob(s, p, n):
x = 1.0 - p
a = n - s
b = s + 1
c = a + b - 1
prob = 0.0
for j in xrange(a, c + 1):
prob += factorial(c) / (factorial(j)*factorial(c-j)) \
* x**j * (1 - x)**(c-j)
return prob
>>> prob(20, 0.3, 100)
0.016462853241869437
>>> 1-prob(40-1, 0.3, 100)
0.020988576003924564
Normal Estimate, good for large n
import math
def erf(z):
t = 1.0 / (1.0 + 0.5 * abs(z))
# use Horner's method
ans = 1 - t * math.exp( -z*z - 1.26551223 +
t * ( 1.00002368 +
t * ( 0.37409196 +
t * ( 0.09678418 +
t * (-0.18628806 +
t * ( 0.27886807 +
t * (-1.13520398 +
t * ( 1.48851587 +
t * (-0.82215223 +
t * ( 0.17087277))))))))))
if z >= 0.0:
return ans
else:
return -ans
def normal_estimate(s, p, n):
u = n * p
o = (u * (1-p)) ** 0.5
return 0.5 * (1 + erf((s-u)/(o*2**0.5)))
>>> normal_estimate(20, 0.3, 100)
0.014548164531920815
>>> 1-normal_estimate(40-1, 0.3, 100)
0.024767304545069813
Poisson Estimate: Good for large n and small p
import math
def poisson(s,p,n):
L = n*p
sum = 0
for i in xrange(0, s+1):
sum += L**i/factorial(i)
return sum*math.e**(-L)
>>> poisson(20, 0.3, 100)
0.013411150012837811
>>> 1-poisson(40-1, 0.3, 100)
0.046253037645840323
I was on a project where we needed to be able to calculate the binomial CDF in an environment that didn't have a factorial or gamma function defined. It took me a few weeks, but I ended up coming up with the following algorithm which calculates the CDF exactly (i.e. no approximation necessary). Python is basically as good as pseudocode, right?
import numpy as np
def binomial_cdf(x,n,p):
cdf = 0
b = 0
for k in range(x+1):
if k > 0:
b += + np.log(n-k+1) - np.log(k)
log_pmf_k = b + k * np.log(p) + (n-k) * np.log(1-p)
cdf += np.exp(log_pmf_k)
return cdf
Performance scales with x. For small values of x, this solution is about an order of magnitude faster than scipy.stats.binom.cdf, with similar performance at around x=10,000.
I won't go into a full derivation of this algorithm because stackoverflow doesn't support MathJax, but the thrust of it is first identifying the following equivalence:
For all k > 0, sp.misc.comb(n,k) == np.prod([(n-k+1)/k for k in range(1,k+1)])
Which we can rewrite as:
sp.misc.comb(n,k) == sp.misc.comb(n,k-1) * (n-k+1)/k
or in log space:
np.log( sp.misc.comb(n,k) ) == np.log(sp.misc.comb(n,k-1)) + np.log(n-k+1) - np.log(k)
Because the CDF is a summation of PMFs, we can use this formulation to calculate the binomial coefficient (the log of which is b in the function above) for PMF_{x=i} from the coefficient we calculated for PMF_{x=i-1}. This means we can do everything inside a single loop using accumulators, and we don't need to calculate any factorials!
The reason most of the calculations are done in log space is to improve the numerical stability of the polynomial terms, i.e. p^x and (1-p)^(1-x) have the potential to be extremely large or extremely small, which can cause computational errors.
EDIT: Is this a novel algorithm? I've been poking around on and off since before I posted this, and I'm increasingly wondering if I should write this up more formally and submit it to a journal.
I think you want to evaluate the incomplete beta function.
There's a nice implementation using a continued fraction representation in "Numerical Recipes In C", chapter 6: 'Special Functions'.
I can't totally vouch for the efficiency, but Scipy has a module for this
from scipy.stats.distributions import binom
binom.cdf(successes, attempts, chance_of_success_per_attempt)
An efficient and, more importantly, numerical stable algorithm exists in the domain of Bezier Curves used in Computer Aided Design. It is called de Casteljau's algorithm used to evaluate the Bernstein Polynomials used to define Bezier Curves.
I believe that I am only allowed one link per answer so start with Wikipedia - Bernstein Polynomials
Notice the very close relationship between the Binomial Distribution and the Bernstein Polynomials. Then click through to the link on de Casteljau's algorithm.
Lets say I know the probability of throwing a heads with a particular coin is P.
What is the probability of me throwing
the coin T times and getting at least
S heads?
Set n = T
Set beta[i] = 0 for i = 0, ... S - 1
Set beta[i] = 1 for i = S, ... T
Set t = p
Evaluate B(t) using de Casteljau
or at most S heads?
Set n = T
Set beta[i] = 1 for i = 0, ... S
Set beta[i] = 0 for i = S + 1, ... T
Set t = p
Evaluate B(t) using de Casteljau
Open source code probably exists already. NURBS Curves (Non-Uniform Rational B-spline Curves) are a generalization of Bezier Curves and are widely used in CAD. Try openNurbs (the license is very liberal) or failing that Open CASCADE (a somewhat less liberal and opaque license). Both toolkits are in C++, though, IIRC, .NET bindings exist.
If you are using Python, no need to code it yourself. Scipy got you covered:
from scipy.stats import binom
# probability that you get 20 or less successes out of 100, when p=0.3
binom.cdf(20, 100, 0.3)
>>> 0.016462853241869434
# probability that you get exactly 20 successes out of 100, when p=0.3
binom.pmf(20, 100, 0.3)
>>> 0.0075756449257260777
From the portion of your question "getting at least S heads" you want the cummulative binomial distribution function. See http://en.wikipedia.org/wiki/Binomial_distribution for the equation, which is described as being in terms of the "regularized incomplete beta function" (as already answered). If you just want to calculate the answer without having to implement the entire solution yourself, the GNU Scientific Library provides the function: gsl_cdf_binomial_P and gsl_cdf_binomial_Q.
The DCDFLIB Project has C# functions (wrappers around C code) to evaluate many CDF functions, including the binomial distribution. You can find the original C and FORTRAN code here. This code is well tested and accurate.
If you want to write your own code to avoid being dependent on an external library, you could use the normal approximation to the binomial mentioned in other answers. Here are some notes on how good the approximation is under various circumstances. If you go that route and need code to compute the normal CDF, here's Python code for doing that. It's only about a dozen lines of code and could easily be ported to any other language. But if you want high accuracy and efficient code, you're better off using third party code like DCDFLIB. Several man-years went into producing that library.
Try this one, used in GMP. Another reference is this.
import numpy as np
np.random.seed(1)
x=np.random.binomial(20,0.6,10000) #20 flips of coin,probability of
heads percentage and 10000 times
done.
sum(x>12)/len(x)
The output is 41% of times we got 12 heads.
How can I convert a uniform distribution (as most random number generators produce, e.g. between 0.0 and 1.0) into a normal distribution? What if I want a mean and standard deviation of my choosing?
There are plenty of methods:
Do not use Box Muller. Especially if you draw many gaussian numbers. Box Muller yields a result which is clamped between -6 and 6 (assuming double precision. Things worsen with floats.). And it is really less efficient than other available methods.
Ziggurat is fine, but needs a table lookup (and some platform-specific tweaking due to cache size issues)
Ratio-of-uniforms is my favorite, only a few addition/multiplications and a log 1/50th of the time (eg. look there).
Inverting the CDF is efficient (and overlooked, why ?), you have fast implementations of it available if you search google. It is mandatory for Quasi-Random numbers.
The Ziggurat algorithm is pretty efficient for this, although the Box-Muller transform is easier to implement from scratch (and not crazy slow).
Changing the distribution of any function to another involves using the inverse of the function you want.
In other words, if you aim for a specific probability function p(x) you get the distribution by integrating over it -> d(x) = integral(p(x)) and use its inverse: Inv(d(x)). Now use the random probability function (which have uniform distribution) and cast the result value through the function Inv(d(x)). You should get random values cast with distribution according to the function you chose.
This is the generic math approach - by using it you can now choose any probability or distribution function you have as long as it have inverse or good inverse approximation.
Hope this helped and thanks for the small remark about using the distribution and not the probability itself.
Here is a javascript implementation using the polar form of the Box-Muller transformation.
/*
* Returns member of set with a given mean and standard deviation
* mean: mean
* standard deviation: std_dev
*/
function createMemberInNormalDistribution(mean,std_dev){
return mean + (gaussRandom()*std_dev);
}
/*
* Returns random number in normal distribution centering on 0.
* ~95% of numbers returned should fall between -2 and 2
* ie within two standard deviations
*/
function gaussRandom() {
var u = 2*Math.random()-1;
var v = 2*Math.random()-1;
var r = u*u + v*v;
/*if outside interval [0,1] start over*/
if(r == 0 || r >= 1) return gaussRandom();
var c = Math.sqrt(-2*Math.log(r)/r);
return u*c;
/* todo: optimize this algorithm by caching (v*c)
* and returning next time gaussRandom() is called.
* left out for simplicity */
}
Where R1, R2 are random uniform numbers:
NORMAL DISTRIBUTION, with SD of 1:
sqrt(-2*log(R1))*cos(2*pi*R2)
This is exact... no need to do all those slow loops!
Reference: dspguide.com/ch2/6.htm
Use the central limit theorem wikipedia entry mathworld entry to your advantage.
Generate n of the uniformly distributed numbers, sum them, subtract n*0.5 and you have the output of an approximately normal distribution with mean equal to 0 and variance equal to (1/12) * (1/sqrt(N)) (see wikipedia on uniform distributions for that last one)
n=10 gives you something half decent fast. If you want something more than half decent go for tylers solution (as noted in the wikipedia entry on normal distributions)
I would use Box-Muller. Two things about this:
You end up with two values per iteration
Typically, you cache one value and return the other. On the next call for a sample, you return the cached value.
Box-Muller gives a Z-score
You have to then scale the Z-score by the standard deviation and add the mean to get the full value in the normal distribution.
It seems incredible that I could add something to this after eight years, but for the case of Java I would like to point readers to the Random.nextGaussian() method, which generates a Gaussian distribution with mean 0.0 and standard deviation 1.0 for you.
A simple addition and/or multiplication will change the mean and standard deviation to your needs.
The standard Python library module random has what you want:
normalvariate(mu, sigma)
Normal distribution. mu is the mean, and sigma is the standard deviation.
For the algorithm itself, take a look at the function in random.py in the Python library.
The manual entry is here
This is a Matlab implementation using the polar form of the Box-Muller transformation:
Function randn_box_muller.m:
function [values] = randn_box_muller(n, mean, std_dev)
if nargin == 1
mean = 0;
std_dev = 1;
end
r = gaussRandomN(n);
values = r.*std_dev - mean;
end
function [values] = gaussRandomN(n)
[u, v, r] = gaussRandomNValid(n);
c = sqrt(-2*log(r)./r);
values = u.*c;
end
function [u, v, r] = gaussRandomNValid(n)
r = zeros(n, 1);
u = zeros(n, 1);
v = zeros(n, 1);
filter = r==0 | r>=1;
% if outside interval [0,1] start over
while n ~= 0
u(filter) = 2*rand(n, 1)-1;
v(filter) = 2*rand(n, 1)-1;
r(filter) = u(filter).*u(filter) + v(filter).*v(filter);
filter = r==0 | r>=1;
n = size(r(filter),1);
end
end
And invoking histfit(randn_box_muller(10000000),100); this is the result:
Obviously it is really inefficient compared with the Matlab built-in randn.
This is my JavaScript implementation of Algorithm P (Polar method for normal deviates) from Section 3.4.1 of Donald Knuth's book The Art of Computer Programming:
function normal_random(mean,stddev)
{
var V1
var V2
var S
do{
var U1 = Math.random() // return uniform distributed in [0,1[
var U2 = Math.random()
V1 = 2*U1-1
V2 = 2*U2-1
S = V1*V1+V2*V2
}while(S >= 1)
if(S===0) return 0
return mean+stddev*(V1*Math.sqrt(-2*Math.log(S)/S))
}
I thing you should try this in EXCEL: =norminv(rand();0;1). This will product the random numbers which should be normally distributed with the zero mean and unite variance. "0" can be supplied with any value, so that the numbers will be of desired mean, and by changing "1", you will get the variance equal to the square of your input.
For example: =norminv(rand();50;3) will yield to the normally distributed numbers with MEAN = 50 VARIANCE = 9.
Q How can I convert a uniform distribution (as most random number generators produce, e.g. between 0.0 and 1.0) into a normal distribution?
For software implementation I know couple random generator names which give you a pseudo uniform random sequence in [0,1] (Mersenne Twister, Linear Congruate Generator). Let's call it U(x)
It is exist mathematical area which called probibility theory.
First thing: If you want to model r.v. with integral distribution F then you can try just to evaluate F^-1(U(x)). In pr.theory it was proved that such r.v. will have integral distribution F.
Step 2 can be appliable to generate r.v.~F without usage of any counting methods when F^-1 can be derived analytically without problems. (e.g. exp.distribution)
To model normal distribution you can cacculate y1*cos(y2), where y1~is uniform in[0,2pi]. and y2 is the relei distribution.
Q: What if I want a mean and standard deviation of my choosing?
You can calculate sigma*N(0,1)+m.
It can be shown that such shifting and scaling lead to N(m,sigma)
I have the following code which maybe could help:
set.seed(123)
n <- 1000
u <- runif(n) #creates U
x <- -log(u)
y <- runif(n, max=u*sqrt((2*exp(1))/pi)) #create Y
z <- ifelse (y < dnorm(x)/2, -x, NA)
z <- ifelse ((y > dnorm(x)/2) & (y < dnorm(x)), x, z)
z <- z[!is.na(z)]
It is also easier to use the implemented function rnorm() since it is faster than writing a random number generator for the normal distribution. See the following code as prove
n <- length(z)
t0 <- Sys.time()
z <- rnorm(n)
t1 <- Sys.time()
t1-t0
function distRandom(){
do{
x=random(DISTRIBUTION_DOMAIN);
}while(random(DISTRIBUTION_RANGE)>=distributionFunction(x));
return x;
}