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Generate random doubles out of the whole positive range of doubles

for testing a specific math function I implemented I need to generate a lot of doubles in the full range of doubles > 0. So the random generated value should be between 2^−52 × 2^−1022 and (2−(2^−52)) × 2^1023. I tried using
ThreadLocalRandom.current().nextDouble(origin, bound)
but it only gives values close to 1e300.
I tested it with
public void testRandomDouble() {
double current;
for (int j = 0; j < 10; j++) {
double min = Double.POSITIVE_INFINITY;
for (int i = 0; i < 100_000_000; i++) {
current = ThreadLocalRandom.current().nextDouble(Double.MIN_VALUE, Double.MAX_VALUE);
if (current < min) {
min = current;
}
}
System.out.println(min);
}
}
generating the output
1.2100736287390257E300
1.2292284466505449E300
1.4318629398915128E299
6.922983256724938E299
1.3927453080775622E300
4.8454315085367987E300
1.4899199386555683E299
3.7592835763389994E299
2.0561053862668256E300
1.6268118313101214E299
Even when recompiling and rerunning the test (so the local thread has a different seed) results in approx. the same output. I didn't find anything about this behavior online. What am I missing?

There are ten times as many values in the range 1e300 as 1e299, and ten times as many values in the range 1e299 as in 298. This shouldn't necessarily be surprising! You should expect 90% of your values to be e300, 99% to be e299 or e300, etc. If you want a uniform distribution over the values that can possibly be held in a double, not a uniform distribution over the number line, you will need a very different algorithm. That would probably look something like
double d;
do {
d = Double.longBitsToDouble(random.nextLong());
} while (Double.isNaN(d) || Double.isInfinite(d) || d <= 0);
return d;

Java method to calculate the square root via minuend and subtrahend and print each step

I am tasked with creating a program in java which calculates the square root of a double and goes through each step of calculating it manually. The requirements are:
split the number into number pairs including the decimal point (1234.67 -> 12 34 67) to prepare for subtraction. If the number is uneven, a zero must populate (234.67 -> 02 34 67)
Print each pair (each pair is a minuend), one at a time, into the console and have the console show the subtraction. Subtrahend starts at 1 and so long as the result >= 0, the subtrahend increases by 2.
The count of subtrahends is the first number of the final square root output, the count of subtrahends from the second round is the second number of the square root output, etc.
From the first subtrahend round, take the remainder and join it to the second number pair, this is the new minuend for the second round of subtraction
Calculate the second subtrahend in round two by doubling the first number of the square root output and adding 1 in the first digit position
Repeat step 2, increasing by 2 each time
Step 5 and 6 repeat until two decimal places are reached
My question is with the number pairs in step 1 and getting the subsequent subtrahends after step 3 as a number to calculate. We are given the following visual:
My current thought is to put the double into a string and then tell java that each number pair is a number. I have a method created which creates a string from a double, but I am still missing how to incorporate the decimal place numbers. From my C class, I remember multiplying decimals by 100 to "store" the decimal numbers before converting them back later with another division by 100. I'm sure there is a java library that is able to do this but we are specifically not allowed to use them.
I think I should be able to continue on with the rest of the problem once I get past this point of splitting the number into number pairs inclusive of the decimals.
This is also my first stack post so if you have any tips on how to better write questions for future posts that would be helpful as well.
This is my current array method to store a given double into an array:
public static void printArray(int [] a) //printer helper method
{
for(int i = 0; i < a.length; i++)
{
System.out.print(a[i]);
}
}
public static void stringDigits (double n) //begin string method
{
int a [] = new int [15];
int i = 0;
int stringLength = 0;
while(n > 1)
{
a[i] = (int) (n % 10);
n = n / 10;
i++;
}
for(int j = 0; a[j] != 0; j++)
{
System.out.print(a[j]);
if(a[j] != 0)
{
stringLength++;
}
}
System.out.println("");
System.out.println(stringLength);
int[] numbersArray = new int[stringLength];
int g = 0;
for(int k = a.length-1; g < numbersArray.length; k--)
{
if(a[k] > 0)
{
numbersArray[g] = a[k];
g++;
}
}
System.out.println("");
printArray(numbersArray);
}
I've tried at first to store the value of the double into an int[] a array so that I can then select the numbers in pairs and then somehow combine them back into numbers. So if the array is {1,2,3,4,5,6} my next idea is to get java to convert a[0] + a[1] into the number 12 to prepare for the subtraction step.
This link looks close but does anyone know why the numbers are "10l" and "100l" etc? I've tested some of the answers and they dont produce the proper squareroot compared to the sqrt function from the math library.
Create a program that calculates the square root of a number without using Math.sqrt

Get a random number within a range with a bias

Hello i am trying to make a method to generate a random number within a range
where it can take a Bias that will make the number more likely to be higher/lower depending on the bias.
To do this currently i was using this
public int randIntWeightedLow(int max, int min, int rolls){
int rValue = 100;
for (int i = 0; i < rolls ; i++) {
int rand = randInt(min, max);
if (rand < rValue ){
rValue = rand;
}
}
return rValue;
}
This works okay by giving me a number in the range and the more rolls i add the likely the number will be low. However the problem i am running in to is that the there is a big difference between having 3 rolls and 4 rolls.
I am loking to have somthing like
public void randomIntWithBias(int min, int max, float bias){
}
Where giving a negative bias would make the number be low more often and
a positive bias make the number be higher more often but still keeping the number in the random of the min and max.
Currently to generate a random number i am using
public int randInt(final int n1, final int n2) {
if (n1 == n2) {
return n1;
}
final int min = n1 > n2 ? n2 : n1;
final int max = n1 > n2 ? n1 : n2;
return rand.nextInt(max - min + 1) + min;
}
I am new to java and coding in general so any help would be greatly appreciated.

Ok, here is quick sketch how it could be done.
First, I propose to use Apache commons java library, it has sampling for integers
with different probabilities already implemented. We need Enumerated Integer Distribution.
Second, two parameters to make distribution look linear, p0 and delta.
For kth value relative probability would be p0 + k*delta. For delta positive
larger numbers will be more probable, for delta negative smaller numbers will be
more probable, delta=0 equal to uniform sampling.
Code (my Java is rusty, please bear with me)
import org.apache.commons.math3.distribution.EnumeratedIntegerDistribution;
public int randomIntWithBias(int min, int max, double p0, double delta){
if (p0 < 0.0)
throw new Exception("Negative initial probability");
int N = max - min + 1; // total number of items to sample
double[] p = new double[N]; // probabilities
int[] items = new int[N]; // items
double sum = 0.0; // total probabilities summed
for(int k = 0; k != N; ++k) { // fill arrays
p[k] = p0 + k*delta;
sum += p[k];
items[k] = min + k;
}
if (delta < 0.0) { // when delta negative we could get negative probabilities
if (p[N-1] < 0.0) // check only last probability
throw new Exception("Negative probability");
}
for(int k = 0; k != N; ++k) { // Normalize probabilities
p[k] /= sum;
}
EnumeratedIntegerDistribution rng = new EnumeratedIntegerDistribution(items, p);
return rng.sample();
}
That's the gist of the idea, code could be (and should be) optimized and cleaned.
UPDATE
Of course, instead of linear bias function you could put in, say, quadratic one.
General quadratic function has three parameters - pass them on, fill in a similar way array of probabilities, normalize, sample

A efficient binomial random number generator code in Java

The relevant question is: Algorithm to generate Poisson and binomial random numbers?
I just take her description for the Binomial random number:
For example, consider binomial random numbers. A binomial random
number is the number of heads in N tosses of a coin with probability p
of a heads on any single toss. If you generate N uniform random
numbers on the interval (0,1) and count the number less than p, then
the count is a binomial random number with parameters N and p.
There is a trivial solution in Algorithm to generate Poisson and binomial random numbers? through using iterations:
public static int getBinomial(int n, double p) {
int x = 0;
for(int i = 0; i < n; i++) {
if(Math.random() < p)
x++;
}
return x;
}
However, my purpose of pursing a binomial random number generator is just to avoid the inefficient loops (i from 0 to n). My n could be very large. And p is often very small.
A toy example of my case could be: n=1*10^6, p=1*10^(-7).
The n could range from 1*10^3 to 1*10^10.

If you have small p values, you'll like this one better than the naive implementation you cited. It still loops, but the expected number of iterations is O(np) so it's pretty fast for small p values. If you're working with large p values, replace p with q = 1-p and subtract the return value from n. Clearly, it will be at its worst when p = q = 0.5.
public static int getBinomial(int n, double p) {
double log_q = Math.log(1.0 - p);
int x = 0;
double sum = 0;
for(;;) {
sum += Math.log(Math.random()) / (n - x);
if(sum < log_q) {
return x;
}
x++;
}
}
The implementation is a variant of Luc Devroye's "Second Waiting Time Method" on page 522 of his text "Non-Uniform Random Variate Generation."
There are faster methods based on acceptance/rejection techniques, but they are substantially more complex to implement.

I could imagine one way to speed it up by a constant factor (e.g. 4).
After 4 throws you will toss a head 0,1,2,3 or 4.
The probabilities for it are something like [0.6561, 0.2916, 0.0486, 0.0036, 0.0001].
Now you can generate one number random number and simulate 4 original throws. If that's not clear how I can elaborate a little more.
This way after some original pre-calculation you can speedup the process almost 4 times. The only requirement for it to be precise is that the granularity of your random generator is at least p^4.

Java Biasing Random Numbers in a Triangular Array

This question is an extension of Java- Math.random(): Selecting an element of a 13 by 13 triangular array. I am selecting two numbers at random (0-12 inclusive) and I wanted the values to be equal.
But now, since this is a multiplication game, I want a way to bias the results so certain combinations come up more frequently (like if the Player does worse for 12x8, I want it to come up more frequently). Eventually, I would like to bias towards any of the 91 combinations, but once I get this down, that should not be hard.
My Thoughts: Add some int n to the triangular number and Random.nextInt(91 + n) to bias the results toward a combination.
private int[] triLessThan(int x, int[] bias) { // I'm thinking a 91 element array, 0 for no bias, positive for bias towards
int i = 0;
int last = 0;
while (true) {
int sum = 0;
for (int a = 0; a < i * (i + 2)/2; a++){
sum += bias[a]
}
int triangle = i * (i + 1) / 2;
if (triangle + sum > x){
int[] toReturn = {last,i};
return toReturn;
}
last = triangle;
i++;
}
}
At the random number roll:
int sum = sumOfArray(bias); // bias is the array;
int roll = random.nextInt(91 + sum);
int[] triNum = triLessThan(roll);
int num1 = triNum[1];
int num2 = roll - triNum[0]; //now split into parts and make bias[] add chances to one number.
where sumOfArray just finds the sum (that formula is easy). Will this work?
Edit: Using Floris's idea:
At random number roll:
int[] bias = {1,1,1,...,1,1,1} // 91 elements
int roll = random.nextInt(sumOfBias());
int num1 = roll;
int num2 = 0;
while (roll > 0){
roll -= bias[num2];
num2++;
}
num1 = (int) (Math.sqrt(8 * num2 + 1) - 1)/2;
num2 -= num1 * (num1 + 1) / 2;

You already know how to convert a number between 0 and 91 and turn it into a roll (from the answer to your previous question). I would suggest that you create an array of N elements, where N >> 91. Fill the first 91 elements with 0...90, and set a counter A to 91. Now choose a number between 0 and A, pick the corresponding element from the array, and convert to a multiplication problem. If the answer is wrong, append the number of the problem to the end of the array, and increment A by one.
This will create an array in which the frequencies of sampling will represent the number of times a problem was solved incorrectly - but it doesn't ever lower the frequency again if the problem is solved correctly the next time it is asked.
An alternative and better solution, and one that is a little closer to yours (but distinct) creates an array of 91 frequencies - each initially set to 1 - and keeps track of the sum (initially 91). But now, when you choose a random number (between 0 and sum) you traverse the array until the cumulative sum is greater then your random number - the number of the bin is the roll you choose, and you convert that with the formula derived earlier. If the answer is wrong you increment the bin and update the sum; if it is right, you decrement the sum but never to a value less than one, and update the sum. Repeat.
This should give you exactly what you are asking: given an array of 91 numbers ("bins"), randomly select a bin in such a way that the probability of that bin is proportional to the value in it. Return the index of the bin (which can be turned into the combination of numbers using the method you had before). This function is called with the bin (frequency) array as the first parameter, and the cumulative sum as the second. You look up where the cumulative sum of the first n elements first exceeds a random number scaled by the sum of the frequencies:
private int chooseBin(float[] freq, float fsum) {
// given an array of frequencies (probabilities) freq
// and the sum of this array, fsum
// choose a random number between 0 and 90
// such that if this function is called many times
// the frequency with which each value is observed converges
// on the frequencies in freq
float x, cs=0; // x stores random value, cs is cumulative sum
int ii=-1; // variable that increments until random value is found
x = Math.rand();
while(cs < x*fsum && ii<90) {
// increment cumulative sum until it's bigger than fraction x of sum
ii++;
cs += freq[ii];
}
return ii;
}
I confirmed that it gives me a histogram (blue bars) that looks exactly like the probability distribution that I fed it (red line):
(note - this was plotted with matlab so X goes from 1 to 91, not from 0 to 90).
Here is another idea (this is not really answering the question, but it's potentially even more interesting):
You can skew your probability of choosing a particular problem by sampling something other than a uniform distribution. For example, the square of a uniformly sampled random variate will favor smaller numbers. This gives us an interesting possibility:
First, shuffle your 91 numbers into a random order
Next, pick a number from a non-uniform distribution (one that favors smaller numbers). Since the numbers were randomly shuffled, they are in fact equally likely to be chosen. But now here's the trick: if the problem (represented by the number picked) is solved correctly, you move the problem number "to the top of the stack", where it is least likely to be chosen again. If the player gets it wrong, it is moved to the bottom of the stack, where it is most likely to be chosen again. Over time, difficult problems move to the bottom of the stack.
You can create random distributions with different degrees of skew using a variation of
roll = (int)(91*(asin(Math.rand()*a)/asin(a)))
As you make a closer to 1, the function tends to favor lower numbers with almost zero probability of higher numbers:
I believe the following code sections do what I described:
private int[] chooseProblem(float bias, int[] currentShuffle) {
// if bias == 0, we choose from uniform distribution
// for 0 < bias <= 1, we choose from increasingly biased distribution
// for bias > 1, we choose from uniform distribution
// array currentShuffle contains the numbers 0..90, initially in shuffled order
// when a problem is solved correctly it is moved to the top of the pile
// when it is wrong, it is moved to the bottom.
// return value contains number1, number2, and the current position of the problem in the list
int problem, problemIndex;
if(bias < 0 || bias > 1) bias = 0;
if(bias == 0) {
problem = random.nextInt(91);
problemIndex = problem;
}
else {
float x = asin(Math.random()*bias)/asin(bias);
problemIndex = Math.floor(91*x);
problem = currentShuffle[problemIndex];
}
// now convert "problem number" into two numbers:
int first, last;
first = (int)((Math.sqrt(8*problem + 1)-1)/2);
last = problem - first * (first+1) / 2;
// and return the result:
return {first, last, problemIndex};
}
private void shuffleProblems(int[] currentShuffle, int upDown) {
// when upDown==0, return a randomly shuffled array
// when upDown < 0, (wrong answer) move element[-upDown] to zero
// when upDown > 0, (correct answer) move element[upDown] to last position
// note - if problem 0 is answered incorrectly, don't call this routine!
int ii, temp, swap;
if(upDown == 0) {
// first an ordered list:
for(ii=0;ii<91;ii++) {
currentShuffle[ii]=ii;
}
// now shuffle it:
for(ii=0;ii<91;ii++) {
temp = currentShuffle[ii];
swap = ii + random.nextInt(91-ii);
currentShuffle[ii]=currentShuffle[swap];
currentShuffle[swap]=temp;
}
return;
}
if(upDown < 0) {
temp = currentShuffle[-upDown];
for(ii = -upDown; ii>0; ii--) {
currentShuffle[ii]=currentShuffle[ii-1];
}
currentShuffle[0] = temp;
}
else {
temp = currentShuffle[upDown];
for(ii = upDown; ii<90; ii++) {
currentShuffle[ii]=currentShuffle[ii+1];
}
currentShuffle[90] = temp;
}
return;
}
// main problem posing loop:
int[] currentShuffle = new int[91];
int[] newProblem;
int keepGoing = 1;
// initial shuffle:
shuffleProblems( currentShuffle, 0); // initial shuffle
while(keepGoing) {
newProblem = chooseProblem(bias, currentShuffle);
// pose the problem, get the answer
if(wrong) {
if(newProblem > 0) shuffleProblems( currentShuffle, -newProblem[2]);
}
else shuffleProblems( currentShuffle, newProblem[2]);
// decide if you keep going...
}