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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

Monte Carlo Simulation

I'm a student in a Java Programming class. My problem deals with an interpretation of the Monte Carlo Simulation. I'm supposed to find the probability that three quarters or three pennies will be picked out of a purse that has 3 quarters and 3 pennies. Once a coin is picked it is not replaced. The probability should be 0.1XXXXXXX. I keep getting 0 or 1 for my answer. This is what i have so far.
public class CoinPurse {
public static void main(String[] args) {
System.out.print("Probability of Drawing 3 coins of the Same Type - ");
System.out.println(coinPurseSimulation(100));
}
/**
Runs numTrials trials of a Monte Carlo simulation of drawing
3 coins out of a purse containing 3 pennies and 3 quarters.
Coins are not replaced once drawn.
#param numTrials - the number of times the method will attempt to draw 3 coins
#returns a double - the fraction of times 3 coins of the same type were drawn.
*/
public static double coinPurseSimulation(int numTrials) {
final int P = 1;
final int Q = 2;
int [] purse = {Q, Q, Q, P, P, P};
int [] drawCoins = new int[3];
for (int draw = 0; draw < 3; draw ++) {
int index = (int)(Math.random() * purse.length);
drawCoins[draw] = purse[index];
int [] newPurse = new int[purse.length-1];
int j = 0;
for (int i =0; i < purse.length; i++) {
if (i == index) {
continue;
}
newPurse[j] = purse[i];
j++;
}
purse = newPurse;
}
double number = 0.0;
double result = 0.0;
for (int i = 0; i < numTrials; i++) {
result++;
for (int j = 0; j < numTrials;j++) {
if(drawCoins[0] == drawCoins [1] && drawCoins[1] == drawCoins[2]) {
number++;
}
}
}
return number/result;
}
}

The reason you only ever get 0 or 1 is that you only draw (or pick) coins from the purse once, but you then test that draw numTrials * numTrials times. You have two loops (with indices i and j) iterating numTrials time - your logic is a little messed up there.
You can put the first loop (for drawing coins) within a second loop (for running trials) and your code will work. I've put a minimal refactor below (using your code as closely as possible), with two comments afterwards that might help you some more.
public class CoinPurse
{
public static void main(String[] args)
{
System.out.print("Probability of Drawing 3 coins of the Same Type - ");
System.out.println(coinPurseSimulation(100));
}
/**
* Runs numTrials trials of a Monte Carlo simulation of drawing 3 coins out
* of a purse containing 3 pennies and 3 quarters. Coins are not replaced
* once drawn.
*
* #param numTrials
* - the number of times the method will attempt to draw 3 coins
* #returns a double - the fraction of times 3 coins of the same type were
* drawn.
*/
public static double coinPurseSimulation(int numTrials)
{
final int P = 1;
final int Q = 2;
double number = 0.0;
double result = 0.0;
// Changed your loop index to t to avoid conflict with i in your draw
// loop
for (int t = 0; t < numTrials; t++)
{
result++;
// Moved your draw without replacement code here
int[] purse =
{ Q, Q, Q, P, P, P };
int[] drawCoins = new int[3];
for (int draw = 0; draw < 3; draw++)
{
int index = (int) (Math.random() * purse.length);
drawCoins[draw] = purse[index];
int[] newPurse = new int[purse.length - 1];
int j = 0;
for (int i = 0; i < purse.length; i++)
{
if (i == index)
{
continue;
}
newPurse[j] = purse[i];
j++;
}
purse = newPurse;
}
// Deleted the loop with index j - you don't need to test the same
// combination numTrials times...
if (drawCoins[0] == drawCoins[1] && drawCoins[1] == drawCoins[2])
{
number++;
}
}
return number / result;
}
}
Picking coins code
I have some comments on your routing for drawing coins:
It works correctly
It is rather cumbersome
It would have been easier for you to spot the problem if you had broken this bit of code into a separate method.
I'm going to address 3 and then 2.
Break the drawing code out into a method
private static int[] pickCoins(int[] purse, int numPicks)
{
//A little error check
if (numPicks > purse.length)
{
System.err.println("Can't pick " + numPicks +
" coins from a purse with only " + purse.length + " coins!");
}
int[] samples = new int[numPicks];
// Your sampling code here
return samples;
}
You can now simply call from within your second loop i.e.
drawCoins = pickCoins(purse, 3);
Sampling algorithm
#pjs's answer recommends using Collections.shuffle() and then taking the first 3 coins in your collection (e.g. an ArrayList). This is a good suggestion, but I'm guessing you haven't been introduced to Collections yet, and may not be 'allowed' to use them. If you are - do use them. If not (as I assume), you might want to think about better ways to randomly draw n items from an r length array without replacement.
One (well accepted) way is the Fisher-Yates shuffle and its derivatives. In effect it involves picking randomly from the unpicked subset of an array.
In Java - an working example could be as follows - it works by moving picked coins to the "end" of the purse and picking only from the first maxInd unpicked coins.
private static int[] pickCoins(int[] purse, int numCoins)
{
int[] samples = new int[numCoins];
int maxInd = purse.length - 1;
for (int i = 0; i < numCoins; i++)
{
int index = (int) (Math.random() * maxInd);
int draw = purse[index];
samples[i] = draw;
// swap the already drawn sample with the one at maxInd and decrement maxInd
purse[index] = purse[maxInd];
purse[maxInd] = draw;
maxInd -= 1;
}
return samples;
}
Expected results
You say your expected result is 0.1XXXXXXX. As you're learning Monte Carlo simulation - you might need to think about that a little more. The expected result depends on how many trials you do.
First, in this simple example, you can consider the analytic (or in some sense exact) result. Consider the procedure:
You draw your first coin - it doesn't matter which one it is
Whichever coin it was, there are 2 left in the bag that are the same - the probability of picking one of those is 2 / 5
If you picked one of the matching coins in step 2, there is now 1 matching coin left in the bag. The probability of picking that is 1 / 4
So, the probability of getting 3 matching coins (of either denomination) is 2 / 5 * 1 / 4 == 2 / 20 == 0.1
Your Monte Carlo programme is trying to estimate that probability. You would expect it to converge on 0.1 given sufficient estimates (i.e. with numTrials high enough). It won't always give a value equal to, or even starting with, 0.1. With sufficient number of trials, it's likely to give something starting 0.09 or 0.1. However, if numTrials == 1, it will give either 0 or 1, because it will draw once and the draw will either match or not. If numTrials == 2, the results can only be 0, 0.5 or 1 and so on.
One of the lessons of doing Monte Carlo simulation to estimate probabilities is to have a sufficiently high sample count to get a good estimate. That in turn depends on the accuracy you want - you can use your code to investigate this once it's working.

You need to move the loop where you generate draws down into the numTrials loop. The way you've written it you're making a single draw, and then checking that one result numTrials times.
I haven't checked the logic for your draw carefully, but that's because I'd recommend a different (and much simpler) approach. Use Collections.shuffle() on your set of quarters and pennies, and check the first three elements after each shuffle.
If done correctly, the answer should be 2 * (3/6) * (2/5) * (1/4), which is 0.1.

Calculating Least Common Multiple (how do I know when there isn't one?)

The code below was my first attempt at a LCM (lowest common multiple) calculator with a user interface (UI code not shown) written months ago. I know there are simpler ways to write this, but I'd like help understanding why sometimes THIS specific code is not finding a common multiple (with most number sets it works fine).
When a user inputs almost any number set, the app spits out the correct LCM. But when the number set 1,234 / 2,345 / 5,432 / 4,321 is used, the app initially was stopping when x hit 536,870,912. This was because the result of x * mult was a number that couldn't be held by the int primitive. After changing x to a double and casting result = (int) (mult * x), the code continues to function as expected but seems to increment x indefinitely.
public static void compare(){
result = 0;
int mult = 0;
double x = 1;
int[] nums = UserInterface.getNums();
// finds highest number in user-input set
for(int i = 0; i < nums.length; i ++){
if (nums[i] > mult) mult = nums[i];
}
// finds lowest common multiple
for(int i = 0; i < nums.length;){
if((mult * x) % nums[i] == 0){
result = (int) (mult * x);
i++;
}
else{
result = 0;
x++;
i = 0;
}
}
}

We know the LCM of your test set must be less than or equal to 67,920,681,416,560.
In java the int datatype has a max value of 2^31-1 = 2,147,483,647 so you are obviously going to get an overflow. You can change your code to use long throughout this has a max value of 2^64-1=18,446,744,073,709,551,615 so it should be sufficient for your calculation. If you need bigger values then look at the BigInteger class.
In javascript things are more complicated. All numbers are floating point so you loose accuracy. This probably mean the condition
if((mult * x) % nums[i] == 0)
is never satisfied so your loop never quits.
Your algorithm is very basic, there are much better algorithms out there, elclanrs has one above and see https://en.wikipedia.org/wiki/Least_common_multiple for some hints.
Also you should change the title of the question. As it stands it make no sense as any set of numbers must have a LCM.

Reduce treatment time of the FFT

I'm currently working on Java for Android. I try to implement the FFT in order to realize a kind of viewer of the frequencies.
Actually I was able to do it, but the display is not fluid at all.
I added some traces in order to check the treatment time of each part of my code, and the fact is that the FFT takes about 300ms to be applied on my complex array, that owns 4096 elements. And I need it to take less than 100ms, as my thread (that displays the frequencies) is refreshed every 100ms. I reduced the initial array in order that the FFT results own only 1028 elements, and it works, but the result is deprecated.
Does someone have an idea ?
I used the default fft.java and Complex.java classes that can be found on the internet.
For information, my code computing the FFT is the following :
int bytesPerSample = 2;
Complex[] x = new Complex[bufferSize/2] ;
for (int index = 0 ; index < bufferReadResult - bytesPerSample + 1; index += bytesPerSample)
{
// 16BITS = 2BYTES
float asFloat = Float.intBitsToFloat(asInt);
double sample = 0;
for (int b = 0; b < bytesPerSample; b++) {
int v = buffer[index + b];
if (b < bytesPerSample - 1 || bytesPerSample == 1) {
v &= 0xFF;
}
sample += v << (b * 8);
}
double sample32 = 100 * (sample / 32768.0); // don't know the use of this compute...
x[index/bytesPerSample] = new Complex(sample32, 0);
}
Complex[] tx = new Complex[1024]; // size = 2048
///// reduction of the size of the signal in order to improve the fft traitment time
for (int i = 0; i < x.length/4; i++)
{
tx[i] = new Complex(x[i*4].re(), 0);
}
// Signal retrieval thanks to the FFT
fftRes = FFT.fft(tx);

I don't know Java, but you're way of converting between your input data and an array of complex values seems very convoluted. You're building two arrays of complex data where only one is necessary.
Also it smells like your complex real and imaginary values are doubles. That's way over the top for what you need, and ARMs are veeeery slow at double arithmetic anyway. Is there a complex class based on single precision floats?
Thirdly you're performing a complex fft on real data by filling the imaginary part of your complexes with zero. Whilst the result will be correct it is twice as much work straight off (unless the routine is clever enough to spot that, which I doubt). If possible perform a real fft on your data and save half your time.
And then as Simon says there's the whole issue of avoiding garbage collection and memory allocation.
Also it looks like your FFT has no preparatory step. This mean that the routine FFT.fft() is calculating the complex exponentials every time. The longest part of the FFT calculation is working out the complex exponentials, which is a shame because for any given FFT length the exponentials are constants. They don't depend on your input data at all. In the real time world we use FFT routines where we calculate the exponentials once at the start of the program and then the actual fft itself takes that const array as one of its inputs. Don't know if your FFT class can do something similar.
If you do end up going to something like FFTW then you're going to have to get used to calling C code from your Java. Also make sure you get a version that supports (I think) NEON, ARM's answer to SSE, AVX and Altivec. It's worth ploughing through their release notes to check. Also I strongly suspect that FFTW will only be able to offer a significant speed up if you ask it to perform an FFT on single precision floats, not doubles.
Google luck!
--Edit--
I meant of course 'good luck'. Give me a real keyboard quick, these touchscreen ones are unreliable...

First, thanks for all your answers.
I followed them and made two test :
first one, I replace the double used in my Complex class by float. The result is just a bit better, but not enough.
then I've rewroten the fft method in order not to use Complex anymore, but a two-dimensional float array instead. For each row of this array, the first column contains the real part, and the second one the imaginary part.
I also changed my code in order to instanciate the float array only once, on the onCreate method.
And the result... is worst !! Now it takes a little bit more than 500ms instead of 300ms.
I don't know what to do now.
You can find below the initial fft fonction, and then the one I've re-wroten.
Thanks for your help.
// compute the FFT of x[], assuming its length is a power of 2
public static Complex[] fft(Complex[] x) {
int N = x.length;
// base case
if (N == 1) return new Complex[] { x[0] };
// radix 2 Cooley-Tukey FFT
if (N % 2 != 0) { throw new RuntimeException("N is not a power of 2 : " + N); }
// fft of even terms
Complex[] even = new Complex[N/2];
for (int k = 0; k < N/2; k++) {
even[k] = x[2*k];
}
Complex[] q = fft(even);
// fft of odd terms
Complex[] odd = even; // reuse the array
for (int k = 0; k < N/2; k++) {
odd[k] = x[2*k + 1];
}
Complex[] r = fft(odd);
// combine
Complex[] y = new Complex[N];
for (int k = 0; k < N/2; k++) {
double kth = -2 * k * Math.PI / N;
Complex wk = new Complex(Math.cos(kth), Math.sin(kth));
y[k] = q[k].plus(wk.times(r[k]));
y[k + N/2] = q[k].minus(wk.times(r[k]));
}
return y;
}
public static float[][] fftf(float[][] x) {
/**
* x[][0] = real part
* x[][1] = imaginary part
*/
int N = x.length;
// base case
if (N == 1) return new float[][] { x[0] };
// radix 2 Cooley-Tukey FFT
if (N % 2 != 0) { throw new RuntimeException("N is not a power of 2 : " + N); }
// fft of even terms
float[][] even = new float[N/2][2];
for (int k = 0; k < N/2; k++) {
even[k] = x[2*k];
}
float[][] q = fftf(even);
// fft of odd terms
float[][] odd = even; // reuse the array
for (int k = 0; k < N/2; k++) {
odd[k] = x[2*k + 1];
}
float[][] r = fftf(odd);
// combine
float[][] y = new float[N][2];
double kth, wkcos, wksin ;
for (int k = 0; k < N/2; k++) {
kth = -2 * k * Math.PI / N;
//Complex wk = new Complex(Math.cos(kth), Math.sin(kth));
wkcos = Math.cos(kth) ; // real part
wksin = Math.sin(kth) ; // imaginary part
// y[k] = q[k].plus(wk.times(r[k]));
y[k][0] = (float) (q[k][0] + wkcos * r[k][0] - wksin * r[k][1]);
y[k][1] = (float) (q[k][1] + wkcos * r[k][1] + wksin * r[k][0]);
// y[k + N/2] = q[k].minus(wk.times(r[k]));
y[k + N/2][0] = (float) (q[k][0] - (wkcos * r[k][0] - wksin * r[k][1]));
y[k + N/2][1] = (float) (q[k][1] - (wkcos * r[k][1] + wksin * r[k][0]));
}
return y;
}

actually I think I don't understand everything.
First, about Math.cos and Math.sin : how do you want me not to compute it each time ? Do you mean that I should instanciate the whole values only once (e.g store it in an array) and use them for each compute ?
Second, about the N % 2, indeed it's not very useful, I could make the test before the call of the function.
Third, about Simon's advice : I mixed what he said and what you said, that's why I've replaced the Complex by a two-dimensional float[][]. If that was not what he suggested, then what was it ?
At least, I'm not a FFT expert, so what do you mean by making a "real FFT" ? Do you mean that my imaginary part is useless ? If so, I'm not sure, because later in my code, I compute the magnitude of each frequence, so sqrt(real[i]*real[i] + imag[i]*imag[i]). And I think that my imaginary part is not equal to zero...
thanks !

Interview: Find the whole cubes between range of two Integers

I just gave a coding interview on codility
I was asked the to implement the following, but i was not able to finish it in 20 minutes, now I am here to get ideas form this community
Write a function public int whole_cubes_count ( int A,int B ) where it should return whole cubes within the range
For example if A=8 and B=65, all the possible cubes in the range are 2^3 =8 , 3^3 =27 and 4^3=64, so the function should return count 3
I was not able to figure out how to identify a number as whole cube. How do I solve this problem?
A and B can have range from [-20000 to 20000]
This is what I tried
import java.util.Scanner;
class Solution1 {
public int whole_cubes_count ( int A,int B ) {
int count =0;
while(A<=B)
{
double v = Math.pow(A, 1 / 3); // << What goes here?
System.out.println(v);
if (v<=B)
{
count=count+1;
}
A =A +1;
}
return count ;
}
public static void main(String[] args)
{
System.out.println("Enter 1st Number");
Scanner scan = new Scanner(System.in);
int s1 = scan.nextInt();
System.out.println("Enter 2nd Number");
//Scanner scan = new Scanner(System.in);
int s2 = scan.nextInt();
Solution1 n = new Solution1();
System.out.println(n.whole_cubes_count (s1,s2));
}
}

Down and dirty, that's what I say.
If you only have 20 minutes, then they shouldn't expect super-optimized code. So don't even try. Play to the constraints of the system which say only +20,000 to -20,000 as the range. You know the cube values have to be within 27, since 27 * 27 * 27 = 19683.
public int whole_cubes_count(int a, int b) {
int count = 0;
int cube;
for (int x = -27; x <= 27; x++) {
cube = x * x * x;
if ((cube >= a) && (cube <= b))
count++;
}
return count;
}

For the positive cubes:
i = 1
while i^3 < max
++i
Similarly for the negative cubes but with an absolute value in the comparison.
To make this more general, you need to find the value of i where i^3 >= min, in the case that both min and max are positive. A similar solution works if both min and max are negative.

Well, it can be computed with O(1) complexity, we will need to find the largest cube that fits into the range, and the smallest one. All those that are between will obviously also be inside.
def n_cubes(A, B):
a_cr = int(math.ceil(cube_root(A)))
b_cr = int(math.floor(cube_root(B)))
if b_cr >= a_cr:
return b_cr - a_cr + 1
return 0
just make sure your cube_root returns integers for actual cubes. Complete solution as gist https://gist.github.com/tymofij/9035744

int countNoOfCubes(int a, int b) {
int count = 0;
for (int startsCube = (int) Math.ceil(Math.cbrt(a)); Math.pow(
startsCube, 3.0) <= b; startsCube++) {
count++;
}
return count;
}

The solution suggested by #Tim is faster than the one provided by #Erick, especially when A...B range increased.
Let me quote the ground from github here:
"one can notice that x³ > y³ for any x > y. (that is called monotonic function)
therefore for any x that lies in ∛A ≤ x ≤ ∛B, cube would fit: A ≤ x³ ≤ B
So to get number of cubes which lie within A..B, you can simply count number of integers between ∛A and ∛B. And number of integers between two numbers is their difference."
It seems perfectly correct, isn't it? It works for any power, not only for cube.
Here is my port of cube_root method for java:
/*
* make sure your cube_root returns integers for actual cubes
*/
static double cubeRoot(int x) {
//negative number cannot be raised to a fractional power
double res = Math.copySign(Math.pow(Math.abs(x), (1.0d/3)) , x);
long rounded_res = symmetricRound(res);
if (rounded_res * rounded_res * rounded_res == x)
return rounded_res;
else
return res;
}
private static long symmetricRound( double d ) {
return d < 0 ? - Math.round( -d ) : Math.round( d );
}
I am aware of Math.cbrt in java but with Math.pow approach it is easy to generalize the solution for other exponents.