How do I generate random numbers using Dart?
Use Random class from dart:math:
import 'dart:math';
main() {
var rng = Random();
for (var i = 0; i < 10; i++) {
print(rng.nextInt(100));
}
}
This code was tested with the Dart VM and dart2js, as of the time of this writing.
You can achieve it via Random class object random.nextInt(max), which is in dart:math library. The nextInt() method requires a max limit. The random number starts from 0 and the max limit itself is exclusive.
import 'dart:math';
Random random = new Random();
int randomNumber = random.nextInt(100); // from 0 upto 99 included
If you want to add the min limit, add the min limit to the result
int randomNumber = random.nextInt(90) + 10; // from 10 upto 99 included
try this, you can control the min/max value :
NOTE that you need to import dart math library.
import 'dart:math';
void main() {
int random(int min, int max) {
return min + Random().nextInt(max - min);
}
print(random(5, 20)); // Output : 19, 5, 15.. (5 -> 19, 20 is not included)
}
Here's a snippet for generating a list of random numbers
import 'dart:math';
main() {
var rng = new Random();
var l = new List.generate(12, (_) => rng.nextInt(100));
}
This will generate a list of 12 integers from 0 to 99 (inclusive).
A secure random API was just added to dart:math
new Random.secure()
dart:math
Random added a secure constructor returning a cryptographically secure
random generator which reads from the entropy source provided by the
embedder for every generated random value.
which delegates to window.crypto.getRandomValues() in the browser and to the OS (like urandom on the server)
If you need cryptographically-secure random numbers (e.g. for encryption), and you're in a browser, you can use the DOM cryptography API:
int random() {
final ary = new Int32Array(1);
window.crypto.getRandomValues(ary);
return ary[0];
}
This works in Dartium, Chrome, and Firefox, but likely not in other browsers as this is an experimental API.
Let me answer this question with a practical example in the form of a simple dice rolling app that calls 1 of 6 dice face images randomly to the screen when tapped.
first declare a variable that generates random numbers (don't forget to import dart.math). Then declare a variable that parses the initial random number within constraints between 1 and 6 as an Integer.
Both variables are static private in order to be initialized once.This is is not a huge deal but would be good practice if you had to initialize a whole bunch of random numbers.
static var _random = new Random();
static var _diceface = _random.nextInt(6) +1 ;
Now create a Gesture detection widget with a ClipRRect as a child to return one of the six dice face images to the screen when tapped.
GestureDetector(
onTap: () {
setState(() {
_diceface = _random.nextInt(6) +1 ;
});
},
child: ClipRRect(
clipBehavior: Clip.hardEdge,
borderRadius: BorderRadius.circular(100.8),
child: Image(
image: AssetImage('images/diceface$_diceface.png'),
fit: BoxFit.cover,
),
)
),
A new random number is generated each time you tap the screen and that number is referenced to select which dice face image is chosen.
Dice rolling app using random numbers in dart
its worked for me new Random().nextInt(100); // MAX = number
it will give 0 to 99 random number
Eample::
import 'dart:math';
int MAX = 100;
print(new Random().nextInt(MAX));`
Not able to comment because I just created this account, but I wanted to make sure to point out that #eggrobot78's solution works, but it is exclusive in dart so it doesn't include the last number. If you change the last line to "r = min + rnd.nextInt(max - min + 1);", then it should include the last number as well.
Explanation:
max = 5;
min = 3;
Random rnd = new Random();
r = min + rnd.nextInt(max - min);
//max - min is 2
//nextInt is exclusive so nextInt will return 0 through 1
//3 is added so the line will give a number between 3 and 4
//if you add the "+ 1" then it will return a number between 3 and 5
you can generate by simply in this way
there is a class named Random();
you can use that and genrate random numbers
Random objectname = Random();
int number = objectname.nextInt(100);
// it will generate random number within 100.
An alternative solution could be using the following code DRandom. This class should be used with a seed. It provides a familiar interface to what you would expect in .NET, it was ported from mono's Random.cs. This code may not be cryptography safe and has not been statistically tested.
Just wrote this little class for generating Normal Random numbers... it was a decent starting point for the checking I need to do. (These sets will distribute on a "bell" shaped
curve.) The seed will be set randomly, but if you want to be able to re-generate a set you can just pass some specific seed and the same set will generate.
Have fun...
class RandomNormal {
num _min, _max, _sum;
int _nEle, _seed, _hLim;
Random _random;
List _rNAr;
//getter
List get randomNumberAr => _rNAr;
num _randomN() {
int r0 = _random.nextInt(_hLim);
int r1 = _random.nextInt(_hLim);
int r2 = _random.nextInt(_hLim);
int r3 = _random.nextInt(_hLim);
num rslt = _min + (r0 + r1 + r2 + r3) / 4000.0; //Add the OS back in...
_sum += rslt; //#DEBUG ONLY
return( rslt );
}
RandomNormal(this._nEle, this._min, this._max, [this._seed = null]) {
if (_seed == null ) {
Random r = new Random();
_seed = r.nextInt(1000);
}
_hLim = (_max - _min).ceil() * 1000;
_random = new Random(_seed);
_rNAr = [];
_sum = 0;//#DEBUG ONLY
h2("RandomNormal with k: ${_nEle}, Seed: ${_seed}, Min: ${_min}, Max: ${_max}");//#DEBUG ONLY
for(int n = 0; n < _nEle; n++ ){
num randomN = _randomN();
//p("randomN = ${randomN}");
LIST_add( _rNAr, randomN );
}
h3("Mean = ${_sum/_nEle}");//#DEBUG ONLY
}
}
new RandomNormal(1000, 80, 120);
new RandomNormal(1000, 80, 120);
new RandomNormal(1000, 80, 120);
new RandomNormal(1000, 80, 120);
new RandomNormal(1000, 80, 120);
new RandomNormal(1000, 80, 120);
new RandomNormal(1000, 80, 120);
Then you can just use it like this to check the mean of sets of 1000 nums generated between a low and high limit. The values are stored in the class so they can be accessed after instantiation.
_swarmii
For me the easiest way is to do:
import 'dart:math';
Random rnd = new Random();
r = min + rnd.nextInt(max - min);
//where min and max should be specified.
Thanks to #adam-singer explanation in here.
This method generates random integer. Both minimum and maximum are inclusive.
Make sure you add this line to your file: import 'dart:math'; Then you can add and use this method:
int randomInt(int min, int max) {
return Random().nextInt(max - min + 1) + min;
}
So if you call randomInt(-10,10) it will return a number between -10 and 10 (including -10 and 10).
Use Dart Generators, that is used to produce a sequence of number or values.
main(){
print("Sequence Number");
oddnum(10).forEach(print);
}
Iterable<int> oddnum(int num) sync*{
int k=num;
while(k>=0){
if(k%2==1){
yield k;
}
k--;
}
}
Use class Random() from 'dart:math' library.
import 'dart:math';
void main() {
int max = 10;
int RandomNumber = Random().nextInt(max);
print(RandomNumber);
}
This should generate and print a random number from 0 to 9.
one line solution you can directly call all the functions with constructor as well.
import 'dart:math';
print(Random().nextInt(100));
This works for me
import 'dart:math';
void main() {
Random random = Random();
int randomNumber = random.nextInt(5) + 1;
print(randomNumber);
}
Change the number five for possible results you want
Related
here is the code for generating random numbers,but I am getting duplicate numbers,how can I overcome this.
void getnumbers()
{
Random r = new Random();
int[] trubyte = new int[4];
for (var x = 0; x < 4; ++x)
{
trubyte[x] = r.Next(1, 5);
}
b1.Content = trubyte[0];
b2.Content = trubyte[1];
b3.Content = trubyte[2];
b4.Content = trubyte[3];
}
Just get another random number if the method returns one that you already have.
void getnumbers()
{
Random r = new Random();
int num;
var trubyte = new List<int>();
for (var x = 0; x < 4; ++x)
{
do
{
num = r.Next(1, 5);
} while(trubyte.Contains(num));
trubyte[x] = num;
}
b1.Content = trubyte[0];
b2.Content = trubyte[1];
b3.Content = trubyte[2];
b4.Content = trubyte[3];
}
I'm using List instead of an array just because it offers the Contains method right away, not any other special reason.
This is not efficient if you want to generate a big list of random, unrepeated numbers (it's O(n^2) in the worst case) but for 4 numbers it's more than enough ;)
A random number generator function can return duplicates, because the output is random.
If you are using an RNG to generate numbers which must be unique, you will need to verify that they have not already been generated before using them.
Can't you use something like this [0] on Windows Mobile? It seems more practical than writing your own RNG.
0: http://msdn.microsoft.com/en-us/library/system.security.cryptography.randomnumbergenerator(v=vs.90).aspx
You have to do it by yourself, that means checking if a number was already generated.
You can do it like gjulianm said, but it is a long list of numbers, say 1000 you would be wasting a lot of time. So if you want a randomized list of 1000 you could proceed the following way
Initialize an array trubyte of size 1000 with trubyte[0]=1,trubyte[1]=2 and so on...
Initialize a variable arraysize=1000
run a loop 1000 times in which first extract a random number k btw 0-(arraysize-1). Your random number is a[k] which you can separately in a list. Now swap trubyte[k] with trubyte[arraysize]. And finally decrease the arraysize by one.
Another way, if you don't want the numbers while in the loop is just to use the changed list after the execution of loop
void getnumbers(){
Random r = new Random();
int num;
int[] trubyte = new int[1000];
int finalList[] = new int[1000]
for (int x = 0; x < 1000; ++x)
{
trubyte[x]=x+1;
}
int arraysize=1000;
for (var x = 0; x < 1000; ++x)
{
int k=r.Next(0, arraysize);
finalList[x]=trubyte[k];
trubyte[k]=trubyte[arraysize-1];
arraysize--;
}
//use the finalList
}
we can use dictionary instead of hash-set in windows phone application.
below is the code for generating distinct random numbers.
static int[] GetRandomNumbersNonrepeat(int noOfRandomNumbers, int maxValue)
{
Dictionary<int, int> randomnumbers = new Dictionary<int, int>();
while (randomnumbers.Count < maxValue)
{
Random r = new Random();
int rnum = r.Next(1, maxValue+1);
if (!randomnumbers.ContainsValue(rnum))
{
randomnumbers.Add(randomnumbers.Count + 1, rnum);
}
}
int[] rnums = randomnumbers.Values.ToArray<int>();
return rnums;
}
Say I have 4 possible results and the probabilities of each result appearing are
1 = 10%
2 = 20%
3 = 30%
4 = 40%
I'd like to write a method like GetRandomValue which if called 1000 times would return
1 x 100 times
2 x 200 times
3 x 300 times
4 x 400 times
Whats the name of an algorithm which would produce such results?
in your case you can generate a random number (int) within 1..10 and if it's 1 then select 1, if it's between 2-3 select 2 and if it's between 4..6 select 3 and if is between 7..10 select 4.
In all if you have some probabilities which sum to 1, you can have a random number within (0,1) distribute your generated result to related value (I simplified in your case within 1..10).
To get a random number you would use the Random class of .Net.
Something like the following would accomplish what you requested:
public class MyRandom
{
private Random m_rand = new Random();
public int GetNextValue()
{
// Gets a random value between 0-9 with equal probability
// and converts it to a number between 1-4 with the probablities requested.
switch (m_rand.Next(0, 9))
{
case 0:
return 1;
case 1: case 2:
return 2;
case 3: case 4: case 5:
return 3;
default:
return 4;
}
}
}
If you just want those probabilities in the long run, you can just get values by randomly selecting one element from the array {1,2,2,3,3,3,4,4,4,4}.
If you however need to retrieve exactly 1000 elements, in those specific quantities, you can try something like this (not C#, but shouldn't be a problem):
import java.util.Random;
import java.util.*;
class Thing{
Random r = new Random();
ArrayList<Integer> numbers=new ArrayList<Integer>();
ArrayList<Integer> counts=new ArrayList<Integer>();
int totalCount;
public void set(int i, int count){
numbers.add(i);
counts.add(count);
totalCount+=count;
}
public int getValue(){
if (totalCount==0)
throw new IllegalStateException();
double pos = r.nextDouble();
double z = 0;
int index = 0;
//we select elements using their remaining counts for probabilities
for (; index<counts.size(); index++){
z += counts.get(index) / ((double)totalCount);
if (pos<z)
break;
}
int result = numbers.get(index);
counts.set( index , counts.get(index)-1);
if (counts.get(index)==0){
counts.remove(index);
numbers.remove(index);
}
totalCount--;
return result;
}
}
class Test{
public static void main(String []args){
Thing t = new Thing(){{
set(1,100);
set(2,200);
set(3,300);
set(4,400);
}};
int[]hist=new int[4];
for (int i=0;i<1000;i++){
int value = t.getValue();
System.out.print(value);
hist[value-1]++;
}
System.out.println();
double sum=0;
for (int i=0;i<4;i++) sum+=hist[i];
for (int i=0;i<4;i++)
System.out.printf("%d: %d values, %f%%\n",i+1,hist[i], (100*hist[i]/sum));
}
}
hi i am using Random class for getting random numbers but my requirement is once it generate one no that should not be repeate again pls help me.
Keep a list of the generated numbers and check this list before returning the next random.
Since you have not specified a language, I'll use C#
List<int> generated = new List<int>;
public int Next()
{
int r;
do { r = Random.Next() } while generated.Contains(r);
generated.Add(r);
return r;
}
The following C# code shows how to obtain 7 random cards with no duplicates. It is the most efficient method to use when your random number range is between 1 and 64 and are integers:
ulong Card, SevenCardHand;
int CardLoop;
const int CardsInDeck = 52;
Random RandObj = new Random(Seed);
for (CardLoop = 0; CardLoop < 7; CardLoop++)
{
do
{
Card = (1UL << RandObj.Next(CardsInDeck));
} while ((SevenCardHand & Card) != 0);
SevenCardHand |= Card;
}
If the random number range is greater than 64, then the next most efficient way to get random numbers without any duplicates is as follows from this C# code:
const int MaxNums = 1000;
int[] OutBuf = new int[MaxNums];
int MaxInt = 250000; // Reps the largest random number that should be returned.
int Loop, Val;
// Init the OutBuf with random numbers between 1 and MaxInt, which is 250,000.
BitArray BA = new BitArray(MaxInt + 1);
for (Loop = 0; Loop < MaxNums; Loop++)
{
// Avoid duplicate numbers.
for (; ; )
{
Val = RandObj.Next(MaxInt + 1);
if (BA.Get(Val))
continue;
OutBuf[Loop] = Val;
BA.Set(Val, true);
break;
}
}
The drawback with this technique is that it tends to use more memory, but it should be significantly faster than other approaches since it does not have to look through a large container each time a random number is obtained.
here is my Linq code to generate a list of random numbers which contains 10 numbers ranging from 0 to 20
Random rand = new Random();
var randomSeq = Enumerable.Repeat(0, 10).Select(i => rand.Next(0,20));
Result:
6
19
18
7
18
12
12
9
2
18
as you can see i have three 18s and two 12s..
I have tried to use Distinct() function, but it will not fill up the list (e.g only fill up 8 out of 10 numbers)
Question: How can I generate unique number (i.e non repeatable numbers )
Many thanks
You want to generate a random permutation of the numbers 0 to 19 and pick 10 of these numbers. The standard algorithm for generating a random permutation is Fisher-Yates shuffle. After generating a random permutation you can just pick the first 10 numbers.
It is not to hard to come up with an ad-hoc algorithm like repeatedly choosing a new number if a collision occured but they usually fail to have good statistical properties, have nondeterministic runtime or don't even guarantee termination in the worst case.
Note that this solution is no good choice if the numbers are of different order. Generating a permuation of the numbers below a million only to pick ten is not the smartest thing one can do.
UPDATE
I just realized that you can just stop the algorithm after generating the first ten elements of the permutation - there is no need to build the whole permutation.
In functional programming it is usual to create infinite sequences. It might sound a little bizarre at first but it can be very usefull at some situations. Supose you have an extention as such:
public static class EnumerableExtentions
{
public static IEnumerable<T> Infinite<T>(Func<int, T> generator)
{
int count = 0;
checked {
while (true)
yield return generator(count++);
}
}
}
I can use it to create infinite sequences like:
var oddNumbers = EnumerableExtentions.Infinite(n => 2*n + 1)
That is an infinite sequence of all odd numbers. I could take only the first 10, for example:
oddNumbers.Take(10);
would yield:
1 3 5 7 9 11 13 15 17 19
Because of the defered execution, we don´t get a StackOverflowException (you gotta be carefull though).
The same principle can be used to create an infinite random sequence, distinct it and then taking the first 10:
var r = new Random();
var randomNumbers = EnumerableExtentions
.Infinite(i=> r.Next (0, 20))
.Distinct()
.Take(10);
If you need, you can make an OrderBy(s=>s) at the end.
At LINQ exchange, they discuss a method of randomly reordering a list with LINQ and give a code example which will generate a random permutation of the numbers you want.
They say (paraphrasing, and adapted for this problem):
Randomly Sort a List Array With LINQ OrderBy
// create and populate the original list with 20 elements
List<int> MyList = new List<int>(20);
for (int i = 0; i < 20; i++)
MyList.Add(i);
// use System.GUID to generate a new GUID for each item in the list
List<int> RandomList = MyList.OrderBy(x => System.Guid.NewGuid()).ToList();
LINQ OrderBy will then sort the array by the list of GUID's returned.
Now you can just take the first 10 elements of the list, and you've got your solution.
They note that using the System.Guid.NewGuid() yields the same distribution spread as the Fisher-Yates shuffle algorithm, and this way you won't have to actually implement the algorithm yourself.
Why not do:
Enumerable.Range(0, 20)
.OrderBy(x => Guid.NewGuid().GetHashCode())
.Distinct()
.Take(10)
.ToArray();
How about using a utility enumerable method:
static IEnumerable<int> RandomNumbersBetween(int min, int max)
{
int availableNumbers = (max - min) + 1 ;
int yieldedNumbers = 0;
Random rand = new Random();
Dictionary<int, object> used = new Dictionary<int, object>();
while (true)
{
int n = rand.Next(min, max+1); //Random.Next max value is exclusive, so add one
if (!used.ContainsKey(n))
{
yield return n;
used.Add(n, null);
if (++yieldedNumbers == availableNumbers)
yield break;
}
}
}
Because it returns IEnumerable, you can use it with LINQ and IEnumerable extension methods:
RandomNumbersBetween(0, 20).Take(10)
Or maybe take odd numbers only:
RandomNumbersBetween(1, 1000).Where(i => i%2 == 1).Take(100)
Et cetera.
Edit:
Note that this solution has terrible performance characteristics if you are trying to generate a full set of random numbers between min and max.
However it works efficiently if you want to generate, say 10 random numbers between 0 and 20, or even better, between 0 and 1000.
In worst-case scenario it can also take (max - min) space.
Just create a list of sequential valid numbers. Then generate a random index from this list and return (and remove from list) the number at the index.
static class Excensions
{
public static T PopAt<T>(this List<T> list, int index)
{
T ret = list[index];
list.RemoveAt(index);
return ret;
}
}
class Program
{
static void Main()
{
Random rng = new Random();
int length = 10; //sequence length
int limit = 20; //maximum value
var avail = Enumerable.Range(0, limit).ToList();
var seq = from i in Enumerable.Range(0, length)
select avail.PopAt(rng.Next(avail.Count));
}
}
store the generated result in an array, so anytime you generate e new number check if it has been generated before, if yes generate another one, otherwise take the number and save it in the array
Using a custom RepeatUntil extension and relying on closures:
using System;
using System.Collections.Generic;
using System.Linq;
namespace ConsoleApplication1
{
public static class CoolExtensions
{
public static IEnumerable<TResult> RepeatUntil<TResult>( TResult element, Func<bool> condition )
{
while (!condition())
yield return element;
}
}
class Program
{
static void Main( string[] args )
{
Random rand = new Random();
HashSet<int> numbers = new HashSet<int>();
var randomSeq = CoolExtensions.RepeatUntil( 0, () => numbers.Count >= 10).Select( i => rand.Next( 0, 20 ) ).Select( x => numbers.Add(x));
// just used to evaluate the sequence
randomSeq.ToList();
foreach (int number in numbers)
Console.WriteLine( number );
Console.ReadLine();
}
}
}
Why not order by a random? like this
var rnd = new Random();
var randomSeq = Enumerable.Range(1,20).OrderBy(r => rnd.NextDouble()).Take(10).ToList();
Can you do something like this?
Random rand = new Random();
var randomSeq = Enumerable.Range(0, 20).OrderBy(i => rand.Next(0,20)).Take(10);
This question already has answers here:
Roulette Selection in Genetic Algorithms
(14 answers)
Closed 7 years ago.
Can anyone provide some pseudo code for a roulette selection function? How would I implement this: I don't really understand how to read this math notation.I want General algorithm to this.
The other answers seem to be assuming that you are trying to implement a roulette game. I think that you are asking about roulette wheel selection in evolutionary algorithms.
Here is some Java code that implements roulette wheel selection.
Assume you have 10 items to choose from and you choose by generating a random number between 0 and 1. You divide the range 0 to 1 up into ten non-overlapping segments, each proportional to the fitness of one of the ten items. For example, this might look like this:
0 - 0.3 is item 1
0.3 - 0.4 is item 2
0.4 - 0.5 is item 3
0.5 - 0.57 is item 4
0.57 - 0.63 is item 5
0.63 - 0.68 is item 6
0.68 - 0.8 is item 7
0.8 - 0.85 is item 8
0.85 - 0.98 is item 9
0.98 - 1 is item 10
This is your roulette wheel. Your random number between 0 and 1 is your spin. If the random number is 0.46, then the chosen item is item 3. If it's 0.92, then it's item 9.
Here is a bit of python code:
def roulette_select(population, fitnesses, num):
""" Roulette selection, implemented according to:
<http://stackoverflow.com/questions/177271/roulette
-selection-in-genetic-algorithms/177278#177278>
"""
total_fitness = float(sum(fitnesses))
rel_fitness = [f/total_fitness for f in fitnesses]
# Generate probability intervals for each individual
probs = [sum(rel_fitness[:i+1]) for i in range(len(rel_fitness))]
# Draw new population
new_population = []
for n in xrange(num):
r = rand()
for (i, individual) in enumerate(population):
if r <= probs[i]:
new_population.append(individual)
break
return new_population
First, generate an array of the percentages you assigned, let's say p[1..n]
and assume the total is the sum of all the percentages.
Then get a random number between 1 to total, let's say r
Now, the algorithm in lua:
local c = 0
for i = 1,n do
c = c + p[i]
if r <= c then
return i
end
end
There are 2 steps to this: First create an array with all the values on the wheel. This can be a 2 dimensional array with colour as well as number, or you can choose to add 100 to red numbers.
Then simply generate a random number between 0 or 1 (depending on whether your language starts numbering array indexes from 0 or 1) and the last element in your array.
Most languages have built-in random number functions. In VB and VBScript the function is RND(). In Javascript it is Math.random()
Fetch the value from that position in the array and you have your random roulette number.
Final note: don't forget to seed your random number generator or you will get the same sequence of draws every time you run the program.
Here is a really quick way to do it using stream selection in Java. It selects the indices of an array using the values as weights. No cumulative weights needed due to the mathematical properties.
static int selectRandomWeighted(double[] wts, Random rnd) {
int selected = 0;
double total = wts[0];
for( int i = 1; i < wts.length; i++ ) {
total += wts[i];
if( rnd.nextDouble() <= (wts[i] / total)) selected = i;
}
return selected;
}
This could be further improved using Kahan summation or reading through the doubles as an iterable if the array was too big to initialize at once.
I wanted the same and so created this self-contained Roulette class. You give it a series of weights (in the form of a double array), and it will simply return an index from that array according to a weighted random pick.
I created a class because you can get a big speed up by only doing the cumulative additions once via the constructor. It's C# code, but enjoy the C like speed and simplicity!
class Roulette
{
double[] c;
double total;
Random random;
public Roulette(double[] n) {
random = new Random();
total = 0;
c = new double[n.Length+1];
c[0] = 0;
// Create cumulative values for later:
for (int i = 0; i < n.Length; i++) {
c[i+1] = c[i] + n[i];
total += n[i];
}
}
public int spin() {
double r = random.NextDouble() * total; // Create a random number between 0 and 1 and times by the total we calculated earlier.
//int j; for (j = 0; j < c.Length; j++) if (c[j] > r) break; return j-1; // Don't use this - it's slower than the binary search below.
//// Binary search for efficiency. Objective is to find index of the number just above r:
int a = 0;
int b = c.Length - 1;
while (b - a > 1) {
int mid = (a + b) / 2;
if (c[mid] > r) b = mid;
else a = mid;
}
return a;
}
}
The initial weights are up to you. Maybe it could be the fitness of each member, or a value inversely proportional to the member's position in the "top 50". E.g.: 1st place = 1.0 weighting, 2nd place = 0.5, 3rd place = 0.333, 4th place = 0.25 weighting etc. etc.
Well, for an American Roulette wheel, you're going to need to generate a random integer between 1 and 38. There are 36 numbers, a 0, and a 00.
One of the big things to consider, though, is that in American roulette, their are many different bets that can be made. A single bet can cover 1, 2, 3, 4, 5, 6, two different 12s, or 18. You may wish to create a list of lists where each number has additional flages to simplify that, or do it all in the programming.
If I were implementing it in Python, I would just create a Tuple of 0, 00, and 1 through 36 and use random.choice() for each spin.
This assumes some class "Classifier" which just has a String condition, String message, and double strength. Just follow the logic.
-- Paul
public static List<Classifier> rouletteSelection(int classifiers) {
List<Classifier> classifierList = new LinkedList<Classifier>();
double strengthSum = 0.0;
double probabilitySum = 0.0;
// add up the strengths of the map
Set<String> keySet = ClassifierMap.CLASSIFIER_MAP.keySet();
for (String key : keySet) {
/* used for debug to make sure wheel is working.
if (strengthSum == 0.0) {
ClassifierMap.CLASSIFIER_MAP.get(key).setStrength(8000.0);
}
*/
Classifier classifier = ClassifierMap.CLASSIFIER_MAP.get(key);
double strength = classifier.getStrength();
strengthSum = strengthSum + strength;
}
System.out.println("strengthSum: " + strengthSum);
// compute the total probability. this will be 1.00 or close to it.
for (String key : keySet) {
Classifier classifier = ClassifierMap.CLASSIFIER_MAP.get(key);
double probability = (classifier.getStrength() / strengthSum);
probabilitySum = probabilitySum + probability;
}
System.out.println("probabilitySum: " + probabilitySum);
while (classifierList.size() < classifiers) {
boolean winnerFound = false;
double rouletteRandom = random.nextDouble();
double rouletteSum = 0.0;
for (String key : keySet) {
Classifier classifier = ClassifierMap.CLASSIFIER_MAP.get(key);
double probability = (classifier.getStrength() / strengthSum);
rouletteSum = rouletteSum + probability;
if (rouletteSum > rouletteRandom && (winnerFound == false)) {
System.out.println("Winner found: " + probability);
classifierList.add(classifier);
winnerFound = true;
}
}
}
return classifierList;
}
You can use a data structure like this:
Map<A, B> roulette_wheel_schema = new LinkedHashMap<A, B>()
where A is an integer that represents a pocket of the roulette wheel, and B is an index that identifies a chromosome in the population. The number of pockets is proportional to the fitness proportionate of each chromosome:
number of pockets = (fitness proportionate) · (scale factor)
Then we generate a random between 0 and the size of the selection schema and with this random number we get the index of the chromosome from the roulette.
We calculate the relative error between the fitness proportionate of each chromosome and the probability of being selected by the selection scheme.
The method getRouletteWheel returns the selection scheme based on previous data structure.
private Map<Integer, Integer> getRouletteWheel(
ArrayList<Chromosome_fitnessProportionate> chromosomes,
int precision) {
/*
* The number of pockets on the wheel
*
* number of pockets in roulette_wheel_schema = probability ·
* (10^precision)
*/
Map<Integer, Integer> roulette_wheel_schema = new LinkedHashMap<Integer, Integer>();
double fitness_proportionate = 0.0D;
double pockets = 0.0D;
int key_counter = -1;
double scale_factor = Math
.pow(new Double(10.0D), new Double(precision));
for (int index_cromosome = 0; index_cromosome < chromosomes.size(); index_cromosome++){
Chromosome_fitnessProportionate chromosome = chromosomes
.get(index_cromosome);
fitness_proportionate = chromosome.getFitness_proportionate();
fitness_proportionate *= scale_factor;
pockets = Math.rint(fitness_proportionate);
System.out.println("... " + index_cromosome + " : " + pockets);
for (int j = 0; j < pockets; j++) {
roulette_wheel_schema.put(Integer.valueOf(++key_counter),
Integer.valueOf(index_cromosome));
}
}
return roulette_wheel_schema;
}
I have worked out a Java code similar to that of Dan Dyer (referenced earlier). My roulette-wheel, however, selects a single element based on a probability vector (input) and returns the index of the selected element.
Having said that, the following code is more appropriate if the selection size is unitary and if you do not assume how the probabilities are calculated and zero probability value is allowed. The code is self-contained and includes a test with 20 wheel spins (to run).
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Random;
import java.util.logging.Level;
import java.util.logging.Logger;
/**
* Roulette-wheel Test version.
* Features a probability vector input with possibly null probability values.
* Appropriate for adaptive operator selection such as Probability Matching
* or Adaptive Pursuit, (Dynamic) Multi-armed Bandit.
* #version October 2015.
* #author Hakim Mitiche
*/
public class RouletteWheel {
/**
* Selects an element probabilistically.
* #param wheelProbabilities elements probability vector.
* #param rng random generator object
* #return selected element index
* #throws java.lang.Exception
*/
public int select(List<Double> wheelProbabilities, Random rng)
throws Exception{
double[] cumulativeProba = new double[wheelProbabilities.size()];
cumulativeProba[0] = wheelProbabilities.get(0);
for (int i = 1; i < wheelProbabilities.size(); i++)
{
double proba = wheelProbabilities.get(i);
cumulativeProba[i] = cumulativeProba[i - 1] + proba;
}
int last = wheelProbabilities.size()-1;
if (cumulativeProba[last] != 1.0)
{
throw new Exception("The probabilities does not sum up to one ("
+ "sum="+cumulativeProba[last]);
}
double r = rng.nextDouble();
int selected = Arrays.binarySearch(cumulativeProba, r);
if (selected < 0)
{
/* Convert negative insertion point to array index.
to find the correct cumulative proba range index.
*/
selected = Math.abs(selected + 1);
}
/* skip indexes of elements with Zero probability,
go backward to matching index*/
int i = selected;
while (wheelProbabilities.get(i) == 0.0){
System.out.print(i+" selected, correction");
i--;
if (i<0) i=last;
}
selected = i;
return selected;
}
public static void main(String[] args){
RouletteWheel rw = new RouletteWheel();
int rept = 20;
List<Double> P = new ArrayList<>(4);
P.add(0.2);
P.add(0.1);
P.add(0.6);
P.add(0.1);
Random rng = new Random();
for (int i = 0 ; i < rept; i++){
try {
int s = rw.select(P, rng);
System.out.println("Element selected "+s+ ", P(s)="+P.get(s));
} catch (Exception ex) {
Logger.getLogger(RouletteWheel.class.getName()).log(Level.SEVERE, null, ex);
}
}
P.clear();
P.add(0.2);
P.add(0.0);
P.add(0.5);
P.add(0.0);
P.add(0.1);
P.add(0.2);
//rng = new Random();
for (int i = 0 ; i < rept; i++){
try {
int s = rw.select(P, rng);
System.out.println("Element selected "+s+ ", P(s)="+P.get(s));
} catch (Exception ex) {
Logger.getLogger(RouletteWheel.class.getName()).log(Level.SEVERE, null, ex);
}
}
}
/**
* {#inheritDoc}
* #return
*/
#Override
public String toString()
{
return "Roulette Wheel Selection";
}
}
Below an execution sample for a proba vector P=[0.2,0.1,0.6,0.1],
WheelElements = [0,1,2,3]:
Element selected 3, P(s)=0.1
Element selected 2, P(s)=0.6
Element selected 3, P(s)=0.1
Element selected 2, P(s)=0.6
Element selected 1, P(s)=0.1
Element selected 2, P(s)=0.6
Element selected 3, P(s)=0.1
Element selected 2, P(s)=0.6
Element selected 2, P(s)=0.6
Element selected 2, P(s)=0.6
Element selected 2, P(s)=0.6
Element selected 2, P(s)=0.6
Element selected 3, P(s)=0.1
Element selected 2, P(s)=0.6
Element selected 2, P(s)=0.6
Element selected 2, P(s)=0.6
Element selected 0, P(s)=0.2
Element selected 2, P(s)=0.6
Element selected 2, P(s)=0.6
Element selected 2, P(s)=0.6
The code also tests a roulette wheel with zero probability.
I am afraid that anybody using the in built random number generator in all programming languages must be aware that the number generated is not 100% random.So should be used with caution.
Random Number Generator pseudo code
add one to a sequential counter
get the current value of the sequential counter
add the counter value by the computer tick count or some other small interval timer value
optionally add addition numbers, like a number from an external piece of hardware like a plasma generator or some other type of somewhat random phenomena
divide the result by a very big prime number
359334085968622831041960188598043661065388726959079837 for example
get some digits from the far right of the decimal point of the result
use these digits as a random number
Use the random number digits to create random numbers between 1 and 38 (or 37 European) for roulette.