How do I generate a random int
value in a specific range?
The following methods have bugs related to integer overflow:
randomNum = minimum + (int)(Math.random() * maximum);
// Bug: `randomNum` can be bigger than `maximum`.
Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum = minimum + i;
// Bug: `randomNum` can be smaller than `minimum`.
In Java 1.7 or later, the standard way to do this is as follows:
import java.util.concurrent.ThreadLocalRandom;
// nextInt is normally exclusive of the top value,
// so add 1 to make it inclusive
int randomNum = ThreadLocalRandom.current().nextInt(min, max + 1);
See the relevant JavaDoc. This approach has the advantage of not needing to explicitly initialize a java.util.Random instance, which can be a source of confusion and error if used inappropriately.
However, conversely there is no way to explicitly set the seed so it can be difficult to reproduce results in situations where that is useful such as testing or saving game states or similar. In those situations, the pre-Java 1.7 technique shown below can be used.
Before Java 1.7, the standard way to do this is as follows:
import java.util.Random;
/**
* Returns a pseudo-random number between min and max, inclusive.
* The difference between min and max can be at most
* <code>Integer.MAX_VALUE - 1</code>.
*
* @param min Minimum value
* @param max Maximum value. Must be greater than min.
* @return Integer between min and max, inclusive.
* @see java.util.Random#nextInt(int)
*/
public static int randInt(int min, int max) {
// NOTE: This will (intentionally) not run as written so that folks
// copy-pasting have to think about how to initialize their
// Random instance. Initialization of the Random instance is outside
// the main scope of the question, but some decent options are to have
// a field that is initialized once and then re-used as needed or to
// use ThreadLocalRandom (if using at least Java 1.7).
//
// In particular, do NOT do 'Random rand = new Random()' here or you
// will get not very good / not very random results.
Random rand;
// nextInt is normally exclusive of the top value,
// so add 1 to make it inclusive
int randomNum = rand.nextInt((max - min) + 1) + min;
return randomNum;
}
See the relevant JavaDoc. In practice, the java.util.Random class is often preferable to java.lang.Math.random().
In particular, there is no need to reinvent the random integer generation wheel when there is a straightforward API within the standard library to accomplish the task.
Answered 2023-09-20 20:13:01
max
value is Integer.MAX_VALUE
it is possible to overflow ,resulting into a java.lang.IllegalArgumentException
. You can try with : randInt(0, Integer.MAX_VALUE)
. Also, if nextInt((max-min) + 1)
returns the most high value (quite rare, I assume) won't it overflow again( supposing min and max are high enough values)? How to deal with this kind of situations? - anyone ThreadLocalRandom
solution even if you can't specify the seed or you don't need a thread safe method, since the API is more intuitive if min != 0
. - anyone RandomGenerator
, "It is recommended that multithreaded applications use either ThreadLocalRandom
or (preferably) pseudorandom number generators that implement the RandomGenerator.SplittableGenerator
or RandomGenerator.JumpableGenerator
interface." See: stackoverflow.com/a/70529176/1108305. - anyone Note that this approach is more biased and less efficient than a nextInt
approach, https://stackoverflow.com/a/738651/360211
One standard pattern for accomplishing this is:
Min + (int)(Math.random() * ((Max - Min) + 1))
The Java Math library function Math.random() generates a double value in the range [0,1)
. Notice this range does not include the 1.
In order to get a specific range of values first, you need to multiply by the magnitude of the range of values you want covered.
Math.random() * ( Max - Min )
This returns a value in the range [0,Max-Min)
, where 'Max-Min' is not included.
For example, if you want [5,10)
, you need to cover five integer values so you use
Math.random() * 5
This would return a value in the range [0,5)
, where 5 is not included.
Now you need to shift this range up to the range that you are targeting. You do this by adding the Min value.
Min + (Math.random() * (Max - Min))
You now will get a value in the range [Min,Max)
. Following our example, that means [5,10)
:
5 + (Math.random() * (10 - 5))
But, this still doesn't include Max
and you are getting a double value. In order to get the Max
value included, you need to add 1 to your range parameter (Max - Min)
and then truncate the decimal part by casting to an int. This is accomplished via:
Min + (int)(Math.random() * ((Max - Min) + 1))
And there you have it. A random integer value in the range [Min,Max]
, or per the example [5,10]
:
5 + (int)(Math.random() * ((10 - 5) + 1))
Answered 2023-09-20 20:13:01
Use:
Random ran = new Random();
int x = ran.nextInt(6) + 5;
The integer x
is now the random number that has a possible outcome of 5-10
.
Answered 2023-09-20 20:13:01
With java-8 they introduced the method ints(int randomNumberOrigin, int randomNumberBound)
in the Random
class.
For example if you want to generate five random integers (or a single one) in the range [0, 10], just do:
Random r = new Random();
int[] fiveRandomNumbers = r.ints(5, 0, 11).toArray();
int randomNumber = r.ints(1, 0, 11).findFirst().getAsInt();
The first parameter indicates just the size of the IntStream
generated (which is the overloaded method of the one that produces an unlimited IntStream
).
If you need to do multiple separate calls, you can create an infinite primitive iterator from the stream:
public final class IntRandomNumberGenerator {
private PrimitiveIterator.OfInt randomIterator;
/**
* Initialize a new random number generator that generates
* random numbers in the range [min, max]
* @param min - the min value (inclusive)
* @param max - the max value (inclusive)
*/
public IntRandomNumberGenerator(int min, int max) {
randomIterator = new Random().ints(min, max + 1).iterator();
}
/**
* Returns a random number in the range (min, max)
* @return a random number in the range (min, max)
*/
public int nextInt() {
return randomIterator.nextInt();
}
}
You can also do it for double
and long
values.
Answered 2023-09-20 20:13:01
You can edit your second code example to:
Random rn = new Random();
int range = maximum - minimum + 1;
int randomNum = rn.nextInt(range) + minimum;
Answered 2023-09-20 20:13:01
Just a small modification of your first solution would suffice.
Random rand = new Random();
randomNum = minimum + rand.nextInt((maximum - minimum) + 1);
See more here for implementation of Random
Answered 2023-09-20 20:13:01
ThreadLocalRandom
equivalent of class java.util.Random
for multithreaded environment. Generating a random number is carried out locally in each of the threads. So we have a better performance by reducing the conflicts.
int rand = ThreadLocalRandom.current().nextInt(x,y);
x
,y
- intervals e.g. (1,10)
Answered 2023-09-20 20:13:01
The Math.Random
class in Java is 0-based. So, if you write something like this:
Random rand = new Random();
int x = rand.nextInt(10);
x
will be between 0-9
inclusive.
So, given the following array of 25
items, the code to generate a random number between 0
(the base of the array) and array.length
would be:
String[] i = new String[25];
Random rand = new Random();
int index = 0;
index = rand.nextInt( i.length );
Since i.length
will return 25
, the nextInt( i.length )
will return a number between the range of 0-24
. The other option is going with Math.Random
which works in the same way.
index = (int) Math.floor(Math.random() * i.length);
For a better understanding, check out forum post Random Intervals (archive.org).
Answered 2023-09-20 20:13:01
index
variable will not affect the result of the random number. You can choose to initialize it any way you would like without having to worry about changing the outcome. Hope this helps. - anyone int index = rand.nextInt(i.Length);
- anyone int index; \n index = rand...
if one is fond of declarations and assignments on different lines. Some coding standards are more stringent (and without apparent purpose) than others. - anyone It can be done by simply doing the statement:
Randomizer.generate(0, 10); // Minimum of zero and maximum of ten
Below is its source code.
public class Randomizer {
public static int generate(int min, int max) {
return min + (int)(Math.random() * ((max - min) + 1));
}
}
It is just clean and simple.
Answered 2023-09-20 20:13:01
Forgive me for being fastidious, but the solution suggested by the majority, i.e., min + rng.nextInt(max - min + 1))
, seems perilous due to the fact that:
rng.nextInt(n)
cannot reach Integer.MAX_VALUE
.(max - min)
may cause overflow when min
is negative.A foolproof solution would return correct results for any min <= max
within [Integer.MIN_VALUE
, Integer.MAX_VALUE
]. Consider the following naive implementation:
int nextIntInRange(int min, int max, Random rng) {
if (min > max) {
throw new IllegalArgumentException("Cannot draw random int from invalid range [" + min + ", " + max + "].");
}
int diff = max - min;
if (diff >= 0 && diff != Integer.MAX_VALUE) {
return (min + rng.nextInt(diff + 1));
}
int i;
do {
i = rng.nextInt();
} while (i < min || i > max);
return i;
}
Although inefficient, note that the probability of success in the while
loop will always be 50% or higher.
Answered 2023-09-20 20:13:01
I wonder if any of the random number generating methods provided by an Apache Commons Math library would fit the bill.
For example: RandomDataGenerator.nextInt
or RandomDataGenerator.nextLong
Answered 2023-09-20 20:13:01
I use this:
/**
* @param min - The minimum.
* @param max - The maximum.
* @return A random double between these numbers (inclusive the minimum and maximum).
*/
public static double getRandom(double min, double max) {
return (Math.random() * (max + 1 - min)) + min;
}
You can cast it to an Integer if you want.
Answered 2023-09-20 20:13:01
new Random
(check the JavaDoc): "Creates a new random number generator. This constructor sets the seed of the random number generator to a value very likely to be distinct from any other invocation of this constructor." Very likely might just involve using the current time as seed. If that time uses milliseconds then current computers are fast enough to generate the same number. But besides that 2147483647 is Integer.MAX_VALUE
; the output obviously depends on the input, which you haven't specified. - anyone As of Java 7, you should no longer use Random. For most uses, the random number generator of choice is now ThreadLocalRandom.
For fork join pools and parallel streams, use SplittableRandom.
Joshua Bloch. Effective Java. Third Edition.
For fork join pools and parallel streams, use SplittableRandom (it implements SplittableGenerator interface, see Java 17 notes below) that is usually faster, has a better statistical independence and uniformity properties in comparison with Random.
To generate a random int
in the range [0, 1_000]:
int n = new SplittableRandom().nextInt(0, 1_001);
To generate a random int[100]
array of values in the range [0, 1_000]:
int[] a = new SplittableRandom().ints(100, 0, 1_001).parallel().toArray();
To return a Stream of random values:
IntStream stream = new SplittableRandom().ints(100, 0, 1_001);
It is recommended that multithreaded applications use either ThreadLocalRandom or (preferably) pseudorandom number generators that implement the RandomGenerator.SplittableGenerator or RandomGenerator.JumpableGenerator interface.
RandomGenerator1 documentation.
1 - RandomGenerator is a modern interface that was introduced in Java 17. It is a common protocol for objects that generate random or pseudorandom sequences of numbers and has more features and methods than Random.
Answered 2023-09-20 20:13:01
.parallel()
? It seems to me like generating a 100 random numbers would be too trivial to warrant parallelism. - anyone parallel
processing). By the way, for array of 1_000_000
elements, the parallel
version was 2 times faster on my machine in comparison with sequential. - anyone Let us take an example.
Suppose I wish to generate a number between 5-10:
int max = 10;
int min = 5;
int diff = max - min;
Random rn = new Random();
int i = rn.nextInt(diff + 1);
i += min;
System.out.print("The Random Number is " + i);
Let us understand this...
Initialize max with highest value and min with the lowest value.
Now, we need to determine how many possible values can be obtained. For this example, it would be:
5, 6, 7, 8, 9, 10
So, count of this would be max - min + 1.
i.e. 10 - 5 + 1 = 6
The random number will generate a number between 0-5.
i.e. 0, 1, 2, 3, 4, 5
Adding the min value to the random number would produce:
5, 6, 7, 8, 9, 10
Hence we obtain the desired range.
Answered 2023-09-20 20:13:01
Generate a random number for the difference of min and max by using the nextint(n) method and then add min number to the result:
Random rn = new Random();
int result = rn.nextInt(max - min + 1) + min;
System.out.println(result);
Answered 2023-09-20 20:13:01
To generate a random number "in between two numbers", use the following code:
Random r = new Random();
int lowerBound = 1;
int upperBound = 11;
int result = r.nextInt(upperBound-lowerBound) + lowerBound;
This gives you a random number in between 1 (inclusive) and 11 (exclusive), so initialize the upperBound value by adding 1. For example, if you want to generate random number between 1 to 10 then initialize the upperBound number with 11 instead of 10.
Answered 2023-09-20 20:13:01
Just use the Random class:
Random ran = new Random();
// Assumes max and min are non-negative.
int randomInt = min + ran.nextInt(max - min + 1);
Answered 2023-09-20 20:13:01
These methods might be convenient to use:
This method will return a random number between the provided minimum and maximum value:
public static int getRandomNumberBetween(int min, int max) {
Random foo = new Random();
int randomNumber = foo.nextInt(max - min) + min;
if (randomNumber == min) {
// Since the random number is between the min and max values, simply add 1
return min + 1;
} else {
return randomNumber;
}
}
and this method will return a random number from the provided minimum and maximum value (so the generated number could also be the minimum or maximum number):
public static int getRandomNumberFrom(int min, int max) {
Random foo = new Random();
int randomNumber = foo.nextInt((max + 1) - min) + min;
return randomNumber;
}
Answered 2023-09-20 20:13:01
// Since the random number is between the min and max values, simply add 1
. Why? Doesn't min count? Usually the range is [min, max) where min is included and max is excluded. Wrong answer, voted down. - anyone min + 1
will be twice as likely than the other number to be the result of getRandomNumberBetween
! - anyone In case of rolling a dice it would be random number between 1 to 6 (not 0 to 6), so:
face = 1 + randomNumbers.nextInt(6);
Answered 2023-09-20 20:13:01
int random = minimum + Double.valueOf(Math.random()*(maximum-minimum )).intValue();
Or take a look to RandomUtils from Apache Commons.
Answered 2023-09-20 20:13:01
Double.valueOf(Math.random()*(maximum-minimun)).intValue()
is quite an obfuscated (and inefficient) way to say (int)(Math.random()*(maximum-minimun))
… - anyone You can achieve that concisely in Java 8:
Random random = new Random();
int max = 10;
int min = 5;
int totalNumber = 10;
IntStream stream = random.ints(totalNumber, min, max);
stream.forEach(System.out::println);
Answered 2023-09-20 20:13:01
Here's a helpful class to generate random ints
in a range with any combination of inclusive/exclusive bounds:
import java.util.Random;
public class RandomRange extends Random {
public int nextIncInc(int min, int max) {
return nextInt(max - min + 1) + min;
}
public int nextExcInc(int min, int max) {
return nextInt(max - min) + 1 + min;
}
public int nextExcExc(int min, int max) {
return nextInt(max - min - 1) + 1 + min;
}
public int nextIncExc(int min, int max) {
return nextInt(max - min) + min;
}
}
Answered 2023-09-20 20:13:01
Another option is just using Apache Commons:
import org.apache.commons.math.random.RandomData;
import org.apache.commons.math.random.RandomDataImpl;
public void method() {
RandomData randomData = new RandomDataImpl();
int number = randomData.nextInt(5, 10);
// ...
}
Answered 2023-09-20 20:13:01
I found this example Generate random numbers :
This example generates random integers in a specific range.
import java.util.Random;
/** Generate random integers in a certain range. */
public final class RandomRange {
public static final void main(String... aArgs){
log("Generating random integers in the range 1..10.");
int START = 1;
int END = 10;
Random random = new Random();
for (int idx = 1; idx <= 10; ++idx){
showRandomInteger(START, END, random);
}
log("Done.");
}
private static void showRandomInteger(int aStart, int aEnd, Random aRandom){
if ( aStart > aEnd ) {
throw new IllegalArgumentException("Start cannot exceed End.");
}
//get the range, casting to long to avoid overflow problems
long range = (long)aEnd - (long)aStart + 1;
// compute a fraction of the range, 0 <= frac < range
long fraction = (long)(range * aRandom.nextDouble());
int randomNumber = (int)(fraction + aStart);
log("Generated : " + randomNumber);
}
private static void log(String aMessage){
System.out.println(aMessage);
}
}
An example run of this class :
Generating random integers in the range 1..10.
Generated : 9
Generated : 3
Generated : 3
Generated : 9
Generated : 4
Generated : 1
Generated : 3
Generated : 9
Generated : 10
Generated : 10
Done.
Answered 2023-09-20 20:13:01
public static Random RANDOM = new Random(System.nanoTime());
public static final float random(final float pMin, final float pMax) {
return pMin + RANDOM.nextFloat() * (pMax - pMin);
}
Answered 2023-09-20 20:13:01
Here is a simple sample that shows how to generate random number from closed [min, max]
range, while min <= max is true
You can reuse it as field in hole class, also having all Random.class
methods in one place
Results example:
RandomUtils random = new RandomUtils();
random.nextInt(0, 0); // returns 0
random.nextInt(10, 10); // returns 10
random.nextInt(-10, 10); // returns numbers from -10 to 10 (-10, -9....9, 10)
random.nextInt(10, -10); // throws assert
Sources:
import junit.framework.Assert;
import java.util.Random;
public class RandomUtils extends Random {
/**
* @param min generated value. Can't be > then max
* @param max generated value
* @return values in closed range [min, max].
*/
public int nextInt(int min, int max) {
Assert.assertFalse("min can't be > then max; values:[" + min + ", " + max + "]", min > max);
if (min == max) {
return max;
}
return nextInt(max - min + 1) + min;
}
}
Answered 2023-09-20 20:13:01
It's better to use SecureRandom rather than just Random.
public static int generateRandomInteger(int min, int max) {
SecureRandom rand = new SecureRandom();
rand.setSeed(new Date().getTime());
int randomNum = rand.nextInt((max - min) + 1) + min;
return randomNum;
}
Answered 2023-09-20 20:13:01
private static int SecureRandom rand = new SecureRandom();
2: static {
3: rand.setSeed(...);
4: }
- anyone SecureRandom
, it will be seeded by the system. Directly calling setSeed
is very dangerous, it may replace the (really random) seed with the date. And that will certainly not result in a SecureRandom
, as anybody can guess the time and try and seed their own SecureRandom
instance with that information. - anyone