Related
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.
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))
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.
Use:
minValue + rn.nextInt(maxValue - minValue + 1)
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.
You can edit your second code example to:
Random rn = new Random();
int range = maximum - minimum + 1;
int randomNum = rn.nextInt(range) + minimum;
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
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)
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).
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.
File Randomizer.java
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.
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.
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
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.
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.
Starting from Java 8
For fork join pools and parallel streams, use SplittableRandom 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);
rand.nextInt((max+1) - min) + min;
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.
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);
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.
Just use the Random class:
Random ran = new Random();
// Assumes max and min are non-negative.
int randomInt = min + ran.nextInt(max - min + 1);
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;
}
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);
int random = minimum + Double.valueOf(Math.random()*(maximum-minimum )).intValue();
Or take a look to RandomUtils from Apache Commons.
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);
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;
}
}
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);
// ...
}
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.
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);
}
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;
}
}
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;
}
rand.nextInt((max+1) - min) + min;
This is working fine.
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.
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))
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.
Use:
minValue + rn.nextInt(maxValue - minValue + 1)
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.
You can edit your second code example to:
Random rn = new Random();
int range = maximum - minimum + 1;
int randomNum = rn.nextInt(range) + minimum;
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
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)
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).
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.
File Randomizer.java
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.
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.
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
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.
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.
Starting from Java 8
For fork join pools and parallel streams, use SplittableRandom 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);
rand.nextInt((max+1) - min) + min;
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.
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);
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.
Just use the Random class:
Random ran = new Random();
// Assumes max and min are non-negative.
int randomInt = min + ran.nextInt(max - min + 1);
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;
}
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);
int random = minimum + Double.valueOf(Math.random()*(maximum-minimum )).intValue();
Or take a look to RandomUtils from Apache Commons.
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);
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;
}
}
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);
// ...
}
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.
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);
}
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;
}
}
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;
}
rand.nextInt((max+1) - min) + min;
This is working fine.
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.
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))
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.
Use:
minValue + rn.nextInt(maxValue - minValue + 1)
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.
You can edit your second code example to:
Random rn = new Random();
int range = maximum - minimum + 1;
int randomNum = rn.nextInt(range) + minimum;
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
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)
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).
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.
File Randomizer.java
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.
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.
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
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.
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.
Starting from Java 8
For fork join pools and parallel streams, use SplittableRandom 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);
rand.nextInt((max+1) - min) + min;
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.
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);
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.
Just use the Random class:
Random ran = new Random();
// Assumes max and min are non-negative.
int randomInt = min + ran.nextInt(max - min + 1);
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;
}
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);
int random = minimum + Double.valueOf(Math.random()*(maximum-minimum )).intValue();
Or take a look to RandomUtils from Apache Commons.
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);
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;
}
}
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);
// ...
}
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.
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);
}
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;
}
}
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;
}
rand.nextInt((max+1) - min) + min;
This is working fine.
I have following code:
quesPart1 = ran.nextInt((numbersBetween - 2) + 1) + 2;
quesPart2 = ran.nextInt((numbersBetween - 2) + 1) + 2;
if(quesPart2 > quesPart1)
{
int placeHolder = quesPart1;
quesPart1 = quesPart2;
quesPart2 = placeHolder;
}
//if first part is even
if(quesPart1 % 2 == 0)
{
if(quesPart2 % 2 != 0)
{
--quesPart2;
}
}
else
{
if(quesPart2 % 2 == 0)
{
++quesPart2;
}
}
Above code make sure that if quesPart1 is greater than quesPart2 and both are even or both are odd numbers. Now i want to get only random numbers which are also divisible by one another. Like if i divide quesPart1 by quesPart2 i get integer not decimal number. Any ideas how i can do that without adding too much complexity to above code.
You can do something like:
int div = quesPart1 / quesPart2;
quesPart1 = div * quesPart2;
add this code at the bottom of your code.
Like if i divide quesPart1 by quesPart2 i get integer not decimal number.
Keep it simple: generate random numbers and take their product. Example:
quesPart2 = ran.nextInt(UPPER_BOUND);
int temp = ran.nextInt(UPPER_BOUND);
questPart1 = temp * quesPart2;
Specifying the range, as in the original question, is left an an exercise to the reader. (What, you didn't think I was going to do all the thinking for you, did you? ;-)
Look into the modulus operator, a % b. It returns the left over amount when a is divided by b. When b cleanly divides into a, such that there is no decimal part, a % b will be zero.
In order to generate a number that is divisible by another, given two random numbers, a and b, simply multiply a by b. This will give you c, a number that is a multiple of both a and b, and therefore dividable by both cleanly without remainder.
I have come up with this simple function and a do while loop that is easy to implement.
// This is a simple function to set the min and max integers you want
const getRandomIntInclusive = (min, max) => {
min = Math.ceil(min);
max = Math.floor(max);
return Math.floor(Math.random() * (max - min + 1)) + min;
}
//Define some variables variables
let firstNo = 0
let secondNo = 0
let isDivisible = 0;
//generate random ints until first number is divisible to second number
do {
//get random int between 1-9 for the first and second integer
firstNo = getRandomIntInclusive(1, 9)
secondNo = getRandomIntInclusive(1, 9)
isDivisible = firstNo % secondNo; //Check if it's fully divisible
}
while (isDivisible != 0) //Run until it is fully divisible
To generate Random numbers in java you can use ran.nextInt() or please refer to this link to see how to generate random numbers.
store those 2 random numbers (as num1 and num2).
To verify whether the solution after dividing num1 and num2 is integer or not, use this method:
sol = num1 / num2
if (sol == (int)sol)
{
... //true if the solution is an integer
}
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Given a range of integers, how do I generate a random integer divisible by 5 in that range?
I'm using Java
just generate a regular random integer and multiply it by 5!
details: generate a random integer in [0, n) where n is the number of multiples of 5 in your range, then multiply it by 5 and add the lowest multiple to it.
one-liner: System.out.println(rnd.nextInt(max / 5 - (min + 4) / 5 + 1) * 5 + (min + 4) / 5 * 5); (assuming non-negative and valid arguments)
credits: lowest multiple expression (min + 4) / 5 * 5 from here and expression simplified a bit based on #Thomas's (imo currently incorrect) answer
This question calls for a multiple of five in a range, not number in the period of five in the range.
This solution handles negatives and range validity.
// because Java's % operator doesn't do what one might expect with negatives
int lbound = (min+4) - (((min+4) % 5) + 5) % 5;
int ubound = max - (((max % 5) + 5) % 5);
if (lbound > ubound) {
// do something about the range error
}
if (lbound == ubound) {
return lbound;
}
int range = ((ubound - lbound)/5) + 1;
return ((int)(Math.random() * range) * 5) + lbound;
First create a Random, and round low and high to the nearest higher/lower multiple of 5 respectively:
Random r = new Random();
low = ((low+4)/5)*5; // next multiple of 5
high = (high/5)*5; // previous multiple of 5
This may make low > high, which is infeasible, so don't proceed any further; or it make may make low == high, which may be of no interest whatsover, so you may want to test for that. The code below works correctly either way, because of the +1 and -1: generate a random number in {low..high}
int randomPart = r.nextInt(high-low+1)+low-1;
Then round it upwards to a multiple of 5. The prior shenanigans with low and high assure it is in range:
int nextInt = ((randomPart+4)/5)*5;
This method first computes how many numbers divisible by 5 are in the given range. It picks a number between 0 and that count at random, and translates that random number back into the given range by multiplying it with 5 and adding it to the lower bound.
Note that both lowerBound and upperBound are inclusive.
public static int getRandomDivisibleByFive(int lowerBound, int upperBound) {
if (lowerBound > 0) lowerBound += 4;
if (upperBound < 0) upperBound -= 4;
lowerBound /= 5;
upperBound /= 5;
int n = upperBound - lowerBound + 1;
if (n < 1) {
throw new IllegalArgumentException("Range too small");
}
return 5 * (lowerBound + new Random().nextInt(n));
}
Picks a random number between your values and then tests if it is divisible by div. If it is it returns that value otherwise it will have to do at max div-1 iterations to get to a number divisible by div.
In your situation call rBetweenGenerator(min, max, 5)
public int rBetweenGenerator(int min, int max, int div)
{
int res = min + ((new Random()).nextInt(max - min + 1))
for(int i = res; i < res + div; i++)
{
if( i % div == 0 )
{
return i;
}
} return -1; //error
}