I've been looking for a simple Java algorithm to generate a pseudo-random alpha-numeric string. In my situation it would be used as a unique session/key identifier that would "likely" be unique over 500K+ generation (my needs don't really require anything much more sophisticated).
Ideally, I would be able to specify a length depending on my uniqueness needs. For example, a generated string of length 12 might look something like "AEYGF7K0DM1X".
Algorithm
To generate a random string, concatenate characters drawn randomly from the set of acceptable symbols until the string reaches the desired length.
Implementation
Here's some fairly simple and very flexible code for generating random identifiers. Read the information that follows for important application notes.
public class RandomString {
/**
* Generate a random string.
*/
public String nextString() {
for (int idx = 0; idx < buf.length; ++idx)
buf[idx] = symbols[random.nextInt(symbols.length)];
return new String(buf);
}
public static final String upper = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
public static final String lower = upper.toLowerCase(Locale.ROOT);
public static final String digits = "0123456789";
public static final String alphanum = upper + lower + digits;
private final Random random;
private final char[] symbols;
private final char[] buf;
public RandomString(int length, Random random, String symbols) {
if (length < 1) throw new IllegalArgumentException();
if (symbols.length() < 2) throw new IllegalArgumentException();
this.random = Objects.requireNonNull(random);
this.symbols = symbols.toCharArray();
this.buf = new char[length];
}
/**
* Create an alphanumeric string generator.
*/
public RandomString(int length, Random random) {
this(length, random, alphanum);
}
/**
* Create an alphanumeric strings from a secure generator.
*/
public RandomString(int length) {
this(length, new SecureRandom());
}
/**
* Create session identifiers.
*/
public RandomString() {
this(21);
}
}
Usage examples
Create an insecure generator for 8-character identifiers:
RandomString gen = new RandomString(8, ThreadLocalRandom.current());
Create a secure generator for session identifiers:
RandomString session = new RandomString();
Create a generator with easy-to-read codes for printing. The strings are longer than full alphanumeric strings to compensate for using fewer symbols:
String easy = RandomString.digits + "ACEFGHJKLMNPQRUVWXYabcdefhijkprstuvwx";
RandomString tickets = new RandomString(23, new SecureRandom(), easy);
Use as session identifiers
Generating session identifiers that are likely to be unique is not good enough, or you could just use a simple counter. Attackers hijack sessions when predictable identifiers are used.
There is tension between length and security. Shorter identifiers are easier to guess, because there are fewer possibilities. But longer identifiers consume more storage and bandwidth. A larger set of symbols helps, but might cause encoding problems if identifiers are included in URLs or re-entered by hand.
The underlying source of randomness, or entropy, for session identifiers should come from a random number generator designed for cryptography. However, initializing these generators can sometimes be computationally expensive or slow, so effort should be made to re-use them when possible.
Use as object identifiers
Not every application requires security. Random assignment can be an efficient way for multiple entities to generate identifiers in a shared space without any coordination or partitioning. Coordination can be slow, especially in a clustered or distributed environment, and splitting up a space causes problems when entities end up with shares that are too small or too big.
Identifiers generated without taking measures to make them unpredictable should be protected by other means if an attacker might be able to view and manipulate them, as happens in most web applications. There should be a separate authorization system that protects objects whose identifier can be guessed by an attacker without access permission.
Care must be also be taken to use identifiers that are long enough to make collisions unlikely given the anticipated total number of identifiers. This is referred to as "the birthday paradox." The probability of a collision, p, is approximately n2/(2qx), where n is the number of identifiers actually generated, q is the number of distinct symbols in the alphabet, and x is the length of the identifiers. This should be a very small number, like 2‑50 or less.
Working this out shows that the chance of collision among 500k 15-character identifiers is about 2‑52, which is probably less likely than undetected errors from cosmic rays, etc.
Comparison with UUIDs
According to their specification, UUIDs are not designed to be unpredictable, and should not be used as session identifiers.
UUIDs in their standard format take a lot of space: 36 characters for only 122 bits of entropy. (Not all bits of a "random" UUID are selected randomly.) A randomly chosen alphanumeric string packs more entropy in just 21 characters.
UUIDs are not flexible; they have a standardized structure and layout. This is their chief virtue as well as their main weakness. When collaborating with an outside party, the standardization offered by UUIDs may be helpful. For purely internal use, they can be inefficient.
Java supplies a way of doing this directly. If you don't want the dashes, they are easy to strip out. Just use uuid.replace("-", "")
import java.util.UUID;
public class randomStringGenerator {
public static void main(String[] args) {
System.out.println(generateString());
}
public static String generateString() {
String uuid = UUID.randomUUID().toString();
return "uuid = " + uuid;
}
}
Output
uuid = 2d7428a6-b58c-4008-8575-f05549f16316
static final String AB = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
static SecureRandom rnd = new SecureRandom();
String randomString(int len){
StringBuilder sb = new StringBuilder(len);
for(int i = 0; i < len; i++)
sb.append(AB.charAt(rnd.nextInt(AB.length())));
return sb.toString();
}
If you're happy to use Apache classes, you could use org.apache.commons.text.RandomStringGenerator (Apache Commons Text).
Example:
RandomStringGenerator randomStringGenerator =
new RandomStringGenerator.Builder()
.withinRange('0', 'z')
.filteredBy(CharacterPredicates.LETTERS, CharacterPredicates.DIGITS)
.build();
randomStringGenerator.generate(12); // toUpperCase() if you want
Since Apache Commons Lang 3.6, RandomStringUtils is deprecated.
You can use an Apache Commons library for this, RandomStringUtils:
RandomStringUtils.randomAlphanumeric(20).toUpperCase();
In one line:
Long.toHexString(Double.doubleToLongBits(Math.random()));
Source: Java - generating a random string
This is easily achievable without any external libraries.
1. Cryptographic Pseudo Random Data Generation (PRNG)
First you need a cryptographic PRNG. Java has SecureRandom for that and typically uses the best entropy source on the machine (e.g. /dev/random). Read more here.
SecureRandom rnd = new SecureRandom();
byte[] token = new byte[byteLength];
rnd.nextBytes(token);
Note: SecureRandom is the slowest, but most secure way in Java of generating random bytes. I do however recommend not considering performance here since it usually has no real impact on your application unless you have to generate millions of tokens per second.
2. Required Space of Possible Values
Next you have to decide "how unique" your token needs to be. The whole and only point of considering entropy is to make sure that the system can resist brute force attacks: the space of possible values must be so large that any attacker could only try a negligible proportion of the values in non-ludicrous time1.
Unique identifiers such as random UUID have 122 bit of entropy (i.e., 2^122 = 5.3x10^36) - the chance of collision is "*(...) for there to be a one in a billion chance of duplication, 103 trillion version 4 UUIDs must be generated2". We will choose 128 bits since it fits exactly into 16 bytes and is seen as highly sufficient for being unique for basically every, but the most extreme, use cases and you don't have to think about duplicates. Here is a simple comparison table of entropy including simple analysis of the birthday problem.
For simple requirements, 8 or 12 byte length might suffice, but with 16 bytes you are on the "safe side".
And that's basically it. The last thing is to think about encoding so it can be represented as a printable text (read, a String).
3. Binary to Text Encoding
Typical encodings include:
Base64 every character encodes 6 bit, creating a 33% overhead. Fortunately there are standard implementations in Java 8+ and Android. With older Java you can use any of the numerous third-party libraries. If you want your tokens to be URL safe use the URL-safe version of RFC4648 (which usually is supported by most implementations). Example encoding 16 bytes with padding: XfJhfv3C0P6ag7y9VQxSbw==
Base32 every character encodes 5 bit, creating a 40% overhead. This will use A-Z and 2-7, making it reasonably space efficient while being case-insensitive alpha-numeric. There isn't any standard implementation in the JDK. Example encoding 16 bytes without padding: WUPIL5DQTZGMF4D3NX5L7LNFOY
Base16 (hexadecimal) every character encodes four bit, requiring two characters per byte (i.e., 16 bytes create a string of length 32). Therefore hexadecimal is less space efficient than Base32, but it is safe to use in most cases (URL) since it only uses 0-9 and A to F. Example encoding 16 bytes: 4fa3dd0f57cb3bf331441ed285b27735. See a Stack Overflow discussion about converting to hexadecimal here.
Additional encodings like Base85 and the exotic Base122 exist with better/worse space efficiency. You can create your own encoding (which basically most answers in this thread do), but I would advise against it, if you don't have very specific requirements. See more encoding schemes in the Wikipedia article.
4. Summary and Example
Use SecureRandom
Use at least 16 bytes (2^128) of possible values
Encode according to your requirements (usually hex or base32 if you need it to be alpha-numeric)
Don't
... use your home brew encoding: better maintainable and readable for others if they see what standard encoding you use instead of weird for loops creating characters at a time.
... use UUID: it has no guarantees on randomness; you are wasting 6 bits of entropy and have a verbose string representation
Example: Hexadecimal Token Generator
public static String generateRandomHexToken(int byteLength) {
SecureRandom secureRandom = new SecureRandom();
byte[] token = new byte[byteLength];
secureRandom.nextBytes(token);
return new BigInteger(1, token).toString(16); // Hexadecimal encoding
}
//generateRandomHexToken(16) -> 2189df7475e96aa3982dbeab266497cd
Example: Base64 Token Generator (URL Safe)
public static String generateRandomBase64Token(int byteLength) {
SecureRandom secureRandom = new SecureRandom();
byte[] token = new byte[byteLength];
secureRandom.nextBytes(token);
return Base64.getUrlEncoder().withoutPadding().encodeToString(token); //base64 encoding
}
//generateRandomBase64Token(16) -> EEcCCAYuUcQk7IuzdaPzrg
Example: Java CLI Tool
If you want a ready-to-use CLI tool you may use dice:
Example: Related issue - Protect Your Current Ids
If you already have an id you can use (e.g., a synthetic long in your entity), but don't want to publish the internal value, you can use this library to encrypt it and obfuscate it: https://github.com/patrickfav/id-mask
IdMask<Long> idMask = IdMasks.forLongIds(Config.builder(key).build());
String maskedId = idMask.mask(id);
// Example: NPSBolhMyabUBdTyanrbqT8
long originalId = idMask.unmask(maskedId);
Using Dollar should be as simple as:
// "0123456789" + "ABCDE...Z"
String validCharacters = $('0', '9').join() + $('A', 'Z').join();
String randomString(int length) {
return $(validCharacters).shuffle().slice(length).toString();
}
#Test
public void buildFiveRandomStrings() {
for (int i : $(5)) {
System.out.println(randomString(12));
}
}
It outputs something like this:
DKL1SBH9UJWC
JH7P0IT21EA5
5DTI72EO6SFU
HQUMJTEBNF7Y
1HCR6SKYWGT7
Here it is in Java:
import static java.lang.Math.round;
import static java.lang.Math.random;
import static java.lang.Math.pow;
import static java.lang.Math.abs;
import static java.lang.Math.min;
import static org.apache.commons.lang.StringUtils.leftPad
public class RandomAlphaNum {
public static String gen(int length) {
StringBuffer sb = new StringBuffer();
for (int i = length; i > 0; i -= 12) {
int n = min(12, abs(i));
sb.append(leftPad(Long.toString(round(random() * pow(36, n)), 36), n, '0'));
}
return sb.toString();
}
}
Here's a sample run:
scala> RandomAlphaNum.gen(42)
res3: java.lang.String = uja6snx21bswf9t89s00bxssu8g6qlu16ffzqaxxoy
A short and easy solution, but it uses only lowercase and numerics:
Random r = new java.util.Random ();
String s = Long.toString (r.nextLong () & Long.MAX_VALUE, 36);
The size is about 12 digits to base 36 and can't be improved further, that way. Of course you can append multiple instances.
Surprising, no one here has suggested it, but:
import java.util.UUID
UUID.randomUUID().toString();
Easy.
The benefit of this is UUIDs are nice, long, and guaranteed to be almost impossible to collide.
Wikipedia has a good explanation of it:
" ...only after generating 1 billion UUIDs every second for the next 100 years, the probability of creating just one duplicate would be about 50%."
The first four bits are the version type and two for the variant, so you get 122 bits of random. So if you want to, you can truncate from the end to reduce the size of the UUID. It's not recommended, but you still have loads of randomness, enough for your 500k records easy.
An alternative in Java 8 is:
static final Random random = new Random(); // Or SecureRandom
static final int startChar = (int) '!';
static final int endChar = (int) '~';
static String randomString(final int maxLength) {
final int length = random.nextInt(maxLength + 1);
return random.ints(length, startChar, endChar + 1)
.collect(StringBuilder::new, StringBuilder::appendCodePoint, StringBuilder::append)
.toString();
}
public static String generateSessionKey(int length){
String alphabet =
new String("0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"); // 9
int n = alphabet.length(); // 10
String result = new String();
Random r = new Random(); // 11
for (int i=0; i<length; i++) // 12
result = result + alphabet.charAt(r.nextInt(n)); //13
return result;
}
import java.util.Random;
public class passGen{
// Version 1.0
private static final String dCase = "abcdefghijklmnopqrstuvwxyz";
private static final String uCase = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
private static final String sChar = "!##$%^&*";
private static final String intChar = "0123456789";
private static Random r = new Random();
private static StringBuilder pass = new StringBuilder();
public static void main (String[] args) {
System.out.println ("Generating pass...");
while (pass.length () != 16){
int rPick = r.nextInt(4);
if (rPick == 0){
int spot = r.nextInt(26);
pass.append(dCase.charAt(spot));
} else if (rPick == 1) {
int spot = r.nextInt(26);
pass.append(uCase.charAt(spot));
} else if (rPick == 2) {
int spot = r.nextInt(8);
pass.append(sChar.charAt(spot));
} else {
int spot = r.nextInt(10);
pass.append(intChar.charAt(spot));
}
}
System.out.println ("Generated Pass: " + pass.toString());
}
}
This just adds the password into the string and... yeah, it works well. Check it out... It is very simple; I wrote it.
Using UUIDs is insecure, because parts of the UUID aren't random at all. The procedure of erickson is very neat, but it does not create strings of the same length. The following snippet should be sufficient:
/*
* The random generator used by this class to create random keys.
* In a holder class to defer initialization until needed.
*/
private static class RandomHolder {
static final Random random = new SecureRandom();
public static String randomKey(int length) {
return String.format("%"+length+"s", new BigInteger(length*5/*base 32,2^5*/, random)
.toString(32)).replace('\u0020', '0');
}
}
Why choose length*5? Let's assume the simple case of a random string of length 1, so one random character. To get a random character containing all digits 0-9 and characters a-z, we would need a random number between 0 and 35 to get one of each character.
BigInteger provides a constructor to generate a random number, uniformly distributed over the range 0 to (2^numBits - 1). Unfortunately 35 is not a number which can be received by 2^numBits - 1.
So we have two options: Either go with 2^5-1=31 or 2^6-1=63. If we would choose 2^6 we would get a lot of "unnecessary" / "longer" numbers. Therefore 2^5 is the better option, even if we lose four characters (w-z). To now generate a string of a certain length, we can simply use a 2^(length*numBits)-1 number. The last problem, if we want a string with a certain length, random could generate a small number, so the length is not met, so we have to pad the string to its required length prepending zeros.
I found this solution that generates a random hex encoded string. The provided unit test seems to hold up to my primary use case. Although, it is slightly more complex than some of the other answers provided.
/**
* Generate a random hex encoded string token of the specified length
*
* #param length
* #return random hex string
*/
public static synchronized String generateUniqueToken(Integer length){
byte random[] = new byte[length];
Random randomGenerator = new Random();
StringBuffer buffer = new StringBuffer();
randomGenerator.nextBytes(random);
for (int j = 0; j < random.length; j++) {
byte b1 = (byte) ((random[j] & 0xf0) >> 4);
byte b2 = (byte) (random[j] & 0x0f);
if (b1 < 10)
buffer.append((char) ('0' + b1));
else
buffer.append((char) ('A' + (b1 - 10)));
if (b2 < 10)
buffer.append((char) ('0' + b2));
else
buffer.append((char) ('A' + (b2 - 10)));
}
return (buffer.toString());
}
#Test
public void testGenerateUniqueToken(){
Set set = new HashSet();
String token = null;
int size = 16;
/* Seems like we should be able to generate 500K tokens
* without a duplicate
*/
for (int i=0; i<500000; i++){
token = Utility.generateUniqueToken(size);
if (token.length() != size * 2){
fail("Incorrect length");
} else if (set.contains(token)) {
fail("Duplicate token generated");
} else{
set.add(token);
}
}
}
Change String characters as per as your requirements.
String is immutable. Here StringBuilder.append is more efficient than string concatenation.
public static String getRandomString(int length) {
final String characters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJLMNOPQRSTUVWXYZ1234567890!##$%^&*()_+";
StringBuilder result = new StringBuilder();
while(length > 0) {
Random rand = new Random();
result.append(characters.charAt(rand.nextInt(characters.length())));
length--;
}
return result.toString();
}
import java.util.Date;
import java.util.Random;
public class RandomGenerator {
private static Random random = new Random((new Date()).getTime());
public static String generateRandomString(int length) {
char[] values = {'a','b','c','d','e','f','g','h','i','j',
'k','l','m','n','o','p','q','r','s','t',
'u','v','w','x','y','z','0','1','2','3',
'4','5','6','7','8','9'};
String out = "";
for (int i=0;i<length;i++) {
int idx=random.nextInt(values.length);
out += values[idx];
}
return out;
}
}
I don't really like any of these answers regarding a "simple" solution :S
I would go for a simple ;), pure Java, one liner (entropy is based on random string length and the given character set):
public String randomString(int length, String characterSet) {
return IntStream.range(0, length).map(i -> new SecureRandom().nextInt(characterSet.length())).mapToObj(randomInt -> characterSet.substring(randomInt, randomInt + 1)).collect(Collectors.joining());
}
#Test
public void buildFiveRandomStrings() {
for (int q = 0; q < 5; q++) {
System.out.println(randomString(10, "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789")); // The character set can basically be anything
}
}
Or (a bit more readable old way)
public String randomString(int length, String characterSet) {
StringBuilder sb = new StringBuilder(); // Consider using StringBuffer if needed
for (int i = 0; i < length; i++) {
int randomInt = new SecureRandom().nextInt(characterSet.length());
sb.append(characterSet.substring(randomInt, randomInt + 1));
}
return sb.toString();
}
#Test
public void buildFiveRandomStrings() {
for (int q = 0; q < 5; q++) {
System.out.println(randomString(10, "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789")); // The character set can basically be anything
}
}
But on the other hand you could also go with UUID which has a pretty good entropy:
UUID.randomUUID().toString().replace("-", "")
I'm using a library from Apache Commons to generate an alphanumeric string:
import org.apache.commons.lang3.RandomStringUtils;
String keyLength = 20;
RandomStringUtils.randomAlphanumeric(keylength);
It's fast and simple!
You mention "simple", but just in case anyone else is looking for something that meets more stringent security requirements, you might want to take a look at jpwgen. jpwgen is modeled after pwgen in Unix, and is very configurable.
import java.util.*;
import javax.swing.*;
public class alphanumeric {
public static void main(String args[]) {
String nval, lenval;
int n, len;
nval = JOptionPane.showInputDialog("Enter number of codes you require: ");
n = Integer.parseInt(nval);
lenval = JOptionPane.showInputDialog("Enter code length you require: ");
len = Integer.parseInt(lenval);
find(n, len);
}
public static void find(int n, int length) {
String str1 = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
StringBuilder sb = new StringBuilder(length);
Random r = new Random();
System.out.println("\n\t Unique codes are \n\n");
for(int i=0; i<n; i++) {
for(int j=0; j<length; j++) {
sb.append(str1.charAt(r.nextInt(str1.length())));
}
System.out.println(" " + sb.toString());
sb.delete(0, length);
}
}
}
Here is the one-liner by abacus-common:
String.valueOf(CharStream.random('0', 'z').filter(c -> N.isLetterOrDigit(c)).limit(12).toArray())
Random doesn't mean it must be unique. To get unique strings, use:
N.uuid() // E.g.: "e812e749-cf4c-4959-8ee1-57829a69a80f". length is 36.
N.guid() // E.g.: "0678ce04e18945559ba82ddeccaabfcd". length is 32 without '-'
You can use the following code, if your password mandatory contains numbers and alphabetic special characters:
private static final String NUMBERS = "0123456789";
private static final String UPPER_ALPHABETS = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
private static final String LOWER_ALPHABETS = "abcdefghijklmnopqrstuvwxyz";
private static final String SPECIALCHARACTERS = "##$%&*";
private static final int MINLENGTHOFPASSWORD = 8;
public static String getRandomPassword() {
StringBuilder password = new StringBuilder();
int j = 0;
for (int i = 0; i < MINLENGTHOFPASSWORD; i++) {
password.append(getRandomPasswordCharacters(j));
j++;
if (j == 3) {
j = 0;
}
}
return password.toString();
}
private static String getRandomPasswordCharacters(int pos) {
Random randomNum = new Random();
StringBuilder randomChar = new StringBuilder();
switch (pos) {
case 0:
randomChar.append(NUMBERS.charAt(randomNum.nextInt(NUMBERS.length() - 1)));
break;
case 1:
randomChar.append(UPPER_ALPHABETS.charAt(randomNum.nextInt(UPPER_ALPHABETS.length() - 1)));
break;
case 2:
randomChar.append(SPECIALCHARACTERS.charAt(randomNum.nextInt(SPECIALCHARACTERS.length() - 1)));
break;
case 3:
randomChar.append(LOWER_ALPHABETS.charAt(randomNum.nextInt(LOWER_ALPHABETS.length() - 1)));
break;
}
return randomChar.toString();
}
You can use the UUID class with its getLeastSignificantBits() message to get 64 bit of random data, and then convert it to a radix 36 number (i.e. a string consisting of 0-9,A-Z):
Long.toString(Math.abs( UUID.randomUUID().getLeastSignificantBits(), 36));
This yields a string up to 13 characters long. We use Math.abs() to make sure there isn't a minus sign sneaking in.
Here it is a Scala solution:
(for (i <- 0 until rnd.nextInt(64)) yield {
('0' + rnd.nextInt(64)).asInstanceOf[Char]
}) mkString("")
Using an Apache Commons library, it can be done in one line:
import org.apache.commons.lang.RandomStringUtils;
RandomStringUtils.randomAlphanumeric(64);
Documentation
public static String randomSeriesForThreeCharacter() {
Random r = new Random();
String value = "";
char random_Char ;
for(int i=0; i<10; i++)
{
random_Char = (char) (48 + r.nextInt(74));
value = value + random_char;
}
return value;
}
I think this is the smallest solution here, or nearly one of the smallest:
public String generateRandomString(int length) {
String randomString = "";
final char[] chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz01234567890".toCharArray();
final Random random = new Random();
for (int i = 0; i < length; i++) {
randomString = randomString + chars[random.nextInt(chars.length)];
}
return randomString;
}
The code works just fine. If you are using this method, I recommend you to use more than 10 characters. A collision happens at 5 characters / 30362 iterations. This took 9 seconds.
public class Utils {
private final Random RANDOM = new SecureRandom();
private final String ALPHABET = "0123456789QWERTYUIOPASDFGHJKLZXCVBNMqwertyuiopasdfghjklzxcvbnm";
private String generateRandomString(int length) {
StringBuffer buffer = new StringBuffer(length);
for (int i = 0; i < length; i++) {
buffer.append(ALPHABET.charAt(RANDOM.nextInt(ALPHABET.length())));
}
return new String(buffer);
}
}
I encountered a problem while coding and I can't seem to find where I messed up or even why I get a wrong result.
First, let me explain the task.
It's about "Yijing Hexagram Symbols".
The left one is the original and the right one is the result that my code should give me.
Basically every "hexagram" contains 6 lines that can be either diveded or not.
So there are a total of
2^6 = 64 possible "hexagrams"
The task is to calculate and code a methode to print all possible combinations.
Thats what I have so far :
public class test {
public String toBin (int zahl) {
if(zahl ==0) return "0";
if (zahl ==1 ) return "1";
return ""+(toBin( zahl/2)+(zahl%2));
}
public void show (String s) {
for (char c : s.toCharArray()){
if (c == '1'){
System.out.println("--- ---");
}
if(c=='0'){
System.out.println("-------");
}
}
}
public void ausgeben (){
for(int i = 0 ; i < 64; i++) {
show (toBin(i));
}
}
}
The problem is, when I test the 'show'-methode with "10" I get 3 lines and not 2 as intended.
public class runner {
public static void main(String[] args){
test a = new test();
a.ausgeben();
a.show("10");
}
}
Another problem I've encoutered is, that since I'm converting to binary i sometimes have not enough lines because for example 10 in binary is 0001010 but the first "0" are missing. How can I implement them in an easy way without changing much ?
I am fairly new to all this so if I didn't explain anything enough or made any mistakes feel free to tell me.
You may find it easier if you use the Integer.toBinaryString method combined with the String.format and String.replace methods.
String binary = String.format("%6s", Integer.toBinaryString(zahl)).replace(' ', '0');
This converts the number to binary, formats it in a field six spaces wide (with leading spaces as necessary), and then replaces the spaces with '0'.
Well, there are many ways to pad a string with zeros, or create a binary string that is already padded with zeros.
For example, you could do something like:
public String padToSix( String binStr ) {
return "000000".substring( 0, 5 - binStr.length() ) + binStr;
}
This would check how long your string is, and take as many zeros are needed to fill it up to six from the "000000" string.
Or you could simply replace your conversion method (which is recursive, and that's not really necessary) with one that specializes in six-digit numbers:
public static String toBin (int zahl) {
char[] digits = { '0','0','0','0','0','0' };
int currDigitIndex = 5;
while ( currDigitIndex >= 0 && zahl > 0 ) {
digits[currDigitIndex] += (zahl % 2);
currDigitIndex--;
zahl /= 2;
}
return new String(digits);
}
This one modifies the character array ( which initially has only zeros ) from the right to the left. It adds the value of the current bit to the character at the given place. '0' + 0 is '0', and '0' + 1 is '1'. Because you know in advance that you have six digits, you can start from the right and go to the left. If your number has only four digits, well, the two digits we haven't touched will be '0' because that's how the character array was initialized.
There are really a lot of methods to achieve the same thing.
Your problem reduces to printing all binary strings of length 6. I would go with this code snippet:
String format = "%06d";
for(int i = 0; i < 64; i++)
{
show(String.format(format, Integer.valueOf(Integer.toBinaryString(i))));
System.out.println();
}
If you don't wish to print leading zeros, replace String.format(..) with Integer.toBinaryString(i).
Okay, I implemented this SO question to my code: Return True or False Randomly
But, I have strange behavior: I need to run ten instances simultaneously, where every instance returns true or false just once per run. And surprisingly, no matter what I do, every time i get just false
Is there something to improve the method so I can have at least roughly 50% chance to get true?
To make it more understandable: I have my application builded to JAR file which is then run via batch command
java -jar my-program.jar
pause
Content of the program - to make it as simple as possible:
public class myProgram{
public static boolean getRandomBoolean() {
return Math.random() < 0.5;
// I tried another approaches here, still the same result
}
public static void main(String[] args) {
System.out.println(getRandomBoolean());
}
}
If I open 10 command lines and run it, I get false as result every time...
I recommend using Random.nextBoolean()
That being said, Math.random() < 0.5 as you have used works too. Here's the behavior on my machine:
$ cat myProgram.java
public class myProgram{
public static boolean getRandomBoolean() {
return Math.random() < 0.5;
//I tried another approaches here, still the same result
}
public static void main(String[] args) {
System.out.println(getRandomBoolean());
}
}
$ javac myProgram.java
$ java myProgram ; java myProgram; java myProgram; java myProgram
true
false
false
true
Needless to say, there are no guarantees for getting different values each time. In your case however, I suspect that
A) you're not working with the code you think you are, (like editing the wrong file)
B) you havn't compiled your different attempts when testing, or
C) you're working with some non-standard broken implementation.
Have you tried looking at the Java Documentation?
Returns the next pseudorandom, uniformly distributed boolean value from this random number generator's sequence ... the values true and false are produced with (approximately) equal probability.
For example:
import java.util.Random;
Random random = new Random();
random.nextBoolean();
You could also try nextBoolean()-Method
Here is an example: http://www.tutorialspoint.com/java/util/random_nextboolean.htm
Java 8: Use random generator isolated to the current thread: ThreadLocalRandom nextBoolean()
Like the global Random generator used by the Math class, a ThreadLocalRandom is initialized with an internally generated seed that may not otherwise be modified. When applicable, use of ThreadLocalRandom rather than shared Random objects in concurrent programs will typically encounter much less overhead and contention.
java.util.concurrent.ThreadLocalRandom.current().nextBoolean();
Why not use the Random class, which has a method nextBoolean:
import java.util.Random;
/** Generate 10 random booleans. */
public final class MyProgram {
public static final void main(String... args){
Random randomGenerator = new Random();
for (int idx = 1; idx <= 10; ++idx){
boolean randomBool = randomGenerator.nextBoolean();
System.out.println("Generated : " + randomBool);
}
}
}
You can use the following for an unbiased result:
Random random = new Random();
//For 50% chance of true
boolean chance50oftrue = (random.nextInt(2) == 0) ? true : false;
Note: random.nextInt(2) means that the number 2 is the bound. the counting starts at 0. So we have 2 possible numbers (0 and 1) and hence the probability is 50%!
If you want to give more probability to your result to be true (or false) you can adjust the above as following!
Random random = new Random();
//For 50% chance of true
boolean chance50oftrue = (random.nextInt(2) == 0) ? true : false;
//For 25% chance of true
boolean chance25oftrue = (random.nextInt(4) == 0) ? true : false;
//For 40% chance of true
boolean chance40oftrue = (random.nextInt(5) < 2) ? true : false;
The easiest way to initialize a random number generator is to use the parameterless constructor, for example
Random generator = new Random();
However, in using this constructor you should recognize that algorithmic random number generators are not truly random, they are really algorithms that generate a fixed but random-looking sequence of numbers.
You can make it appear more 'random' by giving the Random constructor the 'seed' parameter, which you can dynamically built by for example using system time in milliseconds (which will always be different)
you could get your clock() value and check if it is odd or even. I dont know if it is %50 of true
And you can custom-create your random function:
static double s=System.nanoTime();//in the instantiating of main applet
public static double randoom()
{
s=(double)(((555555555* s+ 444444)%100000)/(double)100000);
return s;
}
numbers 55555.. and 444.. are the big numbers to get a wide range function
please ignore that skype icon :D
You can also make two random integers and verify if they are the same, this gives you more control over the probabilities.
Random rand = new Random();
Declare a range to manage random probability.
In this example, there is a 50% chance of being true.
int range = 2;
Generate 2 random integers.
int a = rand.nextInt(range);
int b = rand.nextInt(range);
Then simply compare return the value.
return a == b;
I also have a class you can use.
RandomRange.java
Words in a text are always a source of randomness. Given a certain word, nothing can be inferred about the next word. For each word, we can take the ASCII codes of its letters, add those codes to form a number. The parity of this number is a good candidate for a random boolean.
Possible drawbacks:
this strategy is based upon using a text file as a source for the words. At some point,
the end of the file will be reached. However, you can estimate how many times you are expected to call the randomBoolean()
function from your app. If you will need to call it about 1 million times, then a text file with 1 million words will be enough.
As a correction, you can use a stream of data from a live source like an online newspaper.
using some statistical analysis of the common phrases and idioms in a language, one can estimate the next word in a phrase,
given the first words of the phrase, with some degree of accuracy. But statistically, these cases are rare, when we can accuratelly
predict the next word. So, in most cases, the next word is independent on the previous words.
package p01;
import java.io.File;
import java.nio.file.Files;
import java.nio.file.Paths;
public class Main {
String words[];
int currentIndex=0;
public static String readFileAsString()throws Exception
{
String data = "";
File file = new File("the_comedy_of_errors");
//System.out.println(file.exists());
data = new String(Files.readAllBytes(Paths.get(file.getName())));
return data;
}
public void init() throws Exception
{
String data = readFileAsString();
words = data.split("\\t| |,|\\.|'|\\r|\\n|:");
}
public String getNextWord() throws Exception
{
if(currentIndex>words.length-1)
throw new Exception("out of words; reached end of file");
String currentWord = words[currentIndex];
currentIndex++;
while(currentWord.isEmpty())
{
currentWord = words[currentIndex];
currentIndex++;
}
return currentWord;
}
public boolean getNextRandom() throws Exception
{
String nextWord = getNextWord();
int asciiSum = 0;
for (int i = 0; i < nextWord.length(); i++){
char c = nextWord.charAt(i);
asciiSum = asciiSum + (int) c;
}
System.out.println(nextWord+"-"+asciiSum);
return (asciiSum%2==1) ;
}
public static void main(String args[]) throws Exception
{
Main m = new Main();
m.init();
while(true)
{
System.out.println(m.getNextRandom());
Thread.sleep(100);
}
}
}
In Eclipse, in the root of my project, there is a file called 'the_comedy_of_errors' (no extension) - created with File> New > File , where I pasted some content from here: http://shakespeare.mit.edu/comedy_errors/comedy_errors.1.1.html
For a flexible boolean randomizer:
public static rbin(bias){
bias = bias || 50;
return(Math.random() * 100 <= bias);
/*The bias argument is optional but will allow you to put some weight
on the TRUE side. The higher the bias number, the more likely it is
true.*/
}
Make sure to use numbers 0 - 100 or you might lower the bias and get more common false values.
PS: I do not know anything about Java other than it has a few features in common with JavaScript. I used my JavaScript knowledge plus my inferring power to construct this code. Expect my answer to not be functional. Y'all can edit this answer to fix any issues I am not aware of.