Develop Reference

Selecting random ROUND number without being non-even distribution

So I want to make a number selector in JS that select a ROUND number between 1 and 100. Like selecting a winning ticket for example.
I found out I could use a function and then Math.round, like this
function sort(min,max) {
return Math.round(Math.randon() * (max - min) + min)
}
console.log(sort(1,100))
But then I readed Math.round will make a non-even distribution, with max and min values less likely to roll than the others.
So how can I solve this one? Getting a ROUND number between 1 and 100 with even distribution?

Sort is not a good name for random number picker. Anyway, looks pretty even to me (if you use floor)
function rand_int(min, max) {
return Math.floor(Math.random() * (max - min + 1) + min)
}
var bucket = {}
for (var i = 0; i < 1000000; i++) {
var x = rand_int(1, 10)
bucket[x] = (bucket[x] || 0) + 1
}
console.log(bucket)

if x bigger than y then return x smaller than y javascript

I'am trying to make a simple calculation generator in prompt window. But it should not generate a negative number as answer, such as: x-y=-answer
My code so far:
CodepenLink
How it should look like
function myFunction() {
var randomNumber = Math.floor(Math.random() * 10);
var nxtRandomNumber = Math.floor(Math.random() * 10);
var question = prompt("What is:"+ randomNumber+ " minus " + nxtRandomNumber);
if(nxtRandomNumber > randomNumber){
return ;
}
var result = Number(randomNumber) - Number(nxtRandomNumber);
document.getElementById("demo").innerHTML = "Number:" + answer + " was right!";
}
<p>Push the button to start </p>
<button onclick="myFunction()">Calculate!</button>
<p id="demo"></p>

make sure when you calculate nxtRandomNumber to use the value of randomNumber as a minimum.
So if randomNumber is 5, nextRandomNumber should be calculated to compute a number at least equal to 'randomNumber'
for ex:
nextRandomNumber = Math.floor(Math.random() * 10) + randomNumber;

Theres a trick to generate random number between 2 values.
var nxtRandomNumber = Math.floor(Math.random() * (10 - randomNumber) + randomNumber);
From mdn
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/random
This example returns a random number between the specified values. The returned value is no lower than (and may possibly equal) min, and is less than (and not equal) max.>>>

Javascript: Generate a random number within a range using crypto.getRandomValues

I understand you can generate a random number in JavaScript within a range using this function:
function getRandomInt (min, max) {
return Math.floor(Math.random() * (max - min + 1)) + min;
}
Courtesy of Ionuț G. Stan here.
What I want to know is if you can generate a better random number in a range using crypto.getRandomValues() instead of Math.random(). I would like to be able to generate a number between 0 and 10 inclusive, or 0 - 1, or even 10 - 5000 inclusive.
You'll note Math.random() produces a number like: 0.8565239671015732.
The getRandomValues API might return something like:
231 with Uint8Array(1)
54328 with Uint16Array(1)
355282741 with Uint32Array(1).
So how to translate that back to a decimal number so I can keep with the same range algorithm above? Or do I need a new algorithm?
Here's the code I tried but it doesn't work too well.
function getRandomInt(min, max) {
// Create byte array and fill with 1 random number
var byteArray = new Uint8Array(1);
window.crypto.getRandomValues(byteArray);
// Convert to decimal
var randomNum = '0.' + byteArray[0].toString();
// Get number in range
randomNum = Math.floor(randomNum * (max - min + 1)) + min;
return randomNum;
}
At the low end (range 0 - 1) it returns more 0's than 1's. What's the best way to do it with getRandomValues()?
Many thanks

IMHO, the easiest way to generate a random number in a [min..max] range with window.crypto.getRandomValues() is described here.
An ECMAScript 2015-syntax code, in case the link is TL;TR:
function getRandomIntInclusive(min, max) {
const randomBuffer = new Uint32Array(1);
window.crypto.getRandomValues(randomBuffer);
let randomNumber = randomBuffer[0] / (0xffffffff + 1);
min = Math.ceil(min);
max = Math.floor(max);
return Math.floor(randomNumber * (max - min + 1)) + min;
}

The easiest way is probably by rejection sampling (see http://en.wikipedia.org/wiki/Rejection_sampling). For example, assuming that max - min is less than 256:
function getRandomInt(min, max) {
// Create byte array and fill with 1 random number
var byteArray = new Uint8Array(1);
window.crypto.getRandomValues(byteArray);
var range = max - min + 1;
var max_range = 256;
if (byteArray[0] >= Math.floor(max_range / range) * range)
return getRandomInt(min, max);
return min + (byteArray[0] % range);
}

Many of these answers are going to produce biased results. Here's an unbiased solution.
function random(min, max) {
const range = max - min + 1
const bytes_needed = Math.ceil(Math.log2(range) / 8)
const cutoff = Math.floor((256 ** bytes_needed) / range) * range
const bytes = new Uint8Array(bytes_needed)
let value
do {
crypto.getRandomValues(bytes)
value = bytes.reduce((acc, x, n) => acc + x * 256 ** n, 0)
} while (value >= cutoff)
return min + value % range
}

If you are using Node.js, it is safer to use the cryptographically secure pseudorandom crypto.randomInt. Don't go write this kind of sensitive methods if you don't know what you are doing and without peer review.
Official documentation
crypto.randomInt([min, ]max[, callback])
Added in: v14.10.0, v12.19.0
min <integer> Start of random range (inclusive). Default: 0.
max <integer> End of random range (exclusive).
callback <Function> function(err, n) {}.
Return a random integer n such that min <= n < max. This implementation avoids modulo bias.
The range (max - min) must be less than 2^48. min and max must be safe integers.
If the callback function is not provided, the random integer is generated synchronously.
// Asynchronous
crypto.randomInt(3, (err, n) => {
if (err) throw err;
console.log(`Random number chosen from (0, 1, 2): ${n}`);
});
// Synchronous
const n = crypto.randomInt(3);
console.log(`Random number chosen from (0, 1, 2): ${n}`);
// With `min` argument
const n = crypto.randomInt(1, 7);
console.log(`The dice rolled: ${n}`);

Necromancing.
Well, this is easy to solve.
Consider random number in ranges without crypto-random:
// Returns a random number between min (inclusive) and max (exclusive)
function getRandomArbitrary(min, max) {
return Math.random() * (max - min) + min;
}
/**
* Returns a random integer between min (inclusive) and max (inclusive).
* The value is no lower than min (or the next integer greater than min
* if min isn't an integer) and no greater than max (or the next integer
* lower than max if max isn't an integer).
* Using Math.round() will give you a non-uniform distribution!
*/
function getRandomInt(min, max) {
min = Math.ceil(min);
max = Math.floor(max);
return Math.floor(Math.random() * (max - min + 1)) + min;
}
So all you need to do is replace Math.random with a random from crypt.
So what does Math.random do ?
According to MDN, the Math.random() function returns a floating-point, pseudo-random number in the range 0 to less than 1 (inclusive of 0, but not 1)
So we need a crypto-random number >= 0 and < 1 (not <=).
So, we need a non-negative (aka. UNSIGNED) integer from getRandomValues.
How do we do this?
Simple:
Instead of getting an integer, and then doing Math.abs, we just get an UInt:
var randomBuffer = new Int8Array(4); // Int8Array = byte, 1 int = 4 byte = 32 bit
window.crypto.getRandomValues(randomBuffer);
var dataView = new DataView(array.buffer);
var uint = dataView.getUint32();
The shorthand version of which is
var randomBuffer = new Uint32Array(1);
(window.crypto || window.msCrypto).getRandomValues(randomBuffer);
var uint = randomBuffer[0];
Now all we need to do is divide uint by uint32.MaxValue (aka 0xFFFFFFFF) to get a floating-point number. And because we cannot have 1 in the result-set, we need to divide by (uint32.MaxValue+1) to ensure the result is < 1.
Dividing by (UInt32.MaxValue + 1) works because a JavaScript integer is a 64-bit floating-point number internally, so it is not limited at 32 bit.
function cryptoRand()
{
var array = new Int8Array(4);
(window.crypto || window.msCrypto).getRandomValues(array);
var dataView = new DataView(array.buffer);
var uint = dataView.getUint32();
var f = uint / (0xffffffff + 1); // 0xFFFFFFFF = uint32.MaxValue (+1 because Math.random is inclusive of 0, but not 1)
return f;
}
the shorthand of which is
function cryptoRand()
{
const randomBuffer = new Uint32Array(1);
(window.crypto || window.msCrypto).getRandomValues(randomBuffer);
return ( randomBuffer[0] / (0xffffffff + 1) );
}
Now all you need to do is replace Math.random() with cryptoRand() in the above functions.
Note that if crypto.getRandomValues uses the Windows-CryptoAPI on Windows to get the random bytes, you should not consider these values a truly cryptographically secure source of entropy.

Rando.js uses crypto.getRandomValues to basically do this for you
console.log(rando(5, 10));
<script src="https://randojs.com/2.0.0.js"></script>
This is carved out of the source code if you want to look behind the curtain:
var cryptoRandom = () => {
try {
var cryptoRandoms, cryptoRandomSlices = [],
cryptoRandom;
while ((cryptoRandom = "." + cryptoRandomSlices.join("")).length < 30) {
cryptoRandoms = (window.crypto || window.msCrypto).getRandomValues(new Uint32Array(5));
for (var i = 0; i < cryptoRandoms.length; i++) {
var cryptoRandomSlice = cryptoRandoms[i].toString().slice(1, -1);
if (cryptoRandomSlice.length > 0) cryptoRandomSlices[cryptoRandomSlices.length] = cryptoRandomSlice;
}
}
return Number(cryptoRandom);
} catch (e) {
return Math.random();
}
};
var min = 5;
var max = 10;
if (min > max) var temp = max, max = min, min = temp;
min = Math.floor(min), max = Math.floor(max);
console.log( Math.floor(cryptoRandom() * (max - min + 1) + min) );

Read this if you're concerned about the randomness of your number:
If you use a 6 sided dice to generate a random number 1 thru 5, what do you do when you land on 6? There's two strategies:
Re-roll until you get a 1 thru 5. This maintains the randomness, but creates extra work.
Map the 6 to one of the numbers you do want, like 5. This is less work, but now you skewed your distribution and are going to get extra 5s.
Strategy 1 is the "rejection sampling" mentioned by #arghbleargh and used in their answer and a few other answers.
Strategy 2 is what #Chris_F is referring to as producing biased results.
So understand that all solutions to the original post's question require mapping from one pseudo-random distribution of numbers to another distribution with a different number of 'buckets'.
Strategy 2 is probably fine because:
With strategy 2, as long as you are taking the modulo then no resulting number will be more than 2x as likely as any other number. So it is not significantly easier to guess than strategy 1.
And as long as your source distribution is MUCH bigger than your target distribution, the skew in randomness will be negligible unless you're running a Monte Carlo simulation or something (which you probably wouldn't be doing in JavaScript to begin with, or at least you wouldn't be using the crypto library for that).
Math.random() uses strategy 2, maps from a ~52 bit number (2^52 unique numbers), though some environments use less precision (see here).

JQuery create a random 16 digit number possible?

As the title says ... is it possible to create a random 16 digit number with jquery?

Just use:
Math.floor(Math.random()*1E16)
EDIT :
Note that there is about a 1/10 chance of a lower number of digits. If Math.random() generates something like 0.0942104924071337 then 0.0942104924071337 * 1E16 is 0942104924071337 which evaluates to 942104924071337; a 15 digit number.
The only way to 100% guarantee that the number is 16 digits in length is to have it be formed as a string. Using this method I would recommend #rjmunro's answer:
number = (Math.random()+' ').substring(2,10)+(Math.random()+' ').substring(2,10);

Not with jQuery, no, but you can do it with plain javascript.
If you want exactly 16 digits (possibly including leading 0s), I would start with Math.random(), convert to a string, pick 8 digits, and concatenate 2 runs together.
number = (Math.random() + '').substring(2,10)
+ (Math.random() + '').substring(2,10);

No, use JAVASCRIPT!
jQuery is not some magic genie.

This is a task which is much better suited for raw javascript. For example
var str = '';
var i;
for (i = 0; i < 16; i++) {
var number = Math.floor(Math.random() * 10) % 10;
str += number;
}

I just tried with #rjmunro 's answer.
Unfortunately, it does generate string less than 16digits,
but very rare, approxly once in 10 million times.
Here is my testing code, runs in nodejs:
'use strict';
var fs = require('fs');
var totalTimes = 100000000;
var times = totalTimes;
var fileName;
var writeStream;
while (times > 0) {
var key = (Math.random() + ' ').substring(2,10) + (Math.random() + ' ').substring(2,10);
times --;
if (key.length !== 16) {
var msg = 'a flaw key gened: ' + key + '\n';
// create a log file at first time
if (!fileName) {
fileName = 'log/flaw_key_' + new Date() + '.txt';
}
writeStream = fs.createWriteStream(fileName);
writeStream.write(msg);
writeStream.end();
}
if (times === 0) {
console.log(totalTimes + ' times key gened');
}
}
Also #Dimitri Mikadze 's answer generate less length string as well, so I eventually adopt a way with some concept of his solution:
function getRandomInt(min, max) {
return Math.floor(Math.random() * (max - min + 1)) + min;
}
/**
* Gen random digits string in specific length
* #param {Int} length of string
*
* #return {String}
*
*/
function genString(length) {
var times = length;
var key = '';
while (times > 0) {
times --;
key += getRandomInt(0, 9);
}
return key;
}
genString(16); // a 16 digits string

u can use this function to generate random digits, just pass minimum and maximum parameters
function randomInt(min, max) {
return Math.floor(Math.random() * (max - min + 1) + min);
}
random 16 digit, usage
randomInt(0, 9999999999999999);

I know this question is old but this simple function will guarantee a 16 (or however many you want) character string every time without the 10% failure rate of other solutions. Can change it to a number if you need to.
function generate() {
let string = ""
while (string.length < 16) {
let number = Math.floor(Math.random() * 10).toString()
string += number
}
return string
}

I think this way is more beautiful:
const generateFixedLengthNumberInString = length =>
[...Array(length).keys()].reduce(
previousValue =>
previousValue + String(Math.floor(Math.random() * 10) % 10),
);
console.log(generateFixedLengthNumberInString(16))
// prints "0587139224228340"

JavaScript, Generate a Random Number that is 9 numbers in length

I'm looking for an efficient, elegant way to generate a JavaScript variable that is 9 digits in length:
Example: 323760488

You could generate 9 random digits and concatenate them all together.
Or, you could call random() and multiply the result by 1000000000:
Math.floor(Math.random() * 1000000000);
Since Math.random() generates a random double precision number between 0 and 1, you will have enough digits of precision to still have randomness in your least significant place.
If you want to ensure that your number starts with a nonzero digit, try:
Math.floor(100000000 + Math.random() * 900000000);
Or pad with zeros:
function LeftPadWithZeros(number, length)
{
var str = '' + number;
while (str.length < length) {
str = '0' + str;
}
return str;
}
Or pad using this inline 'trick'.

why don't just extract digits from the Math.random() string representation?
Math.random().toString().slice(2,11);
/*
Math.random() -> 0.12345678901234
.toString() -> "0.12345678901234"
.slice(2,11) -> "123456789"
*/
(requirement is that every javascript implementation Math.random()'s precision is at least 9 decimal places)

Also...
function getRandom(length) {
return Math.floor(Math.pow(10, length-1) + Math.random() * 9 * Math.pow(10, length-1));
}
getRandom(9) => 234664534

Three methods I've found in order of efficiency:
(Test machine running Firefox 7.0 Win XP)
parseInt(Math.random()*1000000000, 10)
1 million iterations: ~626ms. By far the fastest - parseInt is a native function vs calling the Math library again. NOTE: See below.
Math.floor(Math.random()*1000000000)
1 million iterations: ~1005ms. Two function calls.
String(Math.random()).substring(2,11)
1 million iterations: ~2997ms. Three function calls.
And also...
parseInt(Math.random()*1000000000)
1 million iterations: ~362ms.
NOTE: parseInt is usually noted as unsafe to use without radix parameter. See https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/parseInt or google "JavaScript: The Good Parts". However, it seems the parameter passed to parseInt will never begin with '0' or '0x' since the input is first multiplied by 1000000000. YMMV.

Math.random().toFixed(length).split('.')[1]
Using toFixed alows you to set the length longer than the default (seems to generate 15-16 digits after the decimal. ToFixed will let you get more digits if you need them.

In one line(ish):
var len = 10;
parseInt((Math.random() * 9 + 1) * Math.pow(10,len-1), 10);
Steps:
We generate a random number that fulfil 1 ≤ x < 10.
Then, we multiply by Math.pow(10,len-1) (number with a length len).
Finally, parseInt() to remove decimals.

Thought I would take a stab at your question. When I ran the following code it worked for me.
<script type="text/javascript">
function getRandomInt(min, max) {
return Math.floor(Math.random() * (max - min)) + min;
} //The maximum is exclusive and the minimum is inclusive
$(document).ready(function() {
$("#random-button").on("click", function() {
var randomNumber = getRandomInt(100000000, 999999999);
$("#random-number").html(randomNumber);
});
</script>

Does this already have enough answers?
I guess not. So, this should reliably provide a number with 9 digits, even if Math.random() decides to return something like 0.000235436:
Math.floor((Math.random() + Math.floor(Math.random()*9)+1) * Math.pow(10, 8))

Screen scrape this page:
9 random numbers

function rand(len){var x='';
for(var i=0;i<len;i++){x+=Math.floor(Math.random() * 10);}
return x;
}
rand(9);

If you mean to generate random telephone number, then they usually are forbidden to start with zero.
That is why you should combine few methods:
Math.floor(Math.random()*8+1)+Math.random().toString().slice(2,10);
this will generate random in between 100 000 000 to 999 999 999
With other methods I had a little trouble to get reliable results as leading zeroes was somehow a problem.

I know the answer is old, but I want to share this way to generate integers or float numbers from 0 to n. Note that the position of the point (float case) is random between the boundaries. The number is an string because the limitation of the MAX_SAFE_INTEGER that is now 9007199254740991
Math.hRandom = function(positions, float = false) {
var number = "";
var point = -1;
if (float) point = Math.floor(Math.random() * positions) + 1;
for (let i = 0; i < positions; i++) {
if (i == point) number += ".";
number += Math.floor(Math.random() * 10);
}
return number;
}
//integer random number 9 numbers
console.log(Math.hRandom(9));
//float random number from 0 to 9e1000 with 1000 numbers.
console.log(Math.hRandom(1000, true));

function randomCod(){
let code = "";
let chars = 'abcdefghijlmnopqrstuvxwz';
let numbers = '0123456789';
let specialCaracter = '/{}$%&#*/()!-=?<>';
for(let i = 4; i > 1; i--){
let random = Math.floor(Math.random() * 99999).toString();
code += specialCaracter[random.substring(i, i-1)] + ((parseInt(random.substring(i, i-1)) % 2 == 0) ? (chars[random.substring(i, i-1)].toUpperCase()) : (chars[random.substring(i, i+1)])) + (numbers[random.substring(i, i-1)]);
}
code = (code.indexOf("undefined") > -1 || code.indexOf("NaN") > -1) ? randomCod() : code;
return code;
}

With max exclusive: Math.floor(Math.random() * max);
With max inclusive: Math.round(Math.random() * max);

To generate a number string with length n, thanks to #nvitaterna, I came up with this:
1 + Math.floor(Math.random() * 9) + Math.random().toFixed(n - 1).split('.')[1]
It prevents first digit to be zero.
It can generate string with length ~ 50 each time you call it.

var number = Math.floor(Math.random() * 900000000) + 100000000

var number = Math.floor(Math.random()*899999999 + 100000000)

For a number of 10 characters
Math.floor(Math.random() * 9000000000) + 1000000000
From https://gist.github.com/lpf23/9762508
This answer is intended for people who are looking to generate a 10 digit number (without a country code)