Random number with fixed length - javascript

I want to generate a random integer number with 0-9 numbers and with length = 5. I try this:
function genRand(min,max) {
for (var i = 1; i <= 5; i++) {
var range = max - min + 1;
return Math.floor(Math.random()*range) + min;
}
}
and call:
genRand(0,9);
But it always returns 1 number, not 5 (
Help please!

function genRand() {
return Math.floor(Math.random()*89999+10000);
}

return exits the function on the first loop.

The smallest 5 digit number is 10000, the largest is 99999, or 10000+89999.
Return a random number between 0 and 89999, and add it to the minimum.
var ran5=10000+Math.round(Math.floor()*90000)
Math.floor rounds down, and Math.random is greater than or equal to 0 and less than 1.

Here's a more generalized version of Nate B's answer:
function rand(digits) {
return Math.floor(Math.random()*parseInt('8' + '9'.repeat(digits-1))+parseInt('1' + '0'.repeat(digits-1)));
}

To get a 5 digit random number generate random numbers between the range (10000, 99999). Or generate randomly 5 single digits and paste them.
EDIT
The process you have shown simply will generate one number and return to the caller. The think which might work is (pseudo code) :
int sum = 0;
int m = 1;
for (i=0;i<5;i++)
{
sum = sum + m * random (0, 9);
/* or */
sum = sum * m + random (0, 9);
m = m * 10;
}
Or better generate 5 digit random numbers with rand (10000, 99999)

Related

Why does it take so many attempts to generate a random value between 1 and 10000?

I have the following code that generates an initial random number between 1 and 10000, then repeatedly generates a second random number until it matches the first:
let upper = 10000;
let randomNumber = getRandomNumber(upper);
let guess;
let attempt = 0;
function getRandomNumber(upper) {
return Math.floor(Math.random() * upper) + 1;
}
while (guess !== randomNumber) {
guess = getRandomNumber(upper);
attempt += 1;
}
document.write('The randomNumber is ' + randomNumber);
document.write(' it took' + attempt);
I am confused at (attempt) variables. Why is it that the computer took this many attempts to get the randomNumber? Also, it didn't put attempt in the loop condition.
Just to give you a start. This is what your code does:
// define the maximum of randomly generated number. range = 0 - 10.000
let upper = 10000;
// generate a random number out of the range 0-10.000
let randomNumber = getRandomNumber(upper);
// predefine variable guess
let guess;
// set a counter to 0
let attempt = 0;
// generate and return a random number out of the range from 0 to `upper`
function getRandomNumber(upper) {
return Math.floor(Math.random() * upper) + 1;
}
// loop until guess equals randomNumber
while (guess !== randomNumber) {
// generate a new random number and assign it to the variable guess
guess = getRandomNumber(upper);
// increase the counter by 1
attempt += 1;
}
// output the initially generated number
document.write('The randomNumber is ' + randomNumber);
// output the number of repetitions
document.write(' it took' + attempt);
So, once again. You generate a random number at start. And then you repeat generating another random number until this second random number matches the first. As you don't set any limits e.g. "each random number can only appear once" or "no more than 10.000 tries" your program might need millions of tries until the number matches, because you have a range of 10.000 possible numbers and they might repeat a hundreds of times each before the match is finally there.
Try to optimize your program by limiting the number of tries to 10.000. And you could your computer just let count upwards from 0 to 10.000 instead of guessing with a randomly generated number.
When I repeatedly run your snippet, I am seeing your code take anywhere from a few thousand to a few tens of thousands of repetitions to get a given value for a random number sampled between 1 and 10000. But this is not surprising -- it is expected.
Assuming your getRandomNumber(upper) function does indeed return a number between 1 and upper with a uniform distribution, the expected probability that the number returned will not be the initial, given value randomNumber is:
1 - (1/upper)
And the chance that the first N generated numbers will not include the given value is:
(1 - (1/upper)) ^ N
And so the chance P that the first N generated numbers will include given value is:
P = 1 - (1 - (1/upper)) ^ N
Thus the following formula gives the number of repetitions you will need to make to generate your initial value with a given probability P:
N = ln(1.0 - P) / ln(1.0 - (1.0/upper))
Using this formula, there is only a 50% chance of getting randomValue after 6932 repetitions, and a 95% chance after 29956 repetitions.
let upper = 10000;
function numberOfRepetitionsToGetValueWithRequiredProbability(upper, P) {
return Math.ceil(Math.log(1.0 - P) / Math.log(1.0 - (1.0/upper)))
}
function printNumberOfRepetitionsToGetValueWithRequiredProbability(upper, P) {
document.write('The number of tries to get a given value between 1 and ' + upper + ' with a ' + P + ' probability: ' + numberOfRepetitionsToGetValueWithRequiredProbability(upper, P) + ".<br>");
}
var probabilities = [0.10, 0.20, 0.30, 0.40, 0.50, 0.60, 0.70, 0.80, 0.90, 0.95, 0.99, 0.9999];
probabilities.forEach((p) => printNumberOfRepetitionsToGetValueWithRequiredProbability(upper, p));
This is entirely consistent with the observed behavior of your code. And of course, assuming Math.random() is truly random (which it isn't, it's only pseudorandom, according to the docs) there is always going to be a vanishingly small probability of never encountering your initial value no matter how many repetitions you make.

Prettify numbers to their closest number with zeros at the end in javascript

I have some chunks as following example:
// lowest and highest values of chunk arrays
[
[0, 945710.3843175517],
[945710.3843175517, 2268727.9557668166],
[2268727.9557668166, 14965451.25314727],
[14965451.25314727, 17890252.39415521],
[17890252.39415521, 3501296406.880383]
]
what I want to get from these chunks is something like this:
< 1.000.000
1.000.000 - 3.000.000
3.000.000 - 15.000.000
15.000.000 - 18.000.000
> 10.000.000
I will use these new numbers as a legend of an informative map.
I use a function to achieve this goal named as roundClosestLegendNumber for each value.
All numbers are positive numbers and there is no maximum limitation.
roundClosestLegendNumber(5) \\ should give 10
roundClosestLegendNumber(94) \\ should give 100
roundClosestLegendNumber(125) \\ should give 200
roundClosestLegendNumber(945710.3843175517) \\ should give 1000000
roundClosestLegendNumber(14965451.25314727) \\ should give 15000000
roundClosestLegendNumber(17890252.39415521) \\ should give 18000000
// and so on
I changed the comparison for multiplier from 10 * to 100 * to get a precision of 2 digits before the series of 0s.
var roundClosestLegendNumber = function(number) {
if(number < 10) {
return number;
}
var multiplier = 10;
while(number >= 100 * multiplier) {
multiplier = 10 * multiplier;
}
count = 1;
while(number > multiplier * count) {
count++;
}
return multiplier * count;
}
You could use significant figures, set sf (I've set mine to 3) then only the first 3 numbers are displayed.
function sigFigure(v) {
let sf = 3;
if (v >= Math.pow(10,sf)) {
let number = v.toPrecision(sf);
let numbers = number.split("e+")
return parseInt((numbers[0]*Math.pow(10,numbers[1])).toFixed(0));
} else {
return v
}
}
var array = [0, 945710.38431755, 2268727.9557668166, 14965451.25314727, 17890252.39415521, 3501296406.880383];
console.log(array.map(sigFigure));

factorial with trailing zeros, but without calculating factorial

I'm calculating the trailing zeros of a factorial. My solution is to calculate the factorial then determine how many trailing zeros it has. As you can imagine this isn't very scalable. How can I solve this without calculating the factorial?
I've found these pages on SO:
Trailing zeroes in a Factorial
Calculating the factorial without trailing zeros efficiently?
However, neither are in Javascript. If you downvote this question please let me know why. Thank-you for your time and feedback.
My solution:
function zeros(n) {
var result = [];
var count = 0;
for (var i = 1; i <= n; i++) {
result.push(i);
} //generating range for factorial function
var factorial = result.reduce(function(acc, el) {
return acc * el;
}, 1); //calculating factorial
factorial = factorial.toString().split('');
for (var j = factorial.length - 1; j > 0; j--) {
if (parseInt(factorial[j]) === 0) {
count += 1;
} else {
break;
}
} //counting trailing zeros
return count;
}
Knowing the number of trailing zeroes in a number comes down to knowing how many times it can be divided by 10, i.e. by both 5 and 2.
With factorial numbers that is quite easy to count:
f! = 1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16. ... .f
^ ^ ^
The places where a factor 5 gets into the final product are marked. It is clear that factors of 2 occur more often, so the count of factors of 5 are determining the number of trailing zeroes.
Now, when the factor 25 occurs, it should be counted for 2; likewise 125 should count for 3 factors of 5, etc.
You can cover for that with a loop like this:
function zeros(n) {
var result = 0;
while (n = Math.floor(n / 5)) result += n;
return result;
}
public static void main(String[] args) {
int n=23;
String fact= factorial(BigInteger.valueOf(23)).toString();
System.out.format("Factorial value of %d is %s\n", n,fact);
int len=fact.length();
//Check end with zeros
if(fact.matches(".*0*$")){
String[] su=fact.split("0*$");
//Split the pattern from whole string
System.out.println(Arrays.toString(fact.split("0*$")));
//Subtract from the total length
System.out.println("Count of trailing zeros "+(len-su[0].length()));
}
}
public static BigInteger factorial(BigInteger n) {
if (n.equals(BigInteger.ONE) || n.equals(BigInteger.ZERO)) {
return BigInteger.ONE;
}
return n.multiply(factorial(n.subtract(BigInteger.ONE)));
}
You don't really need to calculate the factorial product to count the trailing zeroes.
Here a sample to count the number of trailing zeroes in n!
temp = 5;
zeroes = 0;
//counting the sum of multiples of 5,5^2,5^3....present in n!
while(n>=temp){
fives = n/temp;
zeroes = zeroes + fives;
temp = temp*5;
}
printf("%d",zeroes);
Note that each multiple of 5 in the factorial product will contribute 1 to the number of trailing zeros. On top of this, each multiple of 25 will contribute an additional 1 to the number of trailing zeros. Then, each multiple of 125 will contribute another 1 to the number of trailing zeros, and so on.
Here's a great link to understand the concept behind this: https://brilliant.org/wiki/trailing-number-of-zeros/
I came across this algorithm somewhere on here can not remember now, but it looks like this,
def zeros(n)
return 0 if n.zero?
k = (Math.log(n)/Math.log(5)).to_i
m = 5**k
n*(m-1)/(4*m)
end
This very effiecient as it does not need a loop.
You can further optimize it to look like this.
def zeros(n)
return 0 if n.zero?
n*(n-1)/(4*n)
end
A javascript translation of this will be.
function zeros(n) {
if (n == 0) return 0;
return n * (n-1)/(4*n);
}
Note that this algorithm is correct till about n >= 1000000000, in which case the return value has an error margin of +1, and this error margin increases by +1 every n * 10000.

Understanding formula for generating random number in interval [duplicate]

How can I generate random whole numbers between two specified variables in JavaScript, e.g. x = 4 and y = 8 would output any of 4, 5, 6, 7, 8?
There are some examples on the Mozilla Developer Network page:
/**
* 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;
}
Here's the logic behind it. It's a simple rule of three:
Math.random() returns a Number between 0 (inclusive) and 1 (exclusive). So we have an interval like this:
[0 .................................... 1)
Now, we'd like a number between min (inclusive) and max (exclusive):
[0 .................................... 1)
[min .................................. max)
We can use the Math.random to get the correspondent in the [min, max) interval. But, first we should factor a little bit the problem by subtracting min from the second interval:
[0 .................................... 1)
[min - min ............................ max - min)
This gives:
[0 .................................... 1)
[0 .................................... max - min)
We may now apply Math.random and then calculate the correspondent. Let's choose a random number:
Math.random()
|
[0 .................................... 1)
[0 .................................... max - min)
|
x (what we need)
So, in order to find x, we would do:
x = Math.random() * (max - min);
Don't forget to add min back, so that we get a number in the [min, max) interval:
x = Math.random() * (max - min) + min;
That was the first function from MDN. The second one, returns an integer between min and max, both inclusive.
Now for getting integers, you could use round, ceil or floor.
You could use Math.round(Math.random() * (max - min)) + min, this however gives a non-even distribution. Both, min and max only have approximately half the chance to roll:
min...min+0.5...min+1...min+1.5 ... max-0.5....max
└───┬───┘└────────┬───────┘└───── ... ─────┘└───┬──┘ ← Math.round()
min min+1 max
With max excluded from the interval, it has an even less chance to roll than min.
With Math.floor(Math.random() * (max - min +1)) + min you have a perfectly even distribution.
min... min+1... ... max-1... max.... (max+1 is excluded from interval)
└───┬───┘└───┬───┘└─── ... ┘└───┬───┘└───┬───┘ ← Math.floor()
min min+1 max-1 max
You can't use ceil() and -1 in that equation because max now had a slightly less chance to roll, but you can roll the (unwanted) min-1 result too.
var randomnumber = Math.floor(Math.random() * (maximum - minimum + 1)) + minimum;
Math.random()
Returns an integer random number between min (included) and max (included):
function randomInteger(min, max) {
return Math.floor(Math.random() * (max - min + 1)) + min;
}
Or any random number between min (included) and max (not included):
function randomNumber(min, max) {
return Math.random() * (max - min) + min;
}
Useful examples (integers):
// 0 -> 10
Math.floor(Math.random() * 11);
// 1 -> 10
Math.floor(Math.random() * 10) + 1;
// 5 -> 20
Math.floor(Math.random() * 16) + 5;
// -10 -> (-2)
Math.floor(Math.random() * 9) - 10;
** And always nice to be reminded (Mozilla):
Math.random() does not provide cryptographically secure random
numbers. Do not use them for anything related to security. Use the Web
Crypto API instead, and more precisely the
window.crypto.getRandomValues() method.
Use:
function getRandomizer(bottom, top) {
return function() {
return Math.floor( Math.random() * ( 1 + top - bottom ) ) + bottom;
}
}
Usage:
var rollDie = getRandomizer( 1, 6 );
var results = ""
for ( var i = 0; i<1000; i++ ) {
results += rollDie() + " "; // Make a string filled with 1000 random numbers in the range 1-6.
}
Breakdown:
We are returning a function (borrowing from functional programming) that when called, will return a random integer between the the values bottom and top, inclusive. We say 'inclusive' because we want to include both bottom and top in the range of numbers that can be returned. This way, getRandomizer( 1, 6 ) will return either 1, 2, 3, 4, 5, or 6.
('bottom' is the lower number, and 'top' is the greater number)
Math.random() * ( 1 + top - bottom )
Math.random() returns a random double between 0 and 1, and if we multiply it by one plus the difference between top and bottom, we'll get a double somewhere between 0 and 1+b-a.
Math.floor( Math.random() * ( 1 + top - bottom ) )
Math.floor rounds the number down to the nearest integer. So we now have all the integers between 0 and top-bottom. The 1 looks confusing, but it needs to be there because we are always rounding down, so the top number will never actually be reached without it. The random decimal we generate needs to be in the range 0 to (1+top-bottom) so we can round down and get an integer in the range 0 to top-bottom:
Math.floor( Math.random() * ( 1 + top - bottom ) ) + bottom
The code in the previous example gave us an integer in the range 0 and top-bottom, so all we need to do now is add bottom to that result to get an integer in the range bottom and top inclusive. :D
NOTE: If you pass in a non-integer value or the greater number first you'll get undesirable behavior, but unless anyone requests it I am not going to delve into the argument checking code as it’s rather far from the intent of the original question.
All these solutions are using way too much firepower. You only need to call one function: Math.random();
Math.random() * max | 0;
This returns a random integer between 0 (inclusive) and max (non-inclusive).
Return a random number between 1 and 10:
Math.floor((Math.random()*10) + 1);
Return a random number between 1 and 100:
Math.floor((Math.random()*100) + 1)
function randomRange(min, max) {
return ~~(Math.random() * (max - min + 1)) + min
}
Alternative if you are using Underscore.js you can use
_.random(min, max)
If you need a variable between 0 and max, you can use:
Math.floor(Math.random() * max);
The other answers don't account for the perfectly reasonable parameters of 0 and 1. Instead you should use the round instead of ceil or floor:
function randomNumber(minimum, maximum){
return Math.round( Math.random() * (maximum - minimum) + minimum);
}
console.log(randomNumber(0,1)); # 0 1 1 0 1 0
console.log(randomNumber(5,6)); # 5 6 6 5 5 6
console.log(randomNumber(3,-1)); # 1 3 1 -1 -1 -1
Cryptographically strong
To get a cryptographically strong random integer number in the range [x,y], try:
let cs = (x,y) => x + (y - x + 1)*crypto.getRandomValues(new Uint32Array(1))[0]/2**32 | 0
console.log(cs(4, 8))
Here's what I use to generate random numbers.
function random(min,max) {
return Math.floor((Math.random())*(max-min+1))+min;
}
Math.random() returns a number between 0 (inclusive) and 1 (exclusive). We multiply this number by the range (max-min). This results in a number between 0 (inclusive), and the range.
For example, take random(2,5). We multiply the random number 0≤x<1 by the range (5-2=3), so we now have a number, x where 0≤x<3.
In order to force the function to treat both the max and min as inclusive, we add 1 to our range calculation: Math.random()*(max-min+1). Now, we multiply the random number by the (5-2+1=4), resulting in an number, x, such that 0≤x<4. If we floor this calculation, we get an integer: 0≤x≤3, with an equal likelihood of each result (1/4).
Finally, we need to convert this into an integer between the requested values. Since we already have an integer between 0 and the (max-min), we can simply map the value into the correct range by adding the minimum value. In our example, we add 2 our integer between 0 and 3, resulting in an integer between 2 and 5.
Use this function to get random numbers in a given range:
function rnd(min, max) {
return Math.floor(Math.random()*(max - min + 1) + min);
}
Here is the Microsoft .NET Implementation of the Random class in JavaScript—
var Random = (function () {
function Random(Seed) {
if (!Seed) {
Seed = this.milliseconds();
}
this.SeedArray = [];
for (var i = 0; i < 56; i++)
this.SeedArray.push(0);
var num = (Seed == -2147483648) ? 2147483647 : Math.abs(Seed);
var num2 = 161803398 - num;
this.SeedArray[55] = num2;
var num3 = 1;
for (var i_1 = 1; i_1 < 55; i_1++) {
var num4 = 21 * i_1 % 55;
this.SeedArray[num4] = num3;
num3 = num2 - num3;
if (num3 < 0) {
num3 += 2147483647;
}
num2 = this.SeedArray[num4];
}
for (var j = 1; j < 5; j++) {
for (var k = 1; k < 56; k++) {
this.SeedArray[k] -= this.SeedArray[1 + (k + 30) % 55];
if (this.SeedArray[k] < 0) {
this.SeedArray[k] += 2147483647;
}
}
}
this.inext = 0;
this.inextp = 21;
Seed = 1;
}
Random.prototype.milliseconds = function () {
var str = new Date().valueOf().toString();
return parseInt(str.substr(str.length - 6));
};
Random.prototype.InternalSample = function () {
var num = this.inext;
var num2 = this.inextp;
if (++num >= 56) {
num = 1;
}
if (++num2 >= 56) {
num2 = 1;
}
var num3 = this.SeedArray[num] - this.SeedArray[num2];
if (num3 == 2147483647) {
num3--;
}
if (num3 < 0) {
num3 += 2147483647;
}
this.SeedArray[num] = num3;
this.inext = num;
this.inextp = num2;
return num3;
};
Random.prototype.Sample = function () {
return this.InternalSample() * 4.6566128752457969E-10;
};
Random.prototype.GetSampleForLargeRange = function () {
var num = this.InternalSample();
var flag = this.InternalSample() % 2 == 0;
if (flag) {
num = -num;
}
var num2 = num;
num2 += 2147483646.0;
return num2 / 4294967293.0;
};
Random.prototype.Next = function (minValue, maxValue) {
if (!minValue && !maxValue)
return this.InternalSample();
var num = maxValue - minValue;
if (num <= 2147483647) {
return parseInt((this.Sample() * num + minValue).toFixed(0));
}
return this.GetSampleForLargeRange() * num + minValue;
};
Random.prototype.NextDouble = function () {
return this.Sample();
};
Random.prototype.NextBytes = function (buffer) {
for (var i = 0; i < buffer.length; i++) {
buffer[i] = this.InternalSample() % 256;
}
};
return Random;
}());
Use:
var r = new Random();
var nextInt = r.Next(1, 100); // Returns an integer between range
var nextDbl = r.NextDouble(); // Returns a random decimal
I wanted to explain using an example:
Function to generate random whole numbers in JavaScript within a range of 5 to 25
General Overview:
(i) First convert it to the range - starting from 0.
(ii) Then convert it to your desired range ( which then will be very
easy to complete).
So basically, if you want to generate random whole numbers from 5 to 25 then:
First step: Converting it to range - starting from 0
Subtract "lower/minimum number" from both "max" and "min". i.e
(5-5) - (25-5)
So the range will be:
0-20 ...right?
Step two
Now if you want both numbers inclusive in range - i.e "both 0 and 20", the equation will be:
Mathematical equation: Math.floor((Math.random() * 21))
General equation: Math.floor((Math.random() * (max-min +1)))
Now if we add subtracted/minimum number (i.e., 5) to the range - then automatically we can get range from 0 to 20 => 5 to 25
Step three
Now add the difference you subtracted in equation (i.e., 5) and add "Math.floor" to the whole equation:
Mathematical equation: Math.floor((Math.random() * 21) + 5)
General equation: Math.floor((Math.random() * (max-min +1)) + min)
So finally the function will be:
function randomRange(min, max) {
return Math.floor((Math.random() * (max - min + 1)) + min);
}
After generating a random number using a computer program, it is still considered as a random number if the picked number is a part or the full one of the initial one. But if it was changed, then mathematicians do not accept it as a random number and they can call it a biased number.
But if you are developing a program for a simple task, this will not be a case to consider. But if you are developing a program to generate a random number for a valuable stuff such as lottery program, or gambling game, then your program will be rejected by the management if you are not consider about the above case.
So for those kind of people, here is my suggestion:
Generate a random number using Math.random() (say this n):
Now for [0,10) ==> n*10 (i.e. one digit) and for[10,100) ==> n*100 (i.e., two digits) and so on. Here square bracket indicates that the boundary is inclusive and a round bracket indicates the boundary is exclusive.
Then remove the rest after the decimal point. (i.e., get the floor) - using Math.floor(). This can be done.
If you know how to read the random number table to pick a random number, you know the above process (multiplying by 1, 10, 100 and so on) does not violate the one that I was mentioned at the beginning (because it changes only the place of the decimal point).
Study the following example and develop it to your needs.
If you need a sample [0,9] then the floor of n10 is your answer and if you need [0,99] then the floor of n100 is your answer and so on.
Now let’s enter into your role:
You've asked for numbers in a specific range. (In this case you are biased among that range. By taking a number from [1,6] by roll a die, then you are biased into [1,6], but still it is a random number if and only if the die is unbiased.)
So consider your range ==> [78, 247]
number of elements of the range = 247 - 78 + 1 = 170; (since both the boundaries are inclusive).
/* Method 1: */
var i = 78, j = 247, k = 170, a = [], b = [], c, d, e, f, l = 0;
for(; i <= j; i++){ a.push(i); }
while(l < 170){
c = Math.random()*100; c = Math.floor(c);
d = Math.random()*100; d = Math.floor(d);
b.push(a[c]); e = c + d;
if((b.length != k) && (e < k)){ b.push(a[e]); }
l = b.length;
}
console.log('Method 1:');
console.log(b);
/* Method 2: */
var a, b, c, d = [], l = 0;
while(l < 170){
a = Math.random()*100; a = Math.floor(a);
b = Math.random()*100; b = Math.floor(b);
c = a + b;
if(c <= 247 || c >= 78){ d.push(c); }else{ d.push(a); }
l = d.length;
}
console.log('Method 2:');
console.log(d);
Note: In method one, first I created an array which contains numbers that you need and then randomly put them into another array.
In method two, generate numbers randomly and check those are in the range that you need. Then put it into an array. Here I generated two random numbers and used the total of them to maximize the speed of the program by minimizing the failure rate that obtaining a useful number. However, adding generated numbers will also give some biasedness. So I would recommend my first method to generate random numbers within a specific range.
In both methods, your console will show the result (press F12 in Chrome to open the console).
function getRandomInt(lower, upper)
{
//to create an even sample distribution
return Math.floor(lower + (Math.random() * (upper - lower + 1)));
//to produce an uneven sample distribution
//return Math.round(lower + (Math.random() * (upper - lower)));
//to exclude the max value from the possible values
//return Math.floor(lower + (Math.random() * (upper - lower)));
}
To test this function, and variations of this function, save the below HTML/JavaScript to a file and open with a browser. The code will produce a graph showing the distribution of one million function calls. The code will also record the edge cases, so if the the function produces a value greater than the max, or less than the min, you.will.know.about.it.
<html>
<head>
<script type="text/javascript">
function getRandomInt(lower, upper)
{
//to create an even sample distribution
return Math.floor(lower + (Math.random() * (upper - lower + 1)));
//to produce an uneven sample distribution
//return Math.round(lower + (Math.random() * (upper - lower)));
//to exclude the max value from the possible values
//return Math.floor(lower + (Math.random() * (upper - lower)));
}
var min = -5;
var max = 5;
var array = new Array();
for(var i = 0; i <= (max - min) + 2; i++) {
array.push(0);
}
for(var i = 0; i < 1000000; i++) {
var random = getRandomInt(min, max);
array[random - min + 1]++;
}
var maxSample = 0;
for(var i = 0; i < max - min; i++) {
maxSample = Math.max(maxSample, array[i]);
}
//create a bar graph to show the sample distribution
var maxHeight = 500;
for(var i = 0; i <= (max - min) + 2; i++) {
var sampleHeight = (array[i]/maxSample) * maxHeight;
document.write('<span style="display:inline-block;color:'+(sampleHeight == 0 ? 'black' : 'white')+';background-color:black;height:'+sampleHeight+'px"> [' + (i + min - 1) + ']: '+array[i]+'</span> ');
}
document.write('<hr/>');
</script>
</head>
<body>
</body>
</html>
For a random integer with a range, try:
function random(minimum, maximum) {
var bool = true;
while (bool) {
var number = (Math.floor(Math.random() * maximum + 1) + minimum);
if (number > 20) {
bool = true;
} else {
bool = false;
}
}
return number;
}
Here is a function that generates a random number between min and max, both inclusive.
const randomInt = (max, min) => Math.round(Math.random() * (max - min)) + min;
To get a random number say between 1 and 6, first do:
0.5 + (Math.random() * ((6 - 1) + 1))
This multiplies a random number by 6 and then adds 0.5 to it. Next round the number to a positive integer by doing:
Math.round(0.5 + (Math.random() * ((6 - 1) + 1))
This round the number to the nearest whole number.
Or to make it more understandable do this:
var value = 0.5 + (Math.random() * ((6 - 1) + 1))
var roll = Math.round(value);
return roll;
In general, the code for doing this using variables is:
var value = (Min - 0.5) + (Math.random() * ((Max - Min) + 1))
var roll = Math.round(value);
return roll;
The reason for taking away 0.5 from the minimum value is because using the minimum value alone would allow you to get an integer that was one more than your maximum value. By taking away 0.5 from the minimum value you are essentially preventing the maximum value from being rounded up.
Using the following code, you can generate an array of random numbers, without repeating, in a given range.
function genRandomNumber(how_many_numbers, min, max) {
// Parameters
//
// how_many_numbers: How many numbers you want to
// generate. For example, it is 5.
//
// min (inclusive): Minimum/low value of a range. It
// must be any positive integer, but
// less than max. I.e., 4.
//
// max (inclusive): Maximum value of a range. it must
// be any positive integer. I.e., 50
//
// Return type: array
var random_number = [];
for (var i = 0; i < how_many_numbers; i++) {
var gen_num = parseInt((Math.random() * (max-min+1)) + min);
do {
var is_exist = random_number.indexOf(gen_num);
if (is_exist >= 0) {
gen_num = parseInt((Math.random() * (max-min+1)) + min);
}
else {
random_number.push(gen_num);
is_exist = -2;
}
}
while (is_exist > -1);
}
document.getElementById('box').innerHTML = random_number;
}
Random whole number between lowest and highest:
function randomRange(low, high) {
var range = (high-low);
var random = Math.floor(Math.random()*range);
if (random === 0) {
random += 1;
}
return low + random;
}
It is not the most elegant solution, but something quick.
I found this simple method on W3Schools:
Math.floor((Math.random() * max) + min);
Math.random() is fast and suitable for many purposes, but it's not appropriate if you need cryptographically-secure values (it's not secure), or if you need integers from a completely uniform unbiased distribution (the multiplication approach used in others answers produces certain values slightly more often than others).
In such cases, we can use crypto.getRandomValues() to generate secure integers, and reject any generated values that we can't map uniformly into the target range. This will be slower, but it shouldn't be significant unless you're generating extremely large numbers of values.
To clarify the biased distribution concern, consider the case where we want to generate a value between 1 and 5, but we have a random number generator that produces values between 1 and 16 (a 4-bit value). We want to have the same number of generated values mapping to each output value, but 16 does not evenly divide by 5: it leaves a remainder of 1. So we need to reject 1 of the possible generated values, and only continue when we get one of the 15 lesser values that can be uniformly mapped into our target range. Our behaviour could look like this pseudocode:
Generate a 4-bit integer in the range 1-16.
If we generated 1, 6, or 11 then output 1.
If we generated 2, 7, or 12 then output 2.
If we generated 3, 8, or 13 then output 3.
If we generated 4, 9, or 14 then output 4.
If we generated 5, 10, or 15 then output 5.
If we generated 16 then reject it and try again.
The following code uses similar logic, but generates a 32-bit integer instead, because that's the largest common integer size that can be represented by JavaScript's standard number type. (This could be modified to use BigInts if you need a larger range.) Regardless of the chosen range, the fraction of generated values that are rejected will always be less than 0.5, so the expected number of rejections will always be less than 1.0 and usually close to 0.0; you don't need to worry about it looping forever.
const randomInteger = (min, max) => {
const range = max - min;
const maxGeneratedValue = 0xFFFFFFFF;
const possibleResultValues = range + 1;
const possibleGeneratedValues = maxGeneratedValue + 1;
const remainder = possibleGeneratedValues % possibleResultValues;
const maxUnbiased = maxGeneratedValue - remainder;
if (!Number.isInteger(min) || !Number.isInteger(max) ||
max > Number.MAX_SAFE_INTEGER || min < Number.MIN_SAFE_INTEGER) {
throw new Error('Arguments must be safe integers.');
} else if (range > maxGeneratedValue) {
throw new Error(`Range of ${range} (from ${min} to ${max}) > ${maxGeneratedValue}.`);
} else if (max < min) {
throw new Error(`max (${max}) must be >= min (${min}).`);
} else if (min === max) {
return min;
}
let generated;
do {
generated = crypto.getRandomValues(new Uint32Array(1))[0];
} while (generated > maxUnbiased);
return min + (generated % possibleResultValues);
};
console.log(randomInteger(-8, 8)); // -2
console.log(randomInteger(0, 0)); // 0
console.log(randomInteger(0, 0xFFFFFFFF)); // 944450079
console.log(randomInteger(-1, 0xFFFFFFFF));
// Error: Range of 4294967296 covering -1 to 4294967295 is > 4294967295.
console.log(new Array(12).fill().map(n => randomInteger(8, 12)));
// [11, 8, 8, 11, 10, 8, 8, 12, 12, 12, 9, 9]
Here is an example of a JavaScript function that can generate a random number of any specified length without using Math.random():
function genRandom(length)
{
const t1 = new Date().getMilliseconds();
var min = "1", max = "9";
var result;
var numLength = length;
if (numLength != 0)
{
for (var i = 1; i < numLength; i++)
{
min = min.toString() + "0";
max = max.toString() + "9";
}
}
else
{
min = 0;
max = 0;
return;
}
for (var i = min; i <= max; i++)
{
// Empty Loop
}
const t2 = new Date().getMilliseconds();
console.log(t2);
result = ((max - min)*t1)/t2;
console.log(result);
return result;
}
Use:
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8" />
</head>
<body>
<script>
/*
Assuming that window.crypto.getRandomValues
is available, the real range would be from
0 to 1,998 instead of 0 to 2,000.
See the JavaScript documentation
for an explanation:
https://developer.mozilla.org/en-US/docs/Web/API/RandomSource/getRandomValues
*/
var array = new Uint8Array(2);
window.crypto.getRandomValues(array);
console.log(array[0] + array[1]);
</script>
</body>
</html>
Uint8Array creates an array filled with a number up to three digits which would be a maximum of 999. This code is very short.
This is my take on a random number in a range, as in I wanted to get a random number within a range of base to exponent. E.g., base = 10, exponent = 2, gives a random number from 0 to 100, ideally, and so on.
If it helps using it, here it is:
// Get random number within provided base + exponent
// By Goran Biljetina --> 2012
function isEmpty(value) {
return (typeof value === "undefined" || value === null);
}
var numSeq = new Array();
function add(num, seq) {
var toAdd = new Object();
toAdd.num = num;
toAdd.seq = seq;
numSeq[numSeq.length] = toAdd;
}
function fillNumSeq (num, seq) {
var n;
for(i=0; i<=seq; i++) {
n = Math.pow(num, i);
add(n, i);
}
}
function getRandNum(base, exp) {
if (isEmpty(base)) {
console.log("Specify value for base parameter");
}
if (isEmpty(exp)) {
console.log("Specify value for exponent parameter");
}
fillNumSeq(base, exp);
var emax;
var eseq;
var nseed;
var nspan;
emax = (numSeq.length);
eseq = Math.floor(Math.random()*emax) + 1;
nseed = numSeq[eseq].num;
nspan = Math.floor((Math.random())*(Math.random()*nseed)) + 1;
return Math.floor(Math.random()*nspan) + 1;
}
console.log(getRandNum(10, 20), numSeq);
//Testing:
//getRandNum(-10, 20);
//console.log(getRandNum(-10, 20), numSeq);
//console.log(numSeq);
This I guess, is the most simplified of all the contributions.
maxNum = 8,
minNum = 4
console.log(Math.floor(Math.random() * (maxNum - minNum) + minNum))
console.log(Math.floor(Math.random() * (8 - 4) + 4))
This will log random numbers between 4 and 8 into the console, 4 and 8 inclusive.
Ionuț G. Stan wrote a great answer, but it was a bit too complex for me to grasp. So, I found an even simpler explanation of the same concepts at Math.floor( Math.random () * (max - min + 1)) + min) Explanation by Jason Anello.
Note: The only important thing you should know before reading Jason's explanation is a definition of "truncate". He uses that term when describing Math.floor(). Oxford dictionary defines "truncate" as:
Shorten (something) by cutting off the top or end.
A function called randUpTo that accepts a number and returns a random whole number between 0 and that number:
var randUpTo = function(num) {
return Math.floor(Math.random() * (num - 1) + 0);
};
A function called randBetween that accepts two numbers representing a range and returns a random whole number between those two numbers:
var randBetween = function (min, max) {
return Math.floor(Math.random() * (max - min - 1)) + min;
};
A function called randFromTill that accepts two numbers representing a range and returns a random number between min (inclusive) and max (exclusive)
var randFromTill = function (min, max) {
return Math.random() * (max - min) + min;
};
A function called randFromTo that accepts two numbers representing a range and returns a random integer between min (inclusive) and max (inclusive):
var randFromTo = function (min, max) {
return Math.floor(Math.random() * (max - min + 1)) + min;
};
You can you this code snippet,
let randomNumber = function(first, second) {
let number = Math.floor(Math.random()*Math.floor(second));
while(number < first) {
number = Math.floor(Math.random()*Math.floor(second));
}
return number;
}

Random integer in a certain range excluding one number

I would like get a random number in a range excluding one number (e.g. from 1 to 1000 exclude 577). I searched for a solution, but never solved my issue.
I want something like:
Math.floor((Math.random() * 1000) + 1).exclude(577);
I would like to avoid for loops creating an array as much as possible, because the length is always different (sometimes 1 to 10000, sometimes 685 to 888555444, etc), and the process of generating it could take too much time.
I already tried:
Javascript - Generating Random numbers in a Range, excluding certain numbers
How can I generate a random number within a range but exclude some?
How could I achieve this?
The fastest way to obtain a random integer number in a certain range [a, b], excluding one value c, is to generate it between a and b-1, and then increment it by one if it's higher than or equal to c.
Here's a working function:
function randomExcluded(min, max, excluded) {
var n = Math.floor(Math.random() * (max-min) + min);
if (n >= excluded) n++;
return n;
}
This solution only has a complexity of O(1).
One possibility is not to add 1, and if that number comes out, you assign the last possible value.
For example:
var result = Math.floor((Math.random() * 100000));
if(result==577) result = 100000;
In this way, you will not need to re-launch the random method, but is repeated. And meets the objective of being a random.
As #ebyrob suggested, you can create a function that makes a mapping from a smaller set to the larger set with excluded values by adding 1 for each value that it is larger than or equal to:
// min - integer
// max - integer
// exclusions - array of integers
// - must contain unique integers between min & max
function RandomNumber(min, max, exclusions) {
// As #Fabian pointed out, sorting is necessary
// We use concat to avoid mutating the original array
// See: http://stackoverflow.com/questions/9592740/how-can-you-sort-an-array-without-mutating-the-original-array
var exclusionsSorted = exclusions.concat().sort(function(a, b) {
return a - b
});
var logicalMax = max - exclusionsSorted.length;
var randomNumber = Math.floor(Math.random() * (logicalMax - min + 1)) + min;
for(var i = 0; i < exclusionsSorted.length; i++) {
if (randomNumber >= exclusionsSorted[i]) {
randomNumber++;
}
}
return randomNumber;
}
Example Fiddle
Also, I think #JesusCuesta's answer provides a simpler mapping and is better.
Update: My original answer had many issues with it.
To expand on #Jesus Cuesta's answer:
function RandomNumber(min, max, exclusions) {
var hash = new Object();
for(var i = 0; i < exclusions.length; ++i ) { // TODO: run only once as setup
hash[exclusions[i]] = i + max - exclusions.length;
}
var randomNumber = Math.floor((Math.random() * (max - min - exclusions.length)) + min);
if (hash.hasOwnProperty(randomNumber)) {
randomNumber = hash[randomNumber];
}
return randomNumber;
}
Note: This only works if max - exclusions.length > maximum exclusion. So close.
You could just continue generating the number until you find it suits your needs:
function randomExcluded(start, end, excluded) {
var n = excluded
while (n == excluded)
n = Math.floor((Math.random() * (end-start+1) + start));
return n;
}
myRandom = randomExcluded(1, 10000, 577);
By the way this is not the best solution at all, look at my other answer for a better one!
Generate a random number and if it matches the excluded number then add another random number(-20 to 20)
var max = 99999, min = 1, exclude = 577;
var num = Math.floor(Math.random() * (max - min)) + min ;
while(num == exclude || num > max || num < min ) {
var rand = Math.random() > .5 ? -20 : 20 ;
num += Math.floor((Math.random() * (rand));
}
import random
def rng_generator():
a = random.randint(0, 100)
if a == 577:
rng_generator()
else:
print(a)
#main()
rng_generator()
Exclude the number from calculations:
function toggleRand() {
// demonstration code only.
// this algorithm does NOT produce random numbers.
// return `0` - `576` , `578` - `n`
return [Math.floor((Math.random() * 576) + 1)
,Math.floor(Math.random() * (100000 - 578) + 1)
]
// select "random" index
[Math.random() > .5 ? 0 : 1];
}
console.log(toggleRand());
Alternatively, use String.prototype.replace() with RegExp /^(577)$/ to match number that should be excluded from result; replace with another random number in range [0-99] utilizing new Date().getTime(), isNaN() and String.prototype.slice()
console.log(
+String(Math.floor(Math.random()*(578 - 575) + 575))
.replace(/^(577)$/,String(isNaN("$1")&&new Date().getTime()).slice(-2))
);
Could also use String.prototype.match() to filter results:
console.log(
+String(Math.floor(Math.random()*10))
.replace(/^(5)$/,String(isNaN("$1")&&new Date().getTime()).match(/[^5]/g).slice(-1)[0])
);

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