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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))
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'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.
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;
}
How to determine if a number is tenthousandth, thousandth, hundredth, tens. I need to convert them to their nearest.
For example, 124. since it's hundredth, i will convert it to 200 using
Math.ceil(124/100) * 100
For example, 4560. since it's thousandth, i will convert it to 5000 using
Math.ceil(5000/1000) * 1000
you can use the following way to convert given number into nearest 10
function convertToNearest(num){
var mul = 1;
while(num > (mul * 10))
mul = mul * 10;
return Math.ceil(num/mul) * mul;
}
This solution handles integers (positive AND negative):
function nearest(n) {
if (n === 0)
return 0;
let neg = (n < 0);
if (neg)
n = -n;
let x = 10 ** Math.floor(Math.log10(n));
if (x === 1)
x = 10;
let ans = Math.ceil(n/x) * x;
return (neg) ? -ans : ans;
}
[-230, 0, 1, 8, -18, -6540, -120, -5, 12, 4560, 320, 888, 83293, 2323203]
.forEach(n => console.log('%s\t->\t%s', n, nearest(n)));
I need to loop through a simple array in a "weird" way.
The length of my array is always a odd squared number.
Let's say the length is 49. To make it clearer, my elements are the index of the array. So I have something like:
myArray = [0, 1, 2, 3, 4 ... 47, 48]
So, you can imagine the following square matrix:
0 7 14 21 28 35 42
1 8 15 22 29 36 43
2 9 16 23 30 37 44
3 10 17 24 31 38 45
4 11 18 25 32 39 46
5 12 19 26 33 40 47
6 13 20 27 34 41 48
I have to start with the center of this matrix (i.e. myArray[Math.floor(myArray.length / 2)])
In my example: 24
Then I have to alternate between left and right numbers until I pass through all the row.
In my example, for the first iteration : 17, 31, 10, 38, 3, 45
Once a row is done, I alternate between up then bottom numbers an reiterate the left/right logic for the given row.
In my example, for myArray as input I should loop in the following order:
24, 17, 31, 10, 38, 3, 45,
23, 16, 30, 9, 37, 2, 44,
25, 18, 32, 11, 39, 4, 46,
22, 15, 29, 8, 36, 1, 43,
26, 19, 33, 12, 40, 5, 47,
21, 14, 28, 7, 35, 42, 0,
27, 20, 34, 13, 41, 6, 48
Could you help me the achieved it properly?
Here is what I did so far: https://jsfiddle.net/6qzkk2zx/
I recommend nesting two ES6 generators:
for (let y of outwards(7)) {
for (let x of outwards(7)) {
var i = y*7+x;
use(arr[i]); // for one-dimensional array
use(arr[x][y]) // for two-dimensional array
}
}
function* outwards(n) {
console.assert(n%2 == 1);
var x = (n-1)/2; // starting in the middle
yield x;
for (var i=1; i<n;) {
x -= i++; // jumping to the left
yield x;
x += i++; // and right
yield x;
}
}
Starting from that, you could either manually translate it back to an equivalent ES5 construct, or just let a transpiler do the work for you: Demo.
I think I have it:
https://jsfiddle.net/2dbj68t3/2/
var start = (array.length - 1) / 2;
var square = Math.sqrt(array.length);
newArray.push(array[start]);
for (j = 1; j <= Math.floor((square / 2)); j++) {
newArray.push((array[start - (j * square)]));
newArray.push((array[start + (j * square)]));
}
for (i = 1; i <= Math.floor((square / 2)); i++) {
newArray.push((array[start - i]));
for (j = 1; j <= Math.floor((square / 2)); j++) {
newArray.push((array[start - i - (j * square)]));
newArray.push((array[start - i + (j * square)]));
}
newArray.push((array[start + i]));
for (j = 1; j <= Math.floor((square / 2)); j++) {
newArray.push((array[start + i - (j * square)]));
newArray.push((array[start + i + (j * square)]));
}
}
Another option:
// Define number of elements
var n = 49;
// Calculate center
var start = Math.floor(n / 2);
// Calculate solution
var result = [get_row(start, n)];
for (var l = 1; l < Math.sqrt(n) / 2; l++) {
result.push(get_row(start - l, n));
result.push(get_row(start + l, n));
}
// Print solution
for (var k = 0; k < result.length; k++) {
console.log(result[k]);
}
///////////////////////////////////////////////
function get_row(c, size) {
var a = [];
a.push(c);
for (var i = 1; i < Math.sqrt(size) / 2; i++) {
a.push(c - i * Math.sqrt(size));
a.push(c + i * Math.sqrt(size));
}
return a;
}
Here's another approach using nested loops, single resulting array.
JSfiddle Demo
var num = 7,
middleElement = Math.floor(num * num / 2); // Square number and get the center number
var result = []; // To store resulting array
for(var j = 1; j <= num; j++) {
// This will be the middle element, i.e. 24, 23, 25, 22, 26, 21, 27
// Will be calculated dynamically
element = j % 2 ? middleElement + Math.floor(j / 2) : middleElement - Math.floor(j / 2);
result.push(element); // Add middle element in the resulting array
for(var i = 1; i < num; i++) {
// For next six elements
// Get the number from current number
element = i % 2 ? element - num * i : element + num * i;
result.push(element);
}
}
var num = 7,
middleElement = Math.floor(num * num / 2); // Square number and get the center number
var result = []; // To store resulting array
for(var j = 1; j <= num; j++) {
// This will be the middle element, i.e. 24, 23, 25, 22, 26, 21, 27
// Will be calculated dynamically
element = j % 2 ? middleElement + Math.floor(j / 2) : middleElement - Math.floor(j / 2);
result.push(element); // Add middle element in the resulting array
for(var i = 1; i < num; i++) {
// For next six elements
// Get the number from current number
element = i % 2 ? element - num * i : element + num * i;
result.push(element);
}
}
console.log(result);
Another approach, with an array for the position of the numbers:
var array = [], // 3 2 4 1 5 0 6
length = 7,
lower = 0, upper = length - 1,
result;
while (lower < upper) { // generate array
array.unshift(upper--);
array.unshift(lower++);
}
lower === upper && array.unshift(upper);
result = Array.apply(null, { length: length }).map(function (_, j) {
return Array.apply(null, { length: length }).map(function (_, i) {
return array[i] * length + array[j];
});
});
document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');
Numbers that follows the one you are processing (let's say 24) are always some subtraction of your number (24) by some multiple of the length of the matrix (in your example, this length is 7).
Let subject be the number you are processing and hop the length of the matrix.
So, iterate through the rows, two per iteration, always filling two cells at a time, the first with subject - hop * <counter of iteration>, and the second cell with subject + hop * <counter of iteration>
I hope I understand your algorithm, but this ought to work:
If the length is always an odd square, then your index minima are 0 and maxima are Sqrt(n+1)-1.
var verticalup=false;
var horizontalforward=false;
var initial=myArray[Math.floor(myArray.length / 2)];
var maximum=Math.sqrt(myArray.length+1);
var current={vertical:initial,horizontal:initial};
var continue=true;
while (continue) {
//do something here with the current selection
if (current.horizontal===0) {
verticalup=!verticalup;
current.vertical+=(verticalup?1:-1);
}
horizontalforward=!horizontalforward;
current.horizontal+=(horizontalforward?1:-1);
continue=current.vertical>=0 && current.vertical<maxima && current.horizontal>=0 && current.horizontal<maximum;
}
I think something in the lines of...
var center = Math.floor(myArray.length / 2;
var parity = function(p){
return (p%2)*2-1; //if the number is even, returns -1 otherwise 1
}
//adds to cetner location half on index, signed with its parity, so we get an alternating effect each time index grows bigger
//for 0,1,2,3,4,5,6... will return 0,1,-1,2,-2,3,-3...
var step = function(index){
return (Math.floor((index+1)/2))*parity(index);
}
for (var i=0;i<size;i++){
for (var j=0;j<size;j++){
a[center-step(i)][center-step(j)];
}
}
Basically it utilizes the parity of i and j to determine whether to go up or down, and or left or right from the (center, center) point.
Had a hard time coming up with a concise title for this. I'm sure there are terms for what I want to accomplish and there is no doubt a common algorithm to accomplish what I'm after - I just don't know about them yet.
I need to break up a number into n pieces that are each a multiple of 50. The number is itself a multiple of 50. Here is an example:
Divide 5,000 by 3 and end up with three numbers that are each multiples of 50:
1,650
1,700
1,650
I also would like to have the numbers distributed so that they flip back and forth, here is an example with more numbers to illustrate this:
Divide 5,000 by 7 and end up with 7 numbers that are each multiples of 50:
700
750
700
750
700
700
700
Note that in the above example I'm not worried that the extra 50 is not centered in the series, that is I don't need to have something like this:
700
700
750 <--- note the '50s' are centered
700
750 <--- note the '50s' are centered
700
700
Hopefully I've asked this clearly enough that you understand what I want to accomplish.
Update: Here is the function I'll be using.
var number = 5000;
var n = 7;
var multiple = 50;
var values = getIntDividedIntoMultiple(number, n, multiple)
function getIntDividedIntoMultiple(dividend, divisor, multiple)
{
var values = [];
while (dividend> 0 && divisor > 0)
{
var a = Math.round(dividend/ divisor / multiple) * multiple;
dividend -= a;
divisor--;
values.push(a);
}
return values;
}
var number = 5000;
var n = 7;
var values = [];
while (number > 0 && n > 0) {
var a = Math.floor(number / n / 50) * 50;
number -= a;
n--;
values.push(a);
} // 700 700 700 700 700 750 750
Edit
You can alternate Math.floor and Math.ceil to obtain the desired result:
while (number > 0 && n > 0) {
if (a%2 == 0)
a = Math.floor(number / n / 50) * 50;
else
a = Math.ceil(number / n / 50) * 50;
number -= a;
n--;
values.push(a);
} // 700 750 700 750 700 700 700
// i - an integer multiple of k
// k - an integer
// n - a valid array length
// returns an array of length n containing integer multiples of k
// such that the elements sum to i and the array is sorted,
// contains the minimum number of unique elements necessary to
// satisfy the first condition, the elements chosen are the
// closest together that satisfy the first condition.
function f(i, k, n) {
var minNumber = (((i / k) / n) | 0) * k;
var maxNumber = minNumber + k;
var numMax = (i - (minNumber * n)) / k;
var nums = [];
for (var i = 0; i < n - numMax; ++i) {
nums[i] = minNumber;
}
for (var i = n - numMax; i < n; ++i) {
nums[i] = maxNumber;
}
return nums;
}
So your second example would be
f(5000, 50, 7)
which yields
[700,700,700,700,700,750,750]
Let a be your starting number, k - number of parts you want to divide to.
Suppose, that b = a/n.
Now you want to divide b into k close integer parts.
Take k numbers, each equal to b/k (integer division).
Add 1 to first b%k numbers.
Multiply each number by n.
Example:
a = 5000, n = 50, k = 7.
b = 100
Starting series {14, 14, 14, 14, 14, 14, 14}
Add 1 to first 2 integers {15, 15, 14, 14, 14, 14, 14}.
Multiply by 50 {750, 750, 700, 700, 700, 700, 700}.
Your problem is the same as dividing a number X into N integer pieces that are all within 1 of each other (just multiply everything by 50 after you've found the result). Doing this is easy - set all N numbers to Floor(X/N), then add 1 to X mod N of them.
I see your problem as basically trying to divide a sum of money into near-equal bundles of bills of a certain denomination.
For example, dividing 10,000 dollars into 7 near-equal bundles of 50-dollar bills.
function getBundles(sum, denomination, count, shuffle)
{
var p = Math.floor(sum / denomination);
var q = Math.floor(p / count);
var r = p - q * count;
console.log(r + " of " + ((q + 1) * denomination)
+ " and " + (count - r) + " of " + (q * denomination));
var b = new Array(count);
for (var i = 0; i < count; i++) {
b[i] = (r > 0 && (!shuffle || Math.random() < .5 || count - i == r)
? (--r, q + 1) : q)
* denomination;
}
return b;
}
// Divide 10,000 dollars into 7 near-equal bundles of 50-dollar bills
var bundles = getBundles(10000, 50, 7, true);
console.log("bundles: " + bundles);
Output:
4 of 1450 and 3 of 1400
bundles: 1400,1450,1450,1400,1450,1400,1450
If the last argument shuffle is true, it distributes the extra amount randomly between the bundles.
Here's my take:
public static void main(String[] args) {
System.out.println(toList(divide(50, 5000, 3)));
System.out.println(toList(divide(50, 5000, 7)));
System.out.println(toList(divide(33, 6600, 7)));
}
private static ArrayList<Integer> toList(int[] args) {
ArrayList<Integer> list = new ArrayList<Integer>(args.length);
for (int i : args)
list.add(i);
return list;
}
public static int[] divide(int N, int multiplyOfN, int partsCount) {
if (N <= 0 || multiplyOfN <= N || multiplyOfN % N != 0)
throw new IllegalArgumentException("Invalid args");
int factor = multiplyOfN / N;
if (partsCount > factor)
throw new IllegalArgumentException("Invalid args");
int parts[] = new int[partsCount];
int remainingAdjustments = factor % partsCount;
int base = ((multiplyOfN / partsCount) / N) * N;
for (int i = 0; i < partsCount; i ++) {
parts[i] = (i % 2 == 1 && remainingAdjustments-- > 0) ? base + N : base;
}
return parts;
}
My algorithm provides even distribution of remainder across parts:
function splitValue(value, parts, multiplicity)
{
var result = [];
var currentSum = 0;
for (var i = 0; i < parts; i++)
{
result[i] = Math.round(value * (i + 1) / parts / multiplicity) * multiplicity - currentSum;
currentSum += result[i];
}
return result;
}
For value = 5000, parts = 7, multiplicity = 50 it returns
[ 700, 750, 700, 700, 700, 750, 700 ]
Okay....
I have a lot of uncontrolled numbers i want to round:
51255 -> 55000
25 -> 25
9214 -> 9500
13135 -> 15000
25123 -> 30000
I have tried modifying the numbers as string and counting length....
But is there a simple way using some Math function maybe?
Here's my late answer. Uses no Math methods.
function toN5( x ) {
var i = 5;
while( x >= 100 ) {x/=10; i*=10;}
return ((~~(x/5))+(x%5?1:0)) * i;
}
DEMO: http://jsbin.com/ujamoj/edit#javascript,live
[51255, 24, 25, 26, 9214, 13135, 25123, 1, 9, 0].map( toN5 );
// [55000, 25, 25, 30, 9500, 15000, 30000, 5, 10, 0]
Or this is perhaps a bit cleaner:
function toN5( x ) {
var i = 1;
while( x >= 100 ) {x/=10; i*=10;}
return (x + (5-((x%5)||5))) * i;
}
DEMO: http://jsbin.com/idowan/edit#javascript,live
To break it down:
function toN5( x ) {
// v---we're going to reduce x to the tens place, and for each place
// v reduction, we'll multiply i * 10 to restore x later.
var i = 1;
// as long as x >= 100, divide x by 10, and multiply i by 10.
while( x >= 100 ) {x/=10; i*=10;}
// Now round up to the next 5 by adding to x the difference between 5 and
// the remainder of x/5 (or if the remainder was 0, we substitute 5
// for the remainder, so it is (x + (5 - 5)), which of course equals x).
// So then since we are now in either the tens or ones place, and we've
// rounded to the next 5 (or stayed the same), we multiply by i to restore
// x to its original place.
return (x + (5-((x%5)||5))) * i;
}
Or to avoid logical operators, and just use arithmetic operators, we could do:
return (x + ((5-(x%5))%5)) * i;
And to spread it out a bit:
function toN5( x ) {
var i = 1;
while( x >= 100 ) {
x/=10;
i*=10;
}
var remainder = x % 5;
var distance_to_5 = (5 - remainder) % 5;
return (x + distance_to_5) * i;
}
var numbers = [51255, 25, 9214, 13135, 25123, 3, 6];
function weird_round(a) {
var len = a.toString().length;
var div = len == 1 ? 1 : Math.pow(10, len - 2);
return Math.ceil(a / 5 / div) * div * 5;
}
alert(numbers.map(weird_round));
Also updated for numbers below 10. Won't work properly for negative numbers either, just mention if you need this.
DEMO
I'm not sure why, but I thought it would be fun with regular expressions:
var result = +(number.toString().replace(/([1-9])([0-9])(.+)/, function() {
return Math.ceil(+(arguments[1] + '.' + arguments[2])) * 10 - (+arguments[2] < 5?5:0) + arguments[3].replace(/./g, '0');
}));
Working Demo
with(Math) {
var exp = floor(log(number)/log(10)) - 1;
exp = max(exp,0);
var n = number/pow(10,exp);
var n2 = ceil(n/5) * 5;
var result = n2 * pow(10,exp);
}
http://jsfiddle.net/NvvGf/4/
Caveat: only works for the natural numbers.
function round(number) {
var numberStr = number + "",
max,
i;
if (numberStr[1] > '4') {
numberStr[0] = parseInt(numberStr[0]) + 1;
numberStr[1] = '0';
} else {
numberStr[1] = '5';
}
for (i = 2; max = numberStr.length; i < max; i += 1) {
numberStr += '0';
}
return parseInt(numberStr);
}
Strange coincidence, I wrote something really similar not so long ago!
function iSuckAtNames(n) {
var n = n.toString(), len = n.length, res;
//Check the second number. if it's less than a 5, round down,
//If it's more/equal, round up
//Either way, we'll need to use this:
var res = parseFloat(n[0]) * Math.pow(10, len - 1); //e.g. 5 * 10^4 = 50000
if (n[1] <= 5) {
//we need to add a 5 right before the end!
res += 5 * Math.pow(10, len - 2);
}
else {
//We need another number of that size
res += Math.pow(10, len - 1);
}
return res;
}