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Optimising Javascript

I was given a quiz and I had gotten the answer wrong and It's been bugging me ever since so I thought I'd ask for your thoughts
I needed to optimise the following function
function sumOfEvenNumbers(n) {
var sum = 0;
for(let i = 2; i < n;i++){
if(i % 2 == 0) sum += i;
}
return sum;
}
console.log(sumOfEvenNumbers(5));
I came up with
function sumOfEvenNumbers(n) {
var sum = 0;
while(--n >= 2) sum += n % 2 == 0? n : 0
return sum;
}
console.log(sumOfEvenNumbers(5));
What other ways were there?

It's a bit of a math question. The sum appears to be the sum of an arithmitic sequence with a common difference of 2. The sum is:
sum = N * (last + first) / 2;
where N is the number of the numbers in the sequence, last is the last number of those numbers, and first is the first.
Translated to javascript as:
function sumOfEvenNumbers(n) {
return Math.floor(n / 2) * (n - n % 2 + 2) / 2;
}
Because the number of even numbers between 2 and n is Math.floor(n / 2) and the last even number is n - n % 2 (7 would be 7 - 7 % 2 === 6 and 8 would be 8 - 8 % 2 === 8). and the first is 2.

Sum of n numbers:
var sum = (n * (n+1)) / 2;
Sum of n even numbers:
var m = Math.floor(n/2);
var sum = 2 * (m * (m+1) /2);

You can compute these sums using an arithmetic sum formula in constant time:
// Return sum of positive even numbers < n:
function sumOfEvenNumbers(n) {
n = (n - 1) >> 1;
return n * (n + 1);
}
// Example:
console.log(sumOfEvenNumbers(5));
Above computation avoids modulo and division operators which consume more CPU cycles than multiplication, addition and bit-shifting. Pay attention to the limited range of the bit-shifting operator >> though.
See e.g. http://oeis.org/A002378 for this and other formulas leading to the same result.

First thing is to eliminate the test in the loop:
function sumOfEvenNumbers(n) {
var sum = 0;
var halfN= Math.floor(n/2);
for(let i = 1; i < n/2;i++) {
sum += i;
}
return sum * 2;
}
Then we can observe that is just calculating the sum of all the integers less than a limit - and there is a formula for that (but actually formula is for less-equal a limit).
function sumOfEvenNumbers(n) {
var halfNm1= Math.floor(n/2)-1;
var sum = halfNm1 * (halfNm1+1) / 2;
return sum * 2;
}
And then eliminate the division and multiplication and the unnecessary addition and subtraction:
function sumOfEvenNumbers(n) {
var halfN= Math.floor(n/2);
return (halfN-1) * halfN;
}

Your solution computes in linear (O(N)) time.
If you use a mathematical solution, you can compute it in O(1) time:
function sum(n) {
let half = Math.ceil(n/2)
return half * (half + 1)
}

Because the question is tagged ecmascript-6 :)
const sumEven = x => [...Array(x + 1).keys()].reduce((a, b) => b % 2 === 0 ? a + b : a, 0);
// set max number
console.log(sumEven(10));

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;
}

Need help understanding recursive function example from Eloquent Javascript

function power(base, exponent) {
if (exponent == 0)
return 1;
else
return base * power(base, exponent - 1);
}
I think that I understand the basic principle of recursion, it simply means you are calling the function within the function itself. This can be used to perform a loop of sorts, but what I cant figure out is how the above code actually decides to loop in order to figure out the exponential value of a number. I used function power(2,5) as my argument and the function knew the answer was 32, but how? Does the function loop itself subtracting 1 from the exponent each time, and multiplying base * base until exponent reaches zero? And if thats the case, how does calling the power function within the function accomplish this exactly? And once exponent reaches zero, wouldnt the function then just return 1 and not the correct answer?

I consider each recursive step (the function calling itself) producing a shorter and easier problem.
The easiest problem is power(base, 0), which satisfies exponent == 0 and returns one (any base to the zeroth power is 1).
Then, notice that no matter how large exponent is, it is reducing exponent by one, guaranteeing that it will eventually reach the "easiest" problem where the exponent is zero. It only can't be negative, or else this "base case" is never reached.
So, 2^5, or power(2, 5), becomes 2 * 2^4. And 2^4 = 2 * 2^3. By continuing this expansion, we get 2 * 2 * 2 * 2 * 2 * 1, which equals 32. The 1 represents the case exponent == 0 being true.
The computation has to keep track how many of these multiplications it has accumulated, and once the base case of exponent == 0 is reached, multiply all numbers together. It cannot know in advance with certainty what power(base, exponent-1) will return.

Follow the call pattern.. Let's assume we do power(2,2).. You get this:
power(2,2) -> (exponent != 0) 2 * power(2, 1)
2 * power(2, 1) -> (exponent != 0) 2 * power(2, 0)
2 * 2 * power(2,0) -> (exponent == 0) 1
2 * 2 * 1 = 4
The way it works is basically your call stack, as long as you keep calling sub-methods, your parent doesn't return. So it keeps nesting itself until it hits a concrete # -- in this case, 1, then it goes back up the stack actually doing the *.

This shows intermediate results which may help you to follow the logic:
Each level has its own value of base, exponent, answer.
function power(base, exponent) {
var answer; // local
level = level + 1;
console.log("Entering: power(" + base + ", " + exponent +
") (level " + level + ")");
if (exponent == 0) { // don't recurse any more
answer = 1; }
else { // recurse to get answer
answer = base * power(base, exponent - 1); }
// now return answer
console.log("Leaving: power("+ base + ", " + exponent +
") (level " + level + ") ans=" + answer);
level = level - 1
return answer;
}
var level = 0; // global
console.log("Final answer: " + power(2, 5));

The best way to explain the above recursive is to see what the console returns
function power(base, exponent) {
// termination/base case
if (exponent == 0)
return 1;
// recursive case
else
console.log(base + ':' + exponent)
return base * power(base, exponent - 1);
// 2 * (2, 4) = 4
// 4 * (2, 3) = 8
// 8 * (2, 2) = 16
// 16 *(2, 1) = 32
// 16 *(2, 0) = 1 recursive stops and returns 1
// function calls the last return = 32
}
var result = power(2,10)
console.log(result)
I hope this give you a more visual view of how this recursive works

This also confused heck out of me when i first saw it, after 10 minutes starring at it, it just came to me....nothing magical....
function power(base, exponent) {
if (exponent == 0)
return 1;
else
return base * power(base, exponent - 1);
}
console.log(power(2, 5));
this is how it runs through:
return base * power(base, exponent - 1)
looking at the above line itself, I was lost too. there are no operations(+,-,*,/,%) done to the given parameters 2 and 5, but somehow at the end, console.log just know to produce correct number.
Because it does not return a numeric value, it simply returns base * power(base, exponent -1) itself, when program reaches power(base, exponent -1), it executes it before return happens, until exponent == 0 becomes true.
1st:
return base * power(base, 5 - 1);
2nd:
return base * base * power(base, 4 - 1);
3rd:
return base * base * base * power(base, 3 - 1);
4th:
return base * base * base * base * power(base, 2 - 1);
5th:
return base * base * base * base * base * power(base, 1 - 1);
6th:
return base * base * base * base * base * 1;
because if (exponent == 0) return 1
so:
2*2*2*2*2*1 = 32

Can someone please explain this recursive JS code to calculate exponents?

I can't understand this recursion even though it's a really simple example. When it goes to power(base, exponent - 1); what is that supposed to do? How are things being multiplied when power keeps getting invoked until exponent equals 0?
function power(base, exponent) {
if (exponent === 0) {
return 1;
} else {
return base * power(base, exponent - 1);
}
}

Let's start from the beginning.
Let's say you call power(base, 0). Since exponent is 0, the function returns 1.
Now, let's say you call power(base, 1). Since exponent isn't 0 this time, the function calls power(base, exponent - 1) and multiplies it by base. (That's the key here...it takes the result from the recursive call, and adds its own twist.) Since exponent - 1 = 0, and power(base, 0) is 1, the result is effectively base * 1. Read: base.
Now on to power(base, 2). That ends up being base * power(base, 1). And power(base, 1) is base * power(base, 0). End result: base * (base * 1). Read: base squared.
And so on.
In case it wasn't obvious, by the way, this function will only work with non-negative integer exponents. If exponent is negative, or is even the tiniest bit more or less than a whole number, the function will run "forever". (In reality, you'll more than likely cause a stack overflow, once recursion eats up all of your stack.)
You could fix the function for negative powers with some code like
if (exponent < 0) return 1 / power(base, -exponent);
As for non-integers...there's no good way to solve that other than throwing an exception. Raising a number to a non-integer power makes sense, so you don't want to just truncate the exponent or otherwise pretend they didn't try to do it -- you'd end up returning the wrong answer.

This is similar to Math.pow(); it raises the base argument to the exponent argument.
For example, 2 ^ 4 is 16, so power(2, 4) would return 16. The if() statement checks to see whether the exponent (power) is zero and returns 1 if it is - any number raised to the power 0 equals 1.
The last line
return base * power(base, exponent - 1);
Is a recursive function that calls power() from within itself however many times specified by the value in exponent.
I'll try to explain recursion from the bottom up, or "from the middle" shall we say; it's probably easier to understand.
The bottom most call of power() takes 2 and 1 as it's arguments, and will return 1. This return value is then used in the second up call of power(), so this time the arguments passed are 2 and 2, which outputs 4, and so on until the top-most call to power() is passed 2 and 4 which returns 16.

Using a 2^3 example:
power(2, 3);
calls:
function power(2, 3) {
if (3 === 0) {
return 1;
} else {
return 2 * power(2, 2); //called
}
}
which leads to:
function power(2, 2) {
if (2 === 0) {
return 1;
} else {
return 2 * power(2, 1); //called
}
}
which leads to:
function power(2, 1) {
if (1 === 0) {
return 1;
} else {
return 2 * power(2, 0); //called
}
}
which leads to:
function power(2, 0) {
if (1 === 0) {
return 1; //returned
} else {
return 2 * power(2, -1);
}
}
which leads to:
function power(2, 1) {
if (1 === 0) {
return 1;
} else {
return 2 * 1; //returned
}
}
which leads to:
function power(2, 2) {
if (2 === 0) {
return 1;
} else {
return 2 * 2; //returned
}
}
which leads to:
function power(2, 3) {
if (3 === 0) {
return 1;
} else {
return 2 * 4; //returned
}
}
which ultimately returns 8, which is 2^3.

Assuming the initial call is power(10, 3)...
v-----first power() call returns base * (result of next power() call)
v-----second power() call returns base * (result of next power() call)
v-----third power() call returns base * (result of last power() call)
v------result of last power() call returns 1
(10 * (10 * (10 * (1))))
^-----return 1
^-----return base * 1 (10)
^-----return base * 10 (100)
^-----return base * 100 (1000)
Or go down the left, and up the right. Each line is a subsequent call to power() starting with power(10, 3)...
return base * power(base, 2); // return base * 100 (1000)
return base * power(base, 1); // return base * 10 (100)
return base * power(base, 0); // return base * 1 (10)
return 1; // return 1 (1)

base = 10
power = 3
10 * power(10,2)
10 * 10 * power(10,1)
10 * 10 * 10
maybe ok for positive integers...

Let's try to explain this with some maths.
f(x,y) = x^y # (1) function definition
= x * x * x * ... * x # (2) multiply x with itself y times
= x * (x * x * ... * x) # (3) rewrite using parentheses for clarity
= x * (x^(y-1)) # (4) replace the second part by (1) notation
= x * f(x, y-1) # (5) replace again by using f(x,y) notation according to (1)
f(x,0) = 1 # base case: x^0 = 1
Following this you can see that f(x,y) = x * f(x, y-1).
You can also see where
if (exponent === 0) {
return 1;
}
comes from, namely the base case that something to the 0th power always equals 1: f(x,0) = 1.
That's how this recursion was derived.

Javascript intelligent rounding

I currently need to round numbers up to their nearest major number. (Not sure what the right term is here)
But see an example of what I'm trying to achieve
IE:
13 // 20
349 // 400
5645 // 6000
9892 // 10000
13988 // 20000
93456 // 100000
231516 // 300000
etc. etc.
I have implemented a way of doing this but its so painful and only handles numbers up to a million and if I want it to go higher I need to add more if statements (yeah see how i implmented it :P im not very proud, but brain is stuck)
There must be something out there already but google is not helping me very much probably due to me not knowing the correct term for the kind of rounding i want to do

<script type="text/javascript">
function intelliRound(num) {
var len=(num+'').length;
var fac=Math.pow(10,len-1);
return Math.ceil(num/fac)*fac;
}
alert(intelliRound(13));
alert(intelliRound(349));
alert(intelliRound(5645));
// ...
</script>
See http://jsfiddle.net/fCLjp/

One way;
var a = [13, // 20
349, // 400
5645, // 6000
9892, // 10000
13988, // 20000
93456, // 100000
231516 // 300000
]
for (var i in a) {
var num = a[i];
var scale = Math.pow(10, Math.floor(Math.log(num) / Math.LN10));
print([ num, Math.ceil(num / scale) * scale ])
}
13,20
349,400
5645,6000
9892,10000
13988,20000
93456,100000
231516,300000

The answer from #rabudde works well, but for those that need to handle negative numbers, here's an updated version:
function intelliRound(num) {
var len = (num + '').length;
var result = 0;
if (num < 0) {
var fac = Math.pow(10, len - 2);
result = Math.floor(num / fac) * fac;
}
else {
var fac = Math.pow(10, len - 1);
result = Math.ceil(num / fac) * fac;
}
return result;
}
alert(intelliRound(13));
alert(intelliRound(349));
alert(intelliRound(5645));
alert(intelliRound(-13));
alert(intelliRound(-349));
alert(intelliRound(-5645));

you can use Math.ceil function, as described here:
javascript - ceiling of a dollar amount
to get your numbers right you'll have to divide them by 10 (if they have 2 digits), 100 (if they have 3 digits), and so on...

The intelliRound function from the other answers works well, but break with negative numbers. Here I have extended these solutions to support decimals (e.g. 0.123, -0.987) and non-numbers:
/**
* Function that returns the floor/ceil of a number, to an appropriate magnitude
* #param {number} num - the number you want to round
*
* e.g.
* magnitudeRound(0.13) => 1
* magnitudeRound(13) => 20
* magnitudeRound(349) => 400
* magnitudeRound(9645) => 10000
* magnitudeRound(-3645) => -4000
* magnitudeRound(-149) => -200
*/
function magnitudeRound(num) {
const isValidNumber = typeof num === 'number' && !Number.isNaN(num);
const result = 0;
if (!isValidNumber || num === 0) return result;
const abs = Math.abs(num);
const sign = Math.sign(num);
if (abs > 0 && abs <= 1) return 1 * sign; // percentages on a scale -1 to 1
if (abs > 1 && abs <= 10) return 10 * sign;
const zeroes = `${Math.round(abs)}`.length - 1; // e.g 123 => 2, 4567 => 3
const exponent = 10 ** zeroes; // math floor and ceil only work on integer
const roundingDirection = sign < 0 ? 'floor' : 'ceil';
return Math[roundingDirection](num / exponent) * exponent;
}