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I am new to coding and javascript and was asked, for an assignment, to convert base 10 numbers to a binary base without using specific Javascript built in methods (like alert(a.toString(16))), and I am only allowed to use loops,arrays and functions. This is what i have so far:
var number = prompt("Enter an unsigned base 10 number");
if (number>=0) {
var base = prompt("Enter b for binary, o for octal, or h for hexadecimal");
if (base=="h"||base=="H") {
;
}
So as you can see, I don't have much to go on. I was curious as to what equation or formula I would use to convert the base 10 number, as well as how i'm supposed to show A=10, B=11, C=12 and so forth for a hexadecimal base. Any help would be greatly appreciated!
edit: This is a rather complicated way to do it,
as Alnitak showed me (see discussion below).
It is more a scibble, or the long way by foot.
Short explanation:
If we want to get the binary of the decimal number 10,
we have to try 2^n so that 2^n is still smaller than 10.
For example 2^3 = 8 (that is OK). But 2^4 = 16 (thats too big).
So we have 2^3 and store a 1 for that in an array at index 3.
Now we have to get the rest of 10-2^3, which is 2, and have to
make the same calculation again until we get a difference of zero.
At last we have to reverse the array because its the other way arround.
var a = prompt("Enter an unsigned base 10 number");
var arr = [];
var i = 0;
function decToBin(x) {
y = Math.pow(2, i);
if (y < x) {
arr[i] = 0;
i++;
decToBin(x);
} else if (y > x) {
i--;
newX = (x - Math.pow(2, i));
arr[i] = 1;
i = 0;
decToBin(newX)
} else if (y == x) {
arr[i] = 1;
result = arr.reverse().join();
}
return result;
}
var b = decToBin(a); // var b holds the result
document.write(b);
Is it possible to get a random number between 1-100 and keep the results mainly within the 40-60 range? I mean, it will go out of that range rarely, but I want it to be mainly within that range... Is it possible with JavaScript/jQuery?
Right now I'm just using the basic Math.random() * 100 + 1.
The simplest way would be to generate two random numbers from 0-50 and add them together.
This gives a distribution biased towards 50, in the same way rolling two dice biases towards 7.
In fact, by using a larger number of "dice" (as #Falco suggests), you can make a closer approximation to a bell-curve:
function weightedRandom(max, numDice) {
let num = 0;
for (let i = 0; i < numDice; i++) {
num += Math.random() * (max/numDice);
}
return num;
}
JSFiddle: http://jsfiddle.net/797qhcza/1/
You have some good answers here that give specific solutions; let me describe for you the general solution. The problem is:
I have a source of more-or-less uniformly distributed random numbers between 0 and 1.
I wish to produce a sequence of random numbers that follow a different distribution.
The general solution to this problem is to work out the quantile function of your desired distribution, and then apply the quantile function to the output of your uniform source.
The quantile function is the inverse of the integral of your desired distribution function. The distribution function is the function where the area under a portion of the curve is equal to the probability that the randomly-chosen item will be in that portion.
I give an example of how to do so here:
http://ericlippert.com/2012/02/21/generating-random-non-uniform-data/
The code in there is in C#, but the principles apply to any language; it should be straightforward to adapt the solution to JavaScript.
Taking arrays of numbers, etc. isn't efficient. You should take a mapping which takes a random number between 0 to 100 and maps to the distribution you need. So in your case, you could take f(x)=-(1/25)x2+4x to get a distribution with the most values in the middle of your range.
I might do something like setup a "chance" for the number to be allowed to go "out of bounds". In this example, a 20% chance the number will be 1-100, otherwise, 40-60:
$(function () {
$('button').click(function () {
var outOfBoundsChance = .2;
var num = 0;
if (Math.random() <= outOfBoundsChance) {
num = getRandomInt(1, 100);
} else {
num = getRandomInt(40, 60);
}
$('#out').text(num);
});
function getRandomInt(min, max) {
return Math.floor(Math.random() * (max - min + 1)) + min;
}
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<button>Generate</button>
<div id="out"></div>
fiddle: http://jsfiddle.net/kbv39s9w/
I needed to solve this problem a few years ago and my solution was easier than any of the other answers.
I generated 3 randoms between the bounds and averaged them. This pulls the result towards the centre but leaves it completely possible to reach the extremities.
It looks stupid but you can use rand twice:
var choice = Math.random() * 3;
var result;
if (choice < 2){
result = Math.random() * 20 + 40; //you have 2/3 chance to go there
}
else {
result = Math.random() * 100 + 1;
}
Sure it is possible. Make a random 1-100. If the number is <30 then generate number in range 1-100 if not generate in range 40-60.
There is a lot of different ways to generate such random numbers. One way to do it is to compute the sum of multiple uniformly random numbers. How many random numbers you sum and what their range is will determine how the final distribution will look.
The more numbers you sum up, the more it will be biased towards the center. Using the sum of 1 random number was already proposed in your question, but as you notice is not biased towards the center of the range. Other answers have propose using the sum of 2 random numbers or the sum of 3 random numbers.
You can get even more bias towards the center of the range by taking the sum of more random numbers. At the extreme you could take the sum of 99 random numbers which each were either 0 or 1. That would be a binomial distribution. (Binomial distributions can in some sense be seen as the discrete version of normal distributions). This can still in theory cover the full range, but it has so much bias towards the center that you should never expect to see it reach the endpoints.
This approach means you can tweak just how much bias you want.
What about using something like this:
var loops = 10;
var tries = 10;
var div = $("#results").html(random());
function random() {
var values = "";
for(var i=0; i < loops; i++) {
var numTries = tries;
do {
var num = Math.floor((Math.random() * 100) + 1);
numTries--;
}
while((num < 40 || num >60) && numTries > 1)
values += num + "<br/>";
}
return values;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="results"></div>
The way I've coded it allows you to set a couple of variables:
loops = number of results
tries = number of times the function will try to get a number between 40-60 before it stops running through the while loop
Added bonus: It uses do while!!! Awesomeness at its best
You can write a function that maps random values between [0, 1) to [1, 100] according to weight. Consider this example:
Here, the value 0.95 maps to value between [61, 100].
In fact we have .05 / .1 = 0.5, which, when mapped to [61, 100], yields 81.
Here is the function:
/*
* Function that returns a function that maps random number to value according to map of probability
*/
function createDistributionFunction(data) {
// cache data + some pre-calculations
var cache = [];
var i;
for (i = 0; i < data.length; i++) {
cache[i] = {};
cache[i].valueMin = data[i].values[0];
cache[i].valueMax = data[i].values[1];
cache[i].rangeMin = i === 0 ? 0 : cache[i - 1].rangeMax;
cache[i].rangeMax = cache[i].rangeMin + data[i].weight;
}
return function(random) {
var value;
for (i = 0; i < cache.length; i++) {
// this maps random number to the bracket and the value inside that bracket
if (cache[i].rangeMin <= random && random < cache[i].rangeMax) {
value = (random - cache[i].rangeMin) / (cache[i].rangeMax - cache[i].rangeMin);
value *= cache[i].valueMax - cache[i].valueMin + 1;
value += cache[i].valueMin;
return Math.floor(value);
}
}
};
}
/*
* Example usage
*/
var distributionFunction = createDistributionFunction([
{ weight: 0.1, values: [1, 40] },
{ weight: 0.8, values: [41, 60] },
{ weight: 0.1, values: [61, 100] }
]);
/*
* Test the example and draw results using Google charts API
*/
function testAndDrawResult() {
var counts = [];
var i;
var value;
// run the function in a loop and count the number of occurrences of each value
for (i = 0; i < 10000; i++) {
value = distributionFunction(Math.random());
counts[value] = (counts[value] || 0) + 1;
}
// convert results to datatable and display
var data = new google.visualization.DataTable();
data.addColumn("number", "Value");
data.addColumn("number", "Count");
for (value = 0; value < counts.length; value++) {
if (counts[value] !== undefined) {
data.addRow([value, counts[value]]);
}
}
var chart = new google.visualization.ColumnChart(document.getElementById("chart"));
chart.draw(data);
}
google.load("visualization", "1", { packages: ["corechart"] });
google.setOnLoadCallback(testAndDrawResult);
<script src="https://www.google.com/jsapi"></script>
<div id="chart"></div>
Here's a weighted solution at 3/4 40-60 and 1/4 outside that range.
function weighted() {
var w = 4;
// number 1 to w
var r = Math.floor(Math.random() * w) + 1;
if (r === 1) { // 1/w goes to outside 40-60
var n = Math.floor(Math.random() * 80) + 1;
if (n >= 40 && n <= 60) n += 40;
return n
}
// w-1/w goes to 40-60 range.
return Math.floor(Math.random() * 21) + 40;
}
function test() {
var counts = [];
for (var i = 0; i < 2000; i++) {
var n = weighted();
if (!counts[n]) counts[n] = 0;
counts[n] ++;
}
var output = document.getElementById('output');
var o = "";
for (var i = 1; i <= 100; i++) {
o += i + " - " + (counts[i] | 0) + "\n";
}
output.innerHTML = o;
}
test();
<pre id="output"></pre>
Ok, so I decided to add another answer because I felt like my last answer, as well as most answers here, use some sort of half-statistical way of obtaining a bell-curve type result return. The code I provide below works the same way as when you roll a dice. Therefore, it is hardest to get 1 or 99, but easiest to get 50.
var loops = 10; //Number of numbers generated
var min = 1,
max = 50;
var div = $("#results").html(random());
function random() {
var values = "";
for (var i = 0; i < loops; i++) {
var one = generate();
var two = generate();
var ans = one + two - 1;
var num = values += ans + "<br/>";
}
return values;
}
function generate() {
return Math.floor((Math.random() * (max - min + 1)) + min);
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div id="results"></div>
I'd recommend using the beta distribution to generate a number between 0-1, then scale it up. It's quite flexible and can create many different shapes of distributions.
Here's a quick and dirty sampler:
rbeta = function(alpha, beta) {
var a = 0
for(var i = 0; i < alpha; i++)
a -= Math.log(Math.random())
var b = 0
for(var i = 0; i < beta; i++)
b -= Math.log(Math.random())
return Math.ceil(100 * a / (a+b))
}
var randNum;
// generate random number from 1-5
var freq = Math.floor(Math.random() * (6 - 1) + 1);
// focus on 40-60 if the number is odd (1,3, or 5)
// this should happen %60 of the time
if (freq % 2){
randNum = Math.floor(Math.random() * (60 - 40) + 40);
}
else {
randNum = Math.floor(Math.random() * (100 - 1) + 1);
}
The best solution targeting this very problem is the one proposed by BlueRaja - Danny Pflughoeft but I think a somewhat faster and more general solution is also worth mentioning.
When I have to generate random numbers (strings, coordinate pairs, etc.) satisfying the two requirements of
The result set is quite small. (not larger than 16K numbers)
The result set is discreet. (like integer numbers only)
I usually start by creating an array of numbers (strings, coordinate pairs, etc.) fulfilling the requirement (In your case: an array of numbers containing the more probable ones multiple times.), then choose a random item of that array. This way, you only have to call the expensive random function once per item.
Distribution
5% for [ 0,39]
90% for [40,59]
5% for [60,99]
Solution
var f = Math.random();
if (f < 0.05) return random(0,39);
else if (f < 0.95) return random(40,59);
else return random(60,99);
Generic Solution
random_choose([series(0,39),series(40,59),series(60,99)],[0.05,0.90,0.05]);
function random_choose (collections,probabilities)
{
var acc = 0.00;
var r1 = Math.random();
var r2 = Math.random();
for (var i = 0; i < probabilities.length; i++)
{
acc += probabilities[i];
if (r1 < acc)
return collections[i][Math.floor(r2*collections[i].length)];
}
return (-1);
}
function series(min,max)
{
var i = min; var s = [];
while (s[s.length-1] < max) s[s.length]=i++;
return s;
}
You can use a helper random number to whether generate random numbers in 40-60 or 1-100:
// 90% of random numbers should be between 40 to 60.
var weight_percentage = 90;
var focuse_on_center = ( (Math.random() * 100) < weight_percentage );
if(focuse_on_center)
{
// generate a random number within the 40-60 range.
alert (40 + Math.random() * 20 + 1);
}
else
{
// generate a random number within the 1-100 range.
alert (Math.random() * 100 + 1);
}
If you can use the gaussian function, use it. This function returns normal number with average 0 and sigma 1.
95% of this number are within average +/- 2*sigma. Your average = 50, and sigma = 5 so
randomNumber = 50 + 5*gaussian()
The best way to do that is generating a random number that is distributed equally in a certain set of numbers, and then apply a projection function to the set between 0 and a 100 where the projection is more likely to hit the numbers you want.
Typically the mathematical way of achieving this is plotting a probability function of the numbers you want. We could use the bell curve, but let's for the sake of easier calculation just work with a flipped parabola.
Let's make a parabola such that its roots are at 0 and 100 without skewing it. We get the following equation:
f(x) = -(x-0)(x-100) = -x * (x-100) = -x^2 + 100x
Now, all the area under the curve between 0 and 100 is representative of our first set where we want the numbers generated. There, the generation is completely random. So, all we need to do is find the bounds of our first set.
The lower bound is, of course, 0. The upper bound is the integral of our function at 100, which is
F(x) = -x^3/3 + 50x^2
F(100) = 500,000/3 = 166,666.66666 (let's just use 166,666, because rounding up would make the target out of bounds)
So we know that we need to generate a number somewhere between 0 and 166,666. Then, we simply need to take that number and project it to our second set, which is between 0 and 100.
We know that the random number we generated is some integral of our parabola with an input x between 0 and 100. That means that we simply have to assume that the random number is the result of F(x), and solve for x.
In this case, F(x) is a cubic equation, and in the form F(x) = ax^3 + bx^2 + cx + d = 0, the following statements are true:
a = -1/3
b = 50
c = 0
d = -1 * (your random number)
Solving this for x yields you the actual random number your are looking for, which is guaranteed to be in the [0, 100] range and a much higher likelihood to be close to the center than the edges.
This answer is really good. But I would like to post implementation instructions (I'm not into JavaScript, so I hope you will understand) for different situation.
Assume you have ranges and weights for every range:
ranges - [1, 20], [21, 40], [41, 60], [61, 100]
weights - {1, 2, 100, 5}
Initial Static Information, could be cached:
Sum of all weights (108 in sample)
Range selection boundaries. It basically this formula: Boundary[n] = Boundary[n - 1] + weigh[n - 1] and Boundary[0] = 0. Sample has Boundary = {0, 1, 3, 103, 108}
Number generation:
Generate random number N from range [0, Sum of all weights).
for (i = 0; i < size(Boundary) && N > Boundary[i + 1]; ++i)
Take ith range and generate random number in that range.
Additional note for performance optimizations. Ranges don't have to be ordered neither ascending nor descending order, so for faster range look-up range that has highest weight should go first and one with lowest weight should go last.
So, to be short,
3√(-8) = (-8)1/3
console.log(Math.pow(-8,1/3));
//Should be -2
But when I test it out, it outputs
NaN
Why? Is it a bug or it is expected to be like this in the first place? I am using JavaScript to draw graphs, but this messes up the graph.
You can use this snippet to calculate it. It also works for other powers, e.g. 1/4, 1/5, etc.
function nthroot(x, n) {
try {
var negate = n % 2 == 1 && x < 0;
if(negate)
x = -x;
var possible = Math.pow(x, 1 / n);
n = Math.pow(possible, n);
if(Math.abs(x - n) < 1 && (x > 0 == n > 0))
return negate ? -possible : possible;
} catch(e){}
}
nthroot(-8, 3);
Source: http://gotochriswest.com/blog/2011/05/06/cube-root-an-beyond/
A faster approach for just calculating the cubic root:
Math.cbrt = function(x) {
var sign = x === 0 ? 0 : x > 0 ? 1 : -1;
return sign * Math.pow(Math.abs(x), 1 / 3);
}
Math.cbrt(-8);
Update
To find an integer based cubic root, you can use the following function, inspired by this answer:
// positive-only cubic root approximation
function cbrt(n)
{
var a = n; // note: this is a non optimized assumption
while (a * a * a > n) {
a = Math.floor((2 * a + (n / (a * a))) / 3);
}
return a;
}
It starts with an assumption that converges to the closest integer a for which a^3 <= n. This function can be adjusted in the same way to support a negative base.
There's no bug; you are raising a negative number to a fractional power; hence, the NaN.
The top hit on google for this is from Dr Math the explanation is pretty good. It says for for real numbers (not complex numbers anyway), a negative number raised to a fractional power may not be a real number. The simplest example is probably
-4 ^ (1/2)
which is essentially computing the square root of -4. Even though the cubic root of -8 does have real solutions, I think that most software libraries find it more efficient not to do all the complex arithmetic and return NaN only when the imaginary part is nonzero and give you the nice real answer otherwise.
EDIT
Just to make absolutely clear that NaN is the intended result, see the official ECMAScript 5.1 Specification, Section 15.8.2.13. It says:
If x<0 and x is finite and y is finite and y is not an integer, the result is NaN.
Again, even though SOME instances of raising negative numbers to fractional powers have exactly one real root, many languages just do the NaN thing for all cases of negative numbers to fractional roots.
Please do not think JavaScript is the only such language. C++ does the same thing:
If x is finite negative and y is finite but not an integer value, it causes a domain error.
Two key problems:
Mathematically, there are multiple cubic roots of a negative number: -2, but also 2 complex roots (see cube roots of unity).
Javascript's Math object (and most other standard math libraries) will not do fractional powers of negative numbers. It converts the fractional power to a float before the function receives it, so you are asking the function to compute a floating point power of a negative number, which may or may not have a real solution. So it does the pragmatic thing and refuses to attempt to calculate such a value.
If you want to get the correct answer, you'll need to decide how mathematically correct you want to be, and write those rules into a non-standard implementation of pow.
All library functions are limited to avoid excessive calculation times and unnecessary complexity.
I like the other answers, but how about overriding Math.pow so it would be able to work with all nth roots of negative numbers:
//keep the original method for proxying
Math.pow_ = Math.pow;
//redefine the method
Math.pow = function(_base, _exponent) {
if (_base < 0) {
if (Math.abs(_exponent) < 1) {
//we're calculating nth root of _base, where n === 1/_exponent
if (1 / _exponent % 2 === 0) {
//nth root of a negative number is imaginary when n is even, we could return
//a string like "123i" but this would completely mess up further computation
return NaN;
}/*else if (1 / _exponent % 2 !== 0)*/
//nth root of a negative number when n is odd
return -Math.pow_(Math.abs(_base), _exponent);
}
}/*else if (_base >=0)*/
//run the original method, nothing will go wrong
return Math.pow_(_base, _exponent);
};
Fiddled with some test cases, give me a shout if you spot a bug!
So I see a bunch of methods that revolve around Math.pow(...) which is cool, but based on the wording of the bounty I'm proposing a slightly different approach.
There are several computational approximations for solving roots, some taking quicker steps than others. Ultimately the stopping point comes down to the degree of precision desired(it's really up to you/the problem being solved).
I'm not going to explain the math in fine detail, but the following are implementations of cubed root approximations that passed the target test(bounty test - also added negative range, because of the question title). Each iteration in the loop (see the while(Math.abs(xi-xi0)>precision) loops in each method) gets a step closer to the desired precision. Once precision is achieved a format is applied to the number so it's as precise as the calculation derived from the iteration.
var precision = 0.0000000000001;
function test_cuberoot_fn(fn) {
var tested = 0,
failed = 0;
for (var i = -100; i < 100; i++) {
var root = fn(i*i*i);
if (i !== root) {
console.log(i, root);
failed++;
}
tested++;
}
if (failed) {
console.log("failed %d / %d", failed, tested);
}else{
console.log("Passed test");
}
}
test_cuberoot_fn(newtonMethod);
test_cuberoot_fn(halleysMethod);
Newton's approximation Implementation
function newtonMethod(cube){
if(cube == 0){//only John Skeet and Chuck Norris
return 0; //can divide by zero, we'll have
} //to settle for check and return
var xi = 1;
var xi0 = -1;
while(Math.abs(xi-xi0)>precision){//precision = 0.0000000000001
xi0=xi;
xi = (1/3)*((cube/(xi*xi))+2*xi);
}
return Number(xi.toPrecision(12));
}
Halley's approximation Implementation
note Halley's approximation takes quicker steps to solving the cube, so it's computationally faster than newton's approximation.
function halleysMethod(cube){
if(cube == 0){//only John Skeet and Chuck Norris
return 0; //can divide by zero, we'll have
} //to settle for check and return
var xi = 1;
var xi0 = -1;
while(Math.abs(xi-xi0)>precision){//precision = 0.0000000000001
xi0=xi;
xi = xi*((xi*xi*xi + 2*cube)/(2*xi*xi*xi+cube));
}
return Number(xi.toPrecision(12));
}
It's Working in Chrome Console
function cubeRoot(number) {
var num = number;
var temp = 1;
var inverse = 1 / 3;
if (num < 0) {
num = -num;
temp = -1;
}
var res = Math.pow(num, inverse);
var acc = res - Math.floor(res);
if (acc <= 0.00001)
res = Math.floor(res);
else if (acc >= 0.99999)
res = Math.ceil(res);
return (temp * res);
}
cubeRoot(-64) // -4
cubeRoot(64) // 4
As a heads up, in ES6 there is now a Math.cbrt function.
In my testing in Google chrome it appears to work almost twice as fast as Math.pow. Interestingly I had to add up the results otherwise chrome did a better job of optimizing away the pow function.
//do a performance test on the cube root function from es6
var start=0, end=0, k=0;
start = performance.now();
k=0;
for (var i=0.0; i<10000000.0; i+=1.0)
{
var j = Math.cbrt(i);
//k+=j;
}
end = performance.now();
console.log("cbrt took:" + (end-start),k);
k=0;
start = performance.now();
for (var i=0.0; i<10000000.0; i+=1.0)
{
var j = Math.pow(i,0.33333333);
//k+=j;
}
end = performance.now();
console.log("pow took:" + (end-start),k);
k=0;
start = performance.now();
for (var i=0.0; i<10000000.0; i+=1.0)
{
var j = Math.cbrt(i);
k+=j;
}
end = performance.now();
console.log("cbrt took:" + (end-start),k);
k=0;
start = performance.now();
for (var i=0.0; i<10000000.0; i+=1.0)
{
var j = Math.pow(i,0.33333333);
k+=j;
}
end = performance.now();
console.log("pow took:" + (end-start),k);
Result:
cbrt took:468.28200000163633 0
pow took:77.21999999921536 0
cbrt took:546.8039999977918 1615825909.5248165
pow took:869.1149999940535 1615825826.7510242
//aren't cube roots of negative numbers the same as positive, except for the sign?
Math.cubeRoot= function(n, r){
var sign= (n<0)? -1: 1;
return sign*Math.pow(Math.abs(n), 1/3);
}
Math.cubeRoot(-8)
/* returned value: (Number)
-2
*/
Just want to highlight that in ES6 there is a native cubic root function. So you can just do this (check the support here)
Math.cbrt(-8) will return you -2
this works with negative number and negative exponent:
function nthRoot(x = 0, r = 1) {
if (x < 0) {
if (r % 2 === 1) return -nthRoot(-x, r)
if (r % 2 === -1) return -1 / nthRoot(-x, -r)
}
return x ** (1 / r)
}
examples:
nthRoot( 16, 2) 4
nthRoot( 16, -2) 0.25
nthRoot(-16, 2) NaN
nthRoot(-16, -2) NaN
nthRoot( 27, 3) 3
nthRoot( 27, -3) 0.3333333333333333
nthRoot(-27, 3) -3
nthRoot(-27, -3) -0.3333333333333333
How would I go about rounded a number up the nearest multiple of 3?
i.e.
25 would return 27
1 would return 3
0 would return 3
6 would return 6
if(n > 0)
return Math.ceil(n/3.0) * 3;
else if( n < 0)
return Math.floor(n/3.0) * 3;
else
return 3;
Simply:
3.0*Math.ceil(n/3.0)
?
Here you are!
Number.prototype.roundTo = function(num) {
var resto = this%num;
if (resto <= (num/2)) {
return this-resto;
} else {
return this+num-resto;
}
}
Examples:
y = 236.32;
x = y.roundTo(10);
// results in x = 240
y = 236.32;
x = y.roundTo(5);
// results in x = 235
I'm answering this in psuedocode since I program mainly in SystemVerilog and Vera (ASIC HDL). % represents a modulus function.
round_number_up_to_nearest_divisor = number + ((divisor - (number % divisor)) % divisor)
This works in any case.
The modulus of the number calculates the remainder, subtracting that from the divisor results in the number required to get to the next divisor multiple, then the "magic" occurs. You would think that it's good enough to have the single modulus function, but in the case where the number is an exact multiple of the divisor, it calculates an extra multiple. ie, 24 would return 27. The additional modulus protects against this by making the addition 0.
As mentioned in a comment to the accepted answer, you can just use this:
Math.ceil(x/3)*3
(Even though it does not return 3 when x is 0, because that was likely a mistake by the OP.)
Out of the nine answers posted before this one (that have not been deleted or that do not have such a low score that they are not visible to all users), only the ones by Dean Nicholson (excepting the issue with loss of significance) and beauburrier are correct. The accepted answer gives the wrong result for negative numbers and it adds an exception for 0 to account for what was likely a mistake by the OP. Two other answers round a number to the nearest multiple instead of always rounding up, one more gives the wrong result for negative numbers, and three more even give the wrong result for positive numbers.
This function will round up to the nearest multiple of whatever factor you provide.
It will not round up 0 or numbers which are already multiples.
round_up = function(x,factor){ return x - (x%factor) + (x%factor>0 && factor);}
round_up(25,3)
27
round up(1,3)
3
round_up(0,3)
0
round_up(6,3)
6
The behavior for 0 is not what you asked for, but seems more consistent and useful this way. If you did want to round up 0 though, the following function would do that:
round_up = function(x,factor){ return x - (x%factor) + ( (x%factor>0 || x==0) && factor);}
round_up(25,3)
27
round up(1,3)
3
round_up(0,3)
3
round_up(6,3)
6
Building on #Makram's approach, and incorporating #Adam's subsequent comments, I've modified the original Math.prototype example such that it accurately rounds negative numbers in both zero-centric and unbiased systems:
Number.prototype.mround = function(_mult, _zero) {
var bias = _zero || false;
var base = Math.abs(this);
var mult = Math.abs(_mult);
if (bias == true) {
base = Math.round(base / mult) * _mult;
base = (this<0)?-base:base ;
} else {
base = Math.round(this / _mult) * _mult;
}
return parseFloat(base.toFixed(_mult.precision()));
}
Number.prototype.precision = function() {
if (!isFinite(this)) return 0;
var a = this, e = 1, p = 0;
while (Math.round(a * e) / e !== a) { a *= 10; p++; }
return p;
}
Examples:
(-2).mround(3) returns -3;
(0).mround(3) returns 0;
(2).mround(3) returns 3;
(25.4).mround(3) returns 24;
(15.12).mround(.1) returns 15.1
(n - n mod 3)+3
$(document).ready(function() {
var modulus = 3;
for (i=0; i < 21; i++) {
$("#results").append("<li>" + roundUp(i, modulus) + "</li>")
}
});
function roundUp(number, modulus) {
var remainder = number % modulus;
if (remainder == 0) {
return number;
} else {
return number + modulus - remainder;
}
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
Round up to nearest multiple of 3:
<ul id="results">
</ul>
A more general answer that might help somebody with a more general problem: if you want to round numbers to multiples of a fraction, consider using a library. This is a valid use case in GUI where decimals are typed into input and for instance you want to coerce them to multiples of 0.25, 0.2, 0.5 etc. Then the naive approach won't get you far:
function roundToStep(value, step) {
return Math.round(value / step) * step;
}
console.log(roundToStep(1.005, 0.01)); // 1, and should be 1.01
After hours of trying to write up my own function and looking up npm packages, I decided that Decimal.js gets the job done right away. It even has a toNearest method that does exactly that, and you can choose whether to round up, down, or to closer value (default).
const Decimal = require("decimal.js")
function roundToStep (value, step) {
return new Decimal(value).toNearest(step).toNumber();
}
console.log(roundToStep(1.005, 0.01)); // 1.01
RunKit example
Using remainder operator (modulus):
(n - 1 - (n - 1) % 3) + 3
By the code given below use can change any numbers and you can find any multiple of any number
let numbers = [8,11,15];
let multiple = 3
let result = numbers.map(myFunction);
function myFunction(n){
let answer = Math.round(n/multiple) * multiple ;
if (answer <= 0)
return multiple
else
return answer
}
console.log("Closest Multiple of " + multiple + " is " + result);
if(x%3==0)
return x
else
return ((x/3|0)+1)*3
The JavaScript Math.random() function returns a random value between 0 and 1, automatically seeded based on the current time (similar to Java I believe). However, I don't think there's any way to set you own seed for it.
How can I make a random number generator that I can provide my own seed value for, so that I can have it produce a repeatable sequence of (pseudo)random numbers?
One option is http://davidbau.com/seedrandom which is a seedable RC4-based Math.random() drop-in replacement with nice properties.
If you don't need the seeding capability just use Math.random() and build helper functions around it (eg. randRange(start, end)).
I'm not sure what RNG you're using, but it's best to know and document it so you're aware of its characteristics and limitations.
Like Starkii said, Mersenne Twister is a good PRNG, but it isn't easy to implement. If you want to do it yourself try implementing a LCG - it's very easy, has decent randomness qualities (not as good as Mersenne Twister), and you can use some of the popular constants.
EDIT: consider the great options at this answer for short seedable RNG implementations, including an LCG option.
function RNG(seed) {
// LCG using GCC's constants
this.m = 0x80000000; // 2**31;
this.a = 1103515245;
this.c = 12345;
this.state = seed ? seed : Math.floor(Math.random() * (this.m - 1));
}
RNG.prototype.nextInt = function() {
this.state = (this.a * this.state + this.c) % this.m;
return this.state;
}
RNG.prototype.nextFloat = function() {
// returns in range [0,1]
return this.nextInt() / (this.m - 1);
}
RNG.prototype.nextRange = function(start, end) {
// returns in range [start, end): including start, excluding end
// can't modulu nextInt because of weak randomness in lower bits
var rangeSize = end - start;
var randomUnder1 = this.nextInt() / this.m;
return start + Math.floor(randomUnder1 * rangeSize);
}
RNG.prototype.choice = function(array) {
return array[this.nextRange(0, array.length)];
}
var rng = new RNG(20);
for (var i = 0; i < 10; i++)
console.log(rng.nextRange(10, 50));
var digits = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9'];
for (var i = 0; i < 10; i++)
console.log(rng.choice(digits));
If you want to be able to specify the seed, you just need to replace the calls to getSeconds() and getMinutes(). You could pass in an int and use half of it mod 60 for the seconds value and the other half modulo 60 to give you the other part.
That being said, this method looks like garbage. Doing proper random number generation is very hard. The obvious problem with this is that the random number seed is based on seconds and minutes. To guess the seed and recreate your stream of random numbers only requires trying 3600 different second and minute combinations. It also means that there are only 3600 different possible seeds. This is correctable, but I'd be suspicious of this RNG from the start.
If you want to use a better RNG, try the Mersenne Twister. It is a well tested and fairly robust RNG with a huge orbit and excellent performance.
EDIT: I really should be correct and refer to this as a Pseudo Random Number Generator or PRNG.
"Anyone who uses arithmetic methods to produce random numbers is in a state of sin."
--- John von Neumann
I use a JavaScript port of the Mersenne Twister:
https://gist.github.com/300494
It allows you to set the seed manually. Also, as mentioned in other answers, the Mersenne Twister is a really good PRNG.
The code you listed kind of looks like a Lehmer RNG. If this is the case, then 2147483647 is the largest 32-bit signed integer, 2147483647 is the largest 32-bit prime, and 48271 is a full-period multiplier that is used to generate the numbers.
If this is true, you could modify RandomNumberGenerator to take in an extra parameter seed, and then set this.seed to seed; but you'd have to be careful to make sure the seed would result in a good distribution of random numbers (Lehmer can be weird like that) -- but most seeds will be fine.
The following is a PRNG that may be fed a custom seed. Calling SeedRandom will return a random generator function. SeedRandom can be called with no arguments in order to seed the returned random function with the current time, or it can be called with either 1 or 2 non-negative inters as arguments in order to seed it with those integers. Due to float point accuracy seeding with only 1 value will only allow the generator to be initiated to one of 2^53 different states.
The returned random generator function takes 1 integer argument named limit, the limit must be in the range 1 to 4294965886, the function will return a number in the range 0 to limit-1.
function SeedRandom(state1,state2){
var mod1=4294967087
var mul1=65539
var mod2=4294965887
var mul2=65537
if(typeof state1!="number"){
state1=+new Date()
}
if(typeof state2!="number"){
state2=state1
}
state1=state1%(mod1-1)+1
state2=state2%(mod2-1)+1
function random(limit){
state1=(state1*mul1)%mod1
state2=(state2*mul2)%mod2
if(state1<limit && state2<limit && state1<mod1%limit && state2<mod2%limit){
return random(limit)
}
return (state1+state2)%limit
}
return random
}
Example use:
var generator1=SeedRandom() //Seed with current time
var randomVariable=generator1(7) //Generate one of the numbers [0,1,2,3,4,5,6]
var generator2=SeedRandom(42) //Seed with a specific seed
var fixedVariable=generator2(7) //First value of this generator will always be
//1 because of the specific seed.
This generator exhibit the following properties:
It has approximately 2^64 different possible inner states.
It has a period of approximately 2^63, plenty more than anyone will ever realistically need in a JavaScript program.
Due to the mod values being primes there is no simple pattern in the output, no matter the chosen limit. This is unlike some simpler PRNGs that exhibit some quite systematic patterns.
It discards some results in order to get a perfect distribution no matter the limit.
It is relatively slow, runs around 10 000 000 times per second on my machine.
Bonus: typescript version
If you program in Typescript, I adapted the Mersenne Twister implementation that was brought in Christoph Henkelmann's answer to this thread as a typescript class:
/**
* copied almost directly from Mersenne Twister implementation found in https://gist.github.com/banksean/300494
* all rights reserved to him.
*/
export class Random {
static N = 624;
static M = 397;
static MATRIX_A = 0x9908b0df;
/* constant vector a */
static UPPER_MASK = 0x80000000;
/* most significant w-r bits */
static LOWER_MASK = 0x7fffffff;
/* least significant r bits */
mt = new Array(Random.N);
/* the array for the state vector */
mti = Random.N + 1;
/* mti==N+1 means mt[N] is not initialized */
constructor(seed:number = null) {
if (seed == null) {
seed = new Date().getTime();
}
this.init_genrand(seed);
}
private init_genrand(s:number) {
this.mt[0] = s >>> 0;
for (this.mti = 1; this.mti < Random.N; this.mti++) {
var s = this.mt[this.mti - 1] ^ (this.mt[this.mti - 1] >>> 30);
this.mt[this.mti] = (((((s & 0xffff0000) >>> 16) * 1812433253) << 16) + (s & 0x0000ffff) * 1812433253)
+ this.mti;
/* See Knuth TAOCP Vol2. 3rd Ed. P.106 for multiplier. */
/* In the previous versions, MSBs of the seed affect */
/* only MSBs of the array mt[]. */
/* 2002/01/09 modified by Makoto Matsumoto */
this.mt[this.mti] >>>= 0;
/* for >32 bit machines */
}
}
/**
* generates a random number on [0,0xffffffff]-interval
* #private
*/
private _nextInt32():number {
var y:number;
var mag01 = new Array(0x0, Random.MATRIX_A);
/* mag01[x] = x * MATRIX_A for x=0,1 */
if (this.mti >= Random.N) { /* generate N words at one time */
var kk:number;
if (this.mti == Random.N + 1) /* if init_genrand() has not been called, */
this.init_genrand(5489);
/* a default initial seed is used */
for (kk = 0; kk < Random.N - Random.M; kk++) {
y = (this.mt[kk] & Random.UPPER_MASK) | (this.mt[kk + 1] & Random.LOWER_MASK);
this.mt[kk] = this.mt[kk + Random.M] ^ (y >>> 1) ^ mag01[y & 0x1];
}
for (; kk < Random.N - 1; kk++) {
y = (this.mt[kk] & Random.UPPER_MASK) | (this.mt[kk + 1] & Random.LOWER_MASK);
this.mt[kk] = this.mt[kk + (Random.M - Random.N)] ^ (y >>> 1) ^ mag01[y & 0x1];
}
y = (this.mt[Random.N - 1] & Random.UPPER_MASK) | (this.mt[0] & Random.LOWER_MASK);
this.mt[Random.N - 1] = this.mt[Random.M - 1] ^ (y >>> 1) ^ mag01[y & 0x1];
this.mti = 0;
}
y = this.mt[this.mti++];
/* Tempering */
y ^= (y >>> 11);
y ^= (y << 7) & 0x9d2c5680;
y ^= (y << 15) & 0xefc60000;
y ^= (y >>> 18);
return y >>> 0;
}
/**
* generates an int32 pseudo random number
* #param range: an optional [from, to] range, if not specified the result will be in range [0,0xffffffff]
* #return {number}
*/
nextInt32(range:[number, number] = null):number {
var result = this._nextInt32();
if (range == null) {
return result;
}
return (result % (range[1] - range[0])) + range[0];
}
/**
* generates a random number on [0,0x7fffffff]-interval
*/
nextInt31():number {
return (this._nextInt32() >>> 1);
}
/**
* generates a random number on [0,1]-real-interval
*/
nextNumber():number {
return this._nextInt32() * (1.0 / 4294967295.0);
}
/**
* generates a random number on [0,1) with 53-bit resolution
*/
nextNumber53():number {
var a = this._nextInt32() >>> 5, b = this._nextInt32() >>> 6;
return (a * 67108864.0 + b) * (1.0 / 9007199254740992.0);
}
}
you can than use it as follows:
var random = new Random(132);
random.nextInt32(); //return a pseudo random int32 number
random.nextInt32([10,20]); //return a pseudo random int in range [10,20]
random.nextNumber(); //return a a pseudo random number in range [0,1]
check the source for more methods.
Here is quite an effective but simple javascript PRNG function that I like to use:
// The seed is the base number that the function works off
// The modulo is the highest number that the function can return
function PRNG(seed, modulo) {
str = `${(2**31-1&Math.imul(48271,seed))/2**31}`
.split('')
.slice(-10)
.join('') % modulo
return str
}
I hope this is what you're looking for.
Thank you, #aaaaaaaaaaaa (Accepted Answer)
I really needed a good non-library solution (easier to embed)
so... i made this class to store the seed and allow a Unity-esque "Next" ... but kept the initial Integer based results
class randS {
constructor(seed=null) {
if(seed!=null) {
this.seed = seed;
} else {
this.seed = Date.now()%4645455524863;
}
this.next = this.SeedRandom(this.seed);
this.last = 0;
}
Init(seed=this.seed) {
if (seed = this.seed) {
this.next = this.SeedRandom(this.seed);
} else {
this.seed=seed;
this.next = this.SeedRandom(this.seed);
}
}
SeedRandom(state1,state2){
var mod1=4294967087;
var mod2=4294965887;
var mul1=65539;
var mul2=65537;
if(typeof state1!="number"){
state1=+new Date();
}
if(typeof state2!="number"){
state2=state1;
}
state1=state1%(mod1-1)+1;
state2=state2%(mod2-1)+1;
function random(limit){
state1=(state1*mul1)%mod1;
state2=(state2*mul2)%mod2;
if(state1<limit && state2<limit && state1<mod1%limit && state2<mod2%limit){
this.last = random;
return random(limit);
}
this.last = (state1+state2)%limit;
return (state1+state2)%limit;
}
this.last = random;
return random;
}
}
And then checked it with these... seems to work well with random (but queryable) seed value (a la Minecraft) and even stored the last value returned (if needed)
var rng = new randS(9005646549);
console.log(rng.next(20)+' '+rng.next(20)+' '+rng.next(20)+' '+rng.next(20)+' '+rng.next(20)+' '+rng.next(20)+' '+rng.next(20));
console.log(rng.next(20) + ' ' + rng.next(20) + ' ' + rng.last);
which should output (for everybody)
6 7 8 14 1 12 6
9 1 1
EDIT: I made the init() work if you ever needed to reseed, or were testing values (this was necessary in my context as well)
Note: This code was originally included in the question above. In the interests of keeping the question short and focused, I've moved it to this Community Wiki answer.
I found this code kicking around and it appears to work fine for getting a random number and then using the seed afterward but I'm not quite sure how the logic works (e.g. where the 2345678901, 48271 & 2147483647 numbers came from).
function nextRandomNumber(){
var hi = this.seed / this.Q;
var lo = this.seed % this.Q;
var test = this.A * lo - this.R * hi;
if(test > 0){
this.seed = test;
} else {
this.seed = test + this.M;
}
return (this.seed * this.oneOverM);
}
function RandomNumberGenerator(){
var d = new Date();
this.seed = 2345678901 + (d.getSeconds() * 0xFFFFFF) + (d.getMinutes() * 0xFFFF);
this.A = 48271;
this.M = 2147483647;
this.Q = this.M / this.A;
this.R = this.M % this.A;
this.oneOverM = 1.0 / this.M;
this.next = nextRandomNumber;
return this;
}
function createRandomNumber(Min, Max){
var rand = new RandomNumberGenerator();
return Math.round((Max-Min) * rand.next() + Min);
}
//Thus I can now do:
var letters = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'];
var numbers = ['1','2','3','4','5','6','7','8','9','10'];
var colors = ['red','orange','yellow','green','blue','indigo','violet'];
var first = letters[createRandomNumber(0, letters.length)];
var second = numbers[createRandomNumber(0, numbers.length)];
var third = colors[createRandomNumber(0, colors.length)];
alert("Today's show was brought to you by the letter: " + first + ", the number " + second + ", and the color " + third + "!");
/*
If I could pass my own seed into the createRandomNumber(min, max, seed);
function then I could reproduce a random output later if desired.
*/
OK, here's the solution I settled on.
First you create a seed value using the "newseed()" function. Then you pass the seed value to the "srandom()" function. Lastly, the "srandom()" function returns a pseudo random value between 0 and 1.
The crucial bit is that the seed value is stored inside an array. If it were simply an integer or float, the value would get overwritten each time the function were called, since the values of integers, floats, strings and so forth are stored directly in the stack versus just the pointers as in the case of arrays and other objects. Thus, it's possible for the value of the seed to remain persistent.
Finally, it is possible to define the "srandom()" function such that it is a method of the "Math" object, but I'll leave that up to you to figure out. ;)
Good luck!
JavaScript:
// Global variables used for the seeded random functions, below.
var seedobja = 1103515245
var seedobjc = 12345
var seedobjm = 4294967295 //0x100000000
// Creates a new seed for seeded functions such as srandom().
function newseed(seednum)
{
return [seednum]
}
// Works like Math.random(), except you provide your own seed as the first argument.
function srandom(seedobj)
{
seedobj[0] = (seedobj[0] * seedobja + seedobjc) % seedobjm
return seedobj[0] / (seedobjm - 1)
}
// Store some test values in variables.
var my_seed_value = newseed(230951)
var my_random_value_1 = srandom(my_seed_value)
var my_random_value_2 = srandom(my_seed_value)
var my_random_value_3 = srandom(my_seed_value)
// Print the values to console. Replace "WScript.Echo()" with "alert()" if inside a Web browser.
WScript.Echo(my_random_value_1)
WScript.Echo(my_random_value_2)
WScript.Echo(my_random_value_3)
Lua 4 (my personal target environment):
-- Global variables used for the seeded random functions, below.
seedobja = 1103515.245
seedobjc = 12345
seedobjm = 4294967.295 --0x100000000
-- Creates a new seed for seeded functions such as srandom().
function newseed(seednum)
return {seednum}
end
-- Works like random(), except you provide your own seed as the first argument.
function srandom(seedobj)
seedobj[1] = mod(seedobj[1] * seedobja + seedobjc, seedobjm)
return seedobj[1] / (seedobjm - 1)
end
-- Store some test values in variables.
my_seed_value = newseed(230951)
my_random_value_1 = srandom(my_seed_value)
my_random_value_2 = srandom(my_seed_value)
my_random_value_3 = srandom(my_seed_value)
-- Print the values to console.
print(my_random_value_1)
print(my_random_value_2)
print(my_random_value_3)