在Node.js包含范围之间生成大随机数

Question

So I'm very familiar with the good old

Math.floor(Math.random() * (max - min + 1)) + min;

and this works very nicely with small numbers, however when numbers get larger this quickly becomes biased and only returns numbers one zero below it (for ex. a random number between 0 and 1e100 will almost always (every time I've tested, so several billion times since I used a for loop to generate lots of numbers) return [x]e99 ). And yes I waited the long time for the program to generate that many numbers, twice. By this point, it would be safe to assume that the output is always [x]e99 for all practical uses.

So next I tried this

Math.floor(Math.pow(max - min + 1, Math.random())) + min;

and while that works perfectly for huge ranges it breaks for small ones. So my question is how can do both - be able to generate both small and large random numbers without any bias (or minimal bias to the point of not being noticeable)?

Note: I'm using Decimal.js to handle numbers in the range -1e2043 < x < 1e2043 but since it is the same algorithm I displayed the vanilla JavaScript forms above to prevent confusion. I can take a vanilla answer and convert it to Decimal.js without any trouble so feel free to answer with either.

Note #2: I want to even out the odds of getting large numbers. For example 1e33 should have the same odds as 1e90 in my 0-1e100 example. But at the same time I need to support smaller numbers and ranges.

Answer 1

Your Problem is Precision. That's the reason you use Decimal.js in the first place. Like every other Number in JS, Math.random() supports only 53 bit of precision (Some browser even used to create only the upper 32bit of randomness) . But your value 1e100 would need 333 bit of precision. So the lower 280 bit (~75 decimal places out of 100) are discarded in your formula.

But Decimal.js provides a random() method. Why don't you use that one?

function random(min, max){
    var delta = new Decimal(max).sub(min);
    return Decimal.random( +delta.log(10) ).mul(delta).add(min);
}

Another "problem" why you get so many values with e+99 is probability. For the range 0 .. 1e100 the probabilities to get some exponent are

e+99  => 90%, 
e+98  =>  9%,
e+97  =>  0.9%,
e+96  =>  0.09%,
e+95  =>  0.009%,
e+94  =>  0.0009%,
e+93  =>  0.00009%,
e+92  =>  0.000009%,
e+91  =>  0.0000009%,
e+90  =>  0.00000009%,
and so on

So if you generate ten billion numbers, statistically you'll get a single value up to 1e+90 . That are the odds.

I want to even out those odds for large numbers. 1e33 should have the same odds as 1e90 for example

OK, then let's generate a 10 ^random in the range min ... max .

function random2(min, max){
    var a = +Decimal.log10(min), 
        b = +Decimal.log10(max);
    //trying to deal with zero-values. 
    if(a === -Infinity && b === -Infinity) return 0;  //a random value between 0 and 0 ;)
    if(a === -Infinity) a = Math.min(0, b-53);
    if(b === -Infinity) b = Math.min(0, a-53);

    return Decimal.pow(10, Decimal.random(Math.abs(b-a)).mul(b-a).add(a) );
}

now the exponents are pretty much uniformly distributed, but the values are a bit skewed. Because 10 ¹ to 10 ^1.5 10 .. 33 has the same probability as 10 ^1.5 to 10 ² 34 .. 100

Answer 2

The issue with Math.random() * Math.pow(10, Math.floor(Math.random() * 100)); at smaller numbers is that random ranges [0, 1) , meaning that when calculating the exponent separately one needs to make sure the prefix ranges [1, 10) . Otherwise you want to calculate a number in [1eX, 1eX+1) but have eg 0.1 as prefix and end up in 1eX-1 . Here is an example, maxExp is not 100 but 10 for readability of the output but easily adjustable.

 let maxExp = 10; function differentDistributionRandom() { let exp = Math.floor(Math.random() * (maxExp + 1)) - 1; if (exp < 0) return Math.random(); else return (Math.random() * 9 + 1) * Math.pow(10, exp); } let counts = new Array(maxExp + 1).fill(0).map(e => []); for (let i = 0; i < (maxExp + 1) * 1000; i++) { let x = differentDistributionRandom(); counts[Math.max(0, Math.floor(Math.log10(x)) + 1)].push(x); } counts.forEach((e, i) => { console.log(`E: ${i - 1 < 0 ? "<0" : i - 1}, amount: ${e.length}, example: ${Number.isNaN(e[0]) ? "none" : e[0]}`); }); 

You might see the category <0 here which is hopefully what you wanted (the cutoff point is arbitrary, here [0, 1) has the same probability as [1, 10) as [10, 100) and so on, but [0.01, 0.1) is again less likely than [0.1, 1) )

If you didn't insist on base 10 you could reinterpret the pseudorandom bits from two Math.random calls as Float64 which would give a similar distribution, base 2 :

 function exponentDistribution() { let bits = [Math.random(), Math.random()]; let buffer = new ArrayBuffer(24); let view = new DataView(buffer); view.setFloat64(8, bits[0]); view.setFloat64(16, bits[1]); //alternatively all at once with setInt32 for (let i = 0; i < 4; i++) { view.setInt8(i, view.getInt8(12 + i)); view.setInt8(i + 4, view.getInt8(20 + i)); } return Math.abs(view.getFloat64(0)); } let counts = new Array(11).fill(0).map(e => []); for (let i = 0; i < (1 << 11) * 100; i++) { let x = exponentDistribution(); let exp = Math.floor(Math.log2(x)); if (exp >= -5 && exp <= 5) { counts[exp + 5].push(x); } } counts.forEach((e, i) => { console.log(`E: ${i - 5}, amount: ${e.length}, example: ${Number.isNaN(e[0]) ? "none" : e[0]}`); }); 

This one obviously is bounded by the precision ends of Float64 , there are some uneven parts of the distribution due to some details of IEEE754, eg denorms/subnorms and i did not take care of special values like Infinity . It is rather to be seen as a fun extra, a reminder of the distribution of float values. Note that the loop does 1 << 11 (2048) times a number iterations, which is about the exponent range of Float64 , 11 bit, [-1022, 1023] . That's why in the example each bucket gets approximately said number (100) hits.

Answer 3

You can create the number in increments less than Number.MAX_SAFE_INTEGER , then concatenate the generated numbers to a single string

 const r = () => Math.floor(Math.random() * Number.MAX_SAFE_INTEGER); let N = ""; for (let i = 0; i < 10; i++) N += r(); document.body.appendChild(document.createTextNode(N)); console.log(/e/.test(N));