I need to generate 26 digit numbers with Math.random, but when I use this:
Math.floor(Math.random() * 100000000000000000000000000) + 900000000000000000000000000
I gets 9.544695043285823e+26
I need to generate 26 digit numbers with Math.random, but when I use this:
Math.floor(Math.random() * 100000000000000000000000000) + 900000000000000000000000000
I gets 9.544695043285823e+26
MAX_SAFE_INTEGER
)
– CertainPerformance
Commented
Dec 22, 2018 at 8:48
9.544695043285823e+26
is a darn big number with actually 27 digits, as e+26
means that you should move the decimal separator 26 places to the right.
– Chri.s
Commented
Dec 22, 2018 at 8:57
Modern browsers support BigInt
and bigint
primitive type and we can bine it with a random generated array containing 8 bytes (the sizeof bigint
is 8 bytes (64 bits)).
BigInt
Performance WiseWe can generate a random hex string of 16 characters length and apply it directly to BigInt
:
const hexString = Array(16)
.fill()
.map(() => Math.round(Math.random() * 0xF).toString(16))
.join('');
const randomBigInt = BigInt(`0x${hexString}`);
// randomBigInt will contain a random Bigint
document.querySelector('#generate').addEventListener('click', () => {
const output = [];
let lines = 10;
do {
const hexString = Array(16)
.fill()
.map(() => Math.round(Math.random() * 0xF).toString(16))
.join('');
const number = BigInt(`0x${hexString}`);
output.push(`${
number.toString().padStart(24)
} : 0x${
hexString.padStart(16, '0')
}`);
} while (--lines > 0);
document.querySelector('#numbers').textContent = output.join('\n');
});
<button id="generate">Generate</button>
<pre id="numbers"><pre>
BigInt
from random bytes arrayIf we want to use Uint8Array
or if we want more control over the bits manipulation, we can bine Array.prototype.fill
with Array.prototype.map
to generate an array containing 8 random byte number values (beware this is around 50% slower than the above method):
const randomBytes = Array(8)
.fill()
.map(() => Math.round(Math.random() * 0xFF));
// randomBytes will contain something similar to this:
// [129, 59, 98, 222, 20, 7, 196, 244]
Then we use Array.prototype.reduce
to initialize a BigInt
of zero value and left shift each randum byte value its position X 8 bits and applying bitwise or to the current value of each reduce
iteration:
const randomBigInt = randomBytes
.reduce((n, c, i) => n | BigInt(c) << BigInt(i) * 8n, 0n);
// randomBigInt will contain a random Bigint
BigInt
values
document.querySelector('#generate').addEventListener('click', () => {
const output = [];
let lines = 10;
do {
const number = Array(8)
.fill()
.map(() => Math.round(Math.random() * 0xFF))
.reduce((n, c, i) => n | BigInt(c) << BigInt(i) * 8n, 0n);
output.push(`${
number.toString().padStart(24)
} : 0x${
number.toString(16).padStart(16, '0')
}`);
} while (--lines > 0);
document.querySelector('#numbers').textContent = output.join('\n');
});
<button id="generate">Generate</button>
<pre id="numbers"><pre>
Floating point numbers in JavaScript (and a lot of other languages) can contain only about 15.955 digits without losing precision. For bigger numbers you can look into JS libraries, or concatenate few numbers as strings. For example:
console.log( Math.random().toString().slice(2, 15) + Math.random().toString().slice(2, 15) )
If I use a 6 decillion LG about 72.9 million out of 300 but when I switch to 10 centillion it es like this 10 ^ 999 - 99999 + 26 just like how 9 - 9 - 9 - 9 - googolplex is equal to 0 googolplex