generate random UUIDs

README.md

UUID

Returns a random v4 UUID of the form xxxxxxxx-xxxx-4xxx-yxxx-xxxxxxxxxxxx, where each x is replaced with a random hexadecimal digit from 0 to f, and y is replaced with a random hexadecimal digit from 8 to b.

There's also @LeverOne's approach using iteration, which is one byte shorter.

annotated.js

function b(
  a                  // placeholder
){
  return a           // if the placeholder was passed, return
    ? (              // a random number from 0 to 15
      a ^            // unless b is 8,
      Math.random()  // in which case
      * 16           // a random number from
      >> a/4         // 8 to 11
      ).toString(16) // in hexadecimal
    : (              // or otherwise a concatenated string:
      [1e7] +        // 10000000 +
      -1e3 +         // -1000 +
      -4e3 +         // -4000 +
      -8e3 +         // -80000000 +
      -1e11          // -100000000000,
      ).replace(     // replacing
        /[018]/g,    // zeroes, ones, and eights with
        b            // random hex digits
      )
}

index.js

function b(a){return a?(a^Math.random()*16>>a/4).toString(16):([1e7]+-1e3+-4e3+-8e3+-1e11).replace(/[018]/g,b)}

LICENSE.txt

            DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE
                    Version 2, December 2004

 Copyright (C) 2011 Jed Schmidt <http://jed.is>

 Everyone is permitted to copy and distribute verbatim or modified
 copies of this license document, and changing it is allowed as long
 as the name is changed.

            DO WHAT THE FUCK YOU WANT TO PUBLIC LICENSE
   TERMS AND CONDITIONS FOR COPYING, DISTRIBUTION AND MODIFICATION

  0. You just DO WHAT THE FUCK YOU WANT TO.

package.json

{
  "name": "UUID",

  "description": "Generates random UUIDs",

  "keywords": [
    "UUID",
    "ID",
    "random" 
  ]
}

@jsejcksn you probably should hoist your hex table to the parent scope, to avoid recalculating every time. Also note (though you might not care, depending on your use-case) that I was deliberately avoiding loops in my version for performance reasons - using map makes for fewer lines, but much slower execution. 🙂

@mindplay-dk Thanks, it’s primarily for readability.

For max optimization, one of the other golfed versions could be used, or you could compile your code using an optimizing transpiler.

We can use coercion and exponential notation to get template string and save 7 bytes:
function(){return(""+1e7+-1e3+-4e3+-8e3+-1e11).replace(/1|0/g,function(){return(0|Math.random()*16).toString(16)})}

@ ### #

We have been using this in production on the web, and are seeing a lot of collisions with the versions using Math.random in certain browsers. If you are using any of these snippets on the web, definitely use one that uses crypto.getRandomValues! Even if you tested it on your system and did not see any collisions, some other system/browser combination may implement Math.random in a way that causes easy collisions.

It has been a huge treat to read this thread, so much positivity, creativity, knowledge and collaboration!

Hi, what is the latest version of this function?

The latest version I've got is this:

var uuid = () => ([1e7]+-1e3+-4e3+-8e3+-1e11).replace(/[018]/g,c =>(c^(window.crypto||window.msCrypto).getRandomValues(new Uint8Array(1))[0]&15>>c/4).toString(16));

Is the OP not updating the code anymore? @phyreman how does your function differ?

@RobertoMachorro Just ensuring the random numbers come from a crypto source instead of Math.random(). It's definitely slower than molasses though, lol. I think the OP stopped once it hit the point of basically not being able to golf it much more and a spec-compliant version was posted.

Ok I don't even know how, but I typed in 'var konami' with github copilot and it gave me:

var konami = { // https://gist.github.com/jed/982883
    "up": false,
    "down": false,
    "left": false,
    "right": false,
    "b": false,
    "a": false,
    "start": false,
}

how

Can anyone explain the reasoning behind what seems rather arbitrary, in the replacing only 0s, 1s, and 8s with random hex digits, and not, for example, replacing all numbers with such digits? I'm sure I missed some detail that explains it.

IIRC the spec requires those digits to be certain values.

@jed thanks for the link. I browsed through that spec to see if I could find something that jumps out. I see lots of discussion of bits in fields and such, but nothing that obviously connects digits 0, 1, and 8 to needing to be replaced. In fact, the part you linked to specifically says:

Set all the other bits to randomly (or pseudo-randomly) chosen values.

In any case, appreciate having further info. My understanding is still incomplete, but it's less incomplete now. ;-)

Oh, wait... I think I maybe get more of it now... the code has stuff like 1e3 and 8e3, which are like 1000 and 8000... but it also has 4e3... so basically it's targeting all the 1s, 0s, and 8s for replacement, but leaving the 4 in the middle alone.

That's because it's not part of the spec, it's part of the genius of @subzey. Look at the string that's being replaced

• 0 and 1 (1 is used as the head of a segment to prevent truncation),
• 8, which is special cased with the ^ operation to clamp it to the proper values, but not
• 4, which cannot be replaced because it specifies the version in the spec.

right, you got it. the 1 could be any other digit, really, it's just a placeholder, but the 8 was selected for its bitwise operation properties.

(there's a walkthrough here, fwiw)

@jed thanks! :)

I was perusing the minified source of a website to see how they did something, came across this interesting way to create a GUID, and traced it back to here. I thought it was interestingly minified in the source I was looking at (with an insanely clever minifier!). Made me laugh when I saw that the original was golfed. 😆 A+ on the golfing, BTW! Clever minifier indeed--a clever human minifier. This was really fun to mentally chew on.

For funsies, here's a rough draft on a UUIDv7 generator:

[...Date.now().toString(16).padStart(12,0)].toSpliced(8,0,'-').join``+([-7e3]+-8e3+-1e11).replace(/[018]/g,b=>(b^crypto.rng(1)[0]%16>>b/4).toString(16))