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.

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)})}

subzey, you are a genius. fixing and annotating now.

you also inspired me to cut 10 more bytes out:

function b(a){return a?(0|Math.random()*16).toString(16):(""+1e7+-1e3+-4e3+-8e3+-1e11).replace(/1|0/g,b)}

(0|Math.random()*16).toString(16)

can also be trimmed 4 bytes to

Math.random().toString(16)[2]

if you don't care about IE... thoughts?

@jed, Math.random() may return 0, so Math.random().toString(16)[2] may return undefined.

Also, isn't setting name of function affects the outer scope?

well, with the chance of Math.random returning a zero 1 in 2<<61, seems like the definition of an edge case. more likely that IE barfs on it.

also, we're allowing named functions because we assume that they'll be assigned externally.

@jed, thanks for your clarification about function names, I didn't know that

@subzey, i hope you use this knowledge to bring us some awesome new stuff.

""+1e7+-1e3+-4e3+-8e3+-1e11 --> [1e7]+-1e3+-4e3+-8e3+-1e11
it's а flaw of many of you.

@LeverOne, perfect! Thanks

wow, that's crazy. good work @LeverOne.

What about replacing /1|0/ with /\d/?

the 4 is hardcoded according to the spec, and the 8 is a bit of a hack... technically it should be anything from 8 to b. so replacing those would make it less compliant.

Ah I see, I missed that.

i wonder if we finally have enough bytes to make it compliant, such that the 8 can be 8, 9, a, or b...

It is enough already :)

function c(a,b){return a?(~~b+0|Math.random()*16>>b/4).toString(16):([1e7]+-1e3+-4e3+-8e3+-1e11).replace(/1|0|(8)/g,c)}

Or this, shorter:

function c(a,b){return a?(b|Math.random()*16>>b/4).toString(16):([1e7]+-1e3+-4e3+-8e3+-1e11).replace(/1|0|(8)/g,c)}

thanks again, @subzey. updated.

Line 20 of annotated.js should be 'c', not 'b'.

BTW, you're all insane. I love it! :)

good catch! for those not in the know, @broofa is a celebrity in the world of UUIDs: https://github.com/broofa/node-uuid

Maybe save 4 bytes...?
```function c(a){return a?(a^Math.random()*16>>a/4).toString(16):([1e7]+-1e3+-4e3+-8e3+-1e11).replace(/[108]/g,c)}

what the... how on earth does this work?

There are three possible values 0,1,8 for a.

For case 0: 0^Math.random()*16>>0/4, >>0/4 is actually >>0, just works as Math.floor.
The 0^ here does nothing.

For case 1: 1^Math.random()*16>>1/4, >>1/4 is still >>0.
Although 1^ here toggles the least significant bit, we can still use it to generate random numbers 0-15.

For case 8: 8^Math.random()*16>>8/4, >>8/4 turns out >>2.
The maximum number of Math.random()*16 in binary form is 1111, that means the range is 0000-1111.
By right shifting 2 bits, the maximum number of Math.random()*16>>2 is 0011, and the range, 0000-0011. What it does here is to generate random numbers 0-3.
Finally, because the most significant bit is always 0, 8^ is actually 8|, it is to ensure the range starts from 8. And now we get 8 + {0, 1, 2, 3}.

incredible work, @tsaniel!

it's amazing how this gist has evolved, from

function(){return"00000000-0000-4000-8000-000000000000".replace(/0/g,function(){return(0|Math.random()*16).toString(16)})}

to

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

while hitting all the greatest hits: recursion, coercion of a number into a string through an array (!), operator repurposing...

I love those tricks, especially the string coercion with array, though i love bitwise tricks.
By the way, i am still thinking if we can shorten /[018]/...

@tsaniel - no way, the other way round would be /[^4-]/, which is just as long.

@atk: That's true. I mean if we can do something to use /\d/ instead.

...which would require that the 4 will be left alone and enlarge the equation probably more than the 3 letters we saved.

Maybe that's the end.

I fear it is.

Once again, excellent work, @tsaniel!

Sorry for my late comment, my mobile browser doesn't work well with github

Recursion & coercion vs. loop & math: https://gist.github.com/1308368

just when you thought it couldn't get any shorter, @LeverOne sheds a byte. good job!

@LeverOne is my hero.

damnit jed, i keep finding you everywhere. thanks for this!

In Firefox you could use the single statement function body to shorten it a few bits still:

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

Shorter yet (cheats, ES6, and works only in Firefox!!!), and less leaky:

function()([1e7]+-1e3+-4e3+-8e3+-1e11).replace(/[018]/g,a=>(a^Math.random()*16>>a/4).toString(16))

Hah, since I already used ES6 and there are no variable assignments, might as well go all the way and avoid Firefox-only stuff:

()=>([1e7]+-1e3+-4e3+-8e3+-1e11).replace(/[018]/g,a=>(a^Math.random()*16>>a/4).toString(16))

This is awesome!

Here's a benchmark vs node-uuid: http://jsperf.com/node-uuid-performance/11

I was so rooting for Crazy Gist to win that one :(