243,583,606,221,817,150,598,111,409x more entropy
I'd recommend using crypto.randomBytes. It's not sha1, but for id purposes, it's quicker, and just as "random".
var id = crypto.randomBytes(20).toString('hex');
//=> f26d60305dae929ef8640a75e70dd78ab809cfe9
The resulting string will be twice as long as the random bytes you generate; each byte encoded to hex is 2 characters. 20 bytes will be 40 characters of hex.
Using 20 bytes, we have 256^20 or 1,461,501,637,330,902,918,203,684,832,716,283,019,655,932,542,976 unique output values. This is identical to SHA1's 160-bit (20-byte) possible outputs.
Knowing this, it's not really meaningful for us to shasum our random bytes. It's like rolling a die twice but only accepting the second roll; no matter what, you have 6 possible outcomes each roll, so the first roll is sufficient.
Why is this better?
To understand why this is better, we first have to understand how hashing functions work. Hashing functions (including SHA1) will always generate the same output if the same input is given.
Say we want to generate IDs but our random input is generated by a coin toss. We have "heads" or "tails"
% echo -n "heads" | shasum
c25dda249cdece9d908cc33adcd16aa05e20290f -
% echo -n "tails" | shasum
71ac9eed6a76a285ae035fe84a251d56ae9485a4 -
If "heads" comes up again, the SHA1 output will be the same as it was the first time
% echo -n "heads" | shasum
c25dda249cdece9d908cc33adcd16aa05e20290f -
Ok, so a coin toss is not a great random ID generator because we only have 2 possible outputs.
If we use a standard 6-sided die, we have 6 possible inputs. Guess how many possible SHA1 outputs? 6!
input => (sha1) => output
1 => 356a192b7913b04c54574d18c28d46e6395428ab
2 => da4b9237bacccdf19c0760cab7aec4a8359010b0
3 => 77de68daecd823babbb58edb1c8e14d7106e83bb
4 => 1b6453892473a467d07372d45eb05abc2031647a
5 => ac3478d69a3c81fa62e60f5c3696165a4e5e6ac4
6 => c1dfd96eea8cc2b62785275bca38ac261256e278
It's easy to delude ourselves by thinking just because the output of our function looks very random, that it is very random.
We both agree that a coin toss or a 6-sided die would make a bad random id generator, because our possible SHA1 results (the value we use for the ID) are very few. But what if we use something that has a lot more outputs? Like a timestamp with milliseconds? Or JavaScript's Math.random? Or even a combination of those two?!
Let's compute just how many unique ids we would get ...
The uniqueness of a timestamp with milliseconds
When using (new Date()).valueOf().toString(), you're getting a 13-character number (e.g., 1375369309741). However, since this a sequentially updating number (once per millisecond), the outputs are almost always the same. Let's take a look
for (var i=0; i<10; i++) {
console.log((new Date()).valueOf().toString());
}
console.log("OMG so not random");
// 1375369431838
// 1375369431839
// 1375369431839
// 1375369431839
// 1375369431839
// 1375369431839
// 1375369431839
// 1375369431839
// 1375369431840
// 1375369431840
// OMG so not random
To be fair, for comparison purposes, in a given minute (a generous operation execution time), you will have 60*1000 or 60000 uniques.
The uniqueness of Math.random
Now, when using Math.random, because of the way JavaScript represents 64-bit floating point numbers, you'll get a number with length anywhere between 13 and 24 characters long. A longer result means more digits which means more entropy. First, we need to find out which is the most probable length.
The script below will determine which length is most probable. We do this by generating 1 million random numbers and incrementing a counter based on the .length of each number.
// get distribution
var counts = [], rand, len;
for (var i=0; i<1000000; i++) {
rand = Math.random();
len = String(rand).length;
if (counts[len] === undefined) counts[len] = 0;
counts[len] += 1;
}
// calculate % frequency
var freq = counts.map(function(n) { return n/1000000 *100 });
By dividing each counter by 1 million, we get the probability of the length of number returned from Math.random.
len frequency(%)
------------------
13 0.0004
14 0.0066
15 0.0654
16 0.6768
17 6.6703
18 61.133 <- highest probability
19 28.089 <- second highest probability
20 3.0287
21 0.2989
22 0.0262
23 0.0040
24 0.0004
So, even though it's not entirely true, let's be generous and say you get a 19-character-long random output; 0.1234567890123456789. The first characters will always be 0 and ., so really we're only getting 17 random characters. This leaves us with 10^17 +1 (for possible 0; see notes below) or 100,000,000,000,000,001 uniques.
So how many random inputs can we generate?
Ok, we calculated the number of results for a millisecond timestamp and Math.random
100,000,000,000,000,001 (Math.random)
* 60,000 (timestamp)
-----------------------------
6,000,000,000,000,000,060,000
That's a single 6,000,000,000,000,000,060,000-sided die. Or, to make this number more humanly digestible, this is roughly the same number as
input outputs
------------------------------------------------------------------------------
( 1×) 6,000,000,000,000,000,060,000-sided die 6,000,000,000,000,000,060,000
(28×) 6-sided die 6,140,942,214,464,815,497,21
(72×) 2-sided coins 4,722,366,482,869,645,213,696
Sounds pretty good, right ? Well, let's find out ...
SHA1 produces a 20-byte value, with a possible 256^20 outcomes. So we're really not using SHA1 to it's full potential. Well how much are we using?
node> 6000000000000000060000 / Math.pow(256,20) * 100
A millisecond timestamp and Math.random uses only 4.11e-27 percent of SHA1's 160-bit potential!
generator sha1 potential used
-----------------------------------------------------------------------------
crypto.randomBytes(20) 100%
Date() + Math.random() 0.00000000000000000000000000411%
6-sided die 0.000000000000000000000000000000000000000000000411%
A coin 0.000000000000000000000000000000000000000000000137%
Holy cats, man! Look at all those zeroes. So how much better is crypto.randomBytes(20)? 243,583,606,221,817,150,598,111,409 times better.
Notes about the +1 and frequency of zeroes
If you're wondering about the +1, it's possible for Math.random to return a 0 which means there's 1 more possible unique result we have to account for.
Based on the discussion that happened below, I was curious about the frequency a 0 would come up. Here's a little script, random_zero.js, I made to get some data
#!/usr/bin/env node
var count = 0;
while (Math.random() !== 0) count++;
console.log(count);
Then, I ran it in 4 threads (I have a 4-core processor), appending the output to a file
$ yes | xargs -n 1 -P 4 node random_zero.js >> zeroes.txt
So it turns out that a 0 is not that hard to get. After 100 values were recorded, the average was
1 in 3,164,854,823 randoms is a 0
Cool! More research would be required to know if that number is on-par with a uniform distribution of v8's Math.random implementation
