问题
I am wondering how to generate a GUID given an input string, such that the same input string results in the same GUID (sort of like an MD5 hash). The problem with MD5 hashes is they just guarantee low collision rate, rather than uniqueness. Instead I would like something like this:
guid('v1.0.0') == 1231231231123123123112312312311231231231
guid('v1.0.1') == 6154716581615471658161547165816154716581
guid('v1.0.2') == 1883939319188393931918839393191883939319
How would you go about implementing this sort of thing (ideally in JavaScript)? Is it even possible to do? I am not sure where to start. Things like the uuid module don't take a seed string, and they don't let you use a custom format/alphabet.
I am not looking for the canonical UUID format, but rather a GUID, ideally one made up of just integers.
回答1:
What you would need is define a one-to-one mapping of text strings (such as "v1.0.0") onto 40 digit long strings (such as "123123..."). This is also known as a bijection, although in your case an injection (a simple one-to-one mapping from inputs to outputs, not necessarily onto) may be enough. As you note, hash functions don't necessarily ensure this mapping, but there are other possibilities, such as full-period linear congruential generators (if they take a seed that you can map one-to-one onto input string values), or other reversible functions.
However, if the set of possible input strings is larger than the set of possible output strings, then you can't map all input strings one-to-one with all output strings (without creating duplicates), due to the pigeonhole principle.
For example, you can't generally map all 120-character strings one-to-one with all 40-digit strings unless you restrict the format of the 120-character strings in some way. However, your problem of creating 40-digit output strings can be solved if you can accept limiting input strings to no more than 1040 values (about 132 bits), or if you can otherwise exploit redundancy in the input strings so that they are guaranteed to compress losslessly to 40 decimal digits (about 132 bits) or less, which may or may not be possible. See also this question.
The algorithm involves two steps:
- First, transform the string to a
BigIntby building up the string'scharCodeAt()values similarly to thestringToIntmethod given in another answer. Throw an error if anycharCodeAt()is 0x80 or greater, or if the resultingBigIntis equal to or greater thanBigInt(alphabet_length)**BigInt(output_length). - Then, transform the integer to another string by taking the mod of the
BigIntand the output alphabet's size and replacing each remainder with the corresponding character in the output alphabet, until theBigIntreaches 0.
回答2:
One approach would be to use the method from that answer:
/*
* uuid-timestamp (emitter)
* UUID v4 based on timestamp
*
* Created by tarkh
* tarkh.com (C) 2020
* https://stackoverflow.com/a/63344366/1261825
*/
const uuidEmit = () => {
// Get now time
const n = Date.now();
// Generate random
const r = Math.random(); // <- swap this
// Stringify now time and generate additional random number
const s = String(n) + String(~~(r*9e4)+1e4);
// Form UUID and return it
return `${s.slice(0,8)}-${s.slice(8,12)}-4${s.slice(12,15)}-${[8,9,'a','b'][~~(r*3)]}${s.slice(15,18)}-${s.slice(s.length-12)}`;
};
// Generate 5 UUIDs
console.log(`${uuidEmit()}
${uuidEmit()}
${uuidEmit()}
${uuidEmit()}
${uuidEmit()}`);And simply swap out the Math.random() call to a different random function which can take your seed value. (There are numerous algorithms out there for creating a seedable random method, so I won't try prescribing a particular one).
Most random seeds expect numeric, so you could convert a seed string to an integer by just adding up the character values (multiplying each by 10^position so you'll always get a unique number):
const stringToInt = str =>
Array.prototype.slice.call(str).reduce((result, char, index) => result += char.charCodeAt(0) * (10**(str.length - index)), 0);
console.log(stringToInt("v1.0.0"));
console.log(stringToInt("v1.0.1"));
console.log(stringToInt("v1.0.2"));If you want to generate the same extract string every time, you can take a similar approach to tarkh's uuidEmit() method but get rid of the bits that change:
const strToInt = str =>
Array.prototype.slice.call(str).reduce((result, char, index) => result += char.charCodeAt(0) * (10**(str.length - index)), 0);
const strToId = (str, len = 40) => {
// Generate random
const r = strToInt(str);
// Multiply the number by some things to get it to the right number of digits
const rLen = `${r}`.length; // length of r as a string
// If you want to avoid any chance of collision, you can't provide too long of a string
// If a small chance of collision is okay, you can instead just truncate the string to
// your desired length
if (rLen > len) throw new Error('String too long');
// our string length is n * (r+m) + e = len, so we'll do some math to get n and m
const mMax = 9; // maximum for the exponent, too much longer and it might be represented as an exponent. If you discover "e" showing up in your string, lower this value
let m = Math.floor(Math.min(mMax, len / rLen)); // exponent
let n = Math.floor(len / (m + rLen)); // number of times we repeat r and m
let e = len - (n * (rLen + m)); // extra to pad us to the right length
return (new Array(n)).fill(0).map((_, i) => String(r * (i * 10**m))).join('')
+ String(10**e);
};
console.log(strToId("v1.0.0"));
console.log(strToId("v1.0.1"));
console.log(strToId("v1.0.2"));
console.log(strToId("v1.0.0") === strToId("v1.0.0")); // check they are the same
console.log(strToId("v1.0.0") === strToId("v1.0.1")); // check they are differentNote, this will only work with smaller strings, (probably about 10 characters top) but it should be able to avoid all collisions. You could tweak it to handle larger strings (remove the multiplying bit from stringToInt) but then you risk collisions.
回答3:
I suggest using MD5...
Following the classic birthday problem, all things being equal, the odds of 2 people sharing a birthday out of a group of 23 people is ( see https://en.wikipedia.org/wiki/Birthday_problem )...
For estimating MD5 collisions, I'm going to simplify the birthday problem formula, erring in the favor of predicting a higher chance of a collision...
Note though that whereas in the birthday problem, a collision is a positive result, in the MD5 problem, a collision is a negative result, and therefore providing higher than expected collision odds provides a conservative estimate of the chance of a MD5 collision. Plus this higher predicted chance can in some way be considered a fudge factor for any uneven distribution in the MD5 output, although I do not believe there is anyway to quantify this without a God computer...
An MD5 hash is 16 bytes long, resulting in a range of 256^16 possible values. Assuming that the MD5 algorithm is generally uniform in its results, lets suppose we create one quadrillion (ie, a million billion or 10^15) unique strings to run through the hash algorithm. Then using the modified formula (to ease the collision calculations and to add a conservative fudge factor), the odds of a collision are...
So, after 10^15 or one quadrillion unique input strings, the estimated odds of a hash collision are on par with the odds of winning the Powerball or the Mega Millions Jackpot (which are on order of 1 in ~300,000,000 per https://www.engineeringbigdata.com/odds-winning-powerball-grand-prize-r/ ).
Note too that 256^16 is 340282366920938463463374607431768211456, which is 39 digits, falling within the desired range of 40 digits.
So, suggest using the MD5 hash ( converting to BigInt ), and if you do run into a collision, I will be more than glad to spot you a lottery ticket, just to have a chance to tap into your luck and split the proceeds...
( Note: I used https://keisan.casio.com/calculator for the calculations. )
来源：https://stackoverflow.com/questions/63503924/how-to-generate-a-guid-with-a-custom-alphabet-that-behaves-similar-to-an-md5-ha