Generating non-repeating random numbers in Python

问题:

Ok this is one of those trickier than it sounds questions so I'm turning to stack overflow because I can't think of a good answer. Here is what I want: I need Python to generate a simple a list of numbers from 0 to 1,000,000,000 in random order to be used for serial numbers (using a random number so that you can't tell how many have been assigned or do timing attacks as easily, i.e. guessing the next one that will come up). These numbers are stored in a database table (indexed) along with the information linked to them. The program generating them doesn't run forever so it can't rely on internal state.

No big deal right? Just generate a list of numbers, shove them into an array and use Python "random.shuffle(big_number_array)" and we're done. Problem is I'd like to avoid having to store a list of numbers (and thus read the file, pop one off the top, save the file and close it). I'd rather generate them on the fly. Problem is that the solutions I can think of have problems:

1) Generate a random number and then check if it has already been used. If it has been used generate a new number, check, repeat as needed until I find an unused one. Problem here is that I may get unlucky and generate a lot of used numbers before getting one that is unused. Possible fix: use a very large pool of numbers to reduce the chances of this (but then I end up with silly long numbers).

2) Generate a random number and then check if it has already been used. If it has been used add or subtract one from the number and check again, keep repeating until I hit an unused number. Problem is this is no longer a random number as I have introduced bias (eventually I will get clumps of numbers and you'd be able to predict the next number with a better chance of success).

3) Generate a random number and then check if it has already been used. If it has been used add or subtract another randomly generated random number and check again, problem is we're back to simply generating random numbers and checking as in solution 1.

4) Suck it up and generate the random list and save it, have a daemon put them into a Queue so there are numbers available (and avoid constantly opening and closing a file, batching it instead).

5) Generate much larger random numbers and hash them (i.e. using MD5) to get a smaller numeric value, we should rarely get collisions, but I end up with larger than needed numbers again.

6) Prepend or append time based information to the random number (i.e. unix timestamp) to reduce chances of a collision, again I get larger numbers than I need.

Anyone have any clever ideas that will reduce the chances of a "collision" (i.e. generating a random number that is already taken) but will also allow me to keep the number "small" (i.e. less than a billion (or a thousand million for your europeans =)).

Answer and why I accepted it:

So I will simply go with 1, and hope it's not an issue, however if it is I will go with the deterministic solution of generating all the numbers and storing them so that there is a guarentee of getting a new random number, and I can use "small" numbers (i.e. 9 digits instead of an MD5/etc.).

回答1:

This is a neat problem, and I've been thinking about it for a while (with solutions similar to Sjoerd's), but in the end, here's what I think:

Use your point 1) and stop worrying.

Assuming real randomness, the probability that a random number has already been chosen before is the count of previously chosen numbers divided by the size of your pool, i.e. the maximal number.

If you say you only need a billion numbers, i.e. nine digits: Treat yourself to 3 more digits, so you have 12-digit serial numbers (that's three groups of four digits – nice and readable).

Even when you're close to having chosen a billion numbers previously, the probability that your new number is already taken is still only 0,1%.

Do step 1 and draw again. You can still check for an "infinite" loop, say don't try more than 1000 times or so, and then fallback to adding 1 (or something else).

You'll win the lottery before that fallback ever gets used.

回答2:

You could use Format-Preserving Encryption to encrypt a counter. Your counter just goes from 0 upwards, and the encryption uses a key of your choice to turn it into a seemingly random value of whatever radix and width you want.

Block ciphers normally have a fixed block size of e.g. 64 or 128 bits. But Format-Preserving Encryption allows you to take a standard cipher like AES and make a smaller-width cipher, of whatever radix and width you want (e.g. radix 10, width 9 for the parameters of the question), with an algorithm which is still cryptographically robust.

It is guaranteed to never have collisions (because cryptographic algorithms create a 1:1 mapping). It is also reversible (a 2-way mapping), so you can take the resulting number and get back to the counter value you started with.

AES-FFX is one proposed standard method to achieve this.

I've experimented with some basic Python code for AES-FFX--see Python code here (but note that it doesn't fully comply with the AES-FFX specification). It can e.g. encrypt a counter to a random-looking 7-digit decimal number. E.g.:

0000000   0731134
0000001   6161064
0000002   8899846
0000003   9575678
0000004   3030773
0000005   2748859
0000006   5127539
0000007   1372978
0000008   3830458
0000009   7628602
0000010   6643859
0000011   2563651
0000012   9522955
0000013   9286113
0000014   5543492
0000015   3230955
...       ...

For another example in Python, using another non-AES-FFX (I think) method, see this blog post "How to Generate an Account Number" which does FPE using a Feistel cipher. It generates numbers from 0 to 2^32-1.

回答3:

With some modular arithmic and prime numbers, you can create all numbers between 0 and a big prime, out of order. If you choose your numbers carefully, the next number is hard to guess.

modulo = 87178291199 # prime
incrementor = 17180131327 # relative prime

current = 433494437 # some start value
for i in xrange(1, 100):
    print current
    current = (current + incrementor) % modulo

回答4:

If they don't have to be random, but just not obviously linear (1, 2, 3, 4, ...), then here's a simple algorithm:

Pick two prime numbers. One of them will be the largest number you can generate, so it should be around one billion. The other should be fairly large.

max_value = 795028841
step = 360287471
previous_serial = 0
for i in xrange(0, max_value):
    previous_serial += step
    previous_serial %= max_value
    print "Serial: %09i" % previous_serial

Just store the previous serial each time so you know where you left off. I can't prove mathmatically that this works (been too long since those particular classes), but it's demonstrably correct with smaller primes:

s = set()
with open("test.txt", "w+") as f:
    previous_serial = 0
    for i in xrange(0, 2711):
        previous_serial += 1811
        previous_serial %= 2711
        assert previous_serial not in s
        s.add(previous_serial)

You could also prove it empirically with 9-digit primes, it'd just take a bit more work (or a lot more memory).

This does mean that given a few serial numbers, it'd be possible to figure out what your values are--but with only nine digits, it's not likely that you're going for unguessable numbers anyway.

回答5:

If you don't need something cryptographically secure, but just "sufficiently obfuscated"...

Galois Fields

You could try operations in Galois Fields, e.g. GF(2)³², to map a simple incrementing counter x to a seemingly random serial number y:

x = counter_value
y = some_galois_function(x)

• Multiply by a constant
  • Inverse is to multiply by the reciprocal of the constant
• Raise to a power: xⁿ
• Reciprocal x^-1
  • Special case of raising to power n
  • It is its own inverse
• Exponentiation of a primitive element: a^x
  • Note that this doesn't have an easily-calculated inverse (discrete logarithm)
  • Ensure a is a primitive element, aka generator

Many of these operations have an inverse, which means, given your serial number, you can calculate the original counter value from which it was derived.

As for finding a library for Galois Field for Python... good question. If you don't need speed (which you wouldn't for this) then you could make your own. I haven't tried these:

• NZMATH
• Finite field Python package
• Sage, although it's a whole environment for mathematical computing, much more than just a Python library

Matrix multiplication in GF(2)

Pick a suitable 32×32 invertible matrix in GF(2), and multiply a 32-bit input counter by it. This is conceptually related to LFSR, as described in S.Lott's answer.

CRC

A related possibility is to use a CRC calculation. Based on the remainder of long-division with an irreducible polynomial in GF(2). Python code is readily available for CRCs (crcmod, pycrc), although you might want to pick a different irreducible polynomial than is normally used, for your purposes. I'm a little fuzzy on the theory, but I think a 32-bit CRC should generate a unique value for every possible combination of 4-byte inputs. Check this. It's quite easy to experimentally check this, by feeding the output back into the input, and checking that it produces a complete cycle of length 2³²-1 (zero just maps to zero). You may need to get rid of any initial/final XORs in the CRC algorithm for this check to work.

回答6:

I think you are overestimating the problems with approach 1). Unless you have hard-realtime requirements just checking by random choice terminates rather fast. The probability of needing more than a number of iterations decays exponentially. With 100M numbers outputted (10% fillfactor) you'll have one in billion chance of requiring more than 9 iterations. Even with 50% of numbers taken you'll on average need 2 iterations and have one in a billion chance of requiring more than 30 checks. Or even the extreme case where 99% of the numbers are already taken might still be reasonable - you'll average a 100 iterations and have 1 in a billion change of requiring 2062 iterations

回答7:

The standard Linear Congruential random number generator's seed sequence CANNOT repeat until the full set of numbers from the starting seed value have been generated. Then it MUST repeat precisely.

The internal seed is often large (48 or 64 bits). The generated numbers are smaller (32 bits usually) because the entire set of bits are not random. If you follow the seed values they will form a distinct non-repeating sequence.

The question is essentially one of locating a good seed that generates "enough" numbers. You can pick a seed, and generate numbers until you get back to the starting seed. That's the length of the sequence. It may be millions or billions of numbers.

There are some guidelines in Knuth for picking suitable seeds that will generate very long sequences of unique numbers.

回答8:

You can run 1) without running into the problem of too many wrong random numbers if you just decrease the random interval by one each time.

For this method to work, you will need to save the numbers already given (which you want to do anyway) and also save the quantity of numbers taken.

It is pretty obvious that, after having collected 10 numbers, your pool of possible random numbers will have been decreased by 10. Therefore, you must not choose a number between 1 and 1.000.000 but between 1 an 999.990. Of course this number is not the real number but only an index (unless the 10 numbers collected have been 999.991, 999.992, …); you’d have to count now from 1 omitting all the numbers already collected.

Of course, your algorithm should be smarter than just counting from 1 to 1.000.000 but I hope you understand the method.

I don’t like drawing random numbers until I get one which fits either. It just feels wrong.

回答9:

My solution https://github.com/glushchenko/python-unique-id, i think you should extend matrix for 1,000,000,000 variations and have fun.

回答10:

I'd rethink the problem itself... You don't seem to be doing anything sequential with the numbers... and you've got an index on the column which has them. Do they actually need to be numbers?

Consider a sha hash... you don't actually need the entire thing. Do what git or other url shortening services do, and take first 3/4/5 characters of the hash. Given that each character now has 36 possible values instead of 10, you have 2,176,782,336 combinations instead of 999,999 combinations (for six digits). Combine that with a quick check on whether the combination exists (a pure index query) and a seed like a timestamp + random number and it should do for almost any situation.

回答11:

Do you need this to be cryptographically secure or just hard to guess? How bad are collisions? Because if it needs to be cryptographically strong and have zero collisions, it is, sadly, impossible.

回答12:

I started trying to write an explanation of the approach used below, but just implementing it was easier and more accurate. This approach has the odd behavior that it gets faster the more numbers you've generated. But it works, and it doesn't require you to generate all the numbers in advance.

As a simple optimization, you could easily make this class use a probabilistic algorithm (generate a random number, and if it's not in the set of used numbers add it to the set and return it) at first, keep track of the collision rate, and switch over to the deterministic approach used here once the collision rate gets bad.

import random

class NonRepeatingRandom(object):

    def __init__(self, maxvalue):
        self.maxvalue = maxvalue
        self.used = set()

    def next(self):
        if len(self.used) >= self.maxvalue:
            raise StopIteration
        r = random.randrange(0, self.maxvalue - len(self.used))
        result = 0
        for i in range(1, r+1):
            result += 1
            while result in self.used:
                 result += 1
        self.used.add(result)
        return result

    def __iter__(self):
        return self

    def __getitem__(self):
        raise NotImplemented

    def get_all(self):
        return [i for i in self]

>>> n = NonRepeatingRandom(20)
>>> n.get_all()
[12, 14, 13, 2, 20, 4, 15, 16, 19, 1, 8, 6, 7, 9, 5, 11, 10, 3, 18, 17]

回答13:

If it is enough for you that a casual observer can't guess the next value, you can use things like a linear congruential generator or even a simple linear feedback shift register to generate the values and keep the state in the database in case you need more values. If you use these right, the values won't repeat until the end of the universe. You'll find more ideas in the list of random number generators.

If you think there might be someone who would have a serious interest to guess the next values, you can use a database sequence to count the values you generate and encrypt them with an encryption algorithm or another cryptographically strong perfect has function. However you need to take care that the encryption algorithm isn't easily breakable if one can get hold of a sequence of successive numbers you generated - a simple RSA, for instance, won't do it because of the Franklin-Reiter Related Message Attack.

回答14:

Bit late answer, but I haven't seen this suggested anywhere.

Why not use the uuid module to create globally unique identifiers

回答15:

To generate a list of totally random numbers within a defined threshold, as follows:

plist=list()
length_of_list=100
upbound=1000
lowbound=0
while len(pList)<(length_of_list):
     pList.append(rnd.randint(lowbound,upbound))
     pList=list(set(pList))

回答16:

I bumped into the same problem and opened a question with a different title before getting to this one. My solution is a random sample generator of indexes (i.e. non-repeating numbers) in the interval [0,maximal), called itersample. Here are some usage examples:

import random
generator=itersample(maximal)
another_number=generator.next() # pick the next non-repeating random number

or

import random
generator=itersample(maximal)
for random_number in generator:
    # do something with random_number
    if some_condition: # exit loop when needed
        break

itersample generates non-repeating random integers, storage need is limited to picked numbers, and the time needed to pick n numbers should be (as some tests confirm) O(n log(n)), regardelss of maximal.

Here is the code of itersample:

import random
def itersample(c): # c = upper bound of generated integers
    sampled=[]
    def fsb(a,b): # free spaces before middle of interval a,b
        fsb.idx=a+(b+1-a)/2
        fsb.last=sampled[fsb.idx]-fsb.idx if len(sampled)>0 else 0
        return fsb.last
    while len(sampled)<c:
        sample_index=random.randrange(c-len(sampled))
        a,b=0,len(sampled)-1
        if fsb(a,a)>sample_index:
            yielding=sample_index
            sampled.insert(0,yielding)
            yield yielding
        elif fsb(b,b)<sample_index+1:
            yielding=len(sampled)+sample_index
            sampled.insert(len(sampled),yielding)
            yield yielding
        else: # sample_index falls inside sampled list
            while a+1<b:
                if fsb(a,b)<sample_index+1:
                    a=fsb.idx
                else:
                    b=fsb.idx
            yielding=a+1+sample_index
            sampled.insert(a+1,yielding)
            yield yielding

回答17:

You are stating that you store the numbers in a database.

Wouldn't it then be easier to store all the numbers there, and ask the database for a random unused number? Most databases support such a request.

Examples

MySQL:

SELECT column FROM table
ORDER BY RAND()
LIMIT 1

PostgreSQL:

SELECT column FROM table
ORDER BY RANDOM()
LIMIT 1