smart way to generate unique random number

问题:

i want to generate a sequence of unique random numbers in the range of 00000001 to 99999999.

So the first one might be 00001010, the second 40002928 etc.

The easy way is to generate a random number and store it in the database, and every next time do it again and check in the database if the number already exists and if so, generate a new one, check it again, etc. But that doesn't look right, i could be regenerating a number maybe 100 times if the number of generated items gets large.

Is there a smarter way?

EDIT as allways i forgot to say WHY i wanted this, and it will probably make things clearer and maybe get an alternative, and it is: we want to generate an ordernumber for a booking, so we could just use 000001, 000002 etc. But we don't want to give the competitors a clue of how much orders are created (because it's not a high volume market, and we don't want them to know if we are on order 30 after 2 months or at order 100. So we want to have an order number which is random (yet unique)

回答1:

You could build a table with all the possible numbers in it, give the record a 'used' field.

1. Select all records that have not been 'used'
2. Pick a random number (r) between 1 and record count
3. Take record number r
4. Get your 'random value' from the record
5. Set the 'used' flag and update the db.

That should be more efficient than picking random numbers, querying the database and repeat until not found as that's just begging for an eternity for the last few values.

回答2:

You can use either an Linear Congruential Generator (LCG) or Linear Feedback Shift Register (LFSR). Google or wikipedia for more info.

Both can, with the right parameters, operate on a 'full-cycle' (or 'full period') basis so that they will generate a 'psuedo-random number' only once in a single period, and generate all numbers within the range. Both are 'weak' generators, so no good for cyptography, but perhaps 'good enough' for apparent randomness. You may have to constrain the period to work within your 'decimal' maximum as having 'binary' periods is necessary.

Update: I should add that it is not necessary to pre-calculate or pre-store previous values in any way, you only need to keep the previous seed-value (single int) and calculate 'on-demand' the next number in the sequence. Of course you can save a chain of pre-calculated numbers to your DB if desired, but it isn't necessary.

回答3:

How about creating a set all of possible numbers and simply randomising the order? You could then just pick the next number from the tail.

Each number appears only once in the set, and when you want a new one it has already been generated, so the overhead is tiny at the point at which you want one. You could do this in memory or the database of your choice. You'll just need a sensible locking strategy for pulling the next available number.

回答4:

Use Pseudo-random Number Generators.

For example - Linear Congruential Random Number Generator

(if increment and n are coprime, then code will generate all numbers from 0 to n-1):

    int seed = 1, increment = 3;
    int n = 10;

    int x = seed;
    for(int i = 0; i < n; i++)
    {
        x = (x + increment) % n;
        Console.WriteLine(x);
    }

Output: 4 7 0 3 6 9 2 5 8 1

Basic Random Number Generators

Mersenne Twister

回答5:

Using this algorithm might be suitable, though it's memory consuming: http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle Put the numbers in the array from 1 to 99999999 and do the shuffle.

回答6:

For the extremely limited size of your numbers no you cannot expect uniqueness for any type of random generation.

You are generating a 32bit integer, whereas to reach uniqueness you need a much larger number in terms around 128bit which is the size GUIDs use which are guaranteed to always be globally unique.

回答7:

In case you happen to have access to a library and you want to dig into and understand the issue well, take a look at

The Art of Computer Programming, Volume 2: Seminumerical Algorithms

by Donald E. Knuth. Chapter 3 is all about random numbers.

回答8:

You could just place your numbers in a set. If the size of the set after generation of your N numbers is too small, generate some more.

Do some trial runs. How many numbers do you have to generate on average? Try to find out an optimal solution to the tradeoff "generate too many numbers" / "check too often for duplicates". This optimal is a number M, so that after generating M numbers, your set will likely hold N unique numbers.

Oh, and M can also be calculated: If you need an extra number (your set contains N-1), then the chance of a random number already being in the set is (N-1)/R, with R being the range. I'm going crosseyed here, so you'll have to figure this out yourself (but this kinda stuff is what makes programming fun, no?).

回答9:

You could put a unique constraint on the column that contains the random number, then handle any constraint voilations by regenerating the number. I think this normally indexes the column as well so this would be faster.

You've tagged the question with C#, so I'm guessing you're using C# to generate the random number. Maybe think about getting the database to generate the random number in a stored proc, and return it.

回答10:

You could try giving writing usernames by using a starting number and an incremental number. You start at a number (say, 12000), then, for each account created, the number goes up by the incremental value.

id = startValue + (totalNumberOfAccounts * inctrementalNumber)

If incrementalNumber is a prime value, you should be able to loop around the max account value and not hit another value. This creates the illusion of a random id, but should also have very little conflicts. In the case of a conflicts, you could add a number to increase when there's a conflict, so the above code becomes. We want to handle this case, since, if we encounter one account value that is identical, when we increment, we will bump into another conflict when we increment again.

id = startValue + (totalNumberOfAccounts * inctrementalNumber) + totalConflicts

回答11:

I've had to do something like this before (create a "random looking" number for part of a URL). What I did was create a list of keys randomly generated. Each time it needed a new number it simply randomly selected a number from keys.Count and XOR the key and the given sequence number, then outputted XORed value (in base 62) prefixed with the keys index (in base 62). I also check the output to ensure it does not contain any naught words. If it does simply take the next key and have a second go. Decrypting the number is equally simple (the first digit is the index to the key to use, a simple XOR and you are done).

I like andora's answer if you are generating new numbers and might have used it had I known. However if I was to do this again I would have simply used UUIDs. Most (if not every) platform has a method for generating them and the length is just not an issue for URLs.

回答12:

You could try shuffling the set of possible values then using them sequentially.

回答13:

I like Lazarus's solution, but if you want to avoid effectively pre-allocating the space for every possible number, just store the used numbers in the table, but build an "unused numbers" list in memory by adding all possible numbers to a collection then deleting every one that's present in the database. Then select one of the remaining numbers and use that, adding it to the list in the database, obviously.

But, like I say, I like Lazaru's solution - I think that's your best bet for most scenarios.

回答14:

function getShuffledNumbers(count) {
 var shuffledNumbers = new Array();
 var choices = new Array();
 for (var i = 0; i<count; i++) {
  // choose a number between 1 and amount of numbers remaining
  choices[i] = selectedNumber = Math.ceil(Math.random()*(99999999 - i));
  // Now to figure out the number based on this selection, work backwards until
  // you figure out which choice this number WOULD have been on the first step
  for (var j = 0; j < i; j++) {
   if (choices[i - 1 - j] >= selectedNumber) {
    // This basically says "it was choice number (selectedNumber) on the last step,
    // but if it's greater than or equal to this, it must have been choice number
    // (selectedNumber + 1) on THIS step."
    selectedNumber++;
   }
  }
  shuffledNumbers[i] = selectedNumber;
 }
 return shuffledNumbers;
}

This is as fast a way I could think of and only uses memory as it needs, however if you run it all the way through it will use double as much memory because it has two arrays, choices and shuffledNumbers.

回答15:

Running a linear congruential generator once to generate each number is apt to produce rather feeble results. Running it through a number of iterations which is relatively prime to your base (100,000,000 in this case) will improve it considerably. If before reporting each output from the generator, you run it through one or more additional permutation functions, the final output will still be a duplicate-free permutation of as many numbers as you want (up to 100,000,000) but if the proper functions are chosen the result can be cryptographically strong.

回答16:

1. create and store ind db two shuffled versions(SHUFFLE_1 and SHUFFLE_2) of the interval [0..N), where N=10'000;

2. whenever a new order is created, you assign its id like this:

ORDER_FAKE_INDEX = N*SHUFFLE_1[ORDER_REAL_INDEX / N] + SHUFFLE_2[ORDER_REAL_INDEX % N]

回答17:

I also came with same kind of problem but in C#. I finally solved it. Hope it works for you also.

Suppose I need random number between 0 and some MaxValue and having a Random type object say random.

int n=0;
while(n<MaxValue)
{
   int i=0;
   i=random.Next(n,MaxValue);
   n++;
   Write.Console(i.ToString());
}

回答18:

By fallowing line we can get e.g. 6 non repetitive random numbers for range e.g. 1 to 100.

var randomNumbers = Enumerable.Range(1, 100)
    .OrderBy(n => Guid.NewGuid())
    .Take(6)
    .OrderBy(n => n);

回答19:

the stupid way: build a table to record, store all the numble first, and them ,every time the numble used, and flag it as "used"

回答20:

System.Random rnd = new System.Random();
IEnumerable<int> numbers = Enumerable.Range(0, 99999999).OrderBy(r => rnd.Next());

This gives a randomly shuffled collection of ints in your range. You can then iterate through the collection in order.

The nice part about this is that you're not actually creating the entire collection in memory.

See comments below - this will generate the entire collection in memory when you iterate to the first element.