Unique random numbers in O(1)?

Question

The problem is this: I'd like to generate unique random numbers between 0 and 1000 that never repeat (I.E. 6 doesn't come out twice), but that doesn't resort to something like an O(N) search of previous values to do it. Is this possible?

Answer 1

You can do this:

1. Create a list, 0..1000.
2. Shuffle the list. (See Fisher-Yates shuffle for a good way to do this.)
3. Return numbers in order from the shuffled list.

So this doesn't require a search of old values each time, but it still requires O(N) for the initial shuffle. But as Nils pointed out in comments, this is amortized O(1).

Answer 2

Another posibility:

You can use an array of flags. And take the next one when it;s already chosen.

But, beware after 1000 calls, the function will never end so you must make a safeguard.

Answer 3

Initialize an array of 1001 integers with the values 0-1000 and set a variable, max, to the current max index of the array (starting with 1000). Pick a random number, r, between 0 and max, swap the number at the position r with the number at position max and return the number now at position max. Decrement max by 1 and continue. When max is 0, set max back to the size of the array - 1 and start again without the need to reinitialize the array.

Update: Although I came up with this method on my own when I answered the question, after some research I realize this is a modified version of Fisher-Yates known as Durstenfeld-Fisher-Yates or Knuth-Fisher-Yates. Since the description may be a little difficult to follow, I have provided an example below (using 11 elements instead of 1001):

Array starts off with 11 elements initialized to array[n] = n, max starts off at 10:

+--+--+--+--+--+--+--+--+--+--+--+
| 0| 1| 2| 3| 4| 5| 6| 7| 8| 9|10|
+--+--+--+--+--+--+--+--+--+--+--+
                                ^
                               max

At each iteration, a random number r is selected between 0 and max, array[r] and array[max] are swapped, the new array[max] is returned, and max is decremented:

max = 10, r = 3
           +--------------------+
           v                    v
+--+--+--+--+--+--+--+--+--+--+--+
| 0| 1| 2|10| 4| 5| 6| 7| 8| 9| 3|
+--+--+--+--+--+--+--+--+--+--+--+

max = 9, r = 7
                       +-----+
                       v     v
+--+--+--+--+--+--+--+--+--+--+--+
| 0| 1| 2|10| 4| 5| 6| 9| 8| 7: 3|
+--+--+--+--+--+--+--+--+--+--+--+

max = 8, r = 1
     +--------------------+
     v                    v
+--+--+--+--+--+--+--+--+--+--+--+
| 0| 8| 2|10| 4| 5| 6| 9| 1: 7| 3|
+--+--+--+--+--+--+--+--+--+--+--+

max = 7, r = 5
                 +-----+
                 v     v
+--+--+--+--+--+--+--+--+--+--+--+
| 0| 8| 2|10| 4| 9| 6| 5: 1| 7| 3|
+--+--+--+--+--+--+--+--+--+--+--+

...

After 11 iterations, all numbers in the array have been selected, max == 0, and the array elements are shuffled:

+--+--+--+--+--+--+--+--+--+--+--+
| 4|10| 8| 6| 2| 0| 9| 5| 1| 7| 3|
+--+--+--+--+--+--+--+--+--+--+--+

At this point, max can be reset to 10 and the process can continue.

Answer 4

You could use A Linear Congruential Generator. Where m (the modulus) would be the nearest prime bigger than 1000. When you get a number out of the range, just get the next one. The sequence will only repeat once all elements have occurred, and you don't have to use a table. Be aware of the disadvantages of this generator though (including lack of randomness).

Answer 5

Use a Maximal Linear Feedback Shift Register.

It's implementable in a few lines of C and at runtime does little more than a couple test/branches, a little addition and bit shifting. It's not random, but it fools most people.

Answer 6

You could use a good pseudo-random number generator with 10 bits and throw away 1001 to 1023 leaving 0 to 1000.

From here we get the design for a 10 bit PRNG..

• 10 bits, feedback polynomial x^10 + x^7 + 1 (period 1023)

• use a Galois LFSR to get fast code

Answer 7

For low numbers like 0...1000, creating a list that contains all the numbers and shuffling it is straight forward. But if the set of numbers to draw from is very large there's another elegant way: You can build a pseudorandom permutation using a key and a cryptographic hash function. See the following C++-ish example pseudo code:

unsigned randperm(string key, unsigned bits, unsigned index) {
  unsigned half1 =  bits    / 2;
  unsigned half2 = (bits+1) / 2;
  unsigned mask1 = (1 << half1) - 1;
  unsigned mask2 = (1 << half2) - 1;
  for (int round=0; round<5; ++round) {
    unsigned temp = (index >> half1);
    temp = (temp << 4) + round;
    index ^= hash( key + "/" + int2str(temp) ) & mask1;
    index = ((index & mask2) << half1) | ((index >> half2) & mask1);
  }
  return index;
}

Here, hash is just some arbitrary pseudo random function that maps a character string to a possibly huge unsigned integer. The function randperm is a permutation of all numbers within 0...pow(2,bits)-1 assuming a fixed key. This follows from the construction because every step that changes the variable index is reversible. This is inspired by a Feistel cipher.

Answer 8

Here's some code I typed up that uses the logic of the first solution. I know this is "language agnostic" but just wanted to present this as an example in C# in case anyone is looking for a quick practical solution.

    // Initialize variables
    Random RandomClass = new Random();
    int RandArrayNum;
    int MaxNumber = 10;
    int LastNumInArray;
    int PickedNumInArray;
    int[] OrderedArray = new int[MaxNumber];      // Ordered Array - set
    int[] ShuffledArray = new int[MaxNumber];     // Shuffled Array - not set

    // Populate the Ordered Array
    for (int i = 0; i < MaxNumber; i++)                  
    {
        OrderedArray[i] = i;
        listBox1.Items.Add(OrderedArray[i]);
    }

    // Execute the Shuffle                
    for (int i = MaxNumber - 1; i > 0; i--)
    {
        RandArrayNum = RandomClass.Next(i + 1);         // Save random #
        ShuffledArray[i] = OrderedArray[RandArrayNum];  // Populting the array in reverse
        LastNumInArray = OrderedArray[i];               // Save Last Number in Test array
        PickedNumInArray = OrderedArray[RandArrayNum];  // Save Picked Random #
        OrderedArray[i] = PickedNumInArray;             // The number is now moved to the back end
        OrderedArray[RandArrayNum] = LastNumInArray;    // The picked number is moved into position
    }

    for (int i = 0; i < MaxNumber; i++)                  
    {
        listBox2.Items.Add(ShuffledArray[i]);
    }