Efficient random permutation of n-set-bits

Question

For the problem of producing a bit-pattern with exactly n set bits, I know of two practical methods, but they both have limitations I\'m not happy with.

Fir

Answer 1

I would use solution number 3, generate the i-th permutation.
But do you need to generate the first i-1 ones?

You can do it a bit faster than that with kind of divide and conquer method proposed here: Returning i-th combination of a bit array and maybe you can improve the solution a bit

Answer 2

Background

From the formula you have given - w! / ((w-n)! * n!) it looks like your problem set has to do with the binomial coefficient which deals with calculating the number of unique combinations and not permutations which deals with duplicates in different positions.

You said:

"There is obviously a third approach of iteratively computing and counting off the legal permutations until a random index is reached, but that's simply a space-for-time trade-off on the first approach, and isn't directly helpful unless there is an efficient way to count off those n permutations.

...

This means that the random number source has been forced to produce bits to distinguish n! different results which are all equivalent. I'd like to know if there's an efficient method to avoid relying on this superfluous randomness. Perhaps by using an algorithm which produces an un-ordered list of bit positions, or by directly computing the nth unique permutation of bits."

So, there is a way to efficiently compute the nth unique combination, or rank, from the k-indexes. The k-indexes refers to a unique combination. For example, lets say that the n choose k case of 4 choose 3 is taken. This means that there are a total of 4 numbers that can be selected (0, 1, 2, 3), which is represented by n, and they are taken in groups of 3, which is represented by k. The total number of unique combinations can be calculated as n! / ((k! * (n-k)!). The rank of zero corresponds to the k-index of (2, 1, 0). Rank one is represented by the k-index group of (3, 1, 0), and so forth.

Solution

There is a formula that can be used to very efficiently translate between a k-index group and the corresponding rank without iteration. Likewise, there is a formula for translating between the rank and corresponding k-index group.

I have written a paper on this formula and how it can be seen from Pascal's Triangle. The paper is called Tablizing The Binomial Coeffieicent.

I have written a C# class which is in the public domain that implements the formula described in the paper. It uses very little memory and can be downloaded from the site. It performs the following tasks:

1. Outputs all the k-indexes in a nice format for any N choose K to a file. The K-indexes can be substituted with more descriptive strings or letters.

2. Converts the k-index to the proper lexicographic index or rank of an entry in the sorted binomial coefficient table. This technique is much faster than older published techniques that rely on iteration. It does this by using a mathematical property inherent in Pascal's Triangle and is very efficient compared to iterating over the entire set.

3. Converts the index in a sorted binomial coefficient table to the corresponding k-index. The technique used is also much faster than older iterative solutions.

4. Uses Mark Dominus method to calculate the binomial coefficient, which is much less likely to overflow and works with larger numbers. This version returns a long value. There is at least one other method that returns an int. Make sure that you use the method that returns a long value.

5. The class is written in .NET C# and provides a way to manage the objects related to the problem (if any) by using a generic list. The constructor of this class takes a bool value called InitTable that when true will create a generic list to hold the objects to be managed. If this value is false, then it will not create the table. The table does not need to be created in order to use the 4 above methods. Accessor methods are provided to access the table.

6. There is an associated test class which shows how to use the class and its methods. It has been extensively tested with at least 2 cases and there are no known bugs.

The following tested example code demonstrates how to use the class and will iterate through each unique combination:

public void Test10Choose5()
{
   String S;
   int Loop;
   int N = 10;  // Total number of elements in the set.
   int K = 5;  // Total number of elements in each group.
   // Create the bin coeff object required to get all
   // the combos for this N choose K combination.
   BinCoeff<int> BC = new BinCoeff<int>(N, K, false);
   int NumCombos = BinCoeff<int>.GetBinCoeff(N, K);
   // The Kindexes array specifies the indexes for a lexigraphic element.
   int[] KIndexes = new int[K];
   StringBuilder SB = new StringBuilder();
   // Loop thru all the combinations for this N choose K case.
   for (int Combo = 0; Combo < NumCombos; Combo++)
   {
      // Get the k-indexes for this combination.  
      BC.GetKIndexes(Combo, KIndexes);
      // Verify that the Kindexes returned can be used to retrive the
      // rank or lexigraphic order of the KIndexes in the table.
      int Val = BC.GetIndex(true, KIndexes);
      if (Val != Combo)
      {
         S = "Val of " + Val.ToString() + " != Combo Value of " + Combo.ToString();
         Console.WriteLine(S);
      }
      SB.Remove(0, SB.Length);
      for (Loop = 0; Loop < K; Loop++)
      {
         SB.Append(KIndexes[Loop].ToString());
         if (Loop < K - 1)
            SB.Append(" ");
      }
      S = "KIndexes = " + SB.ToString();
      Console.WriteLine(S);
   }
}

So, the way to apply the class to your problem is by considering each bit in the word size as the total number of items. This would be n in the n!/((k! (n - k)!) formula. To obtain k, or the group size, simply count the number of bits set to 1. You would have to create a list or array of the class objects for each possible k, which in this case would be 32. Note that the class does not handle N choose N, N choose 0, or N choose 1 so the code would have to check for those cases and return 1 for both the 32 choose 0 case and 32 choose 32 case. For 32 choose 1, it would need to return 32.

If you need to use values not much larger than 32 choose 16 (the worst case for 32 items - yields 601,080,390 unique combinations), then you can use 32 bit integers, which is how the class is currently implemented. If you need to use 64 bit integers, then you will have to convert the class to use 64 bit longs. The largest value that a long can hold is 18,446,744,073,709,551,616 which is 2 ^ 64. The worst case for n choose k when n is 64 is 64 choose 32. 64 choose 32 is 1,832,624,140,942,590,534 - so a long value will work for all 64 choose k cases. If you need numbers bigger than that, then you will probably want to look into using some sort of big integer class. In C#, the .NET framework has a BigInteger class. If you are working in a different language, it should not be hard to port.

If you are looking for a very good PRNG, one of the fastest, lightweight, and high quality output is the Tiny Mersenne Twister or TinyMT for short . I ported the code over to C++ and C#. it can be found here, along with a link to the original author's C code.

Rather than using a shuffling algorithm like Fisher-Yates, you might consider doing something like the following example instead:

// Get 7 random cards.
ulong Card;
ulong SevenCardHand = 0;
for (int CardLoop = 0; CardLoop < 7; CardLoop++)
{
  do
  {
    // The card has a value of between 0 and 51.  So, get a random value and
    // left shift it into the proper bit position.  
    Card = (1UL << RandObj.Next(CardsInDeck));
  } while ((SevenCardHand & Card) != 0);
  SevenCardHand |= Card;
}

The above code is faster than any shuffling algorithm (at least for obtaining a subset of random cards) since it only works on 7 cards instead of 52. It also packs the cards into individual bits within a single 64 bit word. It makes evaluating poker hands much more efficient as well.

As a side, note, the best binomial coefficient calculator I have found that works with very large numbers (it accurately calculated a case that yielded over 15,000 digits in the result) can be found here.

Answer 3

Seems like you want a variant of Floyd's algorithm:

Algorithm to select a single, random combination of values?

Should be especially useful in your case, because the containment test is a simple bitmask operation. This will require only k calls to the RNG. In the code below, I assume you have randint(limit) which produces a uniform random from 0 to limit-1, and that you want k bits set in a 32-bit int:

mask = 0;
for (j = 32 - k; j < 32; ++j) {
    r = randint(j+1);
    b = 1 << r;
    if (mask & b) mask |= (1 << j);
    else mask |= b;
}

How many bits of entropy you need here depends on how randint() is implemented. If k > 16, set it to 32 - k and negate the result.

Your alternative suggestion of generating a single random number representing one combination among the set (mathematicians would call this a rank of the combination) is simpler if you use colex order rather than lexicographic rank. This code, for example:

for (i = k; i >= 1; --i) {
    while ((b = binomial(n, i)) > r) --n;
    buf[i-1] = n;
    r -= b;
}

will fill the array buf[] with indices from 0 to n-1 for the k-combination at colex rank r. In your case, you'd replace buf[i-1] = n with mask |= (1 << n). The binomial() function is binomial coefficient, which I do with a lookup table (see this). That would make the most efficient use of entropy, but I still think Floyd's algorithm would be a better compromise.

Answer 4

[Expanding my comment:] If you only have a little raw entropy available, then use a PRNG to stretch it further. You only need enough raw entropy to seed a PRNG. Use the PRNG to do the actual shuffle, not the raw entropy. For the next shuffle reseed the PRNG with some more raw entropy. That spreads out the raw entropy and makes less of a demand on your entropy source.

If you know exactly the range of numbers you need out of the PRNG, then you can, carefully, set up your own LCG PRNG to cover the appropriate range while needing the minimum entropy to seed it.

ETA: In C++there is a next_permutation() method. Try using that. See std::next_permutation Implementation Explanation for more.

Answer 5

Is this a theory problem or a practical problem?

You could still do the partial shuffle, but keep track of the order of the ones and forget the zeroes. There are log(k!) bits of unused entropy in their final order for your future consumption.

You could also just use the recurrence (n choose k) = (n-1 choose k-1) + (n-1 choose k) directly. Generate a random number between 0 and (n choose k)-1. Call it r. Iterate over all of the bits from the nth to the first. If we have to set j of the i remaining bits, set the ith if r < (i-1 choose j-1) and clear it, subtracting (i-1 choose j-1), otherwise.

Practically, I wouldn't worry about the couple of words of wasted entropy from the partial shuffle; generating a random 32-bit word with 16 bits set costs somewhere between 64 and 80 bits of entropy, and that's entirely acceptable. The growth rate of the required entropy is asymptotically worse than the theoretical bound, so I'd do something different for really big words.

For really big words, you might generate n independent bits that are 1 with probability k/n. This immediately blows your entropy budget (and then some), but it only uses linearly many bits. The number of set bits is tightly concentrated around k, though. For a further expected linear entropy cost, I can fix it up. This approach has much better memory locality than the partial shuffle approach, so I'd probably prefer it in practice.