Compute rank of a combination?

Question

I want to pre-compute some values for each combination in a set of combinations. For example, when choosing 3 numbers from 0 to 12, I\'ll compute some value for each one:

Answer 1

For now, I've reached a compromise: I have a 13x13x13 array which just maps to the index of the combination, taking up 13x13x13x2 bytes = 4 kilobytes (using short ints), plus the normal-sized (13 choose 3) * 2 kilobytes = 572 kilobytes, for a total of 576 kilobytes. Much better than 4 megabytes, and also faster than a rank calculation!

I did this partly cause I couldn't seem to get Moron's answer to work. Also this is more extensible - I have a case where I need combinations with repetitions, and I haven't found a way to compute the rank of those, yet.

Answer 2

What you want are called combinadics. Here's my implementation of this concept, in Python:

def nthresh(k, idx):
  """Finds the largest value m such that C(m, k) <= idx."""
  mk = k
  while ncombs(mk, k) <= idx:
    mk += 1
  return mk - 1


def idx_to_set(k, idx):
  ret = []
  for i in range(k, 0, -1):
    element = nthresh(i, idx)
    ret.append(element)
    idx -= ncombs(element, i)
  return ret


def set_to_idx(input):
  ret = 0
  for k, ck in enumerate(sorted(input)):
    ret += ncombs(ck, k + 1)
  return ret

Answer 3

Use a hash table to store the results. A decent hash function could be something like:

h(x) = (x1*p^(k - 1) + x2*p^(k - 2) + ... + xk*p^0) % pp

Where x1 ... xk are the numbers in your combination (for example (0, 1, 2) has x1 = 0, x2 = 1, x3 = 2) and p and pp are primes.

So you would store Hash[h(0, 1, 2)] = 78 and then you would retrieve it the same way.

Note: the hash table is just an array of size pp, not a dict.

Answer 4

I would suggest a specialised hash table. The hash for a combination should be the exclusive-or of the hashes for the values. Hashes for values are basically random bit-patterns.

You could code the table to cope with collisions, but it should be fairly easy to derive a minimal perfect hash scheme - one where no two three-item combinations give the same hash value, and where the hash-size and table-size are kept to a minimum.

This is basically Zobrist hashing - think of a "move" as adding or removing one item of the combination.

EDIT

The reason to use a hash table is that the lookup performance O(n) where n is the number of items in the combination (assuming no collisions). Calculating lexicographical indexes into the combinations is significantly slower, IIRC.

The downside is obviously the up-front work done to generate the table.

Answer 5

Here is a conceptual answer and a code based on how lex ordering works. (So I guess my answer is like that of "moron", except that I think that he has too few details and his links have too many.) I wrote a function unchoose(n,S) for you that works assuming that S is an ordered list subset of range(n). The idea: Either S contains 0 or it does not. If it does, remove 0 and compute the index for the remaining subset. If it does not, then it comes after the binomial(n-1,k-1) subsets that do contain 0.

def binomial(n,k):
    if n < 0 or k < 0 or k > n: return 0
    b = 1
    for i in xrange(k): b = b*(n-i)/(i+1)
    return b

def unchoose(n,S):
    k = len(S)
    if k == 0 or k == n: return 0
    j = S[0]
    if k == 1: return j
    S = [x-1 for x in S]
    if not j: return unchoose(n-1,S[1:])
    return binomial(n-1,k-1)+unchoose(n-1,S)

def choose(X,k):
    n = len(X)
    if k < 0 or k > n: return []
    if not k: return [[]]
    if k == n: return [X]
    return [X[:1] + S for S in choose(X[1:],k-1)] + choose(X[1:],k)

(n,k) = (13,3)
for S in choose(range(n),k): print unchoose(n,S),S

Now, it is also true that you can cache or hash values of both functions, binomial and unchoose. And what's nice about this is that you can compromise between precomputing everything and precomputing nothing. For instance you can precompute only for len(S) <= 3.

You can also optimize unchoose so that it adds the binomial coefficients with a loop if S[0] > 0, instead of decrementing and using tail recursion.

Answer 6

You can try using the lexicographic index of the combination. Maybe this page will help: http://saliu.com/bbs/messages/348.html

This MSDN page has more details: Generating the mth Lexicographical Element of a Mathematical Combination.

To be a bit more specific:

When treated as a tuple, you can order the combinations lexicographically.

So (0,1,2) < (0,1,3) < (0,1,4) etc.

Say you had the number 0 to n-1 and chose k out of those.

Now if the first element is zero, you know that it is one among the first n-1 choose k-1.

If the first element is 1, then it is one among the next n-2 choose k-1.

This way you can recursively compute the exact position of the given combination in the lexicographic ordering and use that to map it to your number.

This works in reverse too and the MSDN page explains how to do that.