Nth Combination

Question

Is there a direct way of getting the Nth combination of an ordered set of all combinations of nCr?

Example: I have four elements: [6, 4, 2, 1]. All the possible combinations by taking three at a time would be: [[6, 4, 2], [6, 4, 1], [6, 2, 1], [4, 2, 1]].

Is there an algorithm that would give me e.g. the 3rd answer, [6, 2, 1], in the ordered result set, without enumerating all the previous answers?

Answer 1

hi

just a rough sketch: arrange your numbers into upper triangular matrix of tuples:

A(n-1,n-1)   
Aij = [i+1, j-1]

if you traverse matrix row first, you will get combinations in increasing order for two elements. To generalize to three elements, think of your matrix rows as another triangular matrix, rather than a vector. It kind of creates a corner of a cube.

At least this is how I have would approach the problem

let me clarify this, you do not have to store the matrix, you will need to compute index. Let me work out to dimensional example, which you in principle could expand to 20 dimensions(bookkeeping may be atrocious).

ij = (i*i + i)/2 + j // ij is also the combination number
(i,j) = decompose(ij) // from ij one can recover i,j components
I = i // actual first index
J = j + 1 // actual second index

this two-dimensional example works for any number n, and you dont have to tabulate permutations.

Answer 2

One way to do it would be by using properties of bits. This still requires some enumeration, but you wouldn't have to enumerate every set.

For your example, you have 4 numbers in your set. So if you were generating all the possible combinations of 4 numbers, you could enumerate them as follows:

{6, 4, 2, 1}

0000 - {(no numbers in set)}
0001 - {1}
0010 - {2}
0011 - {2, 1}
...
1111 - {6, 4, 2, 1}

See how each "bit" corresponds to "whether that number is in your set"? We see here that there are 16 possibilities (2^4).

So now we can go through and find all of the possibilities that have only 3 bits turned on. This will tell us all of the combinations of "3" that exist:

0111 - {4, 2, 1}
1011 - {6, 2, 1}
1101 - {6, 4, 1}
1110 - {6, 4, 2}

And lets rewrite each of our binary values as decimal values:

0111 = 7
1011 = 11
1101 = 13
1110 = 14

Now that we've done that - well, you said you wanted the "3rd" enumeration. So lets look at the 3rd largest number: 11. Which has the bit pattern 1011. Which corresponds to... {6, 2, 1}

Cool!

Basically, you can use the same concept for any set. So now all we've done is translate the problem from "enumerating all the sets" to "enumerating all of the integers". This might be a lot easier for your problem.

Answer 3

Note you can generate the sequence by recursively generating all combinations with the first element, then all combinations without. In both recursive cases, you drop the first element to get all combinations from n-1 elements. In Python:

def combination(l, r):
    if r == 0:
        yield []
    elif len(l) == r:
        yield l
    else:
        for c in (combination(l[1:], r-1)):
            yield l[0:1]+c
        for c in (combination(l[1:], r)):
            yield c

Any time you're generating a sequence by making a choice like this, you can recursively generate the k_th element by counting how many elements a choice generates and comparing the count to k. If k is less than the count, you make that choice. Otherwise, subtract the count and repeat for the other possible choices you could make at that point. If there are always b choices, you can view this as generating a number in base b. The technique still works if the number of choices varies. In pseudocode (when all choices are always available):

kth(k, choicePoints)
    if choicePoints is empty
        return empty list
    for each choice in head of choicePoints:
        if k < size of choice
            return choice and kth(k, tail of choicePoints)
        else
            k -= size of choice
    signal exception: k is out-of-bounds

This gives you a 0-based index. If you want 1-based, change the comparison to k <= size of choice.

The tricky part (and what is unspecified in the pseudocode) is that the size of a choice depends on previous choices. Note the pseudocode can be used to solve a more general case than the problem.

For this specific problem, there are two choices (b= 2) and the size of the 1st choice (i.e. including the 1^st element) is given by _n-1C_r-1. Here's one implementation (which requires a suitable nCr):

def kthCombination(k, l, r):
    if r == 0:
        return []
    elif len(l) == r:
        return l
    else:
        i=nCr(len(l)-1, r-1)
        if k < i:
            return l[0:1] + kthCombination(k, l[1:], r-1)
        else:
            return kthCombination(k-i, l[1:], r)

If you reverse the order of the choices, you reverse the order of the sequence.

def reverseKthCombination(k, l, r):
    if r == 0:
        return []
    elif len(l) == r:
        return l
    else:
        i=nCr(len(l)-1, r)
        if k < i:
            return reverseKthCombination(k, l[1:], r)
        else:
            return l[0:1] + reverseKthCombination(k-i, l[1:], r-1)

Putting it to use:

>>> l = [6, 4, 2, 1]
>>> [kthCombination(k, [6, 4, 2, 1], 3) for k in range(nCr(len(l), 3)) ]
[[6, 4, 2], [6, 4, 1], [6, 2, 1], [4, 2, 1]]
>>> powOf2s=[2**i for i in range(4,-1,-1)]
>>> [sum(kthCombination(k, powOf2s, 3)) for k in range(nCr(len(powOf2s), 3))]
[28, 26, 25, 22, 21, 19, 14, 13, 11, 7]
>>> [sum(reverseKthCombination(k, powOf2s, 3)) for k in range(nCr(len(powOf2s), 3))]
[7, 11, 13, 14, 19, 21, 22, 25, 26, 28]