(Fast way to) Get a combination given its position in (reverse-)lexicographic order

Solution 1:

To convert a lexicographic position to a combination:

Like the reverse problem discussed in Fast way to get a position of combination (without repetitions), a lexicographic position can be converted to a combination using combinatorial digit place values.

Let us define the tuple $(a,b,c)$ as a binary number $[0 0 0 1 1 1]$ and assign it the lexicographic order ZERO and henceforth treat each combination as a binary number.

Next we define the Least Significant Bit (LSB) and the Most Significant Bit (MSB). To be consistent with ordinary binary number representations, let us define the leftmost position as MSB and the rightmost position as LSB. Because we are picking three objects out of six, each corresponding binary tuple would have three ones and three zeros. Ones represent the objects selected and zeros represent the objects not selected.

Now define the place value to each digit. In ordinary binary numbers, digit place values start at LSB and go to MSB taking values $\{1,2,4,\cdots,2^n,\cdots \}$. Here $n$ is used as the position of the digit from LSB. Likewise, combinatorial place value is defined as $\binom{n}{r}$, where $n$ is the position of the digit from the LSB, following the binary convention. The parameter $r$ is the number of ones to the right of the digit, including its own position.

For example, $n=0$ for LSB and $n=5$ for MSB.

$r=3$ for leftmost one.

$r=1$ for rightmost one.

$r=2$ for the middle one.

Conversion From Lexicographic Position to Binary Tuple:

To convert a lexicographic position $L\_number$ to its corresponding combination, $L\_number$ is compared against the place value of the digit. The comparison starts at MSB. If the place value is less than $L\_number$, the corresponding binary number is set to one and $L\_number$ is decremented by the place value.

If place value $\ge L\_number$

• Place ONE in that position
• Update $L\_number = L\_number - place value$
• Decrement $r$ in $\binom{n}{r}$
• Compare $L\_number$ to place value at next position to right $(n = n - 1)$

If place value $< L\_number$

• Move to next position $(n = n - 1)$

$\textit{Example:}$

Find the combination of three objects from six at the lexicographic place $9$.

$L_n = 9$,

Compare: $\{ \{ L_n = 9\} \geq \binom{5}{3} = 10 \} $, Result: $FALSE$, Combination: $[ 0 . . . . . ]$, $r = 3$, $L_n = 9$

Compare: $\{\{ L_n = 9\}\geq\binom{4}{3} = 4\}$, Result: $TRUE$, Combination: $[ 0 1 . . . . ]$, $r = 3-1 = 2$, $L_n = L_n - 4 = 9-4=5$

Compare: $\{ \{ L_n = 5\}\geq\binom{3}{2} = 3 \} $, Result: $TRUE$, Combination: $[ 0 1 1 . . . ]$, $r = 2-1 = 1$, $L_n = L_n - 3 = 5-3=2$

Compare: $\{\{ L_n = 2\}\geq\binom{2}{1} = 2 \} $, Result: $TRUE$, Combination: $[ 0 1 1 1 . . ]$, $r = 1-1 = 0$, $L_n = L_n - 2 = 2-2=0$

Compare: $\{ \{ L_n = 0\}\geq\binom{1}{0} = 1 \} $, Result: $FALSE$, Combination: $[ 0 1 1 1 0 . ]$, $r = 0$, $L_n = 0$,

Compare: $\{ \{ L_n = 0\}\geq\binom{0}{0} = 1 \} $, Result: $FALSE$, Combination: $[ 0 1 1 1 0 0 ]$, $r = 1-1 = 0$, $L_n = L_n - 2 = 2-2=0$

Since the final answer is $[0 1 1 1 0 0]$, the lexicographic order 9 corresponds to combination $(c,d,e)$.

The following function returns an array of binary values, given the size of the objects $n$, the number of objects to be picked $r$ and the lexicographic order $m$.

function [out]=encode2ts(n,r,m)
%Encodes the input integer 'm' to a constant weight code of n-digits with r-ones
%Most significant digit at highest index.

out = zeros(1,n);
while (n>0)
    if (n>r & r>=0)
        y = nchoosek(n-1,r);
    else
        y = 0;
    end

    if (m>=y)
        m = m - y;
        out(n) = 1;
        r = r - 1;
    else
        out(n) = 0;
    end
    n = n - 1;
 end

Solution 2:

For the preliminaries I have to refer you to my answer to the position-finding problem.

In particular, I will use reverse-lexicographic ordering and zero-based indices because it is simpler. Transforming to one-based indices and positions with lexicographic ordering, as in your example table, can be done by replacing the input position with its distance to $\binom{n}{k}$ and by transforming the output tuple from $(i_0,\ldots,i_{k-1})$ to $(n−i_{k-1},\ldots,n−i_0)$.

To recap: A $k$-index is herein defined to be a $k$-tuple of strictly increasing nonnegative integers. For a $k$-index $I$, its zero-based position (or compressed index) $\operatorname{ordx}(I)$ is defined as the number of $k$-indices that are reverse-lexicographically smaller than $I$. Denoting $I = (i_0,\ldots,i_{k-1})$, we have worked out the explicit formula $$\operatorname{ordx}(I) = \sum_{r=1}^k\binom{i_{r-1}}{r} \tag{1}$$

Using $(1)$ and $$\operatorname{ordx}(I) < \operatorname{ordx}\bigl((0,\ldots,k-2,i_{k-1}+1)\bigr) = \binom{i_{k-1}+1}{k}$$ we can deduce $$\binom{i_{k-1}}{k} \leq \operatorname{ordx}(I) < \binom{i_{k-1}+1}{k} \tag{2}$$

Given the requirement that $k$-index elements be nonnegative and strictly increasing, we also know that $i_{k-1}\geq k-1$. Now $\binom{x}{k}$ is a degree-$k$ polynomial in $x$ with zeros $0,\ldots,k-1$ and positive leading coefficient, so $\binom{x}{k}$ is nonnegative, unbounded, and strictly monotonic for $x\geq k-1$, beginning with $\binom{k-1}{k}=0$, so there exists precisely one solution $i_{k-1}\geq k-1$ for $(2)$.

From $(1)$ and $(2)$ we can figure out an algorithm to recover $i_{k-1}$ and ultimately all of $I$ from $s = \operatorname{ordx}(I)$. Note that it requires explicit knowledge of the tuple length $k$:

Function $\operatorname{xord}(k, s)$:

• Input: tuple length $k$, zero-based position $s$.

• Output: The $k$-index $I$ such that $\operatorname{ordx}(I) = s$.

  1. If $k = 0$, return an empty tuple.
  2. Find $i$ such that $i\geq k-1$ and $b := \binom{i}{k} \leq s < \binom{i+1}{k}$.
  3. Set the $(k-1)$-index $(i_0,\ldots,i_{k-2}) = \operatorname{xord}(k-1, s-b)$.
  4. Return $(i_0,\ldots,i_{k-2},i)$.

The Python implementation below uses loops instead of function call recursion. The search for $i_{k-1}$ with suitable $\binom{i}{k}$ proceeds upward; once found, the remaining $i_r$ are found by downward search. The binomial coefficients are computed on the fly, requiring less than about $2i_{k-1}$ multiplications and divisions in total. The search for $\binom{i}{1} = s$ is shortcut to $i = s$.

In the answer to the question about finding the position $\operatorname{ordx}(I)$, I have also demonstrated a variant named $\operatorname{ords}$ which allows repeated elements, that is, combinations with replacement: Just replace every $i_r$ in the above discussion with $i_r + r$, then the latter forms a strictly increasing sequence even when the former is merely nondecreasing. Code for the corresponding inverse function $\operatorname{sord}$ is given below; and for the sake of brevity, I have implemented xord in terms of sord:

def xord(k, sk):
    """
    Inverse function of ``ordx``, given output tuple size
    and compressed index.
    """
    return [i + r for r,i in enumerate(sord(k, sk))]

Alternatively, you might implement xord like sord below, but with all output assignments of the form idx[r] = j changed to idx[r] = j + r, including the case r = 1.

def sord(k, sk):
    """
    Inverse function of ``ords``, given output tuple size
    and compressed index.
    """
    # Allocate output array. Content does not matter here, only length.
    idx = [0] * k
    if k == 0: return idx
    s = sk
    if k > 1:
        # Find j such that binomial(j+k-1,k) <= s < binomial(j+k,k)
        j = 0
        prev_b = 0
        b = 1
        while b <= s:
            # Maintain invariants: 
            # prev_b == binomial(j+k-1,k), b == binomial(j+k,k)
            prev_b = b
            j += 1
            b *= j + k
            b //= j
        b = prev_b
        # j is now the greatest index occurring in the tuple.
        # From now on, we will search backwards, decreasing j.
        for r in xrange(k-1, 1, -1):
            # b == binomial(j+r,r+1) <= s < binomial(j+r+1,r+1)
            idx[r] = j
            s -= b
            # Update to b = binomial(j+r-1,r)
            b *= r + 1
            b //= j + r
            # Find j such that binomial(j+r-1,r) <= s < binomial(j+r,r)
            while b > s:
                j -= 1
                b *= j
                b //= j + r
        # Simplified handling of r = 1
        # b == binomial(j+r,r+1) <= s < binomial(j+r+1,r+1)
        idx[1] = j
        s -= b
    # Simplified handling of r = 0
    # binomial(j+r,r+1) == s iff j == s
    idx[0] = s
    return idx

If you use fixed-width integer variables, take care that the variable b has enough bits available for the intermediate multiplications.

Verifying that ordx inverts xord can be done with something like:

assert ordx([]) == 0
assert xord(0, 0) == []
for k in xrange(1, 9):
    for s in xrange(10000):
        assert ordx(xord(k, s)) == s