### Ranking and Unranking of Combinations and Permutations

#### by Derrick Stolee

Now that we’ve discussed a nice standard example of canonical deletion, I want to get into more complicated uses. These techniques use more complicated augmentations and deletions, and hence require a little more delicate description of orbit calculations. In order to get to those, we need to be able to store and manipulate subsets of things. Another way to think of “subsets of things” is *combinations*, especially when we know the size of the subset.

Given a set , we will discuss the *power set* of all subsets of , the -subsets , the -tuples , and the -permutations . There are multiple ways to store a given subset, tuple, or permutation, but what is more important is how to store a list of information over the *entire collection* of these objects. My favorite method is to use a *ranking function*, which assigns each object an integer which I can use as an array position. Then, an *unranking function* will invert this process.

This area has a long history, with several different methods of ranking and unranking the objects competing for dominance in several measures of quality. My favorite measure is “readability” and/or “ease of programming” but definitely “computation speed” is important, too. However, sometimes ranking is fast and unranking is slow, sometimes the ranking and unranking is fixed to a certain order, such as lexicographic order.

Today we’ll talk about several ranking and unranking procedures, and the next post will discuss how to use these ranking/unranking functions to compute orbits of these objects for a graph with symmetry.

Since we will be dealing with actual software implementations of these combinatorial objects, we shall use the following conventions:

- .
- For , is the th element of , for .

### Ranking Subsets

Subsets are some of the easiest objects to rank, since there is a very simple bijection between the subsets of and the first integers. For a subset , let be the -bit integer whose th coordinate is the indicator with value 1 if and only if . The best part about this ranking is that the integer rank *is the set, stored directly*. To test membership of in using , simply test if .

Of course, actually ranking all subsets is not recommended for large , but particularly when is near the number of bits in a standard-length integer (32 or 64, depending on your processor/language). Even using that last bit can cause sign errors if you’re not careful!

### Ranking Tuples

To rank tuples, consider the simplest way to count -tuples in : there are ways to fix the th coordinate and ways to extend the tuple to the previous positions (recall ). Therefore, we can rank a tuple by . To unrank the tuple, iteratively pull out the residue modulo and then divide by (using integer division) to find the rank of the remaining tuple of shorter length.

In the special case that , we can assign each block of bits to be the value with the same ranking function but a faster unranking procedure (it avoids integer division). In fact, you could use a similar approach for all , except there will be integers which do not correspond to tuples. This may be trivial when both and are small, but the effect grows exponentially with with base .

### Ranking Combinations

#### Lexicographic Order

The most natural way to order combinations of is lexicographically. This is the same way we order words in a dictionary. So, if for all and . Hence the first -set is and the last is . However, the set is before the set .

This ordering provides a natural ranking procedure:

.

The idea of this ranking is that we go through all of the positions and compute how many -sets start with the first positions of , then have the value , then any -set of elements from . To unrank, we find the positions in increasing order by subtracting these inner sum values, when possible.

unrank_{lex}():while:if:elsereturn

This ranking function is unfortunate that it requires the nested sums (and the binomial requires a loop to calculate, as well). By changing the order slightly, we can improve performance.

#### Co-Lex Order

Using the co-lexicographic ordering of the combinations can simplify a few things. First, the dependence on will be completely removed. Second, the ranking and unranking methods are much simpler. The co-lex order is given by comparing the elements of the sets from largest to smallest, until there is a difference. So, and .

To rank a set in co-lex, we count how many subsets appear before by using the following fact: the value can appear in the th position of only after all -subsets of appear as the first positions of a set, followed by .

To un-rank an index , we must find the most extremal way to write as a sum of binomial coefficients, placing the largest value in the top of the coefficient of largest value.

unrank_{co-lex}():while:while:return

### Ranking Permutations

You know what? I started this post two weeks ago and delayed writing because I’ve never ranked or unranked a permutation. Apparently, it’s not a simple task. See the references below by Er and Myrvold & Ruskey who present a very detailed bibliography on the techniques here. Also, see Glenn Tesler’s presentation on ranking, unranking, and finding sequential permutations.

### Signed Objects

A *signed* object is usually a pairing of a standard (unsigned) object (such as the objects above) and associating a positive or negative value to each element or position in that object. For example, a *signed permutation* is really a pair where is a bijection and is a function. One simple way to rank such objects is to take a tensor product of the rank of the unsigned object with the rank of the set of elements with negative sign. It only matters which object is “first” in this order. So one option is and another is .

### References

J. Arndt, *Matters Computational: Ideas, Algorithms, Source Code*. Available on the FXT page.

T. Johnston, Sorting Signed Permutations Using Cut-And-Paste Operations, *UNL Honors Thesis* (2009).

M. C. Er, Lexicographic Enumeration, Ranking, and Unranking of Permutations of out of Objects, *International Journal of Computer Mathematics* 23(1) (1987).

W. Myrvold, F. Ruskey, Ranking and Unranking Permutations in Linear Time, *Information Processing Letters* (2000).

Thank you so much for this nice article. One minor comment:

In unrankco-lex, right after assigning S[i] there is a line missing that should decrease the variable m by binomial(l-1,i+1).

Thanks for the tip! Yes, I did accidentally leave that out. It is in now.

Hey Derrick, can you tell me this in a simple way? For example I have a way to calculate index (rank) of Permutations without repetitions 5 from 10 elements in the following way:

first index: (1 2 3 4 5) == 0*9!/5! + 0*8!/5! + 0*7!/5! + 0*6!/5! + 0*5!/5!

== 0*3024 + 0*336 + 0*42 + 0*6 + 0*1

== 0

last index: (10 9 8 7 6) == 9*3024 + 8*336 + 7*42 + 6*6 + 5*1

== 30239

Question:

Do you have a way to do the same but with combinations without repetitions? Same elements 5 from 10. First index is 0 and last 250

Thanks!

If you use the co-lex order, then the result will be very similar, but you will use binomial coefficients instead of your n!/r! coefficients. (Also, be sure to sort your combinations!) Also, it helps if your values start at 0, not 1 (if you start at 1, then you will want to replace all instances of “” with “” in the formula below:

Index of .

Thank you for this excellent article. It has helped me immensely.

When implementing unrank_colex in C++, I had to adjust the second while loop condition as follows

while (n_choose_k(l, i + 1) = 0)

It seems like, within the inner while loop, i can at times become less than zero…

Please pardon my comment if such an implementation is already implied by your notation. (I’m fairly new to combinatorial algorithms.)

I think my initial attempt to write the condition was stripped by wordpress formatting

I added i >= 0 as a condition with && so the final inner while loop condition is

while (n_choose_k(l, i + 1) = 0)

Eh, sorry for the above comment spam. I’m seeing other strange issues now, so the issue is likeliest with my code.

Please feel free to nuke the above comments, apologies for the interruption.

Thanks again for the post though, it’s great stuff.