Computing Stirling numbers with limited integers

A couple days ago I wrote a post about a probability problem that involved calculating Stirling numbers. There are two kinds of Stirling numbers, creatively called “Stirling numbers of the first kind” and “Stirling numbers of the second kind.” The second kind come up more often in application, and so when authors say “Stirling numbers” without qualification, they mean the second kind. That’s the convention I’ll use here.

The Stirling number S(n, k) can be computed via the following series, though we will take a different approach for reasons explained below.

S(n,k) = \frac{1}{k!}\sum_{i = 0}^k (-1)^i {k \choose i} (k-i)^n

Problem statement

This made me think of the following problem. Suppose you want to calculate Stirling numbers. You want to use integer arithmetic in order to get exact results.

And suppose you can use integers as large as you like, but you have to specify the maximum integer size before you start calculating. Imagine you have to pay $N to work with N-digit integers, so you want to not use a larger value of N than necessary.

Intermediate results

There are a couple problems with using the equation above. Direct implementation of the equation would first calculate k! S(n, k) first, then divide by k! to get S(n, k). This would be costly because it would require calculating a number k! times bigger than the desired final result. (You could refactor the equation so that there is no division by k! at the end, but this gives an expression that includes non-integer terms.)

In the earlier post, I actually needed to calculate k! S(n, k) rather than S(n, k) and so this wasn’t a problem. This brings up the second problem with the equation above. It’s an alternating series, and it’s not clear whether evaluating the series produces intermediate results that are larger than the final result.


Stirling numbers can be computed recursively using

S(n+1, k) = k S(n,k) + S(n, k-1)

with the base cases S(n, n) = 1 for n ≥ 0 and S(n, 0) = S(0, n) for n > 0. This computes Stirling numbers via a sum of smaller Stirling numbers, and so the intermediate results are no larger than the final result.

So now we’re left with the question of how big Stirling numbers can be. Before computing S(n, k) you need an upper bound on how many digits it will have.

Bounding Stirling numbers

Stirling numbers can be bound in terms of binomial coefficients.

S(n, k) \leq \frac{1}{2} \binom{n}{k}k^{n-k}

This reduces the problem to finding an upper bound on binomial coefficients. I wrote about a simple bound on binomial coefficients here.

Hypergeometric distribution symmetry

One of these days I’d like to read Feller’s probability book slowly. He often says clever things in passing that are easy to miss.

Here’s an example from Feller [1] that I overlooked until I saw it cited elsewhere.

Suppose an urn contains n marbles, n1 red and n2 black. When r marbles are drawn from the urn, the probability that precisely k of the marbles are red follows a hypergeometric distribution.

\frac{\binom{n_1}{k}\binom{n-n_1}{r-k}} {\binom{n}{r}}

Feller mentions that this expression can be rewritten as

 \frac{\binom{r}{k} \binom{n-r}{n_1 - k}} { \binom{n}{n_1} }

What just happened? The roles of the total number of red marbles n1 and the sample size r, have been swapped.

The proof is trivial—expand both expressions using factorials and observe that they’re equal—but that alone doesn’t give much insight. It just shows that two jumbles of symbols represent the same quantity.

The combinatorial interpretation is more interesting and more useful. It says that you have two ways to evaluate the probability: focusing on the sample or focusing on the population of red marbles.

The former perspective says we could multiply the number of ways to choose k marbles from the supply of n1 red marbles and rk black marbles from the supply of nn1 black marbles, then divide by the number of ways to choose a sample of r marbles from an urn of n marbles.

The latter perspective says that rather than partitioning our sample of r marbles, we could think of partitioning the population of red marbles into two groups: those included in our sample and those not in our sample.

I played around with adding color to make it easier to see how the terms move between the two expressions. I colored the number of red marbles red and the sample size blue.

\[ \frac{\binom{{\color{red} n_1}}{k}\binom{n-{\color{red} n_1}}{{\color{blue} r}-k}} {\binom{n}{{\color{blue}r}}} = \frac{\binom{{\color{blue} r}}{k} \binom{n-{\color{blue} r}}{{\color{red} n_1} - k}} { \binom{n}{{\color{red} n_1}} } \]

Moving from left to right, the red variable goes down and the blue variable goes up.

Related posts

[1] William Feller. An Introduction to Probability Theory and Its Applications. Volume 1. Third edition. Page 44.

Shuffle product

riffle shuffle

The shuffle product of two words, w1 and w2, written

w1 Ш w2,

is the set of all words formed by the letters in w1 and w2, preserving the order of each word’s letters. The name comes from the analogy with doing a riffle shuffle of two decks of cards.

For example, bcd Ш ae, the shuffle product of bcd and ae, would be all permutations of abcde in which the consonants appear in alphabetical order and the vowels are also in alphabetical order. So abecd and baecd would be included, but badec would not be because the d and c are in the wrong order.

Side note on Ш

Incidentally, the symbol for shuffle product is the Cyrillic letter sha (Ш, U+0428), the only Cyrillic letter commonly used in mathematics, at least internationally. Presumably Russian mathematicians use other Cyrillic letters, but the only Cyrillic letter an American mathematician, for example, is likely to use is Ш.

The uses of Ш that I’m aware of are the Dirac comb distribution, the Tate–Shafarevich group, and the shuffle product.

Duplicate letters

What is the shuffle product of words containing duplicate letters? For example, what about the shuffle product of bread and crumb? Each word contains an r. The shuffle product, defined above as a set, doesn’t distinguish between the two rs. But another way to define the shuffle product is as a formal sum, with coefficients that count duplicates.

Imagine coloring the letters in abc blue and the letters in cde red. Then abccde and abccde would count as two different possibilities, one with blue c followed by red c, and one the other way around. This term in the formal sum would be 2abccde, the two capturing that there are two ways to arrive at this word.

You could also have duplicate letters within a single word. So in banana, for example, you could imagine coloring each a a different color and coloring the two ns different colors.

Mathematica code

This page gives an implementation of the shuffle product in Mathematica.

shuffleW[s1_, s2_] := 
     Module[{p, tp, ord}, 
          p = Permutations@Join[1 & /@ s1, 0 & /@ s2]\[Transpose];
          tp = BitXor[p, 1];
          ord = Accumulate[p] p + (Accumulate[tp] + Length[s1]) tp;
          Outer[Part, {Join[s1, s2]}, ord, 1][[1]]\[Transpose]]

This code takes two lists of characters and returns a list of lists of characters. You can use this to compute both senses of the shuffle product. For example, let’s compute abc Ш ac.

The Mathematica command

    shuffleW[{a, b, c}, {a, c}]

returns a list of 10 lists:

    {{a, b, c, a, c}, {a, b, a, c, c}, {a, b, a, c, c}, 
     {a, a, b, c, c}, {a, a, b, c, c}, {a, a, c, b, c}, 
     {a, a, b, c, c}, {a, a, b, c, c}, {a, a, c, b, c}, 
     {a, c, a, b, c}}

If we ask for the union of the set above with Union[%] we get

    {{a, a, b, c, c}, {a, a, c, b, c}, {a, b, a, c, c}, 
     {a, b, c, a, c}, {a, c, a, b, c}}

So using the set definition, we could say

abc Ш ac = {aabcc, aacbc, abacc, abcac, acabc}.

Using the formal sum definition we could say

abc Ш ac = 4aabcc + 2aacbc + 2abacc + abcacacabc.

Related posts

Photo by Amol Tyagi on Unsplash

Reverse engineering options

There are 221,184 ways to have a Whopper. You Rule.

This weekend I saw a sign in the window of a Burger King™ that made me think of an interesting problem. If you know the number of possibilities like this, how would you reverse engineer what the options that created the possibilities?

In the example above, there are 211,184 = 213×33 possible answers, and so most likely there are 13 questions that have a yes or no answer, and 3 questions that have 3 possible answers each. But there are other possibilities in principle, a lot of other possibilities.

Maybe there are 15 questions: 1 with 4 possible answers, 12 with 2 possible answers, and 3 with 3 possible answers. Or maybe there are again 15 questions, but one of the questions has 9 answers. If there are 14 questions, maybe one question has 8 answers, or 27 answers, or maybe two questions have 6 answers, … it gets complicated.

More general problem

How would you address this kind of problem in general?

Suppose there is some list of questions Q1, Q2, …, Qk where there are qi possible answers to question Qi. Then the product of all the qi equals the total number of possible responses P to the questions. Note that we don’t know the number of questions k; all we know is the product P. Let’s call the ordered list {q1, q2, …, qk} of number of responses to questions a survey.

There are as many possible surveys as there are multiplicative partitions of P. The number of possible surveys only depends on the exponents in the prime factorization of P, not on the prime factors themselves. It would make no difference if in the example above we had 513×73 possibilities rather than 213×33 possibilities. We’d reason the same way when counting the possible surveys. The number of questions is somewhere between 1 and the sum of the exponents in the prime factorization.

We have to decide how we want to count surveys. For example, if we do have 13 questions with binary answers and 3 questions with trinary answers, does it matter which questions have three answers? If it does, we multiply the number of possibilities by the binomial coefficient C(16, 3) because we choose which of the 16 total questions have three possible answers. And then we have to reason about the case of 15 questions, 14 questions, etc.

Multiplicative partitions are complicated, and its not possible to write down simple expressions for the number of multiplicative partitions of numbers with many prime factors. But in the special case of only two distinct prime factors, i.e.

P = p_1^i \, p_2^j

then there is a relatively simple expression for the total number of possible surveys, assuming order matters [1]:

2^{i+j-1} \sum_{k=0}^j {i \choose k}{j \choose k} 2^{-k}

This works out to 3,760,128 in the example above. Curiously,

3760128 = 221184 × 17.

But note that our result does not depend on the primes in the factorization of P, only on their exponents. There are no p‘s in the summation above, only i‘s and j‘s. As we noted earlier, we’d come to the same count if we’d started with P = 513×73. The factors of 2 and 3 in our count arrive for different reasons than being the primes in the factorization of P.

Related posts

[1] A. Knopfmacher and M. E. Mays. A survey of factorization counting functions, International Journal of Number Theory. Vol. 01, No. 04, pp. 563-581 (2005)

Design of experiments and design theory

Design of experiments is a branch of statistics, and design theory is a branch of combinatorics, and yet they overlap quite a bit.

It’s hard to say precisely what design theory is, but it’s consider with whether objects can be arranged in certain ways, and if so how many ways this can be done. Design theory is pure mathematics, but it is of interest to people working in ares of applied mathematics such as coding theory and statistics.

Here’s a recap of posts I’ve written recently related to design of experiments and design theory.

Design of Experiments

A few weeks ago I wrote about fractional factorial design. Then later I wrote about response surface models. Then a diagram from central composite design, a popular design in response surface methodology, was one the diagrams in a post I wrote about visually similar diagrams from separate areas of application.

I wrote two posts about pitfalls with A/B testing. One shows how play-the-winner sequential testing has the same problems as Condorcet’s voter paradox, with the order of the tests potentially determining the final winner. More seriously, A/B testing cannot detect interaction effects which may be critical.

ANSI and Military Standards

There are several civilian and military standards related to design of experiments. The first of these was MIL-STD-105. The US military has retired this standard in favor of the civilian standard ASQ/ANSI Z1.4 which is virtually identical.

Similarly, the US military standard MIL-STD-414 was replaced by the very similar civilian standard ASQ/ANSI Z1.9. This post looks at the mean-range method for estimating variation which these two standards reference.

Design Theory

I wrote a couple posts on Room squares, one on Room squares in general and one on Thomas Room’s original design now known as a Room square. Room squares are used in tournament designs.

I wrote a couple posts about Costas arrays, an introduction and a post on creating Costas arrays in Mathematica.

Latin Squares

Latin squares and Greco-Latin squares a part of design theory and a part of design of experiments. Here are several posts on Latin and Greco-Latin squares.

The Chicken McNugget Monoid

When McDonalds first introduced Chicken McNuggets, you could buy McNuggets in boxes of 6, 9, or 20. The Chicken McNugget problem is to determine which numbers of McNuggets you can and cannot buy. A number n is a McNugget number if it is possible to buy exactly that many McNuggets (using the original boxes).

Box of 6 Chicken McNuggets

There are only finitely many non-McNugget numbers, and these are listed in OEIS sequence A065003.

Note that if you order eight boxes of 6 nuggets you get 48 nuggets, and so if you order more than eight boxes, in any combination, you get more than 48 nuggets. So the following program will compute all McNugget numbers less than 48.

    ns = set() 
    for i in range(8):
        for j in range(8):
            for k in range(8):
                n = 6*i + 9*j + 20*k
                if n <= 48:
    comp = set(range(48)) - ns

The non-McNugget numbers less than 48 are stored in comp, the set complement of ns. These numbers are

1, 2, 3, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 22, 23, 25, 28, 31, 34, 37, 43.

It turns out 48 and 49 are also McNugget numbers. You can verify this by changing “48” to “50” in the code above and looking at ns. This means that 44, 45, 46, 47, 48, and 49, a run of 6 consecutive numbers are all McNugget numbers, and so you can add boxes of 6 to these numbers to get any greater number. This shows that 43 is the largest non-McNugget number.

The set of McNugget numbers is contains 0, my favorite number of McNuggets, and is closed under addition, and so it forms a monoid.

Update: The next post generalizes this one, giving a little theory and more general code.

Source: Factoring in the Chicken McNugget monoid. arXiv:1709.01606

Efficiently testing a black box

Suppose you have a black box that takes three bits as input and produces one bit as output. You could think of the input bits as positions of toggle switches, and the output bit as a light attached to the box that is either on or off.

Full factorial design

Now suppose that only one combination of 3 bits produces a successful output. There’s one way to set the switches that makes the light turn on. You can find the successful input by brute force if you test all 2³ = 8 possible inputs. In statistical lingo, you are conducting an experiment with a factorial design, i.e. you test all combinations of inputs.

See the Python code below for a text version of this design

In the chart above, each row is an experimental run and each column is a bit. I used – and + rather than 0 and 1 because it is conventional in this topic to use a minus sign to indicate that a factor is not present and a plus sign to indicate that it is present.

Fractional factorial design

Now suppose your black box takes 4 bits as inputs, but only 3 of them matter. One of the bits does nothing, but you don’t know which bit that is. You could use a factorial design again, testing all 24 = 16 possible inputs. But there’s a more clever approach that requires only 8 guesses. In statistical jargon, this is a fractional factorial design.

See the Python code below for a text version of this design

No matter which three bits the output depends on, all combinations of those three bits will be tested. Said another way, if you delete any one of the four columns, the remaining columns contain all combinations of those bits.


Now suppose your black box takes 8 bits. Again only 3 bits matter, but you don’t know which 3. How many runs do you need to do to be certain of finding the successful combination of 3 bits? It’s clear that you need at least 8 runs: if you know that the first three bits are the important ones, for example, you still need 8 runs. And it’s also clear that you could go back to brute force and try all 28 = 256 possible inputs, but the example above raises your hopes that you could get by with less than 256 runs. Could you still get by with 8? That’s too much to hope for, but you could use only 16 runs.

--------, +----+++, -+--+-++, ++--++--, --+-+++-, +-+-+--+, -++--+-+, +++---+-, ---+++-+, +--++-+-, -+-+-++-, ++-+---+, --++--++, +-++-+--, -++++---, ++++++++

Note that since this design works for 8 factors, it also works for fewer factors. If you had 5, 6, or 7 factors, you could use the first 5, 6, or 7 columns of the design above.

This design has some redundancy: every combination of three particular bits is tested twice. This is unfortunate in our setting because we are assuming the black box is deterministic: the right combination of switches will always turn the light on. But what if the right combination of switches probably turns on the light? Then redundancy is a good thing. If there’s an 80% chance that the right combination will work, then there’s a 96% chance that at least one of the two tests of the right combination will work.

Fractional factorial experiment designs are usually used in with the assumption that there are random effects, and so redundancy is a good thing.

You want to test each main effect, i.e. each single bit, and interaction effects, i.e. combinations of bits, such as pairs of bits or triples of bits. But you assume that not all possible interactions are important; otherwise you’d need a full factorial design. You typically hit diminishing return with interactions quickly: pairs of effects are often important, combinations of three effects are less common, and rarely would an experiment consider forth order interactions.

If only main effects and pairs of main effects matter, and you have a moderately large number of factors n, a fractional factorial design can let you use a lot less than 2n runs while giving you as many replications of main and interaction effects as you want.


The following Python code verifies that the designs above have the claimed properties.

    import numpy as np
    from itertools import combinations
    def verify(matrix, k):
        "verify that ever choice of k columns has 2^k unique rows"
        nrows, ncols = matrix.shape
        for (a, b, c) in combinations(range(ncols), k):
            s = set()
            for r in range(nrows):
                s.add((matrix[r,a], matrix[r,b], matrix[r,c]))
            if len(s) != 2**k:
                print("problem with columns", a, b, c)
                print("number of unique elements: ", len(s))
                print("should be", 2**k)
    m = [
        [-1, -1, -1, -1],
        [-1, -1, +1, +1],
        [-1, +1, -1, +1],
        [-1, +1, +1, -1],
        [+1, -1, -1, +1],
        [+1, -1, +1, -1],
        [+1, +1, -1, -1],
        [+1, +1, +1, +1]
    verify(np.matrix(m), 3)
    m = [
        [-1, -1, -1, -1, -1, -1, -1, -1],
        [+1, -1, -1, -1, -1, +1, +1, +1],
        [-1, +1, -1, -1, +1, -1, +1, +1],
        [+1, +1, -1, -1, +1, +1, -1, -1],
        [-1, -1, +1, -1, +1, +1, +1, -1],
        [+1, -1, +1, -1, +1, -1, -1, +1],
        [-1, +1, +1, -1, -1, +1, -1, +1],
        [+1, +1, +1, -1, -1, -1, +1, -1],
        [-1, -1, -1, +1, +1, +1, -1, +1],
        [+1, -1, -1, +1, +1, -1, +1, -1],
        [-1, +1, -1, +1, -1, +1, +1, -1],
        [+1, +1, -1, +1, -1, -1, -1, +1],
        [-1, -1, +1, +1, -1, -1, +1, +1],
        [+1, -1, +1, +1, -1, +1, -1, -1],
        [-1, +1, +1, +1, +1, -1, -1, -1],
        [+1, +1, +1, +1, +1, +1, +1, +1],
    verify(np.matrix(m), 3)

Related services

The original Room square

A few days ago I wrote about Room squares, squares named after Thomas Room. This post will be about Room’s original square.

You could think of a Room square as a tournament design in which the rows represent locations and the columns represent rounds (or vice versa). Every team plays every other team exactly once, and only one pair of teams can play each other at the same location in the same round.

Here is an image creating from Room’s original square using the visualization approach from the earlier post.

This is not a trivial variation on the square in the earlier post. Notice that this square has a block of four contiguous squares in the middle row and middle column. The previous image does not.

Here is Room’s original square, presented as in his half-page announcement in [1] with the rows and columns labeled with binary numbers.

A plain text version of the image above is available below [2].

Room points out an interesting pattern: a cell is filled in if and only if the “dot product” of its coordinate is odd. Think of each coordinate number as three dimensional vector and take the dot product of the vector representing the row and the vector representing the column.

Room didn’t refer to dot products. In his words

(αβγ, λμν) are the coordinates of one of the points marked rs in and only if αλ + βμ + γν is odd.

Related posts

[1] T. G. Room. A new type of magic square. Mathematical Gazette. Volume 39 (1955), p. 307.

[2] Here is Room’s square as a plain text (org-mode) table.

| 001 |  12 |     |  34 |     |  56 |     |  78 |
| 010 |     |  37 |  25 |     |     |  48 |  16 |
| 011 |  47 |  15 |     |     |  38 |  26 |     |
| 100 |     |     |     |  68 |  14 |  57 |  23 |
| 101 |  58 |     |  67 |  24 |     |  13 |     |
| 110 |     |  46 |  18 |  35 |  27 |     |     |
| 111 |  36 |  28 |     |  17 |     |     |  45 |
|     | 001 | 010 | 011 | 100 | 101 | 110 | 111 |

Costas arrays

The famous n queens problem is to find a way to position n queens on a n×n chessboard so that no queen attacks any other. That is, no two queens can be in the same row, the same column, or on the same diagonal. Here’s an example solution:

Costas arrays

In this post we’re going to look at a similar problem, weakening one requirement and adding another. First we’re going to remove the requirement that no two pieces can be on the same diagonal. This turns the n queens problem into the n rooks problem because rooks cannot move diagonally.

Next, imagine running stiff wires from every rook to every other rook. We will require that no two wires have the same length and run in the same direction. in more mathematical terms, we require that the displacement vectors between all the rooks are unique.

A solution to this problem is called a Costas array.  In 1965, J. P. Costas invented what are now called Costas arrays to solve a problem in radar signal processing.

Why do we call a solution a Costas array rather than a Costas matrix? Because a matrix solution can be described by recording for each column the row number of the occupied cell. For example, we could describe the eight queens solution above as

(2, 4, 6, 8, 3, 1, 7, 5).

Here I’m numbering rows from the bottom up, like points in the first quadrant, rather than top to bottom as one does with matrices.

Note that the solution to the eight queens problem above is not a Costas array because some of the displacement vectors between queens are the same. For example, if we number queens from left to right, the displacement vectors between the first and second queens is the same as the displacement vector between the second and third queens.

Visualizing a Costas array

Here’s a visualization of a Costas array of order 5.

It’s clear from the plot that no two dots are in the same row or column, but it’s not obvious that all the connecting lines are different. To see that the lines are different, we move all the tails of the connecting lines to the origin, keeping their length and direction the same.

There are 10 colored lines in the first plot, but at first you may only see 9 in the second plot. That’s because two of the lines have the same direction but different length. If you look closely, you’ll see that there’s a short purple line on top of the long red line. That is, one line runs from the origin to (1, -1) and another from the origin to (4, -4).

Here is a visualization of a Costas array of order 9.

And here are its displacement vectors translated to the origin.

Here is a visualization of a Costas array of order 15.

And here are its displacement vectors translated to the origin.

Generating Costas arrays

There are algorithms for generating some Costas arrays, but not all. Every known algorithm [1] leaves out some solutions, and it is not know whether Costas arrays exist for some values of n.

The Costas arrays above were generated using the Lempel construction algorithm. Given a prime p [2] and a primitive root mod p [3], the following Python code will generate a Costas array of order p – 2.

    p = 11 # prime
    a = 2 # primitive element

    # verify a is a primitive element mod p
    s = {a**i % p for i in range(p)}
    assert( len(s) == p-1 )

    n = p-2
    dots = []

    for i in range(1, n+1):
        for j in range(1, n+1):
            if (pow(a, i, p) + pow(a, j, p)) % p == 1:

Related posts

[1] Strictly speaking, no scalable algorithm will enumerate all Costas arrays. You could enumerate all permutation matrices of order n and test whether each is a Costas array, but this requires generating and testing n! matrices and so is completely impractical for moderately large n.

[2] More generally, the Lempel algorithm can generate a solution of order q-2 where q is a prime power. The code above only works for primes, not prime powers. For prime powers, you have to work with finite fields of order q and the code would be more complicated.

[3] A primitive root for a finite field of order q is a generator of the multiplicative group of the field, i.e. an element x such that every non-zero element of the field is some power of x.

Balanced tournament designs

Suppose you have an even number of teams that you’d like to schedule in a Round Robin tournament. This means each team plays every other team exactly once.

Denote the number of teams as 2n. You’d like each team to play in each round, so you need n locations for the games to be played.

Each game will choose 2 teams from the set of 2n teams, so the number of games will be n(2n – 1). And this means you’ll need 2n – 1 rounds. A tournament design will be a grid with n rows, one for each location, and 2n-1 columns, one for each round.

Let’s pause there for just a second and make sure this is possible. We can certainly assign n(2n – 1) games to a grid of size n by 2n-1. But can we do it in such a way that no team needs to be in two locations at the same time? It turns out we can.

Now we’d like to introduce one more requirement: we’d like to balance the games over the locations. By that we mean we’d like a team to play no more than twice in the same location. A team has to play at least once in each location, but not every team can play twice in each location. If every team played once in each location, we’d have n² games, and if every team played twice in each location we’d have 2n² games, but

n² < n(2n – 1) < 2n²

for n greater than 1. If n = 1, this is all trivial, because in that case we would have two teams. They simply play each other and that’s the tournament!

We can’t have each team play exactly the same number of times in each location, but can we have each team play either once or twice in each location? If so, we call that a balanced tournament design.

Now the question becomes for which values of n can we have a balanced tournament design involving 2n teams. This is called a BTD(n) design.

We said above that this is possible if n = 1: two teams just play each other somewhere. For n = 2, it is not possible to have a balanced tournament design: some team will have to play all three games in the same location.

It is possible to have a balanced tournament design for n = 3. Here’s a solution:

{{af, ce, bc, de, bd}, {be, df, ad, ac, cf}, {cd, ab, ef, bf, ae}}

In fact this is the solution aside from relabeling the teams. That is, given any balanced tournament design involving six teams, there is a way to label the teams so that you have the design above. Said another way, there is one design in BTD(3) up to isomorphism.

There are balanced tournament designs for all positive n except for n = 2. And in general there are a lot of them. There are 47 designs in BTD(4) up to isomorphism [1].

Related posts

[1] The CRC Handbook of Combinatorial Designs. Colbourn and Dinitz editors. CRC Press, 1996.