Round-robin tournaments generated by the Circle Method have maximum carry-over

Abstract

The Circle Method is widely used in the field of sport scheduling to generate schedules for round-robin tournaments. If in such a tournament, team A played team B in its previous match and is now playing team C, team C is said to receive a carry-over effect from team B. The so-called carry-over effect value is a number that can be associated to each round-robin schedule; it represents a degree of unbalancedness of the schedule with respect to carry-over. Here, we prove that, for an even number of teams, the Circle Method generates a schedule with maximum carry-over effect value, answering an open question.

Appendix: Verification of the Bound

We prove the claim that

$$\begin{aligned} \frac{{\left( \ell - 3\right) } n^{2} - {\left( \ell ^{2} - 2 \, \ell - 13\right) } n - 12 \, \ell - 12}{{\left( \ell + 1\right) } n} > 0 \end{aligned}$$

(11)

when $\ell$ lies between 3 and $n-3$. We use the assumption that $n \ge 10$ as well.

Since n and $\ell$ are positive, it is sufficient to check that the numerator is positive:

$$\begin{aligned} (\ell - 3) n^{2} - {\left( \ell ^{2} - 2 \, \ell - 13\right) } n - 12 \, \ell - 12 > 0. \end{aligned}$$

The left hand side is a quadratic polynomial in both n and $\ell$. Given a value for n we want to know for which values of $\ell$ this expression is positive. So we rewrite the previous expression ordered by powers of $\ell$:

$$\begin{aligned} -n \ell ^2+(n^2+2n-12)\ell +(-3n^2+13n -12). \end{aligned}$$

This quadratic polynomial has a strictly negative leading coefficient, so it is positive between its zero points which are given by

$$\begin{aligned} \ell _1 = \frac{n^{2} + 2 \, n - \sqrt{n^{4} - 8 \, n^{3} + 32 \, n^{2} - 96 \, n + 144} - 12}{2 \, n},\\ \ell _2 = \frac{n^{2} + 2 \, n + \sqrt{n^{4} - 8 \, n^{3} + 32 \, n^{2} - 96 \, n + 144} - 12}{2 \, n}. \end{aligned}$$

Now the claim can be formulated as follows: the interval over $\ell$ where expression (11) is positive, contains the interval $[3, n-3]$, or equivalently, we have: $\ell _1 < 3$ and $\ell _2 > n-3$. To prove the first inequality for all values of n, it is sufficient to show that (i) $\ell _1$ is strictly increasing, and (ii) has as limit 3 when n goes to infinity. To calculate this limit, we multiply numerator and denominator by $n^2 + 2 n - 12 + \sqrt{n^4 - 8 n^3 + 32 n^2 - 96 n + 144}$, and get

$$\begin{aligned} \ell _1= & {} \frac{12 n^3 - 52 n^2+ 48 n}{2 n(n^2 + 2 n + \sqrt{n^4 - 8 n^3 + 32 n^2 - 96 n + 144} - 12)}, \\= & {} \frac{6 n^2 - 26 n+ 24 }{n^2 + 2 n + \sqrt{n^4 - 8 n^3 + 32 n^2 - 96 n + 144} - 12}.\\ \end{aligned}$$

Clearly, $\lim \nolimits _{n\rightarrow \infty } \ell _1 = 3$.

To prove that $\ell _1$ is strictly increasing, we consider the derivative of $\ell _1$ with respect to n:

$$\begin{aligned} \ell '_1 = -\frac{n^{4} - 4 \, n^{3} - {\left( n^{2} + 12\right) }\sqrt{n^{4} - 8 \, n^{3} + 32 \, n^{2} - 96 \, n + 144} + 48 \, n - 144}{2 n^{2}\sqrt{n^{4} - 8 \, n^{3} + 32 \, n^{2} - 96 \, n + 144}}. \end{aligned}$$

Notice that $n^{4} - 8 \, n^{3} + 32 \, n^{2} - 96 \, n + 144$ is always positive (indeed, this polynomial has four imaginary roots). So it is sufficient to show that the numerator is positive:

$$\begin{aligned} -n^{4} + 4 n^{3} + (n^{2} + 12)\sqrt{n^{4} - 8 \, n^{3} + 32 \, n^{2} - 96 n + 144} - 48 n + 144 > 0. \end{aligned}$$

(12)

We obtain the following series of equivalent inequalities.

$$\begin{aligned} (n^{2} + 12)\sqrt{n^4 - 8n^3 + 32n^2 - 96n + 144}> & {} n^{4} - 4 n^{3} + 48 n - 144 \Longleftrightarrow \\ (n^{2} + 12)\sqrt{n^4 - 8n^3 + 32n^2 - 96n + 144}> & {} (n^{2} + 12)( n^2 - 4n -12) \Longleftrightarrow \\ \sqrt{n^4 - 8n^3 + 32n^2 - 96n + 144}> & {} n^2 - 4n -12 \Longleftrightarrow \\ n^4 - 8n^3 + 32n^2 - 96n + 144> & {} (n^2 - 4n -12)^2 \Longleftrightarrow \\ 8n(5n-24)+(n^2 - 4n -12)^2> & {} (n^2 - 4n -12)^2 \Longleftrightarrow \\ 8n(5n-24)> & {} 0 \end{aligned}$$

This is true since we assume that $n \ge 10$; it follows that $\ell _1<3$ if $n \ge 10$.

For the other inequality $\ell _2>n-3$, it is sufficient to prove that (i) this inequality holds for $n = 8$, and (ii) that the derivative of $\ell _2$ with respect to n is greater than, or equal to, 1. This condition on the derivative implies that if n increases with 1 that $\ell _2$ increases with at least 1.

One can deduce that for $n=8$ the value of $\ell _2 \approx 8.71058907144937 > n - 3$.

Next, the derivative of $\ell _2$ equals

$$\begin{aligned} \frac{n^{4} - 4 \, n^{3} + {\left( n^{2} + 12\right) }\sqrt{n^{4} - 8 \, n^{3} + 32 \, n^{2} - 96 \, n + 144} + 48 \, n - 144}{2 n^{2} \sqrt{n^{4} - 8 \, n^{3} + 32 \, n^{2} - 96 \, n + 144}}, \end{aligned}$$

and this has to be greater than, or equal to, 1. Thus, the following inequality has to hold

$$\begin{aligned} {n^{4} - 4 \, n^{3} - {\left( n^{2} - 12\right) }\sqrt{n^{4} - 8 \, n^{3} + 32 \, n^{2} - 96 \, n + 144} + 48 \, n - 144} \ge 0. \end{aligned}$$

The calculation to show this is analogous to the previous one. Indeed, we have:

$$\begin{aligned} (n^{2} - 12)\sqrt{n^4 - 8n^3 + 32n^2 - 96n + 144}\le & {} n^{4} - 4n^3 + 48n - 144 \Longleftrightarrow \\ (n^{2} - 12)\sqrt{n^4 - 8n^3 + 32n^2 - 96n + 144}\le & {} (n^{2} - 12)(n^2 - 4n + 12) \Longleftrightarrow \\ \sqrt{n^4 - 8n^3 + 32n^2 - 96n + 144}\le & {} n^2 - 4n + 12 \Longleftrightarrow \\ n^4 - 8n^3 + 32n^2 - 96n + 144\le & {} (n^2 - 4n + 12)^2 \Longleftrightarrow \\ n^4 - 8n^3 + 32n^2 - 96n + 144\le & {} n^4 - 8n^3 + 40n^2 -96n + 144 \Longleftrightarrow \\ 0\le & {} 8n^2 \end{aligned}$$

So the claim that $\ell _2>n-3$, holds for $n \ge 10$.