tournament if and only if $\sum_{i=1}^{k}s_i\geqslant\binom{k}{2}$ for each $k<M$ and $\sum_{i=1}^{M}s_i=\binom{M}{2}$ hold. A tournament is irreducible if and only if all the $M-1$ inequalities above are strict.

In order to find the irreducible components (and hence the condensation) of ${{\cal T}}$, the following very simple algorithm, having complexity $O(M^2)$, can be applied.

1.  Find the smallest $k$ such that $\sum_{i=1}^{k}s_i=\binom{k}{2}$; the sub-tournament ${{\cal T}}_1$ made up of the $k$ nodes with smallest score is an irreducible component of ${{\cal T}}$.

2.  Remove ${{\cal T}}_1$ from ${{\cal T}}$ and repeat Step 1 until no node is left.

Note that if the aim is to find the maximal component ${{{\cal T}}_{\mathrm{MAX}}}$, one can start “from above” and take into account $M-1-s_i$ instead of $s_i$ (i.e. the number of nodes that dominate the $i$-th node), so that only one step is needed. Another algorithm finding the irreducible components of any graph and having complexity $O(M^2)$ is shown in Kocay and Kreher [@Kocay-Kreher].

#### Cycles

A measure of how far a tournament is from being transitive is the number of 3-cycles. If the score sequence of ${{\cal T}}$ is $s_1,s_2,\dotsc,s_M$, then the number of $3$-cycles is at most $\binom{M}{3}-\sum_{i=1}^{M}\binom{s_i}{2}\leqslant
\left\{\begin{array}{ll}\frac{M^3-M}{24} & \mbox{if $2\nmid M$}\\[1pt]
  