  A shortcut of a directed path $v_1 v_2 \cdots v_n$ is an edge $v_iv_j$ with
$j > i+1$. If $j = i+2$ the shortcut is called a hop. If all hops are present,
the path is called hop complete, so the path and its hops form a square of a
path. We prove that every tournament with $n \ge 4$ vertices has a Hamiltonian
path with at least $(4n-10)/7$ hops, and has a hop complete path of order at
least $n^{0.295}$.
  A spanning binary tree of a tournament is a spanning shortcut tree if for
every vertex of the tree, all its left descendants are in-neighbors and all its
right descendants are out-neighbors. It is well-known that every tournament
contains a spanning shortcut tree. The number of shortcuts of a shortcut tree
is the number of shortcuts of its unique induced Hamiltonian path. Let $t(n)$
denote the largest integer such that every tournament with $n$ vertices has a
spanning shortcut tree with at least $t(n)$ shortcuts. We almost determine the
asymptotic growth of $t(n)$ as it is proved that $\Theta(n\log^2n) \ge
t(n)-\frac{1}{2}\binom{n}{2} \ge \Theta(n \log n)$.
