Example 1
The following algorithm will generate a random permutation of the elements $1,2, \ldots, n .$ It is somewhat faster than the one presented in Example $1 \mathrm{a}$ but is such that no position is fixed until the algorithm ends. In this algorithm, $P(i)$ can be interpreted as the element in position $i$ Step 1. Set $k=1$ Step $2 .$ Set $P(1)=1$ Step $3 .$ If $k=n,$ stop. Otherwise, let $k=k+1$ Step 4. Generate a random number $U$ and let
$$\begin{aligned}P(k) &=P([k U]+1) \\P([k U]+1) &=k\end{aligned}$$
Go to step 3
(a) Explain in words what the algorithm is doing.
(b) Show that at iteration $k$ - that is, when the value of $P(k)$ is initially $\operatorname{set}-P(1), P(2), \ldots$
$P(k)$ is a random permutation of $1,2, \ldots, k$
Hint: Use induction and argue that
$$\begin{array}{l}P_{k}\left\{i_{1}, i_{2}, \ldots, i_{j-1}, k, i_{j}, \ldots, i_{k-2}, i\right\} \\
\quad=P_{k-1}\left\{i_{1}, i_{2}, \ldots, i_{j-1}, i, i_{j}, \ldots, i_{k-2}\right\} \frac{1}{k} \\
\quad=\frac{1}{k !} \text { by the induction hypothesis }
\end{array}$$