Question

Consider the following algorithm for generating a random permutation of the elements \(1,2, \ldots, n .\) In this algorithm, \(P(i)\) can be interpreted as the element in position \(i\) Step 1: \(\quad\) Set \(k=1\). Step 2: \(\quad\) 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 set-that \(P(1), P(2), \ldots, P(k)\) is a random permutation of \(1,2, \ldots, k\). Hint: Use induction and argue that $$ \begin{aligned} &P_{k}\left\\{i_{1}, i_{2}, \ldots, i_{j-1}, k, i_{j}, \ldots, i_{k-2}, i\right\\} \\ &\quad=P_{k-1}\left\\{i_{1}, i_{2}, \ldots, i_{j-1}, i, i_{j}, \ldots, i_{k-2}\right\\} \frac{1}{k} \end{aligned} $$ \(=\frac{1}{k !}\) by the induction hypothesis The preceding algorithm can be used even if \(n\) is not initially known.

Short answer

(a) The algorithm generates a random permutation of elements from 1 to n by iteratively swapping elements and their positions based on a random number U. The procedure continues until all elements have been permuted. (b) By using induction, we proved that at iteration k, the set \(P(1), P(2), \ldots, P(k)\) is a random permutation of elements from 1 to k. The base case (k=1) holds true, and for the inductive step (k+1), assuming the induction hypothesis, it is also true as the multiplicative factor \(\frac{1}{(k+1)!}\) is introduced by the random number U.

Step-by-step solution

Part (a): Explanation of the algorithm

The algorithm provided consists of the following steps: 1. Set the initial value of k to 1. 2. Initialize the first position in P with 1 (i.e., P(1) = 1). 3. If k equals n, the process has reached the end and the algorithm terminates. Otherwise, increase k by 1. 4. Generate a random number U, and use it to generate the next element and its position in the sequence: \(P(k) = P(\lfloor kU \rfloor + 1)\) and \(P(\lfloor kU \rfloor + 1) = k\) Then return to Step 3. This algorithm generates a random permutation of the elements from 1 to n. At each iteration, it generates a random number U to determine the position where the current value of k will be placed and swaps the elements at positions k and the chosen position. The algorithm iterates until the entire sequence is generated, terminating when k equals n.

Part (b): Proving random permutation correctness

In order to show that at iteration k, the set \(P(1), P(2), \ldots, P(k)\) is a random permutation of elements \(1,2,\ldots,k\), we will use induction. Inductive hypothesis: For iteration k, we will assume that \(P_{k}\{i_{1}, i_{2}, \ldots, i_{j-1}, k, i_{j}, \ldots, i_{k-2}, i\} = P_{k-1}\{i_{1}, i_{2}, \ldots, i_{j-1}, i, i_{j}, \ldots, i_{k-2}\} \frac{1}{k}\), which equals \(\frac{1}{k!}\) by the induction hypothesis. Base case: For k=1, P(1) = 1. There is only one permutation for the elements 1 to 1, so it holds true that \(\frac{1}{1!} = 1\). Inductive step: Let's consider the iteration k+1. We must show that \(P_{k+1}\{i_1, i_2, \ldots, i_j-1,k+1,i_j,\ldots,i_{k-1}, i\} = P_{k}\{i_{1}, i_{2}, \ldots, i_{j-1}, k, i_{j}, \ldots, i_{k-2}, i\} \frac{1}{k+1}\), which equals to \(\frac{1}{(k+1)!}\). For this step, we generate a random number U, and use it to determine the position and element of \(P(k+1)\): \(P(k+1) = P(\lfloor (k+1)U \rfloor + 1)\) and \(P(\lfloor (k+1)U \rfloor + 1) = k+1\) This will result in one of the \((k+1)\) positions for the element k+1 randomly with equal probability \(\frac{1}{k+1}\), giving us a set \(\left\{1, 2, \ldots, k, k+1 \right\}\) at iteration k+1. Now, by the inductive hypothesis, we have \(P_{k}\{i_{1}, i_{2}, \ldots, i_{j-1}, k, i_{j}, \ldots, i_{k-2}, i\} = \frac{1}{k!}\). Multiplying both sides with \(\frac{1}{k+1}\), we have: \(P_{k}\{i_{1}, i_{2}, \ldots, i_{j-1}, k, i_{j}, \ldots, i_{k-2}, i\} \frac{1}{k+1} = \frac{1}{(k+1)!}\) Thus, by induction, the algorithm generates a random permutation of the elements 1 to n, and at iteration k, the set \(P(1), P(2), \ldots, P(k)\) permutes the elements \(1, 2, \ldots, k\) randomly.