![]() For that, we will generalize the approach we had for the case of k = 1. We now turn to the more general question of computing the expected number of cycles of length k > 1. This means that the expected number of fixed points in a random permutation is 1! Cycles of length k > 1 ![]() Since we pick a permutation uniformly at random, the probability of any number ending at position i is exactly 1/ n. We conclude that we need to compute the probability of having a fixed point at position i. We now observe that for every i = 1, …, n, we haveĮ = 1 Pr + 0 Pr = Pr. Again, by the linearity of expectation, we have Where for every i = 1, …, n, Y( i) is a random variable that is equal to 1 if i is a fixed point, and 0 otherwise. Applying the idea of decomposing to simpler random variables once again, we can write As a reminder, a fixed point is an index i that satisfies σ( i) = i. Thus, we want to compute the expected number of fixed points of a random permutation. Cycles of length 1 correspond to fixed points of the permutation. Let’s start by considering the case k = 1. Thus, we have reduced our original question to the task of computing E, for every k = 1, …, n. In our case, the Linearity of Expectation givesĮ = E + E + … + E. The expected value of a sum of random variables is equal to the sum of their individual expected values, regardless of whether they are independent. We now turn to the extremely useful property known as Linearity of Expectation, which essentially states that It is easy to see that we have the equality Let X( k) denote the number of cycles of length exactly k, for every k = 1, …, n. A natural choice for X in our case is the following. A standard technique when it comes to computing expected values is to write a random variable in terms of simpler random variables. ![]() Let X be the random variable denoting the number of cycles in a random permutation. Linearity of Expectation (again) comes to the rescue Returning to our original question, we would like now to compute the expected number of cycles in a permutation picked uniformly at random from the set of all n! permutations. The above observation immediately implies that there is always at least one cycle in any permutation. For example σ(4) = 6, and thus there is an arrow from the number 4 to the number 6. Note that each arrow corresponds to the result of applying σ to a number. Pick a uniformly random permutation of the first n positive natural numbers, along with its unique cycle decomposition in this case, the cycle decomposition consists of 3 cycles. Therefore, if N is the size of the array and M is the number of elements in the array with its occurrence = 2, then the number of permutations satisfying the condition will be 2 (N – (2 * X) – 1).Or, Why Linearity of Expectation delivers again Introduction.Therefore, there are 2 * 2 = 4 possible permutations. 1 and 3 both have two choices for the left part and right part.Recommended: Please try your approach on. ISRO CS Syllabus for Scientist/Engineer Exam.ISRO CS Original Papers and Official Keys.GATE CS Original Papers and Official Keys.DevOps Engineering - Planning to Production.Python Backend Development with Django(Live).Android App Development with Kotlin(Live).Full Stack Development with React & Node JS(Live).Java Programming - Beginner to Advanced.Data Structure & Algorithm-Self Paced(C++/JAVA).Data Structure & Algorithm Classes (Live).
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