Combinatorial properties of heapsort by Al-Jaber Ah.

By Al-Jaber Ah.

A number of points on the topic of the combinatorial houses of heapsort are mentioned during this thesis. A recursion formulation for the variety of lots fulfilling a given situation among any offsprings with an identical father or mother Is given and a number of other houses of tons are mentioned together with a brand new set of rules to generate the set of all lots of any dimension. additionally during this paintings we outline moment order bushes that have an excellent value within the learn of the complexity of Williams' algorithms to generate a heap. We talk about this type of timber and we end up that the producing functionality of the variety of bushes satisfies a nonlinear differential distinction equation. The numerical computation and the asymptotic growth for a volume with regards to this nonlinear differential distinction equation Is given during this paintings . eventually, we supply an higher sure for the variety of the second one order bushes generated from the set of all lots of dimension N the place N has the shape 2-1 for any confident integer okay.

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N − 1 . k! k − 1 Hint. 9) can be used for the induction step. (b) Give a noninductive combinatorial proof of the formula for f (n, k). , ordered and unordered with repetitions allowed or forbidden. 4. We are selecting k things from an n-set. The rules governing the selections are listed at the top and the left of the figure. As indicated in the figure, these numbers can also be interpreted in terms of placing balls into labeled boxes. The concepts in the Rules of Sum and Product were known in ancient times.

Instead, only its size is given. 1 Using the notation To get a feeling for the notation used to specify a function, it may be helpful to imagine that you have an envelope or box that contains a function. In other words, this envelope contains all the information needed to completely describe the function. Think about what you’re going to see when you open the envelope. ∗ ∗ ∗ Stop and think about this! ∗ ∗ ∗ You might see P = {a, b, c}, g: P → 4, g(a) = 3, g(b) = 1 and g(c) = 4. This tells you that the name of the function is g, the domain of g is P , which is {a, b, c}, and the codomain of g is 4 = {1, 2, 3, 4}.

N − l)! n! ((n − l) − k)! This is the exact answer written in various forms. The exact answer does not give us a good idea of how the probability behaves when the numbers are large. 2): 2 n! ∼ nk e−k /2n (n − k)! and 2 (n − l)! ∼ (n − l)k e−k /2(n−l) . ((n − l) − k)! 3 Thus 2 (n − l)k e−k /2(n−l) P (n, k, l) ∼ = nk e−k2 /2n 1− l n k exp − k2 l 2n(n − l) We need to look at the two factors on the right. Inside the exponential we have compared to n, this is nearly 2 −k l 2n2 . 27 . −k2 l 2n(n−l) .

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