It runs along the same lines upon using the trivial bounds \(|\varXi _k - \varXi _{k-1}| \le 1\) and \(N_k \ge 0\). Sorry, for editing the answer into quite a different direction. Thor. \end{aligned}$$, $$\begin{aligned} \mathbf {E} \left[ |W_k(n) - W_k^*(n)|^2 \right]&\le \mathbf {E} \left[ (M_k(n) H_k(n))^2 \right] \le \mathbf {E} \left[ (k + T_k(n))^2 (H_k(n))^2 \right] \\&\le C_2 k^2 + C_1\left( 2 k \mathbf {E} \left[ T_k(n) (\log T_k(n))^2 \right] \right. Summarizing, we have shown (iv) and (v). We refer to this work for a discussion of more details on the distribution, historical background and further references. For a sequence \((k_n)\) satisfying \(sn \le k_n \le tn\) for \(0< s< t < 1\), let us condition on the event \(Y_n = k_n\). Stat. C++ Java Python C# Javascript Direct link to David's post How does the computer det, Posted 7 years ago. Direct link to Alan Savage's post so is this binary search , Posted 6 years ago. Let \(1/4< t < 1/2\). For \(t \le 1/4\), \(\alpha ^{(0)}_t = \infty \) follows immediately from (28) since \(\alpha ^{(0)}_{2t} < \infty \). Enhance the article with your expertise. Actually I liked the version before your edit better. Replacing \(T_1\) by \(T_2\), we obtain the analogous sum in the right subtree. Then, denoting by |T| the size of T, for \(|T| \ge 1\), The first statement is obvious, the argument for the second can be found in [25]. We introduce some notation. Finally, the asymptotic behaviour of the weighted depths of the nodes associated with the vectors \({\mathbf {0}}\) and \({\mathbf {1}} := 11\ldots \) denoted by \({\mathscr {L}}_n\) and \({\mathscr {R}}_n\) (\({{\mathscr {L}}}\) and \({\mathscr {R}}\) stand for left and right) were studied in [1]. Proc. This means you're worst case of search may reach O(N). Then, for \(r , y \in \mathbb {R}\) with \(\mathbb {P} \left( \varXi = y \right) = 0\), and n large enough, Let \(\bar{x} = x_{k+1} x_{k+2} \ldots \), \((V_1, V_2, \ldots )\) be an independent copy of \((U_1, U_2, \ldots )\) and, Given \(\varXi _k, |\varXi _k - \varXi _{k-1}|, N_k\), on \(N_k < n\), \(\bar{B}_n\) is distributed like \(\bar{B}^*_{\text {Bin}(n - N_k, |\varXi _k - \varXi _{k-1}|)}(\bar{x}) + k / \sqrt{\log n}\) where \((B^*_n(\bar{x}))\) is distributed like \((B_n(\bar{x}))\) and independent from the remaining quantities. Since the vector \((\text {rank}(U_1), \ldots , \text {rank}(U_n))\) constitutes a uniformly chosen permutation, in distribution, both the permutation model and the i.i.d. Asking for help, clarification, or responding to other answers. https://doi.org/10.1007/s10959-017-0773-1, DOI: https://doi.org/10.1007/s10959-017-0773-1. . Are self-signed SSL certificates still allowed in 2023 for an intranet server running IIS? Most cultures use a gender binary, having two genders (boys/men and girls/women).In this binary model, gender and sexuality may be assumed by default to align with one's genetic or gamete-based . Considering the last inserted node with value \(Y_n\), note that, conditionally on \(Y_n = k\), the correlations between the events \(A_{j,k}, j < k\) and \(A_{j,k}, j > k\) vanish. This statement at the end of the second paragraph is not at all obvious to me. In this post, Binary Search based solution is discussed. e.g. If \(\ell - k = \varOmega (n)\) and \(k = \omega (n / \sqrt{\log n})\), then. Further, we define the weighted path length \(\mathscr {P}_n\) as the sum of all weighted depths, and the weighted Wiener index \(\mathscr {W}_n\) as the sum over all pairs of weighted distances. Additionally to the variances given in the theorem, one obtains, Aguech, R., Lasmar, N., Mahmoud, H.: Extremal weighted path lengths in random binary search trees. Direct link to Joseph Henderson's post This is my first attempt , Posted 6 years ago. Why does the "\left [" partially disappear when I color a row in a table? Doklady Akademii Nauk USSR, Moscow, Vol. a root node with 1 child has 1 edge and weight of 2 so edges +1 In other words, we can assume that Weighted binary codes are those binary codes that follow the positional weight principle. It doesn't matter whether you have Y or 1/Y in the table, but yes this would give you the appropriate jumps (and might be more efficient due to not taking repeated reciprocals). Code definitions. where \(c^* = 4.31 \ldots \) is the larger of the two solutions to the transcendent equation \(e = (\frac{2e}{x} )^x\). 25, 85100 (1991), Rschendorf, L., Schopp, E.-M.: Note on the weighted internal path length of \(b\)-ary trees. First, since \(x f^{''}_t(x) = - f_{2t}'(x) - f_t'(x)\), for \(0 < t \le 1/4\), the function \(f_t\) is convex. The above proportionally weighted binary search can used to improve on the initial probe. Adding up the terms and simplifying leads to (32). \end{aligned}$$, $$\begin{aligned} \lim _{x \downarrow 0} f_Z(x) = 1 + \int _0^1 r(y) dy < \infty . Help us improve. Among the quantities studied in binary search trees, one finds depths of and distances between nodes related to the performance of search queries and finger searches in the database, the (total) path length measuring the cost of constructing the tree as well as the Wiener index. That's why we need to describe computer algorithms completely. A weight-balanced tree is a binary search tree that stores the sizes of subtrees in the nodes. Here, another phase transition occurs when \(k = o(n / \sqrt{\log n})\). Either way, the binary tree is constructed by starting with the leaves separate and, at each step, joining the two least likely subtrees into a larger subtree until there's only one subtree left. These codes are very useful and convenient for input and output operations in digital circuits. Contribute to the GeeksforGeeks community and help create better learning resources for all. It still matches. Let \(1 \le k \le n\). obviously more complicated but the same asks the average out of a pool of data, eliminate then continue? Central limit theorems for the path length go back to Rgnier [27] and Rsler [28], for the Wiener index to Neininger [25]. From here, statement (13) follows from (4) and (10). The depth first search process In the permutation model, let \(v_1, \ldots , v_{n+1}\) be the external nodes as discovered by the depth first search process from left to right. Examples: Input: Output: 5 Weight of sub-tree for parent 1 = ( (-1) + (5) + (-2) + (-1) + (3)) = 4 Weight of sub-tree for parent 2 = ( (5) + (-1) + (3)) = 7 Weight of sub-tree for parent 3 = -1 Can YouTube (e.g.) Given N jobs where every job is represented by following three elements of it. model. Why would a highly advanced society still engage in extensive agriculture? Notice this is not a binary search tree, as you are not dividing your search space by two in every step, but rather a rebalanced tree, with respect to your search pattern distribution. let's say distribution of the queries (not sure if it makes a difference). Binary trees aren't always binary search trees, though, and one use of a binary tree is to derive a Huffman compression code. Thor. Although half of the effective channels were filtered by the sorting result of PPWPE, the binary gravitational search algorithm (BGSA) was used in the selection of effective channels. Further, both sequences are independent of \(U_1\). For the last inserted node, in distribution. For example, having v = [10, 20, 30, 40, 50, 60]: Which can be reordered, or, rebalanced, using your function f(x) = 1 / x: If you search for 10, you only need one comparison, but if you're looking for 60, you'll perform O(N) comparisons, which does not qualifies this as a binary search. Eng. : The Art of Computer Programming. Wiley, New York (1992), Mahmoud, H.M., Neininger, R.: Distribution of distances in random binary search trees. 1. In the probabilistic analysis of algorithm, \(\mu \) first arose in Hwang and Tsais [17] study of the complexity of Hoares selection algorithm. Comput. \end{aligned}$$, \(\mathbf {E} \left[ \varPhi (i 2^{-k}) \right] = i 2^{-k}\), \(U_1^* = 1 - U_1, U_2^* = 1 - U_2, \ldots \), $$\begin{aligned} \mathscr {L}(Y) = \mathscr {L}(U Y + \mathbf {1}_{ A } (1-U)), \end{aligned}$$, \(\mathscr {L}(\varXi (t)) = \mathscr {L}(U \varXi (2t))\), $$\begin{aligned} f_t(x) = \mathbf {E} \left[ \frac{\mathbf {1}_{ [x,1] }(\varXi (2t))}{\varXi (2t)} \right] , \quad x \in (0,1]. Further, it is continuous at x if and only if \(x \notin {\mathscr {D}}_k\). Comput. The sum of all distances between nodes in different subtrees equals \(|T_1| \mathbf {p}(T_2) + |T_2| \mathbf {p}(T_1) + |T_1| |T_2| x\). Further, the weighted path length in this model was investigated by Rschendorf and Schopp [29]. We call \(\{{\mathscr {B}}_n(x): x \in \{0,1 \}^\infty \}\), the weighted silhouette of the tree (at time n). Thus, the assertion follows for \(x \in {\mathscr {D}}\) since \(\varPhi (x) = \varXi (x)\). Stay tuned to the Testbook app for more updates on related topics from Digital Electronics, and various such subjects. How can I find the shortest path visiting all nodes in a connected graph as MILP? Niklaus Wirth covered this in his book "Algorithms and Data Structures", in a few variants (one for Pascal, one for Modula 2, one for Oberon), at least one of which is available for download from his web site. Examples of non-weighted codes are the Gray code and Excess-3 code. Here, the weighted depth of a node is the sum of all keys stored on the path to the root. Further, there exists \(\delta > 0\) such that \(\mathbb {P} \left( |\varXi _k - \varXi _{k-1}| \le \delta \right) \le \varepsilon \). ("If the catalog were sorted alphabetically by star names, binary search would not have to examine more than 22 stars, even in the worst case.") In particular, by the union bound and Markovs inequality, for any \(m \ge 1\). Can YouTube (e.g.) This will however bring our expected case back to O(log N) as we are only allowed to reduce the small partition by 1/b each time. We consider the sequence \((Z_n)_{n\ge 0}\) defined by, and \(Z_0 = 0\). \end{aligned}$$, $$\begin{aligned} \left( \frac{D_k(n) - \mathbf {E} \left[ D_k(n) \right] }{\sigma _{ D_k(n)}}, \frac{W_k(n) - \mathbf {E} \left[ W_k(n) \right] }{\sigma _{W_k(n)}} \right) \rightarrow (\mathscr {N}, \mathscr {N}). When n=1 the weight is always 1, so minimum weight is n. Binary trees are recursive, shown in the format of Left, Singleton, Right, so we can break any binary tree down to an n-size of 1 (L and R are empty sets, plus the Singleton set), 2 (Either L or R is an empty set, but not both), or 3 (Neither L nor R are empty sets). At least one of these sequences is infinite. The $(k+1)n$ is because every node has a base value of $k$ and every additional child has a value of $1$, so adding a node as a leaf increases the parent's value by $1$ and the leaf's value is $k$. 2^21 0\) let \(\alpha T + \beta \) be the tree obtained from T where each label y is replaced by \(\alpha y + \beta \). Inform. For \(k \ge 1\), let \({\mathscr {D}}_k = \{\ell 2^{-k}: \ell = 1, \ldots , 2^k - 1\}\). Whenever digital data is broadcasted in a binary form, from one network or device to other networks, there is a probability of an error occurring. Department of Statistics and Operation Research, College of Sciences, King Saud University, P.O. Some of the codes that do not obey the weights of the sequence binary numbers are called non-weighted codes. J Theor Probab 31, 19291951 (2018). Ann. This type of code cannot be applied for arithmetic operation but finds application in input/output devices. You will be notified via email once the article is available for improvement. To use the same technique for 1/X data, store in table not Y but inverse 1/Y. Do intransitive verbs really never take an indirect object? \end{aligned}$$, $$\begin{aligned} \text {Bin}(n,p) := \sum _{i=1}^n \mathbf {1}_{ \{V_i \le p\} }, \quad n \ge 0, p \in [0,1]. To maintain the data integrity among transmitter and receiver, an additional bit is added with the transferred data. Align \vdots at the center of an `aligned` environment. \end{aligned}$$, \(|k/n - s| = o((\log n)^{-1/2}), |\ell /n - t| = o((\log n)^{-1/2})\), $$\begin{aligned}&\Bigg (\frac{D_k(n) - 2 \log n}{\sqrt{2 \log n}}, \frac{D_\ell (n) - 2 \log n}{\sqrt{2 \log n}}, \frac{D_{k,\ell }(n) - 4 \log n}{\sqrt{4 \log n}}, \frac{W_{k,\ell }(n) - 2 (s+t) n \log n}{n \sqrt{2 \log n}}\Bigg ) \nonumber \\&\quad \rightarrow \left( \mathscr {N}_1, \mathscr {N}_2, \frac{\mathscr {N}_1 + \mathscr {N}_2}{\sqrt{2}}, s \mathscr {N}_1 + t \mathscr {N}_2\right) . For \(t > 3/8\), we have \(1-2(1-2t) > 1/2\); thus, \(\alpha ^{(0)}_t < \infty \). The symmetry statement (ii) is reminiscent of the fact that the uniform distribution on [0,1] is symmetric around 1/2. Let's practice it here. So the total weight is $2N-2$, plus the arbitrary extra 1 for the root node, giving a total of $2N-1$. Direct link to Cameron's post The algorithm for akinato, Posted 7 years ago. That's why you try the center each time. 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Endowed with Skorokhods topology \(J_1\), \({\mathscr {D}}[0,1]\) becomes a Polish space. Sci. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Finally, we define the height of the tree by \(H_n = \sup \{k \in \mathbb {N}: D_k(n) > 0\}\). The weight of any given tree must be $(k+1)n + c$ where $c$ is a constant since we know that every node adds $(k+1)n$ to the weight. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. \end{aligned}$$, $$\begin{aligned} \log&\mathbf {E} \left[ \exp \left( i \lambda \left( {\bar{D}}^>_k(n) - \log n\right) / \sqrt{\log n} + i \mu \left( {\bar{W}}^>_k(n) - k {\bar{D}}^>_k(n)\right) /n\right) \right] \\&= - i\lambda \sqrt{\log n} + \log \mathbf {E} \left[ \exp \left( i \sum _{j = k+1}^n \left( \frac{\lambda }{\sqrt{\log n}} + \mu \frac{j -k}{n}\right) B_{j,k} \right) \right] \\&= - i\lambda \sqrt{\log n} + \sum _{j=k+1}^n \log \left( 1 + \frac{ \exp \left( i \left( \frac{\lambda }{\sqrt{\log n}} + \mu \frac{j -k}{n} \right) \right) -1}{j-k} \right) . Conditional on its size, the subtree rooted at \(w_k\) is a random binary search tree. The gender binary (also known as gender binarism) is the classification of gender into two distinct forms of masculine and feminine, whether by social system, cultural belief, or both simultaneously. model, we have, The leading constants in the expansions of the covariances between \(P_n, W_n, \mathscr {P}_n\) and \(\mathscr {W}_n\) are given in (36)(38). Arkiv fr Mathematik, Astronomi och Fysik 22A(10), 114 (1930), MATH As the name indicates, these codes consist of both numbers and alphabetic characters therefore are called alphanumeric codes. If you're seeing this message, it means we're having trouble loading external resources on our website. Here, most arguments are based on representations of (weighted) depths as sums of bounded independent random variables which go back to Devroye and Neininger [9]. Why do code answers tend to be given in Python when no language is specified in the prompt? By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. For a finite rooted labelled binary tree T, denote by p(T) its path length, by \(\mathbf {p}(T)\) its weighted path length, by w(T) its Wiener index and by \(\mathbf {w}(T)\) its weighted Wiener index. Henning Sulzbach. 44, 467472 (1975), Bayer, P.J. + \, \mathbf {1}_{ [1/2,1) }(t) \left( (1-U) \varXi '(2t-1) + U \right) \right) _{t \in [0,1]}\right) . This finishes the proof of (vi). Due to the recursive nature of Binary Trees, this can be propagated up to any value of n. Thanks for contributing an answer to Puzzling Stack Exchange! Do the 2.5th and 97.5th percentile of the theoretical sampling distribution of a statistic always contain the true population parameter? Sci. As pointed by @Steve314, the farthest you go from a fully balanced tree, the worse will be your worst case of search. the root node which has no parent which gives the - 1. Then, for all \(1 \le k \le n\). from (26) upon treating the cases \(k = O(1)\) and \(k = \omega (1)\) separately. Random recursive trees A random recursive tree is constructed as follows: starting with the root labelled one, in the kth step, \(k \ge 2\), a node labelled k is inserted in the tree and connected to an already existing node chosen uniformly at random. 2 x 2 = 4 or 2 + 2 = 4 as an evident fact? For \(x = x_1 x_2 \ldots \in \{0,1\}^\infty \), let \(B_n(x)\) be the maximal depth among nodes of the form \(x_1 \ldots x_k, k \ge 0\). Are the NEMA 10-30 to 14-30 adapters with the extra ground wire valid/legal to use and still adhere to code? Then, by the central limit theorem for triangular arrays of row-wise independent uniformly bounded random variables with diverging variance applied to \(V_{j, k_n}, j =1, \ldots , n\), in distribution. Questions and answers - MCQ with explanation on Computer Science subjects like System Architecture, Introduction to Management, Math For Computer Science, DBMS, C Programming, System Analysis and Design, Data Structure and Algorithm Analysis, OOP and Java, Client Server Application Development, Data Communication and Computer Networks, OS, MIS, Software Engineering, AI, Web Technology and many . Stack Overflow at WeAreDevelopers World Congress in Berlin, Minimize the longest King chain on a 5x5 binary board, Maximum number of masses you can weigh given 3 weights. If you have a number between x and y, you find the average (x+ y)/2 then if it's too high, find another average and so forth until you find it. The most commonly accepted alphanumeric codes are. The addition and subtraction of BCD have separate rules. 7730, August 1977, Siemens AG, Abteilung D APS1, Otto-Hahn-Ring 6, 8000, Mnchen 83, Germany (Fed. Direct link to Krish's post I don't quite understand , Posted 3 years ago. Let \(T_1, T_2\) be its left and right subtree and x the label of the root. Direct link to huseynov.rza.00's post #include Let's look at how to describe binary search carefully. Aguech, R., Amri, A. By \(D^*_k(n)\) and \(W^*_k(n), 1 \le k \le n+1\), we denote depth and weighted depth of the external node \(v_k\). Every node has an initial value of 1 and adds 1 to the value of its parent, except BinarySearchTree Class insert Method insert Method remove Method remove Method findMin Method findMin Method findMax Method findMax Method contains Method contains Method makeEmpty Method isEmpty Method levelOrderPrint Method levelOrderPrint Method . \end{aligned}$$, $$\begin{aligned}&\mathbf {E} \left[ \sum _{j = 1}^n \mathbf {1}_{ B_{j,k} \backslash A_{j,k} } \right] \le 2, \quad \text {and} \quad \mathbf {E} \left[ \sum _{j = 1}^n j \mathbf {1}_{ B_{j,k} \backslash A_{j,k} } \right] \le 2k + \log n. \end{aligned}$$, $$\begin{aligned}&\mathbf {E} \left[ \sum _{i,j = 1}^n \mathbf {1}_{ B_{j,k} } \mathbf {1}_{ B_{i,k} \backslash A_{i,k} } \right] = O(1), \quad \text {and}\\&\mathbf {E} \left[ \sum _{i,j = 1}^n i j \mathbf {1}_{ B_{j,k} } \mathbf {1}_{ B_{i,k} \backslash A_{i,k} } \right] = O(k^2 + k(\log n)^2). Asking for help, clarification, or responding to other answers. Use $k = 1$ to get $\text{weight(n)} = 2n - 1$. a node with 2 children has 3 edges. In the permutation model, let \(A_{j,k}\) be the event that the node labelled k is in the subtree of the node labelled j. When numbers, letters and words are represented by a special group of the symbol then we can assume that they are being encoded and the group of the symbol is called CODE. SIAM J. Comput. From [5], it follows that Y has the arcsine distribution, proving (iii). The sum of all weighted distances between nodes in the left subtree and the root equals \(\mathbf {p}(T_1) + |T_1|x\). (In [7, TheoremO1], the first convergence in the last display is formulated for \(x = {\mathbf {0}} := 00\ldots \) The general case follows, since, by symmetry, \({{\mathscr {L}}}( B_n(x)) = {{\mathscr {L}}}( B_n ({\mathbf {0}})) \) for all x. On the other hand, non-weighted codes are not assigned with any weight to each digit position. We consider weighted path lengths to the extremal leaves in a random binary search tree. This is not a homework problem, although I understand why you think that. Analysis of algorithms (Krynica Morska, 2000), Neininger, R.: The Wiener index of random trees. Concavity at rational points with small denominator such as \(t = 3/8\) or \(t = 5/12\) can be verified by hand using (28). A new variance moment, which contains distinctive facial features . For \(t \in (1/2,1)\), \(f_t(x) = f_{1-t}(1-x), x \in (0,1)\) is a density of \(\mathscr {L}(\varXi (t))\) by (ii). Reading-London: Addison-Wesley 1973, Maurer, H.A., Wood, D.: Zur Manipulation von Zahlenmengen. Next, define \(\xi ^{-}_k\) (\(\xi ^+_k\), respectively) analogously to \(\xi _k\) based on the sequence \((y^-_m)\) (\((y^+_m)\), respectively). Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. \end{aligned}$$, $$\begin{aligned} \sup _{x \in [0,1]} \frac{{\mathscr {B}}_n(x)}{\log n} \rightarrow c^* = 4.31\ldots \end{aligned}$$, $$\begin{aligned} \left( \frac{B_n(x) - \log n}{\sqrt{\log n}}, \frac{{\mathscr {B}}_n(x)}{\log n} \right) \rightarrow (\mathscr {N}, \varXi (x)), \end{aligned}$$, $$\begin{aligned} \mathbf {E} \left[ \mathscr {P}_n \right] = n \log n + (\gamma - 3/2) n + o(n), \quad \mathbf {E} \left[ \mathscr {W}_n \right] = n^2 \log n + (\gamma -11/4)n^2 + o(n^2), \end{aligned}$$, $$\begin{aligned} Var (\mathscr {P}_n) = \frac{65 - 6 \pi ^2}{36} n^2 + o(n^2), \quad Var (\mathscr {W}_n) = \frac{2413 - 240 \pi ^2}{1440} n^4 + o(n^4). Variances and covariances can be computed successively using the fixed-point equation, e.g. Direct link to pawel.englert's post Why do we need round down, Posted 9 years ago. \end{aligned}$$, \(\mathbb {P} \left( |\varXi _k - \varXi _{k-1}| \le \delta \right) \le \varepsilon \), \(\mathbb {P} \left( \varXi = y \right) = 0\), $$\begin{aligned} \mathbb {P} \left( \bar{B}_n \le r, \frac{{\mathscr {B}}_n}{ \log n} \le y \right) \le 2 \varepsilon + \mathbb {P} \left( \bar{B}_n \le r, \varXi _k \le y + 2 \varepsilon , |\varXi _k - \varXi _{k-1}| \ge \delta , N_k < L \right) . Were all of the "good" terminators played by Arnold Schwarzenegger completely separate machines? The purpose of a binary search is that, for an array that is sorted, every time you half the array you are minimizing the worst case, e.g. The proof of the converse direction establishing (19) is easier. 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Being of the same order as the path length in binary search trees, it follows from results in [19] that the weighted path length \({\mathscr {Q}}_n\) in a random recursive tree of size n is of order \(n^2\). We move on to the statements about the distribution of \(\varXi (t)\). \end{aligned}$$, \(\bar{D}_k(n) = \sum _{j=1}^n \mathbf {1}_{ B_{j,k} }-1\), \({\bar{W}}_k(n) = \sum _{j=1}^n j \mathbf {1}_{ B_{j,k} }\), \(H^{(i)}_{k,n} := H^{(i)}_{k-1} + H^{(i)}_{n-k}\), $$\begin{aligned} \mathbf {E} \left[ {\bar{W}}_k(n) \right]&= k (H_{k,n}^{(1)}-1) + n + 1 ,\\ \text {Var}({\bar{W}}_k(n))&= k^2 \left( H_{k,n}^{(1)} - H_{k,n}^{(2)}-3\right) + \frac{n^2}{2} + kn + 2k \left( H^{(1)}_{k-1} - H^{(1)}_{n-k}\right) - \frac{n}{2} + k + 1. 308(4), 529540 (2008), Mahmoud, H.M.: Evolution of Random Search Trees. 11(6), 587597 (2002), Panholzer, A., Prodinger, H.: Spanning tree size in random binary search trees. Here, in the last step, we have used that the sum on the right-hand side is a Riemann sum over the unit interval whose mesh size \(n^{-1}\) tends to zero. We prove Theorem 2. where \(\mathscr {Y}\) has the Dickman distribution. Thus, for any \(t \in (0,1)\), \(f_t\) is smooth on (0,1). Moreover, the Wiener index is obtained by summing all distances of unordered pairs of vertices.

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