Average case analysis of heap building by repeated insertion

TL;DR: It is shown that the average number of swaps required to construct a heap on n keys by Williams’ method of repeated insertion is (! + o(1))n, where the constant ! is about 1.3.

About: This article is published in Journal of Algorithms.The article was published on 1991-01-02 and is currently open access. It has received 19 citations till now. The article focuses on the topics: Heap (data structure).

Summary

1 Introduction

• The heap is a much used and much studied data structure (for example, see Knuth [K]).
• The key observation of this section, Observation 2.1, is the cornerstone of their analysis.

2 Qualitative results for expected times

• And give some results which will be used later to establish bounds on ω.the authors.
• Another is to assume that the keys are n independent random variables, each uniformly distributed on the interval [0,1].
• The authors have now completed part (a) of the proof of Theorem 1.3.
• Let R(n) denote the rank of the nth key among the first n keys.

5 Probability bounds

• In this section the authors shall prove part (c) of Theorem 1.3.
• The authors follow roughly the route taken in section 2 for investigating E[Wn].
• Finally, the authors may use the half of the lemma already proved to handle the sum above, and one further application of inequality (5.1) to handle the last term.

6 Repeated insertion into equi-probable heaps

• The authors consider the simplified approximation of the average case behaviour of Williams’ heap construction in which they repeatedly insert a uniform random key into a uniform random heap.
• The main result of this section is the following.
• The proof of part (c) will then follow from Hoeffding’s inequality (5.1).
• Proof Continuing to mimic previous notation, the authors define x̃k as the aver- age of the expected insertion costs along the kth level Lk, namely ∑2k+1−1 t=2k Ĩ(t)/2k. (See also Lemma 2.6.).

7 Concluding remarks

• The authors have described fairly precisely the average case behaviour of Williams’ method of constructing a heap by repeated insertion.
• The most interesting related open problem is the average case analysis of heapsort (with any of its trickledown variants).
• While some empirical data (e.g. see [K]) and partial theoretical results (e.g. [C]) are known, the problem of determining average numbers of comparisons is open.
• The following series of tables shows the average expected values of the levels of a heap built by Williams’ algorithm, for perfect heaps of sizes 3 to 4095.

Frequently asked questions

Q1. What are the contributions mentioned in the paper "Average case analysis of heap building by repeated insertion" ?

The heap is a much used and much studied data structure ( for example, see Knuth [ K ] ). 

Q2. how can i compute wn in lg n space?

by storing the probability arrays P (n, i, ·) only for nodes i on IP+(n), it is possible to compute E[Wn] in only O(n lg n) space but O(n 3 lg n) time. 

Q3. What is the insertion path of the key?

Observe that during the insertion of some key with R(n) = k, a swap takes place at node i of the insertion path if and only if k ≤ An−1[i]. 

Q4. What is the expected number of swaps along the two links incident with the root?

When bubbling up the nodes in Lk, the expected number of swaps along the two links incident with the root is ∑2k+1−1j=2k 1/j > ln 2. 

Q5. What is the rank of a number x in a set of numbers?

The rank of a number x in a set of numbers is its placement in the ordered set; thus the smallest number has rank 1, the next smallest rank 2, etc. Let An[i] be the rank of A[i] among A[1, . . . , n] after exactly n keys have been inserted. 

Q6. what is the avg. exp. of the array?

Heapsize 31level avg. exp. val.0 0.031250000000000 1 0.092036756202444 2 0.205202826266408 3 0.401927398975604 4 0.703027874420290 

Q7. What is the meaning of array indices?

Here and for the rest of this section, array indices are understood to be integers; any fraction x/y used an array index is actually ⌊x/y⌋. 

Q8. Hence, for k k2 j=k4 l?

Hence for j ≥ k − k2 + 1,Prob{2−jBj,k > 2 j−k+2 + 2−s+1} ≤ (2 lg lg k) exp(−2kk2(lg k)8 ) .Now Σ2 ≤ ∑k−k1j=k−k2+1 2−jBj,k, soProb{Σ2 > 2 −k1+3 + (k2 − k1)2 −s+1} ≤ (k2 − k1)(2 lg lg k) exp(− 2kk2(lg k)8 ) . 

Q9. What is the insertion path of nodes t?

For each node t in level Lk (k > 0) let IP (t) = {⌊t/2j⌋ : j = 1, . . . , k} be the set of nodes on the insertion path from node t to the root. 

Q10. What is the method of heap building?

This method requires (2+o(1))n comparisons in the worst case to build a heap with n keys, which is less than Williams’ method takes on average.