On the number of heaps and the cost of heap construction

Transcription

1 On the number of heaps and the cost of heap construction Hsien-Kuei Hwang Institute of Statistical Science Academia Sinica Taipei 5 Taiwan Jean-Marc Steyaert LIX École Polytechnique 928 Palaiseau France November 2, 200 Abstract Heaps constitute a well-known data structure allowing the implementation of an efficient O(n log n sorting algorithm as well as the design of fast priority queues. Although heaps have been known for long, their combinatorial properties are still partially worked out: exact summation formulae have been stated, but most of the asymptotic behaviors are still unknown. In this paper, we present a number of general (not restricting to special subsequences asymptotic results that give insight on the difficulties encountered in the asymptotic study of the number of heaps of a given size and of the cost of heap construction. In particular we exhibit the influence of arithmetic functions in the apparently chaotic behavior of these quantities. It is also shown that the distribution function of the cost of heap construction using Floyd s algorithm and other variants is asymptotically normal. Introduction Heap is an elementary data structure often used in applications concerned with priority queues and partial (as well as total ordering. It first appeared in Williams Heapsort algorithm [30], which happened to be the first in-place O(n log n sorting algorithm. Besides its original applications to sorting, heap has wide applications in algorithm design, see Aho et al. []. It serves as the prototype, both conceptually and in actual implementations, of many complex data structures in computational geometry (see Preparata and Shamos [25] and in graph-theoretical problems (see Noltemeier [2], Mehlhorn and Tsakalidis [20]. A (max heap is an array with elements a j, j n, satisfying the path-monotonic property: a j a j/2 for j = 2, 3,..., n, where x denotes the integral part of x. It can be viewed as a binary tree in which the value of each element is not smaller than that of its children. The construction of a heap from an arbitrary set of n keys can be performed efficiently in linear time using Floyd s algorithm (see [7, 5.2.3], but it is easy to see that the exact number of operations used varies according to the nature of the permutation on the input keys. Likewise, the number of heaps of a given size n is not a steadily growing function of n, but proves to have a chaotic rate of growth. Several authors have proposed partial answers to these problems. Knuth states the basic Most results of this paper appeared as Rapport de Recherche, LIX/RR/93/07, Ecole polytechnique, 993; only quite recently did we revise the paper. This work was partially supported by the ESPRIT Basic Research Action No. 74 (ALCOM II and by the AQSI research action of the French Ministry for Research and Space.

2 recurrence relation for the number of heaps, and gives an explicit solution as a product; he also shows that the expected cost used by Floyd s algorithm to construct a heap is linear; Doberkat [5] derives the probability generating functions for the number of exchanges and comparisons used by Floyd s algorithm, he also finds the asymptotics of the first two moments; Sprugnoli [27] examines in more detail the average case behavior of Floyd s heap-construction algorithm [7] for general values of the heap size. The method of Doberkat is via probability generating functions and that of Sprugnoli relies largely on refining the results on special sequences. We propose a new approach to the asymptotic study of the number of heaps and the cost of heap construction. It is based on explicit decomposition of the solution to the basic recurrence characteristic of heap structures. The decomposition is in terms of some digital sums (sums expressible using the binary digits of n in base 2. From this canonical representation, elementary asymptotic methods then give the asymptotic behavior of the quantity in question. This paper is organized as follows. We state the basic recurrence together with some notations in the next section. The general solution, in terms of digital sums, of this recurrence is presented in Section 3. Then we apply this result to determine an asymptotic expression for the number of heaps and the cost (mean and variance of Floyd s heap-construction algorithm. Our results on the mean and the variance improve upon that of Doberkat [5] and that of Sprugnoli [27]. In the final section, we establish the asymptotic normality of the cost used by Floyd s algorithm under the uniform permutation model. 2 The basic divide-and-conquer recurrence A characteristic property of a heap, when viewing as a binary tree, is that at least one of the two subtrees of the root node is complete (i.e., it contains 2 k elements for some positive integer k, and then the size of the other subtree is at least half and at most twice this number. Furthermore, this property recursively applies to each node. Given a heap H of size N with left and right subtrees L and R of respective sizes L and R, an additive cost function ϕ on heaps is defined by a relation ϕ[h] = τ[h] + ϕ[l] + ϕ[r], ( for some cost function τ. Summing over all heaps of size N, which obviously decompose into subheaps of the same sizes L and R, we thus define a function f that satisfies the same recurrence on sizes, t being the sum function for τ. Since at least one of L or R is complete, the relation ( can be written into a more precise form as follows. For positive integer k and a given sequence {t n } n f 2 k +j = t 2 k +j + f 0 = 0, f = t, { f2 k + f 2 k +j, if 0 j < 2 k, f 2 k + f j, if 2 k j < 2 k, which is referred to as the (additive heap recurrence (see [27], [3, Ch. 3]. Observe that we can incorporate the two cases into one by writing f n = f 2 log 2 2n/3 + f n 2 log 2 2n/3 + t n (n 2, (2 where 2 log 2 2n/3 represents the unique power of two lying between n/2 and 2n/3. Also the associated generating functions satisfy f(z = t n z n + ( f z 2 k z 3 2k z 3 2k + (z 2k + z 2k f j z j, n k k 2 k j<2 k 2

3 where f(z = n f nz n. Here and in the following, H n denotes the total number of ways to rearrange the integers {, 2,..., n} into a heap. It is obvious that H n satisfies the multiplicative heap recurrence: ( 2 k + j 2 H 2 k +j = k H 2 k H 2 k +j, if 0 j < 2 k, ( 2 k + j 2 k H 2 k H j, if 2 k j < 2 k, since the root has to contain the largest key and the remaining elements can be assigned independently to the left and right subheaps. The sequence {H n } n 2 =, 2, 3, 8, 20, 80, 20, 896, 3360, 9200, 79200, , , , , , , increases very fast and is the A05697 entry in Sloane s On-Line Encyclopedia of Integer Sequences ( njas/sequences. It is more convenient to work with h n = log(n!/h n since h n satisfies (2 with t n = log n. Knuth [7, p. 54] expresses the number of heaps of size n via the product of the sizes of all its subtrees. Denoting by s i, i =, 2,..., n, these sizes, the number H n writes: H n = n! i n s. i The s i s are generally of the form 2 k, for some positive integer k, except for the nodes lying on the special path whose ranks (in hierarchical order are of the form 2 j (+{n/2 j }, where {x} denotes the fractional part of x. If we consider the (backward difference of f n : ϕ n = f n = f n f n, then we obtain, { ϕ2 ϕ 2 k +j = τ 2 k +j + k +j, 0 j < 2 k, ϕ j, 2 k j < 2 k, (k, with the initial condition ϕ 0 = 0, where τ n = t n. Equivalently, if we write n = (b L... b 0 2 in base 2, then this recurrence can be re-written as ϕ n = ϕ (bl...b 0 2 = τ (bl...b ϕ (bl 2...b 0 2 = τ (bj...b j L Before solving (2, we note that there is another very similar type of recurrences [0] { φ2 k + φ φ 2 k +j = τ 2 k +j + 2 k +j, if 0 j 2 k ; φ 2 k + φ j, if 2 k j 2 k, (3 which occurs as the solution of the following equation φ n = τ n + min (φ j + φ n j (n 2, j n/2 (with φ given when the sequence {τ n } n 0 is strictly concave, namely, τ n+2 2τ n+ + τ n < 0 for all n 0. 3

4 On the other hand, when the sequence {τ n } n 0 is strictly convex, the solution φ n satisfies [0] φ n = φ n/2 + φ n/2 + τ n (n 2. (4 This type of recurrences occurs very often in the analysis of algorithms and was systematically treated by an analytic approach by Flajolet and Golin [6]; see also [5]. Recurrences of the types (2 and (4 are typical representatives of the divide-and-conquer recurrences that one naturally encounters from its designing principle: A n = A x(n + A y(n + B n (n n 0 >, with some initial conditions, where B n is some given sequence (called toll function, A n is certain cost measure related to the algorithm in question, x(n and y(n usually satisfy 0 < x(n, y(n < n, their values depending upon the dividing strategy. Besides (2 and (4, let us mention another recurrence recently studied by Panny and Prodinger [22] for the analysis of bottom-up mergesort algorithm: A n = A x(n + A n x(n + B n (n 2, (5 with x(n = 2 log 2 n, where y denotes the least integer y. A typical phenomenon of the behaviors of divide-and-conquer recurrences is that they often involve certain periodic fluctuations depending on the instance size. Intuitively, this is due to the fact that in each dividing step, the sizes of the two subproblems are not necessarily identical. Consequently, the cumulative (recursive effect renders the global behavior less smooth, or even chaotic. Such a fluctuating behavior relies sensitively upon the dividing strategy. Roughly, the more balanced the two sequences x(n and y(n are, the more smooth the resulting oscillating behavior is. For example, the recurrence (4 should be expected to involve oscillating functions that are more smooth than those of (2 and (5. A detailed comparison realizing these arguments for the two recurrences (5 and 4 can be found in [22] and in [4]. Notations. n is a positive integer, and n = (b L b L...b 0 2, the binary representation of n, where L = log 2 n and b L =. n j = (b j...b 0 2 for j =, 2,..., L; n 0 =. v 2 (n denotes the exponent of 2 in the prime decomposition of n; ν(n denotes the number of -digits in the binary representation of n. All limits are taken to be n. 3 Explicit formula To solve the heap recurrence (2 explicitly, we observe that when n = 2 k+, the recurrence is essentially linear: f 2 k+ = t 2 k+ + 2f 2 k, which can be solved easily by iteration. From this, we can find the solution for the sequences {2 k }, {2 k + 2 k },.... But this process does not lead readily to a general solution; see Sprugnoli [27]. Hence, we begin with another approach. We first extend a classical counting argument for Young tableaux of a given shape (see [7, 5..4, 5.2.3] and give an explicit solution to the general recurrence (2. 4

5 Lemma. For n, the solution f n of the heap recurrence (2 is given by f n = ( n n 2 j 2 j t 2 j + t nj, (6 j L for any given sequence {t n } n. 0 j L Proof. As in [7], we call the nodes lying on the path from to n special nodes. Let Σ and Σ 2 denote the first sum and the second sum, respectively. Then Σ counts the weights of nonspecial nodes under the cost function t n, and Σ 2 counts similarly the weights of special nodes. Nonspecial nodes are always perfectly balanced, so their sizes are of the form 2 j. Now any node of rank β = (c r...c 0 2 lying in the range n/2 j < β < n/2 j has as subtree size s β = 0 i<j 2i = 2 j, since s β is the number of positive integers n whose binary representation is of the form (c r...c 0 2, where is any 0- string. Hence, the number of nonspecial subtrees of size 2 j is n n 2 j 2 j. On the other hand, the number of positive integers n whose binary representation has the form (b L...b j for = 0,,..., j is 0 i<j 2i + (b j... b = (b j... b 0 2. Hence, the special subtree sizes are n, n L,..., n 0 = n j. This completes the proof. Note that Σ can be re-written in the form Σ = n 2 j (t 2 j+ t 2 j j L and, by using n j = 2 j ( + {n/2 j }, 0 j L Σ 2 = 0 j L j L+ Similarly, the solution of the recurrence (3 is given by (φ 0 = 0 φ n = ( n n 2 j 2 j τ 2 j + τ nj. t 2 j + t n, (7 t 2 j (+{n/2 j }. (8 0 j L As an easy illustrative example, take t n = an + b with a > 0, we obtain readily 2 f n = anl + (a + bn a2 L+ + a(l + 2 = an log 2 n + F (log 2 nn + O(log n, where F is a continuous periodic function of period : F (u = a + b a{u} a2 {u} = a 2 + b a log 2 + a log 2 k Z\{0} e 2kπiu χ k (χ k + (χ k = 2kπi/ log 2, u R, By the basic property of heap: if a node β has rank (c r...c 0 2, then its left and right children have respective ranks (c r...c 00 2 and (c r...c We could imagine the mergesort variant on heaps, namely: in order to sort a heap, sort the left and right subheaps (divide and then merge (conquer the two sorted heaps by using, for instance, a linear merge; then the number of comparisons f n used in the worst case is described by (2 with t n = n 2 for n 2 and t = 0. So that if we do not take into account the additional space needed, then this algorithm takes f n + C n = nl 2 L+ n/2 + L {n/2} + C n comparisons to sort n elements, where C n is the number of comparisons used to construct a heap from a permutation. As far as the number of comparisons is concerned, this would be the fastest heapsort in the worst case, although it destroys the basic characteristic of heaps. 5

6 the series being absolutely convergent. This simple example reflects the general pattern of solutions to (2. An interesting consequence of Lemma is the following necessary and sufficient condition for the asymptotic linearity of f n. Lemma 2. Assume that f n satisfies the heap recurrence (2. Then f n is asymptotically linear: f n cn iff t n = o(n, and t 2 j 2 j <. (9 The constant c is given by c = j t 2 j 2 j. j L j This result says that without loss of generality, we can, under the hypotheses of Lemma 2, consider only the special sequence {f 2 k } k, as far as the dominant asymptotics is concerned. Proof. Assume that (9 holds. By (6 and t n = o(n, we have f n = ( n 2 j n 2 j t 2 j + ({ n } { n } 2 j 2 j t 2 j + t nj j L j L 0 j L = n t 2 j 2 j + o 2 j + o ( { n } 2 j + 2 j = n j L j L j L t 2 j 2 j + o(n. (0 Since the sum j L t 2 j 2 j converges (by assumption, we obtain f n cn. Note that the expression (0 holds as long as t n = o(n. On the other hand, assume that f n = c 0 n + o(n for some constant c 0. Then by definition (2 t n = f n f 2 log 2 2n/3 f n 2 log 2 2n/3 = c 0 n c 0 2 log 2 2n/3 c 0 (n 2 log 2n/3 2 + o(n = o(n. This implies, by (0 and f n c 0 n, that c 0 = c = t 2 j 2 j <. j This proves the necessity part and the lemma. As to recurrence (3, the constant c in (9 is modified to be c = j 0 τ 2 j2 j, under the same conditions. Note that c may be zero since we did not impose that t n 0. 4 The number of heaps In this section, we derive an asymptotic expression for the number of heaps H n of size n. To this end, we apply Lemma to the sequence h n = log(n!/h n. The extremal and average order of the arithmetic functions involved are also discussed. 6

7 4. The asymptotic behavior of H n First of all, since h n satisfies (2 with t n = log n, Lemma 2 gives, h n n log(2 j 2 j = n 2 log j log( 2 j = n. j j Let α = 2 log 2+ j 2 j log( 2 j. In order to obtain an asymptotically equivalent approximation for H n, we need to determine h n up to the constant term. To this aim, we examine each sum in (6 in more detail. Again, by formula (8, the contribution to h n by the sum Σ with t n = log n is given by Σ 2 = j log 2 + ω(n, 0 j L where ω(n := log( + {n/2 j } 0 j L oscillates between O(log n and O(. Consider now Σ in (7 with t n = log n. Simple algebraic manipulations lead to Σ = j L n 2 j log 2j+ 2 j = α n ν(n log 2 0 j L+ where q = j log( 2 j is a constant and is bounded for all n. Thus r(n = j j L+ log(2 j j log 2 q r(n 2 L + O ( n 2, { n 2 j } log 2 j 2 j (n, h n = Σ + Σ 2 = α n ν(n log 2 + ω(n (L + log 2 q r(n 2 L + O ( n 2. ( Returning to H n, we have H n = n! e hn = n! 2 L++ν(n e α n+q+r(n ω(n ( + 2 L + O ( n 2. For convenience, set Q = e q = j ( 2 j = and R(n = e r(n = j ( 2 j 2 j {n/2 j }. Since ν(n = j 0 {n/2j } = 0 j L {n/2j } + n/2 L, we conclude, by Stirling s formula, our main asymptotic result for H n. 7

8 Theorem. The number of heaps of size n satisfies the asymptotic expression H n = 2Q ( 2πΠ(log 2 nr(n n n+ 3 2 e (α+n + 2 L+ + 2n + O ( n 2, where α = 2 log 2 + j 2 j log( 2 j and Π(log 2 n = 2 ν(n {log 2 n} e ω(n = 2 2{log 2 n} {log 2 n} Note that this is not an usual asymptotic formula since log n log H n n n+3/2 e (α+n 0 j L 2 {n/2j } + {n/2 j }. (2 varies between 0 and log 4 (0/9 (see (3 below. In particular, if n = 2 L+, then H n = 4 ( 2πe α n n+ 3 2 e αn n + 3 4n + O ( n 2 ; If n = 2 L, then H n = 4Q ( 2π n n+ 3 2 e αn n + 7 2n + O ( n 2 ; But when n = 3 (4m+ (namely, n = ( for m 0, we obtain L = 2m and where b = log(0/9/2 and κ = H n = 20Q π κn n+ 3 2 b/ log 2 e αn n ( + O ( n ( 9 log4 3 ( 4 j 3 ( 4 j 0 j ( 4 j ( j ( 2 4 j. 3 (+ 2 4 j Remark. By the definition of h n and (, the information-theoretic lower bound for the number of comparisons to construct a heap of size n is given by h n log 2 = log 2 n! H n = α ω(n n ν(n L + log 2 log 2 + O(, The constant α/ log 2 is approximately equal to ; see [9]. 4.2 Extremal order We now derive bounds for functions R(n and Π(u. Taking logarithms and making use of the inequalities 0 {n/2 j } 2 j, we have, for all n, R(n exp 2 j log( 2 j = j 8

9 Figure : The function Π plotted against log 2 n for n from 32 to 024; the upper and lower bounds as predicted by (3 are also shown. Direct expansion of the function log ( ( 2 j /( 2 j leads to log R(n = l 2 l l(2 l+ 0 j L so that the oscillating behavior of log R(n is completely determined by the reversed weighted digital sums 0 j L b j2 lj. Proposition. For all n, the inequalities hold, where b = log(0/9/2 and c is some constant. b j 2 lj, 0 < c n b/ log 2 Π(log 2 n 2 (3 Proof. Observe first that by (2 and the elementary inequalities we obtain the upper and the lower bounds e log 2 2 2x x 2 (0 x, e 2x log x (0 x, Π(log 2 n e(log 2 ( e 2 log 2 L e(log 2n +/ log 2+log log 2/ log 2. Such a lower bound obtained by considering only the maximum of each term (as a continuous function is, however, too crude and unattainable. 9

10 To prove the better lower bound in (3, we consider the function ϖ(n := ψ ( {n/2 j }, where ψ(x := log( + x x log 2, j L so that Π(log 2 n = 2 2{log 2 n} {log n} 2 e ϖ(n. Note that the function ψ(x attains the maximum value at ( log 2/ log satisfies the recurrence ( n 2 ϖ(n = ϖ(n 2 L L + ψ. log 2 (n 2 L <j L 2 j Also ϖ(n then so that We prove that lim sup n ϖ(n L = b = 2 (ψ(/3 + ψ(2/3 = 2 log 0 9. (4 The proof consists of mainly two parts: First, we prove that if n = ( = 4L/2+ 3 L 2 b ϖ(n L 2 b + b, lim sup n ϖ(n L Second, if n is not of the form (5, then n can be written as (L even, (5 b. (6 n = (00 0 } 0 {{ 0} 2, (7 λ where λ 0, λ and is a binary sequence. We prove by induction that ϖ(n (L + b (n, (8 which together with (6 proves (4. Consider first the case when n is of the form (5. Then ϖ(n = { ( ψ 3 ( j + ψ 3 2 } 3 4 j. j L/2 By using the elementary inequalities (by concavity ( b ψ(/3 x/2 + ψ(2/3 x 2b + log 2 x 2 40 (0 x /4, (9 0

11 we obtain L 2 b ϖ(n L ( 2 b + log < L 2 b + ( log j L/2 4 j < L 2 b + b, for even L. This proves (6. We now consider the case when n is of the form (7. Since ψ(0 = ψ( = 0, the initial cases n 4 are easily checked. We assume that n 5. Then ϖ(n can be decomposed as (writing y = (0. 2 ( ϖ(n = ψ 2 + y j L 2s 4 + j s+ when λ = 0, for some s 0; and ϖ(n = ( ψ 2 j + y 2 j j<λ j s { ψ 0 j L 2s λ ψ({2 j y}, { ( ψ j + y ( j + ψ j + y } 4 j ( ( 3 2λ λ 4 j + y 2 2 λ+ 4 j + ψ 3 2λ λ 4 j + y } 2 λ 4 j ψ({2 j y}, when λ 2, for some s 0. Consider first the case λ = 0. By induction, max ψ({2 j y} (L 2s 3b. 0 y 0 j L 2s 4 It remains to show that for s 0 ( max 0 y ψ 2 + y + { ( ψ j + y ( j + ψ j + y } 4 j j s+ (2s + 3b, which follows from the elementary inequalities { ( max ψ 0 y j + y 2 4 j and + ψ ( j + y } 4 j 2b (j 2, max {ψ(/2 + y/2 + ψ(3/4 + y/4 + ψ(3/8 + y/8} 3b. 0 y

12 Next consider the case λ = 2. Again by induction max ψ({2 j y} (L 2s λb, and we need to show that where for brevity A s (y := ψ 0 y 0 j L 2s λ max A s(y (2s + 3b, 0 y ( 2 + y + { ( ψ j + y ( j + ψ j + y } 4 j. j s+ The case s = 0 is easily checked numerically; indeed, we have max A s(y 3b y Assume that s. Then by applying the inequality (9 max A s(y max A { ( 0(y + max ψ 0 y 0 y 0 y j + y 2 4 j 2 j s+ ( b sb + log 2 max y (2s + 3b < (2s + 3b. 2 j s+ ( 2 + ψ j + y } 4 j 5/3 y 4 j The remaining case λ > 2 is treated as follows. Observe first that we have, by the monotonicity of ψ, ( max ψ 0 y 2 j + y 2 j (λ 3ψ(/4 (λ 3b. 3 j<λ Then for λ 3 { ( max ψ 0 y 2 + y ( + ψ y 4 { ( max ψ 0 y 2 + y + ψ 2 4b 0.04; and, by (9, max 0 y j s { ( ψ ( + ψ ( 4 + y 4 3 2λ λ 4 j + y 3 2λ λ + y + ψ 2 λ+ 4 j ( λ+ 3 2sb + log λ 2sb , 2 2 λ+ ( y 6 ( 2 + ψ + ψ 3 2λ λ + y } 2 ( λ y } 8 ( 2 + ψ 3 2λ λ 4 j + y } 2 λ 4 j

13 for λ 3. This proves (8 for λ 3 and the proposition. Remarks. The proof does not reveal why repeating the pattern 0 yields the best possible upper bound for ϖ(n/l. Roughly, this is because the average of ψ(/3 and ψ(2/3 is somehow the most balanced choice. Here is another rough explanation (by calculations. Write first x = {n/2 L }. Then ϖ(n = 0 j<l ψ({2j x}. The function b ψ(x is positive for 0 x 0.27 and 0.62 x <. So if we assume that (8 holds, then, by induction, we are left with the interval [0.27, 0.62] for x. Then 2b ψ(x ψ({2x} > 0 for /3 < x 0.62, and we are left with the smaller interval [0.27, /3]. Continuing this way, we see that /3 is a limiting point in some sense, which means that n is of the form (5. An alternative proof of (4 is sketched as follows. Let x = x [0, and x j = {2 j x}. We show that for all x [0, and m (ψ(x j b K := 3ψ(/2. j m It suffices to show that there is an interval I [0, such that (i for all x I there is an m 0 such that x m0 + I, (ψ(x j b 0, and (ψ(x j b K/3, j m 0 j m for all m < m 0 ; and (ii for all x I ψ(x b K/3, and x 2, x 3 I. For ϖ(n, we take I = [0, /4] [/3, and check the above conditions case by case. 4.3 Average order Although the function ϖ(n oscillates between O( and O(log n, its sum function has a much smoother behavior. We determine its average order, namely n m n ω(m. Note that we have the Fourier series log( + {x} {x} log 2 = 3 2 log 2 + k 0 e 2kπix 2kπi convergent for all x, where Ei(, z is the integral exponential defined by Ei(, z = e zw w dw (Ei(, 4kπi Ei(, 2kπi, ( arg(z < π, the integration being taken along any path in the z-plan with a cut along the negative real axis. Now summing all such series for j =, 2,..., L, we obtain ( 3 ϖ(n = 2 log 2 L + Ei(, 4kπi Ei(, 2kπi e 2kπni/2j, 2kπi k 0 j L which is nothing but a mere translation of ϖ(n into trigonometric sums. Yet the formula still says something about the average order of ϖ(n: ( 3 ϖ(m = n 2 log 2 log 2 n + O(, m n which can be obtained by the following Ergodic-type result. 3

14 Lemma 3. Let ϕ(x be any real, continuous function on [0, ] and differentiable in (0,. Define φ(m = 0 j log 2 m ϕ({m/2j }. Then n m n ( φ(m = 0 ϕ(x dx log 2 n + O(. In other words, the lemma says that the average order of the function φ(m is equal to log 2 n times the mean value of the function ϕ in [0, ]. Proof. (Sketch Let Φ(n := m n φ(m. Then Φ(n = n+ φ(x dx n+ (φ(x φ( x dx + O(. The first integral on the right-hand side is easily seen to be n+ ( φ(x dx = ϕ(x dx n log 2 n + O(n. The second integral is O(n by the differentiability of ϕ(x. If we take ϕ(x = x, we obtain, in view of the relation ν(n = 0 j L {n/2j } + n/2 L, n m n 0 ν(m = 2 log 2 n + O(; and in general n m n ν q (m = q + log 2 n + O( (q > 0, where ν q (n := 0 j L {n/2j } q. The implied constant in the O-term depends on q. Other examples are given in the next section. 5 The cost of constructing heaps In this section, we again apply the solution (6 of the heap recurrence to refine previous analyses on the mean and the variance of the cost used by Floyd s algorithm to construct a heap from a random sequence. Moreover, the asymptotic normality of the cost is also established. Our methods apply equally to other variants of Floyd s algorithm. Let us first recall the algorithm. It is divide-and-conquer in nature. Briefly, to construct a heap of size n, construct the left and the right subheaps, respectively, by the same procedure and then find the proper place in one of the two subheaps for the root element (by comparisons and keyexchanges. We assume, throughout this section, that a uniform probability measure is assumed on the set of permutations of size n, n. Knuth [7, p. 55] showed that, given a random permutation of size n, Floyd s algorithm preserves randomness in each recursive step. This means that given a random permutation of n elements, the output obtained by Floyd s algorithm is a random heap of size n, namely, all heaps of size n are equally likely. 4

15 Thus, with the notations of (, the probability generating function P n (z for some cost measure of the algorithm like the number of exchanges running over all heaps of size n satisfies (see (2 P 0 (z = P n (z = P 2 log 2 2n/3 (zp n 2 log 2 2n/3 (zq n(z (n, (20 where Q n (z is the probability generating function for the cost of placing the root element of a heap of size n into one of its two subheaps. Note that Q n (z is always a polynomial for all finite n. From (6, we have the explicit representation P n (z = Q n (z j L Q 2 j (z n/2j n/2 j Q nj (z (n. Our analysis below applies to all versions of heap-construction algorithm using divide-and-conquer paradigm such that the randomness is preserved in each conquering step, the randomness is preserved. For concreteness, we consider in detail the number of exchanges used by Floyd s algorithm. Note that if ξ n is a random variable distributed according to P n (z, namely, E(z ξn = P n (z, then E(ξ n satisfies (2 with t n = Q n(, and Var(ξ n satisfies (2 with t n = Q n(+q n( Q n( 2. Cumulants of higher order obey the same type of recurrences. We derive in 5.3 an effective version (namely, with convergence rate of a theorem of Haigh [] on the asymptotic normality of sums of independent random variables. Then we apply this theorem to probability generating functions of the form (20. Roughly, our theorem states that if the maximum degree M n of the polynomial Q n (z is small compared to the standard deviation σ n of the random variable ξ n (more precisely, M n /σ n 0, then the random variable ξ n is asymptotically normal in the sense of convergence in distribution. This fact is well-known since Laplace from the early development of Probability Theory. From now on, given a random permutation π n of size n, let ξ n denote the number of exchanges used to construct a heap from π n using Floyd s algorithm [7, 5.2.3]. 5. The mean of ξ n From [7, 5.2.3], the mean number of exchanges E(ξ n satisfies the heap recurrence (2 with t n = n j n log 2 j = L + (L + 2/n 2 L+ /n, the average height of a node in a random heap of size n (see [7, p. 55]. Applying formulae (7 and (8, we get the following result that is more general than that of Sprugnoli [27]; see also [3, Ch. 3]. Theorem 2. The expected number of exchanges E(ξ n used by Floyd s heap construction algorithm satisfies ( log n E(ξ n = µn + ϖ (n + ϖ 2 (n + O (n, (2 n where µ = 2 + j j(2j = , ϖ (n is defined by ϖ (n = j L + {n/2 j } 2 + {n/2 j }, 5

16 and ϖ 2 (n = O( is given by ϖ 2 (n = n 2 L j 2 j + j j ( + {n/2 j } + { n } j2 j 2 j + 2 j (2 j (2 j+. j j j Furthermore, ϖ (n satisfies lim inf n ϖ (n L =, and lim sup n ϖ (n L = By Lemma 4, the average order of the arithmetic function ϖ (n is (3/2 2 log 2 log 2 n + O(. Proof. (Sketch With the exact formulae (7 and (8, simple manipulations using Maple lead to (2. The inequalities for ϖ (n are derived by similar arguments as in Proposition (both approaches apply. In particular, if n = 2 L+, then ( log n E(ξ n = µn L + µ + O ; n if n = 2 L, then ( log n E(ξ n = µn L + µ + O. n But if n = 3 (4m+, we have E(ξ n = µn 7 20 L + O(, In general, for n = (2 (m+d /(2 d, d 2, we have E(ξ n = µn + K d L + O(, where K d = d(2 d 0 j<d (2 d j 2 d + 2 j. 5.2 The variance of ξ n We now consider the variance of ξ n. Since the variance of a sum of independent random variables is equal to the sum of the variance of individual random variable, the variance of ξ n satisfies (2 with t n = n j n = 6 2L n 4L+ n 2 log 2 j 2 n j n log 2 j 2 L2 n + 2L+2 L n 2 4 L n 6 n + 2L+3 n 2 L2 n 2 4 L n 2 4 n 2, by partial summation. With the help of Maple, we obtain the following result, improving that of Doberkat [5]. 6

17 Figure 2: The function (E(ξ n µn/l plotted against {log 2 n} for n from 2 to 024; the upper bound as predicted by Theorem 2 is also shown. Theorem 3. The variance of the number of exchanges satisfies the asymptotic expression ( log Var (ξ n = σ 2 2 n n + ϖ 3 (n + ϖ 4 (n + O (n, n where σ 2 = 2 j j2 (2 j 2 = , ϖ 3 (n is defined by and satisfies ϖ 3 (n = 2 0 j L {n/2 j } {n/2 j } 2 ( + {n/2 j } 2, and ϖ 4 (n = O(: lim inf n ϖ 3 (n L = 0, and lim sup n ϖ 3 (n L = , j 2 2 j ϖ 4 (n = (2 j 2 + j j { n 2 j } 2 j (j 2 + 4j j+ (2j j (j 2 2j (2 j 2 (2 j+ 2. The average order of ϖ 3 (n is (6 log 2 4 log 2 n + O(. Note that, unlike the two functions ϖ(n/l and ϖ (n/l whose maximum values are achieved when n = (00 0 2, the maximum of ϖ 3 /L is (asymptotically attained when n is of the form n = ( , 7

18 so that the ratio / is equal to 5 (ψ 3(5/3 + ψ 3 (0/3 + ψ 3 (20/3 + ψ 3 (9/3 + ψ 3 (8/3, where ψ 3 (x = 2(x x 2 /( + x 2. The main difference is that the underlying function ψ 3 (x attains its maximum value /4 at x = /3 and not between /3 and /2. The proof follows the same arguments used in the proof of Proposition ; we omit the proof here since it is very involved and uninteresting. 5.3 Asymptotic normality From the special form (20 satisfied by the probability generating function P n (z of ξ n, we show that ξ n is asymptotically normal in the sense of convergence in distribution. Theorem 4. The distribution functions of the random variables (ξ n µn/(σ n converge to the standard normal distribution: ( sup P ξn µn <x< σ n < x x ( e t2 /2 dt log n 2π = O, (22 n where the two constants µ and σ 2 are defined in Theorems 2 and 3, respectively. These results are special cases of the following lemma refined from that of Haigh [], where we explicitly characterize the convergence rate to the normal law by a direct approach based upon characteristic functions and the Berry-Esseen smoothing inequality (see [23]. Lemma 4. Let {X n } n be a sequence of random variables taking only non-negative integral values with mean µ n and variance σn. 2 Suppose that the probability generating function P n (z of Ω n can be decomposed as P n (z = P nj (z, j k n for some sequence {k n } n, where the P nj (z are polynomials such that (i each P nj (z is itself a probability generating function of some random variable, say, X nj ( j k n ; and (ii M n σ n 0, (23 where M n = max j kn deg P nj (z. Then the distribution of X n is asymptotically Gaussian: ( sup P Xn µ n < x x ( e t2 /2 dt 2π = O Mn. (24 <x< σ n Proof. First of all, we have µ n = j k n µ nj and σn 2 = j k n σnj 2, where µ nj and σnj 2 denote, respectively, the mean and the variance of X nj. Note that condition (23 implies that σ n as n. Let ϕ nj (t = P nj (e it/σn and define the random variable Xn = (X n µ n /σ n with characteristic function ϕ n (t; then ϕ n (t = e µnit/σn P n (e it/σn. Set T n = ησ n /M n, where η > 0 is a sufficiently small constant. By the expansion log ϕ nj (t = µ ( nj it σ2 nj σ n 2σn 2 t 2 E( Xnj µ nj + O σn 3 t 3 ( j k n, t T n, 8 σ n

19 we obtain ( log ϕ n (t = t2 2 + O Mn t 3 σ n ( t T n, since σ 3 n E( X nj µ nj 3 2 max X j kn nj σ n j k n = 2M n σ n. Using e w w e w, for all complex w, we have ( ( ( ϕ n (t e t2 /2 Mn = O t 3 exp t2 σ n 2 + O Mn t 3 σ n ( Mn = O t 3 e t2 /2 ( t T n, σ n by choosing η sufficiently small. Thus we obtain Tn ϕ n (t e t2 /2 t dt = O T n ( Mn σ n t 3 e t2 /2 dt = O By the Berry-Esseen smoothing inequality [23, p. 09], we have ( sup P Xn µ n < x x ( e t2 /2 dt <x< σ n 2π = O Tn + Tn T n ( Mn σ n. ϕ n (t e t2 /2 t dt, and the result (24 follows. This simple lemma can be used to derive the asymptotic normality (with convergence rate for many combinatorial quantities; see [4]. Proof of Theorem 4. By (20 and Lemma 5, we need only check that the degree of the polynomials Q n (z in (20 is small compared to the standard deviation of ξ n, which is O( n. The special form of Q n (z given in [5] has no importance here. Since Q n (z is the probability generating function for inserting the root into one of the two subheaps that are of height O(log n, it is obvious that the degree of Q n (z is O(log n. From this, an application of Lemma 5 yields sup <x< P ( ξ n E(ξ n Var(ξn < x x 2π e t2 /2 dt = O ( Mn In view of Theorems 2 and 3, the above equation is asymptotically equivalent to (22. Our approach also applies to the number of comparisons used by Floyd s algorithm and similar results as Theorems 2 4 hold. Here is a more general rule. The cost (the number of key-exchanges or the number of comparisons used to construct a heap from a random permutation of n elements by algorithms using divide-and-conquer paradigm such that the randomness is preserved in each conquer step is asymptotically normal in the sense of convergence in distribution. 9 σ n.

20 It is merely a rule since a formal statement of a precise version would be too heavy. In particular, this rule applies to the heap construction algorithms in [3, 9, 2, 29], the basic ideas of improvement being more or less due to Floyd. Our approach also gives in most cases more precise quantitative results in the form of (22. In connection with this, it should be pointed out that the original on-line algorithm proposed by Williams [30] to construct a heap is not linear in the worst case and that the randomness is not guaranteed (see [24] in each step. The average case analysis of its behavior is more difficult; see [2, 8, 2]. Likewise, a precise analysis of the expected behavior of heapsort is very involved since successive deletions destroy the initial random character. Recent progress in this direction can be found in [26]. 6 Conclusion Analysis in distribution of algorithms has received much attention recently and several tools have been now well developed; see Mahmoud [8], Szpankowski [28]. Specializing to divide-and-conquer algorithms that preserve randomness in each conquer step, we have seen that the limiting distribution of the cost of heap-related algorithms is normal in most cases. For algorithms leading to the recurrence (4, we have, in general, probability generating functions of the form { pn (z = p n/2 (zp n/2 (zq n (z, if n 2; p 0 (z = p (z =. This is studied in detail in [6] in the case of mergesort where the limiting distribution is still of a Gaussian type. Since divide-and-conquer is widely used in algorithmics, the general methods presented in this paper together with those in [6, 6] might be applied to the analysis in distribution of other algorithms. We have seen that digital sums of the form j L ψ({n/2j } appear naturally in the analysis of heap recurrences. We studied the extremal properties of such sums for three special cases: ψ(x = log( + x x log 2, ψ(x = ( + x 2 /( + x and ψ(x = 2(x x 2 /( + x 2 and the behavior of the last case is very different from those of the first two cases. In general, the extremal problems for arbitrary ψ(x satisfying some regularity conditions are very difficult. Acknowledgements The first author thanks Tsung-Hsi Tsai for helpful suggestions. References [] A. V. Aho, J. E. Hopcroft and J. D. Ullman, The Design and Analysis of Computer Algorithms, Addison-Wesley, MA, 974. [2] B. Bollabás and I. Simon, Repeated random insertion into a priority queue structure, Journal of Algorithms, 6 ( [3] S. Carlsson, Average-case results on heapsort, BIT, 27 (

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